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Chapter 11: Problem 22

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 is 0.

Step by step solution

01

Timoshenko Beam Theory Equations

The Timoshenko Beam Theory equations account for both shear deformation and bending. The main resulting equation for the bending moment is given by: \(M(x) = EI \frac{d^2w}{dx^2} +kG A_s \frac{d^2v}{dx^2}\) Here, \(M(x)\) is the bending moment, \(w\) is the vertical displacement of the beam, and \(v\) is the rotation along the length of the beam. \(A_s\) represents the shear area of the cross-sectional beam, and \(k\) is the shear correction factor.
02

Apply Boundary Conditions

For a simply supported beam, the boundary conditions are as follows: 1. At the left end of the beam (\(x=0\)), the vertical displacement and the bending moment are both equal to zero. \(w(0) = M(0) = 0\) 2. At the right end of the beam (\(x=L\)), the vertical displacement and the bending moment are also both equal to zero. \(w(L) = M(L) = 0\) Next, we will find the general solution for both \(w\) and \(v\).
03

Solve for General Solutions

Now, let's plug the distributed load function \(p(x, t) = q_{0} \sin \Omega t\) into the Timoshenko beam theory equation and solve for \(w\) and \(v\). We will get the general solutions: \(w(x) = \frac{q_0}{EI} (\frac{x^4}{24} - \frac{L^3}{6} x + \frac{L^4}{24}) \sin \Omega t\) \(v(x) = \frac{q_0}{6kG EI} (x^3 - L^3) \sin \Omega t\)
04

Calculate Bending Moment at Center

At the center of the span, \(x = \frac{L}{2}\). We'll now substitute this value and the given constants into the bending moment equation: \(M(\frac{L}{2}) = EI \frac{d^2w}{dx^2}|_{x = L/2} +kG A_s \frac{d^2v}{dx^2}|_{x = L/2}\) Substituting the obtained general expressions for \(w\) and \(v\) into this equation, and taking the second derivatives with respect to \(x\), we get: \(M(\frac{L}{2})=EI(\frac{q_0 L^2}{8} - \frac{q_0L^2}{8}) + kGA_s(\frac{q_0L^2}{24E}-\frac{q_0L^2}{24E})\) After simplification, we get: \(M(\frac{L}{2})=0\) So the bending moment at the center of the span is zero.

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