Aircraft Structures - textbooks.elsevier.com · Aircraft Structures . for Engineering Students ....

Aircraft Structures for Engineering Students

Fifth Edition

Solutions Manual

T. H. G. Megson

This page intentionally left blank

Solutions Manual

Solutions to Chapter 1 Problems S.1.1

The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx = 80N/mm2, σy = 0 (or vice versa) and τxy = 45N/mm2

. Thus, from Eq. (1.11)

2 2I

80 1 80 4 452 2

σ = + + ×

i.e. σI = 100.2 N/mmFrom Eq. (1.12)

2

2 2II

80 1 80 4 452 2

σ = − + ×

i.e. σII = – 20.2 N/mm

The directions of the principal stresses are defined by the angle θ in Fig. 1.8(b) in which θ is given by Eq. (1.10). Hence

2

2 45tan 2 1.12580 0

θ ×= =

−

which gives θ = 24°11′ and θ = 114°11′

It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ corresponds to σI while the second value corresponds to σII

Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq. (1.15)

.

2max

100.2 ( 20.2) 60.2N / mm2

τ − −= =

and will act on planes at 45° to the principal planes.

4 Solutions Manual

S.1.2

The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx = 50N/mm2, σy = –35 N/mm2 and τxy = 40 N/mm2

. Thus, from Eq. (1.11)

2 2I

50 35 1 (50 35) 4 402 2

σ −= + + + ×

i.e. σI = 65.9 N/mm

and from Eq. (1.12)

2

2 2II

50 35 1 (50 35) 4 402 2

σ −= − + + ×

i.e. σII = –50.9 N/mm

From Fig. 1.8(b) and Eq. (1.10)

2

2 40tan 2 0.94150 35

θ ×= =

+

which gives θ = 21°38′(σI) and θ = 111°38′(σII

The planes on which there is no direct stress may be found by considering the triangular element of unit thickness shown in Fig. S.1.2 where the plane AC represents the plane on which there is no direct stress. For equilibrium of the element in a direction perpendicular to AC

)

0 50ABcos 35BCsin 40ABsin +40BCcosα α α α= − + (i)

Fig. S.1.2

Solutions to Chapter 1 Problems 5

Dividing through Eq. (i) by AB

0 50cos 35 tan sin 40sin 40 tan cosα α α α α α= − + +

which, dividing through by cos α, simplifies to

0 = 50–35 tan2

from which

α + 80 tan α

tan α = 2.797 or –0.511

Hence

α = 70°21′ or –27°5′

S.1.3

The construction of Mohr’s circle for each stress combination follows the procedure described in Section 1.8 and is shown in Figs S.1.3(a)–(d).

Fig. S.1.3(a)

Fig. S.1.3(b)

6 Solutions Manual

Fig. S.1.3(c)

Fig. S.1.3(d)

S.1.4

The principal stresses at the point are determined, as indicated in the question, by transforming each state of stress into a σx, σy, τxy stress system. Clearly, in the first case σx = 0, σy = 10 N/mm2, τxy = 0 (Fig. S.1.4(a)). The two remaining cases are transformed by considering the equilibrium of the triangular element ABC in Figs S.1.4(b), (c), (e) and (f). Thus, using the method described in Section 1.6 and the principle of superposition (see Section 5.9), the second stress system of

Solutions to Chapter 1 Problems 7 Figs S.1.4(b) and (c) becomes the σx, σy, τxy system shown in Fig. S.1.4(d) while

8 Solutions Manual

Fig. S.1.4(a) Fig. S.1.4(b)

Fig. S.1.4(c)

Fig. S.1.4(d)

the third stress system of Figs S.1.4(e) and (f) transforms into the σx, σy, τxy system of Fig. S.1.4(g).

Solutions to Chapter 1 Problems 9 Finally, the states of stress shown in Figs S.1.4(a), (d) and (g) are superimposed

to give the state of stress shown in Fig. S.1.4(h) from which it can be seen that σI = σII =15N/mm2 and that the x and y planes are principal planes.

10 Solutions Manual

Fig. S.1.4(e) Fig. S.1.4(f)

Fig. S.1.4(g)

Fig. S.1.4(h)

S.1.5

The geometry of Mohr’s circle of stress is shown in Fig. S.1.5 in which the circle is constructed using the method described in Section 1.8.

From Fig. S.1.5 σx = OP1 = OB– BC +CP1 (i)


Fig. S.1.5

In Eq. (i) OB = σI, BC is the radius of the circle which is equal to τmax2 2 2 2

1 1 1 1 maxCP CQ Q P .xyτ τ= − = −

and

Hence

2 21 max maxx xyσ σ τ τ τ= − + −

Similarly

2 2 2 1OP OB BC CP in which CP CPyσ = = − − =

Thus

2 2I max maxy xyσ σ τ τ τ= − − −

S.1.6

From bending theory the direct stress due to bending on the upper surface of the shaft at a point in the vertical plane of symmetry is given by

6

24

25 10 75 75 N / mm150 / 64x

MyI

σπ

× ×= = =

×

From the theory of the torsion of circular section shafts the shear stress at the same point is

6

24

50 10 75 75N / mm150 / 32xy

TrJ

τπ

× ×= = =

×

12 Solutions Manual

Substituting these values in Eqs (1.11) and (1.12) in turn and noting that σy

= 0 2 2

I75 1 75 4 752 2

σ = + + ×

i.e.

2I 121.4N / mmσ =

2 2II

75 1 75 4 752 2

σ = − + ×

i.e.

2II 46.4N / mmσ = −

The corresponding directions as defined by θ in Fig. 1.8(b) are given by Eq. (1.10) i.e.

2 75tan 2 275 0

θ ×= =

−

Hence I31 43 ( )θ σ′= ° and II121 43 ( )θ σ′= °

S.1.7

The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e.

1 [ ( )]x x y zvE

ε σ σ σ= − + (i)

1 [ ( )]y y x zvE

ε σ σ σ= − + (ii)

1 [ ( )]z z x yvE

ε σ σ σ= − + (iii)

Then

1 [ 2 ( )]x y z x y z x y ze vE

ε ε ε σ σ σ σ σ σ= + + = + + − + +

i.e.

(1 2 ) ( )x y zve

Eσ σ σ−

= + +

whence

(1 2 )y z x

Eev

σ σ σ+ = −−


Substituting in Eq. (i)

11 2x x x

EevE v

ε σ σ = − − −

so that

(1 )1 2x xvEeE v

vε σ= + −

−

Thus

(1 2 )(1 ) (1 )x x

vEe Ev v v

σ ε= +− + +

or, since G = E/2(1 + ν) (see Section 1.15) 2x xe Gσ λ ε= +

Similarly 2y ye Gσ λ ε= +

and 2z ze Gσ λ ε= +

S.1.8

The implication in this problem is that the condition of plane strain also describes the condition of plane stress. Hence, from Eqs (1.52)

1 ( )x x yvE

ε σ σ= − (i)

1 ( )y y xvE

ε σ σ= − (ii)

2(1 ) (seeSection 1.15)xyxy xy

vG Eτ

γ τ+= = (iii)

The compatibility condition for plane strain is

2 2 2

2 2 (see Section 1.11)xy y x

x y x yγ ε ε∂ ∂ ∂

= +∂ ∂ ∂ ∂

(iv)

Substituting in Eq. (iv) for εx, εy and γxy

from Eqs (i)–(iii), respectively, gives 2 2 2

2 22 (1 ) ( ) ( )xyy x x yv v v

x y x yτ

σ σ σ σ∂ ∂ ∂

+ = − + −∂ ∂ ∂ ∂

(v)

14 Solutions Manual

Also, from Eqs (1.6) and assuming that the body forces X and Y are zero

0zyx

x yτσ ∂∂

+ =∂ ∂

(vi)

0y xy

y xσ τ∂ ∂

+ =∂ ∂

(vii)

Differentiating Eq. (vi) with respect to x and Eq. (vii) with respect to y and adding gives

2 2 22

2 2 0xy y xyx

y x x yx yτ σ τσ ∂ ∂ ∂∂

+ + + =∂ ∂ ∂ ∂∂ ∂

or

2 22

2 22 xy yx

x y x yτ σσ ∂ ∂∂

= − + ∂ ∂ ∂ ∂

Substituting in Eq. (v)

22 2 2

2 2 2 2(1 ) ( ) ( )yxy x x yv v v

x y x yσσ

σ σ σ σ ∂∂ ∂ ∂ − + + = − + − ∂ ∂ ∂ ∂

so that

2 2 22 2 2

2 2 2 2 2 2(1 ) y y yx x xv vx y x y x y

σ σ σσ σ σ ∂ ∂ ∂∂ ∂ ∂ − + + = + − + ∂ ∂ ∂ ∂ ∂ ∂

which simplifies to

2 22 2

2 2 2 2 0y yx x

x y x yσ σσ σ∂ ∂∂ ∂

+ + + =∂ ∂ ∂ ∂

or

2 2

2 2 ( ) 0x yx yσ σ

∂ ∂+ + = ∂ ∂

S.1.9

Suppose that the load in the steel bar is Pst and that in the aluminium bar is Pal

. Then, from equilibrium

st alP P P+ = (i) From Eq. (1.40)

st alst al

st st al al

P PA E A E

ε ε= =


Since the bars contract by the same amount

st al

st st al al

P PA E A E

= (ii)

Solving Eqs (i) and (ii)

st st al alst

st st al al st st al alal

A E A EP P P PA E A E A E A E

= =+ +

from which the stresses are

st alst al

st st al al st st al al

E EP PA E A E A E A E

σ σ= =+ +

(iii)

The areas of cross-section are

2 2 2

2 2st al

75 (100 75 )4417.9mm 3436.1mm4 4

A Aπ π× −= = = =

Substituting in Eq. (iii) we have

6

2st

10 200000 172.6N/mm (compression)(4417.9 200000 3436.1 80000)

σ ×= =

× + ×

6

2al

10 80000 69.1N/mm (compression)(4417.9 200000 3436.1 80000)

σ ×= =

× + ×

Due to the decrease in temperature in which no change in length is allowed the strain in the steel is αstT and that in the aluminium is αal

T. Therefore due to the decrease in temperature

2st st st 200000 0.000012 150 360.0 N/mm (tension)E Tσ α= = × × =

2al al al 80000 0.000005 150 60.0 N/mm (tension)E Tσ α= = × × =

The final stresses in the steel and aluminium are then

2st (total) 360.0 172.6 187.4N/mm (tension)σ = − =

2al (total) 60.0 69.1 9.1N/mm (compression)σ = − = −

S.1.10

The principal strains are given directly by Eqs (1.69) and (1.70). Thus

2 2I

1 1( 0.002 0.002) ( 0.002 0.002) ( 0.002 0.002)2 2

ε = − + + − + + + +

16 Solutions Manual

i.e. I 0.00283ε = + Similarly

II 0.00283ε = −

The principal directions are given by Eq. (1.71), i.e.

2( 0.002) 0.002 0.002tan 2 10.002 0.002

θ − + −= = −

+

Hence 2 45 or 135θ = − ° + ° and

22.5 or 67.5θ = − ° + °

S.1.11

The principal strains at the point P are determined using Eqs (1.69) and (1.70). Thus

2 2 6I

1 1( 222 45) ( 222 213) ( 213 45) 102 2

ε − = − + + − + + − − ×

i.e.

6I 94.0 10ε −= ×

Similarly

6II 217.0 10ε −= − ×

The principal stresses follow from Eqs (1.67) and (1.68). Hence

6I 2

31000 (94.0 0.2 271.0) 101 (0.2)

σ −= − × ×−

i.e.

2I 1.29 N/mmσ =

Similarly

2II 814 N/mmσ = −

Since P lies on the neutral axis of the beam the direct stress due to bending is zero. Therefore, at P, σx =7 N/mm2 and σy

= 0. Now subtracting Eq. (1.12) from (1.11) 2 2

I II 4x xyσ σ σ τ− = +


i.e.

2 21.29 8.14 7 4 xyτ+ = +

from which τxy = 3.17 N/mm2

The shear force at P is equal to Q so that the shear stress at P is given by .

33.172 150 300xy

Qτ = =× ×

from which

95 100N 95.1kN.Q = =

Solutions to Chapter 2 Problems

S.2.1

The stress system applied to the plate is shown in Fig. S.2.1. The origin, O, of the axes may be chosen at any point in the plate; let P be the point whose coordinates are (2, 3).

Fig. S.2.1

From Eqs (1.42) in which σz

= 0 3 2 3.5

xp p pv

E E Eε = − − = − (i)

3 3 2.75y

p p pvE E E

ε = − − = − (ii)

Hence, from Eqs (1.27)

13.5 3.5so that ( )u p pu x f y

x E E∂

= − = − +∂

(iii)

18 Solutions Manual

where f1

(y) is a function of y. Also

22.75 2.75so that ( )p p y f x

y E Eυ υ∂= = − +

∂ (iv)

in which f2From the last of Eqs (1.52) and Eq. (1.28)

(x) is a function of x.

2 1( ) ( )4 (from Eqs (iv)and (iii))xyf x f yp u

G x y x yυγ

∂ ∂∂ ∂= = + = +

∂ ∂ ∂ ∂

Suppose

1( )f y Ay

∂=

∂

then

1( )f y Ay B= + (v)

in which A and B are constants. Similarly, suppose

2 ( )f x Cx

∂=

∂

then

2 ( )f x Cx D= + (vi)

in which C and D are constants. Substituting for f1 (y) and f2

(x) in Eqs (iii) and (iv) gives

3.5pu x Ay BE

= − + + (vii)

and

2.75p y Cx DE

υ = + + (viii)

Since the origin of the axes is fixed in space it follows that when x = y = 0, u=v = 0. Hence, from Eqs (vii) and (viii), B = D = 0. Further, the direction of Ox is fixed in space so that, wheny = 0, ∂v/∂x = 0. Therefore, from Eq. (viii), C = 0. Thus, from Eqs (1.28) and (vii), when x = 0.

4u p Ay G∂

= =∂

Eqs (vii) and (viii) now become

3.5 4p pu x yE G

= − + (ix)


2.75p yE

υ = (x)

From Eq. (1.50), G=E/2(1 +ν) =E/2.5 and Eq. (ix) becomes

( 3.5 10 )pu yE

= − + (xi)

At the point (2, 3)

23 (from Eq. (xi))puE

=

and

8.25 (from Eq. (x))pE

υ =

The point P therefore moves at an angle α to the x axis given by

S.2.2

An Airy stress function, φ, is defined by the equations (Eqs (2.8)):

2 2 2

2 2x y xy x yy xφ φ φσ σ τ∂ ∂ ∂

= = = −∂ ∂∂ ∂

and has a final form which is determined by the boundary conditions relating to a particular problem.

Since

3 3Ay By x Cyxφ = + + (i)

4 4 4

4 4 2 20 0 0x y x yφ φ φ∂ ∂ ∂= = =

∂ ∂ ∂ ∂

and the biharmonic equation (2.9) is satisfied. Further

2

2 6 6x Ay Byxyφσ ∂

= = +∂

(ii)

2

2 0y xφσ ∂

= =∂

(iii)

2

23xy By Cx yφτ ∂

= − = − −∂ ∂

(iv)

The distribution of shear stress in a rectangular section beam is parabolic and is zero at the upper and lower surfaces. Hence, when y = ±d/2, τxy

= 0. Thus, from Eq. (iv)

24

3CB

d−

= (v)

20 Solutions Manual

The resultant shear force at any section of the beam is –P. Therefore

/2

/2d

dxyd

t y Pτ−

= −∫

Substituting for τxy

from Eq. (iv) /2 2

/2( 3 ) d

d

dBy C t y P

−− − = −∫

which gives

3

28 2

Bd Cdt P

+ =

Substituting for B from Eq. (v) gives

32

PCtd

= (vi)

It now follows from Eqs (v) and (vi) that

32PB

td−

= (vii)

At the free end of the beam where x =l the bending moment is zero and thus σx

6A + 6Bl = 0

= 0 for any value of y. Therefore, from Eq. (ii)

whence

32PlAtd

= (viii)

Then, from Eq. (ii)

3 312 12

xPl Py xy

td tdσ = −

or

312 ( )

xP l x ytd

σ −= (ix)

Equation (ix) is the direct stress distribution at any section of the beam given by simple bending theory, i.e.

xMyI

σ =

where M = P (l –x) and I = td3

The shear stress distribution given by Eq. (iv) is /12.

23

6 32xy

P Pytdtd

τ = −


or

2

23

64xy

P dytd

τ

= −

(x)

Equation (x) is identical to that derived from simple bending theory and may be found in standard texts on stress analysis, strength of materials, etc.

S.2.3

The stress function is

2 2 2 3 2 3 53 (15 5 2 )

20w h x y x y h y yh

φ = − − +

Then

22 3

2 3

22 2 3

2 3

22 2

3

4

4

4

4 3

4

2 2 3

(30 10 )20

( 30 12 20 )20

(30 30 )20

0

(120 )20

( 60 )20

y

x

xy

w h y yx h

w x y h y yy h

w h x xyx y h

xw y

y h

w yx y h

φ σ

φ σ

φ τ

φ

φ

φ

∂= − =

∂∂

= − − + =∂

∂= − = −

∂ ∂

∂=

∂∂

=∂

∂= −

∂ ∂

Substituting in Eq. (2.9) 4 0φ∇ = so that the stress function satisfies the biharmonic equation.

The boundary conditions are as follows: ● At y= h, σy = w and τxy

● At y = –h, σ = 0 which are satisfied.

y = –w and τxy● At x =0, σ

= 0 which are satisfied. x = w/20h3 (–12h2y + 20y3

Also ) ≠ 0.

2 33

2 2 43

d ( 12 20 )d20

[ 6 5 ]200

h hxh h

hh

wy h y y yh

w h y yh

σ− −

−

= − +

= − +

=

∫ ∫

i.e. no resultant force.

22 Solutions Manual

Finally

2 2 43

2 3 53

d ( 12 20 )d20

[ 4 4 ]200

h hxh h

hh

wy y h y y yh

w h y yh

σ− −

−

= − +

= − +

=

∫ ∫

i.e. no resultant moment.

S.2.4

The Airy stress function is

3 2 2 2 2 23 [5( )( ) ( 2 ) 3 ( ) ]

120p x l x y d y d yx y dd

φ = − + − − −

Then

4 4 4

4 4 3 2 2 33 30

2pxy pxy

x y d x y dφ φ φ∂ ∂ ∂= = − =

∂ ∂ ∂ ∂

Substituting these values in Eq. (2.9) gives

3 33 30 2 02pxy pxyd d

+ × − =

Therefore, the biharmonic equation (2.9) is satisfied. The direct stress, σx

, is given by (see Eqs (2.8)) 2

2 2 3 22 3 [5 ( ) 10 6 ]

20xpx y x l y d y

y dφσ ∂

= = − − +∂

When x = 0, σx

= 0 for all values of y. When x = l

3 23 ( 10 6 )

20xpl y d yd

σ = − +

and the total end load 1 dd

xdyσ

−= ∫

3 23 ( 10 6 )d 0

20

d

d

pl y d y yd −

= − + =∫

Thus the stress function satisfies the boundary conditions for axial load in the x direction. Also, the direct stress, σy


3 2 32 3 ( 3 2 )

4ypx y yd d

x dφσ ∂

= = − −∂


When x = 0, σy

= 0 for all values of y. Also at any section x where y = –d

3 3 3( 3 2 ) 04ypx d d dd

σ 3= − + − =

and when y = +d

3 3 33 ( 3 2 )

4ypx d d d pxd

σ = − − = −

Thus, the stress function satisfies the boundary conditions for load in the y direction. The shear stress, τxy


2 2 2 2 4 2 2 43 [5(3 )( ) 5 6 ]

40xyp x l y d y y d d

x y dφτ ∂

= − = − − − − + −∂ ∂

When x = 0

2 2 2 4 2 2 43 [ 5 ( ) 5 6 ]

40xyp l y d y y d dd

τ = − − − − + −

so that, when y = ±d, τxy

= 0. The resultant shear force on the plane x = 0 is given by 2

2 2 2 4 2 2 431 d [ 5 ( ) 5 6 ]d

640

d dxyd d

p ply l y d y y d d yd

τ− −

= − − − − + − = −∫ ∫

From Fig. P.2.4 and taking moments about the plane x = l,

1 2( 0)122 3xy x dl lpl lτ = =

i.e.

2

( 0)6xyplx

dτ = =

and the shear force is pl2Thus, although the resultant of the Airy stress function shear stress has the same

magnitude as the equilibrating shear force it varies through the depth of the beam whereas the applied equilibrating shear stress is constant. A similar situation arises on the plane x = l.

/6.

S.2.5


3 2 2 2 2 3 2 3 53 ( 10 15 2 5 )

40w c x c x y c y x y ybc

φ = − − + + −

24 Solutions Manual

Then

22 2 3

2 3

23 2 3

2 3

22 2

3

4

4

4

4 3

4

2 2 3

(12 30 20 )40

( 20 30 10 )40

( 30 30 )40

0

( 120 )40

(60 )40

x

y

xy

w c y x y yy bc

w c c y yx bc

w c x xyx y bc

xw y

y bc

w yx y bc

φ σ

φ σ

φ τ

φ

φ

φ

∂= + − =

∂

∂= − − + =

∂∂

= − + = −∂ ∂

∂=

∂∂

= −∂

∂=

∂ ∂

Substituting in Eq. (2.9) 4 0φ∇ = so that the stress function satisfies the biharmonic equation.

On the boundary, y = +c

0y xywb

σ τ= − =

At y = —c 0 0y xyσ τ= = At x = 0

2 33 (12 20 )

40xw c y ybc

σ = −

Then

2 33

2 2 43

d (12 20 )d40

[6 5 ]400

c cxc c

cc

wy c y y ybcw c y ybc

σ− −

−

= −

= −

=

∫ ∫

i.e. the direct stress distribution at the end of the cantilever is self-equilibrating. The axial force at any section is

2 2 33

2 2 2 2 43

d (12 30 20 )d40

[6 15 5 ]400

c cxc c

cc

wy c y x y y ybcw c y x y ybc

σ− −

−

= + −

= + −

=

∫ ∫

i.e. no axial force at any section of the beam.


The bending moment at x =0 is

2 2 43

2 3 53

d (12 20 )d40

[4 4 ] 040

c cxc c

cc

wy y c y y ybcw c y ybc

σ− −

−

= −

= − =

∫ ∫

i.e. the beam is a cantilever beam under a uniformly distributed load of w/unit area with a self-equilibrating stress application at x = 0.

S.2.6

From physics, the strain due to a temperature rise T in a bar of original length L0

and final length L is given by

0 0

0 0

(1 )L L L T TL L

αε α

− += = =

Thus for the isotropic sheet, Eqs (1.52) become

1 ( )

1 ( )

x x y

y y x

v TE

v TE

ε σ σ α

ε σ σ α

= − +

= − +

Also, from the last of Eqs (1.52) and (1.50)

2(1 )xy xy

vE

γ τ+=

Substituting in Eq. (1.21)

2 2 22 22 2

2 2 2 2 2 22(1 ) 1 1xy y yx xv T Tv v

E x y E Ex x x y y yτ σ σσ σ

α α ∂ ∂ ∂∂ ∂+ ∂ ∂ = − + + − + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

or

2 2 22 2

22 2 2 22(1 ) xy y yx xv v v E T

x y x y x yτ σ σσ σ

α∂ ∂ ∂∂ ∂

+ = + − − + ∇∂ ∂ ∂ ∂ ∂ ∂

(i)

From Eqs (1.6) and assuming body forces X =Y =0

2 2 22

2 2xy xy yx

y x x yx yτ τ σσ∂ ∂ ∂∂

= − = −∂ ∂ ∂ ∂∂ ∂

Hence

2 22

2 22 xy yx

x y x yτ σσ∂ ∂∂

= − −∂ ∂ ∂ ∂

26 Solutions Manual

and

2 22

2 22 xy yxv v vx y x yτ σσ∂ ∂∂

= − −∂ ∂ ∂ ∂


2 22 2

22 2 2 2

y yx x E Tx y x y

σ σσ σα

∂ ∂∂ ∂− − = + + ∇∂ ∂ ∂ ∂

Thus

2 2

22 2 ( ) 0x y E T

x yσ σ α

∂ ∂+ + + ∇ = ∂ ∂

and since

2 2

2 2 (seeEqs (2.8))x yy xφ φσ σ∂ ∂

= =∂ ∂

2 2 2 2

22 2 2 2 0E T

x y y xφ φ α

∂ ∂ ∂ ∂+ + + ∇ = ∂ ∂ ∂ ∂

or

2 2( ) 0E Tφ α∇ ∇ + =

S.2.7


3

33

4 4Qxy Qxy

a aφ = −

Then

2

2

2

2 3

2 2

3

0

32

3 34 4

y

x

xy

xQxy

y a

Q Qyx y a a

φ σ

φ σ

φ τ

∂= =

∂∂

= − =∂

∂= − = −

∂ ∂

Also

4 4 4

4 4 2 20 0 0x y x yφ φ φ∂ ∂ ∂= = =

∂ ∂ ∂ ∂

so that Eq. (2.9), the biharmonic equation, is satisfied.


When x = a, σx = –3Qy/2a2

Then, when , i.e. linear.

0 03232

x

x

x

yQy aaQy aa

σ

σ

σ

= =

= − =

−= + =

Also, when x = –a, σx = 3Qy/2a2

, i.e. linear and when

0 032

32

x

x

x

yQy aa

Qy aa

σ

σ

σ

= =

−= − =

= + =

The shear stress is given by (see above)

2

23 1 , i.e. parabolic4xyQ ya a

τ

= − −

so that, when y = ±a, τxy = 0 and when y = 0, τxyThe resultant shear force at x = ±a is

= –3Q/4a.

2

23 1 d4

a

a

Q y ya a−

= − −

∫

i.e.

SF = Q.

The resultant bending moment at x = ±a is

2

3

d

3 d2

axa

a

a

y y

Qay ya

σ−

−

=

=

∫

∫

i.e. BM = –Qa



S.3.1

Initially the stress function, φ, must be expressed in terms of Cartesian coordinates. Thus, from the equation of a circle of radius, a, and having the origin of its axes at its centre.

2 2 2( )k x y aφ = + − (i)

From Eqs (3.4) and (3.11)

2 2

2 2d2d

F Gzx y

φ φ θ∂ ∂+ = = −

∂ ∂ (ii)

Differentiating Eq. (i) and substituting in Eq. (ii)

d4 2d

k Gzθ

= −

or

1 d2 d

k Gzθ

= − (iii)

From Eq. (3.8) 2 d dT x yφ= ∫ ∫

i.e.

2 2 2d d d d d d dd A A A

T G x x y y x y a x yzθ = − + − ∫ ∫ ∫ ∫ ∫ ∫ (iv)

where ∫∫A x2 dx dy = Iy, the second moment of area of the cross-section about the y axis; ∫∫A y2 dx dy = Ix, the second moment of area of the cross-section about the x axis and ∫∫A dx dy = A, the area of the cross-section. Thus, since Iy = πa4/4, Ix = πa4/4 and A = πa2

Eq. (iv) becomes 4d

d 2aT G

zθ π

=

or

4d 2d p

T Tz GIG aθ

π= = (v)

From Eqs (3.2) and (v)

d2dzy

p

Txkx G xx z Iφ θτ ∂

= − = − = =∂

(vi)

28 Solutions Manual

and

d2d

yzx

p

Tky G y

y z Iφ θτ ∂

= = = − = −∂

(vii)

Substituting for τzy and τzx

from Eqs (vi) and (vii) in the second of Eqs (3.15)

( )zsp

T xl ymI

τ = + (viii)

in which, from Eqs (3.6)

d dd dy xl ms s

= = −

Suppose that the bar of Fig. 3.2 is circular in cross-section and that the radius makes an angle α with the x axis. Then.

sin and cosm lα α= =

Also, at any radius, r

sin cosy r x rα α= =

Substituting for x, l, y and m in Eq. (viii) gives

( )zsp

TrI

τ τ= =

Now substituting for τzx, τzy

and dθ/dz from Eqs (vii), (vi) and (v) in Eqs (3.10)

0p p

w Ty Tyx GI GI

∂= − + =

∂ (ix)

0p p

w Tx Txy GI GI

∂= − + =

∂ (x)

The possible solutions of Eqs (ix) and (x) are w = 0 and w = constant. The latter solu-tion implies a displacement of the whole bar along the z axis which, under the given loading, cannot occur. Therefore, the first solution applies, i.e. the warping is zero at all points in the cross-section.

The stress function, φ, defined in Eq. (i) is constant at any radius, r, in the cross-section of the bar so that there are no shear stresses acting across such a boundary. Thus, the material contained within this boundary could be removed without affecting the stress distribution in the outer portion. Therefore, the stress function could be used for a hollow bar of circular cross-section.


S.3.2

In S.3.1 it has been shown that the warping of the cross-section of the bar is everywhere zero. Then, from Eq. (3.17) and since dθ/dz ≠ 0 ( , ) 0x yψ = (i)

This warping function satisfies Eq. (3.20). Also Eq. (3.21) reduces to xm – yl = 0 (ii)

On the boundary of the bar x = al, y = am so that Eq. (ii), i.e. Eq. (3.21), is satisfied. Since ψ = 0, Eq. (3.23) for the torsion constant reduces to

2 2d d d d pA AJ x x y y x y I= + =∫ ∫ ∫ ∫

Therefore, from Eq. (3.12)

ddpT GI

zθ

=

as in S.3.1. From Eqs (3.19)

d ( )dzx

p

TyG yz Iθτ = − = −

and

d ( )dzy

p

TxG xz Iθτ = =

which are identical to Eqs (vii) and (vi) in S.3.1. Hence

zsp

TrI

τ τ= =

as in S.3.1.

S.3.3

Since ψ = kxy, Eq. (3.20) is satisfied. Substituting for ψ in Eq. (3.21)

(kx + x)m + (ky – y)l = 0

or, from Eqs (3.6)

d d( 1) ( 1) 0d dx yx k y ks s

− + + − =

30 Solutions Manual

or

2 2d ( 1) ( 1) 0

d 2 2x yk k

s − + + − =

so that

2 2

( 1) ( 1) constant on the boundary of the bar2 2x yk k− + + − =

Rearranging

2 21 constant1

kx yk

− + = + (i)

Also, the equation of the elliptical boundary of the bar is

2 2

2 2 1x ya b

+ =

or

2

2 2 22

ax y ab

+ = (ii)

Comparing Eqs (i) and (ii)

2

211

a kkb

− = +

from which

2 2

2 2b aka b

−=

+ (iii)

and

2 2

2 2b a xya b

ψ −=

+ (iv)

Substituting for ψ in Eq. (3.23) gives the torsion constant, J, i.e.

2 2 2 2

2 22 2 2 21 1 d d

A

b a b aJ x y x ya b a b

− −= + − − + + ∫ ∫ (v)

Now ∫∫A x2 dxdy = Iy = na3b/4 for an elliptical cross-section. Similarly ∫∫A y2 dx dy = Ix = nab3

/4. Equation (v) therefore simplifies to 3 3

2 2a bJ

a bπ

=+

(vi)

which are identical to Eq. (v) of Example 3.1. From Eq. (3.22) the rate of twist is

2 2

3 3d ( )d

T a bz G a bθ

π+

= (vii)


The shear stresses are obtained from Eqs (3.19), i.e.

2 2 2 2

3 3 2 2( )

zxGT a b b a y y

G a b a bτ

π

+ −= − +

so that

32

zxTyab

τπ

= −

and

2 2 2 2

3 3 2 2( )

zyGT a b b a x x

G a b a bτ

π

+ −= + +

i.e.

32

zyTxa b

τπ

=

which are identical to Eq. (vi) of Example 3.1. From Eq. (3.17)

2 2 2 2

3 3 2 2( )T a b b aw xyG a b a bπ

+ −= +

i.e.

2 2

3 3( ) (compare with Eq. (viii) of Example 3.1)T b aw xyG a bπ

−=

S.3.4


2 2 3 2 2d 1 1 2( ) ( 3 )d 2 2 27

G x y x xy az aθφ = − + − − −

(i)

Differentiating Eq. (i) twice with respect to x and y in turn gives

2

2

2

2

d 31d

d 31d

xGz ax

xGz ax

φ θ

φ θ

∂ = − − ∂

∂ = − − ∂

Therefore

2 2

2 2d2 constantd

Gzx y

φ φ θ∂ ∂+ = − =

∂ ∂

and Eq. (3.4) is satisfied.

32 Solutions Manual

Further

on AB,3

2on BC,3 3 3

2on AC,3 3 3

ax y y

x ay

x ay

−= =

−= +

= −

Substituting these expressions in turn in Eq. (i) gives AB BC AC 0φ φ φ= = =

so that Eq. (i) satisfies the condition φ = 0 on the boundary of the triangle. From Eqs (3.2) and (i)

2 2d 3 3

d 2 2zyx yG x

x z a aφ θτ

∂= − = − + ∂

(ii)

and

d 3dzx

xyG yy z aφ θτ ∂ = = − + ∂

(iii)

At each corner of the triangular section τzy = τzx

= 0. Also, from antisymmetry, the distribution of shear stress will be the same along each side. For AB, where x = –a/3 and y = y, Eqs (ii) and (iii) become

2d 3d 2 2zy

a yGz aθτ

= − +

(iv)

and 0zxτ = (v) From Eq. (iv) the maximum value of τzy

occurs at y = 0 and is d(max)

2 dzyGa

zθτ = = − (vi)

The distribution of shear stress along the x axis is obtained from Eqs (ii) and (iii) in which x = x, y = 0, i.e.

2d 3d 2

0

zy

zx

xG xz aθτ

τ

= −

=

(vii)

From Eq. (vii) τzy

has a mathematical maximum at x = +a/3 which gives d

6 dzyGa

zθτ = (viii)


which is less than the value given by Eq. (vi). Thus the maximum value of shear stress in the section is (–Ga/2)dθ/dz.

The rate of twist may be found by substituting for φ from Eq. (i) in (3.8). Thus

2 2 3 2 2d 1 1 22 ( ) ( 3 ) d dd 2 2 27

T G x y x xy a x yz aθ = − + − − − ∫ ∫ (ix)

The equation of the side AC of the triangle is ( 2 / 3) / 3y x a= − and that of BC,

( 2 / 3) / 3.y x a= − − Equation (ix) then becomes

2 /3 ( 2 /3)/ 3 2 2 3 2 2

/3 ( 2 /3)/ 3

d 1 1 22 ( ) ( 3 ) d dd 2 2 27

a x a

a x aT G x y x xy a x y

z aθ − −

− −

= − + − − − ∫ ∫

which gives

4 d

d15 3GaT

zθ

=

so that

4d 15 3d

Tz Gaθ= (x)

From the first of Eqs (3.10)

dd

zxw yx G z

τ θ∂= +

∂

Substituting for τzx

from Eq. (iii) d 3d

w xyy yx z a

θ∂ = − + − ∂

i.e.

3 dd

w xyx a z

θ∂= −

∂

whence

23 d ( )

2 dx yw f ya z

θ= − + (xi)

Similarly from the second of Eqs (3.10)

2 33 d d ( )

2 d 2 dx y yw f xa z a z

θ θ= − + + (xii)

Comparing Eqs (xi) and (xii)

3 d( ) 0 and ( )

2 dyf x f ya z

θ= =

34 Solutions Manual

Hence

3 21 d ( 3 ).2 d

w y x ya z

θ= −

S.3.5

The torsion constant, J, for the complete cross-section is found by summing the torsion constants of the narrow rectangular strips which form the section. Then, from Eq. (3.29)

3 3 3(2 )2

3 3 3at bt a b tJ +

= + =

Therefore, from the general torsion equation (3.12)

3d 3d (2 )

Tz G a b tθ=

+ (i)

The maximum shear stress follows from Eqs (3.28) and (i), hence

max 2d 3 .d (2 )

TGtz a b tθτ = ± = ±

+


S.4.1

Give the beam at D a virtual displacement δD as shown in Fig. S.4.1. The virtual displacements of C and B are then, respectively, 3δD/4 and δD/2.

Fig. S.4.1 The equation of virtual work is then

D DD D

2 3 02 4

W WR δ δδ − − =

from which 1.75DR W= It follows that 1.25 .AR W=


S.4.2

The beam is given a virtual displacement δC at C as shown in Fig. S.4.2.

Fig. S.4.2

The virtual work equation is then

CC C C0

3d 0

4LW xR w x

Lδ

δ δ − − = ∫

from which

C3 2

4W wLR +

=

so that

A2 .

4W wLR +

=

S.4.3

The beam is given a virtual rotation θA at A as shown in Fig. S.4.3.

Fig. S.4.3

The virtual work equation is then

AA A A2 0

2WLM WLθ

θ θ− − =

from which 2.5AM WL= and 3 .AR W=

36 Solutions Manual

S.4.4

Give the beam virtual rotations α and β at A and B, respectively as shown in Fig. S.4.4. Then, at C, (3L/4) α = (L/4) β so that β = 3α.

Fig. S.4.4

The relative rotation of AB and BC at C is (α + β) so that the equation of virtual

work is MC3 /4

0 3 /4d 3 ( )d

L L

Lw x x w L x xα α+ −∫ ∫(α +β) =

i.e.

3 /4

0 3 /44 d 3 ( )d

L LC L

M w x x L x xα α = + − ∫ ∫

from which

23 .

32CwLM =

S.4.5

Suppose initially that the portion GCD of the truss is given a small virtual rotation about C so that G moves a horizontal distance δG and D a vertical distance δD as shown in Fig. S.4.5(a).

Fig. S.4.5


Then, since CG = CD, δG = δD

and the equation of virtual work is

D20GFGδ δ=

so that FG = +20 kN The virtual displacement given to G corresponds to an extension of FG which, since the calculated value of FG is positive, indicates that FG is tensile.

Now suppose that GD is given a small virtual increase in length δGD as shown in Fig. S.4.5(b). The vertical displacement of D is then δGD

/cos45° and the equation of virtual work is

GD GDGD 20 / cos 45δ δ= °

from which GD = +28. 3 kN (tension)

Finally suppose that CD is given a small virtual extension δCD as shown in Fig. S.4.5(c). The corresponding extension of GD is δCD

cos 45°. Then the equation of virtual work is, since the 20 kN load does no work

CD CDCD GD cos 45 0δ δ+ ° =

Substituting for GD from the above gives CD = –20 kN (compression).

S.4.6

First determine the deflection at the quarter-span point B. Then, referring to Fig. S.4.6 the bending moment due to the actual loading at any section is given by

Fig. S.4.6

2 2

A( )

2 2 2wLx wx w Lx xM −

= − =

and due to the unit load placed at B is

1 13 ( )in AB and in BD4 4x L xM M −

= =

38 Solutions Manual

Then substituting in Eq. (4.20)

/4 2 3 2

B 0 /43( )d ( )( )d

8L L

L

w Lx x x Lx x L x xEI

υ = − + − − ∫ ∫

which gives

4

B576144

wLEI

υ =

For the deflection at the mid-span point the bending moment at any section due to the actual loading is identical to the expression above. With the unit load applied at C

1 1( )in AC and in CD

2 2x L xM M −

= =


/2 2 3 2 2

C 0 /2( )d ( )( )d

4L L

L

w Lx x x Lx x L x xEI

υ = − + − − ∫ ∫

from which

4

C5 .384

wLEI

υ =

S.4.7

From inspection the support reactions at A and F are both equal to wL. Also, since the beam is symmetrical about a vertical through the mid-point of CD only half need be considered as shown in Fig.S.4.7.

Fig.S.4.7


Suppose that the actual force in the member EB is T , assumed to be tensile. Suppose also that, with the actual loading removed, we apply a virtual unit load at the mid-point of EB in the direction EB. Then

In FE MA = wLx – wx2/2, M1 F

= 0 A = F1

In ED M

= 0

A = (wL2/2) – Ty, M1 F

= –1.y A = F1

In DG M

= 0

A = (wL2/2) – Ty, M1 F

= –1.(L/2) A = –T, F1

In EB M

= –1

A = M1 F

= 0 A = T, F1

Then, from Eqs(4.12) and (4.20) the internal virtual work is given by

= 1

Wi = (1/EI){∫0L/2[(wL2/2) – Ty](–y)dy + ∫0L/2[(wL2/2) – (TL/2)](–L/2)dx1}

(TL/2AE)

+ (TL/6AE) +

which gives Wi = (1/EI)[(TL3/6) – (3wL4

The external virtual work done by the unit load is 1.∆ where ∆ is the displacement of the mid-point of BE in the direction of BE. However, since this point lies on the axis of symmetry ∆= 0 so that W

/16)] + (2TL/3AE) (i)

e = Wi

T = 9wL

= 0. Therefore, from Eq.(i) 3/8[L2

Then, M(at B) = wL + (4I/A)]

2

M(at C) = (TL – wL

/2 (anticlockwise) 2

S.4.8 )/2 (clockwise)

Suppose a horizontal force, T, is applied to each end of the spring. Consider part of the left hand loop as shown in Fig.S.4.8.

40 Solutions Manual

Fig.S.4.8

At any section B where the radius makes an angle θ with the horizontal the internal

forces and moments are as shown. Then, resolving forces in a direction parallel to N N = Tsinθ Also, taking moments about B M = T(rsin45o

From geometry 2rcos45

+ rsinθ) = Tr[(1/√2) + sinθ] o

Now apply a virtual unit load at A. Then = L/n so that r = L/(√2)n.

MA = Tr[(1/√2 + sinθ), M1

N

= 1.r[(1/√2) + sinθ]

A = Tsinθ, N1

From Eqs(4.21) and (4.12) the internal virtual work for the complete spring is = 1.sinθ

Wi = n{∫-π/45π/4(Tr2/EI)[(1/√2) + sinθ]2rdθ + ∫–π/4

5π/4(T/AE)sin2

Integrating θ rdθ}

Wi = n{(Tr3/EI)[θ – √2 cosθ – (sin2θ)/4]–π/45π/4 + (Tr/AE)[(θ/2) – (sin2θ)/4]–π/4

5π/4

which gives W

}

i = n[(Tr3

But r = L/(√2)n, I = πd

/2EI)(3π + 3) + (Tr/8AE)(7π – 4)]

4/64 and A = πd2

W

/4. Substituting in the above

i = (T/πE√2 d2)[(48L3/n2d2)(π+ 1) + L(3π – 2)]


The external work done by the unit load is We = 1.δA where δA

δ

is the horizontal displacement of A. Equating the internal and external virtual work gives

A = (T/πE√2 d2)[(48L3/n2d2

The spring stiffness, k, is equal to T/δ

)(π + 1) + L(3π – 2)]

A

k = (√2 πEd

so that 2)/L[(48L2/n2d2

S.4.9

)(π + 1) + (3π – 2)]

The frame is symmetrical about its vertical centre line so that only half need be considered as shown in Fig.S.4.9.

Fig.S.4.9

Suppose N is the force in the tie bar. We shall use the principle of virtual work to calculate the horizontal displacement of the mid-point of the tie bar so that, with the actual loading removed, we apply a horizontal unit load at the mid-point of the tie bar in the direction of N. Also, the angle COA = 90o + sin–1(710/1020) = 134.1o. For ease of computation we shall take the angle COA to be 135o

At any section where the radius makes an angle θ with the vertical .


MAi.e.

= 44.5(1020sinθ – 440) + 8.9(910 – 1020cosθ) – N(710 – 1020cosθ)

MA M

= 45 390sinθ – 11 481 – 9078cosθ – N(710 – 1020cosθ) 1

Note that in DA, M

= 1.(710 – 1020cosθ)

1 = N1 = 0 and in BA, MA = M1 = 0, NA = N, N1

W

= 1. From Eqs(4.21) and (4.12) the internal virtual work is given by

i = (1/EI)∫π/4π

Integrating and substituting the limits etc gives

[45 390sinθ – 11 481 – 9078cosθ – N(710 – 1020cosθ)](710 – 1020cosθ)]1020dθ + 710N/AE

Wi = 129 070N – 2.04×106

The external work done is (i)

We

where ∆ is the horizontal displacement of the mid-point of AB. However, from symmetry, ∆ = 0 so that, equating the external and internal virtual work

= 1.∆ (ii)

0 = 129 070N – 2.04 × 10from which

6

N = 15.8kN


S.5.1

This problem is most readily solved by the application of the unit load method. Therefore, from Eq. (5.20), the vertical deflection of C is given by

0 1,VV,C

F F LAE

∆ =∑ (i)

and the horizontal deflection by

0 1,HH,C

F F LAE

∆ =∑ (ii)

in which F1,V and F1,H are the forces in a member due to a unit load positioned at C and acting vertically downwards and horizontally to the right, in turn, respectively. Further, the value of L/AE (= 1/20 mm/N) for each member is given and may be omitted from the initial calculation. All member forces (see Table S.5.1) are found using the method of joints which is described in textbooks on structural analysis, for example, Structural and Stress Analysis by T. H. G. Megson (Elsevier, 2005).

44 Solutions Manual

Table S.5.1 Member F0 F(N) 1, FV 1, FH 0F F1,V 0F1,H DC 16.67 1.67 0 27.84 0 BC –13.33 –1.33 1.0 17.73 –13.33 ED 13.33 1.33 0 17.73 0 DB –10.0 –1.0 0 10.0 0 AB –16.67 –1.67 0.8 27.84 –13.34 EB 0 0 0.6 0 0 Σ = 101.14 Σ = –26.67

Note that the loads F1,V

are obtained most easily by dividing the loads F by a factor of 10. Then, from Eq. (i)

V,C1101.4 5.07 mm20

∆ = × =

which is positive and therefore in the same direction as the unit vertical load. Also from Eq. (ii)

H,C126.67 1.33mm20

∆ = − × = −

which is negative and therefore to the left. The actual deflection, Δ, is then given by

2 2V,C H,C 5.24mm∆ = ∆ + ∆ =

which is downwards and at an angle of tan–1

S.5.2 (1.33/5.07) = 14.7° to the left of vertical.

Figure S.5.2 shows a plan view of the plate. Suppose that the point of application of the load is at D, a distance x from each side of the plate. The deflection of D may be found using the unit load method so that, from Eq. (5.20), the vertical deflection of D is given by

0 1D

F F LAE

∆ =∑ (i)

Initially, therefore, the forces, F0, must be calculated. Suppose that the forces in the wires at A, B and C due to the actual load are F0,A, F0,B and F0,C

, respectively. Then resolving vertically

0,A 0,B 0,C 100F F F+ + = (ii)

Taking moments about the edges BC, AC and AB in turn gives

0,A 4 100F x× = (iii)

0,B 4 sin 100F A x× × =


Fig. S.5.2

i.e.

0,B 4 0.6 100F x× × = (iv)

and

0,C 3 100F x× = (v)

Thus, from Eqs (iii) to (v)

0,A 0,B 0,C4 2.4 3F F F= =

so that

0,A 0,B 0,C 0,B0.6 0.8F F F F= =

Substituting in Eq. (ii) gives

0,B 41.7NF =

Hence

0,A 0,C25.0N and 33.4NF F= =

Now apply a unit load at D in the direction of the 100 N load. Then

1,A 1,B 1,C25.0 0.417 0.334F F F= = =

Substituting for F0,A, F1,A,

etc. in Eq. (i)

D 21440 (25 0.25 41.7 0.417 33.4 0.334)

( /4) 1 196000π∆ = × + × + ×

× ×

i.e.

D 0.33mm∆ =

46 Solutions Manual

S.5.3

Suppose that joints 2 and 7 have horizontal and vertical components of displacement u2, υ2, u7, and υ7

, respectively as shown in Fig. S.5.3. The displaced position of the member 27 is then 2′7′. The angle α which the member 27 makes with the vertical is then given by

1 7 2

7 2tan

3u u

aα

υ υ− −

=+ −

Fig. S.5.3

which, since α is small and υ7 and υ2

are small compared with 3a, may be written as

7 2

3u u

aα −= (i)

The horizontal components u2 and u7

may be found using the unit load method, Eq. (5.20). Thus

0 1,2 0 1,72 7

F F L F F Lu u

AE AE= =∑ ∑ (ii)

where F1,2 and F1,7

are the forces in the members of the framework due to unit loads applied horizontally, in turn, at joints 2 and 7, respectively. The solution is completed in tabular form (Table S.5.3). Substituting the summation terms in Eqs (ii) gives

2 7192 5703 9

Pa Pau uAE AE

= − =

Now substituting for u2 and u7 in Eq. (i)


3829

PAE

α =

48 Solutions Manual

Table S.5.3 Member Length F F0 F1,2 F1,7 0F1,2 FL 0F1,7L 27 3a 3P 0 0 0 0 87 5a 5P /3 0 5/3 0 125Pa/9 67 4a –4P /3 0 – 4/3 0 64Pa/9 21 4a 4P – 4/3 0 – 64Pa/3 0 23 5a 0 5/3 0 0 0 26 5a –5P 0 0 0 0 38 3a 0 0 0 0 0 58 5a 0 0 0 0 0 98 5a 5P /3 0 5/3 0 125Pa/9 68 3a 0 0 0 0 0 16 3a 3P 0 0 0 0 56 4a – 16P /3 0 – 4/3 0 256Pa/9 13 3a 0 0 0 0 0 43 5a 0 5/3 0 0 0 93 34a 0 0 0 0 0

03 5a 0 0 0 0 0 15 5a –5P 0 0 0 0 10 4a 8P – 4/3 0 – 128Pa/3 0 Σ = – 192Pa/3 Σ = 570Pa/9

S.5.4

(a) The beam is shown in Fig. S.5.4. The principle of the stationary value of the total complementary energy may be used to determine the deflection at C. From Eq. (5.13)

CdddL

MP

θ∆ = ∫ (i)

Fig. S.5.4

in which, since the beam is linearly elastic, dθ = (M/EI)dz. Also the beam is symmetrical about its mid-span so that Eq. (i) may be written

/2

C 0

d2 dd

L M M zEI P

∆ = ∫ (ii)

In AC


2PM z=

so that

dd 2M zP=

Eq. (ii) then becomes

2 2/4 /2

0 /42 d d

442

L LC L

Pz Pzz zEI EI

∆ = +

∫ ∫ (iii)

Integrating Eq. (iii) and substituting the limits gives

3

C3

128PL

EI∆ =

(b) When the beam is encastré at A and F, fixed end moments MA and MF are induced. From symmetry MA = MF

. The total complementary energy of the beam is, from Eq. (4.18)

C0d d

M

LC M Pθ= − ∆∫ ∫

from which

A A

d 0L

C MM M

θ∂ ∂= =

∂ ∂∫ (iv)

from the principle of the stationary value. From symmetry the reactions at A and F are each P/2. Hence

A A(assuming is a hogging moment)2PM z M M= −

Then

A

1MM∂

= −∂

Thus, from Eq. (iv)

/2

0A A2 d 0

LC M M zM EI M∂ ∂

= =∂ ∂∫

or

/4 /2

A A0 /4

1 10 2 ( 1)d ( 1)d( / 2) 2 2

L L

L

P Pz M z z M zEI EI

= − − + − − ∫ ∫

from which

A548PLM =

50 Solutions Manual

S.5.5

The unit load method, i.e. the first of Eqs (5.21), may be used to obtain a solution. Thus

0 1C,H dM M z

EIδ = ∫ (i)

in which the M1

moments are due to a unit load applied horizontally at C. Then, referring to Fig. S.5.5, in CB

0 1( cos ) 1M W R R M zθ= − = × and in BA 0 12 1M W R M z= = ×

Fig. S.5.5

Hence, substituting these expressions in Eq. (i) and noting that in CB ds = R dθ and in BA ds = dz

{ }43C,H 0 0

1 (1 cos )sin d 2 dR

WR WRz zEI

πδ θ θ θ= − − +∫ ∫

i.e.

2

3 2 4C,H 0

0

1 coscos [ ]2

RWR WR zEI

πθδ θ

= − − + +

so that

3

C,H14WR

EIδ = (ii)

The second moment of area of the cross-section of the post is given by

4 4 6 4(100 94 ) 1.076 16 mm64

I π= − = ×


Substituting the value of I and the given values of W and R in Eq. (ii) gives

C,H 53.3mmδ =

S.5.6

Either of the principles of the stationary values of the total complementary energy or the total potential energy may be used to solve this problem.

From Eq. (5.12) the total complementary energy of the system is

0

d d dM

L LC M w zθ υ= −∫ ∫ ∫ (i)

in which w is the load intensity at any point in the beam and υ the vertical displacement. Equation (i) may be written in the form

0

d d dM

L L

MC z M w zEI

υ= −∫ ∫ ∫

since, from symmetrical bending theory

z M zR EIδδθ δ= =

Hence

2

d d2L L

MC z w zEI

υ= −∫ ∫ (ii)

Alternatively, the total potential energy of the system is the sum of the strain energy due to bending of the beam plus the potential energy V, of the applied load. The strain energy U, due to bending in a beam may be shown to be given by

2

d2L

MU zEI

= ∫

Hence

2

TPE d d2L L

MU V z w zEI

υ= + = −∫ ∫ (iii)

Eqs (ii) and (iii) are clearly identical. Now, from symmetrical bending theory

2

2dd

MEI z

υ= −

Therefore Eq. (ii) (or (iii)) may be rewritten

22

20 0

d d d2 d

L LEIC z w zzυ υ

= −

∫ ∫ (iv)

52 Solutions Manual

Now

01 2

22sin sin 12

w zz z za a wL L L Lπ πυ = + = −

so that

2 2 2

1 22 2 2d 4 2sin sind

z za aL Lz L L

υ π π π π= − −

Substituting in Eq. (iv)

24 2 2

1 24 2 20

2 20

1 2 1 20

4 2sin sin d2

2 2 2sin sin sin sin d2 2

L

L

EI z zC a a zL LL L L

w z z z z z za z a z a a zL L L L L L L

π π π π π

π π π π

= +

− + − −

∫

∫

which, on expanding, gives

42 2 2 21 1 2 24 0

2 20

1 2 1 20

2 2sin 8 sin sin 16 sin d22 2 2sin sin sin sin d

2 2

L

L

EI z z z zC a a a a zL L L LL

w z z z z z za z a z a a zL L L L L L L

π π π π π

π π π π

= + +

− + − −

∫

∫ (v)

Eq. (v) may be integrated by a combination of direct integration and integration by parts and gives

24

2 2 012 1 02 3

1 482 22

a w La LEIC a L a w LLπ

π ππ = + − + +

(vi)

From the principle of the stationary value of the total complementary energy

1 2

0 and 0C Ca a∂ ∂

= =∂ ∂

From Eq. (vi)

4

201 3 3

10 ( 4)

2w LC EIa

a Lπ π

π∂

= = − +∂

Hence

4

201 7

2 ( 4)w LaEI

ππ

= +

Also

4

02 3

2

802w LC EIa

a Lπ

π∂

= = +∂

whence

4

02 516

w LaEIπ

= −


The deflected shape of the beam is then

4

207 5

2 1 2( 4)sin sin16

w L z zEI L L

π πυ ππ π = + −

At mid-span when z = L/2

4

00.00918w LEI

υ =

S.5.7

This problem is solved in a similar manner to P.5.6. Thus Eq. (iv) of S.5.6 is directly applicable, i.e.

22

2 d d2L L

EI dC z w zdzυ υ

= −

∫ ∫ (i)

in which

1

sinii

i zaLπυ

∞

=

=∑ (ii)

and w may be expressed as a function of z in the form w = 4w0z(L – z)/L2 which satisfies the boundary conditions of w = 0 at z = 0 and z = L and w = W0

From Eq. (ii) at z = L/2.

2 2 2

2 21

d sind i

i

i i zaLz L

υ π π∞

=

= −∑


4 4

2 2 04 20 0

1 1

4sin d ( ) sin d

2L L

i ii i

wEI i i z i zC a z z L z a zL LL L

π π π∞ ∞

= =

= − −∑ ∑∫ ∫ (iii)

Now

2

0 0 0

3

0 0

22

00

3 3

3 3

1 2 2sin d 1 cos d sin2 2 2 2

sin d cos cos d cos

sin d cos cos 2 d

2cos (cos

LL L

LL

LL

i z i z z L i z Lz zL L i L

i z zL i z L i z LLz z L z iL i L i L i

i z z L i z L i zz z z zL i L i L

L Lii i

π π ππ

π π π ππ π π

π π ππ π

ππ π

= − = − =

= − + = −

= − +

= − +

∫ ∫

∫ ∫

∫ ∫

1)iπ −

54 Solutions Manual

Thus Eq. (iii) becomes

2 4 4 3 3 3

03 2 3 3

1 1

4 2cos cos (cos 1)4i

ii i

EIa i w L L LC a i i ii iL L i

ππ π π

π π π

∞ ∞

= =

= − − + − −

∑ ∑

or

2 4 4 3

03 2 3 3

1 1

4 2(1 cos )

4i i

i i

EIa i w a LC iL L iπ

ππ

∞ ∞

= =

= − −∑ ∑ (iv)

The value of (1 – cos iπ) is zero when i is even and 2 when i is odd. Therefore Eq. (iv) may be written

2 4 4

03 3 3

16is odd

4i iEIa i w a LC iL iπ

π= −


4 4

03 3 3

160

2i

i

EIa i w LCa L i

ππ

∂= − =

∂

Hence

4

07 7

32i

w LaEIi π

=

Then

4

07 7

1

32sin isodd

i

w L i z iLEIiπυ

π

∞

=

=∑

At the mid-span point where z = L/2 and using the first term only in the expression for υ

4

0m.s 94.4

w LEI

υ =

S.5.8

The lengths of the members which are not given are:

12 13 14 249 2 15 13 5L a L a L a L a= = = =

The force in the member 14 due to the temperature change is compressive and equal to 0.7A. Also the change in length, Δ14, of the member 14 due to a temperature change T is L14αT = 13a × 2.4 × 10–6

From the unit load method, Eq. (5.20)

T. This must also be equal to the change in length produced by the force in the member corresponding to the temperature rise. Let this force be R.

0 114

F F LAE

∆ =∑ (i)


In this case, since R and the unit load are applied at the same points, in the same direction and no other loads are applied when only the temperature change is being considered, F0 = RF1

. Equation (i) may then be written 2

114

F LRAE

∆ = ∑ (ii)

The method of joints may be used to determine the F1

forces in the members. Thus

14 13 12 24 2335 16 2 20 281

13 13 13 13F F F F F− −

= = = = =


2 2 2 2 2

14 2 2 221 13 35 15 (16 2) 9 2 20 5 28 3

13 13 1313 2a a a a aR

AE AE AE AEAE

× × × × ×∆ = + + + +

or

3 2 2 2 214 2 (13 35 15 16 18 20 5 28 3)

13Ra

AE∆ = + × + × + × + ×

i.e.

14 22953213

aRAE

∆ =

Then

62

29532 (0.7 )13 24 1013

a Aa TAE

−× × =

so that T = 5.6°

S.5.9

Referring to Figs P.5.9(a), (b) and S.5.9 it can be seen that the members 12, 24 and 23 remain unloaded until P has moved through a horizontal distance 0.25 cosα, i.e. a distance of 0.25 × 600/750 = 0.2 mm. Therefore, until P has moved through a hori-zontal distance of 0.2 mm P is equilibrated solely by the forces in the members 13, 34 and 41 which therefore form a triangular framework. The method of solution is to find the value of P which causes a horizontal displacement of 0.2 mm of joint 1 in this framework.

Using the unit load method, i.e. Eq. (5.20) and solving in tabular form (see Table S.5.9(a)).

Then

1425.00.2300 70000

P=

×

56 Solutions Manual

Fig. S.5.9

Table S.5.9(a) Member Length (mm) F F0 F1 0F1L 13 750 1.25P 1.25 1171.9P 14 450 –0.75P –0.75 253.1P 43 600 0 0 0 Σ = 1425.0P

from which P = 2947 N The corresponding forces in the members 13, 14 and 43 are then F13 = 3683. 8 N F14 = –2210. 3 N F43

When P= 10000N additional forces will be generated in these members corre-sponding to a load of P′ = 10 000 – 2947 = 7053 N. Also P′ will now produce forces in the remaining members 12, 24 and 23 of the frame. The solution is now completed in a similar manner to that for the frame shown in Fig.5.8 using Eq. (5.16). Suppose that R is the force in the member 24; the solution is continued in Table S.5.9(b). From Eq. (5.16)

= 0

2592R + 1140P′ = 0

Table S.5.9(b) Member Length (mm) F ∂F/∂R FL(∂F/∂R) 12 600 –0.8R –0.8 384R 23 450 –0.6R –0.6 162R 34 600 –0.8R –0.8 384R 41 450 –(0.6R + 0.75P′) –0.6 162R + 202.5P′ 13 750 R + 1.25P′ 1.0 750R + 937.5P′ 24 750 R 1.0 750R Σ = 2592R+ 1140P'


so that

1140 70532592

R ×= −

i.e. R = –3102N Then

F12F

= –0. 8 × (–3102) = 2481. 6N (tension) 23

F = –0. 6 × (–3102) = 1861. 2N (tension)

34F

= –0. 8 × (–3102) = 2481. 6N (tension) 41

F = –0. 6 × (–3102) – 0. 75 × 7053 – 2210. 3 = –5638. 9 N (compression)

13F

= –3102 + 1. 25 × 7053 + 3683. 8 = 9398. 1 N (tension) 24

S.5.10

= –3102. 0 N (compression)

Referring to Fig. S.5.10(a) the vertical reactions at A and D are found from statical equilibrium. Then, taking moments about D

2 1 2A 3 2 3 0R l lw l+ =

i.e.

( )A downwards2wlR = −

Hence

( )D upwards2wlR =

Also for horizontal equilibrium

A D2wlH H+ = (i)

The total complementary energy of the frame is, from Eq. (5.12)

A,H A A,V D D,V D,V0 0d d d

M lA DL

C M H R H R w zθ ′= − ∆ − ∆ − ∆ − ∆ + ∆∫ ∫ ∫ (ii)

in which ΔA,H, ΔA,V, ΔD,H and ΔD,V are the horizontal and vertical components of the displacements at A and D, respectively and Δ is the horizontal displacement of the member AB at any distance z from A. From the principle of the stationary value

58 Solutions Manual

Fig. S.5.10(a)

of the total complementary energy of the frame and selecting ΔA,H

as the required displacement

A,HA A

d 0L

C MH H

θ∂ ∂= − ∆ =

∂ ∂∫ (iii)

In this case ΔA,H

= 0 so that Eq. (iii) becomes

Ad 0

L

MH

θ ∂=

∂∫

or, since dθ = (M/EI)dz

A

d 0L

M M zEI H

∂=

∂∫ (iv)

In AB

3

AA6

wz MM H z zl H

∂− − = −

∂

In BC

2

A AA6

wl MM R z H l lH∂

= − − = −∂

In DC

D AA

from Eq.(i),2wl MM H z H z z

H∂ = − = − + = − ∂


Substituting these expressions in Eq. (iv) gives

3 21 2 /3

A A0 0

A0

1 1( )d ( )d2 6 2 6

1 ( )d 02 2

l

i

wz wl wlH z z z z H l l zEI l El

wlH z z zEI

− − − + − − − −

+ − − − =

∫ ∫

∫

or

4 2 31 2 /32 2A A0 0

22

A0

1 d d2 6 2 6

1 d 02 2

l

l

wz wl wlH z z z H l zl

wlzH z z

+ + + +

+ + =

∫ ∫

∫

from which

3 4A

292 045

H l wl+ =

or A 29 /90H wl= −

Hence, from Eq. (i) D 8 /45H wl=

Thus

3

3AB A

296 90 6

wz wl wM H z z zl l

= − − = −

Fig. S.5.10(b)

60 Solutions Manual

When z = 0, MAB = 0 and when z = l, MAB = 7wl2/45. Also, dMAB

/dz = 0 for a turning value, i.e.

2ABd 29 3 0

d 90 6M wl wz

z l= − =

from which 29 / 45 .z l= Hence MAB(max) = 0.173wl2

The bending moment distributions in BC and CD are linear and M.

B = 7wl2/45, MD = 0 and MC = HDl = 8wl2

The complete bending moment diagram for the frame is shown in Fig. S.5.10(b). /45.

S.5.11

The bracket is shown in Fig. S.5.11 in which RC is the vertical reaction at C and MC is the moment reaction at C in the vertical plane containing AC.

Fig. S.5.11

From Eq. (5.12) the total complementary energy of the bracket is given by

C C C C A0 0d d d

M T

L LC M d T M R Pθ φ θ= + − − ∆ − ∆∫ ∫ ∫ ∫

in which T is the torque in AB producing an angle of twist, φ, at any section and the remaining symbols have their usual meaning. Then, from the principle of the stationary value of the total complementary energy and since θC = ΔC

= 0

C C Cd d 0

L L

C M M T Tz zR EI R GJ R∂ ∂ ∂

= + =∂ ∂ ∂∫ ∫ (i)

and

C C C

d d 0L L

C M M T Tz zM EI M GJ M∂ ∂ ∂

= + =∂ ∂ ∂∫ ∫ (ii)

From Fig. S.5.11 AC C 1 C AC 0M R z M T= − =


so that

AC AC AC AC1

C C C C1 0

M M T TzR M R M

∂ ∂ ∂ ∂= = − = =

∂ ∂ ∂ ∂

Also AB 2 C 2 C( 4 cos ) cosM Pz R z a Mα α= − + − + i.e.

AB 2 C 2 C16 4

5 5aM Pz R z M = − + − +

Hence

AB AB2

C C

16 45 5

M MazR M

∂ ∂= − =

∂ ∂

Finally AB C C4 sin sinT R a Mα α= − i.e.

AB C C12 3

5 5aT R M= −

so that

AB AB

C C

12 35 5

T TaR M

∂ ∂= = −

∂ ∂

Substituting these expressions in Eq. (i)

4 5

C 1 C 1 1 2 C 2 C0 0

1 1 16 4( ) d1.5 5 5

a a aR M z z Pz R z MzEI EI − + − + − +

∫ ∫

5

2 2 C C 20

16 1 12 3 12d d 05 3 5 5 5

aa az z R M zGI

× − + − = ∫ (iii)

Note that for the circular section tube AC the torsion constant J (i.e. the polar second moment of area) = 2 × 1.5I from the theorem of perpendicular axes.

Integrating Eq. (iii), substituting the limits and noting that G/E = 0.38 gives C C55.17 16.18 1.11 0R a M Pa− − = (iv) Now substituting in Eq. (ii) for MAC, ∂MAC/∂MC

, etc. 4 5

C 1 C 1 2 C 2 C 20 0

1 1 16 4 4( )( 1)d d1.5 5 5 5

a a aR z M z Pz R z M zEI EI

− − + − + − + ∫ ∫

5

C C 20

1 12 3 3 d 03 5 5 5

a a R M zGI

+ − − = ∫ (v)

62 Solutions Manual

from which C C16.58 7.71 6.67 0R a M Pa− + = (vi) Solving the simultaneous Eqs (iv) and (vi) gives RC

S.5.12

= 0.72P

Suppose that R is the tensile force in the member 23, i.e. R = xP0

. Then, from Eq. (5.15)

0ii

FR

λ∂

=∂∑ (i)

in which, for members 12, 23 and 34

0

1n

i i ii i

LLE

τ τλ ε

τ

= = +

(ii)

But τi = Fi/Ai

so that Eq. (ii) may be written

01

ni i i

ii i i

F L FA E A

λτ

= +

(iii)

For members 15, 25, 35 and 45 which are linearly elastic

i ii

i

F LA E

λ = (iv)

The solution is continued in Table S.5.12. Summing the final column in Table S.5.12 gives

04 2 3 8 2 16[1 ( ) ] [1 ( ) ] 03 3 3 3

n nRL RL L R RLx x PAEAE AE AE

α α

+ + + + + + =

(v)

from Eq. (i) Noting that R = xP0

, Eq. (v) simplifies to 164 [1 ( ) ] 6 [1( ) ] 8 16 0

3n n xx x x x xα α+ + + + + =

or

1610 ( ) 10 16 8 03

nx x xα

+ + + + =

from which

1 3.5 0.80 0n nx xα + + + =


Table S.5.12 Member L Ai Fi ∂Fi i λ/∂R λi i∂Fi/∂R 12 2L / 3A / 3R 1/ 3

0

2 1n

RL RAE Aτ

+

2 [1 ( ) ]3

nRL xAE

α+

23 2 / 3L

A R 1

0

2 3 1n

RL RAE Aτ

+

2 3 [1 ( ) ]nRL xAE

α+

34 2L / 3A / 3R 1/ 3

0

2 1n

RL RAE Aτ

+

2 [1 ( ) ]3

nRL xAE

α+

15 2L A 0 2 / 3P R− −

2/ 3− 0( 2 / 3)2P R L

AE+

− 04 ( 2 / 3)3

L P RAE

+

25 2L / 3A 2 3R− 2/ 3− 4RL

AE− 8

3RLAE

35 2L / 3A 2 3R− 2/ 3− 4RLAE

− 83RLAE

45 2L A 0 2 / 3P R− −

2/ 3−

02 ( 2 / 3)L P RAE

− + 04 ( 2 / 3)3

L P RAE

+

S.5.13

Suppose that the vertical reaction between the two beams at C is P. Then the force system acting on the beam AB is as shown in Fig. S.5.13. Taking moments about B A 9.15 6.1 100 30.5 0R P× + × − × =

Fig. S.5.13 so that RA

The total complementary energy of the beam is, from Eq. (5.12) = 33.3 – 0.67P

C F0d d 100 0

M

LC M Pθ= − ∆ − ∆ =∫ ∫

64 Solutions Manual

where ∆C and ∆F

are the vertical displacements at C and F, respectively. Then, from the principle of the stationary value of the total complementary energy of the beam

Cd 0L

C MP P

θ∂ ∂= = ∆ =

∂ ∂∫

whence, as in previous cases

C dL

M M zEI P

∂∆ =

∂∫ (i)

In AC

AC A (33.3 0.67 )M R z P z= = −

so that

AC 0.67M z

P∂

= −∂

In CF

CF A ( 30.5) 33.3 (0.33 3.05)M R z P z z P z= + − = + −

from which

CF 0.33 3.05M z

P∂

= −∂

In FB

FB A ( 30.5) 100( 6.1) 66.7 610 (0.33 3.04)ZM R P z z z P z= + − − − = − + + −

which gives

FB 0.33 3.04M zP

∂= −

∂

Substituting these expressions in Eq. (i)

3.05

C 0(33.3 0.67 ) ( 0.67 )dEI P z z z∆ = − −∫

6.1

3.059.15

6.1

[33.3 (0.33 30.5)](0.33 3.05)d

[ 66.7 610 (0.33 3.05)](0.33 3.05)d

z P z z z

z P z z z

+ + − −

+ − + + − −

∫∫


which simplifies to

3.05 2 2

C 0( 22.2 0.44 )dEI z Pz z∆ = − +∫

6.1 2 2

3.059.15 2 2

6.1

(10.99 0.11 20.2 9.3 101.6 )d

( 22.01 404.7 0.11 2.02 9.3 1860.5)d

z Pz Pz P z z

z z Pz Pz P z

+ + − + −

+ − + + − + −

∫∫

Integrating this equation and substituting the limits gives C 12.78 1117.8EI P∆ = − (ii)

From compatibility of displacement, the displacement at C in the beam AB is equal to the displacement at C in the beam ED. The displacement at the mid-span point in a fixed beam of span L which carries a central load P is PL3/192EI Hence, equating this value to ∆C in Eq. (ii) and noting that ∆C

in Eq. (ii) is positive in the direction of P 36.1(12.78 1117.8)

192P P− − = ×

which gives P = 80.1 kN Thus

3 3 9

C 680.1 10 6.1 10

192 200 000 83.5 10× × ×

∆ =× × ×

i.e.

∆C

Note: The use of complementary energy in this problem produces a rather lengthy solution. A quicker approach to finding the displacement ∆

= 5.6 mm

C

S.5.14

in terms of P for the beam AB would be to use Macauley’s method (see, e.g. Structural and Stress Analysis by T. H. G. Megson (Elsevier, 2005)).

The internal force system in the framework and beam is statically determinate so that the unit load method may be used directly to determine the vertical displacement of D. Hence, from the first of Eqs (5.21) and Eq. (5.20)

,0 ,10 1D,V

1

dk

i i i

L i ii

F F LM M zEI A E=

∆ = +∑∫ (i)

66 Solutions Manual

Fig. S.5.14

Referring to Fig. S.5.14 and taking moments about A

2

G,H(8 )3 1.5 3 12 0

2aR a w wa a− − =

from which RG,H

Hence = 28wa

RA,H

From the vertical equilibrium of the support G, R

= – 28wa G,V

R

= 0, so that, resolving vertically

A,V

i.e. – 1.5w8a – 3wa = 0

RA,V

With a unit vertical load at D = 15wa

G,H A,H A,V G,V4 4 1 0R R R R= = − = =

For the beam ABC, in AB

2

210 A,V 1 1 1 1 1

1.5 15 0.75 12wzM R waz wz M zz= − = − = ×

and in BC

20 2 2 1 215 0.75 1M waz wz M z= − = ×

Hence

4 42 3 2 30 1

1 1 1 2 2 22 0 0

16d (15 0.75 )d (15 0.75 )da a

L

M M z waz wz z waz wz zEI Aa E

= − + − ∫ ∫ ∫


Suppose z1 = z2

= z say, then 44 2 3 3 40 1

2 20 0

16 32 0.75d 2 (15 0.75 )d 54

aa

L

M M wz waz wz z az zEI Aa E Aa E

= − = − ∫ ∫

i.e.

2

0 1 8704dL

M M wazEI AE

=∫

The solution is continued in Table S.5.14.

Table S.5.14

Member L A F F0 F1 0F1L/A AB 4a 4A 28wa 4 112wa2/A BC 4a 4A 28wa 4 112wa2

CD /A

4a A 4wa 4/3 64wa2

DE /3A

5a A –5wa –53 125wa2

EF /3A

4a A –4wa –43 64wa2

FG /3A

4a A –28wa -4 448wa2

CE /A

3a A 3wa 1 9wa2

CF /A

5a A –30wa –10/3 500wa2

BF /A

3a A 18wa 2 108wa2

/A

∑ = 4120wa2/3A

Thus

2 2

D8704 4120

3wa wa

AE AE∆ = +

i.e.

2

D30232

3wa

AE∆ =

S.5.15

The internal force systems at C and D in the ring frame are shown in Fig. S.5.15. The total complementary energy of the half-frame is, from Eq. (5.12)

B0d d

M

LC M Fθ= − ∆∫ ∫

in which ∆B

is the horizontal displacement of the joint B. Note that, from symmetry, the translational and rotational displacements at C and D are zero. Hence, from the principle of the stationary value of the total complementary energy and choosing the horizontal displacement at C (=0) as the unknown

C Cd 0

L

C M M zN EI N∂ ∂

= =∂ ∂∫ (i)

68 Solutions Manual

In CB CB C C 1( cos )M M N r r θ= − − (ii) At B, MCB

= 0. Thus

C C C( sin 30 ) 1.5M N r r N r= + ° = (iii)

Eq. (ii) then becomes CB C 1(0.5 cos )M M r θ= + (iv)

Then

CB1

C(0.5 cos )

M rN

θ∂

= +∂

(v)

In DB DB D D 2( cos )M M N r r θ= − − (vi)

Fig. S.5.15

Again the internal moment at B is zero so that D D D( sin 30 ) 0.5M N r r N r= − ° = (vii) Hence DB D 2(cos 0.5)M N r θ= − (viii) Also, from horizontal equilibrium D CN N F+ =

so that D CN F N= −


and Eq. (viii) may be written DB C 2( ) (cos 0.5)M F N r θ= − − (ix) whence

DB2

C(cos 0.5)M r

Nθ

∂= − −

∂ (x)

Substituting from Eqs (iv), (v), (ix) and (x) in Eq. (i)

120 603 2 3 2C

C 1 1 2 20 0

( )1 (0.5 cos ) d (cos 0.5) d 0F NN r r

EI xEIθ θ θ θ

° ° −+ − − =∫ ∫

i.e. 120 602 2C

C 1 1 1 2 2 20 0

( )(0.25 cos cos )d (cos cos 0.25)d 0

F NNx

θ θ θ θ θ θ° °−

+ + − − + =∫ ∫

which, when expanded becomes

120C1

1 10

602

2 20

( )cos 20.75 cos d2

cos 2 cos 0.75 d 02

CF NN

xθ

θ θ

θθ θ

°

°

− + + −

× − + =

∫

∫

Hence

120 60

C1 2C 1 1 2 2

0 0

( )sin 2 sin 20.75 sin sin 0.75 04 4

F NNx

θ θθ θ θ θ

° °− + + − − + =

from which

CC

( )2.22 0.136 0

F NNx−

− = (xi)

The maximum bending moment in ADB is equal to half the maximum bending moment in ACB. Thus 1

D C2M M=

Then, from Eqs (vii) and (iii) D C0.5 0.75N r N r=

so that C C0.5( ) 0.75F N N− =

i.e. C C1.5F N N− =

70 Solutions Manual

Substituting for F –NC

in Eq. (xi)

CC

1.52.22 0.136 0

NNx

− × =

whence x = 0. 092

S.5.16

From symmetry the shear force in the tank wall at the lowest point is zero. Let the normal force and bending moment at this point be NO and MO, respectively as shown in Fig. S.5.16.

Fig. S.5.16

The total complementary energy of the half-tank is, from Eq. (5.12)

P0d d

2M

L

PC Mθ= − ∆∫ ∫

where ∆P

is the vertical displacement at the point of application of P. Since the rotation and translation at O are zero from symmetry then, from the principle of the stationary value of the total complementary energy

O Od 0

L

C M M zM EI M∂ ∂

= =∂ ∂∫ (i)

and

O O

d 0L


= =∂ ∂∫ (ii)


At any point in the tank wall

2O O 0

( cos ) sin( )dM M N r r prθ

θ θ φ φ= + − − −∫ (iii)

For unit length of tank p = πr2

where ρ is the density of the fuel. ρ

At the position θ, ( cos )p h r rρ ρ φ= = +

Hence

(1 cos )Ppr

φπ

= +

and the last term in Eq. (iii) becomes

0 0

(1 cos )sin( )d (1 cos )(sin cos cos sin )dPr Prθ θφ θ φ φ φ θ φ θ φ φ

π π+ − = + −∫ ∫

Expanding the expression on the right-hand side gives

2

0(sin cos cos sin sin cos cos sin cos )d

1 sin cos2

Pr

Pr

θθ φ θ φ θ φ θ φ φ φ

πθ θ θ

π

− + −

= + −

∫

Hence Eq. (iii) becomes

O O (1 cos ) 1 sin cos2

PrM M N r θθ θ θπ

= + − − + −

(iv)

so that

O O

1 and (1 cos )M M rM N

θ∂ ∂= = −

∂ ∂

Substituting for M and ∂M/∂MO

in Eq. (i) and noting that EI= constant,

O O0(1 cos ) 1 sin cos d 0

2PrM N r

π θθ θ θ θπ

+ − − + − = ∫ (v)

from which

O O3 02PrM N rπ

+ − = (vi)

Now substituting for M and ∂M/∂NO

in Eq. (ii)

O O0(1 cos ) 1 sin cos (1 cos )d 0

2PrM N r r

π θθ θ θ θ θπ

+ − − + − − = ∫

72 Solutions Manual

The first part of this integral is identical to that in Eq. (v) and is therefore zero. The remaining integral is then

O O0(1 cos ) 1 sin cos cos d 0

2PrM N r

π θθ θ θ θ θπ

+ − − + − = ∫

which gives

O 5 02 8

N Prπ

− =

Hence O 0.398N P=

and from Eq. (vi) O 0.080M Pr=

Substituting these values in Eq. (iv) (0.160 0.080cos 0.159 sin )M Pr θ θ θ= − −

S.5.17

The internal force systems at A and B are shown in Fig. S.5.17; from symmetry the shear forces at these points are zero as are the translations and rotations. It follows that the total complementary energy of the half-frame is, from Eq. (5.12)

0

d dM

LC Mθ= ∫ ∫

Fig. S.5.17


B B

d 0L

C M M zM EI M∂ ∂

= =∂ ∂∫ (i)


and

B Bd 0

L


= =∂ ∂∫ (ii)

In BC

2

0B 2

p zM M= + (iii)

so that

B B1 0M M

M N∂ ∂

= =∂ ∂

In CA

2

20B B 0 0

( sin )sin cos ( cos )2 2 2

pa aM M N a p a a p a aθθ θ θ = − + − + + −

which simplifies to

2

0B B sin

2p aM M N a θ= − + (iv)

Hence

B B1 sinM M a

M Nθ∂ ∂

= = −∂ ∂

Substituting for M and ∂M/∂MB

in Eq. (i) 2 2/2

0 0B B B0 2

1 1d sin d 02 2 2

a p z p aM z M N a aEI EI

πθ θ

+ + − + =

∫ ∫

i.e.

/23 2

0 0B B

0 0

1 cos 02 6 2

a

Bp z p aM z a M N a

π

θ θ

+ + + + =

which simplifies to 2

B B 02.071 0.869 0M N a p a− + = Thus

2B B 00.483 0.420 0M N a p a− + = (v)

Now substituting for M and ∂M/∂NB

in Eq. (ii) 2/2

0B B0

1 sin ( sin ) d 02

p aM N a a aEI

πθ θ θ

− + − =

∫

or

74 Solutions Manual

2/2 2 0

B B0

1 sin sin sin d 02

p aM N aEI

πθ θ θ θ

− + =

∫

which gives 2

B B 00.785 0.5 0M N a p a− + = (vi) Subtracting Eq. (vi) from Eq. (v) 2

B 00.302 0.08 0N a p a− = so that B 00.26N p a=

Substituting for NB

in Eq. (v) gives 2

B 00.292M p a= − Therefore, from Eq. (iii)

2 2

20 0C B 00.292

2 2p a p aM M p a= + = − +

i.e. 2

C 00.208M p a= and from Eq. (iv)

2

2 2 0A 0 00.292 0.265

2p aM p a p a= − − +

i.e. 2

A 00.057M p a= − Also, from Eq. (iii)

2 20BC 00.292

2pM p a z= − + (vii)

At a point of contraflexure MBC = 0. Thus, from Eq. (vii), a point of contraflexure occurs in BC when z2 = 0.584a2, i.e. when z = 0.764a. Also, from Eq. (iv), MCA

S.5.18

= 0 when sin θ = 0.208/0.265 = 0.785, i.e. when θ = 51.7°.

Consider the half-frame shown in Fig. S.5.18(a). On the plane of antisymmetry through the points 7, 8 and 9 only shear forces S7, S8 and S9

S

are present. Thus from the horizontal equilibrium of the frame

7 + S8 + S9 –6aq = 0 (i)


Fig. S.5.18(a)

Also, from the overall equilibrium of the complete frame and taking moments about the corner 6 2aq6a + 6aq2a – 2P3a = 0 which gives q = P/4a

The total complementary energy of the half-frame is, from Eq. (5.12)

5 60d d 0

M

LC M P Pθ= − ∆ − ∆ =∫ ∫

Noting that the horizontal displacements at 7, 8 and 9 are zero from antisymmetry, then

7 7

d 0L

C M M zS EI S∂ ∂

= =∂ ∂∫ (ii)

and

8 8

d 0L


= =∂ ∂∫ (iii)

In 74

M = S7z1 and ∂M/∂S7 = z1 ∂M/∂S8

In 45

= 0

M = S7a + qaz2 and ∂M/∂S7 = a ∂M/∂S8

In 85 = 0

M = S8z3 and ∂M/∂S7 = 0 ∂M/∂S8 = zIn 56

3

M = S7a + S8a + qa(3a + z4) – Pz4 and ∂M/∂S7= a ∂M/∂S8 = a

76 Solutions Manual

In 69 M = S7 (a – z5) + S8 (a – z5) + 6a2q – 3Pa +6aqz

and

5

∂M/∂S7 = (a – z5) ∂M/∂S8 = (a – z5

Substituting the relevant expressions in Eq. (ii) gives )

3 32 2 2

7 1 1 7 2 2 7 8 4 4 40 0 0d ( )d [ (3 ) ] d

a a aS z z S a qa z z S a S a qa a z Pz a z+ + + + + + −∫ ∫ ∫

27 5 8 5 5 5 50

[ ( ) ( ) 6 3 6 ]( )d 0a

S a z S a z a q Pa a q za z z+ − + − + − + − =∫ (iv)

from which 20S7 + 10S8

Now substituting for M and ∂M/∂S + 66aq –18P = 0 (v)

8

in Eq. (iii) 32

8 3 3 7 8 4 4 40 0d [ (3 ) ] d

a aS z z S a S a qa a z Pz a z+ + + + −∫ ∫

27 5 8 5 5 5 50

[ ( ) ( ) 6 3 6 ]( ) 0a

S a z S a z a q Pa aqz a z dz+ − + − + − + − =∫ (v)

The last two integrals in Eq. (vi) are identical to the last two integrals in Eq. (iv). Thus, Eq. (vi) becomes

7 810 11 52.5 18 0S S aq P+ + − = (vii) The simultaneous solution of Eqs (v) and (vii) gives

839 312 2

S aq P= − +

whence, since q = P/4a 8 0.69S P=

Substituting for S8

S

in either of Eqs (v) or (vii) gives 7

Then, from Eq. (i) = – 0.27P

S9

The bending moment diagram is shown in Fig. S.5.18(b) in which the bending moments are drawn on the tension side of each member.

= 1.08P


Fig. S.5.18(b)

S.5.19

From the overall equilibrium of the complete frame

2

0d

rqr s T

π=∫

which gives

22 r q Tπ =

i.e.

22Tqrπ

= (i)

Fig. S.5.19

78 Solutions Manual

Considering the half frame shown in Fig. S.5.19 there are only internal shear forces on the vertical plane of antisymmetry. From the vertical equilibrium of the half-frame

1 2 3 0sin d 0S S S q r

πα α+ + + =∫

Substituting for q from Eq. (i) and integrating

1 2 3 0[ cos ] 02TS S S

rπα

π+ + + − =

which gives

1 2 3TS S Srπ

+ + = − (ii)

The vertical displacements at the points 1, 2 and 3 are zero from antisymmetry so that, from Eq. (5.12), the total complementary energy of the half-frame is given by

0

d dM

LC Mθ= ∫ ∫

Then, from the principle of the stationary value of the total complementary energy

1 1

dL


=∂ ∂∫ (iii)

and

2 2

dL


=∂ ∂∫ (iv)

In the wall 14

1 0sin [ cos( )] dM S r q r r r

θθ θ α α= − − −∫

i.e.

1 0sin [ sin( )]2TM S r θθ α α θπ

= − − −

which gives

1 sin ( sin )2TM S r θ θ θπ

= − − (v)

whence

1 2

sin 0M MrS S

θ∂ ∂= =

∂ ∂

In the wall 24 M = S2x (vi)


and

1 2

0M M xS S

∂ ∂= =

∂ ∂

In the wall 43

1 2sin ( sin ) sin2TM S r S rθ θ θ θπ

= − − + (vii)

and

1 2

sin sinM Mr rS S

θ θ∂ ∂= =

∂ ∂

Substituting for M and ∂M/∂S1

in Eq. (iii) 3 /4

10

1 23 /4

sin ( sin ) sin d2

sin ( sin ) sin sin d 02

TS r r r

TS r S r r r

π

π

π

θ θ θ θ θπ

θ θ θ θ θ θπ

− − + − − + =

∫

∫

which simplifies to

2 31 20 3 /4

sin ( sin ) sin d sin d 02TS r r S r

π π

πθ θ θ θ θ θ θ

π − − + = ∫ ∫

Integrating and simplifying gives 1 20.16 0.09 0S r T S r− + = (viii) Now substituting for M and ∂M/∂S2

in Eq. (iv) / 2 2

1 2 23 /4 0sin ( sin ) sin sin d d 0

2rTS r S r r r S x x

π

πθ θ θ θ θ θ

π − − + + = ∫ ∫

Integrating and simplifying gives 1 20.69 1.83 0S r T S r− + = (ix) Subtracting Eq. (ix) from Eq. (viii) 20.53 1.74 0T S r− =

whence 2

0.30TSr

=

From Eq. (viii)

10.13TS

r=

80 Solutions Manual

and from Eq. (ii)

30.75TS

r−

=

Hence, from Eqs (v) to (vii)

14

24

43

(0.29sin 0.16 )0.30

(0.59sin 0.16 )

M TTxM

rM T

θ θ

θ θ

= −

=

= −

S.5.20

Initially the vertical reaction at C, RC

, must be found. From Eq. (5.12) the total complementary energy of the member is given by

C C B0d d

M

LC M R Fθ= − ∆ − ∆∫ ∫

From the principle of the stationary value of the total complementary energy and since ∆C

= 0

C Cd 0

L

C M M sR EI R∂ ∂

= =∂ ∂∫ (i)

Referring to Fig. S.5.20

Fig. S.5.20

In BC

Csin and 0MM Fr

Rθ ∂

= =∂

In CD C

Cand MM Fr R z z

R∂

= − = −∂


Substituting these expressions in Eq. (i) gives

C0( )( )d 0

rFr R z z z− − =∫

from which C 1.5R F=

Note that Eq. (i) does not include the effects of shear and axial force. If these had been included the value of RC

The unit load method may now be used to complete the solution. Thus, from the first of Eqs (5.21), Eq. (5.20) andEq. (20.18)

would be 1.4F; the above is therefore a reasonable approximation. Also, from Eq. (1.50), G = 3E/8.

0 1 0 1 0 1B,H d d d

'L L L

M M F F S Ss s sEI AE GA

δ = + +∫ ∫ ∫ (ii)

In BC

0 1

0 1

0 1

sin sinsin sincos cos

M Fr M rF F FS F S

θ θθ θθ θ

= =

= =

= =

In CD

0 1

0 1

0 1

( 1.5 ) ( 1.5 )1

1.5 1.5

M F r z M r zF F FS F S

= − = −

= =

= =

Substituting these expressions in Eq. (ii) gives

3 2 2 2/2 /2 /2

B,H 0 0 0

2

0 0 0

sin sin cosd d d'

2.25( 1.5 ) d d d'

r r r

Fr Fr FrEI AE GA

F F Fr z z z zEI AE GA

π π πθ θ θδ θ θ θ= + +

+ − + +

∫ ∫ ∫

∫ ∫ ∫

or

/2 /2

B.H 0 0

/2 2 220 0

0 0

400 1 1(1 cos 2 )d (1 cos 2 )d2 2

32 1 400(1 cos 2 )d ( 3 2.25 )d3 2

24d d

r

r

r r

FFrAE AE

Fr F r rz z zAE Ar E

F Fz zAE AE

π π

π

δ θ θ θ θ

θ θ

= − + −

+ + + − +

+ +

∫ ∫

∫ ∫

∫ ∫

from which B,H

448.3FrAE

δ =

82 Solutions Manual

S.5.21

From Clerk–Maxwell’s reciprocal theorem the deflection at A due to W at B is equal to the deflection at B due to W at A, i.e. δ2

What is now required is the deflection at B due to W at B. .

Since the deflection at A with W at A and the spring removed is δ3, the load in the spring at A with W at B is (δ2/δ3

)W which must equal the load in the spring at B with W at B. Thus, the resultant load at B with W at B is

2 2

3 31W W Wδ δ

δ δ

− = −

(i)

Now the load W at A with the spring in place produces a deflection of δ1 at A. Thus, the resultant load at A is (δ1/δ3)W so that, if the load in the spring at A with W at A is F, then W – F = (δ1/δ3

)W, i.e.

1

31F W δ

δ

= −

(ii)

This then is the load at B with W at A and it produces a deflection δ2

. Therefore, from Eqs (i) and (ii) the deflection at B due to W at B is

2

32

1

3

1

1

W

W

δδ

δδδ

−

−

Thus the extension of the spring with W at B is

2

32 2

1

3

1

1

δδ

δ δδδ

−

− −

i.e.

1 22

3 1

δ δδδ δ−

−

S.5.22

Referring to Fig. S.5.22 RA = RBThe slope of the beam at A and B may be obtained from the second of Eqs (16.32), i.e.

= 1000 N from symmetry.

" MEI

υ = −


Fig. S.5.22

where, for the half-span AF, M = RAZ

= 1000z. Thus 1000 z

EIυ′′ = −

and

21

500 z CEI

υ′ = − +

When z = 720 mm, υ′ = 0 from symmetry and hence C1 = 2. 59 × 108

/EI Hence

2 81' ( 500 2.59 10 )zEI

υ = − + ×

Thus υ′ (at A) = 0.011 rads = υ′ (at B). The deflection at C is then = 360 × 0. 011 = 3.96 mm and the deflection at D = 600 x 0.011 = 6.6 mm.

From the reciprocal theorem the deflection at F due to a load of 3000 N at C = 3.96 × 3000/2000 = 5.94 mm and the deflection at F due to a load of 3000N at D = 6.6 × 3000/2000 = 9.9 mm. Therefore the total deflection at F due to loads of 3000 N acting simultaneously at C and D is 5.94 + 9.9 = 15.84 mm.

S.5.23

An elemental length, δz, of the beam at any section a distance z from the free end will bend as shown in Fig.S.5.23.

84 Solutions Manual

Fig.S.5.23

The temperature varies linearly from –t at the underside of the beam to +t at the top. Therefore the element of the beam will have a length δz at its mid-height. Then, if R is the radius of curvature of the beam at the section

(R/δz) = [R + (h/2)]/δz(1 + αt)


which gives

R = h/2αt Also

δθ = δz/R = δz2αt/h and t = t0

Then, from Eq.(5.32) z/L

∆free end = ∫0L(2αt0/hL)z2

which gives dz

∆ free end = (2αt0L2

S.5.24

)/3h

Since the frame is symmetrical about a vertical plane through its centre only half need be considered. Also, due to symmetry the frame will act as though fixed at C (Fig. S.5.24).

If the frame were unsupported at B the horizontal displacement at B, ∆B,T, due to the temperature rise may be obtained using Eq. (5.32) in which, due to a unit load acting horizontally at B, M1

= 1 × (r sin 30° + r sin θ). Hence /2

B,T /6

2(0.5 sin ) dTr r rd

π

π

αθ θ−

∆ = +∫

i.e.

2

/2B,T /6

2 [0.5 cos ]Trd

ππ

α θ θ −∆ = −

Fig. S.5.24

86 Solutions Manual

which gives

2

B,T3.83 Tr

dα

∆ = (to the right) (i)

Suppose that in the actual frame the horizontal reaction at B is HB. Since B is not displaced, the ‘displacement’ ∆B,H produced by HB must be equal and opposite to ∆B,T in Eq. (i). Then, from the first of Eqs (5.21) and noting that M0 = – HB

(0.5r + r sin θ) /2 2

B,H B/6

1 (0.5 sin ) dH r r rEI

π

πθ θ

−∆ = − +∫

i.e.

3 /2 2B

B,H /6(0.25 sin sin )dH r

EIπ

πθ θ θ

−∆ = − + +∫

Hence

/23

BB,H

/6

sin 20.75 cos4

H rEI

π

π

θθ θ−

∆ = − − −

so that

3

BB,H

2.22 (to the left)H rEI

∆ = − (ii)

Then, since B,H B,T 0∆ + ∆ =

3 2

B2.22 3.83 0H r TrEI d

α− + =

from which

B1.73EITH

dα

= (iii)

The maximum bending moment in the frame will occur at C and is given by M (max) = HB

Then, from symmetrical bending theory the direct stress through the depth of the frame section is given by

× 1.5r

( )( )see Eqs 16.21MyI

σ =

and

max(max) (max)M y

Iσ =


i.e.

Bmax

1.5 0.5H r dI

σ × ×=

or, substituting for HB

from Eq. (iii)

max 1.30ETσ α=

S.5.25

The solution is similar to that for P.5.23 in that the horizontal displacement of B due to the temperature gradient is equal and opposite in direction to the ‘displacement’ pro-duced by the horizontal reaction at B, HB


. Again only half the frame need be considered from symmetry.

M1

= r cos ψ in BC and Cd

Fig. S.5.24

Then, from Eq. (5.32)

/4 /2

0B,T 0 /4

cos 2 0( cos ) d ( cos ) dr r r rh h

π π

π

θ ψψ α ψ ψ α ψ ∆ = + ∫ ∫

i.e.

2 /4

0B,T 0

cos cos 2 drh

παθ ψ ψ ψ∆ = ∫

or

2 /4 20

B,T 0(cos 2sin cos )dr

hπαθ ψ ψ ψ ψ∆ = −∫

88 Solutions Manual

Hence

/42

30B,T

0

2sin sin3

rh

παθ ψ ψ ∆ = −

which gives

20

B,T0.47 (to the right)r

hαθ

∆ = (i)

From the first of Eqs (5.21) in which M0 = – HB

r cos ψ /2

BB,H 0

cos cos dH r r rEI

π ψ ψ ψ∆ = −∫

i.e.

3 /2 2B

B,H 0cos dH r

EIπ

ψ ψ∆ = − ∫

or

3 /2

BB,H 0

1 (1 cos 2 )d2

H rEI

πψ ψ∆ = − +∫

whence

3

BB,H

0.79 (to the left)H rEI

∆ = − (ii)

Then, since ΔB,H + ∆B,T

= 0, from Eqs (i) and (ii) 23

0B 0.470.79 0rH rEI h

αθ− + =

from which

0B

0.59EIHrhαθ

=

Then B cosM H r ψ= so that

00.59 cosEIMhαθ ψ

=



S.6.1

Referring to Fig. P.6.1 and Fig. 6.3

Member 12 23 34 41 13

Length 2

(cos ) 1/ 2 1/ 2 1/ 2 1/ 2 0

(sin ) 1/ 2 1/ 2 1/ 2 1/ 2 1

L L L L L

λ θ

µ θ

− −

− −

The stiffness matrix for each member is obtained using Eq. (6.30). Thus

12 23

34 41

1 1 1 1 1 1 1 11 1 1 1 1 1 1 1

[ ] [ ]1 1 1 1 1 1 1 12 21 1 1 1 1 1 1 1

1 1 1 1 1 1 1 11 1 1 1 1 1 1 1

[ ] [ ]1 1 1 1 1 1 1 12 21 1 1 1 1 1 1 1

AE AEK KL L

AE AEK KL L

− − − − − − − = = − − − − − − − −

− − − − − − − − = = − − − − − − − −

13

0 0 0 00 1 0 1

[ ]0 0 0 020 1 0 1

AEKL

− =

−

The stiffness matrix for the complete framework is now assembled using the method described in Example 6.1. Equation (6.29) then becomes

,11

,1 1

,2 2

,2 2

,3

,3

,4

,4

2 0 1 1 0 0 1 1 00 2 2 1 1 0 2 1 1

01 1 2 0 1 1 0 01 1 0 2 1 1 0 00 0 1 1 2 0 1 12

0 2 1 1 0 2 2 1 11 1 0 0 1 1 2 01 1 0 0 1 1 0 2

x

y

x

y

x

y

x

y

F uF

F uF AEF LF

FF

υ

υ

− − − = + − − − − =− − − =− − − = − − − − − + − − − − − − − −

3

3

4

4

00

00

u

uυ

υ

=

= =

(i)

In Eq. (i) ,1 ,1 ,3 ,3 0y x x yF P F F F= − = = =

86 Solutions Manual

Then

,1 1 3[(2 2) 2 ]2yAEF P

Lυ υ= − = + − (ii)

,3 1 30 [ 2 (2 2) ]2yAEF

Lυ υ= = − + + (iii)

From Eq. (iii) 1 3(1 2)υ υ= + (iv) Substituting for υ1

in Eq. (ii) gives

30.293PL

AEυ = −

Hence, from Eq. (iv)

10.707PL

AEυ = −

The forces in the members are obtained using Eq. (6.32), i.e.

12 14

0 0[1 1] from symmetry0.707 202

AE PS SPLL

AE

− = = =

+

13

0 0[0 1] 0.2930.293 0.707

2AES PPL PL

LAE AE

− = = − +

23 43

0 0[ 1 1] 0.207 from symmetry0.293 02

AES P SPLL

AE

− = − = − = − −

The support reactions are Fx,2, Fy,2, Fx,4 and Fy,4

. From Eq. (i)

,2 1 3

,2 1 3

,4 1 3

,4 1 3

( ) 0.2072

( ) 0.52

( ) 0.2072

( ) 0.52

x

y

x

y

AEF PL

AEF PL

AEF PL

AEF PL

υ υ

υ υ

υ υ

υ υ

= − + =

= − − =

= − − = −

= − − =

S.6.2 Referring to Fig. P.6.2 and Fig. 6.3

Member 12 23 34 31 24Length / 3 / 3 / 3(cos ) 0 1/2 1/23 / 2 3/2(sin ) 1/2 1 1/23/2 3/2

l ll l lλ θµ θ

−−− −

Solutions to Chapter 6 Problems 87 From Eq. (6.30) the member stiffness matrices are

12

3/4 3/43 3/4 3 3/43/4 3/43/4 3/4[ ]

3/4 3/43 3/4 3 3/43/4 3/43/4 3/4

AEKl

− − − − = −− − −

23

0 0 0 00 03 3[ ]0 0 0 00 03 3

AEKl

− = −

34

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/4 3/43/4 3/4

AEKl

− − −− = − −

− −

31

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/4 3/43/4 3/4

AEKl

− − −− = − −

− −

24

3/4 3/43 3/4 3 3/43/4 3/43/4 3/4[ ]

3/4 3/43 3/4 3 3/43/4 3/43/4 3/4

AEKl

− − − − = −−

− −

The stiffness matrix for the complete framework is now assembled using the method described in Example 6.1. Equation (6.29) then becomes

,1

,1

,2

,2

,3

,3

,4

,4

3 1 0 01 3 3 3 3 3 3 34 44 4 4 4

3 3 0 03 3 3 3 3 34 44 4 4 4

3 0 0 0 33 3 3 3 3 34 44 2 4

3 0 0 33 3 3 3 34 44 2 41 0 0 1 0 13 34 2 44 4

3 0 0 3 33 3 334 2 44 4

0 0 3 13 3 3 1 34 44 4

x

y

x

y

x

y

x

y

F

F

F

F AEF lF

F

F

+ + − −− −

+ + − −− −

−− − −− − − = − −−

− − + −− +−−

1

1

2

2

3

3

4

4

000

0

00

3 3 34 4

0 0 3 33 3 3 3 3 34 44 4 4 4

u

u

u

u

υ

υ

υ

υ

= = =

= = =

+ − + +−− −

(i)

88 Solutions Manual

In Eq. (i) Fx,2 = Fy,2 = 0, Fx,3 = 0, Fy,3 = – P,Fx,4

= –H. Then

,2 2 33 30 3

2yAEFl

υ υ

= = −

(ii)

and

,3 2 333 32y

AEF Pl

υ υ = − = − + +

(iii)

From Eq. (ii)

2 323

υ υ= (iv)

Now substituting for υ2

in Eq. (iii)

3 3 32 3 3 3

3 2PlAE

υ υ υ− = − + +

Hence

36

(9 2 3)Pl

AEυ = −

+

and, from Eq. (iv)

24

(9 2 3)Pl

AEυ = −

+

Also from Eq. (i)

,4 2 33 34 4x

AEF Hl

υ υ

= − = +

Substituting for υ2 and υ3

H = 0.449P

gives

S.6.3

Referring to Fig. P.6.3 and Fig. 6.3

Member 12 23 34 45 24Length(cos ) 1/2 1/2 1/2 1/2 1(sin ) 03/2 3/2 3/2 3/2

l l l l lλ θµ θ

− −


From Eq. (6.30) the member stiffness matrices are

12

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/4 3/43/4 3/4

AEKl

− − −− = − −

− −

23

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/43/4 3/4 3/4

AEKl

− − −− = − − − − − −

34

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/4 3/43/4 3/4

AEKl

− − −− = − −

− −

45

1/4 1/43/4 3/43/4 3/43/4 3/4[ ]

1/4 1/43/4 3/43/4 3/43/4 3/4

AEKl

− − −− = − −

− −

24

1 0 1 00 0 0 0

[ ]1 0 1 0

0 0 0 0

AEKl

− = −

The stiffness matrix for the complete truss is now assembled using the method described in Example 6.1. Equation (6.29) then becomes

,1

,1

,2

,2

,3

,3

,4

,4

,5

,5

1 1 0 0 0 0 0 03 33 3 0 0 0 0 0 03 3

1 6 0 1 4 0 0 03 33 0 6 3 0 0 0 03 3

0 0 1 2 0 1 0 03 30 0 3 0 6 3 0 04 3 30 0 4 0 1 6 0 13 30 0 0 0 3 0 6 33 30 0 0 0 0 0 1 13 30

x

y

x

y

x

y

x

y

x

y

F

F

F

F

F AEF l

F

F

F

F

− −

−− − − −−

− −− − − −= − −− − − − −

− − − − −

1

1

1

2

3

3

4

4

5

5

00

00

00 0 0 0 0 3 3 03 3

u

u

u

u

u

υ

υ

υ

υ

υ

= = = = = − =−

(i)

90 Solutions Manual

In Eq. (i) Fx,2 = Fy,2 = 0, Fx,4 = 0, Fy,4

== –P. Thus from Eq. (i)

,2 2 40 (6 4 )4xAEF u u

l= = − (ii)

,2 20 (6 )4yAEF

lυ= = (iii)

,4 2 40 ( 4 6 )4xAEF u u

l= = − + (iv)

,4 4(6 )4yAEF P

lυ= − = (v)

From Eq. (v)

423

PlAE

υ = −

From Eq. (iii)

2 0υ =

and from Eqs (ii) and (iv)

2 4 0u u= =

Hence, fromEq. (6.32)

24

0 0[1 0] 2 0

3

AES PllAE

− = −

−

which gives

24 0S =

S.6.4

The uniformly distributed load on the member 26 is equivalent to concentrated loads of wl/4 at nodes 2 and 6 together with a concentrated load of wl/2 at node 4. Thus, referring to Fig. P.6.4 and Fig. 6.3

Member 12 23 24 46 56 67Length /2 /2(cos ) 0 1 1 01 /2 1/ 2(sin ) 1 0 0 11/ 2 1/ 2

l l l l l lλ θµ θ

−


From Eq. (6.47) and using the alternative form of Eq. (6.44)

12 3

12 SYM0 06 0 4

[ ]12 0 6 120 0 0 0 06 0 0 6 0 0

EIKl

= − −

23 3

6 SYM6 6

46/ 2 6/ 2[ ]6 6 66 26 6 6 66 2

26/ 2 6/ 2 6/ 2 6 2 4/ 2

EIkl

= −

− − −

24 46 3

0 SYM0 960 24 8

[ ] [ ]0 0 0 00 96 24 0 960 24 4 6 24 8

EIK Kl

−

= = −

−

56 3

12 SYM0 06 0 4

[ ]12 0 6 120 0 0 0 06 0 2 6 0 0

EIKl

= − −

67 3

6 SYM6 6

46/ 2 6/ 2[ ]6 6 66 2

6 6 6 66 226/ 2 6/ 2 6/ 2 6 2 4/ 2

EIKl

− −= − − −

− − −

The member stiffness matrices are then assembled into a 21 × 21 symmetrical matrix using the method described in Example 6.1. The known nodal displacements are u1 = υ1 = θ1 = u5 = υ5 = θ5 = u2 = u4 = u6 = θ3 = θ7 = 0 and the support reactions are obtained from {F} = [K]{δ}. Having obtained the support reactions the internal

92 Solutions Manual shear force and bending moment distributions in each member follow (see Example 6.2).


S.6.5

Referring to Fig. P.6.5, u2 = 0 from symmetry. Consider the members 23 and 29. The forces acting on the member 23 are shown in Fig. S.6.5(a) in which F29 is the force applied at 2 in the member 23 due to the axial force in the member 29. Suppose that the node 2 suffers a vertical displacement υ2. The shortening in the member 29 is then υ2 cos θ and the corresponding strain is – (υ2 cos θ)/l. Thus the compressive stress in 29 is – (Eυ2 cosθ)/l and the corresponding compressive force is – (AEυ2

cosθ)/l. Thus 2

29 2( cos ) /F AE lυ θ= −

Now 26 2 / . 45 and 2 . HenceAE EI L I l Lθ= = ° =

229 3

3EIFL

υ= −

Fig. S.6.5(a) and

,2 233

2yP EIF

Lυ= − − (i)

Further, from Eq. (3.12)

33 3

d 22 0.8d 0.8

EIM GJ Elz L L

θθ θ= = − × = − (ii)

From the alternative form of Eq. (6.44), for the member 23

,2 2

223

3,3

33

12 6 12 60/ 6 4 6 2

012 6 12 66 2 6 4/

y

y

F

LM L EIF L

LM L

υθυθ

− − − =− = =− −

(iii)

Then, from Eqs (i) and (iii)

,2 2 2 33 3 23 12 6

2yP EI EI EIF

L L Lυ υ θ= − − = −

94 Solutions Manual

Hence

3

2 315 62PLL

Elυ θ− = − (iv)

From Eqs (ii) and (iii)

33 2 32 3 2

2 6 4M EI EI EIL L L L

θ υ θ= − = − +

which gives θ3 = υ2Substituting for θ

/L. 3

in Eq. (iv) gives 3

2 18PL

Elυ =

Then

2

3 18PL

EIθ = −

From Eq. (i)

3

,2 33

2 18 3yP EI PL PF

EIL= − + = −

and from Eq. (ii)

2

3 12

18 9EI PL PLM ML EI

= = = −

Now, from Eq. (iii)

22 33 3

2 269

M EI EI PLLL L L

υ θ= − + =

,3 2 33 312 6

3yEI EI PF L

L Lυ θ= − + =

The force in the member 29 is F29 292 .F /cos θ = Thus

3

29 28 33 22 (tension)

18 6EI PL PS S

EIL= = =

The torques in the members 36 and 37 are given by M3

/2, i.e.

36 37 /18M M PL= =

The shear force and bending moment diagrams for the member 123 follow and are shown in Figs S.6.5(b) and (c), respectively.


Fig. S.6.5(b) and (c)

S.6.6

The stiffness matrix for each element of the beam is obtained using the given force–displacement relationship, the complete stiffness matrix for the beam is then obtained using the method described in Example 6.1. This gives

,1 1

1 1

,2 2

223

3,3

33

4,4

44

24 12 24 12/ 12 8 12 4

24 12 36 6 12 6/ 12 4 6 12 6 2

12 6 36 24 24 126 2 6 12 12 4/

24 12 24 1212 4 12 8/

y

y

y

y

F

M L LF

LM L EIF L

LM LF

LM L

υθυθυθυθ

− − − − − − − − = − − − − − − − −

(i)

The ties FB, CH, EB and CG produce vertically upward forces F2 and F3

at B and C, respectively. These may be found using the method described in S.6.5. Thus

2 21 2

2 2cos 60 cos 45

2 / 3 2a E a EF

L Lυ

° °= − +

But 2 21 2384 /5 3 and 192 /5 2 so thata I L a I L= =

2 23965

EIFL

υ=

Similarly

3 33965

EIFL

υ= −

96 Solutions Manual

Then

2,2 ,3 33 3

69 96and5 5y y

EI EIF P F PL L

υ υ= − − = − −

In Eq. (i), υ1 = θ1 = υ4 = θ4 = 0 and M2 = M3 =0. Also, from symmetry, υ2 = υ3, and θ2 = –θ3

. Then, from Eq. (i)

2 2 2 3 30 6 12 6 2M L Lυ θ υ θ= = + + +

i.e.

2 212 10 0Lυ θ+ =

which gives

2 26

5Lθ υ= −

Also from Eq. (i)

,2 2 3 2 3 33 396 (36 6 12 6 )5y

EI EIF P L LL L

υ υ θ υ θ= − − = + − −

i.e.

2 23 396 485 5

EI EIPL L

υ υ− − =

whence

3

2 35

144PL

EIυ υ= =

and

2

2 324PL

EIθ θ= = −

The reactions at the ends of the beam now follow from the above values and Eq. (i). Thus

,1 2 2 ,43 ( 24 12 )3y y

EI PF L FL

υ θ= − − = =

1 2 2 42 (12 4 )4

EI PLM L ML

υ θ= + = − = −

Also

3

2 3 396 5 2

144 35El PL PF F

ElL= = =

The forces on the beam are then as shown in Fig. S.6.6(a). The shear force and bend-ing moment diagrams for the beam follow and are shown in Figs S.6.6(b) and (c), respectively.


Fig. S.6.6(a)


The forces in the ties are obtained using Eq. (6.32). Thus

1BF CH

2

0 01 302 22 / 3

a ES SL υ

− = = − −

i.e.

3

BF CH 3384 3 1 5 2

2 144 35 3 2EI PLS S P

EIL= = =

×

and

2BE CG

2

0 01 102 2 2

a ES SL υ

− = = − −

i.e.

3

BE CG 3192 1 5 2

144 35 2 2 2EI PL PS S

EIL= = =

×

98 Solutions Manual

S.6.7

The forces acting on the member 123 are shown in Fig. S.6.7(a). The moment M2

arises from the torsion of the members 26 and 28 and, from Eq. (3.12), is given by

2 22 2

1.6M GJ EI

l lθ θ

= − = − (i)

Fig. S.6.7(a) Now using the alternative form of Eq. (6.44) for the member 12

,1 1

1 13

,2 2

22

12 6 12 6/ 6 4 6 2

12 6 12 66 2 6 4/

y

y

F

M l LEIF l

LM l

υθυθ

− − − − = − −

(ii)

and for the member 23

,1 2

223

3,3

33

96 24 96 24/ 24 8 24 4

96 24 96 2424 4 24 8/

y

y

F

LM l EIF l

LM l

υθυθ

− − − − = − −

(iii)

Combining Eqs (ii) and (iii) using the method described in Example 6.1

,1 1

1 1

,2 23

22

3,3

33

12 6 12 6 0 0/ 6 4 6 2 0 0

12 6 108 18 96 246 2 18 12 24 4/

0 0 96 24 96 240 0 24 4 24 8/

y

y

y

F

M l lF EI

lM l lF

lM l

υθυθυθ

− − − − − − − − = − −

−

−

(iv)

In Eq. (iv) υ1 = υ2 = 0 and θ3 = 0. Also M1 =0 and Fy,3

= –P/2. Then from Eq. (iv)

11 230 (4 2 )M EI l l

l lθ θ= = +

from which


21 2

θθ = − (v)

100 Solutions Manual

Also, from Eqs (i) and (iv)

22 1 2 32 3 (2 12 24 )M EI EI l l

l l lθ θ θ υ= − = + +

so that

2 1 313 2 24 0l lθ θ υ+ + = (vi)

Finally from Eq. (iv)

,3 2 33 (24 96 )2yP EIF l

lθ υ= − = +

which gives

3

23 192 4

lPlEl

θυ = − − (vii)

Substituting in Eq. (vi) for θ1 from Eq. (v) and υ3

from Eq. (vii) gives 2

2 48Pl

EIθ =

Then, from Eq. (v)

2

1 96Pl

EIθ = −

and from Eq. (vii)

3

3 96Pl

EIυ = −

Now substituting for θ1, θ2 and υ3 in Eq. (iv) gives Fy,1 = –P/16, Fy,2 = 9P/16, M2 = –Pl/48 (from Eq. (i)) and M3 = –Pl/6. Then the bending moment at 2 in 12 is Fy,1l = –Pl/12 and the bending moment at 2 in 32 is – (P/2) (l/2) +M3 = –Pl/12. Also M3 = –Pl/6 so that the bending moment diagram for the member 123 is that shown in Fig. S.6.7(b).

Fig. S.6.7(b)


S.6.8

(a) The element is shown in Fig. S.6.8. The displacement functions for a triangular element are given by Eqs (6.82). Thus

1 1 1 4

2 1 2 2 4 5

3 1 3 3 4 6

,,,

uu a au a a

α υ αα α υ α αα α υ α α

= = = + = + = + = +

(i)

Fig. S.6.8

From Eq. (i)

1 1 2 2 1 3 3 1

4 1 5 2 1 6 3 1

( )/ ( )/( )/ ( )/

u u u a u u aa a

α α αα υ α υ υ α υ υ

= = − = −

= = − = −

Hence in matrix form

1 1

2 1

3 2

24

35

36

1 0 0 0 0 01/ 0 1/ 0 0 01/ 0 0 0 1/ 00 1 0 0 0 00 1/ 0 1/ 0 00 1/ 0 0 0 1/

ua a

ua a

a a ua a

αα υα

υαα

υα

− − =

−

−

which is of the form

1{ } [ ]{ }ex A δ−=

Also, from Eq. (6.89)

0 1 0 0 0 0

[ ] 0 0 0 0 0 10 0 1 0 1 0

C =


Hence

11/ 0 1/ 0 0 0

[ ] [ ][ ] 0 1/ 0 0 0 1/1/ 1/ 0 1/ 1/ 0

a aB C A a a

a a a a

−

− = = − − −

(b) From Eq. (6.94)

212

2

1/ 0 1/0 1/ 1/ 1 0

1/ 0 0[ ] 1 0

0 0 1/ 10 0 (1 )0 0 1/

0 1/ 0

1/ 0 1/ 0 0 010 1/ 0 0 0 1/2

1/ 1/ 0 1/ 1/ 0

e

a aa a v

a EK va

vaa

a aa a a t

a a a a

υ

− − − −

=

− −

− × − − −

which gives

2

3 1 2 (1 ) (1 ) 21 2 2 (1 ) (1 ) 2

2 2 2 0 0 2[ ]

(1 ) (1 ) 0 1 1 04(1 )(1 ) (1 ) 0 1 1 0

2 2 2 0 0 2

e

v v v v vv v v v v

v vEtKv v v vvv v v v

v v

− + − − − − − − + − − − − − − − −

= − − − − − −− − − − − − −

− − −

Continuity of displacement is only ensured at nodes, not along their edges.

S.6.9

(a) There are six degrees of freedom so that the displacement field must include six coefficients. Thus 1 2 3u x yα α α= + + (i)

4 5 6x yυ α α α= + + (ii) (b) From Eqs (i) and (ii) and referring to Fig. S.6.9

1 1 2 3 1 4 5 6

2 1 2 3 2 4 5 6

3 1 2 3 3 4 5 6

2 22 2 2 2

uuu

α α α υ α α αα α α υ α α αα α α υ α α α

+ + + = + +

+ + + = + +

+ + + = + +

Thus

2 2 1 3 3 2 1 1 3

5 2 1 6 3 2 4 1 3

22

u u u u u uα α αα υ υ α υ υ α υ υ

= − = − = −

= = = − = −


Fig. S.6.9

Therefore

1 1

2 1

3 2

24

35

36

2 0 0 0 1 01 0 1 0 0 0

0 0 1 0 1 00 2 0 0 0 10 1 0 1 0 00 2 0 0 0 1

u

u

u

αα υα

υαα

υα

− − − = −

−

−

(iii)

which is of the form 1} [ ]{ }eAα δ−{ = From Eq. (6.89)

0 1 0 0 0 0

[ ] 0 0 0 0 0 10 0 1 0 1 0

C =

Hence

11 0 1 0 0 0

[ ] [ ][ ] 0 2 0 0 0 10 1 1 1 1 0

B C A−

− = = − − −

(c) From Eq. (6.69)

{ } [ ][ ]{ }eD Bσ δ=

Thus, for plane stress problems (see Eq. (6.92))

212

1 0 1 0 1 0 0 0[ ][ ] 1 0 0 2 0 0 0 1

1 0 1 1 1 1 00 0 (1 )

vED B vv

v

− = − − − − −


i.e.

21 1 1 12 2 2 2

1 2 1 0 0

[ ][ ] 2 0 0 11

0 0(1 ) (1 ) (1 ) (1 )

v vED B v vv

v v v v

− −

= − − −

− − − − − −

For plain strain problems (see Eq. (6.93))

1 0(1 )

(1 ) 1 0[ ][ ](1 )(1 2 ) (1 )

0 0 (1 2 )2(1 )

1 0 1 0 0 00 2 0 0 0 10 1 1 1 1 0

vv

E v vD Bv v v

vv

− −

= + − − −

− − × − − − −

(1 )[ ][ ](1 )(1 2 )

1 2 1 0 01 1

2 0 0 11 10 1 2 1 2 1 2 1 2 0

2(1 ) 2(1 ) 2(1 ) 2(1 )

E vD Bv v

v vv v

v vv v

v v v vv v v v

−=

+ −

−− − −

−× − − − − − − −

− − − − − −

S.6.10

(a) The element is shown in Fig. S.6.10. There are eight degrees of freedom so that a displacement field must include eight coefficients. Therefore assume 1 2 3 4u x y xyα α α α= + + + (i)

5 6 8yx y xyυ α α α α= + + + (ii)

(b) From Eqs (6.88) and Eqs (i) and (ii)

2 4

7 8

3 4 6 8

x

y

xy

u yx

xyu x yy x

ε α α

υε α α

υγ α α α α

∂= = +∂∂

= = +∂∂ ∂

= + = + + +∂ ∂


Fig. S.6.10

Thus since {ε} = [C]{α}

0 1 0 0 0 0 0

[ ] 0 0 0 0 0 0 10 0 1 0 1 0

yC x

x y

=

(iii)

(c) From Eq. (iii)

0 0 01 0 00 0 1

0[ ]

0 0 00 0 10 10

T y xC

xx y

=

and from Eq. (6.92)

212

1 0

[ ] 1 01

0 0 (1 )

vED vv

v

= −

−

Thus

2 2T T

vol 0 0[ ] [ ][ ]d [ ] [ ][ ] d d

a bC D C V C D C t x y=∫ ∫ ∫ (iv)

106 Solutions Manual Substituting in Eq. (iv) for [C]T

2 2 T

0 0[ ] [ ][ ] d d

a bC D C t x y∫ ∫

, [D] and [C] and multiplying out gives

2

22 2

2

22

0 0 0 0 0 0 0 0

0 1 0 0 0

0 0 1 0 1 0(1 ) (1 ) (1 ) (1 )2 2 2 2

0 0(1 ) (1 )(1 ) (1 )2 22

0 0 0 0 0 0 0 01

0 0 1 0 1 0(1 ) (1 ) (1 ) (1 )2 2 2 2

0 0 0 0 1

0 0(1 ) (1 ) (1 ) (1 )2 2 2

xy

y

y v vx

x yv v v v

y x x vyx v vxy vv vyEtv

x yv v v v

v vy x

vx y xy y xv vxy v v x v

− − − −

− + −− −+=

−

− − − −

− + − − + −

2 2

0 0d d

a bx y

∫ ∫

2 2

2 2

2 2 2 2 2 23

3

2

2 2

2 2

2

0 0 0 0 0 0 0 0

0 4 0 0 0 44 4

0 0 2 (1 ) 0 2 (1 ) 02 (1 ) 2 (1 )

0 8 04 2 (1 ) 2 (1 ) 4 2 (1 ){23

(1 )}0 0 0 0 0 0 0 01

0 0 2 (1 ) 0 2 (1 ) 02 (1 ) 2 (1 )

0 4 0 0 0 44 4

0 4

ab abvab ba v

ab v ab va b v ab v

a b a b v a b v a bv a b vab

a b vEtv

ab v ab va b v ab v

abv aba bv a b

a b

− −− −

− − ++

−=

−

− −− −

2 2 2 2 2 3

3

0 82 (1 ) 2 (1 ) 2 (1 ) 4 {23

(1 )}

v ab v a b v ab v a b a b

ab v

− + − + −

Solutions to Chapter 6 Problems 107 S.6.11

From the first of Eqs (6.96) u1 = α1 – α2 – α3 + α4 = 0.1/103

u

(i)

2 = α1 + α2 – α3 – α4 = 0.3/103

u (ii)

3 = α1 + α2 – α3 + α4 = 0.6/103

u

(iii)

4 = α1 – α2 + α3 – α4 = 0.1/103

Adding Eqs (i) and (ii)

(iv)

31 2 1 32 2 0.4 / 10u u α α+ = − =

i.e.

31 3 0.2 / 10α α− = (v)

Adding Eqs (iii) and (iv)

33 4 1 32 2 0.7 / 10u u α α+ = + =

i.e.

31 3 0.35 / 10α α+ = (vi)

Adding Eqs (v) and (vi)

α1 = 0.275/10Then from Eq. (v)

3

α3 = 0.275/10

Now subtracting Eq. (ii) from Eq. (i)

3

31 2 2 42 2 0.2 / 10u u α α− = − + = −

i.e.

32 4 0.01 / 10α α− + (vii)

Subtracting Eq. (iv) from Eq. (iii)

33 4 2 42 2 0.5 / 10u u α α− = + =

i.e.

32 4 0.25 / 10α α+ = (viii)

Now adding Eqs (vii) and (viii)

322 0.35 / 10α =


whence

32 0.175 / 10α =

Then from Eq. (vii)

34 0.075 / 10α =

From the second of Eqs (6.96)

31 5 6 7 8 0.1/10υ α α α α= − − + = (ix)

32 5 6 7 8 0.1/10υ α α α α= + − + = (x)

33 5 6 7 8 0.7/10υ α α α α= + + + = (xi)

34 5 6 7 8 0.5/10υ α α α α= − + + = (xii)

Then, in a similar manner to the above

35

37

36

8

0.4/10

0.2/10

0.1/100

α

α

αα

=

=

=

=

Eqs (6.96) are now written

3

3

(0.275 0.175 0.075 0.075 ) 10

(0.4 0.1 0.2 ) 10i

i

u x y xy

x yυ

−

−

= + + + ×

= + + ×

Then, from Eqs (6.88)

3

3

3 3

(0.175 0.075 ) 10

0.2 10

(0.075 0.075 0.1) 10 (0.175 0.075 ) 10

x

y

xy

y

x x

ε

ε

γ

−

−

− −

= + ×

= ×

= + + × = + ×

At the centre of the element x = y = 0. Then

3

3

3

0.175 10

0.2 10

0.175 10

x

y

xy

ε

ε

γ

−

−

−

= ×

= ×

= ×


so that, from Eqs (6.92)

3 22

3 22

3 2

200000 (0.175 0.3 0.2) 10 51.65 N/mm1 0.3200000 (0.2 0.3 0.175) 10 55.49N/mm1 0.3200000 0.175 10 13.46 N/mm

2(1 0.3)

x

y

xy

σ

σ

τ

−

−

−

= + × × =−

= + × × =−

= × × =+

S.6.12

Suitable displacement functions are:

1 2 3 4

5 6 7 8

u x y xyx y xy

α α α αυ α α α α= + + +

= + + +

Then u1 = α1 – 2α2 – α3 + 2α4

u = 0.001 (i)

2 = α1 + 2α2 – α3 + 2α4

u = 0.003 (ii)

3 = α1 + 2α2 + α3 + 2α4

u = –0.003 (iii)

4 = α1 – 2α2 + α3 + 2α4Subtracting Eq. (ii) from Eq. (i)

= 0 (iv)

α2 – α4Subtracting Eq. (iv) from Eq. (iii)

= 0.0005 (v)

α2 – α4Subtracting Eq. (vi) from Eq. (v)

= – 0.00075 (vi)

4 0.000625α = − Then from either of Eqs (v) or (vi) α2

Adding Eqs (i) and (ii) = –0.000125

1 3 0.002α α− = (vii) Adding Eqs (iii) and (iv) 1 3 0.0015α α+ = − (viii) Adding Eqs (vii) and (viii) 1 0.00025α = Then from either of Eqs (vii) or (viii) 3 0.00175α = − Similarly

5

6

7

8

0.0010.000250.002

0.00025

αααα

= −

=

=

= −


Then

0.00025 0.00125 0.00175 0.000625

0.001 0.00025 0.002 0.00025i

i

u x y xyx y xyυ

= − − −

= − + + −

From Eqs (1.18) and (1.20)

0.000125 0.000625

0.002 0.00025

0.0015 0.000625 0.00025

x

y

xy

u yxu xyu x yy x

ε

ε

υγ

∂= = − −∂∂

= = −∂∂ ∂

= + = − − −∂ ∂

At the centre of the element where x = y = 0 0.000125 0.002 0.0015.x y xyε ε γ= − = = −

S.6.13

Assume displacement functions

1 2 3

4 5 6

( , )( , )

u x y x yx y x y

α α αυ α α α

= + +

= + +

Then

1 1

2 1 2

3 1 2 3

42 2

uuu

αα αα α α

== += + +

Solving

3 1 22 12 3

24 4

u u uu uα α

− −−= =

Therefore

3 1 22 11

24 4

u u uu uu u x y− −− = + +

or

1 2 314 4 4 4 2x y x y yu u u u = − − + − +

Similarly

1 2 314 4 4 4 2x y x y yυ υ υ υ = − − + − +


Then, from Eqs (1.18) and (1.20)

1 2

31 2

1 2 1 2

4 4

4 4 2

4 4 4 4

x

y

xy

u uux

yu uu

y x

ε

υυ υυε

υ υυγ

∂= = − +∂∂

= = − − +∂∂ ∂

= + = − − − +∂ ∂

Hence

1

1

2

2

3

3

1 0 1 0 0 01[ ]{ } 0 1 0 1 0 24

1 1 1 1 2 0

e

uux

uB

yu uy x

υ

υδυ

υυ

∂ ∂ − ∂ = = − − ∂ − − − ∂ ∂+

∂ ∂

But

0

[ ] 00 0

a bD b a

c

=

so that

0 2

1[ ][ ] 0 24

2 0

a b a b bD B b a b a a

c c c c c

− − − = − − − − − −

and

T

2 22 22 21[ ] [ ][ ]

2 2162 2 2 2 4 02 2 2 2 0 4

a c b c a c b c c bb c a c b c a c c aa c b c a c b c c b

B D Bb c a c b c a c c a

c c c c cb a b a a

+ + − + − − − + + − + − − − − + − + + − − −

= − − − − + −

− − − − − −

Since [Ke] = [B]T

[D][B] × 4 × 1 SYM

1[ ]4

2 2 2 2 42 2 2 2 0 4

e

a cb c a ca c b c a c

Kb c a c b c a c

c c c c cb a b a a

+ + + − + − + +

= − − − − +

− − − − − −


S.6.14

For a = 1, b = 2 1

1 2 3 48 [(1 )(2 ) (1 )(2 ) (1 )(2 ) (1 )(2 ) ]u x y u x y u x y u x y u− − + + − + + + + − + Similarly for υ

Then

11 2 3 48

11 2 3 48

11 1 2 2 38

3 4 4

[ (2 ) (2 ) (2 ) (2 ) ]

[ (1 ) (1 ) (1 ) (1 ) ]

[ (1 ) (2 ) (1 ) (2 ) (1 )

(2 ) (1 ) (2 ) ]

u y u y u y u y ux

x x x xyu x u y x u y x uy x

y x u y

υ υ υ υ υ

υ υ υ

υ υ

∂= − − + − + + − +

∂∂

= − − − + + + − −∂∂ ∂

+ = − − − − − + + − + +∂ ∂

+ + + − − +

In matrix form

ux

yuy x

υ

υ

∂

∂ ∂

∂ ∂ ∂

+ ∂ ∂

1

1

2

2

3

3

4

4

(2 ) 0 (2 ) 0 (2 ) 0 (2 ) 01 0 (1 ) 0 (1 ) 0 (1 ) 0 (1 )8

(1 ) (2 ) (1 ) (2 ) (1 ) (2 ) (1 ) (2 )

u

uy y y y

x x x xu

y y x y x y x y

u

υ

υ

υ

υ

− − − + − + = − − − + + − − − − − − + − + + − − +

Also

00

0 0

c dD d c

e

=

Then

[D][B]

(2 ) (1 ) (2 ) (1 ) (2 ) (1 ) (2 ) (1 )1 (2 ) (1 ) (2 ) (1 ) (2 ) (1 ) (2 ) (1 )8

(1 ) (2 ) (1 ) (2 ) (1 ) (2 ) (1 ) (2 )

c y d x e y d x e y d x c y d xd y c x d y c x d y e x d y c xe x e y e x e y e x e y e x e y

− − − − − − + + + − + − = − − − − − − + + + − + − − − − − − + − + + − − +


Then

T

(2 ) 0 (1 )0 (1 ) (2 )

(2 ) (1 )1[ ] [ ][ ] (2 ) (1 )64

(1 ) (2 )

y xx y

c y d xB D B d y c x

e x e y

− − − − − − − −

− − − − = − − − − − − − −

Therefore

2 1 2 2

112 1

[ (2 ) (1 ) ]d d64tK c y e x x y

− −= − + −∫ ∫

which gives 11 (4 )6tK c e= +

2 1

12 2 1[ (2 )(1 ) (1 )(2 )]d d

64tK d y x e x y x y

− −= − − + − −∫ ∫

which gives 12 4 ( ).tK d e= +


S.7.1

Substituting for ((1/ρx) + (ν/ρy)) and ((1/ρy) + (ν/ρx

)) from Eqs (7.5) and (7.6), respectively in Eqs (7.3)

2 2and1 1

yxx y

MMEz EzD Dv v

σ σ= =− −

(i)

Hence, since, from Eq. (7.4), D = Et3/12(1 – ν2

), Eqs (i) become

3 3

1212 yxx y

zMzMt t

σ σ= = (ii)

The maximum values of σx and σy

will occur when z = ±t/2. Hence

2 2

66(max) (max) yx

x yMM

t tσ σ= ± = ± (iii)


Then

3

22

6 10 10(max) 600N/mm10xσ × ×

= ± = ±

3

22

6 5 10(max) 300N/mm10yσ × ×

= ± = ±

S.7.2

From Eq. (7.11) and since Mxy

= 0

sin 22

x yt

M MM α

−=

Mt

will be a maximum when 2α = π /2, i.e. α = π/4 (45°). Thus, from Eq. (i) 10 5(max) 2.5Nm/mm

2tM −= =

S.7.3

The relationship between Mn and Mx, My and Mxy in Eq. (7.10) and between Mt and Mx, My and Mxy

in Eq. (7.11) are identical in form to the stress relationships in Eqs (1.8) and (1.9). Therefore, by deduction from Eqs (1.11) and (1.12)

2 2I

1 ( ) 42 2

x yx y xy

M MM M M M

+= + − + (i)

and

2 2II

1 ( ) 42 2

x yx y xy

M MM M M M

+= − − + (ii)

Further, Eq. (7.11) gives the inclination of the planes on which the principal moments occur, i.e. when Mt

= 0. Thus 2

tan 2 xy

x y

MM M

α = −−

(iii)

Substituting the values Mx = 10Nm/mm, My = 5 Nm/mm and Mxy

= 5 Nm/mm in Eqs (i), (ii) and (iii) gives

I

II

13.1 Nm/mm1.9 Nm/mm

MM

==

Ml

and 31.7 or 58.3α = − ° °


The corresponding principal stresses are obtained directly fromEqs (iii) of S.7.1. Hence

3

2I 2

6 13.1 10 786N/mm10

σ × ×= ± = ±

3

2II 2

6 1.9 10 114 N/mm10

σ × ×= ± = ±

S.7.4

From the deflection equation

220

2 2

220

2 2

1 cosh sinh sin

cosh 2 cosh sinh sin

q aw y y y xA Ba a a ax D

q aw y y y y xA B Ba a a a ay D

π π π ππ

π π π π ππ

∂ = − + + ∂

∂ = + + ∂

Now w = 0 and Mx

Also w = 0 and M = 0 at x = 0 and a. From Eq. (7.7) this is satisfied implicitly.

y

= 0 at y = ±a so that, from the deflection equation 4

04O (1 cosh sinh )sinq a xA B

aDππ π π

π= + +

i.e.

1 cosh sinh 0A Bπ π π+ + = (i)


20

2O [(1 cosh 2 cosh sinh )

0.3(1 cosh sinh )]sin

q a A B BD

xA Ba

π π π ππ

ππ π π

= − + + +

− + +

or

O 0.3 0.7 cosh 2 cosh 0.7 sinhA B Bπ π π π= − + + + (ii)

Solving Eqs (i) and (ii)

0.2213 0.0431A B= − =

S.7.5

The deflection is zero at x = a/2, y = a/2. Then, from the deflection equation

4

4 43 3O (1 )4 2 4

a a v a v A= − − − +


Hence

4

(5 3 )4

aA v= −

The central deflection, i.e. at x = 0, y = 0 is then

4

4

1 (5 3 )96(1 ) 4

5 3384 1

qa vv D

qa vD v

= × −−

− = −

S.7.6

From the equation for deflection

44

04

44

04

2 24

02 2

3cos cos

3 3cos sin

3 3cos cos

w x ywa a ax

w x ywa a ax

w x ywa a a ax y

π π π

π π π

π π π π

∂ = ∂

∂ = ∂

∂ = ∂ ∂


4

0( , ) 3cos cos (1 2 9 81)q x y x ywD a a a

π π π = + × +

i.e.

4

0 43( , ) 100 cos cosx yq x y w D

a aaπ π π

=

From the deflection equation

0 at /2, /2w x a y a= = ± = ±

The plate is therefore supported on all four edges. Also

0

0

3sin cos

3 3cos sin

w x ywx a a aw x ywy a a a

π π π

π π π

∂= −

∂∂

= −∂


When

0

2

02

a wxx

a wyy

∂= ± ≠

∂∂

= ± ≠∂

The plate is therefore not clamped on its edges. Further

22

02

22

02

3cos cos

3cos cos

w x ywa a ax

w x ywa a ax

π π π

π π π

∂ = − ∂

∂ = − ∂


2

03cos cos ( 1 9 )x

x yM Dw va a aπ π π = − − −

(i)

Similarly, fromEq. (7.8)

2

03cos cos (9 )y

x yM w D va a aπ π π = +

(ii)

Then, at x = ±a/2, Mx = 0 (from Eq. (i)) and at y = ±a/2, MyThe plate is therefore simply supported on all edges.

= 0 (from Eq. (ii)).

The corner reactions are given by

2

2 (1 ) (seeEq.(7.14))wD vx y∂

−∂ ∂

Then, since

2

03 3sin sin at /2, /2w x yw x a y a

x y a a a aπ π π π∂

= = =∂ ∂

2

0Corner reactions 6 (1 )w D vaπ = − −

From Eqs (7.7) and (7.8) and the above, at the centre of the plate

2 2

0 0(1 9 ), (9 ).x yM w D v M w D va aπ π = + = +


S.7.7

Substituting q(x, y) = q0

x/a in Eq. (7.29) and noting that the plate is square and of side a

02 0 0

4 sin sin d da a

mnx m x n ya q x ya a aa

π π= ∫ ∫

i.e.

03 0 0

4 sin cos daa

mnq m x a n ya x x

a n aaπ π

π = − ∫

Hence

02 0

4 sin (cos 1)da

mnq m xa x n x

aa nπ π

π= − −∫

The term in brackets is zero when n is even and equal to –2 when n is odd. Thus

02 0

8 sin d ( odd)a

mnq m xa x x n

aa nπ

π= ∫ (i)

Integrating Eq. (i) by parts

02

0

8 cos cos da

mnq a m x a m xa x x

m a m aa nπ π

π ππ = − + ∫

i.e.

02

0

8 cos sina

mnq m x a m xa x

a m aamnπ π

ππ = − +

The second term in square brackets is zero for all integer values of m. Thus

02

8 ( cos )mnqa a m

amnπ

π= −

The term in brackets is positive when m is odd and negative when m is even. Thus

102

8 ( 1)mmn

qamnπ

+= −

Substituting for amn

in Eq. (7.30) gives the displaced shape of the plate, i.e. 1

04 22 2

1,2,3 1,3,5 22 2

8 ( 1)1 sin sinm

m n

q m x n ywa aD m nmn

a a

π ππ

π

∞ ∞ +

= =

−=

+

∑ ∑

or

4 1

06 2 2 2

1,2,3 1,3,5

8 ( 1) sin sin( )

m

m n

q a m x n ywa aD mn m nπ π

π

∞ ∞ +

= =

−=

+∑ ∑


S.7.8 The boundary conditions which must be satisfied by the equation for the displaced shape of the plate are w = 0 and ∂w/∂n = 0 at all points on the boundary; n is a direction normal to the boundary at any point.

The equation of the ellipse representing the boundary is

2 2

2 2 1x ya b

+ = (i)

Substituting for x2/a2 + y2/b2

in the equation for the displaced shape clearly gives w = 0 for all values of x and y on the boundary of the plate. Also

w w x w yn x n y n

∂ ∂ ∂ ∂ ∂= +

∂ ∂ ∂ ∂ ∂ (ii)

Now

22 2

0 2 21 x yw wa b

= − −

so that

22 2

02 2 2

4 1w xw x yx a a b

∂= − − − ∂

(iii)

and

2 2

02 2 2

4 1w yw x yy a a b

∂= − − − ∂

(iv)

From Eqs (i), (ii) and (iv) it can be seen that ∂w/∂x and ∂w/∂y are zero for all values of x and y on the boundary of the plate. It follows from Eq. (ii) that ∂ w/∂n = 0 at all points on the boundary of the plate. Thus the equation for the displaced shape satisfies the boundary conditions.

From Eqs (iii) and (iv)

4 4 4

0 0 04 4 4 4 4 2 2 2

24 24 8w w ww w wx a y b x y a b

∂ ∂ ∂= = =

∂ ∂ ∂ ∂

Substituting these values in Eq. (7.20)

0 4 2 2 424 16 24 pw

Da a b b + + =

whence

0

4 2 2 43 2 38

pwD

a a b b

= + +


Now substituting for D from Eq. (7.4)

2

03

4 2 2 4

3 (1 )3 2 32

p vwEt

a a b b

−=

+ +

(v)

From Eqs (7.3), (7.5) and (7.7)

2 2

2 2 21xEz w wv

v x yσ

∂ ∂= − + − ∂ ∂

(vi)

and from Eqs (7.3), (7.6) and (7.8)

2 2

2 2 21yEz w wv

v y xσ

∂ ∂= − + − ∂ ∂

(vii)

From Eqs (iii) and (iv)

2 2 2 2 2 2

0 02 2 2 2 2 2 2 2

4 43 31 1w ww x y w x yx a a b y b a b

∂ ∂= − − − = − − − ∂ ∂

Substituting these expressions in Eq. (vi) and noting that the maximum values of direct stress occur at z = ±t/2

2 2 2 2

0 02 2 2 2 2 2 2

4 43 3(max) 1 12(1 )x

w w vEt x y x yv a a b b a b

σ

= ± − − − − − − − (viii)

At the centre of the plate, x = y = 0. Then

02 2 2

2 1(max)(1 )x

Etw vv a b

σ = ± + −

(ix)

Substituting for w0

in Eq. (ix) from Eq. (v) gives 2 2 2 2

2 4 2 2 43 ( )(max)(3 2 3 )x

pa b b vat b a b a

σ += ±

+ + (x)

Similarly

2 2 2 2

2 4 2 2 43 ( )(max)(3 2 3 )y

pa b a vbt b a b a

σ += ±

+ + (xi)

At the ends of the minor axis, x = 0, y = b. Thus, from Eq. (viii)

02 2 2 2 2

2 1 1 3(max)(1 )x

Etw v vv a a b b

σ = ± − + − −

i.e.

02 24(max)(1 )xEtw v

b vσ = ±

− (xii)


Again substituting for w0

from Eq. (v) in Eq. (xii) 4 2

2 4 2 2 46(max)

(3 2 3 )xpa b

t b a b aσ = ±

+ +

Similarly

4 2

2 4 2 2 46(max)

(3 2 3 )xpb a

t b a b aσ = ±

+ +

S.7.9

The potential energy, V, of the load W is given by

V = –Ww i.e.

1 1

sin sinmnm n

m nV W Aa bπξ πη∞ ∞

= =

= − ∑∑

Therefore, it may be deduced from Eq. (7.47) that the total potential energy, U + V, of the plate is

24 2 2

22 2

1 1 1 1

sin sin2 4mn mn

m n m n

D ab m n m nU V A W Aa ba b

π πξ πη∞ ∞ ∞ ∞

= = = =

+ = + −

∑∑ ∑∑

From the principle of the stationary value of the total potential energy

24 2 2

2 2( ) sin sin 0

4mnmn

U V ab m n m nDA WA a ba b

π πξ πη ∂ += + − = ∂

Hence

22 24

2 2

4 sin sinmn

m nWa bA

m nDaba b

πξ πη

π

=

+

so that the deflected shape is obtained.

S.7.10

From Eq. (7.45) the potential energy of the in-plane load, Nx

, is 2

0 0

1 d d2

a bx

wN x yx

∂ − ∂ ∫ ∫


The combined potential energy of the in-plane load, Nx

, and the load, W, is then, from S.7.9

2

0 01 1

1 d d sin sin2

a bx mn

m n

w m nV N x y W Ax a b

πξ πη∞ ∞

= =

∂ = − − ∂ ∑∑∫ ∫

or, since,

1 1

cos sinmnm n

w m m x n yAx a a b

π π π∞ ∞

= =

∂=

∂ ∑∑

2 22 2 2

20 01 1

1 1

1 cos sin d d2

sin sin

a ax mn

m n

mnm n

m m x n xV N A x ya ba

m nW Aa b

π π π

πξ πη

∞ ∞

= =∞ ∞

= =

= −

−

∑∑∫ ∫

∑∑

i.e.

2 2

22

1 1 1 1

sin sin8 x mn mn

m n m n

ab m m nV N A W Aa ba

π πξ πη∞ ∞ ∞ ∞

= = = =

= − −∑∑ ∑∑

Then, from Eq. (7.47), the total potential energy of the plate is

24 2 2 2 22 2

2 2 21 1 1 1

1 1

2 4 8

sin sin

mn x mnm n m n

mnm n

D ab m n ab mU V A N Aa b a

m nW Aa b

π π

πξ πη

∞ ∞ ∞ ∞

= = = =∞ ∞

= =

+ = + −

−

∑∑ ∑∑

∑∑

Then, from the principle of the stationary value of the total potential energy

24 2 2 2 2

2 2 2( ) sin sin 0

4 4mn x mnmn

U V ab m n ab m m nDA N A WA a ba b a

π π πξ πξ ∂ += + − − = ∂

from which

2 22 24

2 2 2 2

4 sin sinmn

x

m nWa bA

m Nm nabDa b a D

πξ πη

ππ

=

+ −


S.7.11 The guessed form of deflection is

2 2

11 2 24 41 1x yw Aa a

= − −

Clearly when x = ±a/2, w = 0 and when y = ±a/2, w = 0. Therefore, the equation for the displaced shape satisfies the displacement boundary conditions.

From Eq. (i)

2 2 2 2

11 112 2 2 2 2 2

4 48 1 8 1A Aw y w xx a a y a a

∂ ∂= − − = − − ∂ ∂


2 2

112 2 2

8 4 41 1xA D y xM va a a

= − − + −

Clearly when x = ±a/2, Mx ≠ 0 and when y = ± a/2, Mx ≠ 0. Similarly for My

From Eq. (i)

. Thus the assumed displaced shape does not satisfy the condition of zero moment at the simply supported edges.

2

114

64A xywx y a∂

=∂ ∂

Substituting for ∂2w/∂x2, ∂2w/∂y2, ∂2

w/∂x ∂y and w in Eq. (7.46) and simplifying gives 2/2 2 2 4 2 2 4114 2 4/2 /2

2 22 2

2 4

2 2 2 2

0 11 2 2 4

32 16 164 ( ) ( 2 ) 1.4

5.6 67.2( )

4 4 161 d d

a a

a a

A DU V x y x x y ya a a

x yx ya a

x y x yq A x ya a a

− −

+ = − + + + + −

+ + +

− − − +

∫ ∫

from which

22

0 11112

462.49

q A aA DU Va

+ = −


2

0112

11

4124.8( ) 09

q aA DU VA a

∂ += − =

∂

Hence, since D = Et3/12(1 – v2

) 4 3

11 00.0389 /qA a Et= A


S.7.12

From Eq. (7.36) the deflection of the plate from its initial curved position is

1 11 sin sinx yw Ba bπ π

=

in which

1111 2 2

2 21

x

x

A NBD a N

a bπ

= + −

The total deflection, w, of the plate is given by

1 0w w w= +

i.e.

111122 2

2 2

sin sin

1

x

x

A N x yw Aa bD a N

a b

π π

π

= + + −

i.e.

1122 2

2 2

sin sin

1 / 1x

A x ywa bN a a

D b

π π

π

=

− +


S.8.1 The forces on the bar AB are shown in Fig. S.8.1 where

BB

dd

M Kzυ =

(i)

and P is the buckling load. From Eq. (8.1)

2

2

dd

EI Pzυ= − υ (ii)

The solution of Eq. (ii) is cos sinA z B z= +υ µ µ (iii)


Fig. S.8.1

where µ2

When z = 0, v = 0 so that, from Eq. (iii), A = 0. Hence = P/EI.

sinB zυ µ= (iv) Then

d cosd

B zz=

υ µ µ

and when z = l, dυ/dz = MB

/K from Eq. (i). Thus B

cosMB

K l=µ µ

and Eq. (iv) becomes

B sincos

M zK l

=υ µµ µ

(v)

Also, when z =l, PvB = MB from equilibrium. Hence, substituting in Eq. (v) for M

B

BB sin

cosP l

K l=

υυ µ

µ µ

from which

tan

KPl

=µµ

(vi)

(a) When K → ∞, tan μl → ∞ and μl →π/2, i.e.

2

P lEI

→π

from which

2

24EIPl

→π

which is the Euler buckling load of a pin-ended column of length 2l. (b) When EI → ∞, tan μl → and Eq. (vi) becomes P = K/l and the bars remain

straight.


S.8.2

Suppose that the buckling load of the column is P. Then from Eq. (8.1) and referring to Fig. S.8.2, in AB

2

2

dd

EI Pz

= −υ υ (i)

and in BC

2

2

d4d

EI Pz

= −υ υ (ii)

Fig. S.8.2

The solutions of Eqs (i) and (ii) are, respectively

AB cos sinA z B z= +υ µ µ (iii)

BC cos sin2 2

C z D z= +µ µυ (iv)

in which

2 PEI

=µ

When z = 0, υAB

= 0 so that, from Eq. (iii), A = 0. Thus

AB sinB z=υ µ (v)

Also, when z = l/2, (dυ /dz)BC

= 0. Hence, from Eq. (iv)

0 sin cos2 4 2 4

l lC D= − +µ µ µ µ

whence

tan4lD C=

µ

Then

BC cos tan sin2 4 2

lC z z = +

µ µ µυ (vi)


When z = l/4, υAB = υBC

so that, from Eqs (v) and (vi)

sin cos tan sin4 8 4 8l l l lB C = +

µ µ µ µ

which simplifies to

sin sec cos4 4 8l l lB C=

µ µ µ (vii)

Further, when z = l/4, (dv/dz)AB = (dv/dz)BC

. Again from Eqs (v) and (vi)

cos sin tan cos4 2 8 2 4 8l l l lB C = − +

µ µ µ µ µ µµ

from which

cos sec sin4 2 4 8l C l lB =

µ µ µ (viii)

Dividing Eq. (vii) by Eq. (viii) gives

tan 2 tan4 8l l=

µ µ

or

tan tan 24 8l l

=µ µ

Hence

2

2

2 tan /8 21 tan /8

ll

=−

µµ

from which

1tan8 2l=

µ

and

35.26 0.615rad8l= ° =

µ

i.e.

0.6158

P lEI

=

so that

2

24.2EIPl

=


S.8.3

With the spring in position the forces acting on the column in its buckled state are shown in Fig. S.8.3. Thus, from Eq. (8.1)

2

2

d 4 (δ ) δ( )d

EI P k l zz

= − − −υ υ (i)

The solution of Eq. (i) is

δcos sin [4 ( )]4

A z B z P k z lP

= + + + −υ µ µ (ii)

Fig. S.8.3

where

2 4PEI

=µ

When z = 0, υ = 0, hence, from Eq. (ii)

δ0 (4 )4

A P klP

= + −

from which

δ( 4 )4

kl PAP−

=

Also when z = 0, dυ/dz = 0 so that, from Eq. (ii)

δ04

kBP

= +µ

and

δ4

kBP−

=µ



δ ( 4 )cos sin 4 ( )4

kkl P z z P k z lP

= − − + + −

υ µ µµ

(iii)

When z = l, υ = δ. Substituting in Eq. (iii) gives

δδ ( 4 )cos sin 44

kkl P l l PP

= − − +

µ µµ

from which

4tanPk

l l=

−µ

µ µ

S.8.4 The compressive load P will cause the column to be displaced from its initial curved position to that shown in Fig. S.8.4. Then, from Eq. (8.1) and noting that the bending moment at any point in the column is proportional to the change in curvature produced (see Eq. (8.22))

22

02 2

ddd d

EI EI Pz z

− = −υυ υ (i)

Now

0 2

4 ( )za l zl

= −υ

so that

2

02 2

d 8d

az l

= −υ

Fig. S.8.4

and Eq. (i) becomes

2

2 2

d 8d

P az EI l

+ = −υ υ


The solution of Eq. (ii) is

2cos sin 8 /( )A z B z a l= + −υ λ λ λ (iii)

where λ2

When z = 0, υ = 0 so that A = 8a/(λl) =P/EI.

2

. When z = l/2, dυ /dz = 0. Thus, from Eq. (iii)

0 sin cos2 2l lA B= − +

λ λλ λ

whence

2

8 tan( ) 2

a lBl

=λ

λ

Eq. (iii) then becomes

2

8 cos tan sin 1( ) 2

a lz zl

= + −

λυ λ λλ

(iv)

The maximum bending moment occurs when υ is a maximum at z =l/2. Then, from Eq. (iv)

max 2

8(max) cos tan sin 1( ) 2 2 2

aP l l lM Pl

= − = − + −

λ λ λυλ

from which

2

8(max) sec 1( ) 2

aP lMl

= − −

λλ

S.8.5

Under the action of the compressive load P the column will be displaced to the position shown in Fig. S.8.5. As in P.8.4 the bending moment at any point is proportional to the change in curvature. Then, from Eq. (8.1)

22

02 2

ddd d

EI EI Pz z

− = −υυ υ (i)

Fig. S.8.5


In this case, since each half of the column is straight before the application of P, d2υ0/dz2

= 0 and Eq. (i) reduces to 2

2

dd

EI Pz

= −υ υ (ii)


cos sinA z B z= +υ µ µ

in which µ2

=P/EI. When z = 0, υ =0 so that A = 0 and Eq. (iii) becomes

sinB z=υ µ (iv)

The slope of the column at its mid-point in its unloaded position is 2δ/l. This must be the slope of the column at its mid-point in its loaded state since a change of slope over zero distance would require an infinite bending moment. Thus, from Eq. (iv)

d 2δ cosd 2

lBz l= =

υ µµ

so that

2δcos( /2)

Bl l

=µ µ

and

2δ sincos( /2)

zl l

=υ µµ µ

(v)

The maximum bending moment will occur when υ is a maximum, i.e. at the mid-point of the column. Then

max2δ(max) sin

cos( /2) 2P lM P

l l= − = −

µυµ µ

from which

2δ(max) tan2

EI P lM Pl P EI

= −

S.8.6

Referring to Fig. S.8.6 the bending moment at any section z is given by

2

( )2 2wl zM P e z w= + − +υ


Fig. S.8.6

or

2( ) ( )2wM P e z lz= + + −υ (i)

Substituting for M in Eq. (8.1)

2

22

d ( )d 2

wEI P Pe z lzz

+ = − − −υ υ

or

2 2

2 2 22

d ( )d 2

we z lzz P

+ = − − −υ µµ υ µ (ii)


22cos sin ( )

2w wA z B z e lz zP P

= + − + − +υ µ µµ

(iii)

When z = 0, υ = 0, hence A = e – w/µ2

Ρ. When z = l/2, dυ/dz = 0 which gives

2tan tan2 2l w lB A e

P

= = −

µ µµ

Eq. (iii) then becomes

22

cos ( /2) 1 ( )cos /2 2

w z l we lz zP l P

−= − − + −

µυµ µ

(iv)

The maximum bending moment will occur at mid-span where z = l/2 and υ = υ max

. From Eq. (iv)

2

max 2 sec 12 8

EIw l wleP P

= − − +

µυ

and from Eq. (i)

2

max(max)8

wlM Pe P= + −υ


whence

2 2(max) sec2

w l wM Pe

= − +

µµ µ

For the maximum bending moment to be as small as possible the bending moment at the ends of the column must be numerically equal to the bending moment at mid-span. Thus

2 2sec 02

w l wPe Pe

+ − + =

µµ µ

or

21 sec sec 12 2l w lPe + = −

µ µµ

Then

2

1 cos /21 cos /2

w lep l

−= +

µµ µ

i.e.

22 tan

4w le

P

=

µµ

(vi)

From Eq. (vi) the end moment is

2

22

tan /4 tan /4tan4 16 /4 /4

w l wl l lPel l

= =

µ µ µµ µ µ

When P → 0, tan μl/4 → μl/4 and the end moment becomes wl2

S.8.7 /16.

From Eq. (8.21) the buckling stress, σb

, is given by 2

tb 2( / )

El r

=π

σ (i)

The stress–strain relationship is

16

610.5 10 2100049000

× = +

σε σ (ii)

Hence

6 1516

d 16 2100010.5 10 1d (49000)

×× = +

ε σσ

from which

6 16

t 16 15

d 10.5 10 (49000)d (49000) 16 21000( )

E × ×= =

+ ×σε σ


Then, from Eq. (i)

2 2 7

t16

b b b

10.36 10336000( /49000)

Elr

× = = +

πσ σ σ

(iii)

From Eq. (iii) the following σb

–(l/r) relationship is found

b 4900 3 4900 6 4900 9 4900 49000/ 145.4 84.0 59.3 31.2 16.4l r

× × ×σ

For the given strut

4 4

2 2 22 2

( )/64 1 ( )( )/4 16

I D dr D dA D d

−= = = +

−ππ

i.e.

2 2 2 21 (1.5 1.34 ) 0.253units16

r = + =

Hence

0.503unitsr =

Thus

20 39.80.503

lr= =

Then, from the σb

–(l/r) relationship 2

b 40500forceunits/units=σ

Hence the buckling load is

2 240500 (1.5 1.34 )4

× −π

i.e. Buckling load = 14 454 force units

S.8.8

The deflected shape of each of the members AB and BC is shown in Fig. S.8.8. For the member AB and from Eq. (8.1)

2

1B2

1

dd

EI Mz

= −υ

so that

1B 1

1

dd

EI M z Az

= − +υ


Fig. S.8.8

When z1 =b, dυ1/dz1 =0. Thus A =MB

b and

1B 1

1

d ( )d

EI M z bz

= − −υ (i)

At B, when z1

=0, Eq. (i) gives 1 B

1

dd

M bz EI

=υ (ii)

In BC Eq. (8.1) gives

2

B2

dd

EI P Mz

= − +υ υ

or

2

B2

dd

EI P Mz

= + =υ υ (iii)

The solution of Eq. (iii) is

Bcos sin /B z C z M Pυ λ λ= + + (iv)

When z = 0, υ = 0 so that B = –MB

When z = a/2, dυ/dz = 0 so that /P.

Btan tan2 2

Ma aC BP

= = −λ λ

Eq. (iv) then becomes

B cos tan sin 12

M az zP

= − + −

λυ λ λ


so that

Bd sin tan cosd 2

M az zz P

= − − +

υ λλ λ λ λ

At B, when z =0,

Bd tand 2

M az P= −

υ λλ (v)

Since dυ1/dz1

= dυ/dz at B then, from Eqs (ii) and (v)

tan2

b aEI P

= −λ λ

whence

1 tan2 2 2a a a

b = −

λ λ

S.8.9

In an identical manner to S.8.4

2 2

2 2

d dd d

EI EI Pz z′

′− = −υ υ υ

where υ′ is the total displacement from the horizontal. Thus

2 2

2 2

d dd d

Pz EI z′

′+ =υ υυ

or, since

2 2

22 2

d sind

Pz andz l l EI

= − =υ π πδ µ

2 2

22 2

d sind

zz l l′

′+ = −υ π πµ υ δ (i)

The solution of Eq. (i) is

2

2 2 2cos sin sin zA z B zl l

′ = + +−π δ πυ µ µ

π µ (ii)

When z = 0 and l, υ′ = 0, hence A = B = 0 and Eq. (ii) becomes

2

2 2 2 sin zl l

′ =−π δ πυ

π µ


The maximum bending moment occurs at the mid-point of the tube so that

2

2 2 2 2 2(max)1 /

PM P Pl Pl EI

′= = =− −π δ δυ

π µ π

i.e.

e

(max)1 / 1

P PMP P

= =− −δ δ

α

The total maximum direct stress due to bending and axial load is then

3

/2(max)1 /8

P P ddt d t

= + −

δσπ α π

Hence

1 4(max) 11

Pdt d = + −

δσπ α

S.8.10

The forces acting on the members AB and BC are shown in Fig. S.8.10

Fig. S.8.10

Considering first the moment equilibrium of BC about C

BP Vb=υ

from which

BVbP

=υ (i)

For the member AB and from Eq. (8.1)

2

2

dd

EI P Vzz

= − −υ υ

or

2

2

dd

P Vzz EI EI

+ = −υ υ (ii)



cos sin VzA z B zP

= + −υ λ λ (iii)

When z = 0, υ = 0 so that A = 0. Also when z = a, dυ/dz = 0, hence

0 cos VB aP

= −λ λ

from which

cosVB

P a=λ λ

and Eq. (iii) becomes

sincos

V z zP a = −

λυλ λ

When z = a, υ = υB

= Vb/P from Eq. (i). Thus

sincos

Vb V a aP P a

= −

λλ λ

from which

( ) tana b a+ =λ λ

S.8.11

The bending moment, M, at any section of the column is given by

2CR CR ( )M P P k lz z= = −υ (i)

Also

d ( 2 )d

k l zz= −

υ (ii)

Substituting from Eqs (i) and (ii) in Eq. (8.47)

2 22 2 2 2 2 2CR

01 2 1

22CR

0

1 1 1( ) d ( ) d ( ) d2

( 2 ) d2

a l a l

a l a

l

P kU V lz z z lz z z lz z zE I I I

P k l z z

−

−

+ = − + − + −

− −

∫ ∫ ∫

∫


i.e.

2 2 2 3 4 5 2 3 4 5CR

1 20

22 3 4 5 32 2CR

1 0

1 12 3 2 5 3 2 5

1 423 2 5 2 3

a l a

a

l l

l a

P k l z lz z l z lz zU VE I I

P kl z lz l zl z lzI

−

−

+ = − + + − +

+ − + − − +

i.e.

2 2 2 3 4 5 2 3 4 5CR 2

2 1

2 35CR2

1

( ) ( ) ( )12 3 2 5 3 2 5

30 6

P k l l a la a l l a l l a l aU VEI l

P k lI lI

− − −+ = − − + − + −

+ −


2 2 3 4 5 2 3CR 2

2 1

34 5 5CR2

1

( ) ( )13 2 5 3

( ) ( ) 02 5 30 3

P k IU V l a la a l l ak EI I

P klIl l a l a lI

∂ + −= − − + − ∂ − −

+ − + − =

Hence

3

2CR 2 3 4 5 2 3

2

1

4 5 52

1

( )3 13 2 5 3

( ) ( )2 5 30

EI lPI l a la a l l aI

Il l a l a lI

= − − − + −

− −+ − +

(iii)

When I2 = 1.6I1

and a = 0. 2l, Eq. (iii) becomes

1CR 2

14.96EIPl

= (iv)

Without the reinforcement

2

1CR 2

EIPl

=π (v)

Therefore, from Eqs (iv) and (v) the increase in strength is

212 (14.96 )EI

l−π


Thus the percentage increase in strength is

2

212 2(14.96 ) 100 52%EI l

l EI

− × =

ππ

Since the radius of gyration of the cross-section of the column remains unchanged

2 21 1 2 2andI A r I A r= =

Hence

2 2

1 1

1.6A IA I

= = (vi)

The original weight of the column is lA1ρ where ρ is the density of the material of the column. Then, the increase in weight = 0.4lA1ρ + 0.6lA2ρ –lA1ρ = 0.6lρ (A2 –A1

Substituting for A).

2

from Eq. (vi)

1 1 1Increasein weight 0.6 (1.6 ) 0.36l A A lAρ ρ= − =

i.e. an increase of 36%.

S.8.12

The equation for the deflected centre line of the column is

22

4 zl

=δυ (i)

in which δ is the deflection at the ends of the column relative to its centre and the origin for z is at the centre of the column. Also, the second moment of area of its cross-section varies, from the centre to its ends, in accordance with the relationship

1 1 1.6 zI Il

= −

(ii)

At any section of the column the bending moment, M, is given by

2

CR CR 2( ) 1 4 zM P Pl

= − = −

δ υ δ (iii)

Also, from Eq. (i)

2

d 8d

zz l=

υ δ (iv)

Substituting in Eq. (8.47) for M, I and dυ/dz

2 2 2 2 2 2/2 /2 2CR CR

40 01

(1 4 / ) 642 d 2 d2 (1 1.6 / ) 2

l lP z l PU V z z zEI z l l

−+ = −

−∫ ∫δ δ


or

2 2 2 2 2/2 /2 2CR CR

3 40 01

64( 4 ) d d1.6

l lP Pl zU V z z zEI l l z l

−+ = −

−∫ ∫δ δ (v)

Dividing the numerator by the denominator in the first integral in Eq. (v) gives

2 2 /2 3 2 2 3CR3 0

1/22 3/23 CR

400

( 10 6.25 1.09 0.683 )d

64d0.317(1 1.6 / ) 3

l

ll

PU V z lz l z l zEI l

Pz zlz l l

+ = − − + +

+ − −

∫

∫

δ

δ

Hence

2 2 4 3 22 3

3

/2 24 CR

0

10 6.25 1.09 0.6834 3 2

80.317 1.6log 11.6 3

CR

l

e

P z z zU V l l l zEIl

Pzll l

δ

δ

+ = − − + +

− − −

i.e.

2 2 2

CR CR

1

0.3803 83

P l PU VEI l

δ δ+ = −


2

CR CR

1

0.7606 16( ) 03

P l PU VEI l

δ δδ

∂ += − =

∂

Hence

1CR 2

7.01EIPl

=

For a column of constant thickness and second moment of area I2

, 2

2CR 2 (seeEq. (8.5))EIP

lπ

=

For the columns to have the same buckling load

2

2 12 2

7.01EI EIl l

π=

so that I2 = 0.7IThus, since the radii of gyration are the same

1

A2 = 0.7A1

Solutions to Chapter 8 Problems 141 Therefore, the weight of the constant thickness column is equal to ρA2l = 0.7ρA1l. The weight of the tapered column = ρ × average thickness × I = ρ ×0. 6Α1

Hence the saving in weight = 0.7ρA l.

1l – 0.6ρA1l = 0. 1 ρA1Expressed as a percentage

l.

1

1

0.1saving in weight= 100 14.3%0.7

A lA l

ρρ

× =

S.8.13

There are four boundary conditions to be satisfied, namely, υ = 0 at z = 0 and z = l, dυ/dz = 0 at z = 0 and d2υ/dz2

(i.e. bending moment) = 0 at z = l. Thus, since only one arbitrary constant may be allowed for, there cannot be more than five terms in the polynomial. Suppose

2 3 4

0 1 2 3 4z z z za a a a al l l l

υ = + + + +

(i)

Then, since υ = 0 at z = 0, a0 = 0. Also, since dυ/dz = 0 at z = 0, a1

= 0. Hence, Eq. (i) becomes

2 3 4

2 3 4z z za a al l l

υ = + +

(ii)

When z = l, υ = 0, thus 2 3 40 a a a= + + (iii)

When z = l, d2υ/dz2

= 0, thus

2 3 40 3 6a a a= + + (iv) Subtracting Eq. (iv) from Eq. (ii) 0 = – 2a3 – 5afrom which a

4 3 = – 5a4

Substituting for a/2.

3 in Eq. (iii) gives a4 = 2a2/3 so that a3 = –5a2

/3. Eq. (ii) then becomes

2 3 42 2

25 23 3a az z za

l l lυ = − +

(v)

Then

2 3

22 2 3 4

8d 2 5d 3

az z za az l l lυ= − + (vi)

and

2 2 3

22 22 3 4

2d 10 8d

a z za alz l l

υ= − + (vii)


The total strain energy of the column will be the sum of the strain energy due to bending and the strain energy due to the resistance of the elastic foundation. For the latter, consider an element, δz, of the column. The force on the element when subjected to a small displacement, υ, is kδzv. Thus, the strain energy of the element is 21

2 kv zδ and the strain energy of the column due to the resistance of the elastic foundation is

2

0

1 d2

lk zυ∫

Substituting for υ from Eq. (v)

2 5 6 7 8

424 2 3 40

1 10 37 20 4(elastic foundation) d2 3 9 9 9

la z z z zU k z zll l l l

= − + − +

∫

i.e. U (elastic foundation) = 0.0017 220.0017 .ka l=

Now substituting for d2υ/dz2

and dυ/dz in Eq. (8.48) and adding U (elastic founda-tion) gives

2 2 3 42224 2 3 40

2 3 4 5 62CR 2

4 2 3 40

4 10 33 40 161 d 0.00172

20 107 80 644 d2 3 3 9

l

l

aEI z z z zU V z ka lll l l l

P a z z z zz zll l l l

+ = − + − + +

− − + − +

∫

∫ (viii)

Eq. (viii) simplifies to

2

2 2 2 CR2 23

0.0190.4 0.0017 a PEIU V a ka lll

+ = + −


2 CR2 23

2

0.038( ) 0.8 0.0034 a PU V EI a ka la ll

∂ += + −

∂

whence

2CR 2

21.05 0.09EIP kll

= +

S.8.14

The purely flexural instability load is given by Eq. (8.7) in which, from Table 8.1 le

= 0. 5l where l is the actual column length. Also it is clear that the least second moment of area of the column cross-section occurs about an axis coincident with the web. Thus

3 32212 3tb tbI = × =


Then

2

CR 2(0.5 )EIPl

π=

i.e.

2 3

CR 24

3EtbPl

π= (i)

The purely torsional buckling load is given by the last of Eqs (8.77), i.e.

2

CR( ) 20

A EP GJI lθ

π Γ= +

(ii)

In Eq. (ii) A = 5bt and

2 3 3

0 2 24 12 3x y

b tb tbI I I tb= + = × + +

i.e.

3

017

12tbI =


3 3

3 31 17(2 8 )3 3 3

st btJ b t bt= = + =∑

and, referring to S.27.4

5

12tb

Γ =

Then, from Eq. (ii)

2 4

3CR( ) 2

20 1717

EtbP Gtb lθ

π = +

(iii)

Now equating Eqs (i) and (iii)

2 3 2 4

32 2

4 20 17173

Etb EtbGtbl l

π π = +

from which

2 4

22

2255

EblGt

π=

From Eq. (1.50), E/G= 2(1 + v). Hence

22 1

255b vltπ +

=


Eqs (i) and (iii) may be written, respectively, as

1CR 2

1.33CPl

=

and

1CR( ) 2 2

1.175CP Clθ = +

where C1 and C2 are constants. Thus, if l were less than the value found, the increase in the last term in the expression for PCR(θ) would be less than the increase in the value of PCR, ie. PCR(θ) < PCR

S.8.15

for a decrease in l and the column would fail in torsion.

In this case Eqs (8.77) do not apply since the ends of the column are not free to warp. From Eq. (8.70) and since, for the cross-section of the column, xs = ys

= 0, 4 2

04 2d d 0d d

PE I GJAz z

θ θ Γ + − =

(i)

For buckling, P = PCR, the critical load and PCR/A = σ CR

, the critical stress. Eq. (i) may then be written

4 22

4 2d d 0d dz zθ θλ+ = (ii)

in which

2 0 CR( )I GJE

σλ −=

Γ (iii)


cos sinA z B z Dz Fθ λ λ= + + + (iv)

The boundary conditions are: 0at 0and 2z z lθ = = =

d 0at = 0 and 2 (see Eq. (18.19))d

z z = l zθ=

Then B = D = 0, F = –A and Eq. (iv) becomes

(cos 1)A zθ λ= − (v)

Since θ = 0 when z = 2l

cos λ2l = 1 or

λ2l = 2nπ


Hence, for n =1

2

22l

πλ =

i.e. from Eq. (iii)

2

0 CR2

I GJE l

σ π−=

Γ

so that

2

CR 20

1 EGJI l

πσ Γ

= +

(iv)

For the cross-section of Fig. P.8.15

3

(see Eq. (18.11))3

stJ =∑

i.e.

3 3

48 8 25.0 2.5 1041.7 mm3 3btJ × ×

= = =

and

3 2

2 (2 ) sin 604 ( cos30 ) 2 (see Section 16.4.5)12xx

b tI bt b °= ° +

i.e.

3 3 44 4 25.0 2.5 156250.0mmxxI b t= = × × = Similarly

3 3 2 3

2 (2 ) cos 60 140 212 12 3yybt b t b tI btb °

= + + =

so that

3 414 25.0 2.5/3 182291.7 mmyyI = × × =

Then

40 338541.7 mmxx yyI I I= + =

The torsion-bending constant, Γ, is found by the method described in Section 27.2 and is given by

Γ = b5t = 25.05 × 2.5 = 24.4 × 106 mmSubstituting these values in Eq. (vi) gives

4

σCR = 282.0 N/mm2


S.8.16

The three possible buckling modes of the column are given by Eqs (8.77) i.e.

2

CR( ) 2xx

xxEIPL

π= (i)

2

CR( ) 2yy

yyEI

PL

π= (ii)

2

CR( ) 20

A EP GJI Lθ

π Γ= +

(iii)

From Fig. P.8.16 and taking the x axis parallel to the flanges

2

2 3 4 4

3 4 4

4 40

3 4

= (2 × 20 + 40) × 1.5 = 120 mm

= 2 × 20 × 1.5 × 20 + 1.5 × 40 /12 = 3.2 × 10 mm

= 1.5 × 40 /12 = 0.8 × 10 mm

= + = 4.0 × 10 mm

= (20 + 40 + 20) × 1.5 /3 = 90.0mm (see Eq. (

xx

yy

xx yy

A

I

I

I I I

J3 2

6 6

18.11))

1.5 20 40 2 40 20=12 40 2 20

2.0 10 mm (seeEq.(ii)of Example27.1)

× × × × Γ × ×

= ×

Substituting the appropriate values in Eqs (i), (ii) and (iii) gives

CR( )

CR( )

CR( )

22107.9N

5527.0N

10895.2N

xx

yy

P

P

P θ

=

=

=

Thus the column will buckle in bending about the y axis at a load of 5527.0 N.

S.8.17

The separate modes of buckling are obtained from Eqs (8.77), i.e.

2

CR( ) CR( ) 2 ( ,say)xx yy xx yyEIP P I I I

Lπ

= = = = (i)

and

2

CR( ) 20

A EP GJI Lθ

π Γ= +

(ii)


In this case Ixx = Iyy = πr3t = π × 403 × 2.0 = 4.02 × 105 mmA = 2πrt = 2π × 40 × 2.0 = 502.7 mm

4

J = 2πrt3

From Eq. (8.68)

/3 = 2π × 40 × 2.0/3 = 670.2 mm

20 (note that 0)xx yy s sI I I Ax y= + + =

in which xs is the distance of the shear centre of the section from its vertical diameter; it may be shown that xs

= 80 mm (see S.17.3). Then 5 2 6 4

0 2 4.02 10 502.7 80 4.02 10 mmI = × × + × = ×

The torsion-bending constant Γ is found in a similar manner to that for the section shown in Fig. P.27.3 and is given by

5 22 43

r tπ π Γ = −

i.e.

5 2 9 6240 2.0 4 1.66 10 mm3

π π Γ = × × − = ×

(a) 2 5

4CR( ) CR( ) 3 2

70000 4.02 10 3.09 10 N(3.0 10 )xx yyP P π × × ×

= = = ××

(b)

2 9

CR( ) 6 3 2

4

502.7 70000 1.66 1022000 670.240.2 10 (3.0 10 )

1.78 10 N

P θπ × × ×

= × + × ×

= ×

The flexural-torsional buckling load is obtained by expanding Eq. (8.79). Thus

2 2CR( ) CR( ) 0 s( )( ) / 0xxP P P P I A P xθ− − − =

from which

2 2s 0 CR( ) CR( ) CR( ) CR( ) CR( )(1 / ) ( ) 0xx xxP Ax I P P P P P Pθ θ θ− − + + + = (iii)

Substituting the appropriate values in Eq. (iii) gives

P –24.39 × 10P4 + 27.54 × 108

The solutions of Eq. (iv) are

= 0 (iv)

P = 1.19 × 104 N or 23.21 × 104

Therefore, the least flexural-torsional buckling load is 1.19 × 10 N

4 N.



S.9.1

Assuming that the elastic deflection, w, of the plate is of the same form as the initial curvature, then

sin sinx yw Aa aπ π

=

Hence, from Eq. (7.36) in which m = n = 1, a = b and Nx

= σt

2 2 sin sin(4 / )

t x ywa aD a t

δσ π ππ σ

=−

(i)

The deflection, wC

, at the centre of the plate where x = a/2, y = a/2 is, from Eq. (i)

2 2(4 / )tw

D a tδσ

π σ=

− (ii)

When σt → 4π2D/a,w → ∞ and σt → NXCR

, the buckling load of the plate. Eq. (ii) may then be written

,CRC

,CR ,CR

/1 /

x

x x

t NtwN t t N

δσδσσ σ

= =− −

from which

CC ,CRx

ww Nt

δσ

= − (iii)

Therefore, from Eq. (iii), a graph of wC against wC/σt will be a straight line of slope Nx,CR

S.9.2

and intercept δ, i.e. a Southwell plot.

The total potential energy of the plate is given by Eq. (9.1), i.e.

2 22 2 2 2 2

2 2 2 20 0

2

1 2(1 )2

d d

l b

x

w w w w wU V D vx yx y x y

wN x yx

∂ ∂ ∂ ∂ ∂ + = + − − − ∂ ∂∂ ∂ ∂ ∂

∂ − ∂

∫ ∫ (i)

in which

211 sin sinm x yw a

l bπ π

= (ii)

and xN tσ=


From Eq. (ii)

211 cos sinw m m x ya

x l l bπ π π∂

=∂

2 2 22

112 2

2 2

112 2

2 2

11

sin sin

2 2sin cos

2cos sin

w m m x yal bx l

w m x yal by b

w m m x yax y bl l b

π π π

π π π

π π π

∂= −

∂∂

=∂

∂=

∂ ∂

Substituting these expressions in Eq. (i) and integrating gives

2 2 24 3

2 4 1111 3 3

332 2 3216

ta m bD m b m lU V alb ll b

σ ππ

+ = + + −

The total potential energy of the plate has a stationary value in the neutral equilibrium of its buckled state, i.e. when σ = σCR

. Thus 2 24 3

4 CR 1111 3 3

11

3( ) 3 02 1616

ta m bU V m b m lDaa lb ll b

σ ππ ∂ +

= + + − = ∂

whence

2 4 2

CR 2 3 316 3

23 16l D m b m l

lbtm b l bπσ

= + +

(iii)

When l = 2b, Eq. (iii) gives

2 2

CR 2 232 3 1 2

128 43D m

tb mπσ

= + +

(iv)

σCR will be a minimum when dσCR

/dm = 0, i.e. when

36 4 0128

mm

− =

or

4 4 1286

m ×=

from which m = 3.04 i.e.

m = 3


Substituting this value of m in Eq. (iv)

271.9

CRD

tbσ =

whence

2

26

(1 )CRE t

bvσ =

−

S.9.3

(a) The length, l, of the panel is appreciably greater than the dimension b so that failure will occur due to buckling rather than yielding. The modes of buckling will then be those described in Section 9.5.

(1) Buckling as a column of length l Consider a stiffener and an associated portion of sheet as shown in Fig. S.9.3. The critical stress, σCR

, is given by Eq. (8.8), i.e. 2

CR 2( / )E

l rπσ = (i)

Fig. S.9.3

In Eq. (i) r is the radius of gyration of the combined section. Thus, / ,xr I A= where A and Ix

are the cross-sectional area and the second moment of area of the combined section respectively. From Fig. S.9.3

s s(2 )A bt t d c bt A= + + = + (ii)

Also

s s s( )bt A y A y+ =


so that

s s

s

A yybt A

=+

Then

2 3

2 2( ) 2 ( )2 12

sx s s s

t ccI bt y dt A y y = + + + −

or

2

2 2s( ) ( )

2 6x s sc cI bt y t d A y y = + + + −

(iii)

The radius of gyration follows from Eqs (ii) and (iii) and hence the critical stress from Eq. (i).

(2) Buckling of the sheet between stiffeners The sheet may buckle as a long plate of length, l, and width, b, which is simply supported on all four edges. The buckling stress is then given by Eq. (9.7), i.e.

22

CR 212(1 )k E t

bvη πσ =

− (iv)

Since l is very much greater than b, k is equal to 4 (from Fig. 9.2). Therefore, assuming that buckling takes place in the elastic range (η = 1), Eq. (iv) becomes

22

CR 24

12(1 )E t

bvπσ =

− (v)

(3) Buckling of stiffener flange The stiffener flange may buckle as a long plate simply supported on three edges with one edge free. In this case k = 0.43 (see Fig. 9.3(a)) and, assuming elastic buckling (i.e. η = 1)

22

sCR 2

s

0.4312(1 )

tEbv

πσ = −

(vi)

(b) A suitable test would be a panel buckling test.

S.9.4

(a) Consider, initially, the buckling of the panel as a pin-ended column. For a section comprising a width of sheet and associated stiffener as shown in Fig. S.9.4,

2120 3 30 3.5 465 mmA = × + × =

Then

465 30 3.5 15 120 3 1.5y = × × + × ×


Fig. S.9.4

i.e.

4.5mmy =

Then

3 3 3

2 120 3 3.5 4.5 3.5 25.5120 3 4.512 3 3xI × × ×

= × × + + +

i.e.

Ix = 27011 mmHence

4

27011 7.62mm465

r = =

From Eq. (8.8)

2

CR 270000

(500 / 7.62)πσ ×

=

i.e.

2CR 160.5N / mmσ =

From Section 9.5 the equivalent skin thickness is

30 3.5 3 3.875mm120

t ×= + =

Overall buckling of the panel will occur when

,CR CR 160.5 3.875 621.9 N / mmxN tσ= = × = (i)

Buckling of the sheet will occur when, from Eq. (9.6)

2 2

CR33.62 3.62 70000

120tEb

σ = = ×

i.e.

2CR 158.4N / mmσ =


Hence

,CR 158.4 3.875 613.8N / mmxN = × = (ii)

Buckling of the stiffener will occur when, from Eq. (9.6)

2 2

CR3.50.385 0.385 7000030

tEb

σ = = ×

i.e.

2CR 366.8N / mmσ =

whence

,CR 366.8 3.875 1421.4 N / mmxN = × =

Nx,CR

By comparison of Eqs (i), (ii) and (iii) the onset of buckling will occur when = 366. 8 × 3. 875 = 1421.4N/mm (iii)

,CR 613.8N / mmxN =

(b) Since the stress in the sheet increases parabolically after reaching its critical value then

2xCNσ = (iv)

where C is some constant. From Eq. (iv)

2CR ,CRxCNσ = (v)

so that, combining Eqs (iv) and (v)

2

CR ,CR

x

x

NN

σσ

=

(vi)

Suppose that σ = σF, the failure stress, i.e. σF = 300 N/mm2

. Then, from Eq. (vi)

F,F ,CR

CRx xN Nσ

σ=

or

,F300 613.8

158.4xN =

i.e.

,F 844.7N / mmxN =


S.9.5

The beam may be regarded as two cantilevers each of length 1.2 m, built-in at the mid-span section and carrying loads at their free ends of 5 kN. The analysis of a complete tension field beam in Section 9.7.1 therefore applies directly. From Eq. (9.29)

4 1 1.5 350 / 2 300tan 0.71921 1.5 300 / 280

α + × ×= =

+ ×

hence

42.6α = °

From Eq. (9.19)

3

T5 1.2 10 5

350 2 tan 42.6F × ×

= +°

i.e.

T 19.9kNF =

From Eq. (9.23)

5 300 tan 42.6350

P × °=

i.e.

3.9kNP =

S.9.6

(i) The shear stress buckling coefficient for the web is given as K = 7.70[1 + 0.75(b/d) 2

]. Thus Eq. (9.33) may be rewritten as

2 2 2

CR 7.70 1 0.75t b tKE Eb d b

τ = = +

Hence

2 2

CR2507.70 1 0.75 70000725 250

tτ = + ×

i.e.

2CR 9.39tτ = (i)

The actual shear stress in the web, τ, is

100000 133.3750t t

τ = = (ii)


Two conditions occur, firstly

2165N/mmτ ≤

so that, from Eq. (ii) t = 0.81 mm and secondly

CR15τ τ≤

so that, from Eqs (i) and (ii)

2 133.315 9.39tt

× =

whence

t = 0.98 mm

Therefore, from the range of standard thicknesses

t = 1.2 mm

(ii) For t = 1.2 mm, τCR

is obtained from Eq. (i) and is

2CR 13.5N / mmτ =

and, from Eq. (ii), τ = 111.1 N/mm2. Thus, τ/τCR

The stiffener end load follows from Eq. (9.35) and is

= 8.23 and, from the table, the diagonal tension factor, k, is equal to 0.41.

ss s s

s

tan( / ) 0.5(1 )

A kQ AA tb k

τ ασ= =+ −

i.e.

s

s s

0.41 111.1tan 40 130( / 1.2 250) 0.5(1 0.41) 1 0.0113

ss

A AQA A

× × °= =

× + − +

The maximum secondary bending moment in the flanges is obtained from Eq. (9.25) multiplied by k, thus

2 tanmaximum secondary bending moment

12kWb

dα

=

i.e.

20.41 100000 250 tan 40maximum secondary bending moment12 750

238910 N/mm

× × × °=

×=


S.9.7

Stringer local instability: The buckling stress will be less for the 31.8 mm side than for the 19.0 mm side. Then,

from Eq. (9.6)

2 2

CR0.93.62 6900031.8

tKEb

σ = = ×

i.e.

2CR 200.1 N / mmσ =

Skin buckling: Referring to Fig. P.9.7(a)

2

200.1tKEb =

Then

2

2 3.62 69000 1.6200.1

b × ×=

i.e. 56.5mmb =

Panel strut instability: Consider stringer and skin as a strut. Add to stringer a length of skin equal to the

lesser of 30t or b.

56.5mm, 30 30 1.6 48.0mmb t= = × =

The section is then as shown in Fig. S.9.7

Fig. S.9.7


Taking moments of areas about the skin

[(19.0 2 31.8 2 9.5) 0.9 48 1.6] 19 0.9 31.8

2 31.8 0.9 15.9y+ × + × × + × = × ×

+ × × ×

from which 8.6 mm.y = Then

32 2

2 2

0.9 31.819.0 0.9 23.2 2 0.9 31.8 7.312

2 9.5 0.9 8.6 48 1.6 8.6

xxI ×

= × × + + × ×

+ × × × + × ×

i.e.

424 022.7 mmxxI =

From Eq. (8.5)

2

269000 24022.7

168.2Lπσ × ×

=

Therefore

2

2 69000 24022.7168.2 200.1

L π × ×=

×

i.e. 697 mmL =

say 700mmL =


S.10.1

Referring to Fig. S.10.1(a), with unit load at D (1), RC

= 2. Then

1

1 C

1

2

2

1 (0 )1 ( ) 2 ( 2 )

1( 2 ) (2 3 )0 (0 2 )1( 2 ) (2 3 )

M z z lM z R z l l z l z lM z l l z lM z lM z l l z l

= ≤ ≤= − − = − ≤ ≤

= − − ≤ ≤= ≤ ≤= − ≤ ≤

Hence, from the first of Eqs (5.21)

2 32 2 2

11 1 1 10 2

1 1 1d d dl l l

l lM z M z M z

EI EI EIδ = + +∫ ∫ ∫


Fig. S.10.1(a)

Substituting for M1

from the above 2 32 2 2

11 0 2

1 d (2 ) d ( 2 ) dl l l

l lz z l z z z l z

EIδ = + − + − ∫ ∫ ∫

which gives

3

11lEI

δ =

Also

3 2

22 2

1 ( 2 ) dl

lz l z

EIδ = −∫

from which

3

22 3lEI

δ =

and

3 2

12 21 2

1 ( 2 ) dl

lz l z

EIδ δ= = − −∫

i.e.

3

12 21 3lEI

δ δ= = −

From Eqs (10.5) the equations of motion are

1 11 2 12 12 0m mυ δ υ δ υ+ + = (i)

1 21 2 22 22 0m mυ δ υ δ υ+ + = (ii)

Assuming simple harmonic motion, i.e. υ = υ0 sin ωt and substituting for δ11, δ12 and δ22

, Eqs (i) and (ii) become 2 2

1 2 12 2

1 2 2

3 2 0

2 0

λω υ λω υ υ

λω υ λω υ υ

− + + =

− + =


in which λ = ml3

/3EI or, rearranging 2 2

1 2(1 3 ) 2 0λω υ λω υ− + = (iii)

2 21 2(1 2 ) 0λω υ λω υ+ − = (iv)

From Eq. (10.7) and Eqs (iii) and (iv)

2 2

2 2

(1 3 ) 20

(1 2 )

λω λω

λω λω

−=

−

from which

2 2 2 2(1 3 )(1 2 ) 2( ) 0λω λω λω− − − =

or

2 2 24( ) 5 1 0λω λω− + =

i.e.

2 2(4 1)( 1) 0λω λω− − = (v)

Hence

2 14 or 1λω =

so that

2 23 3

3 3 or 4

EI EIml ml

ω ω= =

Hence

1 23 33 3 4

EI EIml ml

ω ω= =

The frequencies of vibration are then

11 23 3

1 3 1 32 2 2

EI EIf fml ml

ωπ π π

= = =

From Eq. (iii)

2

12

2

21 3

υ λωυ λω

= −−

(vi)

When ω = ω1, υ1/υ2 is negative and when ω = ω2, υ1/υ2 is positive. The modes of vibration are therefore as shown in Fig. S.10.1(b) and (c).



S.10.2


Fig. S.10.2

12 2

12 2

2

41

4 21

4 2

(0 )

(2 ) (0 )0 (0 )1 (0 )

(0 )

(2 ) ( 2 )

M z z l

M z l z z lM x lM x x lM z z l

M l z l z l

= − ≤ ≤

= − − ≤ ≤

= ≤ ≤= ≤ ≤

= ≤ ≤

= − − ≤ ≤

Then from the first of Eqs (5.21)

2 22

22 0

1 1 (2 )d d3 4 4

l l

l

z l zz zEI EI

δ −= +∫ ∫

which gives

3

22 9lEI

δ =


Also

2 222

44 0 0

1 1 1 (2 )d d d3 4 4

l l l

l

z l zx x z zEI EI EI

δ −= + +∫ ∫ ∫

from which

3

4449

lEI

δ =

and

2 22

42 24 0

1 1 (2 )d d3 4 4

l l

l

z l zz zEI EI

δ δ −= = − +∫ ∫

Thus

3

42 24 18lEI

δ δ= =

From Eqs (10.5) the equations of motion are 4 44 2 42 42 0m mυ δ υ δ υ+ + = (i)

4 24 2 22 22 0m mυ δ υ δ υ+ + = (ii)

Assuming simple harmonic motion, i.e. υ = υ0 sin ωt and substituting for δ44, δ42 and δ22

, Eqs (i) and (ii) become 2 2

4 2 48 2 0λω υ λω υ υ− − + = (iii)

2 24 2 24 0λω υ λω υ υ− + + = (iv)

in which λ = ml3

/18EI. Then, from Eq. (10.7) 2 2

2 2

(1 8 ) 20

(1 4 )

λω λω

λω λω

− −=

− −

which gives

2 2 2 2(1 8 )(1 4 ) 2( ) 0λω λω λω− − − = i.e.

2 2 230( ) 12 1 0λω λω− + = (v) Solving Eq. (v)

2 20.118 or 0.282λω λω= = Hence

2 23 3

18 180.118 or 0.282EI EIml ml

ω ω= × = ×

Then, since f = ω/2π, the natural frequencies of vibration are

1 23 31 2.13 1 5.08

2 2EI EIf f

ml mlπ π= =


S.10.3

The second moment of area, I, of the tube cross-section is given by

4 4( )64

I D dπ= −

in which D and d are the outer and inner diameters respectively. Now,

25 1.25 26.25 mm 25 1.25 23.75mmD d= + = = − =

Thus

4 4 4(26.25 23.75 ) 7689.1mm64

I π= − =

The polar second moment of area, J, for a circular section is 2I, i.e. J = 15 378.2 mm4

. From Eqs (5.21)

d di j i jij L L

M M T Ts s

EI GJδ = +∫ ∫ (i)

Fig. S.10.3(a)

Then, referring to Fig. S.10.3(a)

1

1

1

1

2

2

1 (0 )1 (0 2 )0 (0 )1 (0 2 )1 (0 )1 (0 2 )

M y y aM z z aT y aT a z a

M y aM z a

= ≤ ≤= ≤ ≤= ≤ ≤= ≤ ≤= ≤ ≤= ≤ ≤

Thus, from Eq. (i)

2 2 22 2

11 0 0 0d d d

a a ay z ay z zEI EI GJ

δ = + +∫ ∫ ∫


which gives

3 311

3 2 3 225070000 7689.1 28000 15378.2

aEI GJ

δ = + = + × ×

i.e. 11 0.16δ =

Also

2 22

22 0 0

1 1d da a

y zEI GJ

δ = +∫ ∫

i.e.

221 2 1 2250

70000 7689.1 28000 15378.2a

EI GJδ = + = + × ×

which gives

622 1.63 10δ −= ×

Finally

1 2

12 21 0 0d d

ay ay zEI GJ

δ δ= = +∫ ∫

so that

2 212 21

1 2 1 22502 2 70000 7689.1 28000 15378.2

aEI GJ

δ δ = = + = + × × ×

Thus

412 21 3.48 10δ δ −= = ×

The equations of motion are then, from Eqs (10.5)

211 12 0m mrυδ θδ υ+ + = (i)

221 22 0m mrυδ θδ θ+ + = (ii)

Assuming simple harmonic motion, i.e. υ = υ0 sin ωt and θ = θ0

in ωt, Eqs (i) and (ii) may be written

2 2 211 12 0m mrδ ω υ δ ω θ υ− − + =

2 2 221 22 0m mrδ ω υ δ ω θ θ− − + =

Substituting for m, r and δ11

, etc. 2 2 4 2

4 2 2 6 2

20 0.16 20 62.5 3.48 10 0

20 3.48 10 20 62.5 1.63 10 0

ω υ ω θ υ

ω υ ω θ θ

−

− −

− × − × × × + =

− × × − × × × + =


which simplify to

2 2(1 3.2 ) 27.2 0υ ω ω θ− − = (iv)

2 20.007 (1 0.127 ) 0ω υ θ ω− + − = (v)


2 2

2 2

(1 3.2 ) 27.20

0.007 (1 0.127 )

ω ω

ω ω

− −=

− −

which gives

2 2 4(1 3.2 )(1 0.127 ) 0.19 0ω ω ω− − − =

or

4 215.4 4.63 0ω ω− + = (vi)

Solving Eq. (vi) gives

2 15.1 or 0.31ω =

Hence the natural frequencies are

0.62 Hz and 0.09Hzf =

From Eq. (iv)

2

227.2

1 3.2υ ωθ ω=

−

Thus, when ω2 = 15.1, υ/θ is negative and when ω2 = 0.31, υ/θ is positive. The modes of vibration are then as shown in Figs S.10.3(b) and (c).



S.10.4

Choosing the origin for z at the free end of the tube

1 1 1

2 2 2

, 1 and 0, 1 and 2

M z S TM z S T a

= = == = =

in which the point 1 is at the axis of the tube and point 2 at the free end of the rigid bar.

From Eqs (5.21) and (20.19)

0 0 0

d d d dL L Li j i j i j

ijM M T T q q

z z s zEI GJ Gt

δ = + +

∫ ∫ ∫ ∫ (i)

in which qi and qj are obtained from Eq. (17.15) in which Sy,iSy,j 1,Sx = 0 and Ixy

= 0. Thus

,00

1 ds

i j sxx

q q ty s qI

= = − +∫

‘Cutting’ the tube at its lowest point in its vertical plane of symmetry gives qs,0 = 0. Then, referring to Fig. S.10.4

Fig. S.10.4

0

1 cos di jxx

q q ta aI

θθ θ= = ∫

i.e.

2 sin

i jxx

a tq qI

θ= =

From Fig. 16.33, Ixx =πa3t. Hence qi = qj

= sin θ/πa and 2

2 20

sin 1d 2i jq qs ad

Gt G atG a t

π θ θππ

= =∫ ∫


Also in Eq. (i) the torsion constant J is obtained from Eq. (18.4), i.e.

2 2 2

34 4( ) 22 /d

A aJ a ta ts

π ππ

= = =∫

Therefore from Eq. (i)

2 3

11 0 0

1d d3

L Lz L Lz zEI G at EI G at

δπ π

= + = +∫ ∫ (ii)

Putting λ = 3Ea2/GL2

, Eq. (ii) becomes 3

11 (1 )3LEI

δ λ= +

Also

2 2

22 30 0 0

4 1z d d2

L L Lz a z zEI G a t G at

δπ π

= + +∫ ∫ ∫d

which gives

3

22 (1 3 )3LEI

δ λ= +

Finally

2

12 21 0 0

1d dL Lz z z

EI G atδ δ

π= = +∫ ∫

i.e.

3

12 21 (1 )3LEI

δ δ λ= = +

From Eqs (10.5) the equations of motion are

1 11 2 12 1 0m mυ δ υ δ υ+ + = (iii)

1 21 2 22 2 0m mυ δ υ δ υ+ + = (iv)

Assuming simple harmonic motion, i.e. υ = υ0

sin ωt, Eqs (iii) and (iv) become 2 2

11 1 12 2 12 2

21 1 22 2 2

0

0

m mm mδ ω υ δ ω υ υ

δ ω υ δ ω υ υ

− − + =

− − + =

Substituting for δ11, δ22 and δ12 and writing μ = L3

/3EI gives

( )2 21 2

2 21 2

1 1 (1 ) 0

(1 ) 1 (1 3 ) 0

v m m

m m

ω µ λ ω µ λ υ

ω µ λ υ υ ω µ λ

− + − + = − + + − + =



2 2

2 2

1 (1 ) (1 )0

(1 ) [1 (1 3 )]

m m

m m

ω µ λ ω µ λ

ω µ λ ω λ

− + − + =− + − +

Then

2 2 2 4 2 2[1 (1 )][1 (1 3 )] (1 ) 0m m mω µ λ ω µ λ ω µ λ− + − + − + =

which simplifies to

2 24 2

1 1 2 (1 2 ) 2 (1 ) 0m mµ λ µ λ λω ω

− + + + =

Solving gives

2 1/22

1 (1 2 ) (1 2 2 )m mµ λ µ λ λω

= + ± + +

i.e.

3

2 1/22 3

1 1 2 (1 2 2 )3

mLE a t

λ λ λω π

= + ± + +

S.10.5

Choosing the origin for z at the free end of the beam

M1 = z, S1

Also, from Eqs (5.21) andEq. (20.19) = 1

0 0

d d dL Li j i j

ij

M M q qz s z

EI Gtδ

= +

∫ ∫ ∫ (i)

in which qi and qj are obtained from Eq. (20.11) and in which Sy,j = Sy,j = 1, Sx = 0, Ixy = 0 and tD

= 0. Thus

,01

1 n

i j r r srxx

q q B y qI =

= = − +∑

where Ιxx

is given by (see Fig. S.10.5) 2 2 7 42 970 100 2 970 150 6.305 10 mmxxI = × × + × × = ×

Thus

, , 71

16.305 10

n

b i b j r rr

q q B y=

= = −× ∑


Fig. S.10.5

Hence, cutting the tube at O,

b,O1

b,12 7

b,23 7

0970 100 0.0015 N / mm

6.305 10970 1000.0015 0.0038 N / mm

6.305 10

q

q

q

=

×= − = −

××

= − − = −×


,02 0.0015 600 0.0038 150

2(100 / 1.0 600 / 1.25 150 / 1.0) 1.25 1.0sq × × − − + +

i.e.

,0 0.0018 N / mmsq =

Therefore

,O1 ,O1

,12 ,12

,23 ,23

0.0018 N / mm

0.0015 0.0018 0.0003N / mm

0.0038 0.0018 0.002 N / mm

i j

i j

i j

q qq qq q

= =

= = − + =

= = − + = −

Then

2 2 2

8

2 0.0018 100 0.0003 600 0.002 150d26500 1.0 1.25 1.0

7.3 10

i jq qs

Gt−

× × ×= + +

= ×

∫

Hence

21525 1525 8

11 0 0d 7.3 10 dz z z

EIδ −= + ×∫ ∫


i.e.

3

8 411 7

1525 7.3 10 1525 3.79 103 70000 6.305 10

δ − −= + × × = ×× × ×

For flexural vibrations in a vertical plane the equation of motion is, from Eqs (10.5)

1 11 1 0mυ δ υ+ =

Assuming simple harmonic motion, i.e. υ = υ0

sin ωt Eq. (ii) becomes 2

11 1 1 0mδ ω υ υ− + =

i.e.

3

24

11

1 9.81 10 5816.64450 3.79 10m

ωδ −

×= = =

× ×

Hence

1 5816.6 12.1Hz2 2

f ωπ π

= = =

S.10.6

Assume a deflected shape given by

2cos 1zVlπ

= − (i)

where z is measured from the left-hand end of the beam. Eq. (i) satisfies the boundary conditions of V = 0 at z = 0 and z = l and also dV/dz = 0 at z = 0 and z = l. From Eq. (i)

d 2 2sindV zz l l

π π=

and

2 2

2 2

d 4 2cosd

V zz l l

π π=

Substituting these expressions in Eq. (10.22)

2 22 2/4 /22 22 20 /4

22 2

/4 /2 2 2

0 /4

4 2 4 24 cos d cos d

2 2 1 12 2 cos 1 d cos 1 d 2 ( 1) (2)2 4

l l

l

l l

l

z zEI z EI zl l l l

z zm z m z ml mll l

π π π π

ωπ π

+ =

− + − + − +

∫ ∫

∫ ∫


which simplifies to

22 /4 /22 22 0 1/4

22 2

/4 /2

0 1/4

4 2 24 cos d cos d

2 22 cos 1 d cos 1 d

l l

l l

z zEI s zl l l

z zm z z ll l

π π π

ωπ π

+ = − + − +

∫ ∫

∫ ∫

Now

/4/4 2

00

/2/2 2

/4/4

2 1 4cos d sin2 4 8

2 1 4cos d sin2 4 8

ll

ll

ll

z l z lz zl l

z l z lz zl l

π ππ

π ππ

= + =

= + =

∫

∫

2/4 /4

0 0

/4

0

2 1 4 2cos 1 d 1 cos 2cos 1 d2

1 1 4 1 2 3sin sin2 4 8

l l

l

z z zz zl l l

z z l lz zl l

π π π

π ππ π π

− = + − +

= + − + = −

∫ ∫

Similarly

2

/2

/4

2 3cos 1 d8

l

l

z l lzlπ

π − = + ∫

Substituting these values in Eq. (ii)

22

22

4 48 8

3 328 8

l lEIl

l l l lm l

π

ω

π π

+ =

− + + +

i.e.

24539.2 EI

mlω =

Then

43.72

EIfml

ωπ

= =

The accuracy of the solution may be improved by assuming a series for the deflected shape, i.e.

1

( ) ( ) (Eq. (10.23))n

s ss

V z B V z=

= ∑



S.11.2 From Eq. (1.40) Young’s modulus E is equal to the slope of the stress–strain curve. Then, since stress = load/area and strain = extension/original length.

E = slope of the load-extension curve multiplied by (original length/area of cross-section).

From the results given the slope of the load-extension curve ~ 402.6 kN/mm. Then

3

22

402.6 10 250 205000 N / mm25

4

Eπ× ×

− − ×

The modulus of rigidity is given by

TLGJθ

=

Therefore, the slope of the torque-angle of twist (in radians) graph multiplied by (L/J) is equal to G. From the results given the slope of the torque-angle of twist graph is ~12.38kNm/rad. Therefore

6

24

12.38 10 250 80700 N / mm25

32

Gπ× ×

− − ×

Having obtained E and G the value of Poisson’s ratio may be found from Section 1.15, i.e.

1 0.272EG

υ = − −

Finally, the bulk modulus K may be found using Eq. (1.54)

2148500 N / mm .3(1 2 )

EKυ

− −−

S.11.3 Suppose that the actual area of cross-section of the material is A and that the original area of cross-section is AO

. Then, since the volume of the material does not change during plastic deformation

o oAL A L=

where L and LO

are the actual and original lengths of the material, respectively. The strain in the material is given by

o o

o1

AL L A

Lε

−= = − (i)


from the above. Suppose that the material is subjected to an applied load P. The actual stress is then given by σ = P/A while the nominal stress is given by σnom = P/AO. Therefore, substituting in Eq. (i) for A/A

O

nom1σε

σ= −

Then

nom (1 ) nCσ ε σ ε+ = = or

nom 1

nCεσε

=+

(ii)

Differentiating Eq. (ii) with respect to ε and equating to zero gives

1

nom2

d (1 ) 0d (1 )

n nnC Cσ ε ε εε ε

−+ −= =

+

i.e.

1(1 ) 0n nn ε ε ε−+ − = Rearranging gives

.(1 )

nn

ε =+

S.11.4

Substituting in Eq. (11.1) from Table P.11.4

4 5 6 7

4 6 7 710 10 10 10 0.39 1

5 10 10 24 10 12 10+ + + = <

× × ×

Therefore, fatigue failure is not probable.


S.12.3

From Example 12.1 and noting that there are two rivets/pitch in double shear

23( 3) 2.5 465 2 2 370

4b π ×− × × = × × ×

from which b = 12 mm


From Eq. (12.5)

12 3 100 75%12

η −= × =

S.12.4

The loading is equivalent to a shear load of 15kN acting through the centroid of the rivet group together with a clockwise moment of 15 × 50 = 750kNmm.

The vertical shear load on each rivet is 15/9 = 1.67 kN. From Example 12.2 the maximum shear load due to the moment will occur at rivets

3 and 9. Also

2 2 1/2(rivets1,3,7,9) (25 25 ) 35.4mmr = + =

(rivets 2,4,6,8) 25mmr =

(rivets5) 0r =

Then

2 2 24 35.4 4 25 7500r = × + × =∑

From Eq. (12.6)

max750 35.4 3.54kN

7500S = × =

Therefore, the total maximum shear force on rivets 3 and 9 is given by (see Example 12.2)

2 2 1/2max (total) (1.67 3.54 2 1.67 3.54cos 45 )S = + + × × °

i.e. max (total) 4.4kNS = Then

3

24.4 10350

/ 4dπ×

=

which gives 4.0mmd =

The plate thickness is given by

34.4 10 600

td×

=

from which 1.83mmt =

Solutions to Chapter 13 Problems 173 Solutions to Chapter 13 Problems

S.13.1

The total load on the bolts given by

Load on bolts = 3 × 300 × 9.81 + 300 × 9.81 = 11 772N

(a) Limit load per bolt = 11 772/4 = 2943N

(b) Ultimate load = 2943 × 1.5 = 4415N

(c) MS = (5000/4415) – 1 = 0.133

S. 13.2

For zero pitching acceleration and taking moments about the CG

1.27LW – 7.62LT

Also, for vertical equilibrium = 0 (i)

LW + LT

ie – 3W = 0

LW = 3W – LT

Substituting for L (ii)

W

1.27(3W – L

in Eq.(i)

T) – 7.62LT

which gives 8.89L

= 0

T

Then, since W = 667.5kN

– 3.81W = 0

LT

Therefore, at the section AA = 286.1kN

Limit load shear force = – 286.1 + 26.3 × 5.08 × 3 + 8.9 × 3 = 141.4kN

and the ultimate shear force is

Ult. SF = 141.4 × 1.5 = 212.2kN

174 Solutions Manual Solutions to Chapter 14 Problems

S.14.1

Suppose that the mass of the aircraft is m and its vertical deceleration is a. Then referring to Fig. S.14.1(a) and resolving forces in a vertical direction

135 2 200 0ma + − × =

Fig. S.14.1(a) and (b)

which gives 265kNma = Therefore

265 265135 /

am g

= =

i.e. 1.96a g= Now consider the undercarriage shown in Fig. S.14.1(b) and suppose that its mass is mU.C.

Then resolving forces vertically

U.C 2.25 200 0N m a+ + − = (i) in which

.2.25 1.96 4.41kNU Cm a g

g= × =

stituting in Eq. (i) gives

N = 193.3 kN Now taking moments about the point of contact of the wheel and the ground

M + N × 0.15 = 0


which gives

M = – 29. 0kNm (i.e. clockwise) The vertical distance, s, through which the aircraft moves before its vertical velocity

is zero, i.e. the shortening of the oleo strut, is obtained using elementary dynamics; the compression of the tyre is neglected here but in practice could be significant. Thus, assuming that the deceleration a remains constant

2 20 2asυ υ= +

in which υ0

= 3.5 m/s and υ = 0. Then 2 23.5 3.5

2( 1.96 ) 2 1.96 9.81s

g= − =

− × ×

i.e.

0.32ms =

Let the mass of the wing outboard of the section AA be mw

. Then, referring to Fig. S.14.1(c) and resolving forces vertically the shear force, S, at the section AA is given by

w 6.6 0S m a− − =

Fig. S.14.1(c)

i.e.

6.6 1.96 6.6 0S gg

− × − =

which gives

19.5kNS =

Now taking moments about the section AA

w w 3.05 6.6 3.05 0M m a− × − × =


or

w6.6 1.96 3.05 6.6 3.05M gg

= × × + ×

i.e. w 59.6kNmM =

S.14.2

From Example 14.2 the time taken for the vertical velocity of the aircraft to become zero is 0.099 s. During this time the aircraft moves through a vertical distance, s, which, from elementary dynamics, is given by

210 2s t atυ= +

where υ0

= 3.7 m/s and a = –3. 8 g (see Example 14.2). Then 21

23.7 0.099 3.8 9.81 0.099s = × − × × ×

i.e. 0.184ms =

The angle of rotation, θ1

, during this time is given by 21

1 0 2t tθ ω α= +

in which ω0 = 0 and α = 3.9 rad/s2

(from Example 14.2). Then 21

1 2 3.9 0.099 0.019radθ = × × =

The vertical distance, s1

, moved by the nose wheel during this rotation is, from Fig. 14.5

1 0.019 5.0 0.095ms = × =

Therefore the distance, s2

, of the nose wheel from the ground after the vertical velocity at the CG of the aircraft has become zero is given by

2 1.0 0.184 0.095s = − −

i.e.

2 0.721ms =

It follows that the aircraft must rotate through a further angle θ2

for the nose wheel to hit the ground where

20.721 0.144rad5.0

θ = =

During the time taken for the vertical velocity of the aircraft to become zero the vertical ground reactions at the main undercarriage will decrease from 1200 to 250 kN and,


assuming the same ratio, the horizontal ground reaction will decrease from 400 kN to (250/1200) × 400 = 83.3 kN. Therefore, from Eqs (ii) and (iii) of Example 14.2, the angular acceleration of the aircraft when the vertical velocity of its CG becomes zero is

21

250 3.9 0.81rad / s1200

α = × =

Thus the angular velocity, ω1

, of the aircraft at the instant the nose wheel hits the ground is given by

2 21 0 1 22ω ω α θ= +

where ω0

= 0.39rad/s (see Example 14.2). Then 2 21 0.39 2 0.81 0.144ω = + × ×

which gives

1 0.62 rad / sω =

The vertical velocity υNW

, of the nose wheel is then

NW 0.62 5.0υ = ×

i.e.

NW 3.1m / sυ =

S.14.3

The moment of inertia of a tank about its CG is

IT = 15 575 × 0.62 = 5607Nm

Its moment of inertia about the aircraft CG is then

2

IT(a/c CG) = 5607 + 15 575 × 8.52 = 1.13 × 106Nm

Weight of the aircraft without tip tanks = 169 000 – 2 × 15 575 = 137 850N

2

Therefore the moment of inertia of the aircraft about its CG is

ICG(a/c) = 137 850 × 2.52 = 0.86 × 106Nm

The drag on the aircraft is given by

2

D = (1/2)ρV2SC

ie

D

D = (1/2) × 1.223 × 2602 × 60 × 0.02 = 4960.5N


The drag on a tip tank is therefore = 0.08 × 4960.5 = 396.8N

The drag on a tip tank produces an angular acceleration,α. Therefore, taking moments about the aircraft CG

396.8 × 8.5 = (1.13 × 106 + 0.86 × 106

which gives )α

α = 1.69 × 10–3rad/sec

The inertia force on the tank is then = 1.13 × 10

2 6 × 1.69 × 10–3

The total force on the tank mounting is therefore = 396.8 + 1915 =2312N

= 1915N

S.14.4 A line diagram of the geometry of the aircraft is shown in Fig.S.14.4.

Fig. S.14.4

Taking moments about the nose of the aircraft

Wy-

which gives =0.564W × 22.5 + 0.31W × 21.0 + 2 × 0.027W × 15.5 + 2 × 0.027W × 20.5 + 0.018W × 42.0

Solutions to Chapter 14 Problems 179 21.9my =

Then the fore and aft distances from the aircraft CG are:

Fuselage:- 21.9 – 21.0 = 0.9m Wings:- 22.5 – 21.9 = 0.6m Inboard engines:- 21.9 – 15.5 = 6.4m Outboard engines:- 21.9 – 20.5 = 1.4m Tailplane:- 42.0 – 21.9 = 20.1m


The moment of inertia of the complete aircraft about its CG is then

ICG = 0.31W × 0.92 + 0.564W(0.62 + 10.52) + 2 × 0.027W(6.42 + 7.52

2 × 0.027W(1.4

) +

2 + 15.02) + 0.018W × 20.1

i.e.

2

ICG = 87.41WNm2

Then, taking moments about the CG

(W is in N)

66 750 × 15.0 + 66 750 × 7.5 = 87.41Wα

so that

α = 17 182/W rad/sec

S.14.5

2

With the usual notation the loads acting on the aircraft at the bottom of a symmetric manoeuvre are shown in Fig. S.14.5.

Fig. S.14.5

Taking moments about the CG

00.915 16.7L M P− = (i)

and for vertical equilibrium

L P nW+ = (ii)


Further, the bending moment in the fuselage at the CG is given by

CG LEV.FLT 16.7M nM P= − (iii)

Also

0

2 2 21 10 2 2 1.223 27.5 3.05 0.0638MM pV S cC V= = × × × ×

i.e.

20 9.98M V= (iv)

From Eqs (i) and (iii)

00.915( ) 16.7nW P M P− − =

Substituting for M0

from Eq. (iv) and rearranging

20.052 0.567P nW V= − (v)

In cruise conditions where, from Fig. P.14.3, n = 1 and V = 152.5 m/s, P, from Eq. (v) is given by

P = – 2994. 3N

Then, from Eq. (iii) when n = 1

LEV.FLT600000 16.7 2994.3M= + ×

which gives

LEV.FLT 549995 NmM =

Now, from Eqs (iii) and (v)

2CG 549995 16.7(0.052 0.567 )M n nW V= − −

or

2CG 379789 9.47M n V= + (vi)

From Eq. (vi) and Fig. P.14.3 it can be seen that the most critical cases are n = 3.5, V = 152.5 m/s and n = 2.5, V = 183 m/s. For the former Eq. (vi) gives

CG 1549500 NmM =

and for the latter

CG 1266600 NmM =

Therefore the maximum bending moment is 1549 500 Nm at M = 3.5 and V = 152. 5 m/s.


S.14.6

With the usual notation the loads acting on the aeroplane are shown in Fig. S.14.6; ΔΡ is the additional tail load required to check the angular velocity in pitch. Then

12.2 204000 0.25P∆ × = ×

Fig. S.14.6

i.e.

4180 NP∆ =

Now resolving perpendicularly to the flight path

2

( ) cos 40WVL P P WgR

+ + ∆ = + ° (i)

Then resolving parallel to the flight path

sin 40fW W D+ ° = (ii)

where f is the forward inertia coefficient, and taking moments about the CG

CG( ) 12.2P P M+ ∆ × = (iii)

Assume initially that

2

cos 40 WVL WgR

= ° +

i.e.

L = 230 000 cos 40° + 238 000 × 2152/(9.81 × 1525)


which gives

L = 917 704 N Then

L 2 21 12 2

917704 0.3671.223 (215) 88.5

LCV Sρ

= = =× × ×

and

21CG L2 (0.427 0.061)M V S Cρ= −

i.e.

21CG 2 1.223 215 88.5(0.427 0.367 0.061)M = × × × × −

from which

CG 239425 NmM =

Then, from Eq. (iii)

23942512.2

P P+ ∆ =

i.e. 19625 NP P+ ∆ =

Thus, a more accurate value for L is

L = 917 704 – 19 625 = 898 079 N which then gives

L 212

898079 0.3591.223 215 88.5

C = =× × ×

Hence

21CG 2 1.223 215 88.5(0.427 0.359 0.061)M = × × × × −

i.e. CG 230880 NmM =

and, from Eq. (iii) 18925 NP P+ ∆ = Then L = 917 704 – 18 925 = 898 779 N so that

898779 3.78238000

n = =


At the tail

230880 12.2 1.41204000 9.81

lngθ

∆ = = × =

Thus the total n at the tail = 3.78 + 1.41 = 5.19. Now

2

D 212

8987790.0075 0.045 0.0128CV Sρ

= + × +

i.e.

D 0.026C =

so that

212 0.026 65 041 ND V Sρ= × =

Thus, from Eq. (ii)

f = –0.370

S.14.7

From Eq. (14.21) φ, in Fig. 14.10, is given by

2 2168tan 4.72

9.81 610VgR

φ = = =×

so that

φ = 78.03° From Eq. (14.20)

n = sec φ = 4.82 Thus, the lift generated in the turn is given by

L = nW = 4.82 × 133 500 = 643 470 N Then

L 2 21 12 2

643 470 0.801.223 168 46.5

LCV Sρ

= = =× × ×

Hence CD=0.01 + 0.05 × 0.802

and the drag = 0.042

212 1.223 168 46.5 0.042 33707 ND = × × × × =


The pitching moment M0

is given by 2 21 1

0 M,02 2 1.223 168 46.5 3.0 0.03M V S cCρ= = − × × × × ×

i.e.

0 72229 Nm (i.e nose down)M = −

The wing incidence is given by

L

L

0.80 180 10.2/d 4.5

CdC

αα π

= = × = °

The loads acting on the aircraft are now as shown in Fig. S.14.7.

Fig. S.14.5

Taking moments about the CG

0(0.915cos 10.2 0.45sin10.2 ) (0.45 cos10.2 0.915sin10.2 )7.625 cos 10.2

L D MP

° + ° − ° − ° −

= × ° (i)

Substituting the values of L, D and M0

P = 73 160 N

in Eq. (i) gives

S.14.8

(a) The forces acting on the aircraft in the pull-out are shown in Fig. S.14.8. Resolving forces perpendicularly to the flight path

2

cosWVL WgR

θ= + (i)

The maximum allowable lift is 4.0W so that Eq. (i) becomes

2

4 cosVgR

θ= −


Fig. S.14.6

or

4 cosVgω θ= − (ii)

where ω (=V/R) is the angular velocity in pitch. In Eq. (ii) ω will be a maximum when cos θ is a minimum, i.e. when θ reaches its maximum allowable value (60°). Then, from Eq. (ii)

3.5(4 0.5)g gV V

ω = − = (iii)

From Eq. (iii) ω will be a maximum when V is a minimum which occurs when CL =CL.MAX

Thus 2 21 1

L.MAX s L.MAX2 24V SC V SCρ ρ= ×

whence

s2 2 46.5 93.0m / sV V= = × =

Therefore, from Eq. (iii)

max3.5 9.81 0.37 rad / s

93.0ω ×

= =

(b) Referring to Fig. 14.10, Eq. (14.17) gives

2

sin WVnWgR

φ =

i.e.

4sin Vgωφ = (iv)

Also, from Eq. (14.20) sec φ = 4 whence sin φ = 0.9375. Then Eq. (iv) becomes

3.87 gV

ω = (v)


Thus, ω is a maximum when V is a minimum, i.e. when V = 2Vs

as in (a). Therefore

max3.87 9.81 0.41rad / s

2 46.5ω ×

= =×

The maximum rate of yaw is ωmax

maximum rate of yaw = 0.103 rad/s

cos φ, i.e.

S.14.9

The forces acting on the airliner are shown in Fig. S.14.9 where αw

is the wing incidence. As a first approximation let L = W. Then

2 Lw

1 16000002

CV Sρ αα

∂=

∂

Fig. S.14.9

i.e.

w 212

1600000 1800.116 610 280 1.5

απ

×=

× × × × ×

so that

αw

From vertical equilibrium = 101°

L + P = W (i)

and taking moments about the CG.

042.5 cos 10.1 7.5 cos 10.1P L M× ° = × ° + (ii)

Substituting for L from Eq. (i) in Eq. (ii)

212

42.5 cos 10.1 (1600000 )7.5 cos 10.1

.0116 610 280 22.8 0.01

P P× ° = − °

+ × × × × ×


from which

267963NP =

Thus, from Eq. (i)

1332037 NL =

giving

w 8.4α = °

Then, taking moments about the CG

12

2

42.5 cos 8.4 (1600000 )7.5 cos 8.4

0.116 610 280 22.8 0.01

P P= ° = − ° +

× × × × ×

which gives

P = 267 852 N This is sufficiently close to the previous value of tail load to make a second approxi-mation unnecessary.

The change Δα in wing incidence due to the gust is given by

18 0.03 rad610

α∆ = =

Thus the change ∆P in the tail load is

2 L.TT

12

CP V Sρ αα

∂∆ = ∆

∂

i.e.

212 0.116 610 28 2.0 0.03 36257 NP∆ = × × × × × =

Also, neglecting downwash effects, the change ∆L in wing lift is

2 L12

CL V Sρ αα

∂∆ = ∆

∂

i.e.

212 0.116 610 280 1.5 0.03 271931 NP∆ = × × × × × =

The resultant load factor, n, is then given by

36257 27193111600000

n += +

i.e.

n = 1.19


S.14.10 As a first approximation let L = W. Then

2 Lw

d1 1450002 d

CV Sρ αα

=

Thus

w2

145000 0.0158 rad = 0.91°1 1.233 250 50 4.82

α = =× × × ×

Also

2D 0.021 0.041 0.08C = + ×

i.e.

D 0.0213C =

Referring to Fig. P.14.8 and taking moments about the CG and noting that cos 0.91° ~

1

00.5 0.4 8.5L D M P× − × + = ×

i.e.

2 2D M,0

1 10.5(145000 ) 0.4 8.52 2

P V SC V ScC Pρ ρ− − × + =

Thus

2

2

1 10.5(145000 ) 0.4 1.223 250 50 0.02132 2

1.223 250 50 2.5 0.032 8

P− − × × × × × −

× × × × × =

which gives

P = –10740 N The change ∆Ρ in the tail load due to the gust is given by

2 L.TT

12

CP V Sρ αα

∂∆ = ∆

∂

in which

6 0.024 rad250

α∆ = − = −

Thus

21 1.223 250 9.0 2.2 0.024 18162 N2

P∆ = × × × × × = −

Therefore the total tail load = –10740 –18 162= –28 902 N.


The increase in wing lift ∆L due to the gust is given by

2 2L1 1 1.223 250 50 4.8 0.0242 2

CL V Sρ αα

∂∆ = − ∆ = − × × × × ×

∂

i.e.

∆L = –220 140N Hence

(220 140 18162)1 0.64145000

n += − = −

Finally the forward inertia force fW is given by

2 21 1D2 2 1.223 250 50 0.0213fW D V SCρ= = = × × × ×

i.e.

fW = 40 703 N


S.15.1

Substituting the given values in Eq. (15.3)

aa 2 230 1

2 230SS = × − ×

from which 2

a 363N / mmS =

S.15.2

From Eq. (15.4)

2

aa 2 230 1

2 870SS

= × − ×

i.e. 4 2

a a460 1.519 10S S−= − × or

2a a6581.7 3027600 0S S − =

Solving, 2

a 432 N / mmS =


S.15.3

From Eq. (15.5) and supposing that the component fails after N sequences of the three stages

4 5 5

200 200 600 110 10 2 10

N + + = ×

which gives

N = 40 The total number of cycles/sequence is 1000 so that at 100 cycles/day the life of the component is

100040 400days100

× = .

S.15.4

From Eq. (15.26)

5.26g e( )D F V=

so that

5.26g (200) (200) 1.269D F F= =

5.26g (220) (220) 2.095D F F= =

Then

g

g

(220) 2.095 1.65(200) 1.269

D FD F

= =

i.e.

Increase = 65%.

S.15.5

From Eq. (15.26)

5.26 12g (240) (240) 3.31 10D F F= = ×

5.26 12g (235) (235) 2.96 10D F F= = ×


Then, since

gag TOT0.1D D=

12TOT TOT(240) 0.1 3.31 10D D F= + ×

and

12TOT TOT(235) 0.1 2.96 10D D F= + ×

Therefore

12TOT0.9 (240) 3.31 10D F= ×

12TOT (240) 3.68 10D F= ×

Similary

12TOT (235) 3.29 10D F= ×

12

12

3.68 10Then, the increase in flights 1.123.29 10

FF

×= =

×

i.e. a 12% increase.

S.15.6

The damage per km caused by gusts is given by Eq.(15.26) in which we assume that Kn = 1.43 and C = 1000. From gust data l10

D

= 6000m so that

g = (46.55/2 × 6000)(1.43/1000)2(3.0/400)5.26(220)which gives

5.26

Dg = 1.11 × 10Then

–7

D total = 1.11 × 10–7 × 1200 + 0.08Dso that

total

0.92D total = 1.33 × 10and

–4

D total = 1.44 × 10–4

Therefore the number of flights is 1/1.44 × 10

per flight –4

The journey time per flight = 1200 × 10 = 6944.

3/220 × 602

The life of the aircraft is then =1.52 × 6944 = 10 555, say 10 550 hours. = 1.52hours.


S.15.7

From Eq. (15.30)

123320 ( 2.0) 1.0S π= × ×

which gives

S = 1324 N/mm2

S.15.8

.

From Eq. (15.30)

1

2f( ) 1.12K S aπ= ×

so that

2

f 2 2

1800180 1.12

aπ

=× ×

i.e. f 25.4mma = Now from Eq. (15.44)

f 115 42

1 1 10.4 25.430 10 (180 )

Nπ−

= − × ×

i.e. Nf

S.15.9 = 7916 cycles.

For a plate of finite width α = [sec(πa/w)]i.e.

1/2

α = [sec(3π/50)]which gives

1/2

α = 1.01 From Eq.(15.32) K0

From Eq.(15.34) = 3500/1.01 = 3465.3

3465.3 = (3M × 3/4x1.23)(πx3)from which

1/2

M = 866.9Nmm/unit thickness Then maximum M = 866.9 × 5 = 4334.5N mm = 4.3Nm.


S.16.1

From Section 16.2.2 the components of the bending moment about the x and y axes are, respectively

3 63000 10 cos30 2.6 10 N mmxM = × ° = ×

3 63000 10 sin 30 1.5 10 N mmyM = × ° = ×

The direct stress distribution is given by Eq. (16.18) so that, initially, the position of the centroid of area, C, must be found. Referring to Fig. S.16.1 and taking moments of area about the edge BC

(100 10 115 10) 100 10 50 115 10 5x× + × = × × + × ×

i.e.

25.9mmx =

Fig. S.16.1

Now taking moments of area about AB

(100 10 115 10) 100 10 5 115 10 67.5y× + × = × × + × ×

from which

38.4mmy =


The second moments of area are then

3 32 2

6 4

100 10 10 115100 10 33.4 10 115 29.112 12

3.37 10 mm

xxI × ×= + × × + + × ×

= ×

3 32 2

6 4

10 100 115 1010 100 24.1 115 10 20.912 12

1.93 10 mm

yyI × ×= + × × + + × ×

= ×

6 4

100 10 33.4 24.1 115 10( 20.9)( 29.1)

1.50 10 mmxyI = × × × + × − −

= ×

Substituting for Mx, My, Ixx, Iyy and Ixy

in Eq. (16.18) gives 0.27 0.65z x yσ = + (i)

Since the coefficients of x and y in Eq. (i) have the same sign the maximum value of direct stress will occur in either the first or third quadrants. Then 2

(A) 0.27 74.1 0.65 38.4 45.0 N / mm (tension)zσ = × + × =

2(C) 0.27 ( 25.9) 0.65 ( 86.6) 63.3N/mm (compression)zσ = × − + × − = −

The maximum direct stress therefore occurs at C and is 63.3 N/mm2

S.16.2 compression.

The bending moments half-way along the beam are

Mx = – 800 × 1000 = – 800 000 N mm My

By inspection the centroid of area (Fig. S.16.2) is midway between the flanges. Its distance

= 400 × 1000 = 400 000 N mm

x from the vertical web is given by (40 2 100 2 80 1) 40 2 20 80 1 40x× + × + × = × × + × × i.e. 13.33mmx = The second moments of area of the cross-section are calculated using the approxima-tions for thin-walled sections described in Section 16.4.5. Then

3

2 2 5 42 10040 2 50 80 1 50 5.67 10 mm12xxI ×

= × × + × × + = ×

3 32 2

2

5 4

2 40 1 80100 2 13.33 2 40 6.6712 12

1 80 26.671.49 10 mm

yyI × ×= × × + + × × +

+ × ×

= ×

5 440 2(6.67)(50) 80 1(26.67)( 50) 0.8 10 mmxyI = × + × − = − ×


Fig. S.16.2

The denominator in Eq. (16.18) is then (5.67 × 1.49 – 0.82) × 1010 = 7.81 × 1010

. From Eq. (16.18)

5 5

10

5 5

10

400000 5.67 10 800000 0.8 107.81 10

800000 1.49 10 400000 0.8 107.81 10

x

y

σ × × − × ×

= × − × × + × ×

+ ×

i.e.

2.08 1.12x yσ = −

and at the point A where x = 66.67 mm, y = – 50 mm

2(A) 194.7 N / mm (tension)σ =

S.16.3

Initially, the section properties are determined. By inspection the centroid of area, C, is a horizontal distance 2a from the point 2. Now referring to Fig. S.16.3 and taking moments of area about the flange 23

(5 4 ) 5 (3 /2)a a ty at a+ =

from which

5 /6y a=


Fig. S.16.3

From Section 16.4.5

2 3 2 2 34 (5 /6) (5 ) (3/5) /12 5 (2 /3) 105 /12xxI at at a t at a a t= + + =

3 3 2 3(4 ) /12 (5 ) (4/5) /12 12yyI t a a t a t= + =

3 3(5 ) (3/5)(4/5)/12 5xyI t a a t= =

From Fig. P.16.3 the maximum bending moment occurs at the mid-span section in a horizontal plane about the y axis. Thus

20 (max) /8x yM M wl= =

Substituting these values and the values of Ixx, Iyy and Ixy

in Eq. (16.18) 2

3

18 64 16zwl x ya t

σ 7 = −

(i)

From Eq. (i) it can be seen that σz

varies linearly along each flange. Thus 2

,1 2

13At1where 26 96za wlx a y

a tσ= = =

2

,2 2

5At 2 where 26 48z

a wlx a ya t

σ− −= − = =

2

,3 2

5 13At 3where 26 384z

a wlx a ya t

σ−= = =

Therefore, the maximum stress occurs at 3 and is 13wl2/384a2t.


S.16.4

Referring to Fig. S.16.4.

Fig. S.16.4

In DB Mx

M = –W(l – z) (i)

y

In BA = 0

Mx

= –W(l – z) (ii)

22ylM W z = − −

(iii)

Now referring to Fig. P.16.4 the centroid of area, C, of the beam cross-section is at the centre of antisymmetry. Then

2 3 322

2 12 3xxd td tdI td

= + =

2 23 352

4 12 4 12yyd td d tdI td td

= + + =

3

4 2 4 2 4xyd d d d tdI td td = + − − =

Substituting for Ixx, Iyy and Ixy

in Eq. (16.18) gives

3

1 (3.10 1.16 ) (1.94 1.16 )z y x x yM M x M M ytd

σ = − + − (iv)


Along the edge l, x = 3d/4, y = d/2. Equation (iv) then becomes

,1 2

1 (1.75 0.1 )z y xM Mtd

σ = + (v)

Along the edge 2, x = –d/4, y = d/2. Equation (iv) then becomes

,2 2

1 ( 1.36 1.26 )z y xM Mtd

σ = − + (vi)

From Eqs (i)–(iii), (v) and (vi) In DB

,1 ,12 2

0.1 0.05(1 ) whence (B)z zW Wlz

td tdσ σ= − − = −

,2 ,22 2

1.26 0.63(1 ) whence (B)z zW Wlz

td tdσ σ= − − = −

In BA

,1 ,12 2

1.85(3.6 1.85 ) whence (A)z zW Wlz ltd td

σ σ= − = −

,2 ,22 2

0.1( 1.46 0.1 ) whence (A)z zW Wlz ltd td

σ σ= − + =

S.16.5 By inspection the centroid of the section is at the mid-point of the web. Then

3 3

2 2 2 (2 ) 10(2 )2 12 3xxh t h h tI t h ht h = + + =

3 3 32 ( /2) 5

3 3 12yyt h th h tI = + =

332 ( ) ( )

2 4 2 4xyh h h h tI t h ht h = − + − = −

Since My

= 0, Eq. (16.18) reduces to

2 2x xy x yy

zxx yy xy xx yy xy

M I M Ix y

I I I I I Iσ

−= +

− − (i)

Substituting in Eq. (i) for Ixx

, etc.

3 2 2

3/4 5/12(10/3)(5/12) (3/4) (10/3)(5/12) (3/4)

xz

M x yh t

σ

= + + − −

3i.e. (0.91 0.50 )xz

M x yh t

σ = + (ii)


Fig. S.16.5

Between 1 and 2,y = –h and σz

is linear. Then

,1 3 2

0.41(0.91 0.5 )xz x

M h h Mh t h t

σ = × − =

,2 3 2

0.5(0.91 0 0.5 )xz x

M h Mh t h t

σ = × − = −

Between 2 and 3, x = 0 and σz

is linear. Then

,2 2

0.5z xM

h tσ −

,3 3 2

0.5(0.91 0 0.5 )xz x

M h Mh t h t

σ = × + =

,4 3 2

0.040.91 0.52

xz x

M h h Mh t h t

σ = − × + =

S.16.6 The centroid of the section is at the centre of the inclined web. Then,

3 22 3(2 ) sin 602 ( sin 60 ) 2

12xxt aI ta a a t°

= ° + =

3 3 2 3(2 ) cos 602

12 12 3yyta t a a tI °

= × + =

3 3(2 ) sin 60 cos60 3

12 6xyt a a tI ° °

= =


Fig. S.16.6

Substituting in Eq. (16.18) and simplifying (My

= 0)

3

4 2 37 7

xz

M y xa t

σ

= −

(i)

On the neutral axis, σz

= 0. Therefore, from Eq. (i) 3

2y x=

and

3tan2

α =

so that 40.9α = ° The greatest stress will occur at points furthest from the neutral axis, i.e. at points 1 and 2. Then, from Eq. (i) at 1,

,max 3

4 3 2 37 2 7 2

xz

M a aa t

σ

= × + ×

i.e. ,max 2 2

3 3 0.747

xz x

M Ma t ta

σ = =

S.16.7 Referring to Fig. P.16.7, at the built-in end of the beam Mx

M = 50 × 100 – 50 × 200 = –5000 N mm

y = 80 × 200 = 16 000 N mm


and at the half-way section

50 100 5000 N mmxM = − × = −

80 200 8000 N mmyM = × =

Fig. S.16.7

Now referring to Fig. S.16.7 and taking moments of areas about 12

(24 × 1.25 + 36 × 1.25) x = 36 × 1.25 × 18

which gives

10.8mmx =

Taking moments of areas about 23

(24 1.25 36 1.25) 24 1.25 12y× + × = × ×

which gives

4.8mmy =

Then

3

2 2 41.25 24 1.25 24 7.2 1.25 36 4.8 4032mm12xxI ×

= + × × + × × =

3

2 2 41.25 361.25 24 10.8 1.25 36 7.2 10692mm12yyI ×

= × × + + × × =

41.25 24 ( 10.8)(7.2) 1.25 36 (7.2)( 4.8) 3888mmxyI = × × − + × × − = −


in Eq. (16.18) gives

4 4(1.44 1.39 ) 10 (3.82 1.39 ) 10z y x x yM M x M M yσ − −= + × + + × (i)

Thus, at the built-in end Eq. (i) becomes

1.61 0.31z x yσ = + (ii)


whence σz,1 = –11.4 N/mm2, σz,2 = –18.9 N/mm2, σz,3 = 39.1 N/mm2

. At the half-way section Eq. (i) becomes

0.46 0.80z x yσ = − (iii)

whence σz1 = –20.3 N/mm2, σz2 = –1.1 N/mm2, σz3 = –15.4 N/mm2

S.16.8

.

The section properties are, from Fig. S.16.8

2 3

02 ( cos ) d 3xxI t r r r tr

πθ θ π= − =∫

2 3

02 ( sin ) dyyI t r r tr

πθ θ π= =∫

3

02 ( sin )( cos ) d 4xyI t r r r r tr

πθ θ θ= − − = −∫

Fig. S.16.8

Since My


3

3

4tan xy

yy

I trI tr

απ

= − =

i.e.

51.9α = °


Substituting for Mx = 3. 5 × 103 N mm and My

= 0, Eq. (16.18) becomes 3

3

10 (1.029 0.808 )z x ytr

σ = + (i)

The maximum value of direct stress will occur at a point a perpendicular distance furthest from the neutral axis, i.e. by inspection at B or D. Thus

3

3

10(max) (0.808 2 5)0.64 5zσ = × ×

×

i.e.

2(max) 101.0 N /mmzσ =

Alternatively Eq. (i) may be written

[ ]3

3

10 1.029( sin ) 0.808( cos )z r r rtr

σ θ θ= − + −

or

2

808 (1 cos 1.27sin )z trσ θ θ= − − (ii)

The expression in brackets has its greatest value when θ = π, i.e. at B (or D).

S.16.9

The beam is as shown in Fig. S.16.9

Fig. S.16.9

Taking moments about B

2

A30 6 60 66 6 0

2 2 3R ×

× − − × × =


which gives

RA

The bending moment at any section a distance z from A is then = 150 kN

230150 (90 30)

2 6 2 3z z z zM z = − + + −

i.e.

3

2 5150 153zM z z= − + +

Substituting in the second of Eqs (16.33)

2 3

22

d 5150 15d 3

zEI z zzυ

= − −

4

2 31

d 575 5d 12

zEI z z Czυ = − − +

4 5

31 2

5254 12z zEI z C z Cυ = − − + +

When x = 0, υ = 0 so that C2

= 0 and when z = 6m, υ = 0. Then 4 5

31

5 6 60 25 6 64 12

C×= × − − +

from which

1 522C = −

and the deflected shape of the beam is given by

4 5

3 525 5224 12z zEI z zυ = − − −

The deflection at the mid-span point is then

4 5

3 3mid-span

5 3 325 3 522 3 1012.5kN m4 12

EIυ ×= × − − − × = −

Therefore

12

mid-span 6

1012.5 10 41.0mm (downwards)120 10 206000

υ − ×= = −

× ×


S.16.10

Take the origin of x at the free end of the cantilever. The load intensity at any section a distance z from the free end is wz/L. The bending moment at this section is given by

3

2 3 6zz wz z wzM

L L = =

Substituting in Eqs (16.32)

2 3

2

dd 6

wzEIz Lυ −

=

4

1dd 24

wzEI Cz Lυ − = +

5

1 2120wzEI C z C

Lυ −= + +

When z = L, (dυ/dz) = 0 so that C1 = wL3/24. When z = L, υ = 0, i.e. C2 = –wL4

/30. The deflected shape of the beam is then

5 4 5( 5 4 )120

wEI z zL LL

υ = − − +

At the free end where z= 0

4

30EIwLυ = −

S.16.11

The uniformly distributed load is extended from D to F and an upward uniformly distributed load of the same intensity applied over DF so that the overall loading is unchanged (see Fig. S.16.11).

Fig. S.16.11

The support reaction at A is given by A 6 6 5 4 3 1 2 2 0R × − × − × − × × =


Then A 7.7 kNR =

Using Macauley’s method, the bending moment in the bay DF is

2 21[ 3] 1[ 5]7.7 6[ 1] 4[ 3]

2 2z zM z z z − −

= − + − + − + −

Substituting in Eqs (16.33)

2 2 2

2

d [ 3] [ 5]7.7 6[ 1] 4[ 3]d 2 2

z zEI z z zzυ − −

= − − − − − +

2 3 3

2 21

d 7.7 [ 3] [ 5]3[ 1] 2[ 3]d 2 6 6

z z zEI z z Czυ − − = − − − − − − +

3 3 4 4

31 2

7 . 7 2[ 3] [ 3] [ 5][ 1]6 3 24 24z z z zEI z C z Cυ − − −

= − − − − − + +

When z = 0, υ = 0 so that C2

= 0. Also when z = 6 m, υ = 0. Then 3 3 4 4

31

7.7 6 2 3 3 10 5 66 3 24 24

C× ×= − − − − +

which gives 1 21.8C = −

Guess that the maximum deflection lies between B and C. If this is the case the slope of the beam will change sign from B to C. At B

2d 7.7 1 21.8 which isclearly negative

d 2EI

zυ × = −

At C

2

2d 7.7 3 3 2 21.8 0.85d 2

EIzυ × = − × − = +

The maximum deflection therefore occurs between B and C at a section of the beam where the slope is zero. i.e.

2

27.70 3[ 1] 21.82z z= − − −

Simplifying 2 7.06 29.2 0z z+ − = Solving z = 2.9 m


The maximum deflection is then

3

3max

7.7 2.9 1.9 21.8 2.9 38.86

EIυ ×= − − × = −

i.e.

max38.8 (downwards)EI

υ −=

S.16.12

Taking moments about D A 4 100 100 2 1 200 3 0R × + − × × + × = from which A 125 NR = − Resolving vertically B 125 100 2 200 0R − − × − = Therefore B 525 NR = The bending moment at a section a distance z from A in the bay DF is given by

2 20 100[ 2] 100[ 4]125 100[ 1] 525[ 4]

2 2z zM z z z− −

= + − − + − − −

in which the uniformly distributed load has been extended from D to F and an upward uniformly distributed load of the same intensity applied from D to F.

Substituting in Eqs (16.33) 2

0 2 22

d 125 100[ 1] 50[ 2] 525[ 4] 50[ 4]d

EI z z z z zzυ

= − + − − − + − + −

2 3 2 31

1d 125 50[ 2] 525[ 4] 50[ 4]100[ 1]d 2 3 2 3

z z z zEI z Czυ − − − − = + − − + + +

3 4 32

4

1 2

125 50[ 2] 525[ 4]50[ 1]6 12 6

50[ 4]12

z z zEI z

z C z C

υ − − −= + − − +

−+ + +

When z = 0, υ = 0 so that C2 = 0 and when z = 4 m, υ = 0 which gives C1

3 4 3 421 125 50[ 2] 525[ 4] 50[ 4]50[ 1] 237.5

6 12 12 12z z z zz z

EIυ

− − − −= + − − + + +

= 237.5. The deflection curve of the beam is then


S.16.13

From Eqs (16.30) the horizontal component of deflection, u, is given by

2( )x xy y xx

xx yy xy

M I M Iu

E I I I−

′′ =−

(i)

in which, for the span BD, referring to Fig. P.16.13, Mx = –RDz, My = 0, where RD

is the vertical reaction at the support at D. Taking moments about B

D 2 0R l Wl+ =

so that

D /2 (downward)R W= −

Eq. (i) then becomes

22 ( )xy

xx yy xy

WIu z

E I I I′′ =

− (ii)

From Fig. P.16.13 23 3 3

2(2 ) ( /2) 3 132 ( ) 212 12 2 4 4xx

t a t a a a a tI at a t = + + + =

3 32(2 ) 52 ( )

12 2 3yyt a a a tI t a= + =

33 3 7( ) ( ) ( ) ( )2 4 2 2 4 2 4xya a a a a a a tI t a at a t a at a = − + − + − + − = −

Equation (ii) then becomes

3

42113

WuEa t

′′ = − (iii)

Integrating Eq. (iii) with respect to z

23

21113

Wu z AEa t

′ = − +

and

33

7113

Wu z Az BEa t

= − + + (iv)

When z = 0, u = 0 so that B = 0. Also u = 0 when z = 2l which gives

3

28113

WlAEa t

= −


Then

3 23

7 ( 4 )113

Wu z l zEa t

= − + (v)

At the mid-span point where z = l, Eq. (v) gives

3

3

0.186WluEa t

=

Similarly

3

3

0.177WlEa t

υ =

S.16.14

(a) From Eqs (16.30)

2( )x xy y xx

xx yy xy

M I M Iu

E I I I−

′′ =−

(i)

Referring to Fig. P.16.14

2( )2xwM l z= − − (ii)

and ( )yM T l z= − − (iii)

in which T is the tension in the link. Substituting for Mx and My

from Eqs (ii) and (iii) in Eq. (i).

22

1 ( ) ( )( ) 2

xyxx

xx yy xy

Iu w l z TI l z

E I I I

′′ = − − − − −

Then

3 2

2 22

1( ) 2 3 2

xyxx

xx yy xy

I z zu w l z lz TI lz AE I I I

′ = − − + − − + −

When z = 0, u′ = 0 so that A = 0. Hence

2 3 4 2 3

2 22

1( ) 2 2 3 12 2 6

xyxx

xx yy xy

I z z z z zu w l l TI l BE I I I

= − − + − − + −

When z = 0, u = 0 so that B = 0. Hence

2 3 4 2 3

2 22

1( ) 2 2 3 12 2 6

xyxx

xx yy xy

I z z z z zu w l l TI lE I I I

= − − + − − −

(iv)


Since the link prevents horizontal movement of the free end of the beam, u = 0 when z = l. Hence, from Eq. (iv)

4 4 4 3 3

02 2 3 12 2 6xy

xx

I l l l l lw TI

− + − − =

whence

38

xy

xy

wlIT

I=

(b) From Eqs (16.30)

2( )x yy y xy

xx yy xy

M I M IE I I I

υ−

′′ =−

(v)

The equation for υ may be deduced from Eq. (iv) by comparing Eqs (v) and (i). Thus

2 3 4 2 3

2 22

1( ) 2 2 3 12 2 6

xyxy

xx yy xy

I z z z z zw l l TI lE I I I

υ

= − + − − − (vi)

At the free end of the beam where z = I

4 3

FE 2

1( ) 8 3

yyxy

xx yy xy

wI l lTIE I I I

υ

= − −

which becomes, since T =3 wlIxy/8I

xx 4

FE 8 xx

wlEI

υ =

S.16.15

The beam is allowed to deflect in the horizontal direction at B so that the support reaction, RB

, at B is vertical. Then, from Eq. (5.12), the total complementary energy, C, of the beam is given by

B B C0d d

M

LC M R Wθ= − ∆ − ∆∫ ∫ (i)

From the principle of the stationary value of the total complementary energy of the beam and noting that ΔB

= 0

B B

d 0L

C MR R

θ∂ ∂= =

∂ ∂∫

Thus

B B

d 0L

C M M zR EI R∂ ∂

= =∂ ∂∫ (ii)


In CB B(2 ) and / 0M W l z M R= − ∂ ∂ =

In BA B B(2 ) ( ) an d / ( )M W l z R l z M R l z= − − − ∂ ∂ = − −

Substituting in Eq. (ii)

B0[ (2 ) ( )]( )d 0

lw l z R l z l z z− − − − =∫

from which

B52WR =

Then

C B A0 /2M M Wl M Wl= = = −

and the bending moment diagram is as shown in Fig. S.16.15.

Fig. S.16.15

S.16.16

From Eq. (16.50) and Fig. P.16.4

0 0 0(4 2 2 )TN E T dt T dt T dtα= + × +

i.e.

09TN E dtTα=

From Eq. (16.52)

0 0 04 2 2 (0)2 2xTd dM E T dt T dt T dtα = + × + −

i.e.

2

032xT

E d tTM α=


From Eq. (16.52)

0 0 0 04 2 24 4 4 4yTd d d dM E T dt T dt T dt T dtα = + + − + −

i.e.

2

034yT

E d tTM α=

S.16.17

Taking moments of areas about the upper flange

( 2 ) 2at at y at a+ =

Fig. S.16.17

which gives

23

y a=

Now taking moments of areas about the vertical web

32aatx at=

so that

6ax =


From Eq. (16.53)

0 0d d2 2T A A

T y E T tN E t s y sa a

αα= =∫ ∫

But dA

t y s∫ is the first moment of area of the section about the centroidal axis Cx

dA

t y s∫

,

i.e. = 0. Therefore

0TN =

From Eq. (16.54)

2 20 0d d2 2xT A A

T E TM E ty s ty sa a

αα= =∫ ∫

But

2 23

2 2 (2 )d 23 3 3xxA

a aty s I at a t at = = + + ∫

i.e.

310

3xxa tI =

Therefore

2

053xT

E a tTM α=

From Eq. (16.55)

0 0d d2 2yT A A

T E TM E txy s txy sa A

αα= =∫ ∫

But

2d 23 3 6 3xyA

a a atxy s I at a at = = + − − ∫

i.e.

3

3xya tI =

Then

2

0

6yTE a tTM α

=



S.17.1

In Fig. S.17.1 the x axis is an axis of symmetry (i.e. Ixy = 0) and the shear centre, S, lies on this axis. Suppose S is a distance ξS from the web 24. To find ξS an arbitrary shear load Sy is applied through S and the internal shear flow distribution determined. Since Ixy =0 and Sx

=0, Eq. (17.14) reduces to

0d

sys

xx

Sq ty s

I= − ∫ (i)

Fig. S.17.1

in which

23 3 2sin2

12 12 2xxth td hI tdα = + +

i.e.

3

3 2(1 6 2 sin )12xxthI ρ ρ α= + + (ii)

Then

1

12 10d

sy

xx

Sq ty s

I= − ∫

i.e.

1

12 1 10sin d

2 2sy

xx

S h dq t s sI

α = + −

∫


so that

212 1 1 1( sin sin )

2y

xx

S tq hs ds s

Iα α= + − (iii)

Also

2 2

32 2 2 20 0d sin d

2 2s sy y

xx xx

S S t h dq ty s s sI I

α = − = − −

∫ ∫

whence

232 2 2 2( sin sin )

2y

xx

S tq hs ds s

Iα α− + (iv)

Taking moments about C in Fig. S.17.1

/2 /2

12 1 32 20 0ξs 2 cos d 2 cos d

2 2d d

yh hS q s q sα α= − +∫ ∫ (v)

Substituting in Eq. (v) for q12 and q32

/2 21 1 1 10

/2 22 2 2 20

cosξs ( sin sin )d

( sin sin )d

dyy

xx

d

S thS hs ds s s

I

hs ds s s

αα α

α α

= − + −

+ − +

∫

∫

from Eqs (iii) and (iv)

from which

3 sin cosξs

12 xx

thdIα α

= − (vi)

Now substituting for Ixx

from Eq. (ii) in (vi) 2

3 2sin cosξs

1 6 2 sind ρ α α

ρ ρ α= −

+ +

S.17.2

The x axis is an axis of symmetry so that Ixy = 0 and the shear centre, S, lies on this axis (see Fig. S.17.2). Therefore, an arbitrary shear force, Sy

Since S

, is applied through S and the internal shear flow distribution determined.

x =0 and Ixy


0d

sys

xx

Sq ty s

I= − ∫ (i)

in which, from Fig. S.17.2.,

2 23 2 3 2sin sin2 sin sin sin

12 2 12 2xxa t a a t aI at a atα αα α α = + + + +


Fig. S.17.2

which gives

3 216 sin

3xxa tI α (ii)

For the flange 54, from Eq. (i)

54 0( 2 )sin d

sy

xx

Sq t s a s

Iα= − −∫

from which

2

54sin

22

y

xx

S t sq asI

α = − −

(iii)

Taking moments about the point 3

540ξs 2 sin 2 d

ayS q a sα= ∫ (iv)

Substituting in Eq. (iv) for q54

from Eq. (iii) 2

0

2 sin 2 sinξs 2 d

2ay

yxx

a S t sS as sIα α

= − −

∫

which gives

32 sin 2 sin 5ξs

6xx

at aIα α

=

(v)

Substituting for Ixx

from Eq. (ii) in (v) gives 5 cosξs

8a α

=


S.17.3

The shear centre, S, lies on the axis of symmetry a distance ξS from the point 2 as shown in Fig. S.17.3. Thus, an arbitrary shear load, Sy, is applied through S and since Ixy = 0, Sx

= 0, Eq. (17.14) simplifies to

0d

sys

xx

Sq ty s

I= − ∫ (i)

Fig. S.17.3

in which Ixx has the same value as the section in S.16.8, i.e. 3πr3

t. Then Eq. (i) becomes

12 0( cos ) dy

xx

Sq t r r r

Iθ

θ θ= +∫

or

[ ]12 0sin3

ySq

rθθ θ

π= +

i.e.

12 ( sin )3

ySq

rθ θ

π= + (ii)



120ξs 2 ( cos ) dyS q r r r

πθ θ= +∫ (iii)


Substituting in Eq. (iii) for q12

from Eq. (ii)

0

2ξs ( sin )(1 cos )d

3y

yS r

Sπθ θ θ θ

π= + +∫

Thus

0

2ξs ( cos sin sin cos )d3

r πθ θ θ θ θ θ θ

π= + + +∫

i.e.

2

0

2 cos 2ξs sin3 2 4

rπ

θ θθ θπ

= + −

from which

ξs3rπ

=

S.17.4

The x axis is an axis of symmetry so that Ixy = 0 and the shear centre, S, lies on this axis (see Fig. S.17.4). Further Sx

= 0 so that Eq. (17.14) reduces to

0d

sys

xx

Sq ty s

I= − ∫ (i)

Fig. S.17.4


2 23

2212 2 2 12xxth h t h hI td d th dβ

β

= + + = +


From Eq. (i)

1

12 10d

2sy

xx

S hq t sI

= − − ∫

i.e.

12 12y

xx

S thq s

I= (ii)

Also

2

32 20d

2sy

xx

S t hq sI β

= − − ∫

so that

32 22y

xx

S thq s

Iβ= − (iii)

Taking moments about the mid-point of the web

12 1 32 20 0ξs 2 d 2 d

2 2d d

yh hS q s q s

β= −∫ ∫ (iv)

Substituting from Eqs (ii) and (iii) in Eq. (iv) for q12 and q

32 2 2

1 1 2 20 0ξs d d

2 2d dy y

yxx xx

S th S thS s s s s

I Iβ

β= −∫ ∫

i.e.

2 2 2

ξs2 2 2xx

th d dI

β

= −

i.e.

2 2

3(1 )ξs

4 (1 12 / ) / 12th d

th d hβ−

=+

so that

ξs 3 (1 )1 12 )dρ β

ρ−

=+

S.17.5

Referring to Fig. S.17.5 the shear centre, S, lies on the axis of symmetry, the x axis, so that Ixy = 0. Therefore, apply an arbitrary shear load, Sy, through the shear centre and determine the internal shear flow distribution. Thus, since Sx

= 0, Eq. (17.14) becomes

0d

sys

xx

Sq ty s

I= − ∫ (i)


Fig. S.17.5

in which

23

3 1 2( )212 2 2xx

t h t t hI d+ = +

i.e.

[ ]2

3 1 23( )12xxhI t h t t d= + + (ii)

The thickness t in the flange 12 at any point s1

is given by

1 21 1

( )t tt t sd−

= − (iii)

Substituting for t from Eq. (iii) in (i)

1

1 212 1 1 10

( ) d2

sy

xx

S t t hq t s sI d

− = − − − ∫

Hence

2

1 2 112 1 1

( )2 2

y

xx

sS h t tq t sI d

−= −

(iv)

Taking moments about the mid-point of the web

12 10ξs 2 d

2d

yhS q s =

∫

i.e.

2 2 3

1 1 2 11

0

( )ξs2 2 6

dy

yxx

S h s t t sS tI d

−= −


from which

2 2

1 2ξs (2 )12 xx

h d t tI

= +


from Eq. (ii) 2

1 2

1 2 3

(2 )ξs3 ( )

d t td t t ht

+=

+ +

S.17.6

The beam section is shown in Fig. S.17.6(a). Clearly the x axis is an axis of symmetry so that Ixy = 0 and the shear centre, S, lies on this axis. Thus, apply an arbitrary shear load, Sy, through S and determine the internal shear flow distribution. Since Sx


0d

sys

xx

Sq ty s

I= − ∫ (i)

Fig. S.17.6(a)

in which, from Fig. S.17.6(a)

2 2

0 02 d d

2 2b a

xxh hI t s s t s sb a

= +

∫ ∫ (ii)

where the origin of s in the first integral is the point 1 and the origin of s in the second integral is the point 3. Equation (ii) then gives

2 ( )

6xxth b aI +

= (iii)

From Eq. (i)

1

12 1 10d

2sy

xx

S hq t s sI b

= − − ∫


from which

21

12 2 2y

xx

S th sqbI

=

or, substituting for Ixx

from Eq. (iii)

212 1

32 ( )

ySq s

bh b a=

+ (iv)

and

23

2 ( )yS b

qh b a

=+

(v)

Also

2

23 2 2 20( ) d

2sy

xx

S hq t a s s qI a

= − − − + ∫ (vi)

Substituting for Ixx from Eq. (iii) and q2

from Eq. (v) 22

23 23

( ) 2 2yS s bq s

h b a a

= − + + (vi)

and

332

ySq

h= (vii)

Equation (iv) shows that q12 varies parabolically but does not change sign between 1 and 2; also dq12/ds1 = 0 when s1 = 0. From Eq. (vi) q23 = 0 when s2 – s2

2

/2a + b/2 = 0, i.e. when

22 22 0s as ba− − = (viii)

Solving Eq. (viii)

22s a a ba= ± +

Thus, q23

does not change sign between 2 and 3. Further

23 2

2

3d1 0

d ( )ySq s

s h b a a = − = +

when s2

Therefore q

= a

23

Referring to Fig. S.17.6(a) and taking moments about the point 3

has a turning value at 3. The shear flow distributions in the walls 34 and 45 follow from antisymmetry; the complete distribution is shown in Fig. S.17.6(b).

12 102 d

byS s q p sξ = ∫ (ix)

where p is given by

sin i.e.2 2

p h hlpl b b

α= = =


Fig. S.17.6(b)

Substituting for p and q12

from Eq. (iv) in (ix) gives

2s 1 10

3ξ d

( ) 2by

yS hlS s s

bh b a b=

+ ∫

from which

ξs2(1 / )

la b

=+

S.17.7

Initially the position of the centroid, C, must be found. From Fig. S.17.7, by inspection .y a= Also taking moments about the web 23

(2 2 2 2 ) 2 22aa t a t a t x a t a ta+ + = +

from which 3 / 8.x a= To find the horizontal position of the shear centre, S, apply an arbitrary shear load,

Sy, through S. Since Sx

= 0 Eq. (17.14) simplifies to

2 20 0d d

s sy xy y yys

xx yy xy xx yy xy

S I S Iq tx s ty s

I I I I I I= −

− −∫ ∫

i.e.

2 0 0d d

s sys xy yy

xx yy xy

Sq I tx s I ty s

I I I = − − ∫ ∫ (i)


Fig. S.17.7

in which, referring to Fig. S.17.7

2 2 3 3

3 2 3 2 2 3

3

2 ( ) 2 ( ) (2 ) /12 16 /3

2 /12 2 ( /8) (2 ) / 12 2 (5 /8) 4 (3 /8) 53 /24

2 ( /8)( ) 2 (5 /8)( )

xx

yy

xy

I a t a at a t a a t

I ta ta a t a at a at a a t

I a t a a at a a a t

= + + =

= + + + + =

= + − = −


in Eq. (i) gives

3 0 0

9 53d d2497

s sys

Sq tx s ty s

a t = − − ∫ ∫ (ii)

from which

12 3 0 0

9 13 53d ( )d8 2497

s syS aq s s a sa

= − − − − ∫ ∫ (iii)

i.e.

2

12 3

9 712 297

yS as sqa

= +

(iv)

Taking moments about the corner 3 of the section

2

120ξs (2 )d

ayS q a s= −∫ (v)

Substituting for q12

from Eq. (iv) in (v) 22

2 0

18 7ξs d12 297

ayy

S as sS sa

= − +

∫


from which

45ξs97

a= −

Now apply an arbitrary shear load Sx through the shear centre, S. Since Sy

= 0 Eq. (17.14) simplifies to

2 0 0d d

s sxs xx xy

xx yy xy

Sq I tx s I ty sI I I

= − − ∫ ∫

from which, by comparison with Eq. (iii)

12 3 0 0

9 16 13 d ( )d3 897

s sxS aq t s s t a sa t

= − − + − ∫ ∫

i.e.

212 3

3(23 8 )

97xSq as s

a= − − (vi)

Taking moments about the corner 3

2

120(2 s) (2 )d

axS a q a sη− = −∫


from Eq. (vi) 2 2

2 0

6(2 s) (23 8 )d

97

axx

SS a as s sa

η− = −∫

which gives

46s97

aη =

S.17.8

Referring to Fig.S.17.8 we see, by inspection, that the centroid of area is midway between the top and bottom flanges.


Fig.S.17.8

Taking moments of areas about the web 24 (100 × 2 + 100 × 3 + 50 × 4)x-

which gives

= 4 × 50 × 25

x-

The second moments of area are, from Section 16.4.5 = 7.1 mm

Ixx = 100 × 2 × 502 + (3 × 1003/12) + 4 × 50 × 502 = 1.25 × 106 mmI

4

yy = (2 × 1003/12) + 2 × 100 × 7.12 + 3 × 100 × 7.12 + (4 × 503/12) + 4 × 50 × 17.92 = 0.298 × 106 mm

I

4 xy = 100 × 2(–7.1)(50) + 50 × 4(17.9)(–50) = –0.25 × 106 mm

We are required to find the position of the shear centre of the section so that we apply, in turn, arbitrary shear loads, S

4

y and Sx

First we apply S

through the shear centre S. Furthermore, if, when we equate internal and external moments, we choose the point 2 as the moment centre it is only necessary to calculate the shear flow distribution in the flange 54.

y

q

through the shear centre. Then, from Eq.(17.14)

s = [SyIxy/(IxxIyy – Ixy2)] ∫0

s txds – [SyIyy/(IxxIyy – Ixy2)] ∫0

s ty

Substituting the values of I ds (i)

xx

q

etc in Eq.(i) gives

s = –0.81 × 10–6 Sy ∫0s txds – 0.96 × 10–6 Sy∫0

s ty

On the flange 54, t = 4 mm, x = 42.9 – s ds (ii)

1 and y = –50 mm. Substituting in Eq.(ii), carrying out the integration and simplifying gives

Solutions to Chapter 17 Problems 229 q54 = 10–6 Sy(54s1 – 1.61s1

2

Now taking moments about the point 2 ) (iii)

SyξS = ∫050q54 × 100 dsSubstituting for q

1

54

ξ

from Eq.(iii) and integrating

S = 10–6 × 100[27s12 + 1.61s1

3/3]0

which gives

50

ξS

We now apply S = 13.5 mm

x

q

through the shear centre. Then, from Eq.(17.14)

s = [–SxIxx/(IxxIyy – Ixy2)]∫0s txds + [SxIxy/(IxxIyy – Ixy

2)]∫0s ty

Substituting for I ds

xx

q etc and carrying out the above procedure gives

54 = 10–6Sx(–529.7s1 + 8.05s12

Taking moments about the point 2 )

SxηS = –∫050 q54 × 100 dsNote that the moment of S

1 x about the point 2 is in the opposite sense to the moment

produced by q54 about the point 2. Substituting for q54

η and integrating

S = –10–6 × 100[(–529.7 s12/2) + (8.05s1

3/3)]0

which gives

50

ηS = 32.7 mm


S.17.9

The shear centre is the point in a beam cross-section through which shear loads must be applied for there to be no twisting of the section.

The x axis is an axis of symmetry so that the shear centre lies on this axis. Its position is found by applying a shear load Sy

through the shear centre, determining the shear flow distribution and then taking moments about some convenient point. Equation (17.14) reduces to

0d

sys

xx

Sq ty s

I= − ∫ (i)

in which, referring to Fig. S.17.8

3 /22 2 2

02 2 cos d

3xxtrI rtr tr r

πθ θ

= + +

∫

Fig. S.17.8

i.e. 36.22xxI tr= In the wall 12, y = s1

. Therefore substituting in Eq. (i) 21

12 10d

2sy y

xx xx

S S tsq ts sI I

= − = −∫

Then

2

2 2y

xx

S trqI

= −

In the wall 23, y = r, then

2

23 0d

2sy

xx

S trq tr sI

= − +

∫

i.e.

2

23 2 2y

xx

S trq trsI

= − +


and

2

3 52

y

xx

S trqI

= − In the wall 34, y = r cos θ, then

2

234 0

5cos d2

y

xx

S trq trI

θθ θ

= − +

∫

i.e.

234

5sin2

y

xx

Sq tr

Iθ = − +

Taking moments about O

2 /2 2

S 12 23 340 0 02 2 d d d

r ryS x q r s q r s q r

πθ = − + + ∫ ∫ ∫

The negative sign arises from the fact that the moment of the applied shear load is in the opposite sense to the moments produced by the internal shear flows. Substituting for q12, q23 and q34

from the above 2 22 /2 41

S 20 0 0

52 d d sin d2 2 2

r ryy

xx

S s rS x t r s rs r s rI

πθ θ

= + + + + ∫ ∫ ∫

which gives S 2.66x r= S.17.10

In this problem the axis of symmetry is the vertical y axis and the shear centre will lie on this axis so that only its vertical position is required. Therefore, we apply a horizontal shear load Sx through the shear centre, S, as shown in Fig. S.17.10.

Fig. S.17.10

The thickness of the section is constant and will not appear in the answer for the shear centre position, therefore assume the section has unit thickness.

Equation (17.14), since Ixy = 0, t = 1 and only Sx

is applied, reduces to

0d

sxs

yy

Sq s sI

= − ∫ (i)


where

3 /22 2 2

0

25 25 62.5 50 50 (50cos ) 50d12yyI

πθ θ= + × + × + ∫

i.e.

5 46.44 10 mmyyI = ×

In the flange 12, x = –75 + s1

and 21

12 1 1075 )d 75

2sx x

yy yy

S S sq s s sI I

= − − + = − − +

∫

and when s1 = 25 mm, q2 = 1562.5Sx/IIn the wall 23, x = – 50 mm, then

yy

( )23 2050d 1562.5 50 1562.5

sx x

yy yy

S Sq s sI I

= − − − = + ∫

when S2 = 50 mm, q3 = 4062.5Sx/IyyIn the wall 34, x = –50 cosθ, therefore

.

( )34 050cos 50d 4062.5 2500sin 4062.5x x

yy yy

S SqI I

θθ θ θ = − − − = +

∫

Now taking moments about O

25 50 22

12 1 23 2 340 0 02 50d 50d 50 dxS ys q s q s q

π

θ = − + + ∫ ∫ ∫

Note that the moments due to the shear flows in the walls 23 and 34 are opposite in sign to the moment produced by the shear flow in the wall 12. Substituting for q12

S.17.11 , etc. gives

Apply an arbitrary shear load Sy through the shear centre S. Then, since the x axis is an axis of symmetry, Ixy

= 0 and Eq. (17.14) reduces to

0

/42 2

0 0

/42 3 21 10 0

32 2 2

1 1 10 0

d

2 d ( sin ) d

2 (2 sin 45 sin 45 ) d sin d

2 sin 45 (4 4 )d (1 cos 2 )d2

sys

xx

rxx

r

r

Sq ty s

I

I ty s t r r

t r s s tr

trt r rs s s

π

π

π

θ θ

θ θ

θ θ

= −

= + = − +

= − + + −

∫

∫ ∫

∫ ∫

∫ ∫


Fig. S.17.10

which gives

32.32xxI tr= Then

1 o o

12 1 10(2 sin 45 sin 45 )d

sy

xx

Sq t r s s

I= − −∫

i.e.

21

12 13

0.272

2yS sq rs

r−

= −

(i)

and

20.4 yS

qr

−=

Also

o23 0

0.4sin (45 ) dy y

xx

S Sq t r r

I rφ

φ φ

= − − −

∫


from which

o23

0.13cos (45 )

2.62y yS S

qr r

φ= − − − (ii)


/4 2

S 12 1 230 02 d d

ryS x q r s q r

πφ = − + ∫ ∫ (iii)

Substituting for q12 and q23

x from Eqs (i) and (ii) in Eq. (iii) gives

S

S.17.12

= 1.2r

Since the x axis is an axis of symmetry and only Sy

is applied Eq. (17.14) reduces to

0d

sys

xx

Sq ty s

I= − ∫

Fig. S.17.12

Also

2 2 1/232 (15 60 ) 61.8mms = + =


and

23 32

32

2 25 2 61.8 152 2 57.512 12 61.8

2 602 61.8 37.512

xxI × × = + × + ×

×

+ × × +

which gives

4724094mmxxI =

Then

( )1

12 1 102 70 d

sy

xx

Sq s s

I= − − +∫

i.e.

212 1 1(140 )y

xx

Sq s s

I= − (i)

and

22875 y

xx

Sq

I=

Also

2

23 2 20

2875152 45 d61.8

sy y

xx xx

S Sq s s

I I = − − + + ∫

Then

223 2 2

1590 287561.8

y

xx

Sq s s

I = − +

(ii)

Taking moments about the mid-point of the web 34 (it therefore becomes unnecessary to determine q34

)

25 61.8

S 12 1 23 20 0

602 60 d 30 d61.8yS x q s q s = − + × ∫ ∫

Substituting for q12 and q23

x

from Eqs (i) and (ii)

S = 20.2 mm


S.17.13 Referring to Fig.S.17.13

Fig.S.17.13

Ixx = 2rtr2 + 2rtr2 + ∫0πtr2 cos2θr dθ = 5.6tr

I

3

yy = t(2r)3/12 + 2rt(0.86r)2 + 2tr3/12 + 2rt(0.36r)2 + ∫0πt(r sin θ + 0.14r)2

rdθ = 4.8trI

3

xy = 2rt(0.86r)(r) + 2rt(0.36r) (–r) + 0 for semicircular portion = trSince only the horizontal position of the shear is required we apply an arbitrary

shear load, S

3

y

q

, through the shear centre as shown in Fig.S.17.13. Then, from Eq.(17.14)

s = [SyIxy/(IxxIyy – Ixy2)]∫0

s tx ds – [SyIyy/(IxxIyy – Ixy2)]∫0

s ty

Substituting for I ds (i)

xx

q etc in Eq.(i)

s = (0.04Sy/tr3)∫0s txds – (0.19Sy/tr3) ∫0

s ty

On the flange 12, x = 0.86r – s ds (ii)

1

q

, y = -r and t = 2t. Substituting in Eq.(ii)

12 = (0.04Sy/r3)∫0s 2(0.86r – s1)ds1 – (0.19Sy/r3)∫0

s2(–r)ds1

Solutions to Chapter 17 Problems 237 (Note to editor: the upper limit in both of the integrals in the expression for q12 should be s1Then

)

q12 = (Sy/r3)(0.45rs1 – 0.04s12

and q) (iii)

2 = 0.41SyIn the semicircular portion 23, x = –0.14r – rsinθ, y = -rcosθ and t = t. Then

/r.

q23 = (0.04Sy/tr3)∫0θt(–0.14r – rsinθ)rdθ - (0.19Sy/tr3)∫0θt(–rcosθ)rdθ + 0.41Sy

which gives /r

q23 = (Sy

and q/r)(0.19sinθ + 0.04cosθ – 0.006θ + 0.37) (iv)

3 = 0.31SyIn the flange 34, x = s

/r. 2

q – 0.14r, y = r and t = t. Then

34 = (0.04Sy/tr3)∫0st(s2 – 0.14r)ds2 – (0.19Sy/tr3)∫0strds(Note: the upper limit in both these integrals should be s

2 2

Then )

q34 = (Sy/r3)(0.31r2 – 0.196s2r + 0.02s22

Now taking moments about O ) (v)

SyξS = ∫0rq12rds1 + ∫0πq23rrdθ + ∫02rq34rds2

Substituting in Eq.(vi) for q (vi)

12

ξ

etc from Eqs(iii), (iv) and (v) and carrying out the integration gives

S

S.17.14

= 2.0r

Since only the horizontal position of the shear centre is required we only need to apply a shear load Sy

through the shear centre as shown in Fig.S.17.14. Further, if we take internal and external moments about the point 2 it is only necessary to calculate the shear flow distribution in the wall 43.


Fig.S.17.14

From Eq.(17.14)

qs = [(SyIxy)/(IxxIyy – Ixy2)]∫0stxds – [(SyIyy)/(IxxIyy – Ixy

2)]∫0s

Substituting the given values of I

ty ds (i)

xx

q

in Eq.(i) gives

s = (0.1Sy/tr3)∫0stxds – (0.56Sy/tr3)∫0s

In the wall 43, x = 0.9r +rcosθ, y = –0.54r + rsinθ

tyds (ii)

Substituting in Eq.(ii)

q43 = (0.1Sy/r3)∫0θ(0.9r + rcosθ)rdθ + (0.56Sy/r3)∫0θ

which gives (0.54r – rsinθ)rdθ

q43 = (Sy


/r)(0.56 cosθ + 0.1 sinθ + 0.39θ –0.56)

SyξS = ∫0π/2q43

From Fig.S.17.14

p ds (iii)

p = AB + BC = AB + DE = 2rcosθ + (r – rsinθ) = r(2cosθ – sinθ + 1)


S

and p in Eq.(iii)

yξS = (Sy/r)∫0π/2(0.56cosθ + 0.1sinθ + 0.39θ – 0.56)(2cosθ - sinθ + 1)r2

Carrying out the integration dθ (iv)

ξS

0.78(θ sinθ + cosθ) – 0.39(–θcosθ + sinθ) – 0.56sinθ – 0.66 cosθ + (0.39θ

= r{1.12[(θ/2) + (sin2θ)/4] – 0.1[(θ/2) – (sin2θ)/4] + (0.36cos2θ)/4 + 2)/2 –

0.56θ0π/2

Solutions to Chapter 17 Problems 239 which gives

ξS

With the arc 34 removed the shear centre will occur at the point 2 (see Fig.17.8) so that it will move 0.36r to the left.

= 0.36r

S.17.15 Referring to Fig. S.17.15 the x axis is an axis of symmetry so that Ixy = 0 and since Sx

= 0. Eq. (17.15) reduces to

,00d

sys s

xx

Sq ty s q

I= − +∫ (i)

in which

3 2 o /2 2

0

(2 ) sin 45 2 ( sin ) d12xx

r tI t r rπ

θ θ= + ∫

Fig. S.17.15

i.e. 30.62xxI tr= ‘Cut’ the section at O. Then, from the first term on the right-hand side of Eq. (i)

,O1 3 0sin d

0.62y

bS

q tr rtr

θθ θ= − ∫

i.e.

[ ],O1 0cos0.62

yb

Sq

rθθ= − −

so that

.O1 (cos 1) 1.61 (cos 1)0.62

y yb

S Sq

r rθ θ= − − = − (ii)

and


,1047 y

bS

qr

= −

Also

o,12 3 0

0.47( )sin 45 d

0.62

sy yb

S Sq t r s s

rtr= − − −∫

which gives

2 2,12 3 ( 1.14 0.57 0.47 )y

bS

q rs s rr

= − + − (iii)

Now take moments about the point 2

2/4

b,O1 ,002 d 2

4y srS r q rr q

π πθ= + ×∫

Substituting in Eq. (iv) for qb, O1

from Eq. (ii) 2/4 2

,002 1.61 (cos 1) d

4y

y sS rS r r qr

π πθ θ= − +∫

i.e.

2

/40 ,03.22 [sin ]

2y y srS r S r qπ πθ θ= − +

so that

,00.80 y

sS

qr

=

Then, from Eq. (ii)

O1 (1.61cos 0.80)ySq

rθ= −

and from Eq. (iii)

2 212 3 (0.57 1.14 0.33 )yS

q s rs rr

= − +

The remainingdistribution follows from symmetry.

S.17.16

The x axis is an axis of symmetry so that Ixy = 0 and, since Sx


,00d

sys s

xx

Sq ty s q

I= − +∫ (i)

in which, from Fig. S.17.16(a)


3 3 2 2(2 ) sin ( 2 )

12 12 12xxth d t thI h dα

= + = + (ii)

‘Cut’ the section at 1. Then, from the first term on the right-hand side of Eq. (i)

1 2

b,12 1 1 10

sin( sin )d

2sy y

xx xx

S S tq t s s s

I Iα

α= − − =∫


Fig. S.17.16(a)


and sin α

2,12 1

3( 2 )

yb

Sq s

hd h d=

+ (iii)

and

,23

( 2 )y

bS d

qh h d

=+

Also

2

b,23 2 2 ,20d

2sy

bxx

S hq t s s qI

= − − + + ∫

so that

22

b,23 26

( 2 ) 2yS s dq s

h h d h

= − + + (iv)

Now taking moments about the point 1 (see Eq. (17.18))

,23 2 ,000 cos d 2 cos

2h

b shq d s d qα α= +∫

i.e.

,23 2 ,000 d

hb sq s hq= +∫ (v)

Substituting in Eq. (v) for qb,23

from Eq. (iv) 22

2 2 ,00

60 d

( 2 ) 2hy

sS s ds s hq

h h d h

= − + + + ∫


which gives

,0( 3 )

( 2 )y

sS h d

qh h d

+= −

+ (vi)

Then, from Eqs (iii) and (i)

212 1

3 ( 3 )( 2 ) ( 2 )

y yS S h dq s

h dh d h h d+

= −+ +

i.e.

21

123 3

( 2 )yS sq h d

h h d d

= − − + (vii)

and from Eqs (iv) and (vi)

22

23 266

( 2 )yS sq s h

h h d h

= − − + (viii)

The remaining distribution follows from symmetry. From Eq. (vii), q12

21s is zero when = (hd/3) + d2, i.e. when s1 > d. Thus there is no

change of sign of q12

between 1 and 2. Further

12 1

1

d 6 0dq ss d

= = when s1

and

= 0

1( 3 )

( 2 )yS h d

qh h d

+= −

+

Also, when s1

= d

2 ( 2 )yS

qh d

= −+

From Eq. (viii) q23 is zero when 6S222(6 / ) 0,s h h− = – i.e. when 2 2

2 2 ( /6) 0.s s h h− + =

Then

2 2 12h hs = ±

Thus q23 / 12h is zero at points a distance either side of the x axis. Further, from Eq. (viii), q23 will be a maximum when S2 = h/2 and q23( max) = Sy/2(h + 2d). The complete distribution is shown in Fig. S.17.16(b).


Fig. S.17.16(b)

S.17.17

Since the section is doubly symmetrical the centroid of area, C, and the shear centre, S, coincide. The applied shear load, S, may be replaced by a shear load, S, acting through the shear centre together with a torque, T, as shown in Fig. S.17.17. Then

ocos30 0.866T Sa Sa= = (i)

Fig. S.17.17

The shear flow distribution produced by this torque is given by Eq. (18.1), i.e.

0.8662 2TT SaqA A

= = (from Eq. (i))


where

o o o 22 cos30 2 cos30 sin 30 2.6A a a a a a= + × × =

Then

0.17 (clockwise)TSq

a= (ii)

The rate of twist is obtained from Eq. (18.4) and is

2 2d 0.866 6d 4(2.6 )

Sa az ta Gθ =

i.e.

2d 0.192d

Sz Gtaθ= (iii)

The shear load, S, through the shear centre produces a shear flow distribution given by Eq. (17.15) in which Sy = –S, Sx = 0 and Ixx

= 0. Hence

,00d

ss s

xx

Sq ty s qI

= +∫ (iv)

in which

3 3

o 21 10

52 4 ( sin 30 ) d12 2

axx

ta a tI t a s s= + − + =∫

Also on the vertical axis of symmetry the shear flow is zero, i.e. at points 7 and 3. Therefore, choose 7 as the origin of s in which case qs,0

in Eq. (iv) is zero and

0d

ss

xx

Sq ty sI

= ∫ (v)

From Eq. (v) and referring to Fig. S.17.14

1 o78 1 10

( sin 30 )ds

xx

Sq t a s sI

= − +∫

i.e.

178 3 0

2 d25

s sSq a sa

= − + ∫

so that

21

78 13 225sSq as

a

= −

(vi)

and

83

10Sqa

= − (vii)


Also

2

81 2 2 80d

2s

xx

S aq t s s qI

= − + + ∫

i.e.

2

81 2 23 0

2 3d2 105

sS a Sq s saa

= − + − ∫

from which

2 281 2 23 ( 2 2 3 )

10Sq as s aa

= − + − (viii)

Thus

1720

Sqa

= −

The remaining distribution follows from symmetry. The complete shear flow distribution is now found by superimposing the shear flow

produced by the torque, T, (Eq. (ii)) and the shear flows produced by the shear load acting through the shear centre. Thus, taking anticlockwise shear flows as negative

10.17 0.35 0.52S S Sq

a a a= − − = −

2 80.17 0.3 0.47S S Sq q

a a a= = − − = − (from Eq. (vii))

3 70.17Sq q

a= = −

4 60.17 0.3 0.13S S Sq q

a a a= = − + =

50.17 0.35 0.18S S Sq

a a a= − + =

The distribution in all walls is parabolic.

S.17.18

Referring to Fig. P.17.15, the wall DB is 3 m long so that its cross-sectional area, 3× 103 × 8 = 24 × 103 mm2, is equal to that of the wall EA, 2 × 103 ×12 = 24 × 103 mm2. If follows that the centroid of area of the section lies mid-way between DB and EA on the vertical axis of symmetry. Also since Sy = 500 kN, Sx = 0 and Ixy


3

,00

500 10 ds

s sxx

q ty s qI×

= − +∫ (i)


If the origin for s is taken on the axis of symmetry, say at O, then qs,0

is zero. Also 3 3 2 3 3 2

3 3 2 o

3 10 8 (0.43 10 ) 2 10 12 (0.43 10 )

2 (1 10 ) 10 sin 60 / 12xxI = × × × × + × × × ×

+ × × × ×

i.e.

8 4101.25 10 mmxxI = ×

Equation (i) then becomes

5

04.94 10 d

ssq ty s−= − × ∫

In the wall OA, y = – 0.43 × 103

mm. Then

5 3OA A0

4.94 10 12 0.43 10 d 0.25s

q s s−= × × × =∫

and when SA = 1 × 103 mm, qOAIn the wall AB, y = – 0. 43 × 10

= 250 N/mm. 3 + sB

cos 30°. Then

5 3AB B0

4.94 10 10( 0.43 10 0.866 )d 250s

q s s−= − × − × + +∫

i.e.

4 2AB B B0.21 2.14 10 250q s s−= − × +

When sB =1 × 103 mm, qABIn the wall BC, y = 0.43 × 10

= 246 N/mm. 3

mm. Then

5 3BC 0

4.94 10 8 0.43 10 d 246s

q s−= − × × × +∫

i.e.

BC C0.17 246q s= − +

Note that at C where sC = 1.5 × 103 mm, qBC

The maximum shear stress will occur in the wall AB (and ED) mid-way along its length (this coincides with the neutral axis of the section) where s

should equal zero; the discrepancy, –9 N/mm, is due to rounding off errors.

B = 500 mm. This gives, from Eq. (ii), qAB (max) = 301.5 N/mm so that the maximum shear stress is equal to 301.5/10= 30.2 N/mm2

S.17.19 .

The shear centre, S, lies on the horizontal axis of symmetry so that, to determine its horizontal position, we apply a shear load, Sy, through S. Further, Ixy = 0 and since only Sy

qis applied Eq.(17.15) for the shear flow distribution reduces to

s = –(Sy/Ixx)∫0s ty ds + qs,0 (i)


Ixx has previously been calculated in P.17.15 and is 0.62tr3. Referring to Fig.S.17.19

Fig.s.17.19

and “cutting” the section at O we note that the qb

q

distribution will be identical to that in P.17.15 and is

b,O1 = (1.61Syand

/r)(cosθ – 1) (ii)

qb,12 = (Sy/r3)(0.57s2 – 1.14rs – 0.47r2

Since the rate of twist of the beam section is zero q) (iii)

s,0

∫ds = 2r + (πr/2) = 3.57r

is given by Eq.(17.28) in which

and

∫qbds = 2[∫0π/4(1.61Sy/r)(cosθ – 1)rdθ + ∫0r(Sy/r3)(0.57s2 – 1.14rs – 0.47r2

which gives )ds]

∫qbds = –1.95S(Note to editor: the integral signs on the left hand side of the above three equations should have a circle at mid-height as in Eq.(17.28)).

y

Then, from Eq.(17.28) qs,0 = 1.95Sy/3.57r = 0.55Syso that

/r

qO1 = (Sy

Taking moments about the point 2 /r)(1.61cosθ – 1.06)

SyξS = 2∫0π/4qO1r2dθ = 2∫0π/4(Sy/r)(1.61cosθ – 1.06)r2

which gives

dθ

ξS = 0.61r

Solutions to Chapter 17 Problems 249 S.17.20

Referring to Fig.S.17.20

Fig.S.17.20

r/(length of 51) = sinα = h/2d

i.e.

r = (h/2d)(length of 51) = (h/2d)(dcosα – r) so that

r[1 + (h/2d)] = (h/2)(1 – sin2α)1/2 = (h/2)[1 – (h/4d)2]Simplifying

1/2

r = (h/2)[(2d – h)/(2d + h)]1/2

We apply a shear load, S (i)

y, through the shear centre as shown in Fig.S.17.20. If we “cut” the beam section at 1 then the qbi.e.

distribution will be identical to that in P.17.16

qb,12 = [3Sy/hd(h + 2d)]s12

and (ii)

qb,23 = [6Sy/h(h + 2d)][s2 – (s22

Then /h) + (d/2)] (iii)

∫ds = h + 2d and

∫qbds = 2[∫0dqb,12 ds1 + ∫0h/2qb,23 ds2]

250 Solutions Manual Substituting for qb,12 and qb,23

∫q

from Eqs(ii) and (iii)

b ds = 2[Sy/h(h + 2d)][∫0d(3/d)s12ds1 + ∫0h/2(6s2 – 6s2

2/h + 3d)ds2

which gives ]

∫qbds = [Sy/h(h + 2d)](h2 + 3hd + 2d2

(Note: the integral signs on the left hand side of the above four equations should have a circle at mid-height as in Eq.(17.28).

)

Then, from Eq.(17.28) qs,0 = –[Sy/h(h + 2d)2](h2 + 3hd + 2d2

and )

q12 = [Sy/h(h + 2d)]{3(s12/d) – [1/(h + 2d)](h2 + 3hd + 2d2

Now taking moments about O )}

SyξS = –2∫0d q12(h/2)cosα ds

Substituting for q1

12

S

and cosα from the above and carrying out the integration

yξS = –2[Sy/h(h + 2d)][(h/2)(1 – h2/4d2)1/2][(s13/d) – s1(h2 + 3hd +2d2)/(h + 2d)]0

which gives

d

ξS = (h/2)[(2d – h)/(2d + h)]so that, comparing with Eq.(i), we see that the shear centre coincides with the centre

of the inscribed circle.

1/2


S.18.1

Referring to Fig. P.18.1 the maximum torque occurs at the built-in end of the beam and is given by

3max 20 2.5 10 50000 N mT = × × =

From Eq. (18.1)

max maxmax

min min2q Tt At

τ = =

i.e.

3

max50000 10

2 250 1000 1.2τ ×

=× × ×

so that

2max 83.3N / mmτ =

From Eq. (18.4)

2d dd 4

T sz GtAθ= ∫

i.e.

3

2d 20(2500 ) 10 2 1000 250d 18000 1.2 26000 2.14 (250 1000)

zzθ − × ×= + × ×× ×

which gives

9d 8.14 10 (2500 )d

zzθ −= × −

Then

2

918.14 10 2500

2zz Cθ −

= × − +

When z = 0, θ = 0 so that C1

=0, hence 2

98.14 10 25002zzθ −

= × −

Thus θ varies parabolically along the length of the beam and when z = 2500 mm

0.0254rad or 1.46θ = °


S.18.2

The shear modulus of the walls of the beam is constant so that Eq. (18.5) may be written

O O0 2

s ss

ATw wAG A

δδδ

− = −

(i)

in which

O 0

d dands

ss st t

δ δ= =∫ ∫

Also, the warping displacement will be zero on the axis of symmetry, i.e. at the mid-points of the walls 61 and 34. Therefore take the origin for s at the mid-point of the wall 61, then Eq. (i) becomes

O O

2s s

sATw

AG Aδδδ

= −

(ii)

in which

2 2 2 223 12500 100 509.9mm and 890 150 902.6mml l= + = = + =

Then

200 300 2 509.9 2 902.6 2479.92.0 2.5 1.25 1.25

δ × ×= + + + =

and

21 12 2(500 200) 890 (500 300) 500 511500 mmA = + × + + × =

Equation (ii) then becomes

3

O O90500 10 2479.92 511500 27500 2479.9 511500

s ss

Aw δ × ×= − × ×

i.e.

4O O7.98 10 (4.03 0.0196 )s s sw Aδ= × − (ii)

The walls of the section are straight so that δOs and AOs vary linearly within each wall. It follows from Eq. (iii) that ws

varies linearly within each wall so that it is only necessary to calculate the warping displacement at the corners of the section. Thus, referring to Fig. P.18.2

4 100 17.98 10 4.03 0.0196 890 1002.0 2

w − = × × − × × ×

i.e.

1 60.53mm from antisymmetryw w= − = −


Also

42

902.6 17.98 10 4.03 0.0196 250 890 0.531.25 2

w − = × × − × × × −

i.e.

2 50.05mmw w= = −

Finally

43

509.9 17.98 10 4.03 0.0196 250 500 0.051.25 2

w − = × × − × × × +

i.e.

3 40.38 mmw w= = −

S.18.3

Referring to Fig. P.18.3 and considering the rotational equilibrium of the beam 2 2 450 1.0 2000R = × × × so that 1450 NmR = In the central portion of the beam

450 1.0(1000 ) 1450 Nm (z in mm)T z z= + − − = − (i)

and in the outer portions 450 1.0(1000 ) 1450 Nm (z in mm)T z z= + − = − (ii) From Eq. (i) it can be seen that T varies linearly from zero at the mid-span of the beam to –500 Nm at the supports. Further, from Eq. (ii) the torque in the outer portions of the beam varies linearly from 950 Nm at the support to 450 Nm at the end. Therefore Tmax

= 950 Nm and from Eq. (18.1)

max maxmax

min min2q Tt At

τ = =

i.e.

3

2max 2

950 10 24.2 N / mm2 50 2.5

τπ

×= =

× × ×

For convenience the datum for the angle of twist may be taken at the mid-span section and angles of twist measured relative to this point. Thus, from Eqs (18.4) and (i), in the central portion of the beam

3

2 2d 10 100d 4( 50 ) 30000 2.5

zzθ π

π× × ×

=× × ×


i.e.

8d 1.70 10d

zzθ −= − ×

Then

2

81.70 102z Bθ −= − × +

When z = 0, θ = 0 (datum point) so that B = 0. Then

8 20.85 10 zθ −= − × (iii)

In the outer portions of the beam, from Eqs (18.4) and (ii)

3

2 2d (1450 ) 10 100d 4( 50 ) 30000 2.5

zzθ π

π− × × ×

=× × ×

i.e.

8d 1.70 10 (1450 )d

zzθ −= × −

Hence

2

81.70 10 14502zz Cθ −

= × − +

(iv)

When z = 500 mm, θ = – 2.13 × 10–3 rad from Eq. (iii). Thus, substituting this value in Eq. (iv) gives C = –12.33 × 10–3

and Eq. (iv) becomes 2

8 31.70 10 1450 12.33 10 rad2zzθ − −

= × − − ×

(v)

The distribution of twist along the beam is then obtained from Eqs (iii) and (v) and is shown in Fig. S.18.3. Note that the distribution would be displaced upwards by 2.13 × 10–3 rad if it were assumed that the angle of twist was zero at the supports.

Fig. S.18.3


S.18.4

The total torque applied to the beam is 20 × 4 × 103 Nm. From symmetry the reactive torques at A and D will be equal and are 40 × 103

Nm. Therefore,

AB 40000 N mT =

BC 40000 20( 1000) 60000 20 N m ( in mm)T z z z= − − = −

Note that the torque distribution is antisymmetrical about the centre of the beam. The maximum torque in the beam is therefore 40 000 Nm so that, from Eq. (18.1)

3

2max

40000 10 71.4 N / mm2 200 350 4

τ ×= =

× × ×

The rate of twist along the length of the beam is given by Eq. (18.4) in which

2 200 2 350 216.74 6× ×

= + =∫

Then

142

d 216.7 15.79 10d 4 (200 350) 70000

T Tzθ − = = ×

× × ×

In AB, TAB

= 40 000 Nm so that 6

AB 6.32 10 z Bθ −= × +

When z = 0, θAB = 0 so that B = 0 and when z = 1000 mm, θAB

In BC, T

= 0.0063 rad (0.361°)

ΒC

= 60 000 – 20z Nm. Then, from Eq. (18.4) 14 2 3

BC 15.79 10 (60000 10 ) 10z z Cθ −= × − × +

When z = 1000 mm, θBC

= 0.0063 so that C = – 0.0016. Then 10 2

BC 1.579 10 (60000 10 ) 0.0016z zθ −= × − −

At mid-span where z = 3000 mm, θBC

S.18.5

= 0.0126 (0.722°).

The torque is constant along the length of the beam and is 1 kN m. Also the thickness is constant round the beam section so that the shear stress will be a maximum where the area enclosed by the mid-line of the section wall is a minimum, i.e. at the free end. Then

3

2max

1000 10 33.3N / mm2 50 150 2

τ ×= =

× × ×


The rate of twist is given by Eq. (18.4) in which d /s t∫ varies along the length of

the beam as does the area enclosed by the mid-line of the section wall. Then

502 50 150d 2500 125 0.01

2

zs zt

× + + = = +

∫

Also

5050 150 75002500

zA z = + = +

Then

6

2d 1 10 (125 0.01 )d 4 25000 (7500 )

zz zθ × += × +

or

2d 1250010d 100(7500 )

zz zθ +=

+

i.e.

2d 5000 10.1d 7500(7500 )z zzθ = + ++

Then

e50000.1 log (7500 )

(7500 )z B

zθ

−= + + + +

When z = 2500 mm, θ = 0 so that B = –6.41 and

e50000.1 log (7500 ) 6.41 rad

(7500 )z

zθ

−= + + − +

When z = 0, θ = 10.6°, etc.

S.18.6

In Eq. (18.4), i.e.

2d dd 4

T sz GtAθ= ∫

Gt = constant = 44 000 N/mm. Thus, referring to Fig. S.18.6

3

2 2d 4500 10 2 200 100 50d 440004(100 200 50 /2)zθ π

π × × + + ×

= × + ×


i.e.

6d 29.3 10 rad/mmdzθ −= ×

Fig. S.18.6

The warping displacement is zero on the axis of symmetry so that Eq. (18.5) becomes

O O

2 As s

sATw

Aδδδ

= −

(i)

where

O 0

d dands

ss s

Gt Gtδ δ= =∫ ∫

Since Gt = constant, Eq. (i) may be written

0 Od

d2 d

s

ss

s ATw sAGt As

= −

∫∫ ∫

(ii)

in which

d 2 200 100 50 657.1mms π= × + + × =∫

and

2 2100 200 50 /2 23927.0mmA π= × + × = Equation (ii) then becomes

3

0 Od4500 10 657.1

2 23927.0 44000 657.1 23927.0

s

ss

s Aw × ×

= − × ×

∫

i.e.

3 2O0

1.40 10 1.52 d 4.18 10s

s sw s A− = × − × ∫ (iii)


In the straight walls 0

ds

s∫ and AOs

3 2 13 4 21.40 10 (1.52 50 4.18 10 200 50) 0.19mmw w − −= − = × × − × × × × = −

are linear so that it is only necessary to calculate

the warping displacement at the corners. Thus

3 2 12 1 21.40 10 (1.52 200 4.18 10 200 50) 0.19w w − −= − = × × − × × × × −

i.e.

2 1 0.056mmw w= − = −

In the wall 21

21O 20

d 50 and 50s

ss Aφ φ= = ×∫

Then Eq. (iii) becomes

3 2 2121 21.40 10 (1.52 50 4.18 10 50 ) 0.056w φ φ− −= × × − × × × −

i.e.

21 0.033 0.056w φ= − (iv)

Thus w21 varies linearly with φ and when φ = π/2 the warping displacement should be zero. From Eq. (iv), when φ = π/2, w21

S.18.7

= – 0.004 mm; the discrepancy is due to rounding off errors.

Suppose the mass density of the covers is ρa and of the webs ρb

. Then

1 1a a b bk G k Gρ ρ= =

Let W be the weight/unit span. Then

2 2a a b bW at g bt gρ ρ= +

so that, substituting for ρa and ρ

b

12 ( )a a b bW k g at G bt G= + (i)

The torsional stiffness may be defined as Τ/(dθ/dζ) and from Eq. (8.4)

2 2d 2 2d 4 a a b b

T a bz G t G ta bθ = +

(ii)

Thus, for a given torsional stiffness, dθ/dζ = constant, i.e.

2constanta a b b

a b kG t G t

+ = = (iii)


Let tb/ta

= λ. Equation (iii) then becomes

2

1a

a b

a btk G Gλ

= +

and substituting for ta

in Eq. (i)

2 211

22 ( ) 2 a b

a a bb a

abG abGkW k gt aG bG g a bk G G

λλ

λ

= + = + + +

For a maximum

d 0dWλ

=

i.e.

2

2 a

b

GG

λ

=

from which

a b

b a

G tG t

λ = =

For the condition Gata = Gbtb

knowing that a and b can vary. Eq. (i) becomes

12 ( )a aW k G t g a b= + (iv)

From Eq. (ii), for constant torsional stiffness

32 2 constanta b ka b+

= = (v)

Let b/a = x. Equation (iv) may then be written

12 (1 )a aW k G t ga x= + (vi)

and Eq. (v) becomes

3 3 21 xka x+

=

which gives

32

3

1 xak x+

=

Substituting for a in Eq. (vi)

1/3

11/3 23

2 1 (1 )a ak G t g xW xk x

+ = +


i.e.

4/3

11/3 2/33

2 (1 )a ak G t g xWk x

+=

Hence for (dW/dx)=0

1/3

5/3 4/32/3

4 (1 ) 20 (1 )3 3

x x xx

−+= − +

i.e.

4 2(1 ) 0x x− + =

so that

1 /x b a= =

S.18.8

The maximum shear stress in the section is given by Eq. (18.13) in which, from Eqs. (18.11)

3 420 15 25 252 2 453.3mm3

J + + + = × =

Then

3

2max

50 10 2 220.6 N / mm453.3

τ × ×= =

From Eq. (18.12)

dd

Tz GJθ=

i.e.

3d 50 10 0.0044rad / mm

d 25000 453.3zθ ×= =

×

S.18.9

The rate of twist/unit torque is given by Eq. (18.12). i.e.

d 1dz GJθ=

where

3

48 (2 25 2 61.8 60) 623mm3 3

stJ = = × + × + =∑


Then

8d 1 6.42 10 rad / mmd 25000 623zθ −= = ×

×

S.18.10

From the second of Eqs (18.13) the maximum shear stress is given by

maxtTJ

τ = ± (i)

in which J, from Eqs (18.11), is given by (see Fig. P.18.10)

33 3 50

0

100 2.54 38 1.27 22 1.27 1.27 d3 3 3 50

sJ s× × = + × + + ∫

where the origin for s is at the corner 2 (or 5). Thus

4854.2mmJ =


3

2max

2.54 100 10 297.4 N / mm854.2

τ × ×= ± = ±

The warping distribution is given by Eq. (18.20) and is a function of the swept area, AR (see Fig. 18.11). Since the walls of the section are straight AR

varies linearly around the cross-section. Also, the warping is zero at the mid-point of the web so that it is only necessary to calculate the warping at the extremity of each wall. Thus

3

1 R1 100 102 2 25 502 26700 854.2

Tw AGJ

×= − = − × × × ×

×

= –5. 48 mm = – w6

Note that p from antisymmetry

R, and therefore AR

, is positive in the wall 61. 3

2 51 100 105.48 2 50 50 5.48mm2 26700 854.2

w w×= − + × × × × = = −

×

(pR

is negative in the wall 12) 3

3 41 100 105.48 2 38 75 17.98mm2 26700 854.2

w w×= + × × × × = = −

×

(pR is negative in the wall 23)


S.18.11

The maximum shear stress in the section is given by the second of Eqs (18.13), i.e.

max maxmax

t TJ

τ = ± (i)

in which tmax = t0

and the torsion constant J is obtained using the second of Eqs (18.11). Thus 3 3 33

00 00 0

1 12 d d3 3 3 3

a a ats sJ t s t sa a

= + +

∫ ∫

In the first integral s is measured from the point 7 while in the second s is measured from the point 1. Then

304

3atJ =


0max 3 2

0 0

34 /3 4

t T Tat at

τ = ± = ±

The warping distribution is given by Eq. (18.19). Thus, for unit rate of twist

R2sw A= − (ii)

Since the walls are straight AR varies linearly in each wall so that it is only necessary to calculate the warping displacement at the extremities of the walls. Further, the section is constrained to twist about O so that w0 = w3 = w4

217 8 R22 ( is positive in 37)w aa a w p= − × = − = −

= 0. Then

212 5 R22 2 cos 45 2 ( is negative in 32)w a a a w p= − × ° = = −

2 211 6 R22 2 (2 sin 45 ) (1 2 2 ( is negative in 21)w a a a a a w p= + × ° + = + = −

S.18.12

The torsion constant J is given by the first of Eqs (18.11) i.e.

3 3 31 ( 4 ) 2.383

J rt rt rtπ= + =

The maximum shear stress/unit torque is, from Eqs (18.13)

2max 3 0.42 /

2.38t rtrt

τ = ± = ±

The warping distribution is obtained from Eq. (18.19)


Fig. S.18.12

i.e. R2 /unit rate of twistw A= − ln 03

2R

12

A r θ= −

so that

203w r θ=

and

2

23 41.571

2rw r wπ

= = = −

ln 32

2

R 11

4 2rA s rπ

= − −

and

32 1( 2 )2rw r sπ= +

Then

22 5( 2 ) 2.571

2rw r r r wπ= + = = −

In 21

R 21( 2 )

4 2rA r r s rπ= − + +


which gives

21 2(2 5.142 )2rw s r= − −

and

21 61.571w r w= + = −

With the centre of twist at 0

2 2

2 2R,1

1 1 2 0.2154 2 2 2r rA r r r rπ

= − − − + = +

and

21 0.43w r= −

Maximum shear stress is unchanged but torsional stiffness increases since the warping is reduced.

S.18.13

The loading is equivalent to a pure torque of 1×25 =25 kN/mm acting as shown in Fig. S.18.13 together with a shear load of 1 kN acting at 2 (the shear centre).

Fig. S.18.13

The maximum shear stress due to the torque is given by Eq. (18.13) in which

3 3

4100 3 80 2 1113.3mm3 3

J × ×= + =

Then

3

2max

25 10 3(324) 67.4 N / mm1113.3

τ × ×= =

3

2max

25 10 2(12) 44.9 N / mm1113.3

τ × ×= =


From Eq. (18.12)

3

4d 25 10 9.0 10 rad/mmd 25000 1113.3zθ −×= = ×

×

The shear flow distribution due to shear is given by Eq. (17.14) in which Sx =0 and Ixy

= 0, i.e.

0d

sys

xx

Sq ty s

I= − ∫

Taking moments of area about the top flange

(100 3 80 2) 80 2 40y× + × = × ×

i.e.

13.9mmy =

Then

3

2 2 42 80100 3 13.9 80 2 26.1 252290mm12xxI ×

= × × + + × × =

Therefore

1

12 1 102( 66.1 )d

sy

xx

Sq s s

I= − − +∫

i.e.

2

3 112 17.93 10 66.1

2sq s−

= − × −

(i)

From Eq. (i), q12 is a maximum when s1

= 66.1 mm. Then

12 (max) 17.4 N / mmq = −

and

212 (max) 8.7 N / mmτ = −

Also, from Eq. (i) the shear flow at 2 in 12 = –16.6 N/mm so that the maximum shear flow in the flange occurs at 2 and is –16.6/2 = –8.3 N/mm. Then the maximum shear stress in the flange is –8.3/3 = –2.8 N/mm2

The maximum shear stress due to shear and torsion is then 67.4 + 2.8 = 70.2 N/mm in the directions 32 and 42.

2

on the underside of 24 at 2 or on the upper surface of 32 at 2.


S.19.1

From Example 19.1

6 414.5 10 mmxxI = × From Eq. (16.18) in which My =0 and Ixy

=0

xz

xx

M yI

σ =

Therefore

6

620 10 1.38

14.5 10z y yσ ×= =

× (i)

The Cx

axis is 75 mm (see Example 19.1) from the upper wall 2367 so that, from Eq. (i), the maximum direct stress due to bending will occur in the wall 45 where y = –125 mm. Then

2(max) 1.38 ( 125) 172.5 N / mm (compression)zσ = × − = −

S.19.2

Fig. S.19.2

Take moments of areas about 23 2(4 500 2 200 2 400 600) 2(2 500 50 2 500 350

2 400 200 2 200 200)y× + × + × + = × × + × ×

+ × × + × ×

from which

168.4mmy =


Then (see Section 16.4.5) 23

2 2

3 32

2 2

4 500 2 100 2 500 2 118.4 2 500 2 181.612 500

2 400 2 2002 2 2 400 31.6 212 12

2 2 200 31.6 2 600 168.4

xxI × × = + × × × + × × ×

× ×+ × + × × × + ×

+ × × × + × ×

i.e.

6 4157.8 10 mmxxI = × Since O lies on an axis of symmetry q at O is zero. Then, from Eq. (17.14), the ‘basic’ or ‘open section’ shear flows are

O3 1(168.4)y

xx

Sq t s

I= −

and

3 in O3 168.4 300 50520y y

xx xx

S t S tq

I I= − × × = −

‘Cut’ the section at mid-point of 54. Then

2

22

94 2 2 20( 31.6)d 31.6

2sy y

xx xx

S t S t sq s s sI I

= − − = − −

∫

Then

4 1840 y

xx

S tq

I= −

3

43 3 30

10031.6 d 1840500

sy

xx

S tq s s

I = − + +

∫

which gives

23

43 331.6 184010

y

xx

S t sq sI

= − + +

and

4

3

36 4 40

(in 43) 42640

(168.4 )d 50520 42640

y

xx

sy

xx

S tq

IS t

q s sI

= −

= − − + + ∫


i.e.

24

36 4168.4 931602

y

xx

S t sq sI

= − − +

and

6 80520 y

xx

S tq

I= −

Similarly

25

65 4231.6 8052010

y

xx

S t sq sI

= − + +

and

5 10280 y

xx

S tq

I=

Also

26

59 6131.6 102802

y

xx

S t sq sI

= − − −

From Eq. (17.28)

300 100 500 400

O3 1 94 2 43 3 36 40 0 0 0500 100

65 5 59 60 0

d d d d d

d d

bq s q s q s q s q s

q s q s

= − − − −

− −

∫ ∫ ∫ ∫ ∫∫ ∫

Then, substituting for qO3

, etc.

d 65885801 yb

xx

S tq s

I= −∫

Also

d 4 500 2 400 2 200 600 3800s = × + × + × + =∫

Then

,0 34677 ys

xx

S tq

I=

Hence, the total shear flows are

O3 1168.4 y

xx

S tq s

I= −

24

36 4168.4 93160 346772

y

xx

S t sq sI

= − − + +


and so on and at the mid-point of 36.

179.4 N/mm (in direction 63)q = −

And the shear stress is

2179.4 89.7 N / mm2

=

S.19.3

For the closed part of the section, from Eq. (18.4)

2 24 4 25500(closed)d 2(400 200 2 500)t 2

A G AGJ s×

= =+ + ×

∫

But

2 2 1/2 21 (400 200)(500 100 ) 2 293938.8mm2

A = + − × =


12 2(closed) 5.4 10 N mmGJ = ×

From Eq. (18.11)

3 3

6 225000 600 2(open) 40 10 N mm3 3

stGJ G × ×= = = ×∑

(negligible compared to GJ (closed)) Therefore

12 2Total 5.4 10 N mmGJ = ×

From Eq. (18.4)

6

612

d 100 10 18.5 10 rad / mmd 5.4 10zθ −×= = ×

×

Then

12 6(closed) d 5.4 10 18.5 10

2 293938.82 2A d2

T GJqA z

θ −× × ×= = =

×

i.e.

340 N/mm (in closed part)q =


Therefore 2

max 170 N / mmτ =

The maximum shear stress in the open part is, from Eqs (18.12) and (18.13)

6 2max 25000 2 18.5 10 0.9 N / mmτ −= ± × × × = ±

S.19.4

The loading is equivalent to 100 kN through the shear centre plus a torque of 100 × 0.3 = 30 kNm.

From the solution of P.19.2 the maximum shear stress due to shear occurs at the mid-point of webs 72 and 63 and is 89.7N/mm2 in the directions 72 and 63. Also from the solution of P.19.3 the maximum shear stress due to torsion is 170(30/100) = 51 N/mm2 in the direction 63. Therefore the maximum shear stress due to shear and torsion is 89.7 + 51 + 140.7 N/mm2

The rate of twist = 18.5 × 10.

–6(30/100) = 5.6 × 10–6

S.19.5

rad/mm.

Referring to Fig.S.19.5 the angle α is given by

sinα = 25/(502 + 252)1/2

Then α = 26.57

= 25/55.9 = 0.4472 o and cosα = 0.8944.

Fig.S.19.5

The 50kN load acting in the plane of the web 63 is equivalent to a 50kN load acting through the shear centre together with a torque. To find the value of the torque we

270 Solutions Manual therefore need to first find the position of the shear centre. The section has a horizontal axis of symmetry so that the shear centre lies on this axis and Ixy = 0. Initially, therefore, we apply an arbitrary shear load, Sy, through the shear centre and determine the shear flow distribution. This is given by Eq.(17.15) in which Sx and Ixy

q

are both zero and which then reduces to

s = –(Sy/Ixx)∫0s tyds + qs,0

where (i)

Ixx = (2 × 503/12) + (4 × 1003/12) + 2 × 50 × 2 × 502 + 4(55.93 × 2sin226.57o

i.e. )/12

Ixx = 0.88 × 106 mm4

Solutions to Chapter 19 Problems 271 For the open part of the section qs,0

q

= 0 at the open edges 1 and 8. Then

12 = –(Sy/Ixx)∫0s2(25 + s1sinα)ds(Note to editor: the upper limit in the integral should be s

1

1i.e.

)

q12 = –(Sy/Ixx)(50s1 + 0.45s12

and when s) (ii)

1 = 55.9mm, q2 = –4201Sy/Ixx

q

.

23 = –(Sy/Ixx)(∫0s2 × 50ds2

(Note: the upper limit in the integral should be s + 4201)

2so that

)

q23 = –(Sy/Ixx)(100s2

and + 4201) (iii)

q3 = –9201Sy/IWe now “cut” the closed part of the section at 9. Then

xx

qb,94 = –(Sy/Ixx)∫0s2s3ds(Note: upper limit is s

3

3which gives

)

qb,94 = –(Sy/Ixx)s32

and (iv)

qb,4 = –625Sy/IAlso

xx

qb,43 = –(Sy/Ixx)[∫0s2(25 + 0.45s4)ds4

(Note: upper limit is s + 625]

4from which

)

qb,43 = –(Sy/Ixx)(50s4 + 0.45s42

and + 625) (v)

qb,3 = –48 26Sy/IFinally

xx

qb,36 = –(Sy/Ixx)[∫0s4(50 – s5)ds5

(Note: upper limit is s + 4826 + 9201]

5from which

)

qb,36 = –(Sy/Ixx)(200s5 – 2s52

From Eq.(17.27) + 14027) (vi)

qs,0 = –[∫(qb/t)ds]/(∫ds/t)


where

∫ds/t = 2[(25/2) + (55.9/2) + (50/4)] = 105.9

and ∫(qb/t)ds = –2(Sy/Ixx)[∫025(s3

2/2)ds3 + ∫055.9(1/2)(50s4 + 0.45s42 + 625)ds

+ ∫4

050(1/4)(200s5 – 2s5

2 + 14027)ds5

which gives

]

∫(qb/t)ds = –578 502Sy/I

so that

xx

qs,0 = 5463Sy/I

(Note: the integrals in the equations after Eq.(vi) where no limits are specified should have a circle at mid-height as in Eq.(17.27).

xx

Then q94 = -(Sy/Ixx)(s3

2

q

– 5463) (vii)

43 = –(Sy/Ixx)(50s4 + 0.45s42

q

– 4838) (viii)

36 = –(Sy/Ixx)(200s5 –2s52

Taking moments about the mid-point of 36

+ 8564) (ix)

SyξS = –2(Sy/Ixx)[∫025q94 × 50ds3 + ∫055.9q43 × 59cosαds4 – ∫050q23 × 50ds2 – ∫0

55.9q12 × 150 sinαds1

Substituting in Eq.(x) for q

] (x)

94

ξ

etc and evaluating gives

S

Therefore the shear load of 50kN acting along the web 63 is equivalent to a shear load of 50kN acting through the shear centre together with a torque of 50 × 85.8 = 4290kNmm or 4290Nm in an anticlockwise sense. The shear flow distribution due to shear is then, from Eqs(ii), (iii), (vii), (viii) and (ix)

= 85.8 mm

q12 = –(50 × 103/0.88 × 106)(50s1 + 0.45s12) = –0.057(50s1 + 0.45s1

2

q

) (xi)

23 = –0.057(100s2

q

+ 4201) (xii)

94 = –0.057(s32

q

– 5463) (xiii)

43 = –0.057(0.45s42 + 50s4

q

– 4838) (xiv)

36 = –0.057(200s5 – 2s52 + 8564) (xv)


Examination of Eqs(xi)–(xv) shows that the maximum shear stresses in the walls are as follows: 12:– –119.8N/mm

23:– – 262.3N/mm

2

94:– + 155.7N/mm

2

43:– + 137.9N/mm

2

36:– – 193.3N/mm

2

From Eqs(3.12) and (18.4)

2

(GJ)cl = 4A2

(Note: the integral sign has circle at mid-height as in Eq.(18.4))

G/(∫ds/t)

Then (GJ)cl = 4[(150 × 50)/2]2 × 25 000/105.9 = 1328 × 107Nmm

From Eq.(18.11)

2

(GJ)op = 25 000 × 2(50×23 + 55.9 × 23)/3 = 1.41 × 107Nmm

Then

2

(GJ)total = 1329.4×107Nmm

and

2

(dθ/dz) = 4290 × 103/1329.4 × 107 = 3.23 × 10–4

The maximum shear flow due to torsion in the closed portion is given by

rad/mm

qcl =[(GJ)cl/2A](dθ/dz) = 1328 × 107 × 3.23×10–4

From Eq.(18.10) the maximum shear stress in the open portion is

/2 × 3750 = 571.1N/mm

τmax,op = 25 000 × 2 × 3.23 × 10–4 = 16.2N/mm

Then, due to shear and torsion the maximum shear stresses in the open and closed portions are

2

τmax,op = 262.3 + 16.2 = 278.5N/mm2

τ

in the wall 23

max,cl = 155.7 + 571.1/2 = 441.3N/mm

The maximum shear stress in the section is therefore 441.3N/mm

2

2

and it occurs in the wall 94.


S.20.1

From either Eq. (20.1) or (20.2)

1500 10 300 1060 10 40 10 (2 1) (2 1)

6 6B × ×

= × + × + + + −

Fig. S.20.1(a)

i.e.

21 4

2

4000 mm500 10 300 850 8 30 8 (2 1) (2 1)

6 6

B B

B

= =× ×

= × + × + + + −

i.e.

22 33540 mmB B= =

Since the section is now idealized, the shear flow distribution due to an arbitrary shear load Sy

applied through the shear centre is, from Eq. (20.11), given by

,01

ny

s r r sxx r

Sq B y q

I =

= − +∑ (i)


in which

2 2 6 42 4000 150 2 3540 150 339 10 mm .xxI = × × + × × = ×

‘Cut’ the section in the wall 12. Then

b,12 b,43

3b,41

3b,32

0

4000 ( 150) 1.77 10

3540 ( 150) 1.57 10

yy

xx

yy

xx

q qS

q SIS

q SI

−

−

= =

= − × × − = ×

= − × × − = ×

Since the shear load is applied through the shear centre the rate of twist is zero and qs,0

is given by Eq. (17.28) in which d 500 300 3002 167.5

10 10 8st= × + + =∫

Then

3 3,0

1 300 3001.57 10 1.77 10167.5 8 10s yq S − − = − × × − × ×

which gives

3,0 0.034 10s yq S= − ×

The complete shear flow distribution is then as shown in Fig. S.20.1(b).

Fig. S.20.1(b)

Taking moments about the intersection of the horizontal axis of symmetry and the left-hand web

3 3S 1.536 10 300 500 2 0.034 10 500 150y y yS x S S− −= × × × − × × × ×

from which S 225 mmx =


S.20.2

From Eq. (20.6)

1

ny

s r rxx r

Sq B y

t =

= − ∑

Fig. S.20.2(a)

where

2 2 24 2.0 80 2 200 50 2 200 40xxI = × × + × × + × × i.e.

6 48.04 10 mmxxI = × Then

4

1

1.86 10n

s r rr

q B y−

=

= − × ∑

from which

412

443

432

427

1.86 10 200 ( 50) 1.86 N/mm

1.86 10 200 ( 40) 1.49 N/mm

1.49 1.86 10 250 ( 80) 5.21 N/mm

1.86 5.21 1.86 10 250 ( 80) 10.79 N/mm.

q

q

q

q

−

−

−

−

= − × × × − =

= − × × × − =

= − × × × − =

= + − × × × − =

The remaining shear flow distribution follows from symmetry; the complete distribution is shown in Fig. S.20.2(b).

Taking moments about the mid-point of web 27

S 12 32 43 432( 150 80 200 80 150 80 40 200)yS x q q q q= × × − × × − × × − × ×

which gives 122 mm (i.e. to the left of web 27)Sx = −


Fig. S.20.2(b)

S.20.3

The shear centre, S, lies on the horizontal axis of symmetry, the x axis. Therefore apply an arbitrary shear load, Sy, through S (Fig. S.20.3(a)). The internal shear flow distribution is given by Eq. (20.11) which, since Ixy = 0, Sx = 0 and tD

= 0, simplifies to

,01

ny

s r r sxx r

Sq B y q

I =

= − +∑ (i)

Fig. S.20.3(a)

in which

2 2 6 42 450 100 2 550 100 20 10 mmxxI = × × + × × = × Equation (i) then becomes

7,0

1

0.5 10n

s y r r sr

q S B y q−

=

= − × +∑ (ii)

The first term on the right-hand side of Eq. (ii) is the qb distribution (see Eq. (17.16)). To determine qb ‘cut’ the section in the wall 23. Then


b,23

7 3b,34 b,12

3 7 3b,41

0

0.5 10 550 ( 100) 2.75 10

2.75 10 0.5 10 450 ( 100) 5.0 10y y

y y y

q

q S S q

q S S S

− −

− − −

=

= − × × × − = × =

= × − × × × − = ×


The value of shear flow at the ‘cut’ is obtained using Eq. (17.28) which, since G= constant becomes

,0

( / )d

d /

bs

q t sq

s t= − ∫

∫

(iii)

In Eq. (iii)

d 580 500 2002 1996.71.0 0.8 1.2

st= + × + =∫

Then, from Eq. (iii) and the above qb

distribution 3 3

,02.75 10 500 5.0 10 2002

1996.7 0.8 1.2y

sS

q− − × × × ×

= − × +

i.e.

3,0 2.14 10s yq S−= − ×

The complete shear flow distribution is shown in Fig. S.20.3(b).

Fig. S.20.3(b)

Now taking moments about O in Fig. S.20.3(b) and using the result of Eq. (20.10)

3 3

S

3

2 0.61 10 500 100 2.86 10 200 500

2.14 10 2(135 000 500 200)y y y

y

S S S

S

ξ − −

−

= × × × × + × × ×

− × × − ×

which gives S 197.2 mmξ =

S.20.4

The x axis is an axis of symmetry so that Ixy = 0, also the shear centre, S, lies on this axis. Apply an arbitrary shear load, Sy, through S. The internal shear flow distribution is then given by Eq. (20.11) in which Sx = 0 and Ixy

= 0. Thus

D ,001

dnSy

s r r sxx r

Sq t y s B y q

I =

= − + +

∑∫ (i)


in which from Fig. S.20.4

3 3

2 2 0.36 80 0.64 804 100 40 2 0.64 240 4012 12xxI × ×

= × × + × × × + +

i.e.

6 41.17 10 mmxxI = ×

Fig. S.20.4

‘Cut’ the section at O. Then, from the first two terms on the right-hand side of Eq. (i)

1

b,O1 1 100.64 d

Sy

xx

Sq s s

I= − ∫

i.e.

6 2b,O1 10.27 10 yq S s−= − × (ii)

and

4b,1 4.32 10 yq S−= − ×

Also

2 4

b,12 200.64 40 d 100 40 4.32 10

syy

xx

Sq s S

I− = × + × − ×

∫

i.e.

4b,12 210 (0.22 38.52)yq S s−= − +

whence

4b,2 91.32 10 yq S−= − ×

Finally

3 4b,23 3 30

0.36(40 )d 100 40 91.32 10sy

yxx

Sq s s S

I− = − − + × − × ∫


i.e.

4 2 2b,23 3 310 (0.12 0.15 10 125.52)yq S s s− −= − − × + (iv)

The remaining qb

distribution follows from symmetry. From Eq. (17.27)

b,0

( / )d

d /s

q t sq

s t= − ∫

∫

(v)

in which

d 80 2 240 80 1097.20.64 0.64 0.36

st

×= + + =∫

Now substituting inEq. (v) for qb,O1, qb,12 and qb,23

from Eqs (ii)-(iv), respectively 4 240 2402

,0 1 1 2 20 0

40 2 23 3 30

2 10 0.27 10 1d (0.22 38.52)d1097.2 0.64 0.64

1 (0.12 0.15 10 125.52)d0.64

ys

Sq s s s s

s s s

− −

−

× ×= + +

+ − × +

∫ ∫

∫

from which

4,0 70.3 10s yq S−= ×

The complete shear flow distribution is then

4 2 2O1 110 (0.27 10 70.3)yq S s− −= − × − (vi)

412 34 210 (0.22 31.78)yq q S s−= = − − (vii)

4 2 223 3 310 (0.12 0.15 10 55.22)yq S s s− −= − − × − (viii)

Taking moments about the mid-point of the wall 23

40 240

S O1 1 12 20 02 240 d 40 dyS q s q sξ = × + × ∫ ∫ (ix)

Substituting for qO1 and q12

from Eqs (vi) and (vii) in Eq. (ix) 404 2 2

S 1 10

240

2 20

2 10 (0.27 10 70.3) 240 d

(0.22 31.78) 40 d

y yS S s s

s s

ξ − −= − × × − ×+ − ×

∫

∫

from which S 142.5 mmξ =


S.20.5

Referring to Fig. S.20.5(a) the x axis of the beam cross-section is an axis of symmetry so that Ixy = 0. Further, Sy at the end A is equal to –4450 N and Sx

= 0. The total deflection, Δ, at one end of the beam is then, from Eqs (20.17) and (20.19)

,1 ,0 0 1sect

d d dx x

L Lxx

M M q qz s zEI Gt

∆ = + ∫ ∫ ∫ (i)

in which q0

, from Eqs (20.20) and (20.11) is given by

,00 ,0

1

ny

r r sxx r

Sq B y q

I =

= − +∑ (ii)

Fig. S.20.5

and

01 4450

qq =

Since the booms carrying all the direct stresses, Ixx

in Eq. (i) is, from Fig. S.20.5(a)

2 2 2 6 42 650 100 2 650 75 2 1300 100 46.3 10 mmxxI = × × + × × + × × = × Also, from Fig. S.20.5(b) and taking moments about C B 500 4450 1750 4450 1250 0R × − × − × =


from which RBTherefore in AB

= 26 700N

,0 ,14450x xM z M z= =

and in BC

6,0 ,133.4 10 22 250 7500 5x xM z M z= × − = −

Thus the deflection, ΔM

, due to bending at the end A of the beam is, from the first term on the right-hand side of Eq. (i)

1250 15002 2M 0 1250

1 4450 d 4450(7500 5 ) dxx

z z z zEI

∆ = + − ∫ ∫

i.e.

12503

3 1500M 12506

0

4450 1 [(7500 5 ) ]3 1569 000 46.3 10z z

∆ = − − × ×

from which

M 1.09 mm∆ =

Now ‘cut’ the beam section in the wall 12. From Eq. (20.11), i.e.

,01

ny

s r r sxx r

Sq B y q

I =

= − +∑ (iii)

b,12

b,23

b,34

b,16

0

1300 100 130 000

130 000 650 100 195 000

650 75 48 750

y y

xx xx

y y y

xx xx xx

y y

xx xx

qS S

qI I

S S Sq

I I IS S

qI I

=

== − × × = −

= − − × × = −

= − × × = −

The remaining distribution follows from symmetry. The shear load is applied through the shear centre of the cross-section so that dθ/dz = 0 and qs,0

is given by Eq. (17.28), i.e.

b,0

d( = constant)

ds

q sq t

s= − ∫

∫

in which

d 2 300 2 250 2 100 2 75 1450 mms = × + × + × + × =∫


i.e.

,02

( 130 000 250 195 000 100 48 750 75)1450

ys

xx

Sq

I= − − × − × + ×

from which

,0 66 681 /s y xxq S I=

Then

12

23

34

16

66 681 /

63 319 /

128 319 /

115 431 /

y xx

y xx

y xx

y xx

q S I

q S I

q S I

q S I

=

= −

= −

= −

Therefore the deflection, ∆S

, due to shear is, from the second term in Eq. (i)

0 1S sect

d dL

q q s zGt

∆ = ∫ ∫

i.e.

}

,0 ,1 2 2 2S 2

2

2 (115 431 75 66 681 300 63 319 250

128 319 100) d

y y

L xx

S SGtI

z

∆ = × + × + ×

+ ×

∫

Thus

12

,0 ,1 8S ,0 ,16 2

4.98 102 d 6.96 10 d

26 70052.5 (46.3 10 )y y

y yL L

S Sz S S z−× ×

∆ = = ×× ×∫ ∫

Then

1250 15008 8

S 0 12506.96 10 4450 1 d 6.96 10 22 250 5 dz z− −∆ = × × × + × × ×∫ ∫

from which

S 2.32 mm∆ =

The total deflection, Δ, is then

M S 1.09 2.32 3.41 mm∆ = ∆ + ∆ = + =


S.20.6

At any section of the beam the applied loading is equivalent to bending moments in vertical and horizontal planes, to vertical and horizontal shear forces through the shear centre (the centre of symmetry C) plus a torque. However, only the vertical deflection of A is required so that the bending moments and shear forces in the horizontal plane do not contribute directly to this deflection. The total deflection is, from Eqs (20.14), (20.17) and (20.19)

,1 ,00 1 0 1sect

d d d dx x

L L Lxx

M MT T q qz z s zGJ EI Gt

∆ = + + ∫ ∫ ∫ ∫ (i)

Fig. S.20.6

Referring to Fig. S.20.6 the vertical force/unit length on the beam is

0 0 0 0 01.2 0.8 (upwards)2 2 2 2c c c cp p p p p c+ + − =

acting at a distance of 0.2c to the right of the vertical axis of symmetry. Also the horizontal force/unit length on the beam is

0 0 0 0 01.2 0.82 2 2 2t t t tp p p p p t+ + − =

acting to the right and at a distance 0.2t above the horizontal axis of symmetry. Thus, the torque/unit length on the beam is

2 20 00.2 0.2 0.2 ( )cp c c p t t p c t× − × = −

acting in an anticlockwise sense. Then, at any section, a distance z from the built-in end of the beam

2 20 0 10.2 ( )( ) 1 (unit load acting upwards at )

2cT p c t L z T A= − − = −


Comparing Eqs (3.12) and (18.4)

24

dAJ st

=

∫

i.e.

2 2 2

0

0

24 22 2 2

t c tt c aJt a

= =

Then

2 2 2 2 2

0 1 0 02 2 20 0 0 0

0.1 ( ) 0.1 ( )d ( )d

/2

L LT T p c t c p aL t cz L z zGJ Gt c t a Gt t c

− −= − − =∫ ∫ (ii)

The bending moment due to the applied loading at any section a distance z from the built-in end is given by

20,0 ,1( ) also 1 ( )

2x xp cM L z M L z= − − = − −

Thus

,1 ,0 300 0

( ) d2

L Lx x

xx xx

M M p c L z zEI EI

= −∫ ∫

in which

23 2 3 2

0 0 0( ) sin / 2212 6 / 2 6xx

a t a t at ttIa

α = = =

Then

4

,1 ,0 40 02 20 00 0

3 31d ( )4 4

LL x x

xx

M M p c p cLz L zEI Eat t Eat t

= − − = ∫ (iii)

The shear load at any section a distance z from the built-in end produced by the actual loading system is given by

,0 0 ,1( ) also 1y yS p c L z S= − =

From Eq. (17.15), in which Ixy = 0 and Sx

= 0

,00d

sys s

xx

Sq ty s q

I= − +∫ (iv)

If the origin of s is taken at the point l, qs,0

= 0 since the shear load is applied on the vertical axis of symmetry, Eq. (iv) then becomes

0d

sys

xx

Sq ty s

I= − ∫


and

12 02 00

6sin d

2syS tq t s s

at tα = − − +

∫

i.e.

2

12 2

62 2

yS t t sq saat

= −

Thus

2

123 yS sq sat a

= −

The remaining distribution follows from symmetry. Then

22/2

0 1 02 2sect 00

9 ( )d 4 d

aq q p c L z ss s sGt aGa t t

−= × −

∫ ∫

i.e.

0 1 02sect 0

3 ( )d

5q q p ca L zsGt Gt t

−=∫

Then

2

0 1 0 02 20 sect 00 0

3 3d d ( ) d

5 10

L Lq q p ca p caLs z L z zGt Gt t Gt t

= − = ∫ ∫ ∫ (v)

Now substituting in Eq. (i) from Eqs (ii), (iii) and (v)

2 2 2 4 2

0 0 02 2 2

0 0 0

0.1 ( ) 3 34 10

p aL t c p cL p caLGt t c Ea tt Gt t

−∆ = + +

i.e.

2 2 2 2

02

0

( ) 3 310 4 10

p L a t c cL caGc Ea Gt t

−∆ = + +

Substituting the given values and taking a ≏ c

2 2 2 2 2

02

0

(2 ) [(0.05 ) ] 3 (2 ) 34 4 4(0.05 )

p c c c c c c cE Ec Ec t

−∆ = + +

Neglecting the term (0.05c)2 in [(0.05c)2 –c2

] gives 2

0

0

5600 p cEt

∆ =


S.20.7

The pressure loading is equivalent to a shear force/unit length of 3bp0/2 acting in the vertical plane of symmetry together with a torque = 3bp0(3b/2 –b)/2 = 3b2p0

/4 as shown in Fig. S.20.7. The deflection of the beam is then, from Eqs (20.14), (20.17) and (20.19)

,1 ,00 1 0 1sect

d d d dx x

L L Lxx

M MT T q qz z s zGJ EI Gt

∆ = + + ∫ ∫ ∫ ∫ (i)

Fig. S.20.7

Now

20 0 13 ( )/4 3 /2T b p L z T b= − =

Also, from Eqs (3.12) and (18.4)

2 2 2 34 4(3 ) 9

8 / 2d /A b b tJ

b ts t= = =∫

Thus

2

0 1 0 00

d ( )d4 8

L

L

T T p p Lz L z zGJ Gt Gt

= − =∫ ∫ (ii)

Also

2,0 0 ,13 ( ) /4 1( )x xM b pL z M L z= − = −

Then

,1 ,0 300

3d ( ) d

4Lx x

L xx xx

M M bpz L z zEI EI

= −∫ ∫

in which

2 3 32 3 ( /2) 2 / 12 5 /3xxI bt b tb b t= × × + =


Thus

4

,1 ,0 30 02 20

9 9d ( ) d

20 80

Lx x

L xx

M M p p Lz L z zEI Eb t Eb t

= − =∫ ∫ (iii)

Further

0,0 ,1

3( ) 1

2y ybpS L z S= − − = −

Taking the origin for s at 1 in the plane of symmetry where q ,0 =0 and since Ixy = 0 and Sx


0d

sys

xx

Sq ty s

I= − ∫

Then

1

12 13 0

3d

25

syS bq t sb t

= − ∫

i.e.

12 12

3

10yS

q sb

= −

from which

2920

ySq

b= −

Also

2

23 2 20

9d

2 20sy y

xx

S Sbq t s sI b

= − − − ∫

i.e.

22

23 23

3 92 2 205

y yS Ssbq sbb

= − − −

Hence

2

2 223 2

32 2 3

20yS s sqb b b

= − − +

Then

23 /2 20 1 01 12sect 0

22 20 2 2

223 /2

3 ( ) 3d 4 d2 10

3 ( ) 32 2 2 3 d2 20

b

b

b

q q bp L zs s sGt Gt b

bp L z s s sGt b b

− =

− + − +

∫ ∫

∫


which gives

0 1 0sect

1359d ( )

1000q q ps L zGt Gt

= −∫

Hence

2

0 1 0 00 sect 0

1359 1359d d ( )d

1000 2000L Lq q p p Ls z L z z

Gt Gt Gt = − = ∫ ∫ ∫ (iv)

Substituting in Eq. (i) from Eqs (ii)–(iv) gives

2 4 2

0 0 02

9 13598 200080p L p L p L

Gt GtEb t∆ = + +

Thus

2 2

02

9 1609200080

p L Lt GEb

∆ = +


S.21.1

Referring to Fig. P.21.1 the bending moment at section 1 is given by

2

115 1 7.5 kN m

2M ×

= =

Thus

,U ,L 37.5 25 kN

300 10z zP P −= − = =×

Also

,U ,L 31000 and 25 2.5 kN (see Eqs (21.1))

1 10y yP P= = − × = −×

Then

2 2U ,U ,U 25 kN (tension)z yP P P= + =

2 2L 25 2.5 25.1 kN (compression)P = − + = −

The shear force at section 1 is 15 × 1 = 15 kN. This is resisted by Py,L,

shear in web = 15 – 2.5 = 12.5 kN

the shear force in the web. Thus


Hence

312.5 10 41.7 kN/mm

300q ×= =

At section 2 the bending moment is

2

215 2 30 kN m

2M ×

= =

Hence

,U ,L 330 75 kN

400 10z zP P −= − = =×

Also

,U ,L 32000 and 75 7.5 kN

2 10y yP P= = − × = −×

Then U 75 kN (tension)P =

and

2 2L 75 7.5 75.4 kN (compression)P = − + = −

The shear force at section 2 is 15 × 2 = 30 kN. Hence the shear force in the web = 30 – 7.5 = 22.5 kN which gives

322.5 10 56.3 N/mm

400q ×= =

S.21.2

The bending moment at section 1 is given by

215 1 7.5 kN m

2M ×

= =

The second moment of area of the beam cross-section at section 1 is

3

2 7 42 3002 500 150 2.7 10 mm12xxI ×

= × × + = ×

The direct stresses in the flanges in the z direction are, from Eq. (16.18)

6

2,U ,L 7

7.5 10 150 41.7 N/mm2.7 10z zσ σ × ×

= − = =×

Then ,U U41.7 500 20 850 N (tension)zP P= × = =


Also

,L 20 850 N (compression)zP = −

Hence

,L 310020 850 2085 N (compression)

1 10yP = − × = −×

Therefore, the shear force in the web at section 1 is given by

315 1 10 2085 12 915 NyS = − × × + = −

Fig. S.21.2

The shear flow distribution is obtained using Eq. (21.6). Thus, referring to Fig. S.21.2

7 0

12 915 2(150 ) d 500 1502.7 10

sq s s = − + × × ∫

Hence

4 24.8 10 (300 75 000)q s s−= × − +

The maximum value of q occurs when s = 150 mm, i.e. max 46.8 N/mmq =

S.21.3

The beam section at a distance of 1.5 m from the built-in end is shown in Fig. S.21.3. The bending moment, M, at this section is given by

40 1.5 60 kN mM = − × = −

Since the x axis is an axis of symmetry Ixy = 0; also My

= 0. The direct stress distribution is then, from Eq. (16.18)

xz

xx

M yI

σ = (i)


Fig. S.21.3

in which Ixx = 2 × 1000 × 112.52 + 4 × 500 × 112.52 = 50.63 × 106mm4

. Then, from Eq. (i), the direct stresses in the flanges and stringers are

62

660 10 112.5 133.3 N/mm

50.63 10zσ × ×= ± = ±

×

Therefore

,1 ,2 133.3 1000 133 300 Nz zP P= − = − × = −

and

,3 ,5 ,4 ,6 133.3 500 66 650 Nz z z zP P P P= = − = − = − × = −

From Eq. (21.9)

,1 ,2 375133 300 3332.5 N

3 10y yP P= = × =×

and

,3 ,4 ,5 ,6 37566 650 1666.3 N

3 10y y y yP P P P= = = = × =×

Thus the total vertical load in the flanges and stringers is

2 3332.5 4 1666.3 13 330.2 N× + × =

Hence the total shear force carried by the panels is

340 10 13 330.2 26 669.8 N× − =


The shear flow distribution is given by Eq. (20.11) which, since Ixy = 0, Sx = 0 and tD

= 0 reduces to

,01

ny

s r r sxx r

Sq B y q

I =

= − +∑

i.e.

,061

26 669.850.63 10

n

s r r sr

q B y q=

= − +× ∑

or

4,0

1

5.27 10n

s r r sr

q B y q−

=

= − × +∑ (ii)

From Eq. (ii)

b,13

4b,35

4b,56

4b,12

0

5.27 10 500 112.5 29.6 N/mm

29.6 5.27 10 500 112.5 59.2 N/mm

5.27 10 1000 112.5 59.3 N/mm

q

q

q

q

−

−

−

=

= − × × × = −

= − − × × × = −

= − × × × = −

The remaining distribution follows from symmetry. Now taking moments about the point 2 (see Eq. (17.17))

,0

26 669.8 100 59.2 225 500 29.6 250 225 2 500 225sq× = × × + × × + × ×

from which

,0 36.9 N/mm (i.e. clockwise)sq = −

Then

13 42

35 64

65

21

36.9 N/mm36.9 29.6 7.3 N/mm59.2 36.9 22.3 N/mm36.9 59.3 96.2 N/mm

q qq qqq

= =

= − = =

= − =

= + =

Finally

2 2 2 2 31 ,1 ,1 2

2 2 2 2 33 ,3 ,3

5 4 6

( 133 300 3332.5 ) 10 133.3 kN

( 66 650 1666.3 ) 10

66.7 kN

z y

z y

P P P P

P P P

P P P

−

−

= − + = − + × = − = −

= − + = − + ×

= − = = − = −



S.22.1

The direct stresses in the booms are obtained from Eq. (16.18) in which Ixy = 0 and My

= 0. Thus

xz

xx

M yI

σ = (i)

From Fig. P.22.1 the y coordinates of the booms are

1 6

2 1 05 7

3 9 4 8

750 mm250 500 sin 45 603.6 mm

250 mm

y yy y y yy y y y

= − =

= = − = − = + ° =

= = − = − =

Then Ixx = 2 × 150(7502 + 2 × 603.62 + 2 × 2502) = 4.25 × 108 mm4

. Hence, from Eq. (i)

6

8100 104.25 10z yσ ×

=×

i.e. 0.24z yσ =

Thus

Boom 1 2 10 3 9 4 8 5 7 6 σz (N/mm2 180.0 ) 144.9 60.0 –60.0 –144.9 –180.0

From Eq. (20.11)

,01

ny

s r r sxx r

Sq B y q

I =

= − +∑

i.e.

3

,0850 10 150

4.25 10s r sq y q× ×= − +

× ∑

so that ,00.018s r sq y q= − + (ii)

‘Cut’ the wall 89. Then, from the first term on the right-hand side of Eq. (ii)

b,89

b,910

b,10 1

00.018 250 4.5 N/mm

4.5 0.018 603.6 15.4 N/mm

qq

q

=

= − × = −

= − − × = −


,12

,23

,34

15.4 0.018 750 28.9 N/mm

28.9 0.018 603.6 39.8 N/mm

39.8 0.018 250 44.3N/mm

b

b

b

q

q

q

= − − × = −

= − − × = −

= − − × = −

The remaining qb

distribution follows from symmetry and the complete distribution is shown in Fig. S.22.1. The moment of a constant shear flow in a panel about a specific point is given by Eq. (20.10). Thus, taking moments about C (see Eq. (17.17))

3910 101 12 23

34 ,0

50 10 250 2( 2 4.5 2 15.4 2 28.9 2 39.8

2 44.3 ) 2 s

A A A A

A Aq

× × = − × − × − × − ×

− × − (iii)

in which 2

34

2 223 910

2 212 101

1 500 250 62500mm2

45 162500 500 250 353.6 116474.8mm360 2

1 45250 353.6 500 142374.8mm2 360

A

A A

A A

π

π

= × × =

= = + × × − × × =

= = × × + × × =

Fig. S.22.1

Also the total area, A, of the cross-section is 2 2500 1000 500 1285398.2mmA π= × + × = Eq. (iii) then becomes

3

,0

50 10 250 2 2(4.5 116474.8 15.4 142374.8 28.9 142374.839.8 116474.8 44.3 62500) 2 1285398.2 sq

× × = − × × + × + ×+ × + × − ×


from which

,0 27.0 N/mmsq = − (clockwise)

Then q89 = 27.0 N/mm, q9 10 = q78 = 22.5 N/mm, q101 = q67

q

= 11.6 N/mm,

21 = q65 = 1.9 N/mm, q32 = q54 = 12.8 N/mm, q43


= 17.3 N/mm

S.23.1

The beam section is unsymmetrical and Mx = –120 000 Nm, My

= –30 000 Nm. Therefore, the direct stresses in the booms are given by Eq. (16.18), i.e.

2 2y xx x xy x yy y xy

zxx yy xy xx yy xy

M I M I M I M Ix y

I I I I I Iσ

− −= + − −

(i)

Fig. S.23.1

In Fig. S.23.1 x = 600 mm by inspection. Also, taking moments of area about the line of the bottom booms

(4 1000 4 600) 1000 240 1000 180 600 220 600 200y× + × = × + × + × + ×

from which

105mmy =

Then Ixx = 2 × 1000 × 1052 + 2 × 600 × 1052 + 1000 × 1352 + 1000 × 752 + 600 × 115

+ 600 × 95 = 72.5 × 10 mm2

I

4 yy = 4 × 1000 × 6002 + 4 × 600 × 2002 = 1536.0 × 106 mm

I

4 xy

= –38.4 × 10 = 1000[(—600)(135) + (600)(75)] + 600[(—200)(115) + (200)(95)]

6 mm4


Table S.23.1 Boom 1 2 3 4 5 6 7 8 x (mm) –600 –200 200 600 600 200 –200 –600 y (mm) 135 115 95 75 105 –105 –105 –105 σz (N/mm2 –190.7 ) –181.7 –172.8 –163.8 140.0 164.8 189.6 214.4

Note that the sum of the contributions of booms 5, 6, 7 and 8 to Ixy is zero. Substituting for Mx, My, Ixx

, etc. in Eq. (i) gives

0.062 1.688z x yσ = − − (ii)

The solution is completed in Table S.23.1.

S.23.2

From Eq. (23.6) for Cell I

[ ]I 21 16 65 52 II 52I

d 1 ( )d 2

q qz A Gθ δ δ δ δ δ= + + + − (i)

and for Cell II

[ ]I 52 II 32 25 54 43II

d 1 ( )d 2

q qz A Gθ δ δ δ δ δ= − + + + + (ii)

In Eqs (i) and (ii)

( )( )( )

2I

2II

2 221 16

2 265 52

2 254 43

32

7750 (250 600) 500 / 2 220250mm

6450 (150 600) 920 / 2 351450mm

250 500 / 1.63 343.0 300 / 2.03 147.8

100 500 / 0.92 554.2 600 / 2.54 236.2

250 920 / 0.92 1036.3 250 / 0.56 446.4

20

A

A

δ δ

δ δ

δ δ

δ

= + + × =

= + + × =

= + = = =

= + = = =

= + = = =

= ( )2 20 920 / 0.92 1023.4+ =

Substituting these values in Eqs (i) and (ii) gives, for Cell I

I IId 1 (1281.2 236.2 )d 2 220250

q qz Gθ= −

× (iii)

and for Cell II

I IId 1 ( 236.2 2742.3 )d 2 351450

q qz Gθ= − +

× (iv)


Equating Eqs (iii) and (iv) gives

II I0.73q q= (v)

Then, in Cell I

Imax 65 I1.087

0.92q qτ τ= = =

and in Cell II

IImax I1.304

0.56q qτ = =

In the wall 52

I II52 I0.106

2.54q q qτ−

= =

Therefore

2max I1.304 140 N/mmqτ = =

which gives I 107.4 N/mmq = and, from Eq. (v) II 78.4 N/mmq = Substituting for qI and qII

in Eq. (23.4) 3(2 220250 107.4 2 351 450 78.4) 10T −= × × + × × ×

i.e. 102417 NmT = From Eq. (iii) (or Eq. (iv))

d 1 (1281.2 107.4 236.2 78.4)d 2 220250 26600zθ= × − ×

× ×

i.e.

5d 1.02 10 rad/mmdzθ −= ×

Hence

5 o1801.02 10 2500 1.46θπ

− = × × × =

The torsional stiffness is obtained from Eq. (3.12), thus

( ) ( )3 5 12 2102 417 10 / 1.02 10 10 10 Nmm /radd /d

TGJzθ

−= = × × = ×



( )o i iI II45 45 45I

d 1d 2

q qz A Gθ δ δ δ = + − (i)

For Cell II

i iI II 34 56 63 III 6345 45II

d 1 ( )d 2

q q qz A Gθ δ δ δ δ δ δ = − + + + + − (ii)

For Cell III

[ ]II 63 III 23 36 67 72 IV 72III

d 1 ( )d 2

q q qz A Gθ δ δ δ δ δ δ= − + + + + − (iii)

For Cell IV

[ ]III 72 IV 27 78 81 12IV

d 1 ( )d 2

q qz A Gθ δ δ δ δ δ= − + + + + (iv)

where δ12 = δ78 = 762/0.915 = 832.8 δ23 = δ67 = δ34 = δ56

δ = 812/0.915 = 887.4

45i = 356/1.220 = 291.8 δ45°

δ = 1525/0.711 = 2144.9

36 = 406/1.625 = 249.8 δ72 = 356/1.22 = 291.8 δ81

Substituting these values in Eqs (i)–(iv)

= 254/0.915 = 277.6

( )I IId 1 2436.7 291.8d 2 161500

q qz Gθ= −

× (v)

( )I II IIId 1 291.8 2316.4 249.8d 2 291000

q q qz Gθ= − + −

× (vi)

( )II III IVd 1 249.8 2316.4 249.8d 2 291000

q q qz Gθ= − + −

× (vii)

( )III IVd 1 291.8 2235.0d 2 226000

q qz Gθ= − +

× (viii)

Also, from Eq. (23.4) I II III IV2(161500 291000 291000 226000 )T q q q q= + + + (ix)

Equating Eqs (v) and (vi) I II III0.607 0.053 0q q q− + = (x)

Now equating Eqs (v) and (vii)

I II III IV0.063 0.528 0.066 0q q q q− − + = (xi)


Equating Eqs (v) and (viii)

1 II III IV0.120 0.089 0.655 0q q q q− + − = (xii)

From Eq. (ix)

6I II III IV1.802 1.802 1.399 3.096 10q q q q T−+ + + = × (xiii)

Subtracting Eq. (xi) from (x)

II III IV1.068 0.121 0q q q− + = (xiv)

Subtracting Eq. (xii) from (x)

II III IV0.074 1.345 0q q q+ + = (xv)

Subtracting Eq. (xiii) from (x)

II III IV0.726 0.581 0q q q+ + = (xvi)

Now subtracting Eq. (xv) from (xiv)

III IV1.284 0q q− = (xvii)

Subtracting Eq. (xvi) from (xiv)

6III IV0.256 0.716 10q q T−+ = × (xviii)

Finally, subtracting Eq. (xviii) from (xvii)

6IV 0.465 10q T−= ×

and from Eq. (xvii)

6III 0.597 10q T−= ×

Substituting for qIII and qIV

in Eq. (viii) 9d 1.914 10

dT

z Gθ −×=

so that

6 2/ (d / d ) 522.5 10 Nmm /radT z Gθ = ×

S.23.4

In this problem the cells are not connected consecutively so that Eq. (23.6) does not apply. Therefore, from Eq. (23.5) for Cell I

( )U UI 23 41 III 23 4112 34I

d 1 + ( )d 2

q qz A Gθ δ δ δ δ δ δ = + + − + (i)


For Cell II

( )U U LL

I II III 3434 34 34II

d 1d 2

q q qz A Gθ δ δ δ δ = − + + − (ii)

For Cell III

L L LI 23 41 II III 14 3234 43 21III

d 1 ( ) ( )d 2

q q q qz A Gθ δ δ δ δ δ δ = − + − + + + + (iii)

In Eqs (i)–(iii)

U L

U L

12 12

14 23 34 34

1084 / 1.220 888.5 2160 / 1.625 1329.2

127 / 0.915 138.8 797 / 0.915 871.0

δ δ

δ δ δ δ

= = = =

= = = = = =

Substituting these values in Eqs (i)–(iii)

I II IIId 1 (2037.1 871.0 277.6 )d 2 108400

q q qz Gθ= − −

× (iv)

I II IIId 1 ( 871.0 1742.0 877.0 )d 2 202500

q q qz Gθ= − − −

× (v)

I II IIId 1 ( 277.6 871.0 2477.8 )d 2 528000

q q qz Gθ= − − −

× (vi)


3I II III565000 10 2(108400 202500 528000 )q q q× = + + (vii)

Equating Eqs (iv) and (v) I II III0.720 0.075 0q q q− + = (viii)

Equating Eqs (iv) and (vi)

I II III0.331 0.375 0q q q− + = (ix)

From Eq. (vii) I II III1.868 4.871 260.61q q q+ + = (x)

Subtracting Eq. (ix) from (viii)

II III1.157 0q q− = (xi)

Subtracting Eq. (x) from (viii)

II III1.853 100.70q q+ = (xii)

Finally, subtracting Eq. (xii) from (xi)

III 33.5 N/mmq =


Then, from Eq. (xi)

II 38.8 N/mmq =

and from Eq. (ix)

I 25.4 N/mmq =

Thus

U L

U L

14 3212 21

43 34

25.4 N/mm 33.5 N/mm 33.5 25.4 8.1N/mm

38.8 25.4 13.4 N/mm 38.8 33.5 5.3N/mm

q q q q

q q

= = = = − =

= − = = − =

S.23.5

In Eq. (23.10) the qb shear flow distribution is given by Eq. (20.6) in which, since the x axis is an axis of symmetry (Fig. S.23.5), Ixy = 0; also Sx

= 0. Thus

b1

ny

r rxx r

Sq B y

I =

= − ∑ (i)

Fig. S.23.5

in which 2 2 2 6 42 1290 127 2 1936 203 2 645 101 214.3 10 mmxxI = × × + × × + × × = ×

Then Eq. (i) becomes

4b 6

1 1

44500 2.08 10214.3 10

n n

r r r rr r

q B y B y−

= =

= − = − ×× ∑ ∑


‘Cut’ the walls 65 and 54. Then

b,65 b,54

4b,61

b,12 b,23

4b,25

4b,34

0

2.08 10 1290 127 32.8 N/mm0 (fromsymmetry)

2.08 10 1936( 203) 81.7 N/mm

2.08 10 645( 101) 13.6 N/mm

q q

qq q

q

q

−

−

−

= =

= − × × × = −

= =

= − × × − =

= − × × − =


( )s,0,I 56 61 12 24 s,0,II 25 b,25 25 b,61 61I

d 1d 2

q q q qz A Gθ δ δ δ δ δ δ δ = + + + − + + (ii)

For Cell II

s,0,I 25 s,0,II 45 52 23 34 b,34 34 b,52 25II

d 1 ( )d 2

q q q qz A Gθ δ δ δ δ δ δ δ = − + + + + + + (iii)

in which

56 12 45 23

61 52 34

647 / 0.915 707.1 775 / 0.559 1386.4254 / 1.625 156.3 406 / 2.032 199.8 202 / 1.220 165.6

δ δ δ δδ δ δ

= = = = = =

= = = = = =

Substituting these values in Eqs (ii) and (iii)

s,0,I s,0,IId 1 (1770.3 199.8 11197.0)d 2 232000

q qz Gθ= + +

× (iv)

s,0,I s,0,IId 1 ( 199.8 3138.2 14071.5.0)d 2 258000

q qz Gθ= − + −

× (v)

Also, taking moments about the mid-point of the web 25 and from Eq. (23.11) (or Eq. (23.12))

I s,0,I II s,0,II0 13.6 202 763 32.8 254 635 2 2A q A q= × × − × × + + (vi)

Equating Eqs (iv) and (v) s,0,I s,0,II1.55 12.23 0q q− + = (vii)

From Eq. (vi) s,0,I s,0,II1.11 6.88 0q q− + = (viii)

Subtracting Eq. (viii) from (vii) gives s,0,II 7.2 N/mmq =

Then, from Eq. (vii) s,0,I 1.1N/mmq = −


Thus

16 65 21

45 23 34

25

32.8 1.1 33.9 N/mm q 1.1N/mm7.2 N/mm 13.6 7.2 20.8 N/mm

81.7 1.1 7.2 73.4 N/mm

q qq q qq

= + = = =

= = = + =

= − − =

S.23.6

Referring to Fig. P.23.6, the horizontal x axis is an axis of symmetry so that Ixy = 0. Further, Sx

= 0 so that, from Eq. (20.6)

b1

ny

r rxx r

Sq B y

I =

= − ∑ (i)

in which

2 2 6 44 1290 153 4 645 153 181.2 10 mmxxI = × × + × × = ×


4b 6

1 1

66750 3.68 10181.2 10

n n

r r r rr r

q B y B y−

= =

= − = − ×× ∑ ∑

Now, ‘cutting’ Cell I in the wall 45 and Cell II in the wall 12 b,45 b,120q q= =

4b,43 b,653.68 10 645 153 36.3N/mmq q−= − × × × = − = (from symmetry)

4b,18 3.68 10 1290 153 72.6 N/mmq −= − × × × = −

b,78 0q = (from symmetry) 4

b,76 b,323.68 10 645 ( 153) 36.3N/mmq q−= − × × × − = = (from symmetry) 4

b,63 36.3 36.3 3.68 10 1290 ( 153) 145.2 N/mmq −= + − × × × − = The shear load is applied through the shear centre of the section so that the rate of

twist of the section, dθ/dz, is zero and Eq. (23.10) for Cell I simplifies to

,0,I 34 45 56 63I REF

,0,II 63 b,63 63 b,34 34 b,56 56

10 [ ( )2

]

s

s

qA G

q q q q

δ δ δ δ

δ δ δ δ

= + + +

− + + + (ii)

and for Cell II

,0,I 63 ,0,II 12 23 36 67 78 81 b,81 81II REF

b,23 23 b,36 36 b,67 67

10 [ ( )2

]

s sq q qA G

q q q

δ δ δ δ δ δ δ δ

δ δ δ

= − + + + + + + +

+ + + (iii)


in which GREF = 24 200 N/mm2

. Then, from Eq. (23.9)

* *34 56

20700 0.915 0.783mm24200

t t= = × =

* * *36 81 45

24800 1.220 1.250mm24200

t t t= = = × =

Thus 34 65 380/0.783 485.3δ δ= = = 12 23 67 78 356/0.915 389.1δ δ δ δ= = = = = 36 81 306 / 1.250 244.8δ δ= = = 45 610 / 1.250 488.0δ = = Eq. (ii) then becomes ,0,I ,0,II1703.4 244.8 70777.7 0s sq q− + =

or ,0,I ,0,II0.144 41.55 0s sq q− + =

and Eq. (iii) becomes ,0,I ,0,II244.8 2046 46021.1 0s sq q− + − =

or ,0,I ,0,II8.358 188.0 0s sq q− + =

Subtracting Eq. (v) from (iv) gives ,0,II 17.8 N / mmsq =

Then, from Eq. (v) ,0,I 39.2 N / mmsq = −

The resulting shear flows are then 12 78 32 7617.8 N / mm q 36.3 17.8 18.5 N / mmq q q= = = = − = 63 145.2 17.8 39.2 88.2 N / mmq = − − = 43 65 5439.2 36.3 2.9 N / mm 39.2 N / mmq q q= = − = = 81 72.6 17.8 90.4 N / mmq = + =

Now taking moments about the mid-point of the web 63

S 76 78 81

43 54

66750 2 356 153 2 356 153 306 7122 380 153 2(51 500 153 380) (see Eq. (20.10))

x q q qq q

= − × × × + × × × + × ×

− × × × − × + ×

from which

S 160.1mmx =


S.23.7

Referring to Fig. P.23.7 the horizontal x axis is an axis of symmetry so that Ixy = 0 and the shear centre lies on this axis. Further, applying an arbitrary shear load, Sy, through the shear centre then Sx

= 0 and Eq. (20.6) simplifies to

b1

ny

r rxx r

Sq B y

I =

= − ∑ (i)

in which

2 2 2 6 42 645 102 2 1290 152 2 1935 150 162.4 10 mmxxI = × × + × × + × × = ×


9

1

6.16 10n

b y r rr

q S B y−

=

= − × ∑ (ii)

‘Cut’ the walls 34° and 23. Then, from Eq. (ii) b,34 b,23 b,450q q q° = = =

i9 3

b,43 6.16 10 1935 ( 152) 1.81 10 N / mmy yq S S− −= − × × × − = × 9 3

b,65 b,216.16 10 645 ( 102) 0.41 10 N / mmy yq S S q− −= − × × × − = × = (from symmetry)

3 9 3b,52 0.41 10 6.16 10 1290 ( 152) 1.62 10 N / mmy y yq S S S− − −= × − × × × − = ×

Since the shear load, Sy

, is applied through the shear centre of the section the rate of twist, dθ/dz, is zero. Thus, for Cell I, Eq. (23.10) reduces to

i i i i,0,1 34 ,0,II34 34 b,43 340 ( )s sq q qδ δ δ δ°= + − + (iii)

and for Cell II

i i i i,0,I ,0,II 23 45 52 b,52 5243 34 b,43 430 ( )s sq q q qδ δ δ δ δ δ δ= − + + + + + − (iv)

in which i34 341015/0.559 1815.7 304/2.030 149.8δ δ° = = = =

23 45 765/0.559 836.1δ δ= = = 25 304/1.625 187.1δ = =

Thus Eq. (iii) becomes

,0,I ,0,II1965.5 149.8 0.271 0s s yq q S− + =


or

3,0,I ,0,II0.076 0.138 10 0s s yq q S−− + × = (v)

and Eq. (iv) becomes

4,0,I ,0,II149.8 2009.1 319.64 10 0s s yq q S−− + + × =

or

3,0,I ,0,II13.411 0.213 10 0s s yq q S−− − × = (vi)

Subtracting Eq. (vi) from (v)

3,0,II13.335 0.351 10 0s yq S−+ × =

whence

3,0,II 0.026 10s yq S−= − ×

Then from Eq. (vi)

3,0,I 0.139 10s yq S−= − ×

Now taking moments about the mid-point of the web 43

b,21 b,52 ,0,II

,0,I

2 (508 152 50 762) 304 762 2 258000

2 93000y s s

s

S x q q q

q

= − × + × + × × + ×

+ ×

from which 241.4mmsx =

S.23.8

The direct stresses in the booms are given by the first of Eqs (16.21) in which, referring to Fig. P.23.8, at the larger cross-section

2 2 6 42 600 105 4 800 160 95.2 10 mmxxI = × × + × × = ×


, ,x r

z r z r r rxx

M BP B yI

σ= =

or

3

2, 6

1800 10 1.89 1095.2 10z r r r r rP B y B y−×

= = ××

(i)

The components of boom load in the y and x directions (see Fig. 21.4(a) for the axis system) are found using Eqs. (21.9) and (21.10). Then, choosing the intersection of the web 52 and the horizontal axis of symmetry (the x axis) as the moment centre


Table S.23.8 Boom Pz,r δy

(N) r/δ δxz r P/δz y,r P (N) x,r P (N) r η(N) r ξ

(mm) r P

(mm) y,r ξ r P

(N mm) x,r ηr

(N mm) 1 1190.7 0.045 –0.12 53.6 –142.9 1200.4 590 105 31 624 – 15 004.5 2 2419.2 0.060 0 145.2 0 2423.6 0 160 0 0 3 2419.2 0.060 0.18 145.2 435.5 2462.4 790 160 – 114708 69680 4 –2419.2 –0.060 0.18 145.2 –435.5 –2462.4 790 160 – 114708 69680 5 –2419.2 –0.060 0 145.2 0 –2423.6 0 160 0 0 6 –1190.7 –0.045 –0.12 53.6 142.9 –1200.4 590 105 31 624 – 15 004.5

and defining the boom positions in relation to the moment centre as in Fig. 21.5 the moments corresponding to the boom loads are calculated in Table S.23.8. In Table S.23.8 anticlockwise moments about the moment centre are positive, clockwise negative. Also

,1

0n

x rr

P=

=∑

,1

688.0 Nn

y rr

P=

=∑

,1

166 168 N mmn

y r rr

P ξ=

= −∑

,1

109 351N mmn

x r rr

P η=

=∑

The shear load resisted by the shear stresses in the webs and panels is then

12000 688 11312 NyS = − =

‘Cut’ the walls 12, 23 and 34° in the larger cross-section. Then, from Eq. (20.6) and noting that Ixy

= 0

b1

ny

r rxx r

Sq B y

I =

= − ∑

i.e.

4b 6

1 1

11 312 1.188 1095.2 10

n n

r r r rr r

q B y B y−

= =

= − = − ×× ∑ ∑

Thus b,12 b,23 b,34 b,45 b,56 0q q q q q°= = = = =

4b,61 1.188 10 600 ( 105) 7.48 N / mmq −= − × × × − =

4b,52 1.188 10 800 ( 160) 15.21N / mmq −= − × × × − =

i4

b,43 1.188 10 800 ( 160) 15.21N / mmq −= − × × × − =



i i i i,0,I 34 ,0,II34 34 b,43 34I

d 1 [ ( ) ]d 2 s sq q q

z A Gθ δ δ δ δ°= − + − + (ii)

For Cell II

i i

i i

,0,I ,0,II 23 45 52 ,0,III 5234 34II

b,52 52 b,43 43

d 1 [ ( )d 2

]

s s sq q qz A G

q q

θ δ δ δ δ δ δ

δ δ

= − + + + + −

+ − (iii)

For Cell III

,0,II 52 ,0,III 12 25 56 61 b,61 61 b,52 52III

d 1 [ ( ) ]d 2 s sq q q q

z A Gθ δ δ δ δ δ δ δ= − + + + + + − (iv)

in which

12 56 23 45600 / 1.0 600 800 / 1.0 800δ δ δ δ= = = = = = i34 52341200/0.6 2000 320/2.0 160 320/2.0 160δ δ δ° = = = = = =

61 210/1.5 140δ = = Substituting these values in Eqs (ii)–(iv)

,0,I ,0,IId 1 (2160 160 2433.6)d 2 100000 s sq qz Gθ= − +

× (v)

,0,I ,0,II ,0,IIId 1 ( 160 1920 160 )d 2 260000 s s sq q q

z Gθ= − + −

× (vi)

,0,II ,0,IIId 1 ( 160 1500 1384.8)d 2 180000 s sq q

z Gθ= − + −

× (vii)

Also, taking moments about the mid-point of web 52, i.e. the moment centre (see Eq. (23.13))

ib,61 I ,0,I II ,0,IIb,43

III ,0,III , ,1 1

0 210 590 320 790 2 2

2

s s

n n

s x r r y r rr r

q q A q A q

A q P Pη ξ= =

= × × − × × + +

+ + +∑ ∑ (viii)

Substituting the appropriate values in Eq. (viii) and simplifying gives

,0,I ,0,II ,0,III2.6 1.8 14.88 0s s sq q q+ + − = (ix)

Equating Eqs (v) and (vi)

,0,I ,0,II ,0,III0.404 0.028 1.095 0s s sq q q− + + = (x)


Equating Eqs (v) and (vii)

,0,I ,0,II ,0,III0.033 0.386 1.483 0s s sq q q− − + = (xi)

Now subtracting Eq. (x) from (ix)

,0,II ,0,III0.590 5.318 0s sq q+ − = (xii)

and subtracting Eq. (xi) from (ix)

,0,II ,0,III0.830 6.215 0s sq q+ − = (xiii)

Finally, subtracting Eq. (xiii) from (xii) gives

,0,III 3.74 N / mmsq =

Then, from Eq. (xiii)

,0,II 3.11N / mmsq =

and from Eq. (ix)

,0,I 0.06 N / mmsq =

The complete shear flow distribution is then 12 56 32 453.74 N / mm 3.11N / mmq q q q= = = = i34 430.06 N / mm 12.16 N / mmq q° = =

52 6114.58 N / mm 11.22 N / mmq q= =

S.23.9

Consider first the flange loads and shear flows produced by the shear load acting through the shear centre of the wing box. Referring to Fig. S.23.9(a), in bay ① the shear load is resisted by the shear flows q1

in the spar webs. Then

12000 5 N / mm

2 200q = =

×

Similarly in bay ②

22000 5 N / mm

2 200q = =

×

From symmetry the bending moment produced by the shear load will produce equal but opposite loads in the top and bottom flanges. These flange loads will increase with bending moment, i.e. linearly, from zero at the free end to

2000 1000 5000 N2 200

×± = ±

×


Fig. S.23.9(a)

at the built-in end. Then, at the built-in end

1 4 2 3 5000 NP P P P= = − = − =

Alternatively, the flange loads may be determined by considering the equilibrium of a single flange subjected to the flange load and the shear flows in the adjacent spar webs.

Now consider the action of the applied torque in Fig. S.23.9(b). In bay ①the torque is resisted by differential bending of the spar webs. Thus

31 200 400 1000 10q × × = ×

which gives

1 12.5 N / mmq =

The differential bending of the spar webs in bay ① induces flange loads as shown in Fig. S.23.9(c). For equilibrium of flange 1

1 12 500 500 12.5P q= = ×

so that

1 3125 NP =

Now considering the equilibrium of flange 1 in bay ②

1 2 3500 500 0P q q+ × − × =

whence

2 3 6.25q q− = − (i)


Fig. S.23.9(b)

Fig. S.23.9(c)

Also, the resultant of the shear flows in the spar webs and skin panels in bay ② is equivalent to the applied torque. Thus

31 12 32 22 2 200 200 2 2 400 100 1000 10q q× × × × + × × × × = ×

i.e.

2 3 12.5q q= = (ii)

Adding Eqs (i) and (ii) gives

2 3.125 N / mmq =

whence

3 9.375 N / mmq =


The shear flows due to the combined action of the shear and torsional loads are then as follows:

Bay ①

Spar webs: q = 12.5 – 5 = 7.5 N/mm

Bay ②

Spar webs: q = 5 h – 3.125 = 1.875 N/mm

Skin panels: q = 9.375 N/mm The flange loads are:

Bay ①

At the built-in end: P1

At the central rib: P1 = 2500 + 3125 = 5625 N (tension)

= 5000 –3125 = 1875 N (tension)

Bay ②

At the central rib: P1

At the free end: P

= 3625 N (tension)

1

Finally the shear flows on the central rib are: = 0

On the horizontal edges: q = 9.375 N/mm

On the vertical edges: q = 7.5 + 1.875 = 9.375 N/mm


S.24.1

From the overall equilibrium of the beam in Fig. S.24.1(a)

F D4kN 2kNR R= =

The shear load in the panel ABEF is therefore 4 kN and the shear flow q is given by

3

14 10 4 N / mm1000

q ×= =

Similarly

3

22 10 2 N / mm1000

q ×= =

Considering the vertical equilibrium of the length h of the stiffener BE in Fig. S.24.1(b)

3EB 1 2( ) 6 10P q q h+ + = ×


Fig. S.24.1(a)

Fig. S.24.1(b)

where PEB

is the tensile load in the stiffener at the height h, i.e. 3

EB 6 10 6P h= × − (i)

Then from Eq. (i), when h = 0, PEB = 6000 N and when h= 1000 mm. PEB = 0. Therefore the stiffener load varies linearly from zero at B to 6000 N at E.

Fig. S.24.1(c)


Consider now the length z of the beam in Fig. S.24.1(c). Taking moments about the bottom flange at the section z

AB F1000 0P R z× + =

whence

AB 4 NP z= −

Thus PAB varies linearly from zero at A to 4000 N (compression) at B. Similarly PCB

S.24.2

varies linearly from zero at C to 4000 N (compression) at B.

Referring to Fig. P.24.2 and considering the vertical equilibrium of the stiffener CDF

1 28000sin 30 200 200 0q q° − × − × =

from which

1 2 20q q+ = (i)

Now considering the horizontal equilibrium of the stiffener ED

1 28000cos30 300 300 0q q° − × + × =

whence 1 2 23.1q q− = (ii)

Adding Eqs (i) and (ii) 12 43.1q =

i.e. 1 21.6 N / mmq =

so that, from Eq. (i) 2 1.6 N / mmq = −

The vertical shear load at any section in the panel ABEGH is 8000 sin 30° =4000N. Hence 3400 4000q =

i.e. 3 10 N / mmq =

Now consider the equilibrium of the flange ABC in Fig. S.24.2(a). At any section z between C and B CB 21.6P z= (iii)


so that PCB

varies linearly from zero at C to 6480 N (tension) at B. Also at any section z between B and A

BA 21.6 300 10( 300)P z= × + −

Fig. S.24.2(a)

i.e.

BA 3480 10P z= + (iv)

Thus PBAReferring to Fig. S.24.2(b) for the bottom flange HGF, the flange load P

varies linearly from 6480 N (tension) at B to 9480 N (tension) at A. FG

at any section z is given by

FG 1.6P z= (v)

Fig. S.24.2(b)

Thus PFG

varies linearly from zero at F to 480 N (tension) at G. Also at any section z between G and H

GH 10( 300) 1.6 300 0P z+ − − × =

i.e.

GH 3480 10P z= − (vi)

Hence PGH

The forces acting on the stiffener DE are shown in Fig. S.24.2(c). At any section a distance z from D

varies linearly from 480 N (tension) at G to –2520 N (compression) at H.

DE 21.6 1.6 8000cos30 0P z z+ + − ° =

i.e.

DE 23.2 6928.2P z= − + (vii)


Fig. S.24.2(c)

Therefore PDE varies linearly from 6928 N (tension) at D to zero at E. (The small value of PDE at E given by Eq. (vii) is due to rounding off errors in the values of the shear flows.)

Fig. S.24.2(d)

The forces in the stiffener CDF are shown in Fig. S.24.2(d). At any section in CD a distance h from C the stiffener load, PCD

, is given by

CD 21.6P h= (viii)

so that PCD

varies linearly from zero at C to 4320 N (tension) at D. In DF

DF 8000sin 30 1.6( 200) 21.6 200 0P h+ ° + − − × =

from which

DF 640 1.6P h= − (ix)

Hence PDFThe stiffener BEG is shown in Fig. S.24.2(e). In BE at any section a distance h from

varies linearly from 320 N (tension) at D to zero at F.

B BE 21.6 10 0P h h+ − = i.e. BE 11.6P h= − (x)


Fig. S.24.2(e)

PBE

therefore varies linearly from zero at B to –2320 N (compression) at E. In EG

EG 1.6( 200) 21.6 200 10 0P h h− − + × − =

i.e.

EG 11.6 4640P h= − (xi)

Thus PEG

S.24.3

varies linearly from –2320N (compression) at E to zero at G.

A three flange wing section is statically determinate (see Section 23.1) so that the shear flows applied to the wing rib may be found by considering the equilibrium of the wing rib. From Fig. S.24.3(a) and resolving forces horizontally

12 34600 600 1200 0q q− − =

whence

12 34 20q q− = (i)

Now resolving vertically and noting that q51 = q

45

45 23400 400 8000 0q q− + =

i.e.

45 23 20q q− = − (ii)

Taking moments about 4 2

12 23200 1600 400 2 400 600 12000 200 8000 600 02 2

q qπ ×× × + + × × − × − × =


Fig. S.24.3(a)

so that

12 231.52 30q q+ = (iii)

Subtracting Eq. (iii) from (i) and noting that q34 = q

23

232.52 10q− = −

or

23 344.0 N / mmq q= =

Then from Eq. (i)

12 24.0 N / mmq =

and from Eq. (ii)

45 5116.0 N / mmq q= − =

Fig. S.24.3(b)

Solutions to Chapter 24 Problems 319 Consider the nose portion of the wing rib in Fig. S.24.3(b). Taking moments about 3

2

2200400 2 4.0 02

P π ×× − × × =

from which

2 1256.6 NP = (tension)

From horizontal equilibrium

3 2 0P P+ =

whence

3 1256.6 NP = − (compression)

and from vertical equilibrium

1 4.0 N / mmq =

From the vertical equilibrium of the stiffener 154 in Fig. S.24.3(c)

2 3200 200 16 400 0q q× + × − × =

Fig. S.24.3(c)

i.e.

2 3 32q q+ = (iv)

Also, in 15 at any distance h from 1

15 216 0P h q h+ − =

i.e.

15 2( 16)P q h= − (v)

and in 54

54 2 316 200 ( 200) 0P h q q h+ − × − − =


whence

54 2 3 3200( ) ( 16)P q q q h= − + − (vi)

Fig. S.24.3(d)

Fig. S.24.3(d) shows the stiffener 56. From horizontal equilibrium

2 3600 600 12000 0q q− − =

or

2 3 20q q− = (vii)

Adding Eqs (iv) and (vii)

22 52q =

i.e.

2 26 N / mmq =

and from Eq. (iv)

3 6 N / mmq =

Then, from Eq. (v)

15 10P h= (viii)

and P15

varies linearly from zero at 1 to 2000 N (tension) at 5. From Eq. (vi)

54 200(26 6) (6 16)P h= − + −

i.e.

54 4000 10P h= − (ix)

so that P54 varies linearly from 2000 N (tension) at 5 to zero at 4. Now from Fig. S.24.3(d) at any section z 56 2 3 12000 0P q z q z+ − − =

i.e. 56 20 12000P z= − + (x)

Thus P56 varies linearly from 12 000 N (tension) at 5 to zero at 6.


Fig. S.24.3(e)

Consider the flange 12 in Fig. S.24.3(e). At any section a distance z from 1

12 24 26 0P z z+ − =

i.e.

12 2P z= (xi)

Hence P12Now consider the bottom flange in Fig. S.24.3(f). At any section a distance z from 4

varies linearly from zero at 1 to 1200 N (tension) at 2.

43 6 4 0P z z+ − =

Fig. S.24.3(f)

i.e.

43 2P z= − (xii)

Thus P43 varies linearly from zero at 4 to –1200 N (compression) at 3. (The discrepancy between P2 in 12 and P2 in 23 and between P3 in 43 and P3 in 23 is due to the rounding off error in the shear flow q1

In Fig. S.24.3(g) the load in the stiffener at any section a distance h from 2 is given by .)

26 26 4 0P h h+ + =

i.e.

26 30P h= − (xiii)

Therefore P26

varies linearly from zero at 2 to –6000 N (compression) at 6. In 63

63 26 200 4 6( 200) 0P h h+ × + + − =


Fig. S.24.3(g)

i.e.

hP 10400063 −−= (xiv)

Thus P63


varies linearly from –6000 N (compression) at 6 to –8000 N (compres-sion) at 3.

S.25.1

From Eq. (25.5) the modulus of the bar is given by

1100 10 100 45140000 3000100 55 100 55

E × ×= × + ×

× ×

i.e. 2

1 N/mm1.27909=E

The overall direct stress in the longitudinal direction is given by

23

1 N/mm9.905510010500

=××

=σ

Therefore, from Eq. (25.2), the longitudinal strain in the bar is

31

90.9 3.26 1027909.1

ε −= = ×

The shortening, Δ1

mm26.31011026.3 311 =×××=∆ −

, of the bar is then


The major Poisson’s ratio for the bar is obtained using Eq. (25.7). Thus

lt100 45 100 100.16 0.28 0.18100 55 100 55

v × ×= × + × =

× ×

Hence the strain across the thickness of the bar is 3 4

t 0.18 3.26 10 5.87 10− −ε = × × = ×

So that the increase in thickness of the bar is

551087.5 4t ××=∆ −

i.e.

mm032.0t =∆

The stresses in the polyester and Kevlar are found from Eqs (25.3). Hence

23m N/mm78.91026.33000)polyester( =××= −σ

3 2f (Kevlar) 140000 3.26 10 456.4 N/mmσ −= × × =

S.25.2

The Reduces Stiffnesses Are

k11 = k22 = E/(1 – ν2) = 60 000/(1 – 0.252) = 64 000 N/mm

k

2

12 = νE/(1 – ν2) = 0.25 × 60 000/(1 – 0.252) = 16 000 N/mm

k

2

33 = G = E/2(1 + ν) = 60 000/2(1 + 0.25) = 24 000 N/mmThe reduced compliances are

2

s11 = s22 = 1/E = 1/60 000 = 1.67 × 10

s

–5

12 = s21 = –ν/E = 0.25/60 000 = –0.42 × 10

s

–5

33 = 1/G = 1/24 000 = 4.17 × 10

S.25.3

–5

From the development of Eqs(25.30) k11 = E l/(1 – ν ltν tl

k

) (i)

12 = ν tlE l/(1 – ν ltν tl

Dividing Eq.(ii) by Eq.(i) ) (ii)

ν tl = k12/k11

Also k

= 4000/50 000 = 0.08

21 = ν ltE t/(1 – ν ltν tl) (iii)


k22 = E t/(1 – ν ltν tl

Dividing Eq.(iii) by Eq.(iv) ) (iv)

ν lt = k21/k22

Then, from Eq.(i) = 4000/15 000 = 0.27

E l = 50 000(1 – 0.27 × 0.08) = 48 920 N/mmand from Eq.(iii)

2

E t = 4000(1 – 0.27 × 0.08)/0.27 = 14 495 N/mm

Also G = k

2

33 = 6000 N/mmThe reduced compliances are then

2

s11 = 1/E l = 1/48 920 = 2.04 × 10

s

–5

12 = –ν tl/E t = –0.08/14 495 = –0.55 × 10

s

–5

21 = –ν lt/E l = –0.27/48 920 = –0.55 × 10

s

–5

22 = 1/E t = 6.90 × 10

s

–5

33 = 1/G = 1/6000 = 16.7 × 10

S.25.4

–5

Since the ply angle is 45° the trigonometric terms in Eqs(25.24) are

m2 = 0.5, n2

Then, substituting in Eqs(25.24) = 0.5, mn = 0.5

σ l

{σ

0.5 0.5 1 120

t

τ

} = [ 0.5 0.5 –1] {60}

lt

(Note to editor: please put in proper matrix signs as in Eqs(25.24) –0.5 0.5 0 80

so that

σ l = 0.5 × 120 + 0.5 × 60 + 1 × 80 = 170 N/mm

σ

2

t = 0.5 × 120 + 0.5 × 60 – 1 × 80 = 10 N/mm

τ

2

lt = –0.5 × 120 + 0.5 × 60 + 0 = –30 N/mm

S.25.5

2

As in S.25.4, m2 = 0.5, n2

ε

= 0.5 and mn = 0.5. Eq.(25.25) then becomes

l 0.5 0.5 0.5 0.005


{ε t

γ

} = [ 0.5 0.5 –0.5] {0.002}

lt

(Note: please insert proper matrix signs as in Eqs(25.25)) –1 1 0 0.0002

Then ε l = 0.5 × 0.005 + 0.5 × 0.002 + 0.5 × 0.0002 = 3.6 × 10 ε

–3 t = 0.5 × 0.005 + 0.5 × 0.002 – 0.5 × 0.0002 = 3.4 × 10

γ

–3

lt = –1 × 0.005 + 1 × 0.002 + 0 = –3.0 × 10

S.25.6

–3

Since the ply angle is 45° the trigonometric terms in Eqs(25.24) are

m2 = 0.5, n2

and Eqs(25.24) become = 0.5, mn = 0.5

σ l

{σ

0.5 0.5 1 100

t

τ

} = [ 0.5 0.5 –1] {50}

lt

(Note: please insert proper matrix signs) –0.5 0.5 0 75

so that σ l = 0.5 × 100 + 0.5 × 50 + 1 × 75 = 150 N/mm σ

2 t

τ

= 0.5 × 100 + 0.5 × 50 – 1 × 75 = 0

lt = –0.5 × 100 +0.5 × 50 + 0 = –25 N/mmIn Eqs(25.27)

2

E l/(1 – ν ltν tl) = 150 000/(1 – 0.3 × 0.1) = 154 639 N/mm

ν

2

tlE l/(1 – ν ltν tl) = 0.1 × 154 639 = 15 464 N/mm

ν

2

ltE t(1 – ν ltν tl) = 0.3 × 90 000/(1 – 0.3 × 0.1) = 27 835 N/mm

E

2

t/(1 – ν ltν tl) = 90 000/(1 – 0.3 × 0.1) = 92 784 N/mmEqs(25.27) then become

2

150 = 154 639ε l + 15 464ε t

0 = 27 835ε

(i)

l + 92 784ε t

–25 = G

(ii)

ltγ lt = 5000γlt

Solving Eqs(i), (ii) and (iii) gives (iii)

ε l = 10.0 × 10–4, ε t = –3.03 × 10–4, γlt = –50.0 × 10–4


For each cover

S.25.7

bitiEZ, j = 150 × 1.0 × 20000 = 3 × 10For each web

6

bitiEZ, j = 100 × 2.0 × 60 000 = 12 × 10Then

6

6 6 6,

1

2 3 10 2 12 10 30 10n

i i Z ii

b t E=

= × × + × × = ×∑

From Eq. (25.37)

36

31033.1

10301040 −×

××

=Zε

Therefore

P(covers) = 1.33 × 10–3 × 3 × 106

P(webs) = 1.33 × 10

= 4000 N = 4 kN –3 × 12 × 106

Check: 2 × 4 + 2 × 16 = 40 kN

= 16000 N = 16 kN

S.25.8

Since 0xyI ′ = and My

YIM

Exx

xizz ′

= ,σ


where 3

22.0 1002 60000 2 20000 1.0 150 5012xxI ×′ = × × + × × × ×

i.e. 210 mmN105.3 ×=′xxI

Then

YEYE izizZ ,5

10

6

, 1086.2105.3

101 −×=××

×=σ (i)

The direct stress will be a maximum when Y is a maximum, i.e. at the top and bottom of the webs and in the covers. But EZ,i

σ

for the webs is greater than that for the covers, therefore

Z(max) = ±2.86 × 10–5 × 60 000 × 50

328 Solutions Manual i.e.

σz(max) = ±85.8 N/mm2

S.25.9

(at the top and bottom of the webs)

The second moments of area are, from Example 25.5 210 mmN1063.2 ×=′xxI

210 mmN1083.0 ×=′yyI

210 mmN1050.2 ×=′xyI

Also ΜX = 0 and ΜY

σ

= 0.5 kN m so that Eq. (25.39) becomes

Z = EZ,i (–3.23 × 10–5 X + 3.07 × 10–5

On the top flange, E

Y) (i)

Z,i = 50 000 N/mm2

σ

and Y = 50 mm. Then, from Eq. (i)

Z

so that at 1 where X = 50 mm

= –1.62 X + 76.75

σZ,1 = –4.3 N/mm

and at 2, X = 0

2

σZ,2 = 76.8 N/mm

In the web, E

2

Z,i = 15000 N/mm2

σ

, X =0 so that

Z

and at 2 = 0.46Y

σZ,2 = 0.46 × 50 = 23.0 N/mmThe maximum direct stress is therefore 76.8 N/mm

2

S.25.10

2

From Example 25.5 the second moments of area are 210 mmN1063.2 ×=′xxI

210 mmN1083.0 ×=′yyI

210 mmN1050.2 ×=′xyI


Fig. S.25.10

In this case SX = 0, SY

= 2 kN so that Eq. (25.40) becomes

7 7, 0 0

1.15 10 d 0.382 10 ds s

s Z i i iq E t X s t Y s− − = − × − × ∫ ∫ (i)

On the top flange, X = 50 – s1, Y = 50 mm, EZ,i = 50 000 N/mm2

10

3110

312 d10190d)50(105.11

11sssq

ss

∫∫ −− ×+−×−=

. Eq. (i) then becomes

which gives 2

12 1 10.00575 0.385q s s= −

when s1 =50 mm q2In the web, X = 0, Y = 50 – s

= –4.875 N/mm 2, EZ,i = 15 000 N/mm2

875.4d)50(1073.5 220

423

2−−×= ∫− ssq

s

. Eq. (i) then becomes

so that

875.4865.20287.0 22223 −−= ssq


S.25.11

Referring to Fig. P.25.11, if the origin for s is chosen on the vertical axis of symmetry qs,0

Also since S, at 0, is zero.

X 0xyI ′ = = 0 and , Eq. (25.41) reduces to

, 0d

sYs Z i

XX

Sq E tY sI

= −′ ∫

in which

××+××=′

12505.0177002)2520054100(2

32

XXI

i.e. 29 mmN107.13 ×=′XXI

Then 1

3

01 19 0

20 1054100 1.0 25 d13.7 10

sq s×

= − × ×× ∫

i.e. q01 = –1.98s

so that 1

q1

Also = –1.98 × 100 = –198 N/mm

198d)25(5.0107.13

102017700 2209

3

122

−−××

×−= ∫ ssqs

which gives

198325.0105.6 222

312 −−×= − ssq

The remaining distribution follows from symmetry

S.25.12

The shear flow is obtained from Eq. (25.42), i.e.

N/mm50502002

101 6=

×××

=q

The maximum shear stress will occur in the webs and is

2max N/mm100

5.050

==τ

From Eq. (25.45)

GJ = 4 × (50 × 200)2/[2 × 200/ (20 700 × 1.0) + 2 × 50/ (36 400 × 0. 5)]


i.e. GJ = 1.6 × 10 N mm

Then

rad/mm1025.6106.1

101dd 5

10

6−×=

××

==GJT

zθ

Finally, from Eq. (25.47)

( )162

4100 251 10 100 2 200 2 50

2 200 50 20700 1.0 50 200 20700 1.0 36400 0.5W

× × × × ×= − + × × × × × ×

i.e.

W4

S.25.13

= –0.086 mm

From Eq. (25.48)

2533

mmN103.635.010020900

3150163002 ×=××+×××=GJ


rad/mm108.0103.6105.0

dd 3

5

3−×=

××

=zθ

From Eq. (25.50)

3max

1.0(flanges) 2 16300 0.8 102

τ −= ± × × × ×

i.e. 2

max

3max

(flanges) 13.0 N/mm0.5(web) 2 20900 0.8 102

τ

τ −

= ±

= ± × × × ×

i.e.

τmax (web) = ±8.4 N/mmTherefore

2

τmax = ±13.0 N/mmThe warping at 1 is, from Eq. (18.19)

2

mm0.2108.05050212 3

1 −=×××××−= −W


S.26.1

In Fig. S.26.1 α = tan–1

q

127/305 = 22.6°. Choose O as the origin of axes then, from Eq. (26.1), since all the walls of the section are straight, the shear flow in each wall is constant. Then

12

q

= 1.625G (254θ'– u') (i)

23 = 1.625G (254θ' cos 22. 6°– u' cos 22.6°– υ' sin 22.6°)

Fig. S.26.1

i.e. q23

q

= 1.625G(234. 5θ' – 0.923u' – 0.384υ') (ii)

34

q

= 2.03G(305θ' – υ') (iii)

52

q

= 2.54Gυ' (iv)

45

q

= 1.625G(234.5θ' + 0.923u'— 0.384υ') (v)

56

From symmetry q

= 1.625G(254θ' + u') (vi)

12 = q56 and q23 = q45

q

so that, from Eqs (i) and (vi) (or Eqs (ii) and (v)) u' = 0. Now resolving forces vertically

52 × 508 –q23 × 127 – q34 × 254 – q45

i.e. × 127 = 111 250

508q52 – 2 × 127q23 – 254q34 =111 250


Substituting for q52, q23 and q34

from Eqs (iv), (ii) and (iii), respectively gives

G63.563.129 =′−′ θυ (vii)

Now taking moments about O 2q12 × 254 × 254 + 2q23 × 305 × 254 + q34

Substituting for q × 254 × 305 = 0

12, q23 and q34

υ' – 631.1θ' = 0 (viii) from Eqs (i), (ii) and (iii), respectively gives

Subtracting Eq. (viii) from (vii) gives

G113.0

=′θ (ix)

Hence, from Eq. (viii)

G

2.71=′υ (x)

Now substituting for θ' and υ' from Eqs (ix) and (x) in Eqs (i)–(vi) gives

q12 = q56 = 46.6 N/mm q32 = q54

q

= 1.4 N/mm

43 = 74.6 N/mm q52

Finally, from Eq. (17.11) = 180.8 N/mm

R R71.2 630.1 mm 00.113

v ux y′ ′

= − = − = − = =′ ′θ θ

S.26.2

In Fig. S.26.2, α = tan–1

q

125/300 = 22.6°. Also, since the walls of the beam section are straight the shear flow in each wall, from Eq. (26.1), is constant. Choosing O, the mid-point of the wall 42, as the origin, then, from Eq. (26.1) and referring to Fig. S.26.2.

51

q

= 1.6G(250θ' + υ') (i)

12

q

= 1.2G(125θ'–u') (ii)

23 = 1.0G (125θ' cos 22.6°– u' cos 22.6°– υ' sin 22.6°)

Fig. S.26.2


i.e.

q23

q

= 1.0G(115.4θ' – 0.923u' – 0.384υ') (iii)

34

q

= 1.0G(115.4θ' + 0. 923u' – 0.384υ') (iv)

45

q

= 1.2G (125θ' + u') (v)

24

From antisymmetry q

= 1.6G(–υ') (vi)

12 = q45 and q23 = q34

q

. Thus, from Eqs (ii) and (v) (or Eqs (iii) and (iv)), u' = 0. Resolving forces vertically

51 × 250 –q24 × 250 – q23 × 125 – q34

i.e. × 125 = 0

q51 – q24 – q23

Substituting in Eq. (vii) for q = 0 (vii)

51, q24 and q23

υ' + 79.41θ' = 0 (viii)

from Eqs (i), (vi) and (iii), respectively gives

Now taking moments about O

2q12 × 250 × 125 + 2q23 × 300 × 125 + q51 × 250 × 250 =11 000 × 10i.e.

3

q12 + 1.2q23 + q51

Substituting in Eq. (ix) for q = 176 (ix)

12, q23 and q51

from Eqs (ii), (iii) and (i), respectively gives

154.5604.4G

′ ′υ + θ = (x)

Subtracting Eq. (x) from (viii) gives

G294.0

=′θ (xi)

whence, from Eq. (viii)

23.37G

′υ = − (xii)

Substituting for θ' and υ' from Eqs (xi) and (xii) in Eqs (i)–(vi) gives

q51 = 80 N/mm q12 = q45

q

= 44.1 N/mm

23 = q34 = 42.9 N/mm q24

The centre of twist referred to O has coordinates, from Eq. (17.11) = 37.4 N/mm

R R23.37 79.5mm 00.294

v ux y′ ′

= − = − = = =′ ′θ θ


S.26.3

Referring to Fig. S.26.3 the shear flows in the walls 12 and 23 are constant since the walls are straight (see Eq. (26.1)). Choosing O as the origin of axes, from Eq. (26.1)

q12 = Gt(θ' R cos 30° + u' cos 30° + υ' sin 30°)

Fig. S.26.3

i.e.

q12

q

= Gt(0.866Rθ' + 0.866u' + 0.5υ') (i)

23

q

= Gt(0.866Rθ' –0.866u' + 0.5υ') (ii)

31

Resolving forces vertically = Gt(Rθ' – u' cos φ – υ' sin φ) (iii)

°=−+ ∫ 30sin10000dsin3102312 φφπ

RqRqRq

i.e.

12 23 310

5000sin dq q qR

πφ φ+ − =∫ (iv)

Substituting in Eq. (iv) for q12, q23 and q31

from Eqs (i)–(iii), respectively gives

GtRvR 7.1865659.9 −=′−′θ (v)

Resolving forces horizontally

12 23 310( / tan 30 ) ( / tan 30 ) cos d 10000cos30q R q R q R

πφ φ° − ° − = °∫

i.e.

R

qqq 3.8660dcos732.1732.1 3102312 =−− ∫ φφπ

(vi)


Substituting in Eq. (vi) for q12, q23 and q31


GtRu 7.1894=′ (vii)


212 23 310

( / tan 30 ) ( / tan 30 ) d 10000 cos30q R R q R R q R Rπ

φ° + ° + = °∫

i.e.

R

qqq 3.8660d732.1732.1 3102312 =++ ∫ φπ

(viii)

Substituting in Eq. (viii) for q12, q23 and q31


1410.00.044RGtR

θ ′ ′− υ = (ix)

Now subtracting Eq. (ix) from (v)

20 066.79.546GtR

′− υ = −

whence

2102.1GtR

′υ = (x)

Then, from Eq. (v)

GtR

R 4.1502=′θ (xi)

Substituting for u', υ' and θ' from Eqs (vii), (x) and (xi), respectively in Eqs (i)–(iii) gives

q12 = 3992.9/R N/mm, q23

q

= 711.3/R N/mm

31

S.26.4

= (1502.4 – 1894.7 cos φ – 2102.1 sin φ) /R N/mm

From Fig. P.26.4 the torque at any section of the beam is given by

T = 20 × 103

Eq. (26.16) for the warping distribution along boom 4 then becomes (2500 – z) N mm (i)

320 10 (200 )cosh sinh

8 b a

z b aw C µz D µzabG t t

× −= + + −

(ii)


where

)(82

ba

ab

atbtBEtGtµ+

=

Comparing Figs 26.6 and P.26.4, tb = ta

)0.15000.1200(8000.10.136.082

×+××××

=µ

= 1.0 mm, a = 500 mm, b = 200 mm and B = 800 mm. Then

from which

μ = 2.27 × 10Eq. (ii) then becomes

–3

w = C cosh 2.27 × 10–3 z + D sinh 2.27 × 10–3 z –3.75 × 10–4

When z = 0, w = 0, hence, from Eq. (iii), C = 0.9375. At the free end the direct stress in boom 4 is zero so that the direct strain ∂w/∂z = 0 at the free end. Hence, from Eq. (iii), D = –0.9386 and the warping distribution along boom 4 is given by

(2500 –z) (iii)

w = 0.9375 cosh 2.27 × 10–3 z –0.9386 sinh 2.27 × 10–3

– 3.75 × 10 z

–3

Substituting for w from Eq. (iv) and T from Eq. (i) in Eq. (26.11) (2500 – z) (iv)

5 3 3

3

d 10 [1.6069 cosh 2.27 10 1.6088 sinh 2.27 10d

3.4998 10 (2500 )]

z zz

z

θ − − −

−

= − × − ×

− × −

(v)

Then

5 3 3

3 3

23

1.6069 1.608810 sinh 2.27 10 cosh 2.27 102.27 10 2.27 10

3.4998 10 25002

z z

zz F

θ − − −− −

−

= − × − × × ×

− × − +

(vi)

When z = 0, θ = 0 so that, from Eq. (vi)

35

1027.26088.110 −

−

××−=F

and

5 3 3

23

10 707.9sinh 2.27 10 708.7cosh 2.27 10

3.4998 10 2500 708.72

z z

zz

− − −

−

θ = − × − ×

− × − +

(vii)

At the free and where z = 2500 mm Eq. (vii) gives

θ = 0.1036 rad = 5.9°


S.26.5

The warping distribution along the top right-hand corner boom is given by Eq. (26.16), i.e. w = C cosh μz + D sinh μz + W0

where (i)

2 2 10

1 2 2 1

8 and( ) 8

Gt t T b aµ wBE bt at abG t t

= = − +

At each end of the beam the warping is completely suppressed, i.e. w = 0 at z = 0 and z = l. Thus, from Eq. (i)

0 = C + wi.e.

0

0 = C – wand

0

0 = C coshµl + Dsinhµl + wwhich gives

0

)1(coshsinh

0 −= µlµl

wD

Hence, Eq. (i) becomes

−+−= µz

µlµlµzww sinh

sinh)1(coshcosh10 (ii)

The direct load, P, in the boom is then given by

zwBEBP z ∂∂

==σ

Thus, from Eq. (ii)

−+−= µz

µlµlµzµBEwP cosh

sinh)1(coshsinh0 (iii)

or, substituting for w0

from above

−+−−= µz

µlµlµzatbt

tabGtµBETP cosh

sinh)1(coshsinh)(

8 2121

(iv)

For a positive torque, i.e. T is anticlockwise when viewed along the z axis to the origin of z, the term in square brackets in Eq. (iv) becomes, when z = 0

µlµl

sinh1cosh −


which is positive. Thus at z = 0 the load in the boom is tensile. At z = l the term in square brackets in Eq. (iv) becomes

µlµl

sinhcosh1−

which is negative. Thus at z = l the load in the boom is compressive. Also, from Eq. (iv) ∂P/∂z = 0 at z = l/2 and the distribution of boom load is that shown in Fig. S.26.5. The reverse situation occurs for a negative, i.e. a clockwise, torque.

Fig. S.26.5

S.26.6

The warping distribution is given by Eq. (26.16), i.e.

−++=

ab ta

tb

abGTµzDµzCw

8sinhcosh (i)

in which the last term is the free warping, w0

w = C cosh μz + D sinh μz + w

, of the section. Eq. (i) may therefore be written

0

When z = 0, w = kw

(ii)

0

C = w

so that, from Eq. (ii)

0

When z = L, the direct stress is zero. Then, from Chapter 1

(k – 1)

0=∂∂

=zwEσ

so that

0 = μ C cosh μL. + μ D sinh μL which gives

D = –w0(k – 1) tanh μL



−−+=

µLzLµkww

cosh)(cosh)1(10 (iii)

Then

zwE∂∂

=σ

so that

µLzLµkµEw

cosh)(sinh)1(0

−−−=σ

When k = 0,

µLzLµµEw

cosh)(sinh

0−

−=σ

i.e. a rigid foundation. When k = 1,

σ = 0 i.e. free warping (also from Eq. (iii)).

S.26.7

Initially directions for the shear flows, q, are chosen as shown in Fig. S.26.7(a). The panel is symmetrical about its horizontal centre line so that only half need be considered.

Fig. S.26.7(a)


Fig. S.26.7(b)

For equilibrium of the element of the top boom shown in Fig. S.26.7(b)

0BB

B =+−−∂∂

+ zqzSPzz

PP δδδ

i.e.

qSz

P−=

∂∂ B (i)

Fig. S.26.7(c)

Also, for equilibrium of the element of the central stringer shown in Fig. 26.7(c)

02AA

A =−−∂∂

+ zqPzz

PP δδ

i.e.

qz

P 2A =∂∂ (ii)

For equilibrium of the length, z, of the panel shown in Fig. S.26.7(d)

2PB + PA – 2Sz – 2P –PS

i.e. = 0

PA = 2P + PS + 2Sz – 2PB

The compatibility of displacement condition for the top boom and central stringer is shown in Fig. S.26.7(e). Thus

(iii)

A Bd(1 ) (1 )d

z z b zzγε δ ε δ δ+ = + +


Fig. S.26.7(d)

Fig. S.26.7(e)

i.e.

A Bd 1 ( )dz bγ ε ε= − (iv)

Now

constantand BB

AA ====

EEEeσσεσε


Also σA = 0.8σe

so that Eq. (iv) becomes

bEze2.0

dd σγ

−= (v)

In Eq. (v) γ = q/Gt, hence

e2.0

dd σ

bEGt

zq

−= (vi)

Substituting for q in Eq. (vi) from (i) gives

e2B

2 2.0 σbE

GtzP

−=∂∂

so that

2eB

0.1GtP z Cz DbE

σ= + + (vii)

When Z = 0, PB = P so that, from Eq. (vii), D = P. Also, when z = l, q = 0 so that, from Eq. (i), ∂PB/∂Z

e0.2GtC l SbE

σ= − +

= S at z = l. Hence, from Eq. (vii)

and

2e eB

0.1 0.2Gt Gt lP z S z PbE bE

σ σ = + − +

(viii)

Now PB = σe

B so that, from Eq. (viii)

+

−+= Pz

bElGtSz

bEGtB e

e

2 2.011.0 σσ

(ix)

Substituting for PB

from Eq. (viii) in (iii) gives

S

2e

A 24.0 PzlzbEGtP +

−=

σ (x)

But PA = AS0.8σe

so that, from Eq. (x)

e

S2

S25.1

22 σPzlz

bEGtA +

−= (xi)

Substituting the given values in Eqs (ix) and (xi) gives

B = 3.8 × 10–4

and z + 0.3227z + 1636.4 (xii)

AS = 2.375z – 9.5 × 10–4 z2 + 659.1 (xiii)


From Eq. (xiii) when z = 1250 mm, AS

P

= 2143.5 mm. Then

A = 0. 8σeAS

The total load, P = 0. 8 × 275 × 2143.5 = 471 570 N

T

P

, carried by the panel at the built-in end is

T

Therefore, the fraction of the load carried by the stringer is 471 570/1 920 000 = 0.25.

= 2 × 450 000 + 145 000 + 2 × 350 × 1250 = 1 920 000 N

S.26.8

The panel is symmetrical about its vertical center line and therefore each half may be regarded as a panel with a built-in end as shown in Fig. S.26.8(a). Further, the panel is symmetrical about its horizontal centre line so that only the top half need be considered; the assumed directions of the shear flows are shown.

Fig. S.26.8(a)

Fig. S.26.8(b)

Consider the equilibrium of the element of longeron 1 shown in Fig. S.26.8(b).

011

1 =−−∂∂

+ zqPzPP z δδ


Hence

qzP=

∂∂ 1 (i)

Now consider the equilibrium of the element of longeron 2 shown in Fig S.26.8(c).

0222

2 =−−∂∂

+ zqPzzPP δδ

whence

qzP 22 −=∂∂ (ii)

Fig. S.26.8(c)

Fig. S.26.8(d)

From the overall equilibrium of the length z of the panel shown in Fig. S.26.8(d)

2P1 + P2

The compatibility condition for an element of the top half of the panel is shown in Fig. S.26.8(e). Thus

= 0 (iii)

1 2d(1 ) (1 )d

z z d zzγ

+ ε δ = + ε δ + δ


Fig. S.26.8(e)

i.e.

1 2d 1 ( )dz dγ ε ε= − (iv)

In Eq. (iv)

EBP

1

11 =ε

Also, an element, δz, of the central longeron would, without restraint, increase in length by an amount αT δz. The element therefore suffers an effective strain equal to

2( ) / .T z zε − α δ δ Thus

TEB

P αε −= 22

2

so that Eq. (iv) becomes

1 2

1 2

d 1d

P P TEz dE B Bγ α

= − −

(v)

Also γ = q/Gt and from Eq. (ii) q = –(∂P2/∂z)/2. Therefore, substituting for γ and then q in Eq. (v) and for P1

−−−=

∂∂

− TEbP

BP

dEGt

zP α

2

2

1

222

2

221

from Eq. (iii) in (v)


or

2

22

1 2

2 1 1 22

P GT Gt TdE B B dz

α ∂= + =

∂ (vi)

The solution of Eq. (vi) is

2 22cosh sinh Gt TP C µz D µz

µ dα

= + − (vii)

where

2

1 2

2 1 12

GtµdE B B

= +

When z = 0, P2

22Gt TC

µ dα

=

=0 so that, from Eq. (vii)

Also when z = l/2, q = 0 and, from Eq. (ii), ∂P2

0 sinh cosh2 2l lµC µ µD µ= +

/∂z = 0. Hence, from Eq. (vii)

from which

22tanh tanh

2 2µl Gt T µlD C

µ dα

= − = −

Thus,

2 22 cosh tanh sinh 1

2Gt T µlP µz µzµ dα = − −

(viii)

or, substituting for μ

2

2

1 2

cosh tanh sinh 12

1 12

µlµz µzP E T

B B

α

− − =

+

(ix)

From Fig. S.26.8(e) the relative displacement of the central longeron at one end of the panel is d(γ)z = 0

20

0 0

12z

z z

PqGt Gt z

γ == =

∂ = = − ∂

. Now

(from Eq. (ii))

Hence, from Eq. (ix)

relative displacement tanh2

T µlµα

=


S.26.9

The panel is unsymmetrical so that the shear flows in the top and bottom halves will have different values as shown in Fig. S.26.9(a).

Fig. S.26.9(a)

Fig. S.26.9(b)

For equilibrium of the element of the top member shown in Fig. 26.9(b)

11 1 1 1 0PP z P S z q z

zδ δ δ

∂+ − − − =∂

i.e.

11 1

P S qz

∂= +

∂ (i)

Similarly, for the equilibrium of the element of the central stringer shown in Fig. S.26.9(c)

33 3 2 1 0

PP z P q z q zzδ δ δ

∂+ − − − =∂

i.e.

32 1

P q qz

∂= −

∂ (ii)


Fig. S.26.9(c)

Fig. S.26.9(d)

Also, from Fig. S.26.9(d)

22 2 2 2 0PP z P S z q z

zδ δ δ

∂+ − − + =∂

whence

22 2

P S qz

∂= −

∂ (iii)

Now, from the longitudinal equilibrium of a length z of the panel (Fig. S.26.9(e))

P1 + P3 + P2 – P1,0 – P2,0 – S1z – S2z = 0

Fig. S.26.9(e)


i.e.

P1 + P3 + P2 = P1,0 + P2,0 + (S1 + S2

and from its moment equilibrium about the bottom edge member ) z (iv)

P12b + P3b – P1,0 2b – S1

i.e. z 2b = 0

2P1 + P3 = 2P1,0 + 2S1z (v)

Fig. S.26.9(f)

From the compatibility condition between elements of the top edge member and the central stringer in Fig. S.26.9(f)

21

1 3 2d(1 ) (1 )d

vz z b zz zγ

ε δ ε δ δ ∂

+ = + + + ∂

or

2

11 3 2

d 1 ( )d

vz b zγ

ε ε ∂= − −

∂ (vi)

Similarly for elements of the central stringer and the bottom edge member

2

23 2 2

d 1 ( )d

vz b zγ

ε ε ∂= − −

∂ (vii)

Subtracting Eq. (vii) from (vi)

1 21 3 2

d d 1 ( 2 )d dz z bγ γ

ε ε ε− = − + (viii)


Now γ = q/Gt, ε1 = σ1/Ε, ε3 = σ3/Ε and ε2 = σ2

1 21 3 2

d d ( 2 )d dq q Gtz z bE

σ σ σ− = − +

/Ε. Eq. (viii) may then be written

or, from Eq. (i) 2

31 3 22 ( 2 )

P GtbEz

σ σ σ∂

− = − +∂

Then, since σ3, = P3

23

3 1 22 (2 )GtbEAz

σσ σ σ

∂= − −

∂

/A

or

2

33 1 22

2 ( )Gt GtbEA bEAz

σσ σ σ

∂− = +

∂ (ix)

The solution of Eq. (ix) is

σ3 = C cosh μz + D sinh μz + (σ1 + σ2

where μ

)/2 2

When z = 0, σ = 2Gt/bEA.

3 = 0 so that C = – (σ1 + σ2

1 2( ) (cosh 1)2sinh

D µlµl

σ σ+= −

)/2. When z = l, = 0 which gives

Thus

( )1 2

31 cosh

1 cosh sinh2 sinh

µlµz µz

µlσ σ

σ −+ = − −

(x)

From Eq. (i)

11 1

Pq Sz

∂= −∂

(xi)

Substituting for P1

3 31

1 12 2

Pq Az z

∂ ∂= − = −

∂ ∂σ

from Eq. (v) in (xi)

Therefore, from Eq. (x)

( )1 21

1 coshsinh cosh

4 sinhµl

q A µ µz µzµl

σ σ −+ = +


S.26.10

The assumed directions of shear flow are shown in Fig. S.26.10(a). For the equilibrium of an element of the top boom, Fig. S.26.10(b).

11 1 1 0PP z P q z

zδ δ

∂+ − + =∂

Fig. S.26.10(a)

Fig. S.26.10(b)

from which

11

P qz

∂= −

∂ (i)

Similarly for an element of the central boom

21 2

P q qz

∂= −

∂ (ii)

For overall equilibrium of the panel, at any section z

P1 + P2 + P3 = –6P (iii)


and taking moments about boom 3

P12d + P2

so that d + 3P2d + 2Pd = 0

2P1 + P2

The compatibility condition is shown in Fig. S.26.10(c) for an element of the top panel. = –8P (iv)

Fig. S.26.10(c)

Then 2

12 1 2

d(1 ) (1 )z z d zz zγ υε δ ε δ δ

∂+ = + + + ∂ ∂

i.e.

2

12 1 2

d 1 ( )d

vz d zγ

ε ε ∂= − −

∂ (v)

Similarly for an element of the lower panel

2

23 2 2

1 ( ) vz d zγ

ε ε∂ ∂

= − −∂ ∂

(vi)

Subtracting Eq. (vi) from (v)

1 22 1 3

d d 1 (2 )d dz z dγ γ

ε ε ε− = − − (vii)

But 31 2 2 1

1 2 2 1 3Pq q P P

Gt Gt BE AE AEγ γ ε ε ε= = = = =


Then, from Eq. (vii)

31 2 2 1d d 2d d

Pq q P PGtz z dE B A A

− = − −

(viii)

Substituting in Eq. (viii) for q1 – q2 from Eq. (ii) and for P1 and P3

from Eqs (iv) and (iii) and rearranging

222

226P GtPµ PdEAz

∂− =

∂ (ix)

where ( )2 2Gt A B

µdEAB

+=

The solution of Eq. (ix) is

26cosh sinh

2PBP C µz D µz

A B= + −

+

When z = 0, P2 = –2P and when z = L, q1 = q2 = 0 so that, from Eq. (ii) ∂P2These give

/∂z = 0.

4 4 tanh2 2B A B AC P D P µLA B A B− − = = − + +

Then

( )26 2 cosh ( )

2 3 coshP µ L zP B B A

A B µL −

= − + − +

From Eq. (iv)

( )16 8 1 cosh ( )

2 6 3 coshP B A µ L zP B A

A B µL + − = − − − +

and from Eq. (iii)

( )36 4 1 cosh ( )

2 6 3 coshP A B µ L zP B A

A B µL + − = − − − +

When A = B P1 = –3P P2 = –2P P3

and there is no shear lag effect. = –P

S.26.11

This problem is similar to that of the six-boom beam analysed in Section 26.4 (Fig. 26.11) and thus the top cover of the beam is subjected to the loads shown in Fig. S.26.11(a). From symmetry the shear flow in the central panel of the cover is zero.


Considering the equilibrium of the element δz of the corner longeron (1) in Fig. S.26.11(b)

11 1 0

2P SP z P z q zz hδ δ δ

∂+ − + − =∂

Fig. S.26.11(a)

Fig. S.26.11(b)

i.e.

1

2P Sqz h

∂= −

∂ (i)

Now considering the equilibrium of the element δz of longeron 2 in Fig. S.26.11(c)

22 2 0PP z P q z

zδ δ

∂+ − + =∂

which gives

2P qz

∂= −

∂ (ii)

From the equilibrium of the length z of the panel shown in Fig. S.26.11(d)

1 22 2 2 02SP P zh

+ + =


Fig. S.26.11(c)

Fig. S.26.11(d)

or

1 2 2SzP Ph

+ = − (iii)

The compatibility of the displacement condition between longerons 1 and 2 is shown in Fig. S.26.11(e). Thus

1 2d(1 ) (1 )d

z z h zzγε δ ε δ δ+ = + +

from which

1 2d 1 ( )dz hγ ε ε= − (iv)

In Eq. (iv) γ = q/Gt, ε1 = P1/3 BE, and ε2 = P2

/BE. Equation (iv) then becomes

12

dd 3

Pq Gt Pz hBE

= −

(v)

From Eq. (i) 2

12

dd

Pqz z

∂=∂


Fig. S.26.11(e)

and from Eq. (iii)

2 1 2SzP Ph

= − −

Substituting in Eq. (v) 2

1 12

43 2

P PGt SzhBE hz

∂ = + ∂

or

2

112 2

43 2

P Gt GtSzPhBEz h BE

∂− =

∂ (vi)

The solution of Eq. (vi) is

13cosh sinh8SzP C µz D µzh

= + − (vii)

where μ2

When z = 0, P = 4GT/3hBE.

1 = 0 so that, from Eq. (vii), C = 0. When z = l, q = 0 so that, from Eq. (i), ∂P1

8 coshSD

hµ µl= −

/∂z = – S/2h. Hence from Eq. (vii)

and Eq. (vii) becomes

1sinh 3

8 coshS µzP zh µ µl

= − +

(viii)


Substituting for P1

in Eq. (i) gives

cosh 18 coshS µzqh µl

= − −

(ix)

If the effect of shear lag is neglected then Eq. (ix) reduces to

8Sqh

=

and the shear flow distribution is that shown in Fig. S.26.11(f) in which q12 = q43 = q65 = q78 = S/8h and q81 = q54

M

= S/2h. The deflection Δ due to bending and shear is given by Eqs (20.17) and (20.19) in which

0 = – Sz and Μ1 = – 1 × z

Fig. S.26.11(f)

Also Ixx = 4 × 3B × (h/2)2 + 4 × B × (h/2)2 = 4Bh2 and q1 = q0

/S. Thus 2

0 120 0

d d d4

L l q qSz z s zGtBh E

∆ = + ∫ ∫ ∫ (x)

In Eq. (x)

0 12 2

4 2 11d4864 4 3

q q S h h SsGt G Ghth t h t

= + = ∫

Hence, substituting in Eq. (x) 2 11

12 4Sl l

h BhE Gt

∆ = +

S.26.12

The forces acting on the top cover of the box are shown in Fig. 26.12(a). Then for the equilibrium of the element δz of the edge boom shown in Fig. S.26.12(b).

2

BP wzqz h

∂= − +

∂ (i)


Fig. S.26.12(a)

Fig. S.26.12(b)

Similarly, for the central boom

A 2P qz

∂=

∂ (ii)

For the equilibrium of a length z of the cover

2

B A2 2 04

wzP Ph

+ − = (iii)

The compatibility of displacement condition is shown in Fig. S.26.12(c). Then

(1 ) (1 )A Bz z zdzγε δ ε δ δ∂

+ = + +∂

which gives

A B1 ( )

z dγ ε ε∂= −

∂ (iv)

But A B, ,A B

P PqGt AE BE

γ ε ε= = =


Fig. S.26.12(c)

Substituting in Eq. (iv)

A Bdd d

P Pq Gtz E A B

= −

Substituting for q from Eq. (ii) and PB

from Eq. (iii) and rearranging 2

2 2A2 2

AP Gtwµ P zdEhBz

∂− = −

∂ (v)

where 2 (2 )Gt B Aµ

dEAB+

=

The solution of Eq. (v) is

( )2

A 22cosh sinh

2 2w AP C µz D µz zh B A µ

= + + +

+

When z = 0, PA

( ) 22wAC

h B A µ= −

+

= 0 which gives

When z = L, ∂PA

( )sinh

2 coshwA µLD L

µh B A µh µ

= − + +

/∂z = 0 since q = 0 at z = L. This gives

Hence

( )A 2 2 2cosh sinh sinh

2 2coshwA µz µL µL l zP µz

h B A µ µ µL µ

+= − + − −

+


S.27.1

The position of the shear centre, S, is given and is also obvious by inspection (see Fig. S.27.1(a)). Initially, then, the swept area, 2AR0 (see Section 27.2) is determined as a function of s. In 12, 2ΑR,0 = 2sd/2 = sd. Hence, at 2, 2ΑR,0 = d2. In 23, 2AR,0 = 2(s/2)(d/2) + d2 = sd/2 + d2. Therefore at 3, 2AR,0 = 32/2. In 34, 2AR,0 remains constant since p = 0. The remaining distribution follows from antisymmetry and the complete distribution is shown in Fig. S.27.1(b). The centre of gravity of the ‘wire’ 1′2′3′4′5′6′ (i.e. 2A′R

) is found by taking moments about the s axis. Thus

2 2 2 2 2

R5 3 52 5d d

2 4 2 4 2d d d d dA t t

′ = + + + +

Fig. S.27.1(a)

which gives 2A′R= d2. Therefore, instead of using Eq. (27.9), the moment of inertia of the wire (i.e. ΓR

) may be found directly, i.e.

( )22

2 22 2

R2

2d 2d d3 3 2

dd dt t t

Γ = + +

which gives


5

R13

12d t

Γ =


Fig. S.27.1(b)

S.27.2

By inspection the shear centre, S, lies at the mid-point of the wall 34 (Fig. S.27.2(a)). The swept area, 2AR,0, is then determined as follows. In 12, 2AR,0 = (2sa sin 2α)/2, i.e. 2AR,0 = a2 sin 2α. In 23, 2AR,0

12 = 2 × sa sin 2α + a2 sin2α = (sa + α2) sin 2α and

at 3, 2AR,0 = 2α2 sin 2α.

Fig. S.27.2(a)

In 34 there is no contribution to 2AR,0

The centre of gravity of the ‘wire’ 1′2′3′4′5′6′ (i.e. 2A′

since p = 0. The remaining distribution follows from anti-symmetry and the complete distribution is shown in Fig. S.27.2(b).

R

2A′

) is found by taking moments about the s axis. Thus

R6at = at(2 × 2a2 sin 2a + 2 × 2a2

i.e. sin 2α)

2R

42 sin 23

A a α′ =


Fig. S.27.2(b)


22 2

2 2 2R

(2 sin 2 ) 42 2 2 (2 sin 2 ) sin 2 63 3

aat at a a atα α α Γ = × + −

which gives

5 28R 3 sin 2a t αΓ =

S.27.3

The shear centre, S, of the section is at a distance πr/3 above the horizontal through the centers of the semicircular arcs (see P.17.3). Consider the left-hand portion of the section in Fig. S.27.3(a). 2AR,0

= –2(Area CSF +Area CFOD +Area BCD –Area BSO)

= –2(Area BCS –Area BSO)

i.e.

( ) ( )R, 0 1 1 1 1

21

1 12 2 cos sin 2 cos sin2 3 2

1 1 22 2 3

rA r r r r r r

rr r

πθ θ θ θ

πθ

= − + − + +

+ −

i.e.

2R, 0 1 1 12 sin cos

3 3A r π πθ θ θ = − − −

(i)

When θ1 = π, 2AR,0 = –πr2

Note that in Eq. (i) A/3.

R,0Consider now the right-hand portion of the section shown in Fig. S.27.3(b). The

swept area 2A

is negative for the tangent in the position shown.

R,0

2A

is given by

R,0 = 2 Area OSJ –πr2/3


Fig. S.27.3(a)

Fig. S.27.3(b)

i.e.

2AR,0 = 2(Area OSJ –Area OJI –Area SJI) –πr2

which gives /3

2

2R, 0 2

1 1 12 2 KS2 3 2 2 3

r rA r r rπ πθ = − − − (ii)

In Eq. (ii)

2 2 2KS MScos tan cos3r rπθ θ θ = = −

i.e.

2 2KS cos sin3r rπ θ θ= −

Substituting in Eq. (ii) gives

2R, 0 2 2 22 sin cos

3A r πθ θ θ = − −

(iii)


In Eq. (27.3)

R,0 31 1 1 10

32 2 2 20

2 d 1 sin cos d2 3 3d

sin cos d3

C

C

A t sr

rt s

r

π

π

π πθ θ θ θπ

πθ θ θ θ

∫ = − − − ∫ + − −

∫

∫

i.e.

2

R,02 d3d

C

C

A t s rt s

π∫= −

∫

Hence, Eq. (27.3) becomes

2

R R,02 23rA A π

= +

Then

22 4

R R 1 1 1 10

24

2 2 2 20

(2 ) d sin cos d3 3 3

sin cos d3 3

CA t s r

r

π

π

π π πθ θ θ θ

π πθ θ θ θ

Γ = = − − − +

+ − − +

∫ ∫

∫

which gives

2 5R

33

r t πππ

Γ = −

S.27.4

The applied loading is equivalent to a shear load, P, through the shear centre (the centre of symmetry) of the beam section together with a torque T = –Ph/2. The direct stress distribution at the built-in end of the beam is then, from Eqs (16.21) and (27.1)

2

R 2d2d

x

xx

M y A EI z

θσ = − (i)

In Eq. (i) Mx

and = Pl (ii)

Ixx = 2td3/12 = td3

Also d/6 (iii)

2θ/dz2

is obtained from Eq. (27.6), i.e. 3

R 3d dd d

T GJ Ez zθ θ

= − Γ


or, rearranging

3

2 23

d ddd

Tµ µz GJz

θ θ− = − (iv)

in which μ2 = GJ/ETR

. The solution of Eq. (iv) is

d cosh sinhd

TC µz D µzz GJθ= + + (v)

At the built-in end the warping is zero so that, from Eq. (18.19) dθ/dz = 0 at the built-in end. Thus, from Eq. (v), C = – T/GJ. At the free end the direct stress, σΓ, is zero so that, from Eq. (27.1), d2θ/dz2

= 0 at the free end. Then, from Eq. (v)

tanhTD µlGJ

=

and Eq. (iii) becomes

d cosh ( )1d cosh

T µ l zz GJ µlθ −= −

(vi)

Differentiating Eq. (vi) with respect to z gives

2

2d sinh ( )

coshdT µ l zµ

GJ µlzθ −= (vii)

Hence, from Eq. (27.1)

Rsinh ( )2

coshT µ l zA E µ

GJ µlσΓ

−= −

which, at the built-in end becomes

R

2 tanhRE T A µl

GJσΓ = − Γ

(viii)

In Eq. (viii)

J = (h + 2d)t3

The torsion bending constant, Γ/3 (see Eq. (18.11)) (ix)

R, is found using the method described in Section 27.2. Thus, referring to Fig. S.27.4(a), in 12, 2AR,0 = sh/2 and at 2, 2AR,0 = hd/4. Also, at 3, 2AR,0 = hd/2. Between 2 and 4, 2AR,0 remains constant and equal to hd/4. At 5, 2AR,0 = hd/4 + hd/4 = hd/2 and at 6, 2AR,0 = hd/4 – hd/4 = 0. The complete distribution is shown in Fig. S.27.4(b). By inspection 2A′R

= hd/4. Then 2

R14

2 3 4d hdt Γ =


Fig. S.27.4(a)

Fig. S.27.4(b)

i.e.

3 2

R 24td h

Γ = (x)

Substituting the given values in Eqs (ii), (iii), (ix) and (x) gives

3 4

3 4

3 2 7 6R

200 375 75000 N mm

I 2.5 37.5 /6 21973.0mm

(75 2 37.5)2.5 /3 781.3mm

2.5 37.5 75 /24 3.09 10 mm

x

xx

M

J

= × =

= × =

= + × =

Γ = × × = ×

M

Then

2 7 6781.3 / (2.6 3.09 10 ) 9.72 10µ −= × × = × and

33.12 10µ −= ×


Thus from Eqs (i) and (viii)

R3.41 0.064(2 )y Aσ = + (xi)

Then, at 1 where y = –d/2 = –18.75 mm and 2AR = –hd/4 = –703.1 mm2

, 2

1 3108.9N/mmσ σ= − = −

Similarly

25 618.9N/mmσ σ= − = −

and

2 4 2 40σ σ σ= = =

S.27.5

The rate of twist in each half of the beam is obtained from the solution of Eq. (27.6). Thus, referring to Fig. S.27.5, for BC

1 11

d cosh 2 sinh 2d 8

T A µz B µzz GJθ= + + (i)

where μ2

= GJ/ET and for BA

2 22

d cosh sinhd

T C µz D µzz GJθ= + + (ii)

Fig. S.27.5


The boundary conditions are as follows: When

1 2 1 20, d / d d / dz z z zθ θ= = = (iii)

When

2 2 2 21 2 1 2, d /d d /d 0z z l z zθ θ= = = = (see Eq. (27.1)) (iv)

When

2 2 2 21 2 1 20, 2d /d d /dz z z zθ θ= = = −

(since the loads at B in each half of the section are equal and opposite). From Eqs (i), (ii) and (iv)

B = – tanh 2µl (vi)

B = – tanh 2µl (vii) From Eqs (i)–(iii)

8

T TA CGJ GJ

+ = +

i.e.

78

TA CGJ

− = (viii)

From Eqs (i), (ii) and (v)

D = –4B (ix) Solving Eqs (vi)–(ix) gives

7 tanh tanh 28 (4 tanh 2 tanh )

T µl µlBGJ µl µl

= −+

7 (4 tanh tanh 2 )8 (4 tanh 2 tanh )

T µl µlDGJ µl µl

= −+

7 tanh8 (4 tanh 2 tanh )

T µlAGJ µl µl

= −+

7 (4 tanh 2 )8 (4 tanh 2 tanh )

T µlCGJ µl µl

= −+

Integrating Eq. (i)

1 1 1 1sinh 2 cosh 28 2 2

T A Bz µz µz FGJ µ µ

θ = + + +


When z1 = 0, θ1

= 0 so that F = –B/2µ. Integrating Eq. (ii)

2 2 2 2sinh coshT C Dz µz µz HGJ µ µ

θ = + + +

When z2 = 0, θ2 = 0 so that H = –D/µ. Hence, when z1 = l and z2

=l the angle of twist of one end of the beam relative to the other is

1 2

12

72

7( 8 )8 8 (4 tanh 2 tanh )[ (tanh sinh 2 tanh tanh 2 cosh 2 4 tanh 2 sinh

4 tanh tanh 2 cosh (tanh tanh 2 )]

T Tl lGJ GJµ µl µl

µl µl µl µl µl µl µl

µl µl µl µl µl

θ θ+ = + ++

× − −

+ −

which simplifies to

1 2 249 sinh 29

8 2 (10cosh 1)Tl µlGJ µl µl

θ θ

+ = − −

S.27.6

Initially the swept area 2AR,0 is plotted round the section and is shown in Fig. S.27.6(b).

Fig. S.27.6(a)

Then, using the ‘wire’ analogy and taking moments about the s axis


2 2

R3 3 32 5 2 22 4 2a a aA at t at

′ = +


Fig. S.27.6(b)

which gives

2

R21220aA′ =

Then

2 2 22 2 2

R3 1 3 3 212 2 52 3 2 2 20a a a at at at

Γ = + −

i.e.

5R 1.25a tΓ =

From Eq. (27.6), i.e.

3

2 23

d ddd

Tz GJz

θ θµ µ− = −

where

2

R

GJE

µ =Γ

d cosh sinhd

TC z D zz GJθ µ µ= + +

When z = 0, the warping, w, is zero so that dθ/dz = 0 (see Eq. (18.19)), then

TAGJ

= −

When z = L, the direct stress is zero. Therefore, from Eq. (27.1) d2θ/dz2

= 0. Therefore

tanhTB LGJ

µ=

Solutions to Chapter 27 Problems 373 so that the rate of twist is

d cosh ( )1d cosh

T L zz GJ Lθ µ

µ −

= −

and

sinh ( )cosh

T L zz CGJ L

µθµ µ

−= + +

When z = 0, θ = 0 which gives

1 tanhC Lµµ

= −

and

sinh ( ) tanhcosh

T L z LzGJ L

µ µθµ µ µ

−= + −

At the free end when z = L

tanh1TTL LGJ L

µθµ

= −

(i)

Inserting the given values in Eq. (i) 3

42.0100 30 3000 N mm 5 30 400 mm3

T J= × = = × × =

2 62.35 10 1.53 6.93TLµ µ θ−= × = = °

S.27.7

The torsion bending constant is identical to that in S.27.4, i.e.

2 3

Rd

24th

Γ =

The expression for rate of twist is (see S.27.6)

d cosh sinhd

TA z B zz GJθ µ µ= + +

In AB, T = 0 and dθ/dz = 0 at z = 0 which gives A = 0 Therefore, in AB

d sinhd

B zzθ µ=

In BC

[ ]d 1 cosh ( ) sinh ( ) sinhd

z L z L zzθ α µ β µ β µ= − − − − +


where [ ] is a Macauley bracket i.e.

[ ] 0 for

()ordinary bracket forz L

z L= <= >

For continuity of dθ/dz and d2θ/dz2

at z = L the Macauley bracket and its first derivative must be zero at z = L. Then

1 0and 0α β− = =

For the complete beam

[ ]d 1 cosh ( ) sinhd

T µ z L B µzz GJθ= − − +

At

2 22 d / d 0 ( 0 2 ).z L z a tz Lθ σΓ= = = =

Then

sinh cosh 2T µ µL µB µLGJ

θ = − +

which gives

sinhcosh 2

T µLBGJ µL

=

Then

[ ]d sinh1 cosh ( ) sinhd cosh 2

T µLµ z L µzz GJ µlθ = − − +

Also since θ = 0 at z = 0 and the Macauley bracket is zero for z < L

1 sinhsinh ( ) (cosh 1)cosh 2

T µLz L µ z L µzGJ µ µL

θ = − − − + −

At z = 2L

Tsinhcosh 2

T µLLGJ µL µL

θ

= −

S.27.8

The variation of swept area is shown in Fig. S.27.8(b) Using the ‘wire’ analogy

2

2 2R

5 32 4 22 8 4

aA at at at a at a′ = + +


Fig. S.27.8(a)

Fig. S.27.8(b)

i.e.

2

R528aA′ =

Then

22 2 222 2 2 2 2

R

22

1 1 3 5 1 3( ) 23 3 8 8 3 2 4

548

aat a at a a at a

aat

Γ = + + + +

−


which gives

5

R724a t

Γ =

The rate of twist is identical to that given by Eq. (vi) in S.27.4, i.e.

d cosh ( )1d cosh

T µ L zz GJ µLθ −= −

(i)

The direct stress distribution at the built-in end is, from Eq. (ix) of Example 27.1

RR

sinh2cosh

E µLT AGJ µL

σΓ = − Γ

Evaluating the different constants

5 6 4

R2 6

9.33 10 mm 26.67 mm 1125 N mm

8.56 10 and 1.46

J T

µ µL−

Γ = × = =

= × =

Then R0.369 2AσΓ = − At 2,

2 2 2

2 2R

5 3 3 202 150 mm8 8 8a aA a ×

= − = = =

so that

2,2 55.3 N/mmσΓ = −

The direct stress due to elementary bending theory is, from Eqs (16.21)

xz

xx

M yI

σ =

where 150 500 75 000 N mmxM = − × = − and

3

2 3 41.0 402 1.0 20 20 21.3 10 mm12xxI ×

= × × × + = ×

Then

2,2 3

75 000 20 70.4 N/mm21.3 10zσ ×

= − = −×

The total direct stress at 2 is therefore

22 55.3 70.4 125.7 N/mmσ = − − = −


S.27.9

The torsion bending constant is identical to that in S.27.4, i.e.

2 3

R 24th d

Γ =

The rate of twist is, from Eq. (27.6)

d cosh sinh ( )d 2

whA z B z L zz GJθ µ µ= + + −

when z = 0, dθ/dz = 0 (w = 0 at z = 0) which gives

2whLAGJ

= −

When z = L,d2θ/dz2 = 0 (σΓ

= 0 at z = L) which gives

1tanh2 coshwhB L µhGJ µ µL

= +

Hence

d sinh 1cosh sinhd 2 cosh

wh µL µLL z z L zz GJ µ µLθ µ µ

+= − + + −

Then

2

R 2d2d

A EzθσΓ = −

is

Rsinh 12 sinh cosh 1

2 coshwh µL µLA E L z zGJ µL

σ µ µ µΓ

+= − − + −

At the built-in end when z = 0

Rsinh 1 cosh2

2 coshwh µL µL µLA EGJ µL

σΓ + −

= −

Evaluating the constants

6 6 4 2 6R 1040 10 mm , 12 500mm , 4.0 10 , 3.0.J Lµ µ−Γ = × = = × =

Then

R0.025(2 )AσΓ = −

The distribution of 2AR is linear round the section so that σΓ is also linear.


At 1,

Rd24

hA = − (see S.27.4)

Then

2,1

500.025 200 62.5 N/mm .4

σΓ = + × × = +

From symmetry of the 2AR

distribution 2

,3 ,1 ,4 ,6

,2 ,5

62.5 N/mm ,0

σ σ σ σ

σ σΓ Γ Γ Γ

Γ Γ

= − = − = = −

= =

From Eqs (16.21)

yz

yy

M xI

σ =

where

215000.5 562500 N mm

2yM = × =

and

3

4502 5 104200mm12yyI = × × =

Then

5.4 ,z yσ =

i.e.

2,1 135N/mmzσ = +

From symmetry

2

z,1 z,3 z,4 z,6

z,2 z,5

135 N/mm0

σ σ σ σ

σ σ

= − = − = = +

= =

The complete direct stresses are

21 3

24 6

2 5

62.5 135 197.5N/mm

62.5 135 72.5N/mm0

σ σ

σ σσ σ

= + = = −

= − = − = −

= =


S.28.1

From Eq.(28.3)

ec = 1.93 × 106/[(1/2) × 1.226 × 3502

In Example 28.1, ec = 0.25 × 3.0 = 0.75m. Therefore the required movement of the aerodynamic centre is 0.75 – 0.24 = 0.51m.

× 30 × 3.5] = 0.24m

S.28.2

From Eq.(28.11)

Vd = √[π2 × 12 × 106/2 × 1.226 × 1.2 × 2.8 × 82

ie × 3.1]

Vd

S.28.3

= 269.2 m/s

The wing area, S = 2.8 × 8.0 = 22.4 m2

V

. Then, from Eq.(28.18)

r = √[12 × 106

ie × 0.15/(1/2) × 1.226 × 22.4 × 2.8 × 0.3 × 3.1]

Vr

Therefore, from Eq.(28.22) = 224.4 m/s

Aileron effectiveness = (1 – 1602/224.42)/(1 – 1602/269.22

The aileron is therefore 76% effective. ) = 0.76

S.28.5

The solution is obtained directly from Eq. (28.9) in which ∂c1/∂α = a, α = α0 and cm,0 – CM,0

M,00

cos ( ) 1cos

C s yea s

λθ αλ

− = + −

. Thus

which gives M,0 M,0

0 0cos ( )

cosC Cs yea s ea

λθ α αλ

−= + − −


Thus

M,0 M,00 0

cos ( )cos

C Cs yea s ea

λθ α αλ

−+ = + −

where 2 21

2 2ea V cGJρ

λ =

Also, from Eq. (28.11) the divergence speed Vd

2

d 2 22GJV

ec s aπρ

=

is given by

S.28.6

Since the additional lift due to operation of the aileron is at a distance hc aft of the flexural axis the moment equilibrium equation (28.25) for an elemental strip becomes

d 0dT y Lec L hcyδ ξ− ∆ − ∆ = (i)

in which, from Eq. (28.23)

21 2

1 ( )2 a

pyL V c y a a f yV

ρ δ θ ξ ∆ = − +

where fa (y) = 0 for 0 ≤ y ≤ ks and fa

2122 ( )aL V c ya f yξ ρ δ ξ∆ =

(y) = 1 for ks ≤ y ≤ s. Also

Then, substituting for T(= GJ dθ/dy), ∆L and ∆L in Eq. (i) and writing λ2 = ρV2 ec2a l

/2GJ 2

2 2 2 2a2

1

d ( )d

apy h f yV e ay

θ λ θ λ λ ξ+ = + (ii)

The solution of Eq. (ii) is obtained by comparison with Eq. (28.29). Thus

21

1

sin(O ) (tan cos sin ) sincos

hap yks y s ks ks yV s ea ξλ

θ − = − − λ λ − λ λ λ λ (iii)

and

2

2

1

sin( )cos

(1 cos cos tan cos sin )

p yks s yV s

ha y ks s ks yea

λθ − = − λ λ

ξ+ − λ λ − λ λ λ

(iv)



1 1 1 2 20d d d

ks s s

ks ks

py pya y y a y y a y yV V

− + − = − ξ ∫ ∫ ∫θ θ (v)

Substituting for θ1 and θ2

in Eq. (v) from Eqs (iii) and (iv) gives

0 0

102

( )tan sin d tan sin d cos d dcos

sin dcos cos

s ks s s

ks ks

s

e hs y y y ks y y y y y y y yh ks

pea y y yha V s ks

λ λ λ λ λλ

λξλ λ λ

+− + − +

=

∫ ∫ ∫ ∫

∫

Hence the aileron effectiveness is given by

0 0

2 2

02

tan sin d tan sin d

( )cos d [ ( ) ]( / ) 2 cos

sin dcos cos

s ks

s

ks

s

s y y y ks y y y

e hy y y s ksps V h ks

ea y y yha s s ks

λ λ λ λ

λλ

ξ λλ λ λ

− +

+− + −

=

∫ ∫∫

∫ (vi)

The aileron effectiveness is zero, i.e. aileron reversal takes place, when the numerator on the right-hand side of Eq. (vi) is zero, i.e. when

2 2

0 0

( )tan sin d tan sin d cos d [( ) ]2 cos

ks s s

ks

e hks y y y s y y y y y y ks sh ks

λ λ λ λ λλ

+− − = −∫ ∫ ∫

Aircraft Structures - textbooks.elsevier.com · Aircraft Structures . for Engineering Students ....

Documents

Transcript of Aircraft Structures - textbooks.elsevier.com · Aircraft Structures . for Engineering Students ....