1. Chapter Six Shearing Stresses in Beams and Thin-Walled Members

2. 6.1 Introduction -- In a long beam, the dominating design factor: -- Primary design factor -- Minor design factor -- In a short beam, the dominating design factor: [due to transverse loading]

3. Equation of equilibrium: -- Shear stress  xy is induced by transverse loading. -- In pure bending -- no shear stress

4. Materials weak in shear resistance shear failure could occur.

5. 6.2 Shear on the Horizontal Face of a Beam Element Knowing and We have (6.3)

6. Since Therefore, and (6.4) (6.5) Defining = shear flow = horizontal shear/length here Q = the first moment w.t.to the neutral axis Q = max at y = 0  The same result can be obtained for the lower element C ' D ' D '' C''

7. 6.3 Determination of the Shearing Stresses in a Beam (6.6) = ave. shear stress A = t  x At the N.A. Q = max, but  ave may not be max, because of t

8.  xy = 0 at top and bottom fibers Variation of  xy < 0.8% if b  h/4 -- for narrow rectangular beams

9. 6.4 Shearing Stresses  xy in Common Types of Beams -- for narrow rectangular beams t = b (6.7) Also, Hence,

10. Knowing A = 2bc, it follows (6.9) This is a parabolic equation with @ y = c @ y = 0 -- i.e. the neutral axis At y = 0, (6.10)  This is only true of rectangular cross-section beams.

11. Special cases: American Standard beam (S-beam) or a wide-flange beam (W-beam) -- over section aa’ or bb’ -- Q = about cc’ (6.6) (6.11) For the web: Assuming the entire V is carried by the web, since the flanges carry little shear force:

12. 6.5 Further Discussion of the Distribution of Stresses in a Narrow Rectangular Beam (6.12) (6.13) Plane sections do NOT remain plane – warping takes place, when a beam is subjected to a transverse shear loading

14. 6.6 Longitudinal Shear on a Beam Element of Arbitrary Shape

15. Using similar procedures in Sec. 6.2, we have = shear flow (6.5) (6.4)

16. 6.7 Shearing Stresses n Thin-Walled Members These two equations are valid for thin-walled members: (6.4) (6.6)

17. （ 6.4) （ 6.6) From Sec. 6.2 We have: Therefore,  xz  0

18. (6.6) -- This equation can be applied to a variety of cross sections.