EXAMPLES: BOUNDARY CONDITIONS FOR TORSION
In this section, a few examples are shown, illustrating the boundary conditions for beams in torsion. In all of the following examples l is the length of the beam and x = 0 is the left end of the beam. Figure 1↓ shows the frame of reference for all the examples in this section, where θ is the twist angle about x axis.
Clamped-free beam
Figure 2↓ shows a clamped-free beam under torsion. The boundary condition can be prescribed as θ = 0 at the clamped end. By definition, the forces and moments at a free end are equal to zero. Here, only the twisting moment is of interest so the boundary condition at the free end is T = GJ(∂θ)/(∂x) = 0.
Elastically constrained
Figure 3↓ shows a beam constrained with torsional springs at both ends. Figure 4↓ shows action and reaction twisting moment on the beam and springs at each end of the beam. Considering the conventional positive sign of twisting moment at each end of the beam, one can write the boundary condition as follows:
(1)
x = 0:T = kθ GJ(∂θ)/(∂x) = kθ
(2)
x = l:T = − kθ GJ(∂θ)/(∂x) = − kθ
Intertial Constraint
Figure 5↓ shows a beam with rigid bodies attached at both ends. Figure 6↓ shows action and reaction twisting moments on the beam and rigid body at each end of the beam. Considering the conventional positive sign of twisting moment at each end of the beam, one can write the moment equilibrium equation for rigid bodies and as a result boundary conditions for the beam.
(3)
x = 0:T = Ic(∂2θ)/(∂2t) GJ(∂θ)/(∂x) = Ic(∂2θ)/(∂2t)
(4)
x = l: − T = Ic(∂2θ)/(∂2t) − GJ(∂θ)/(∂x) = Ic(∂2θ)/(∂2t)