What are the conditions of deflection and bending moment in simply supported beam?
Answers
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 Consequently, the the fourth boundary condition is no longer valid, and is typically replaced by the condition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