ShearTransverse-11

ShearTransverse-11 - Transverse Shear/Non-uniform bending...

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Transverse Shear/Non-uniform bending (Chapter 5 in Gere) KML 9/30/09 Shear forces produce non-uniform bending of beams since they generate shear stresses and bending stresses that vary over the length of beams. Curvature is not constant over the beam. Since the shear stress on a differential element has the same magnitude on all four planar faces, the shear stresses generated in the vertical direction are equivalent to the horizontal shear stresses, which will be derived here. The following is a differential beam element with the accompanying moment and bending stress that vary due to non-uniform bending. y Beam element: side view Differential element with bending stress y x σ 1 σ 2 Differential sub-element with horizontal stresses τ y 1 h/2 Right end view of sub-element y z b dA y F 1 F 2 F 3 I y M 1 - = σ I y dM) (M 2 + - = σ dx x M V M + dM V + dV x M + dM σ 1 σ 2 M The magnitudes of the horizontal forces due to 1 σ and 2 σ are: = σ = dA I My dA F 1 1 + = σ = dA I y dM) (M dA F 2 2 For horizontal force equilibrium: = = - = dA y I dM dA I y dM F F F 1 2 3
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The force due to the shear load distributed over the width of the beam (assuming the shear stress is uniformly distributed over the beam width, which holds approximately for b < h): dx b F 3 τ = ( A F τ = , where dx is the length, and b is the width over which τ acts) Equating the expressions for 3 F and solving for shear stress, τ : = = τ
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This note was uploaded on 10/01/2009 for the course MEGR 2144 taught by Professor Sharpe during the Fall '08 term at UNC Charlotte.

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ShearTransverse-11 - Transverse Shear/Non-uniform bending...

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