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What are the conditions of deflection and bending moment in simply supported beam? |
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Answer» Boundary Conditions It is a general mathematical principle that the number of boundary conditions necessary to determine a solution to a differential EQUATION matches the order of the differential equation. The static beam equation is fourth-order (it has a fourth derivative), so each mechanism for supporting the beam should give rise to four boundary conditions. Cantilevered Beams Figure 5: A cantilevered beam. For a cantilevered beam, the boundary conditions are as FOLLOWS: w(0)=0 . This boundary condition SAYS that the base of the beam (at the wall) does not experience any deflection. w'(0)=0 . We also assume that the beam at the wall is horizontal, so that the derivative of the deflection function is zero at that point. w''(L)=0 . This boundary condition models the assumption that there is no bending moment at the FREE end of the cantilever. w'''(L)=0 . This boundary condition models the assumption that there is no shearing force acting at the free end of the beam. If a concentrated force is applied to the free end of the beam (for example, a weight of mass m is hung on the free end), then this induces a shear on the end of the beam. Consequently, the the fourth boundary condition is no longer valid, and is typically replaced by the condition w'''(L)= -mg where g is the acceleration due to GRAVITY (approximately 9.8 m/s^2). We note that we could actually use this boundary condition all the time, since if m=0 , it reduces to the previous case. Simply-Supported Beams |
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