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# Biomech-Bending-Double Integration and Distributed Loads-Lecture

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Bending Under Distributed Loads and the Double Integration Method Introduction to Biomechanics 14:125:208

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Distributed Loads Let’s look at a distributed load R A =R B =1/2 ω L For the shear diagram @A, V A =R A =1/2 ω L Find the shear V at any distance x away from A
Distributed Loads Substitute for V A This is the equation for a line Where V=0 @ L/2 Crosses the line at L/2

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Distributed Loads @A, M=0 so M A =0 This gives us a parabola Max is @ x=L/2 since

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Distributed Loads What about cantilever beams? The lack of a second reaction force will change the diagrams
Distributed Loads Let’s cut the beam at a point C between A and B and draw a free body diagram Sum the forces in the y direction −𝜔𝑥 − 𝑉 = 0

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Distributed Loads Rearrange to solve for shearing force, V 𝑉 = −𝜔𝑥
Distributed Loads Now solve for the moment about C The sum the moments about C = 0 𝜔𝑥 𝑥 2 + 𝑀 = 0 Rearrange, 𝑀 = − 1 2 𝜔𝑥 2

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Distributed Loads The maximum values for the shear and moment occur at B
Distributed Loads Note that in both cases the shear diagram was an oblique straight line and the bending moment diagram was a parabola

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Originally, the line segments ab, a’b’ and cd are the same length, but after bending, the top surface (ab) gets shorter and the bottom (a’b’) longer, and cd stays the same length.
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