Objective:

1. Lecture No. 20 Shearing Stress in Beams Objective: • How to find the value of q or  or at any horizontal level. • How to find shear stress at glue and force carried by nail. • How to find the maximum shear stress due to V at a • given cross section.

2. a Laminated Beams Unglued Lamina Glued Lamina b a b a D B A C y yx yx yx b B x D Fy=0 Fx=0 M=0 C A xy = yx Shearing Stress in Beams • yx = Horizontal shearing stress xy = Transverse shearing stress (vertical shearing stress at the cross section)

3. 200mm A1 30mm A 25mm y1 250mm A2 y2 B 30mm y3 A3 Ref. line 125mm Problem: The beam is subjected to shear force V = 15 kN Find A and B and show these stresses over elements ?

4. y Ax A 0.1353 z 0.1747 B Solution: 1) Locate centroid and draw y – z axis through centroid. 2) Calculate moment of inertia about z – axis (or N.A.) Iz = 0.21818 × 10-3 m4

5. 3) Element A Element B 1.65 MPa 1.99 MPa Note: arrows meet by their heads or tails.

6. h b N.A. N.A. Problem: Plot variation of Q and  as a function of depth. Solution:

7. Possible max location N.A. N.A. N.A. N.A.

8. Horizontal plane Nail a a Vertical plane Glue Remarks: Always think about the internal vertical shear force Vy as a source for creating shear stress (indirectly through the change of moment) in two ┴ planes ; one is the vertical plane where Vy acts and the other is the horizontal plane to keep  Fx=0. These two shear stresses are equally at their intersection as shown below. Note that for pure moment (M=constant) , V=0 accordingly the shear stresses in vertical and horizontal planes are zeros and ,therefore, no slipping and no need for nails or glue.

9. Force carried by nail = q·s s s s = spacing

10. The enD