Prescribed displacement
Figure 4↓ shows a special case. Root of this beam has a prescribed displacement w = a0sin(ωt) and the other end has a clamped boundary condition.
(5)
x = 0
: w = a0sin(ωt) θ = 0
x = l
: w = 0 M = 0
with the Euler-Bernoulli assumption:
(6)
x = 0
: w = a0sin(ωt) θ = 0
x = l
: w = 0 θ = (∂w)/(∂x) = 0
Translational spring reinforcement
Figure 5↓ shows a beam that is reinforced with a translational spring at each end. Figure 6↓ shows action and reaction force on springs and beam. Considering conventional positive sign for shear force at each end of the beam, one can write the boundary conditions as below
(7)
x = 0
: M = 0 F = kw
x = l
: M = 0 F = − kw
with the Euler-Bernoulli assumption:
(8)
x = 0
: M = 0 F = − (∂)/(∂x)⎛⎝EI(∂2w)/(∂x2)⎞⎠ = kwx = 0
x = l
: M = 0 F = − (∂)/(∂x)⎛⎝EI(∂2w)/(∂x2)⎞⎠ = − kwx = l
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