Understanding Horizontal Shear Stress in Beams - Mechanical Engineering Lecture
Dive into the concept of horizontal shear stress in beams through a detailed lecture focusing on a numerical example. Learn how to calculate average shearing stress at each joint of a beam and determine maximum load values based on allowable bending and shear stresses.
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Horizontal Shear Stress in Beam Lecture Number - 06 D.B. Raut Mechanical Engineering SINHGAD COLLEGE OF ENGG ,Vadgaon Strength of Materials
Shear stress in beam: Numerical Beam AB is made of three planks glued together and is subjected in its plane of symmetry to the loading shown. If width of the each glued joint is 20mm determine average shearing stress in each joint at section n-n. The centroid of the section is shown in the diagram and I=8.63 10-6m4 Solution: As loading is symmetric reaction at each end is half of the total load i.e. 1.5kN. The vertical shear stress at n-n is given by, = 1 5 . V kN Strength of Materials
Shear stress in beam: Numerical Strength of Materials
Shear stress in beam: Numerical Shearing stress at top joint (Joint a) is given by VA y bI A = = = a a a where Area of top flange a Distance of centroid of top flange from N.A. I = M.I. about N.A. b = width of the joint y a = = = + = 4 100 20 8 63 10 . 2000 10 22 7 A mm = , y 12.7 = . mm a = a 6 4 20 1500 I mm b , mm V , N Strength of Materials
Shear stress in beam: Numerical 1500 2000 22 7 20 8 63 10 . VA y bI . = = = 0 725 . a a MPa 6 a Thus shearing stress at top joint is 725kPa Strength of Materials
Shear stress in beam: Numerical Shearing stress at bottom joint (Joint b) is given by VA y bI A = = = b b b where Area of bottom flange b Distance of centroid of bottom flange from N.A. I = M.I. about N.A. b = width of the joint y b = = = + = 4 60 20 1200 8 63 10 . 10 78 3 A mm = , y 68.3 = . mm b = b 6 4 20 1500 I mm b , mm V , N Strength of Materials
Shear stress in beam: Numerical 1500 1200 78 3 20 8 63 10 . VA y bI . = = = 0 608 . a a MPa 6 a Thus shearing stress at bottom joint is 608kPa Strength of Materials
Shear stress in beam: Numerical For the loading as shown in the diagram find the max. value of load P if the allowable stress in bending and shear are 100 MPa and 10 MPa respectively Solution: Support reactions TakingmomentsaboutA P R TakingmomentsaboutD + = + = = 2 5 10 15 0 2 5 . P , R PN D D 10 2 = 5 5 0 0 5 . R P P , R PN A A Strength of Materials
Shear stress in beam: Numerical Shear force and bending moment diagrams are shown in the diagram FromSFD = max = shear forceis 1 5 . V max V PN FromBMD = max = Bendingmomentis = 5 5000 M M P N .m P N .mm max M I aboutN A is . . = . . 3 3 bd bd I 12 12 outer inner 3 3 8 10 12 6 8 12 = = = 4 410 7 4 107 10 . . cm 6 4 mm Strength of Materials
Shear stress in beam: Numerical Design for shear ( ( )( 8 5 2 5 52 1 1 2 b cm = + = ) ( ) ( 52 10 20 = ) 6 4 2 = Ay Ay Ay outer outer )( ) = = . = 3 3 3 cm mm mm VAy bI = = all max 52 10 3 1 5 20 4 107 10 10 53 . kN . P = 10 6 . = P Strength of Materials
Shear stress in beam: Numerical Design for bending My I P = = all max 5000 4 107 10 16 428 . 5 = 100 6 . = P kN Therefore, safe load is minimum of maximum load for bending and maximum load for shear which is 10.53 kN Strength of Materials
