A simply supported uniform beam of length \(L\), bending stiffness \(E I\) and mass per unit length \(m\) is subjected to the uniform distributed load \(p(x, t)=q_{0} \sin \Omega t\). Determine the bending moment at the center of the span if the behavior of the beam is predicted using Timoshenko beam theory for a structure where \(r_{G} / L=0.1\) and \(E / k G\) \(=5\).

Short Answer

Expert verified
Answer: The bending moment at the center of the span can be calculated as: \[M\left(\frac{L}{2}\right)= {-q_0 EI \sum_{n=1}^{\infty} \frac{(1 - (-1)^{n})}{(n\pi)^{2}} \left(\frac{n \pi}{L}\right)^{2} \sin\left(\frac{n\pi}{2}\right) } \] Compute and evaluate the sum to obtain the bending moment M(L/2) at the center of the span.

Step by step solution

01

Write down the equation for Timoshenko beam theory

The governing equation for Timoshenko beam theory can be written as: \[ \frac{\partial^{2}w}{\partial t^2}+\frac{ρ_{G}G A}{E I w} \frac{\partial^{4}w}{\partial x^{4}}-\frac{ρ_{G}G A}{E I w} \frac{\partial^{2}}{\partial x^{2}}\left[\frac{\partial^{2}w}{\partial t^{2}}\right]=q_{0} \sin \Omega t \] where w is the transverse displacement, ρG is the mass per unit length, G is the shear modulus, A is the cross-sectional area, E is the Young's modulus, and I is the moment of inertia.
02

Apply the boundary conditions for the simply supported beam

For a simply supported beam, the boundary conditions are: 1. At x = 0 and x = L, w = 0 (simply supported on both ends) 2. At x = 0 and x = L, M = 0 (simply supported on both ends) To apply these boundary conditions to the Timoshenko beam equation, we will express the displacement w in terms of a Fourier series and apply the boundary conditions: \[ w(x, t) = \sum_{n=1}^{\infty} \frac{\sin \left(\frac{n \pi x}{L}\right)}{n \pi} \] \[\sum_{n}^{\infty} \frac{A_n \sin \left(\frac{n \pi x}{L}\right)}{(n \pi)(r_G / L)} \left(\frac{E}{k G}\right) = q_{0} \sin \Omega t \]
03

Solve for the response function

To solve for the response function An, which represents the Fourier coefficients in the displacement series, we will integrate both sides of the equation over the domain 0 < x < L and simplify the expression: \[\int_{0}^{L} \left(\sum_{n}^{\infty} \frac{A_n \sin \left(\frac{n \pi x}{L}\right)}{(n \pi)(r_G / L)} \left(\frac{E}{k G}\right) - q_{0} \sin \Omega t \right) dx = 0 \] \[ \sum_{n=1}^{\infty} \left[ \frac{2A_n \sin \left(\frac{n \pi L}{L}\right)}{(n \pi)^{2}} - \frac{q_{0} (\sin\left(\frac{n \pi L}{L}\right))}{n^{2} \pi^{2}} \right] = 0 \] \[ A_n = \frac{q_{0} (1 - (-1)^{n})}{n^{2} \pi^{2}} \]
04

Calculate the bending moment at the center of the span

To calculate the bending moment M at the center of the span x = L/2, we need to calculate the second derivative of the displacement with respect to x and substitute the expression for An: \[ M(x) = E I \frac{\partial^{2}w}{\partial x^2} \] \[ M\left(\frac{L}{2}\right) = E I \sum_{n=1}^{\infty} A_n \left(-\frac{n \pi}{L}\right)^{2} \sin \left(\frac{n \pi}{2}\right) \] \[M\left(\frac{L}{2}\right)= {-q_0 EI \sum_{n=1}^{\infty} \frac{(1 - (-1)^{n})}{(n\pi)^{2}} \left(\frac{n \pi}{L}\right)^{2} \sin\left(\frac{n\pi}{2}\right) } \] Compute and evaluate the sum, and the bending moment M(L/2) at the center of the span can be obtained.

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