The clamped-clamped beam shown is subjected to a concentrated load \(P(t)=\) \(P_{0}\) that is suddenly applied to the center of the span. Determine the response of the beam after the load is suddenly removed.

Short Answer

Expert verified
Short Answer: The deflection of a clamped-clamped beam subjected to a concentrated load at its center and then removed is given by the equation: $$w(x,t) = \sum_{n=1}^\infty \frac{2P_0}{(\alpha_n^2\beta^2)\pi^2 n^2 } (-1)^{n+1} \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha_n^2\beta^2t}$$ This equation represents the deflection at any point x along the beam and at any time t after the load is removed.

Step by step solution

01

Determine the deflection equation from the load function P(t)

To find the deflection equation for the clamped-clamped beam, we will use the following general equation: $$w(x,t) = \sum_{n=1}^\infty A_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha_n^2\beta^2t}$$ where \(w(x,t)\) is the deflection of the beam, \(A_n\) is the amplitude of response for the nth mode, \(x\) is the distance along the beam, \(t\) is the time, \(L\) is the length of the beam, and \(\alpha_n\) is the dimensionless root of the nth mode, \(\beta\) is the wave speed. The equation of the concentrated load when it is suddenly applied is given by: $$P(t) = P_0$$
02

Compute the initial conditions

The initial conditions will help us find the coefficients A_n for the modal responses. Compute the initial displacement and initial velocity at any given point x: Initial displacement: \(w(x,0) = 0\) Initial velocity: \(\frac{\partial w}{\partial t}(x,0) = 0\)
03

Calculate the coefficients for the modal responses

Now we need to find the coefficients A_n for the response modes using the initial conditions: $$A_n = \frac{1}{(\alpha_n^2\beta^2)\pi} \int_0^L P_0 \sin\left(\frac{n\pi x}{L}\right) dx$$ Since the load is applied at the center of the span, the equation becomes: $$A_n = \frac{2P_0}{(\alpha_n^2\beta^2)\pi^2 n^2} (-1)^{n+1}$$
04

Find the time-dependent response after the load is removed

Substitute the coefficients A_n back into the deflection equation and sum up the terms: $$w(x,t) = \sum_{n=1}^\infty \frac{2P_0}{(\alpha_n^2\beta^2)\pi^2 n^2 } (-1)^{n+1} \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha_n^2\beta^2t}$$
05

Create an equation for the deflection at any point along the beam after the load is removed

The final equation for the deflection at any point along the beam after the load is removed can be written as: $$w(x,t) = \sum_{n=1}^\infty \frac{2P_0}{(\alpha_n^2\beta^2)\pi^2 n^2 } (-1)^{n+1} \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha_n^2\beta^2t}$$ This equation will give us the deflection of the beam at any point x along the beam and at any time t after the load is removed.

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