Bending and Torsion of Thin, Open Sections

If the shear forces and their locations are known, then we can calculate the shear flows assuming there is only bending produced, and from those shear flows (or through use of the sectorial area function) determine the location of the shear center.

Vy

Vz

yd

zd

Then the torsion induced by the shear forces can be calculated:

VyVz

yd

zd

bending + torsion = bending only + torsion only

VyVz

ye

zeVy

Vz

T

( ) ( )z y y y z zT V e d V e d= − + −

and we can solve for the shear stresses generated by T and superimpose them on the bending stresses to get the total shear stresses

As an example consider our previous problem but now let the 1000 lb load act through the vertical web:

10O O O

1000 lb

3 5

=

1000 lb

+0.828

T = 828.2 in-lb

To determine the maximum shearing stress and its location (neglecting stress concentrations), for the torsional part we have

( )( ) ( )( )3 3

4

1 12 8 0.1 10 0.13 3

0.008666

effJ

in

= × +

=

( )( )max

828.2 0.19556

0.008666max

eff

Tt psiJ

τ = = =

Now, the maximum shear stress will occur in the vertical web at the center where the shear flow is a maximum

1000 lb

0.828 max 109 /q lb in=

max max /1090q t

psiτ =

=

bending torsion

T = 828.2 in-lb

max 9556 psiτ =

so at point O on the right side we will have

O O

max 1090 9556 10,646 psiτ = + =

O

8466 psiτ =

Bending and Torsion of a Thin Closed Section(single cell)

Pq

Vz

Vy

dz

dy

Note that the torsion that the shear forces generate will induce a constant shear flow qT as shown. However in the bending induced, the shear flow is only known up to a constant since we have

PBq

Vz

Vy

ez

eyP=

bendingtorsion

+ T

Tq

( ) ( ) ( ) ( )0 z yz y yy z y yz z zz y

B B

V I V I Q V I V I Qq s q

D− + −

= +

and, unlike open sections, we cannot find an end where qB = 0

Thus, the total shear flow will be determined only to within a constant (which we must determine) that is partly due to bending and partly due to torsion.

Bending and Torsion of a Thin Closed Section(single cell)

Pr⊥

q

Ω = area contained within the centerline of the cross section

Vz

Vy

dz

dy

P y y z zC

M V d V d q r ds⊥= + =∑ ∫

( ) ( ) ( )c y zq s q V f s V g s= + +

( )

( )

2

2

yz y yy z

yy zz yz

yz z zz y

yy zz yz

I Q I Qf s

I I I

I Q I Qg s

I I I

−=

−

−=

−

this is the unknown constant shear flow due to both bending and torsion

Placing the q expression into the moment equation gives:

( ) ( )2P c y zC

M q V f s V g s r ds⊥⎡ ⎤= Ω + +⎣ ⎦∑ ∫

Also, recall we have a relationship between the shear flow and the twist/unit length induced by that shear flow

12 C

qdsG t

φ′ =Ω ∫

(1)

(2)

1. If the shear forces and their positions are known, then qc can be found directly from Eq.(1) since the left hand side of that equation is known explicitly and f and g can be found for the given geometry. Then Eq. (2) can be used (since q is now given completely) to find φ'

2. If the shear forces are known but assumed to act through the shear center (whose position is unknown), we can set φ' =0, Vy = 0 and solve Eq.(2) for the unknown qc . Then Eq.(1) gives the location of the shear center, dz , since

z zC

V d q r ds⊥= ∫

( ) ( )c zq s q V g s= +

We can repeat this process by setting φ' =0, Vz = 0 and solving Eq.(2) again for a new qc . The Eq.(1) gives the location of the shear center, dy :

( ) ( )c yq s q V f s= +

y yC

V d q r ds⊥= ∫

Bending and Torsion of a Thin Closed Section(multiple cells)

P

Vz

Vy

dz

dy

As in the case of the torsion of a multiple cell closed section, we need to account for unknown constant shear flows in each cell. This can be accomplished by conceptually decomposing our problem into two simpler problems as shown on the next page.

P

Vz

Vy

dz

dy

q= q1

q=q2

=

q= 0

q=0

+

constant shear flows from constant parts of q due to bending and constant q's due to torsion.

bending shear flows q(s) due to this "open" section

1q 2q

( ) ( )2

1 1 2 21

2 2m

y y z z y zm C

V d V d q q V f s V g s r ds⊥=

⎡ ⎤+ = Ω + Ω + +⎣ ⎦∑ ∫

1

2

1

2

12

12

C

C

q dsG t

q dsG t

φ

φ

′ =Ω

′ =Ω

∫

∫

(1)

(2)

(3)

q= 0

q=0

+1q 2q

1. If the shear forces and their locations are known, then q1 and q2are first found in terms of the unknown φ' from Eqs. (2) and (3). These qm 's are then placed into Eq.(1) which is solved for the unknown φ' . Once φ' is known in this manner, the qm 's are completely determined.

( ) ( )2

1 1 2 21

2 2m

y y z z y zm C

V d V d q q V f s V g s r ds⊥=

⎡ ⎤+ = Ω + Ω + +⎣ ⎦∑ ∫

1

2

1

2

12

12

C

C

q dsG t

q dsG t

φ

φ

′ =Ω

′ =Ω

∫

∫

(1)

(2)

(3)

2. If the shear forces are known but assumed to act through the shear center (whose position is unknown), we can set φ' =0, Vy = 0 and solve Eqs. (2) and (3) for the unknowns q1 and q2 . Then Eq.(1) gives the location of the shear center, dz , since

( )2

1 1 2 21

2 2m

z z zm C

V d q q V g s r ds⊥=

= Ω + Ω + ⎡ ⎤⎣ ⎦∑ ∫

We can repeat this process by setting φ' =0, Vz = 0 and solving Eqs.(2) and (3) again for new values q1 and q2 . Then Eq.(1) gives the location of the shear center, dy :

( )2

1 1 2 21

2 2m

y y ym C

V d q q V f s r ds⊥=

⎡ ⎤= Ω + Ω + ⎣ ⎦∑ ∫

Example; Calculate the shear flows in the walls of the box beam shown. The beam has a uniform wall thickness of t = 0.1 in. all dimensions are in inches

1000 lb

5.21

3.5C

9

12

4y

z

32

1

4

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1 1 1

2 2 2

3 3 3

4 4 1

1.433 18.44 4.074

? ? ?

6.975 24.43 4.689

14.70 15.15 3.318

yy zz yz

yy zz yz

yy zz yz

yy zz yz

I I I

I I I

I I I

I I I

= = = −

= = =

= = = −

= = = −

To find the I’s for 2

5.21

6.79

0.5

512 13

sds

50.513

126.7913

z s

y s

= +

= −

yz

( )

( ) ( )

132 2

0

132

04

0.5 5 /13 0.1

14.41

yyI z dA

s ds

in

=

= +

=

∫

∫

similarly ( )

( )

132 2 4

0

132 4

0

16.41

3.419

zz

yz

I y dA in

I yzdA in

= =

= = −

∫

∫

( )

( )

( )

44

14

4

14

4

1

37.52

74.43

15.50

myy yy

m

mzz zz

m

myz yz

m

I I in

I I in

I I in

=

=

=

= =

= =

= = −

∑

∑

∑

so total areamoments are

1000 , 0z yV lb V= =

( )

( )( )( )( ) ( )

2

2

74.43 15.5 1000

37.52 74.43 15.5

zz y yz z z

yy zz yz

y z

I Q I Q Vq

I I I

Q Q

− −∆ =

−

− −=

−

29.16 6.073y zq Q Q∆ = − −

Now, use this relation for each section

6.79

3.5s

1q

y

z

( ) ( ) ( )( )

( ) ( ) ( )( )

1

2

1

3.5 / 2 0.1

0.35 0.05

6.79 0.1

0.679

y

z

Q zdA s s

s s

Q ydA s

s

= = − −

= − +

= =

=

∫

∫

( ) ( ) [ ]

( ) ( )

1 21

21

11

29.16 0.35 0.05 6.073 0.679

6.082 1.458 /

4 1 /

q s q s s s

q s s lb in

q q lb in

⎡ ⎤= − − + −⎣ ⎦= + −

= +

z

z A

Ay

6.79

1 1q +

y

z

512

s/2s/2

yz 0.5

6.79 12/ 2 130.5 5

/ 2 13

ys

zs

−=

−=

13

66.791350.526

y s

z s

= −

= +

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

2 2

2 2

0.5 5 / 26 0.1 0.05 0.1923

6.79 6 /13 0.1 0.679 0.04615y

z

Q s s s s

Q s s s s

= + = +

= − = −

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

2 2 21

2 21

21

1 29.16 0.05 0.01923 6.073 0.679 0.04615

1 5.58 0.2802 /

13 118.9 /

q s q s s s s

q s q s s lb in

q q lb in

⎡ ⎤ ⎡ ⎤= + − + − −⎣ ⎦ ⎣ ⎦

= + − −

= −

s

1 118.9q −

y5.21

5.5

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

3 2

3

5.5 / 2 0.1 0.55 0.05

5.21 0.1 0.521y

z

Q s s s s

Q s s

= − = −

= − = −

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

3 21

3 21

31

118.9 29.16 0.55 0.05 6.073 0.521

118.9 12.87 1.458 /

9 116.6 /

q s q s s s

q s q s s lb in

q q lb in

⎡ ⎤= − − − − −⎣ ⎦

= − − +

= −

z

s

1 116.6q −

y5.21

z

3.5

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

4

4 2

3.5 0.1 0.35

5.21 / 2 0.1 0.521 0.05y

z

Q s s

Q s s s s

= − = −

= − − = − +

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

4 21

4 21

41 1

116.6 29.16 0.35 6.073 0.521 0.05

116.6 13.37 0.3036 /

12 116.6 160.44 43.72

q s q s s s

q s q s s lb in

q q q

⎡ ⎤= − − − − − +⎣ ⎦

= − + −

= − + − =

1q

(approximately)

total force in each wall

( ) ( )

( ) ( )

( ) ( )

( ) ( )

41 2

1 10

42 2

1 10

43 2

1 10

44 2

1 10

6.082 1.458 4 17.55

1 5.58 0.2802 13 663.7

118.9 12.87 1.458 9 1237

116.6 13.37 0.3036 12 611.4

F q s s ds q lb

F q s s ds q lb

F q s s ds q lb

F q s s ds q lb

= + − = +

= + − − = −

= − − + = −

= − + − = +

∫

∫

∫

∫

19 1237q −

112 611.4q −

14 17.55q +

113 663.7q −

9

12

A

19 1237q −

112 611.4q −

14 17.55q +

113 663.7q −

9

12

A

1000 lb

to find q1 we must equate moment about any point generated by the shear flows to the moment due to the applied load

=

A

( )( ) ( )( )1 1

1

0

2 116.4 9 4 17.55 12 033.92 /

AM

q qq lb in

=

− + + =

=

∑

then the shear flows are explicitly

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

1 2

2 2

3 2

4 2

33.9 6.08 1.46 /

34.9 5.58 0.208 /

85.0 12.9 1.46 /

82.7 13.4 0.304 /

q s s s lb in

q s s s lb in

q s s s lb in

q s s s lb in

= + −

= − −

= − − +

= − + −

-85

-82.7

-113(midpoint)

34.9

33.9

40.2 (midpoint)

-85

34.9

-13.2 (midpoint)

-82.7

33.9

-13.5 (midpoint)

Now, find the location of the shear center for the vertical load

We need to set1 0

2q ds

G tφ′ = =

Ω ∫and find the q1 that meets this requirement

Since t is a constant here we have equivalently

0qds =∫which is just the sum of the forces we already computed

( ) ( ) ( ) ( )1 1 1 14 17.55 13 663.7 9 1237 2 611.4 0q q q q+ + − + − + − =

which gives

1 65.65 /q lb in=

19 1237q −

112 611.4q −

14 17.55q +

113 663.7q −

9

12

A1000 lb

=

A

e

1 65.65 /q lb in=

( )( ) ( )( )1 1

1000

4 17.55 12 12 611.4 9 10004.95

AM e

q q ee in

=

+ + − =

=

∑

If we want to find the vertical location of the shear center we would have to solve this problem for a horizontal applied shear force

1000 lb

= +

1000 lb

4.95

bending torsion

T

shear flow due to torsion (constant around the whole cross section)

( )( ) ( )( )4950 31.7 /

2 2 4 12 5 12 / 2Tq lb in= = =Ω +⎡ ⎤⎣ ⎦

Thus, the q1 in our original problem is due partially to the bending part of the solution, partially due to torsion

1000 lb

= +

1000 lb

4.95

T

1 33.9q = 1 65.5q = 1 31.7q = −