A uniform beam of length L carries a concentrated load wo at x = =(x-4). y(x) = Elday dx4 subject to the given boundary conditions y(0) = 0, y'(0) = 0, y(L) = 0, y'(L) = 0. - WOL +²_ )+((x - 2)²³ ) · ²(x - 1/12 6EI W0_3 12EI The beam is embedded at both ends as shown in the figure below. Use the Laplace transform to solve the differential equation Wo X 1)

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
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Chapter2: Loads On Structures
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### Analysis of a Uniform Beam Under a Concentrated Load

**Problem Statement:**

A uniform beam of length \( L \) carries a concentrated load \( w_0 \) at \( x = \frac{1}{2}L \). The beam is embedded at both ends, as shown in the figure below. We are required to solve the differential equation using the Laplace transform technique.

The differential equation governing this scenario is:

\[ EI \frac{d^4 y}{dx^4} = w_0 \delta \left( x - \frac{1}{2}L \right), \]

where:
- \( EI \) is the flexural rigidity of the beam,
- \( \delta \left( x - \frac{1}{2}L \right) \) is the Dirac delta function.

**Boundary Conditions:**

The beam is subjected to the following boundary conditions:
- \( y(0) = 0 \)
- \( y'(0) = 0 \)
- \( y(L) = 0 \)
- \( y'(L) = 0 \)

**Solution:**

The deflection \( y(x) \) of the beam is given by:

\[ y(x) = \left( \frac{w_0 L}{6EI} x^2 - \frac{w_0}{12EI} x^3 \right) \times \]

\[
+ \left( \frac{w_0}{6EI} \left( x - \frac{L}{2} \right)^3 \right) \cdot u \left( x - \frac{L}{2} \right).
\]

where \( u(x) \) is the unit step function.

**Explanation of the Diagram:**

The diagram illustrates a beam embedded at both ends with length \( L \). The beam is oriented along the x-axis and has a concentrated load \( w_0 \) applied at the midpoint of the beam \( x = \frac{1}{2}L \). 

*Key Components of the Diagram:*
- The beam is shown horizontally with fixed supports (hashed blocks) on both ends.
- The load \( w_0 \) is represented by a downward arrow at the midpoint of the beam.
- The length \( L \) and the coordinate system with \( x \) and \( y \) axes are also indicated.

---

This detailed
Transcribed Image Text:--- ### Analysis of a Uniform Beam Under a Concentrated Load **Problem Statement:** A uniform beam of length \( L \) carries a concentrated load \( w_0 \) at \( x = \frac{1}{2}L \). The beam is embedded at both ends, as shown in the figure below. We are required to solve the differential equation using the Laplace transform technique. The differential equation governing this scenario is: \[ EI \frac{d^4 y}{dx^4} = w_0 \delta \left( x - \frac{1}{2}L \right), \] where: - \( EI \) is the flexural rigidity of the beam, - \( \delta \left( x - \frac{1}{2}L \right) \) is the Dirac delta function. **Boundary Conditions:** The beam is subjected to the following boundary conditions: - \( y(0) = 0 \) - \( y'(0) = 0 \) - \( y(L) = 0 \) - \( y'(L) = 0 \) **Solution:** The deflection \( y(x) \) of the beam is given by: \[ y(x) = \left( \frac{w_0 L}{6EI} x^2 - \frac{w_0}{12EI} x^3 \right) \times \] \[ + \left( \frac{w_0}{6EI} \left( x - \frac{L}{2} \right)^3 \right) \cdot u \left( x - \frac{L}{2} \right). \] where \( u(x) \) is the unit step function. **Explanation of the Diagram:** The diagram illustrates a beam embedded at both ends with length \( L \). The beam is oriented along the x-axis and has a concentrated load \( w_0 \) applied at the midpoint of the beam \( x = \frac{1}{2}L \). *Key Components of the Diagram:* - The beam is shown horizontally with fixed supports (hashed blocks) on both ends. - The load \( w_0 \) is represented by a downward arrow at the midpoint of the beam. - The length \( L \) and the coordinate system with \( x \) and \( y \) axes are also indicated. --- This detailed
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