The simply supported aluminum beam shown is fabricated from a W356 x 122 wide-flange section. 4/3 19⁰0 qo Given: • L = 19.5 m • 9o = 4 kN/m B₁ B₂ = 33 = C₁ = C₂ = ←4/2 Write an elastic curve equation valid for the entire beam in the following form: y(x) = ³₁x³ + B₂ < x-9.75 >4 + 83 <2-16.25 + C₁ + C₂] Place the coordinate system origin (x = 0) at the left end of the beam. Give values for the missing terms, in the units listed. || number (rtol=0.01, atol=1e-05) number (rtol=0.01, atol=1e-05) number (rtol=0.01, atol=1e-05) number (rtol=0.01, atol=1e-05) number (rtol=0.01, atol=1e-05) L number (rtol=0.01, atol=1e-05) number (rtol=0.01, atol=1e-05) kN kN/m kN/m kN.m² kN.m³ Determine the mid-span deflection and the slope at both ends. Ymid = number (rtol=0.01, atol=1e-05) Yleft-end Fright-end mm

Elements Of Electromagnetics
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**HW6.4. Simply Supported Beam with a Uniformly Distributed Load**

The simply supported aluminum beam shown is fabricated from a W356 × 122 wide-flange section.

**Diagram Description:**
The diagram presents a simply supported beam with a uniform distributed load. The beam is represented horizontally with supports at both ends. The load, indicated by downward arrows labeled \( q_0 \), is applied over the right two-thirds of the beam. The beam's total length is \( L \), with the left half labeled \( L/2 \) and the right section divided into thirds, with the uniform load covering \( L/3 \).

**Given:**
- \( L = 19.5 \, \text{m} \)
- \( q_0 = 4 \, \text{kN/m} \)

**Instructions:**

Write an elastic curve equation valid for the entire beam in the following form:

\[ y(x) = \frac{1}{EI} \left[ \beta_1 x^3 + \beta_2 < x-9.75 >^4 + \beta_3 < x-16.25 >^4 + C_1 x + C_2 \right] \]

- Place the coordinate system origin (\( x = 0 \)) at the left end of the beam.
- Provide values for the missing terms in the units listed.

| Parameter | Value (with tolerance) | Unit  |
|-----------|------------------------|-------|
| \(\beta_1\) | number (rtol=0.01, atol=1e-05) | kN |
| \(\beta_2\) | number (rtol=0.01, atol=1e-05) | kN/m |
| \(\beta_3\) | number (rtol=0.01, atol=1e-05) | kN/m |
| \(C_1\) | number (rtol=0.01, atol=1e-05) | kN ⋅ m² |
| \(C_2\) | number (rtol=0.01, atol=1e-05) | kN ⋅ m³ |

**Determine the mid-span deflection and the slope at both ends.**

| Measurement                | Value (with tolerance)   | Unit |
|----------------------------|--------------------------|------|
| \( y_\
Transcribed Image Text:**HW6.4. Simply Supported Beam with a Uniformly Distributed Load** The simply supported aluminum beam shown is fabricated from a W356 × 122 wide-flange section. **Diagram Description:** The diagram presents a simply supported beam with a uniform distributed load. The beam is represented horizontally with supports at both ends. The load, indicated by downward arrows labeled \( q_0 \), is applied over the right two-thirds of the beam. The beam's total length is \( L \), with the left half labeled \( L/2 \) and the right section divided into thirds, with the uniform load covering \( L/3 \). **Given:** - \( L = 19.5 \, \text{m} \) - \( q_0 = 4 \, \text{kN/m} \) **Instructions:** Write an elastic curve equation valid for the entire beam in the following form: \[ y(x) = \frac{1}{EI} \left[ \beta_1 x^3 + \beta_2 < x-9.75 >^4 + \beta_3 < x-16.25 >^4 + C_1 x + C_2 \right] \] - Place the coordinate system origin (\( x = 0 \)) at the left end of the beam. - Provide values for the missing terms in the units listed. | Parameter | Value (with tolerance) | Unit | |-----------|------------------------|-------| | \(\beta_1\) | number (rtol=0.01, atol=1e-05) | kN | | \(\beta_2\) | number (rtol=0.01, atol=1e-05) | kN/m | | \(\beta_3\) | number (rtol=0.01, atol=1e-05) | kN/m | | \(C_1\) | number (rtol=0.01, atol=1e-05) | kN ⋅ m² | | \(C_2\) | number (rtol=0.01, atol=1e-05) | kN ⋅ m³ | **Determine the mid-span deflection and the slope at both ends.** | Measurement | Value (with tolerance) | Unit | |----------------------------|--------------------------|------| | \( y_\
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