AERSP 301Shear of closed section beams

Jose Palacios

Shear of closed section beams

• Consider the closed section beams subjected to shear loads Sx and Sy that cause bending stresses and shear flows

• Equilibrium relation:

y

x

z Sx

Sy

S

Shear of closed section beamsShear of closed section beams

• As with the open-section beam:

• Unlike the open-section case – you start at an open end, s = 0, q = 0 – it is generally not possible to choose an origin for s at which the shear flow is known. Assume that for the origin chosen (s = 0), shear flow has value qs,o (unknown).

Shear of closed section beamsShear of closed section beams

• First two terms on the RHS represent shear flow in an open section beam loaded through its shear center. Call that qb (“basic” shear flow)

• qb is obtained by introducing a cut at some convenient point in the closed section, thereby converting it to an open section

• The value of shear flow at the cut (s = 0) is found by equating applied and internal moments about some convenient moment center.

Shear of closed section beamsShear of closed section beams

• The value of shear flow at the cut (s = 0) is found by equating applied and internal moments about some convenient moment center.

• From the

figure

A – enclosed area

Shear of closed section beamsShear of closed section beams

• If moment center is chosen to coincide with the lines of action of Sx and Sy then:

• Above equation can be used to obtain qs,o

This expression can be used to find the shear center: Needed in problem 2 Hw 3

A

dsqpq

b

s 20,

Twist and warping of shear loaded closed Twist and warping of shear loaded closed section beamssection beams

• Shear loads not applied through the shear center of a closed section beam causes it to twist and warp (warping = out of plane axial displacements)

• Expressions for twist and warping can be derived in terms of shear flow:

RECALL! Center of TwistRECALL! Center of Twist

• Equivalent to pure rotation about some pt. R

(center of twist [for loading such as pure torsion])

• For the point N

• Origin O of axes chosen arbitrarily, and axes undergo disp. u, v,

Twist and warping of shear loaded closed section beams (cont’d)

• Previously

• Integrate w.r.t. s from the origin for s

dx dy2A0s

Twist and warping of shear loaded closed section beams cont…

• Aos area, swept by generator, center at origin of axes, O, from origin for s to any point s.

• Integrating over the entire c/s yields:

Twist and warping of shear loaded closed Twist and warping of shear loaded closed section beamssection beams

• Substituting for and rearranging terms:

• If the origin coincides with center of twist, R, then last two terms on RHS are zero.

Twist and warping of shear loaded closed Twist and warping of shear loaded closed section beams (cont’d)section beams (cont’d)

• In problems with singly or doubly symmetric sections, the origin for s may be taken to coincide with a point of zero warping– This occurs at points where

axis of symmetry crosses a wall of the section

• It is assumed that the direct stress distribution attributed to the axial constraint is proportional to the free warping of the section = constant x w

• Since a pure torque is applied, resultant of any internal direct stress must be zero. Thus

Shear center of closed section beamsShear center of closed section beams

• As with the open section beam, apply a shear load Sy and find the coordinate s of the shear center.

• First you have to determine the shear flow, qs

• To determine qs,o use the condition that the shear load through the shear center produces zero twist

Shear center of closed section Shear center of closed section beamsbeams

• This gives:

• If Gt = const, then:

This expression can also be used to find the shear center

Sample ProblemSample Problem

• For the closed section beam shown, calculate the shear center.

• Y-location of shear center?

• Symmetry? Ixy = ?• What load should be

applied, Sx or Sy?

Sample Problem

0,0 s

S

xx

ys qydst

I

Sq Need to Find Ixx:

s

xx dAyI0

2

8a

6a

10a

S 1

aaah 1068 22

110

8

s

y

y

ta

dsstdsstIa a

xx

3

10

0

17

0 2

2

21

2

1

1152

17

8

10

82

8a

6a

17a S 2

217

8

s

y

Sample ProblemSample Problem

0,0 s

S

xx

ys qydst

I

Sq

2

1

0

222312,

2130 11341,

4017

817

1152

5

2

115210

8

1152

Syb

ySyb

adssaa

Sq

sa

Sdsst

ta

Sq

Shear Flow q1 forS1 = 10ay

BAsy

Evaluate A and B at beginning and end of path

Do you want me to demonstrate this?

Sample ProblemSample Problem

• Integrating

2

2223

0

212312,

40817

4

1152

4017

817

1152

2

aassa

S

adssaa

Sq

y

Syb

The shear flow, q, on walls 23 and 34 follow from symmetry.

So far we have obtained the basic shear flow (like we did for open cross-section)Now we need to find qs,0

0,0 s

S

xx

ys qydst

I

Sq

Sample ProblemSample Problem

• To obtain qs,0 we use:ds

dsqq

b

s 0,

For the given cross-section ads 54

aays dsasasdss

aa

Sq

17

0 22

2221

21

10

030, 40817

4

5

2

115254

2

Evaluating the Integral: 230, 7.58

1152a

a

Sq ys

2

222312 7.188

17

4

1152aass

a

Sq y

22

1341 7.585

2

1152as

a

Sq y

1

42

q12

q41

9a ξs

q12q41

The Shear Flow Becomes:

Sample ProblemSample ProblemTo find the shear center let the clockwise moments (due to q41) be equal to the

Counter clockwise moments (due to q12) and solving for ξs:Sum of the moments around what point?

1

4

q12

q41

9a ξs

2

530

2

17

0

0121

10

0

041 28sin953sin dsaqdsq

a

s

a

s

280

Sample ProblemSample Problem

2

17

0

022

223

1

10

0

02213

28sin97.18817

4

1152

53sin7.585

2

1152

dsaaassa

S

dsasa

S

a

sy

a

sy

0.799

0.4695

a

a

s

ss

33.3

957.21248.362

Shear Center INSIDE the beam cross-section