 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

 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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 ( )

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 2005 Pearson Education South Asia Pte Ltd 7. Transverse Shear 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

 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.

 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)

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

 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)

 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

 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) = 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