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L L/2 x P Figure 1: Schematic of the three-point bend test. 

          L
L/2
x
P
Figure 1: Schematic of the three-point bend test.
L L/2 x P Figure 1: Schematic of the three-point bend test.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Modeling Three-Point Bending Derivation: Consider the three-point bending schematic shown in Figure 1. Using Euler-Bernoulli beam theory, derive the equation for deflection (v), at any position (x) on the beam, for a given load (P). Also, derive the equation for the maximum deflection (Vmax) of the beam. L L/2 P x Figure 1: Schematic of the three-point bend test
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Transcript

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00:01 So, hello everyone, in this question for beam a, m is equals to 3x by 2 plus x.
00:12 Right, and the shear force for beam a, f is equal to tau m by do x.
00:21 That would be 3x plus or we can say 3 by 2, this would be minus 2 sine x.
00:33 Right so the load distribution will be minus do f by do x right so this will be 2 cox so w is 2 co6 this is the load distribution and this is for beam a now coming to b for beam b m is 4 l x plus x plus 6 plus 7 sine x minus 3 e x so the shear force is do m by do x that would be 4 by x plus 7 plus x minus 3 e 4 x right and the load distribution will be minus do f by do x which is 4 by x squared minus 4 by x squared minus 7 x x x minus 3 x x x so this is w for bb.
01:51 Now, the condition of maximizing the pending movement, it is, for maximizing the bending movement, dm by dx should be 0, which means 3 by 2 minus 2 sine x should be 0.
02:10 So, sine x will be 3 by 4...
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