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Beam Bending Dynamics

General Solution

The solution to the equation of motion can be performed by the separation of variables method

(7) w(x, t) = X(x)Y(t)

Substituting this into the Eq. 6↑, yields

(8) (X’’’’)/(X) = − (1)/(β⁴)(Ÿ)/(Y).

The left hand side and right hand side of the above equation have their dependencies (on x and t, respectively) separated. Therefore, each side must equate to a constant α⁴. The resulting differential equations are

(9) X’’’’ − β⁴X = 0, Ÿ + α⁴β⁴Y = 0.

For the non-trivial (α ≠ 0), the time-dependent equation’s general solution can be written as

(10) Y(t) = Asin(ωt) + Bcos(ωt)

By examination

(11) ω = α²β² = α²√((EI)/(m)) = (αl)²√((EI)/(ml⁴)).

The non-trivial general solution to the spatially dependent equation by presuming a solution of the form

(12) X(x) = e^λx

and by substituting this form into the spatially dependent equation results in

(13) λ⁴ − α⁴ = (λ − iα)(λ + iα)(λ − α)(λ + α)0.

By invoking some trigonometric identities, this equation can be rewritten as:

(14) X(x) = D₁sin(αx) + D₂cos(αx) + D₃sinh(αx) + D₄cosh(αx)

which can be rewritten to a slightly more advantageous form

(15) X(x) = E₁[sin(αx) + sinh(αx)] + E₂[sin(αx) − sinh(αx)] + E₃[cos(αx) + cosh(αx)] + E₄[cos(αx) − cosh(αx)]

The constants A and B can be determined from the initial condition (deflection and its derivative). The constants of the above equation E₁, E₂, E₃, E₄ can be evalutated from the boundary conditions.

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