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Beam Bending Example

Consider a uniform cantilever beam of length , bending rigidity EI, and mass per unit length m. Until time t = 0 the beam is undeflected and at rest. At time t = 0 a transverse concentrated load of magnitude Fcos(Ωt) is applied at x = ℓ. Determine the total beam displacement ν(x, t) for time t > 0.

Solution

We assume the solution to be of the form
(1) X(x) =  E1[sin(αx) + sinh(αx)] + E2[sin(αx) − sinh(αx)]  + E3[cos(αx) + cosh(αx)] + E4[cos(αx) − cosh(αx)]
and we then apply the boundary conditions, which, for a clamped-free beam, are
(2) X(0) = X(0) = X′′(ℓ) = X′′′(ℓ) = 0
These correspond to zero displacement and zero slope at the cantilevered end and zero shear and zero bending moment at the free end.
Applying X(0)=0:
(3) X(0) = 0 =  E1[sin(0) + sinh(0)] + E2[sin(0) − sinh(0)]  + E3[cos(0) + cosh(0)] + E4[cos(0) − cosh(0)]
All sin and sinh terms are 0, the cos and cosh terms are 1, so we are left with
(4) 2E3 = 0 E3 = 0
Applying now the X(0):
(5) X(0) = 0 =  E1[cos(0) + cosh(0)] + E2[cos(0) − cosh(0)]  + 0[ − sin(0) + sinh(0)] + E4[ − sin(0) − sinh(0)]
Noting that the E2 and E4 terms disappear, we have that
(6) 2E1 = 0 E1 = 0
Applying now the X′′(ℓ), and leaving out the E1 and E3 terms:
(7) X′′(ℓ) = 0 = E2[ − α2sin(αℓ) − α2sinh(αℓ)] + E4[ − α2cos(αℓ) − α2cosh(αℓ)]
Factoring out  − α2 we get
(8) E2[sin(αℓ) + sinh(αℓ)] + E4[cos(αℓ) + cosh(αℓ)] = 0
Applying now the X′′′(ℓ), and leaving out the E1 and E3 terms:
(9) X′′′(ℓ) = 0 = E2[ − α3cos(αℓ) − α3cosh(αℓ)] + E4[α3sin(αℓ) − α3sinh(αℓ)]
Factoring out  − α3 we get
(10) E2[cos(αℓ) + cosh(αℓ)] + E4[sinh(αℓ) − sin(αℓ)] = 0
We rewrite equations (8) and (10) as follows
(11) sinh(αℓ) + sin(αℓ) cosh(αℓ) + cos(αℓ) cosh(αℓ) + cos(αℓ) sinh(αℓ) − sin(αℓ) E2 E4  =  0 0
In order to find the characteristic equation, we take the determinant of the 2x2 matrix
(12) sinh2(αℓ) − sin2(αℓ) − [cosh(αℓ) + cos(αℓ)]2 = 0 cos(αℓ)cosh(αℓ) + 1 = 0
For i ≥ 5, the solutions to this characteristic equation are essentially those of cos(αℓ), which are
(13) (2i − 1)π ⁄ 2
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