6cex. Deflection Examples • Ex. 6c.1 • Ex. 6c.2 
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The moment is constant over the entire length of the beam. The curvature of the beam's deflection is then: Integrating once gives the equation defining the slope of beam: The constant C_{1} will be found later using the boundary conditions. Integrating a second time gives the deflection of the beam: The constants C_{1} and C_{2} are determined using the boundary conditions (BC) . Since the beam is simply supported:
Due to symmetry of the load and geometry (supports), another BC that be applied in this case is v'(L/2) = 0. Solving for the constants: The deflection of the beam is then:: From the symmetry of the beam we know the maximum deflection is at x = L/2 = 25 in. From the Materials Property Table, E for 6061 aluminum is 10 Msi. The maximum deflection is then:

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