Problem 2 (30 points): The formula for the deflection of the beam shown is: y(x) = Wo (-x5 + 5Lx*- 10L²X3 + 10L3x2) 120LEI Wo = 15 x 107 N L = 2m Omax %3D %3D т N E = 7 x 1011. m2 1 = 6 x 10-4m %3D %3D Use the Newton-Raphson Method to determine the x location along the beam where the deflection is y(x) = 0.14 m. Use a starting guess of xo = 1.2 m. Be sure to show your calculation for the derivative of the function and your calculations you use to fill out the table. You only need to perform two iterations of the method, which means you should fill out the table below. Notice that y is defined as positive in the downward direction, which means that both the function and the target deflection are positive values in the coordinate system. y(xn) y'(xn) Хn+1 Step Xn 2

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**Problem 2 (30 points):** The formula for the deflection of the beam shown is:

\[
y(x) = \frac{\omega_0}{120EI}(-x^5 + 5Lx^4 - 10L^2x^3 + 10L^3x^2)
\]

Given:
- \(\omega_0 = 15 \times 10^7 \, \frac{N}{m}\)
- \(L = 2 \, m\)
- \(E = 7 \times 10^{11} \, \frac{N}{m^2}\)
- \(I = 6 \times 10^{-4} \, m\)

Use the **Newton-Raphson Method** to determine the \(x\) location along the beam where the deflection is \(y(x) = 0.14 \, m\). 

- Use a starting guess of \(x_0 = 1.2 \, m\).
- Be sure to show your calculation for the derivative of the function and your calculations used to fill out the table.
- You only need to perform two iterations of the method, which means you should fill out the table below.

**Note:**
- \(y\) is defined as positive in the downward direction. This means both the function and the target deflection are positive in the coordinate system.

| Step | \(x_n\) | \(y(x_n)\) | \(y'(x_n)\) | \(x_{n+1}\) |
|------|------|-----------|------------|-------------|
| 0    |      |           |            |             |
| 1    |      |           |            |             |
| 2    |      |           |            |             |

**Diagram Explanation:**
The diagram shows the deflection of a beam under a uniform load \(\omega\), with parameters:
- \(l\) representing the length of the beam.
- \(y\) denoting the deflection which is portrayed as curving downwards.
- \(x\) is the horizontal position along the beam.
- \(\delta_{max}\) is indicated as the maximum deflection.
Transcribed Image Text:**Problem 2 (30 points):** The formula for the deflection of the beam shown is: \[ y(x) = \frac{\omega_0}{120EI}(-x^5 + 5Lx^4 - 10L^2x^3 + 10L^3x^2) \] Given: - \(\omega_0 = 15 \times 10^7 \, \frac{N}{m}\) - \(L = 2 \, m\) - \(E = 7 \times 10^{11} \, \frac{N}{m^2}\) - \(I = 6 \times 10^{-4} \, m\) Use the **Newton-Raphson Method** to determine the \(x\) location along the beam where the deflection is \(y(x) = 0.14 \, m\). - Use a starting guess of \(x_0 = 1.2 \, m\). - Be sure to show your calculation for the derivative of the function and your calculations used to fill out the table. - You only need to perform two iterations of the method, which means you should fill out the table below. **Note:** - \(y\) is defined as positive in the downward direction. This means both the function and the target deflection are positive in the coordinate system. | Step | \(x_n\) | \(y(x_n)\) | \(y'(x_n)\) | \(x_{n+1}\) | |------|------|-----------|------------|-------------| | 0 | | | | | | 1 | | | | | | 2 | | | | | **Diagram Explanation:** The diagram shows the deflection of a beam under a uniform load \(\omega\), with parameters: - \(l\) representing the length of the beam. - \(y\) denoting the deflection which is portrayed as curving downwards. - \(x\) is the horizontal position along the beam. - \(\delta_{max}\) is indicated as the maximum deflection.
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