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Beam Bending Example
Table 1 shows the values for αiℓ, (2i − 1)π ⁄ 2, and βi for the first 5 mode shapes
i αiℓ (2i − 1)π ⁄ 2 βi
1 1.87510 1.57080 0.734096
2 4.69409 4.71239 1.01847
3 7.85476 7.85398 0.999224
4 10.9955 10.9956 1.00003
5 14.1372 14.1372 0.999999
From the αiℓ values we determine ωi by
(14) ωi = (αiℓ)2√((EI)/(mℓ4))
Normalizing equation (1) by  − E4i, we see that the mode shapes become
(15) φi = cosh(αix) − cos(αix) − βi[sinh(αix) − sin(αix)]
where
(16) βi =  − (E2i)/(E4i) = (cosh(αiℓ) + cos(αiℓ))/(sinh(αiℓ) + sin(αiℓ))
and
(17) ℓ⌠⌡0φ2i dx = ℓ     φi(ℓ) = 2( − 1)i + 1
Having solved for φi(x) and ωi, we move on and solve for the deflection of the beam as
(18) ν(x, t) = ξi(t)φi(x)

A) Find Ξi

(19) Ξi  = ℓ⌠⌡0Fcos(Ωt)φi(x) dx      = Fcos(Ωt)1(t)φi(ℓ)      = Fcos(Ωt)( − 1)i + 1  for t > 0

B) Using A), determine the general solution of ξi

Since our forcing function is of the form cos(Ωt), our assumed solution is
(20) ξi = Aisin(ωit) + Bicos(ωit) + Cicos(Ωt)

C) Using the initial conditions, solve for the unknown coefficients Ai, Bi, and Ci

For this, we have the initial conditions that
(21) ν(x, 0) = ν̇(x, 0) = 0
solving the displacement initial condition first:
(22) ν(x, 0) = 0  = Aisin(0) + Bicos(0) + Cicos(0)     0  = Bi + Ci     Bi  =  − Ci
For the velocity initial condition:
(23) ν̇(x, t) = 0  = Aiωicos(0) − Biωisin(0) − Cisin(0)      = Aiωi     0  = Ai
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