1. AERSP 301 Shear of closed section beams Jose Palacios

2. Shear of closed section beams Consider the closed section beams subjected to shear loads S x and S y that cause bending stresses and shear flows Equilibrium relation: y x z SxSx SySy S

3. Shear 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 q s,o (unknown).

4. Shear 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 q b (“basic” shear flow) q b 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.

5. Shear 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

6. Shear of closed section beams If moment center is chosen to coincide with the lines of action of S x and S y then: Above equation can be used to obtain q s,o This expression can be used to find the shear center: Needed in problem 2 Hw 3

7. Twist and warping of shear loaded closed section 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:

8. RECALL! 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, 

9. Twist and warping of shear loaded closed section beams (cont’d) Previously Integrate w.r.t. s from the origin for s dxdy 2A 0s

10. Twist and warping of shear loaded closed section beams cont… A os 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:

11. Twist and warping of shear loaded closed section beams Substituting for and rearranging terms: If the origin coincides with center of twist, R, then last two terms on RHS are zero.

12. Twist and warping of shear loaded closed 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

13. Shear center of closed section beams As with the open section beam, apply a shear load S y and find the coordinate  s of the shear center. First you have to determine the shear flow, q s To determine q s,o use the condition that the shear load through the shear center produces zero twist

14. Shear center of closed section beams This gives: If Gt = const, then: This expression can also be used to find the shear center

15. Sample Problem For the closed section beam shown, calculate the shear center. Y-location of shear center? Symmetry? I xy = ? What load should be applied, S x or S y ?

16. Sample Problem Need to Find I xx : 8a 6a 10a S1S1 y

17. 8a 6a 17a S2S2

18. Sample Problem Shear Flow q 1 for S 1 = 10a y Evaluate A and B at beginning and end of path Do you want me to demonstrate this?

19. Sample Problem Integrating 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 q s,0

20. Sample Problem To obtain q s,0 we use: For the given cross-section Evaluating the Integral: 1 4 2 q 12 q 41 9a ξsξs q 12 q 41 The Shear Flow Becomes:

21. Sample Problem To find the shear center let the clockwise moments (due to q 41 ) be equal to the Counter clockwise moments (due to q 12 ) and solving for ξ s : Sum of the moments around what point? 1 4 q 12 q 41 9a ξsξs 2 53 0 28 0

22. Sample Problem 0.799 0.4695 Shear Center INSIDE the beam cross-section