1.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 1 CHAPTER OBJECTIVES Develop a method for finding the shear stress in a beam having a prismatic cross-section and made from homogeneous material that behaves in a linear-elastic manner This method of analysis is limited to special cases of cross-sectional geometry

2.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 2 CHAPTER OUTLINE 1.Shear in Straight Members 2.The Shear Formula 3.Shear Stresses in Beams

3.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 3 7.1 SHEAR IN STRAIGHT MEMBERS Shear V is the result of a transverse shear-stress distribution that acts over the beam’s cross-section. Due to complementary property of shear, associated longitudinal shear stresses also act along longitudinal planes of beam

4.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 4 7.1 SHEAR IN STRAIGHT MEMBERS As shown below, if top and bottom surfaces of each board are smooth and not bonded together, then application of load P will cause the boards to slide relative to one another. However, if boards are bonded together, longitudinal shear stresses will develop and distort cross-section in a complex manner Boards not bonded together Boards bonded together

5.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 5 7.1 SHEAR IN STRAIGHT MEMBERS As shown, when shear V is applied, the non-uniform shear-strain distribution over cross-section will cause it to warp, i.e., not remain in plane. Before deformation After deformation

6.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 6 7.2 THE SHEAR FORMULA By first principles, flexure formula and V = dM/dx, we obtain  = = VQ It Eq. 7-3  = shear stress in member at the point located a distance y’ from the neutral axis. Assumed to be constant and therefore averaged across the width t of member V = internal resultant shear force, determined from method of sections and equations of equilibrium

7.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 7 7.2 THE SHEAR FORMULA I = moment of inertia of entire cross-sectional area computed about the neutral axis t = width of the member’s cross-sectional area, measured at the point where  is to be determined Q = ∫ A’ y dA’ = y’A’, where A’ is the top (or bottom) portion of member’s cross-sectional area, defined from section where t is measured, and y’ is distance of centroid of A’, measured from neutral axis  = = VQ It Eq. 7-3

8.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 8 7.2 THE SHEAR FORMULA The equation derived is called the shear formula Since Eqn 7-3 is derived indirectly from the flexure formula, the material must behave in a linear-elastic manner and have a modulus of elasticity that is the same in tension and in compression Shear stress in composite members can also be obtained using the shear formula To do so, compute Q and I from the transformed section of the member as discussed in section 6.6. Thickness t in formula remains the actual width t of cross-section at the pt where  is to be calculated

9.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 9 7.3 SHEAR STRESSES IN BEAMS Rectangular cross-section Consider beam to have rectangular cross- section of width b and height h as shown. Distribution of shear stress throughout cross-section can be determined by computing shear stress at arbitrary height “ y” from neutral axis, and plotting the function. Hence, Q = y’A’ =  y 2 b 1212 h24h24 ( )

10.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 10 7.3 SHEAR STRESSES IN BEAMS Rectangular cross-section  =  y 2 ( ) 6V bh 3 h24h24 Eq. 7-4 Eqn 7-4 indicates that shear-stress distribution over cross-section is parabolic. After deriving Q and applying the shear formula, we have Shear stress distribution

11.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 11 7.3 SHEAR STRESSES IN BEAMS  max = 1.5 VAVA Eq. 7-5 By comparison,  max is 50% greater than the average shear stress determined from  avg = V/A. At y = 0, we have  =  y 2 ( ) 6V bh 3 h24h24 Eq. 7-4

12.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear EXAMPLE-1 The beam shown in the figure is made of wood and is subjected to a resultant vertical shear force of V = 3 kN. Determine: a)The shear stress in the beam at point P. b)The maximumshear stress in the beam.

13.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear EXAMPLE-1 Section Properties: Moment of inertia about the neutral axis I = = = 16.28x10 6 mm 4 bh 3 12 (100)(125) 3 12 A horizontal section line is drawn through point P and the partial area is shown shaded. Hence, the first moment with respect to the neutral axis is [ 12.5+(1/2)(50) ][ (50)(100) ] mm 3 = 18.75x10 4 mm 4 Part a)

14.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear EXAMPLE-1 Shear stress at point P:  P = = = 0.346 N/mm 2 VQ I t (3x10 3 )(18.75x10 4 ) (16.28x10 6 )(100) = 0.346 MPa

15.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear EXAMPLE-1 Maximum shear stress occurs at the neutral axis. Hence, the first moment with respect to the neutral axis is largest, thus [62.5/2](100)(62.5) = 19.53x10 4 mm 4 Part b) Shear stress maximum:  max = = = 0.36 N/mm 2 VQ I t (3x10 3 )(19.53x10 4 ) (16.28x10 6 )(100) = 0.36 MPa Note that this equivalent to  max = 1.5 = 1.5 = 0.36 N/mm 2 VAVA 3x10 3 (100)(125)

16.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear EXAMPLE-2 The steel rod has radius of 30 mm. If it is subjected to a shear force of V = 25 kN, determine the maximum shear stress. Section Property: Moment of inertia about the neutral axis I = r 4 = (30 4 ) = 0.636x10 6 mm 4 44 44

17.  2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear Maximum shear stress occurs at the neutral axis. Hence, the height of the centroid is r = (30) = 12.73 mm 4343 4343 The first moment with respect to the neutral axis is largest, thus [12.73][  (30 2 )/2] = 18x10 3 mm 4 Shear stress maximum:  max = = = 11.8 N/mm 2 VQ I t (25x10 3 )(18x10 3 ) (0.636x10 6 )(60) = 11.8 MPa EXAMPLE-2 y’y’ _ Neutral axis V