Shear stresses in the bending of thin, unsymmetrical sections

sds

dx t(s)

consider an element of a thin beam, as shown

Vy

Vz

y

z

x

My

Mz

dxds

q tτ=

q dq+

xxxx

d dxdxσσ +

xxσ

dA

( )

0

0

x

xxxx xx

xx

F

dq dq dx qdx dx dA dAdx

ddq dAdx

σσ σ

σ

=

⎛ ⎞+ − + + − =⎜ ⎟⎝ ⎠

= −

∑

If we integrate from s1 to s2

s1

s2

A

( ) ( )

2

1

2 1

sxx

s A

xx

A

ddq dAdx

dq s q s q dAdx

σ

σ

= −

− = Δ = −

∫ ∫

∫

( )1q s

( )2q s

Since

( ) ( )( )2

y zz z yz z yy y yzxx

yy zz yz

M I M I z M I M I y

I I Iσ

+ − +=

−

( )2

y yz zzz yz yy yz

xx

yy zz yz

dM dMdM dMI I z I I ydx dx dx dxd

dx I I Iσ

⎛ ⎞ ⎛ ⎞+ − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=−

zVzVyV− yV−

( ) ( )( )2

z zz y yz z yz y yyxx

yy zz yz

V I V I z V I V I yddx I I Iσ − − −

=−

placing this in our shear flow expression gives

( ) ( )z yz y yy z y yz z zz yV I V I Q V I V I Qq

D− + −

Δ =

where2

yy zz yzD I I I= −

zA

yA

Q ydA

Q zdA

=

=

∫

∫

If we have a symmetrical expression and if 0yV =

z y

yy

V Qq

I−

Δ =

If q(s1) =0 then

( ) ( )1 1q s t sτΔ =

and we obtain z y

yy

V QI t

τ−

=

the minus sign is here since we have taken + zV

Note: in our expression for Δq the q flows "out" from the end of the section under consideration, whether the section is cut from the left or right of the cross section. At the beginning of the section the q flows "in".Examples:

( )1q s1s 1s( )1q s

( )0q( )0q

in both these cases

AA

( ) ( ) ( ) ( )1 0 z yz y yy z y yz z zz yV I V I Q V I V I Q

q q s qD

− + −Δ = − =

Just as in the torsion of closed sections, the shear flows generated by bending must be conserved at a junction. This follows from equilibrium:

1q

3q

2q

xxxx

d dxdxσσ +xxσ

dA

dx

( )

( )

1 2 3

1 2 3

0

0

0

x

xxxx xx

xx

F

dq q q dx dx dA dAdx

dAq q q ddx

σσ σ

σ

=

⎛ ⎞+ + + + − =⎜ ⎟⎝ ⎠

+ + + =

∑

so as 0dA→ 1 2 3 0q q q+ + =

i.e. the net shear flow out of (or into) a junction must be zero

z

yeS

1000 lb

3 5

3 5

10

all dimensions are in inches

A 1000 lb load produces bending but no twisting in the thin open section shown below. Determine the shear flow distributions in this section and the location of the shear center. Assume the thickness of all sections is 0.1 in. All dimensions are measured to the mid planes of the walls.

z

y

53

5

10

3

A B C

DE F

cy

1y

2y

1A

2A

3A

( ) ( )

( )( )( )( )( )( )( ) ( )( )

1 1 3

1 3

2 022 8 0.1 1

2 8 0.1 10 .10.6154

c

y A Ay

A A

in

× +=

+

=+

=

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

4

1 12 8 0.1 8 0.1 5 0.1 1012 12

48.33

yyI

in

⎡ ⎤= × + +⎢ ⎥⎣ ⎦=

negligible

1000 , 0, 0z y yzV lb V I= = =

so we have

z y

yy

V Qq

I−

Δ =

1000 20.6948.33 y yq Q Q−

Δ = = −

for AB

z

y5

( )1q s0q = 1s

( ) ( )( )( )1 10 20.69 0.1 5q q s sΔ = − = − ⎡ ⎤⎣ ⎦

( )1 110.35 /q s s lb in= −

A B

1 31.03 /Bq lb in= −

3

for BC

z

y5

( )2q s0q =2s

( ) ( )( )( )2 20 20.69 0.1 5q q s sΔ = − = − ⎡ ⎤⎣ ⎦

( )2 210.35q s s= −

B C

2 51.75 /Bq lb in= −

5

for BD

y

z5

3s

( )3q s

3Bq

( ) ( )( )( )33 3 320.69 0.1 5 / 2Bq q s q s sΔ = − = − −⎡ ⎤⎣ ⎦

( ) 3 23 3 310.35 1.035Bq s q s s= − +

31.03 51.75

3Bq

3

3

51.75 31.03 0

82.78 /B

B

q

q lb in

+ − =

= −

( ) 23 3 382.78 10.35 1.035q s s s= − − +

At the middle ( )5 109 /q lb in= −

-82.78

-82.78

-109 ( ) 23 3 382.78 10.35 1.035q s s s= − − +

For the lower flange is the same as for the upper flange but with opposite sign so we have

yQ

D F5

+51.75

E D3

+31.03

10

53

the entire shear stressdistribution:

10

53

the total forces carried by each section:

R1

R1

R2

R2

R3

( )

3

1 1 10

5

2 2 20

102

3 3 3 30

10.35 46.58

10.35 129.38

82.78 10.35 1.035 1000

R s ds lb

R s ds lb

R s s ds lb

= =

= =

= + − =

∫

∫

∫

Note: we have used the negative of the shear flow expressions to agree with the directions assumed for the forces shown

10

53

R1

R1

R2

R2

R310

53

to find the shear center

1000 lb

e

We obtained the shear flow distributions under the assumption that there was only bending. These shear flows will generate the shear forces and their moments about a particular point (i.e. the force locations). Thus, we can determine the shear center here by requiring that the moment of the 1000 lb load about any point produce the same moments as the shear flows,

1000OM e=∑

OO

( )( )2 110 10 129.38 46.58 10828

OM R Rin lb

= − = −

= −∑

1000 8280.828

ee in==

location of the shear center by sectorial area

1s

2s

3s

e OS

A C

ED

F

5

5

take the starting point for the integration at O

BD 0 1esω ω= +

B

AC 0 25 5e sω ω= + +

EF 0 35 5e sω ω= − +

0dAω =∫

( ) ( ) ( )5 3 5

0 1 1 0 2 2 0 3 35 5 3

5 5 5 5 0t es ds e s ds e s dsω ω ω− − −

⎡ ⎤+ + + + + − + =⎢ ⎥

⎣ ⎦∫ ∫ ∫

odd

0 0 0

0

10 8 8 0

0

ω ω ω

ω

+ + =

=

so the sectorial area distribution looks like:

+

-

+

-

Thus, 0y dAω =∫automatically and we need only consider

0z dAω =∫

z1s

0z dAω =∫

( ) ( ) ( ) ( )5 3 5

1 1 1 2 2 3 35 5 3

5 5 5 5 5 5 0t es s ds e s ds e s ds+

− − −

⎡ ⎤+ + + − − + =⎢ ⎥

⎣ ⎦∫ ∫ ∫

1z s= 5z = 5z = −

which gives 250 400 400 03

400 0.828400 250 / 3

e e

e in

+ − =

= =+