1

Hints to Problems

Chapter-1

1.18

150 mm 200 mm550 mm

40 kN 10 kN 20 kN 50 kN

Elongation is given by, AEPL=

= ( ) 224.020000050550000301500004000020530)4/(

12 =++ XXXX mm

1.22

Area of column = 3002465604

2 =X mm2; Area of steel = 4072364

4 2 =XX mm2 Area of concrete = 246 300 4072 = 242 228 mm2

We have, c

c

s

s

EE = or

00015000210cs = or cs 14=

s X As + c X Ac = 800 000 14 c X 4072 + c X 242 228 = 800 000 or c = 2.673 MPa s = 2.673 X 14 = 37.429 MPa

Average compressive stress = 248.3300246000800 = MPa

1.23

P

1 2 3 P 1P 2 P 1

P

21

2

On welding, tensile stresses will be developed in 1 and 3 whereas compressive stress in 2. From equilibrium equation, 2 P1 cos = P2 .(assuming negligible change in ) (i) From compatibility equation, 1 = 2 cos

or

=

22

22

11

11

EALP

EALP cos or

=

22

22

11

21

cos EALP

EALP

cos

or 222

22

2

111 cos

=

EALP

LEAP

If all rods are of same material and section,

222

222

21 coscos

=

= P

LAE

AELP

LAEP

From (i), 23

22

cos2 PPLAE =

or

)cos21(cos2

32

3

2

+=

LAEP and

)cos21(cos

32

2

1

+=

LAEP

1.24 Here, Lt= ;

222

226-2

22

222

2

111 cos40120010 X 191200

0008542cos

=

=

EALPXXX

EALPtL

LEAP

From (i), 23

22

222

2

11 cos2 PEALPtL

LEA =

or 23

22

222

2

11 cos2 PEALPtL

LEA =

or ( ) ( )

+

=+

=

311

222

222

3

22

11

32

2

11

2

cos21cos

21

cos2

EAEAL

tLEA

EAEA

tLL

EA

P

1.26 Aa = 80420 =X mm2 , As = 3201620 =X mm2 Total compression in Al = Total tension in steel

32080 XX sa = or sa 4=

c

cccc

s

ssss E

LtLE

LtL =+ or c

cc

s

ss E

tE

t =+ 1.28 Let 3 be the stress in the direction of restraint. Then,

1 = 1/E 3/E (i) 2 = 3/E 1/E (ii)

3

and 3 = 3/E 1/E (iii) If there is no restraint in the third direction, 13 ' E= However, for the given restraint, 133 2

2/' E

==

113 2 EE

= or 13 2 =

Substituting in (i), 1 = ( )212111 222

12

. =

=

EEEE

Modified modulus of elasticity, E = 21

1

22

=E

Chapter-2

2.11 Both stresses are compressive,

tan = 816.0180120 ==

y

x

or = 39.23o

r = 2222 sincos yx +

129598640

23.39sin18023.39cos120 2222

+=+= oo

= 147 MPa compressive 2.14

OL

R

FMCE

S

P

N60

5

10

10.952.5

2.5

2

10.9

5

0.9

4

2.15

(i)Principal stress }4){(21)(

21 22 ++= yxyx

8.5380

1000016002180

})50(4)60100{(21)60100(

21 22

=+=

++=

= 133.8 MPa (tensile) and 26.2 MPa (tensile)

tan2 = yx

2 = 5.2

60100502 =

X

or 2 = 68.2 or 1 = 34.1o and 2 = 34.1o + 90o = 124.1o max = )2.268.133(

21 = 53.8 MPa at 34.1o + 45o =79.1o

2.16

30o240 MPa

180 MPa

A B

C

30o240 MPa

180 MPa

A B

C

120 MPa

120 MPa

207.8 MPa

On plane BC, x = 8.20730cos240 =o MPa 12030sin240 == o MPa The equivalent state of stress on the panes is shown in the figure. Resultant stress on plane AB = 22 120180 + =216.3 MPa

5.1120180tan == or = 56.3o

Principal stress }4){(21)(

21 22 ++= yxyx

8.1208.193

1000025600219.193

})120(4)1808.207{(21)1808.207(

21 22

=+=

++=

5

= 314.6 MPa (tensile) and 73 MPa (tensile)

tan2 = yx

2 = 63.8

1808.2071202 =

X

or 2 = 83.4o or 1 = 41.7o and 2 = 131.7o max = )736.314(

21 = 120.8 MPa at 41.7o + 45o = 86.7o and 176.7o

2.18

pq r

40080200

P

R'

Q

R

CO

28oO

r60o q

60op80

400200

60o 60

E F(a)

(b)

O'

32o

OF = 449 MPa inclined at 28/2 or 14o to 400 MPa stress OE = 251 MPa inclined at 32/2 or 16o to 200 MPa stress 2.19

p q r

1.6 X 10-6

P

R'

QC

O

10oO

r45o

q

45op

45oE F

(a)

O'

10o

45o

3 X10-64 X10-6

1.6 X 10-6

3 X10-64 X10-6

Major principal strain 1= OF = 4.04 X 10-6 at 10/2 or 5o clockwise of plane p

6

Minor principal strain 2 = OE = 1.58 X 10-6 at 95o clockwise of plane p.

Chapter-3 3.13 A = (/4)X242 = 144

Diameter of the reduced section = 24/2 = 12 mm Area of the reduced section = (/4).122 = 36

When the bar is turned down to half the diameter along half of its length, let P be the equivalent load to induce the same maximum stress.

=000205144

60000020536

600X

PXXPX

+ = 0.000 03235 P = kP ..... (Taking k =0.000 03235)

Using the energy equation

W (h+ ) = 21 P.

W (h+ kP) = 21 P. kP

P2 2WP 2Wh/k = 0 (multiplying throughout by 2/k)

3.15 As = ( /4)242 = 144 ; Ab = ( /4)(362 - 242) = 180 Let x = Extension of bar in mm

L

xEss

.= and L

xEbb

.=

Strain energy of the bar LAELxELA

ELxELA

ELA

E bbb

ss

sb

b

bs

s

s

2.2.22 222

2

2222

+=+=

)(222

222

bbssbb

ss AEAE

LxA

LxEA

LxE +=+=

)18000098144000205(30002

2

XXX

x += = 24693 x2 N.mm Potential energy lost by the weight = W(h + x) = 13500 (6 + x) N.mm

Chapter-4

7

4.11

B

(b) SF Diagram

(c) BM Diagram

C A

(a)6m1m 3m

6 kN 8 kN

D

8

36

246

Taking moments about end A = 016986 =+ XXXRb or kNRb 11= ; and kNRa 31114 ==

4.14

40 kN 20 kN40 kN 20 kN

1 m 1 m2 m2 m2 m

A BC D E F

40

40

20

80

20

Taking moments about end A = 2404206202405 XXXXXRb ++=+ or kNRb 40= ; and 804020204040 =+++=aR kN Shearing force diagram Portion AC: Fx = kNRa 40= (constant ) Portion AD: Fx = kN408040 =+ (constant ) Portion DE: Fx = 04040 = (constant )

8

Portion BE: Fx = 20 kN (constant ) Portion BF: Fx = kN204020 =+ (constant ) Bending moment diagram Portion AC: Mx = - x40 (linear )

At A, x = 2 m and Ma = -80 kN.m; Portion AD: Mx = )2(8040 + xx (linear )

At D, x = 4 m and Md = 0 Portion DE: Mx = 0 Portion EB: Mx = - x20 (x from E) (linear )

At B, x = 1 m and MB = -20 kN.m Portion BF: Mx = - )1(4020 + xx (x from E) (linear )

At F, x=2m, Mf = 0

4.17 The loading on the beam is shown in the figure

10 m

w

A BDE

R a R ba 4-a6 mC

Let distance of pier at A be a m from the end D, the other will be at B, (6-a) m from end E. Distance CB = CE BE = 5- (4-a) = 1+ a

Taking moments about B, )1(106 awRa += or 3)1(5 awRa

+= Bending moment

Portion DA: Mx = 22wx (parabolic); Md = 0; Ma = 2

2wa (negative value)

Portion AB: Mx = )(3

)1(52

2

axawwx ++

It is maximum when dM/dx = 0 or 03

)1(5 =++ awwx or 0)1(53 =++ ax or x = 5(1+a)/3 Maximum bending moment =

++++ aaawaw )1(

35

3)1(5

2)]1)(3/5[( 2

= )25(9

)1(5)1(18

25 2 aawaw ++++

= [ ])41055)(1(185 aaaw +++

9

= [ ])5)(1(185 aaw +

= )45(185 2aaw +

The maximum bending moment will be as small as possible if the magnitudes of the

sagging and the hogging bending moment are equal.

4.19

8 kN/m

B

l = 15 m8 m

CA

16 kN

5 m D

25 kN 18.33 kN2.33 kN

51 kN

32 kN

113.9 kN.m 114.9 kN.m

85 kN.m

SF

BM

5.81 m

13.05 m

(a)

(b)

(c) Taking moments about B,

38

288

315

2815101615 XXXXXXRa +=

or 25=aR kN and 8325282316 =

+= XRb kN Intensity of distributed loading at any cross-section at a distance x from A = (w/l).x=wx/l Shear force diagram

Portion AD: Fx = 75.3

25152

82521 22 x

Xxx

lwxRa == (parabolic)

Fa = 25 kN; Fd = 18.33 kN

Portion DB: Fx = 1675.3

252

x ..(parabolic); Fd = 2.33 kN; Fb = -51 kN Between DB, shear force is zero when Fx = 1675.3/25 2 x =0 or x= 5.81 m

10

Portion BC: It is convenient to deal this portion by using variable x from end C.

Fx = 2828

21 22 x

Xxx

lwx == (parabolic); Fc = 0; Fb = 32 kN

Shear force diagram is shown in Fig. 4.21b.

Bending moment diagram

Portion AD: Mx = 25.11

253

.75.3

2532 xxxxx = (cubic); Fa = 0; Fd = 113.9 kN.m

Portion DB: Mx = 25.11

)5(16253xxx (cubic)

Fd(x=5) = 113.9 kN.m; Fb(x=15) = - 85 kN Mmax(x=5.81) =

9.11425.11

81.5)581.5(1681.52525.11

)5(162533

== Xxxx kN.m

Bending moment is zero at 25.11

)5(16253xxx = 0 or 090025.1013 = xx

Solving by trial and error, x = 13.05 m Portion BC: x from end C.

Mx = 632

32 xxx = (cubic) ; Fc = 0; Fb = -85 kN.m 4.20

w'xx' w

x

C

AB

dx

wx

67503600

3600

Let w = rate of loading at the midspan

Then total load = wwX 4632 =

or w47200 = or w = 1800 kN/m Taking C as origin of the parabola, At a distance x from A or x from C,

11

2'' kxw x = , when x =3 m, 1800' =xw kN or k = 200 22 )3(200)1800('200' xworxw xx == or 2)3(2001800 xwx = or 22 2001200)69(2001800 xxxxwx =+= ( ) 1322 3200600360020012003600 CxxdxxxFx ++== At x = 0, 3600=xF , C1 = 0

32006003600

32 xxFx +=

0 1 2 3 4 5 6 Fx 3600 3067 1733 0 -1733 -3067 -3600

2

43

32

3502003600

32006003600 CxxxdxxxFM xx ++=

+==

At x = 0, Mx =0 C2 = 0

3502003600

43 xxxM x +=

0 1 2 3 4 5 6 Mx 0 3417 5867 6750 5867 3417 -3600

4.21

4 kN4 kN

2 m2 m3 m3 m

A BC D

8 kN.m

1

3

4 kN

E

3

8 kN.m5

2

Chapter-5

5.13

12

64444

1045.3664

)200220(64

)( XdDIxx === mm4 3

6

max

104.331110

1045.36 XXyIZ xxx === mm3

Internal area of pipe = 01.0)2.0(4

2 = m2 Weight of water /m run of pipe = 3082.081.9101.0 =XX kN Area of cross-section of pipe = 2

22

3.65974

)200220( mm= Weight of pipe /m run = 0.0065973 X 70 X1 = 0.4618 kN Total weight/m run = 0.3082 + 0.4618 = 0.77 kN 5.14

D

b

d

A B

CD

sinDBC = and cosDAB = ;6

cos.sin6

222 DbdZ == If beam is to be strongest, Z must be maximum.

i.e. 0=ddZ or ( ) 0cossin2.cossin

63

2

=+ D or 22 cos2sin = or 2tan2 = or 2tan = or 3/2sin = or 3/1cos =

DDXBC 816.03/2 == and DDAB 577.03/1 ==

5.16

13

80 mm

20 mmx x

y

y

120 mm

643 10394.11)40(64

12080121 XXXIxx == mm4

10503

5.135002450 =+= XXXRa N ; 90010501950 ==bR N Bending moment between CB,

22

2504506002

500)1(4501050 xxxxxM x +==

For maximum value, 0=dx

dM x or 0500600 = x or x =1.2 m 810)2.1(2504502.1600 2max =+= XM N.m

5.17

2030 30

30 2504/110004/max === XWlM N.m

31907)2030(64

44 == xI mm4 1.2127

1531907

max

===yIZ xxx mm

3; 5.1171.2127

000250/max === ZM MPa For composite section, Moment of inertia about neutral axis,

( ) 4810561520304

319074 222 =

+ X mm4

14

5.18

Neglecting self weight W of the beam, ''.44

'maxmaxmax

max kWWyIly

IlWy

IM =

===

If W is considered,

+=

+=

+=2

'4

.2

'84

'' maxmaxWWky

IlWW

IyWllW

%Error = 1002/'

2/100

2'

'2

'100

'' X

WWWX

WWk

kWWWkX +=

+

+=

or 1002/'

2/ XWW

We += or eWeWW += '2100 or '2)100( eWWe =

or e

eWW

2100' =

5.21

b

t

t'

b/2

t

t'

Modular ratio is 2. The equivalent steel section is shown in the figure. As the neutral axis of the section is to be at the dividing line,

2''.

22. ttbtbt = or 707.0

21

'. ==tt

5.22

2028

If the stress in steel reaches to maximum value, the stress induced in brass

= 1410

75.1150 X = 61.22 MPa

brasssteelr MMM += = 20028720.

3222.61

282028

32150 3

44

=

+

XX N.mm

15

5.25

10 m

x

16080

At a section x-x, at a distance x from the top, side of the square,

xs .10000

8016080 += or )80(125 = sx mm

Moment of inertia about the diagonal or the neutral axis = 12

4s mm4

Bending moment at the section = 1000x =1000 X 125 (s-80) N.mm

3

6

4

3 )80(.2105.1

212/)80(10125

ssXsX

ssX == MPa

For bending stress to be maximum, 0=dsd

Chapter-6

6.12 Area = 2bt + t(2b-2t) = 4bt - 2t2 = 4bt (Neglecting higher values t)

[ ] [ ] tbtbbbtbbtbtbbbIx 33444433 38)4(32)(3212 )22)((12)2( ==== shear stress (maximum) at neutral axis,

zIyAF=

23

22)(

2

222

2 tbtbtbtbttbbtyA =+=+

= neglecting higher powers t.

btF

tbttb

FzIyA

F169

)3/8.(2/3

3

2

=== Average shear stress = F/4bt

25.2=

AverageMax

6.15

16

z

h

b

t

Refer Fig., 3233

101.5075.2755512

555212

605 XXXXXIx =

++= mm4 The shear stress in the flanges,

At a distance z from the tip, zFXXtX

XztFtI

yAF .102.54101.5075.27... 6

3===

Total force in each flange,

FzXFXdzFztXtdzFf 448.025102.54.102.54.

5.27

0

266 =

===

Chapter-7

7. 9

==lxpw

dxdF sin

Integrating, dx

dMClxplF =+

= 1cos

Integrating, 2122

.sin CxClxplM ++

= At x =0, M=0 , C2 = 0; At x =l, M=0 , C1 = 0

Maximum bending moment = 22

pl

Reaction at support (Value of F at x=0) = pl

7.10

P

l/2l

W=wlA BC

17

Taking moments about A, 2/.2/3 lWlPXXlRb += or 2/)3( WPRb += Consider a cross-section at distance x from C,

2

2

2

222

+= lxwlxRPxdx

ydEI b

Integrating, 1322

26222ClxwlxRxP

dxdyEI b +

+=

Integrating again, 21433

.224266

CxClxwlxRxPEIy b ++

+= 7.15

4 kN/m

A BC

8 m

6 kN

D3 m 3 m 3 m

E

2 m

4 kN/m

A

B

C

8 m

6 kND

3 m 3 m 3 m

E

2 m

To apply Macaulays method, uniformly distributed load has to be continuous upto the end of the beam. To compensate the same an upward uniformly distributed load has to be considered from D to E as shown in the figure. Taking moments about A, 5.3341168 XXXXRb += or 5.13=bR kN and 5.45.1318 ==aR kN At any section x from A, )8(5.13

2)5(4

2)2(45.4

22

2

2

+= xxxxdx

ydEI

Integrating, 2

)8(5.136

)5(46

)2(425.4 233

1

2 ++= xxxCxdxdyEI

Integrating again, 6

)8(5.1324

)5(424

)2(4.65.4 344

21

3 +++= xxxCxCxEIy At A, (x=0), y = 0, 02 =C (consider only first part)

At B, (x=8m), y = 0, 24

)58(424

)28(486

85.4044

1

3 ++= CX ; C1 = 22.69 slope and deflection equations are as follows:

18

2)8(5.13

3)5(2

3)2(27.2225.2

2332 ++= xxxx

dxdyEI

6)8(5.13

6)5(

6)2(7.22

65.4 3443 ++= xxxxxEIy

Deflection at E, x=11m

6

)811(5.136

)511(6

)211(117.226

115.4 3443 ++= XXEIy =68.2 kN.m3

mmXX

Xy 7.221015000200

102.686

12

== For maximum deflection at distance x, making slope equal to zero,

mxxxorxx 27.301.6875.6)2(23

)2(27.2225.20 233

2 ==+++=

4.486

)227.3(27.37.226

27.35.4 43 =++= XXEIy kN.m3

mmXX

Xy 1.161015000200

104.486

12

== 7.16

xA B

Mb

l

Ma

MaMb

BM diagram

ymaxzab

A' B'

D'

C

As the intercepts on a given line ( on BD) between the tangents to the elastic curve of a beam at any two points is equal to the net moment taken about that line of the area of the bending moment diagram between the two points divided by EI.

( ) [ ]baabaab MMEIlllMMllM

EIz +=

+= 2

63..

21

2..1

2

Slope of A, [ ] ( )babaa MMEIlMM

lEIl +=+= 2

621.

6

2

19

Now, the difference of slopes between any two points on an elastic curve of a beam is equal to the net area of the bending moment diagram between these two points divided by EI. At C, the section at which deflection is maximum, slope is zero,

Slope of A, ( ) ( )

+=

+= abaabaa MMlxxM

EIxMM

lxxM

EI 2.1

2.1

2

Equating the two,

( ) ( )baabaa MMEIlMM

lxxM

EI+=

+= 2

62.1

2

( ) ( )

( )( ) 03

2.2

23.62

2

22

=++

+=+

ab

ab

ab

a

baaba

MMMMlx

MMlMx

MMlMMxxlM

or ( )( ) 0810382108.

810882 22 =

++XxXXx or 03.277642 =+ xx ; x= 4.07 m

Deflection at A= ( ) ( )

+=

+= abaabaa MMlxM

EIxxxMM

lxxxM

EI23

632.

22..1

2

( ) mmormXXXXXXXX

76.51076.5810108

07.42108310126

07.4 3336

2=

+=

7.18

6 kN/m

A BC

5 m

4 kN.m4 kN.m

4 kN.m4 kN.m -+ wl2/8

As the loading is symmetrical, area of half the bending moment diagram can be considered. Moments of the bending moment diagram about A

= 8384

54

.2

.165.

8.

2.

32 242 MlwlllMlwll =

Deflection at A

= mmmXXXXX

MlwlEI

363.010363.08

54000384

56000510100

18384

51 3246

24

==

=

7.20

20

Let l be the length. Then maximum bending moment is 3000l at the fixed end (the strongest section of 40 mm diameter).

Thus mmlorXXl 1572040)64/(

300075 4 == Let d be the diameter at a distance x from the free end of cantilever. Then for uniform strength throughout,

xdordXXdx 407

2)64/(300075 34 ==

Moments of the bending moment at x about A= xx.3000

Deflection = == 1570

3/4

2

3/4

157

0

24

0

2 .407

291.0.3000.)64/(210000

1.30001 dxxxdxx

dXdxx

EI

l

mmXxXdxxX 265.0)4569(109.573/5

105.96.105.96 6157

0

3/56

157

0

3/26 ==

==

7.22

W

A B

B.M.

4l

Wl

l

4I

EIWl

54

ll

EIWl

45.2.

5

EIWl5

C D

EIWl

54

I

The simply supported beam is shown in the figure a. Reaction at A = Wl/5; Reaction at B = 4Wl/5 Bending moment diagram for the real beam is shown in Fig. b. Now, in the conjugate beam method, this diagram is to be considered as loading diagram and a new bending moment diagram is to be drawn which will give the deflection of the beam. For the same, first we need to find the reaction on the supports.

EIWlRorlXl

EIWllll

EIWllXR bb

2

56.0324.

24.

534.

2.

545 =

+

+=

EIWl

EIWll

EIWll

EIWlRa

22

24.056.024.

52.

54 =+=

Slope at A= shear force at A for conjugate beam

21

= radXX

XXEI

Wl 0014.01020205000

2000600024.024.0 622

== Deflection at midspan = bending moment at C for conjugate beam

mmXX

XEI

WlllXll

EIWllX

EIWl 5.5

10202050002000600047.047.0

35.2.

25.2

45.2.

55.224.0 6

332

===

For maximum deflection, the bending moment is to be maximum or shear force is to be zero. 7.23

W

A B

B.M.

l/2

Wl

l/2

Wl/2

I16I

EIWl2

EIWl

32EI

Wl16

C

Let moment of inertia of portion CB be 4)2/(

64dI =

For portion AC, IdIac 16644 == ,

The cantilever is shown in the figure a. Bending moment diagram for the real beam in Fig. b. the conjugate beam in Fig. c. The fixed end is transformed into a free end and the free end into a fixed end.

Slope at C= shear force at C for conjugate beam = EI

WllEI

WllEI

Wl1283

21.

2.

322.

32

2

=+

Slope at B= shear force at B for conjugate beam = EI

WllEI

WlEI

Wl12819

21.

2.

21283 22 =+

Deflection at C = bending moment at C for conjugate beam

= EI

WlllEI

WlllEI

Wl7685

321.

2.

324.

2.

32

2

=+ Deflection at B = bending moment at B for conjugate beam

22

= EI

WlEI

WlllEI

WlllEI

WlllEI

Wl38423

768)3259(

3.

21.

2.

265

21.

2.

3243.

2.

32

22

=++=++

Chapter-8

8.12

A B

15 kN

6m3m

20 kN

6m

8 kN/m

Fixing moment at each end for uniformly distributed load, 12

2wlM = (Example 8.2) Fixing moment for point load,

22

lbWaMb = and 2

2

lWabM a = . (Example 8.3)

By combination,

6.20712158

156920

1512315 2

2

2

2

2

=++= XXXXXM a kN.m

4.20012158

156920

1512315 2

2

2

2

2

=++= XXXXXMb kN.m Maximum bending moment = 207.6 kN.m

Section modulus = 66

10977.1105

104.200 XX = mm3 or 0.00198 m3 8.13

A B

5 m200 kN

4 m

80 kN/m

C

Fixing moment at each end for uniformly distributed load, 12

2wlM = (Example 8.2) Fixing moment for point load,

22

lbWaMb = and 2

2

lWabM a = . (Example 8.3)

By combination,

5.34212

9809

45200 22

2

=+= XXXM a kN.m (hogging)

23

1.29312

9809

45200 22

2

=+= XXXMb kN.m (hogging) To find reactions, take moments about B,

kNRorX

XXR aa 7.3411.2935.3422980

420092

==+ 3.1787.341200980 == XRb kN

8.15

W

l/2 l/2

A B

C

MaMc =

(Ma+ Mb)/2

Mb

Wl/4

As the slope at A is equal to slope at B =0, net area of the moment diagram must be zero, i.e.

bab

bba

a

baa

EIlWl

EIlWl

EIlMMM

EIlMMM

2.

4.

21

2.

4.

21

2.

22/)(

2.

22/)( +=+++++

or bab

baa

ba IWl

IWl

IlMM

IlMM

22)3()3(

22

+=+++

+=

++

+

babab

baa II

WlII

MII

M 112

3113

++=

+++

ba

ba

ba

baba II

IIWlIIIIMM

3233 (i)

Deflection of A relative to B is zero, so net moments of areas about B must be zero, i.e.

+

+=+

++

++

32.

22.

4.

21

31.

22.

2.

4.

21

6.

21

2.

2)(

4.

2.

2)(

2.

31

221.

2.

2)(

43.

2l

EIlWlll

EIlWll

EIlMM

lEIlMMll

EIlMMl

EIlM

bab

ab

b

ba

a

ab

a

a

24

bab

ab

b

ba

a

ab

a

a

IWl

IWl

IlMM

IlMM

IlMM

IlM

482448)(

16)(

12)(

83 332222 +=++++

+=

+++

+

babbab

bbaaa II

WlIII

MIIII

M211

6121

41

31

121

41

31

23

++=+

++ba

ba

ba

baba II

IIWlIIIIMM

7)2

27)(2 (ii)

From (i) and (ii),

+++=

baba

baaa IIII

IIIWlM14

)3(2 22

and

+++=

baba

babb IIII

IIIWlM14

)3(2 22

8.20

A B C

6 kN/m

5 m4 m

10 kN/m

I 2I

12

31.2525.5

5.625

18.375

30.1

19.9

8.22

AB C

Dw

l2ll The load diagram is shown in Fig. Applying the three moment equation for uniformly distributed loads to the spans AB and BC,

( )

+=+++

12

33

26

4)2(

4222

llEIlwwllXMllMlM accba

In this equation, ,0=aM and 0; == cbc MM (From symmetry of the beam, B and C have same load, so level of C is the

same as of B)

25

The above equation reduces to 22

43

329

lEIwlM ab

= (i)

Also, cbda RRRR == ; (From symmetry of the beam) Thus, badcba RRRRRRW 22 +=+++= or bba RwlRwlR =

= 2

24

and ba R=

(i) becomes, 22

4.3

329

lREIwlM bb

= (ii) Bending moment at B,

=

+=+=

82820

2

2 WRWlWRWlwllRMM bbaab (hogging)

or

=8

3WRlM bb (iii)

From (ii) and (iii)

2

22

4.3

329

812

lREIwlwlRl bb

=

+=+=+==

EIl

lwlEIl

wl

lEI

wl

Rorl

REIwlR bbb 3457

8348

57

43132

57

4.3

3257

3

3

3

4

3

3

++=

+=

+=

=

EIlEIlwl

EIllwl

EIllwllwXRWR ba

34487

8345716

8

3457

824

2

3

3

3

3

3

3

8.28

26

A B6 m 4 m

60kN/mC

+

+

_

180

8032.7

A

197.3

Maximum bending moment for the spans AB and BC treating as simply supported beams,

For span AB, 2708

6608

22

max === XwlM kN.m

For span BC, 1208

4608

22

max === XwlM kN.m For fixing moments, assume the continuous beam ABC to be made up of fixed beams

AB and BC.

For span AB: Fixing moments at A, 18012

66012

22

=== XwlM a kN.m

Fixing moments at B, 18012

66012

22

=== XwlMb kN.m

For span BC: Fixing moments at B, 8012

46012

22

=== XwlMb kN.m

Fixing moments at C, 8012

46012

22

=== XwlMc kN.m In span AB, moments at A and B due to sinking of support B by 8 mm,

2

6lEIM = (Example 8.7, -ve being counter-clockwise)

or 626

103.1090006

8104000002056 XXXXXM == N.mm or -109.3 kN.m Due to sinking of support B by 8 mm, moments at B and C (Refer example 8.7),

2

6lEIM = (-ve being counter-clockwise)

or 626

102460004

8104000002056 XXXXXM == N.mm or 246 kN.m Distribution factors at B:

27

Stiffness factor for AB, 3/26

4 EIEIsba == (as the beam is fixed at A)

and for BC, 4

34

3 EIEIsbc == (as the beam is simply supported at C)

Distribution factor for AB, 178

4/33/23/2 =+=+= EIEI

EIss

skbcab

abba

Distribution factor for BC, 179

4/33/24/3 =+=+= EIEI

EIss

skbcab

bcbc

Distribution factors 16/31 15/31 Fixed end moments Moments due to sinking

-180 180 -109.3 -109.3

-80 80 246 246

Initial moments Release C Carry over

-289.3 70.7

166 326 -326 -163

Net moments Distribute Carry over

-289.3 70.7 -38 -19

3 0 -35.7

Final moments -197.3 32.7 -32.7 0

Chapter-9 9.16 M =WR (1-cos )

[ ] [ ] dEI

RWdRWREIEI

dsMU .cos1.)cos1(2

122 0

232

0

22 ===

Vertical deflection = [ ] ( ) dEIWRd

EIWR

Wu .cos2cos12.cos12

0

23

0

23 ++==

0

3

0

3

sin242sin

22.cos2

22cos112

+++=

+++= EIWRdEIWR

68.3)12/612(000200

150533002

23

333

===

+++=XX

XXXEIWR

EIWR mm

9.8

28

20 kN

20 kN

100 m

m

80mm

P Q

d = 80 mm ; R = 50+40 = 90 mm ; A = 5.502680

42 =X mm2

5.4399080.

1281

1680...

12816 242

2

422 =+=++=

Rddp mm2

As bending moment tends to increase the curvature, it is positive. Resultant stress = Direct stress +Bending stress Stress at outside face (P), y = 290 - 202=88 mm

409040

5.43990.

5.502620000.1

2

2

2

2

2

+=+=

+++= yR

ypR

AW

yRy

pR

ARWR

AW

o 56.22= MPa (tensile) Stress at inside face (Q)

yRy

pR

AW

yRy

pR

ARWR

AW

i =

+= ..1 2

2

2

2

67.584090

4082.105 == X MPa (compressive) Chapter-10

10.12

TNP60

2= , If P is constant, T

N 1

But 16

3dT = , If stress is same,

3dT ; 31dN or 3/11

Nd , 64.4

505000 3/1 =

=

b

a

dd

10.16

aa

aa

ss

ss

JGlT

JGlT ==

Angle of twist per unit length,

29

aa

a

ss

s

JGT

JGT = or 444 60

327.0)60(2

32XXGTX

dXXGT

aa = d =50.5 mm

Also, asast ll

=

7.0 , ast GrGr

=

7.0

As r is same, 4.17.027.0 === XGG

a

st

a

st

As maximum limits are 65 MPa for alloy and 80 MPa for steel, the maximum stress will reach in steel first. 10.18 Maximum shear stress occurs at smallest diameter.

MPaorXX 5.99.16801010

36 ==

( ) ( ) radXX

rrrrrl

GT 0171.0

401

601

101200.

8200031010

'1

'1

'.

3 336

33 =

=

+=

= 0.98o

Chapter-11

11.13

Under axial torque, 3704.012000200

1210020000646444 === XXXX

EdTDn rad

Increase in the number of turns = 059.023704.0 =

Bending stress, 9.11712200003232

33 == XX

dT

MPa Torsional stiffness (torque /rad) = 54

3704.020 = N.m/rad

11.17

4

38Gd

nWD= or 441.220)5.28(8

5.2000808 3

4

3

4

1 ==== XXXX

nDGdWs N/mm

'302.115)'8(8'00080

8 34

3

4

2 dXdXXd

nDGdWs ====

21

21

sssss += or mmdordXdord

dX 136.2''302.1441.2'693.1173.3'302.1441.2'302.1441.23.1 ==++=

The maximum load will be that taken by the spring with smaller diameter.

30

11.18 E =2G(1+) =2G(1+0.3)=2.6G

For a closed-coiled spring, 438

GdnWD

c =

and for an open-coiled spring,

+=

EGdnWD

o

22

4

2 sin2coscos.

8

Now, 02.0=o

io

or 02.0

sin2coscos.

8

8

122

4

2

4

2

=

+

EGd

nWDGd

nWD

975.0sin2cos6.2

cos6.2025.0sin2cos

cos1 2222 =+=+

GGGor

GEE

or 975.02cos6.0

cos6.2975.0)cos1(2cos6.2

cos6.2222 =+=+

or

02cos6.2cos585.0975.02cos6.0

cos6.2 22 =+=+ or

On solving, o2.17= 11.20

'RE

RE

IM

y== (Eq. 11.14)

At proof load, R is infinite and thus E/R = 0

RE

y= or

R000208

2/10630 = or R = 1650 mm or 1.65 m

Let We be the equivalent static load which produces the same maximum stress and deflection as the impact load.

also, 223

nbtlWe=

As the maximum stress is one half of the proof stress, 21080101400.

23

2630

XXXWe=

or 12000=eW N 2.74

000208108010140012000.

83

83

3

3

3

3

===XXX

XEnbt

lWe mm

Chapter-12

31

12.16 Assuming the deflected form under the action of the crippling load be

=

=400

cos12

cos1 xalxay and thus,

400sin

400xa

dxdy =

dxx

adxxadxdxdyl .

2200

cos1

400.

400sin

400.

180

0

2180

0

22

0

2

=

=

22180

0

2

17006.02004002

1200

sin2004002

1 aXaxxa =

=

=

Also, 400

cos400

cos1)( xPaxPaPayaPM =

==

+= 200100

2

2100

01

222

0

2400

cos400

cos

22dx

I

x

dxI

x

EaPdx

EIMl

12.18

( ) 95100445664

44 ==I mm4; ( ) 30044564

22 ==A mm2

317300

951002 ===AIk mm2

Euler load, 1261122200

951000002072

2

2

2

=== XXlEIPe N

Actual load for failure = 126112 X 0.75 = 94584 N

77.49.77sec36.1sec95100000207

945842

2200sec2

sec2

sec ===== oXEI

Pll

+=

+= 77.4317

28130094584

2sec..1 2max X

eXlkye

AP c

MPa

or ( )e421.0136.100290 += or 1 + 0.421e = 2.89 or e = 4.49 mm 12.21 w = 0.8 kN/m = 0.8 N/mm;

( ) 400004064

4 ==I mm4; ( ) 400404

2 ==A mm2

968.140000000200

000362

320022

===

XEIPll rad (=109.7o)

32

( )17.109sec00036

400000002008.012

sec =

= oXXlP

wEIM 66 10215.2)1967.2(105585.0 XXX == N.mm

7.3527.2840000

2010215.2400

00036. 6max ===

XXIyM

AP c

- 381.4 MPa (compressive) and -324 MPa compressive

Chapter-13

13.13

Internal volume of cylinder = 62 107.70314008004

XXX = mm3 180

1028005.4

2===

XX

tpd

c MPa (tensile) and 904 == tpd

l MPa (tensile) Increase in volume, [ ]).(2).().2( lcclcl E

VVv +=+=

[ ] )360450(2050000

1500800)4/()90180(2)18090(1012452

3 =+= XXXXEVX

340360450)360450(205000

107541012506

3 == orXX 306.0=

13.16

Internal volume of shell = 767.15.16

3 =X m3 = 6101767X mm3 For tube:

Volumetric strain = 3 X hoop strain = ( ) ( ) == 14313

tEpd

E (13.9)

Decrease in volume, ( ) ( ) tEVVtEXXV

tEpd /15753.01

41500231

43 ===

For water: Increase in volume, V

KpVv w .. ==

Thus change volume of water VKp

tEV += 1575

or 63 1017672000

2000200

1575104000 XXt

X

+= or t=6.23mm

13.20

33

Maximum circumferential stress in a tube subjected to internal pressure p is at the inner

surface and is pdddd

io

ioc .22

22

+= or pc 57.3150200

15020022

22

=+=

Thus 4457.3 =p or p =12.32 MPa Longitudinal stress =

)(.

22

2

io

i

rrpr

= pXp 2857.1

7510075

22

2

=

)150200(

4

2200002875.122

=X

p or p =12.43 MPa

Maximum pressure =12.43 MPa 13.22

The maximum hoop stress is at d=di, iio

ioc pdd

dd .2222

+=

or 35606055 22

22

Xdd

o

o

+= or )60(

553560 2222 += oo dd

or )60(7601111 2222 += oo dXd or 8.84=od mm Maximum shear stress,

)( 222

io

oi

dddp= or )60(

3550 222

= oo

dd

or 222 7601010 oo dXd = or 5.109=od mm Thickness = 75.24

2605.109 = mm

13.25 (i) Increase of inner diameter of the collar

[ ]E

pdEpdpp

Edd

Ed rc 425.23.0125.2.

150250150250. 22

22

=+=

++=+=

or 000200150425.21.0 pX= or p = 55 MPa

13.27

)(2)2(

33

33

io

ioi

ddddp

+= or 30

)80(280275 33

33

Xd

Xd

o

o

+=

or )80(5802 3333 =+ oo dXd or 33 8074 Xdo = or do = 96.4 mm t = 16.4/2=8.2 mm

34

Chapter-14 14.12 Let the shrinkage pressure between the disc and the shaft at stand still be p.

09.0=iR m and 4.0=oR m Shrink allowance = 08.0 mm

At stand still, the hollow disc acts similar to a thick cylinder subjected to internal pressure (Refer section 13.9).

Shrinkage allowance (initial difference in radii))(

222

2

io

oi

RRERpR= (Eq. 13.31)

or )90400(000206

40090208.0 222

=XpX or p = 86.9 MPa

Maximum hoop stress is at the inner radius and is given by Eq. 13.7.

2.969.86107.1107.1.09.04.009.04.0. 22

22

22

22

===+=

+= XpppRRRR

iio

io MPa

(ii) When the disc rotates and the shrink fit loosens, radial pressure is zero and thus

the radial stress is also zero.

[ ]222 )3()1(4 oi

RR ++= (Eq. 14.17)

14.13

1006030002 == X

At stand still Let the shrinkage pressure between the disc and the shaft at stand still be p.

At stand still, the hollow disc acts similar to a thick cylinder subjected to internal pressure only and thus the results of the same may be used. - For the outer disc, hoop stress at the inner radius (50 mm),

08.0=iR m and 24.0=oR m ppp

RRRR

iio

io 25.1.08.024.008.024.0. 22

22

22

22

=+=

+= N/mm2

Hoop strain = ss Ep

Epp 55.13.025.1 =+

14.15 Let t be the thickness of the disc at radius r. Now as 2/..

22

. reAt = At r = 0, AeA == 2/0.. 22.180 (i)

35

At r = 0.2 m, 2/2.0..22

. = eAt (ii) Dividing (ii) and (i),

2/)04.0(.2/2.0..

222

.8.180

== eA

eAXt

where 1234.0101202

04.0603000275002/04.0 6

22 =

=

XXXXXX

1598 1234.0 == et mm

Chapter-15

15.6

62

3

2

3

108.61)3.01(12

15000200)1(12

XXEtC ===

=

=

2300

108.61324.0

232

22

6

222

2 xXX

xxRC

wxy

Deflections at different cross-sections are tabulated below:

x (mm) 0 60 120 180 240 300 y (mm) 0 0.064 0.24 0.48 0.72 0.82

0.9

0.30.6

r (mm)0 12060 240180 300

Figure shows the profile of the deflected plate.

[ ] [ ]222223 22 )3()1(.83)3()1()1(12.16.1 )2/( xRtwxREtwEtx ++=++= [ ] [ ]2222 3.3117000000667.0)3.03(300)3.01(154.0.83 xxX =++=

[ ] [ ]222223 22 )13()1(83)13()3()1(12.16.1 )2/( xRtwxREtwEtz ++=++= [ ] [ ]2222 9.1000117000667.0)13.03(300)3.01(158 4.03 xxXXX =++=

15.8

36

62

3

2

3

1026.2)28.01(12

5000200)1(12

XXEtC ===

++= )1(log

11log2

21

8

2

xRC

Pxy

++= )1(log

28.0128.01100log2

21

1026.282000

6

2

xXXx

[ ]xxX log773.101021.35 26 =

Chapter-16

16.9

60

80

10

60

y'

AN

10

A'N' y10

(a) (b)

.45

(mm)

Let N-A and N-A be the elastic and plastic neutral axes respectively.

=++=

1070106045107051060

XXXXXXy 26.54 mm

I = 23

23

)54.2645(701012

7010)554.26(106012

1060 +++ XXXXXX =807.76X103 mm4 5000 278383 285833 238540 Yield will start at the bottom edge,

Moment of resistance at first yield, My= yyyX

yI 15110.

)54.2680(1076.807.

3

== In fully plastic state, the neutral axis divides the total area of the beam into two

equal parts.

16.12

STRESS

30

72

12

24

(mm)

280 MPa

Let Wy be the load at first yield at the mid-span,

37

322

10725828067230.

6XXXbhM yy === N.mm or 7258 N.m

Thus, 72584

2.1 =XWy or Wy = 24.2 X103 N or 24.2 kN

For elastoplastic state, moment of resistance,

=

34.

22 ahbM y

=

324

47230280

22

X =9274 X 103 N.mm or 9274 N.m

or 92744

2.1 =XWy or Wy = 30.9 X103 N or 30.9 kN For this required depth of yield, let the yield occur beyond a cross-section at

distance x m from one end of the beam. Then this cross-section will be having the first yield and thus the moment of resistance will be equal to the first yield at the midspan i.e. 7258X103 N.mm.

16.13

120 mm

200mm15mm

15 mm

y

50mm

yyy

150

.

yyy

150)15(

y

150-y

(a) (b) Solution As the load increases and the yield spreads upto 50 mm from the lower edge of web, assume that the top of the flange is still in the elastic state. Let the neutral axis be at a distance y from the top edge. The stress distribution in the beam section will be as shown in Fig. b.

Stress at the top of flange, t = ..150 yy

y

Stress at the bottom of flange, b = ..150

15yy

y

Total tensile stress force 15)15.(2

)15120.(2

++= yX bbt

38

+=

+++=

+

+=

yyy

yyyyy

yy

yy

yy

y

y

y

yy

1505.1181215755.7

1505.16872255.713500900900

)15(150

155.7150

15150

900

2

2

Compressive force = )5.71875(1550)150(15.2

yXXyX yyy =+

16.14

10 mm

100 mm

80mm

AN

10 mm

y

y(a) (b)

y

y

Solution Let the neutral axis be at a distance y from the top edge. The yield starts at the bottom of the web and will cover some portion when the top of the flange will just yield. The stress distribution in the beam section is shown in Fig. b. Let the stress at the bottom of the flange be when the top of flange just yields. Then yy

y .15=

Positive force = yyX yy 545050010)10.(

2)10100.(

2++=++

Negative force = yyyy yXyy 800.1510)280(10.

2+=+

Chapter-17

17.8

A B

C D

E F

6 m

3 m

3 m

50 kN

AB

C D

E F

6 m

3 m

3 m

50 kN

50 kN 50 kN

AB

C D

E F 50 kN

50 kN

50 kN 50 kN

39

To find the reaction at support B, take moments about A,

6506 XXRb = or 50=bR kN (upwards); 50=aR kN (downwards) 50=ahR kN

oABC 6.26)6/3(tan == Joint A: kNFAB 50= (T) )(50 TFAC = kN, Joint B Considering FH = 0, 06.26cos50 =+ oBCF or kNFBC 9.55= (C) Considering FV = 0, 06.26sin9.5550 =+ BDo F or 25=BDF kN (compressive) 17.9

A B

C D

E F

2 m

2 m

2 m

50 kN

2 m

100 kN

H

G

A

B

C D

E F

2 m

2 m

2 m

50 kN

2 m

100 kN

H

G100 kN

50 kN

F EG

F CE

E

G100F GD

F GF =o

FH= 0D

FV= 0D

50 kN

100 kNG100

F GD

F GF =o

FV= 0

To find the reaction at support B, take moments about A, 45041004 XXXRb = or 50=bR kN (upwards); 0=avR

100=ahR kN Joint E: kNFCE 50= (C) 100=EGF kN (C) by judgement Joint F: 0== FDFG FF (C) by judgement Joint G Considering forces along GD,

FGD = 0, 045cos =+ oGEGD FF or 045cos100 =+ oGDF or kNFGD 7.70= (C) Considering FV = 0, 045cos45cos = oGCoGD FF or 7.70=GCF (T)

40

17.11

200 kN

C

D

E

A

F

B

4 m 4 m

2 m

2 m

200 kN

C

D

E

A

F

B

200

447.2

400

200

400

400

400

0

0

447.1200

0 Taking moments about F,

82004 XXFAH = or 400=AHF kN (inwards); 400=FHF kN (outwards) Joint D: 0=EDF as there is no other horizontal member to balance the force. 200=CDF kN (C) tan-1(4/2)= CBE or oCBE 43.63= Joint C: 20043.63cos =oBCF or 2.447=BCF kN (C)

CEo F=43.63sin2.447 or 400=ECF kN (T)

Joint B: BEo F=43.63cos2.447 or 200=BEF kN (T)

ABo F=43.63sin2.447 or 400=ABF kN (C)

Joint A: 0=AEF 0=AVF Joint F: 200=FVF 40043.63sin =oEFF or 1.447=EFF kN (T) 17.12

DCBA

EFG 40 kN

80 kN40 kN

60o 60o

DCBA

EFG 40 kN

80 kN40 kN

40 kN

Taking moments about F, 464.34068024012 XXXXFDV ++= or 21.58=DVF kN; 40=DHF kN

8.612.58120 ==AVF kN Joint D: 2.6721.5860sin == EDoED ForXF kN (C) 6.3360cos2.67 == oCDF kN (T)