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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bending Moment Calculation
The calculation of the bending moment in a beam is a critical aspect of understanding how forces and moments are distributed along the beam's length. In Timoshenko Beam Theory, this calculation is enhanced by considering not only the bending effects but also the influence of shear deformation. The equation used to find the bending moment, \(M(x)\), incorporates both the material bending stiffness \(EI\) and the shear parameters \(kGA_s\). These help account for the additional deformation that occurs beyond simple bending.
  • \( M(x) = EI \frac{d^2w}{dx^2} + kGA_s \frac{d^2v}{dx^2} \)
  • \(w\) is the displacement caused by bending.
  • \(v\) is the rotational displacement due to shear.
  • Shear deformation adds complexity to the calculation by involving \(kGA_s\).
The Timoshenko Beam Theory thus offers a more accurate representation, especially for beams where shear deformation significantly influences the bending moment.
Shear Deformation
Shear deformation occurs when layers within a material slide past each other, as opposed to simply bending. In Timoshenko Beam Theory, this aspect is taken into account, allowing for more precise modeling of beam behavior. This is particularly important for short or deep beams where shear effects can be significant.
  • It is described by the term \(kGA_s \frac{d^2v}{dx^2}\) in the moment equation.
  • \(k\) is the shear correction factor, compensating for cross-sectional shape effects.
  • \(G\) is the shear modulus, representing the material's rigidity against shear.
  • \(A_s\) is the shear area, which is a portion of the cross-sectional area that effectively resists shear deformation.
By including shear deformation in the equation, Timoshenko Beam Theory provides insights into additional stress patterns and deflections that would otherwise be missed using simpler beam theories.
Simply Supported Beam
A simply supported beam is a classical structural element type frequently encountered in engineering. This type of beam is supported at both ends but is free to move vertically or rotate at these supports. In the context of Timoshenko Beam Theory, simply supported beams have specific boundary conditions that influence their deformation and load-carrying behavior.
  • Each end of the beam can pivot, allowing it to take on reactions in the form of vertical forces only.
  • The beam does not experience moments at the supports, which is expressed by setting the boundary conditions \(M(0) = 0\) and \(M(L) = 0\).
  • Deflections \(w(x)\) and shear rotations \(v(x)\) need to satisfy these boundary conditions at the supports.
Simply supported beams are common because they allow for straightforward analysis without the complexity of additional constraints such as fixed or continuous supports.
Boundary Conditions
Boundary conditions are essential constraints applied to solve the differential equations governing beam behavior. They describe the physical conditions at the ends of the beam, which is crucial for solving for displacements and rotations. In Timoshenko Beam Theory, these are more complex.
  • For simply supported beams, they set constraints like \(w(0) = 0\) and \(w(L) = 0\), meaning no vertical displacement at the supports.
  • The moment conditions \(M(0) = 0\) and \(M(L) = 0\) ensure no resistance to rotation at the ends.
  • They help define the shape and behavior of the equations used to calculate displacements \(w(x)\) and rotations \(v(x)\).
By defining these conditions, we allow the dynamic solution of beam deformation under a distributed load, leading to a more realistic model predicting beam behavior.

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Most popular questions from this chapter

A uniform elastic rod of length \(L\), membrane stiffness \(E A\) and mass per unit length \(m\) is attached to a rigid base at its left end, as shown. Determine the steady state motion of the rod if a motor causes the base to undergo the prescribed motion \(\chi_{x}(t)=h_{0} \sin \Omega t\).

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 Rayleigh Beam Theory with \(r_{G} / L=0.1\).

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.

The beam shown in Figure P11.16 is subjected to a concentrated harmonic load at its right end. If the magnitude of the applied load is \(P_{0}\) and its frequency is equal to half of the fundamental frequency of the structure, determine the transverse shear at the left edge of the beam.

A piano wire of length \(L\), cross-sectional area \(A\) and mass density \(\rho\) is tuned to a tension of magnitude \(N_{0}\). Determine the response of the wire if the string is struck at its quarter point by a hammer that imparts an impulse of magnitude \(\mathcal{I}_{0}\)
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