1. Dr. Ali Al-Gadhib 1.How to find the value of q or  or at any horizontal level. 2.How to find shear stress at glue and force carried by nail. 3.How to find the maximum shear stress due to V at a given cross section. Lecture No. 20

2. Dr. Ali Al-Gadhib  yx = Horizontal shearing stress  xy = Transverse shearing stress (vertical shearing stress at the cross section)  xy =  yx x y a b a b a b A B C D Laminated Beams Unglued Lamina Glued Lamina  yx  F x =0  M=0  F y =0 A B C D  yx

3. Dr. Ali Al-Gadhib 30mm 250mm 200mm 25mm A B 125mm A3A3 A2A2 A1A1 y1y1 y2y2 y3y3 Ref. line The beam is subjected to shear force V = 15 kN Find  A and  B and show these stresses over elements ?

4. Dr. Ali Al-Gadhib A B AxAx y z 0.1353 0.1747 1) Locate centroid and draw y – z axis through centroid. 2) Calculate moment of inertia about z – axis (or N.A.) I z = 0.21818 × 10 -3 m 4

5. Dr. Ali Al-Gadhib 3) Element A Element B 1.99 MPa 1.65 MPa Note: arrows meet by their heads or tails.

6. Dr. Ali Al-Gadhib Plot variation of Q and  as a function of depth. h b N.A.

7. Dr. Ali Al-Gadhib N.A. Possible  max location

8. Dr. Ali Al-Gadhib Always think about the internal vertical shear force V y as a source for creating shear stress (indirectly through the change of moment) in two ┴ planes ; one is the vertical plane where V y acts and the other is the horizontal plane to keep  F x =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. Vertical plane Horizontal plane a a Glue Nail

9. Dr. Ali Al-Gadhib s s s = spacing Force carried by nail = q·s

10. Dr. Ali Al-Gadhib