when the beam is horizontal. For W-90 lb and the stiffness of the spring k-313 lb/in, determine the static elongation of the spring (6pring). Le, h-0. [in] W shown, wh

Elements Of Electromagnetics
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Publisher:Sadiku, Matthew N. O.
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**Transcription of the Problem Statement from the Image:**

A weight \( W \) is dropped from a height \( h \) on the mid-span at \( C \) of a simply-supported beam as shown, where the spring always remains vertical and is at its natural length when the beam is horizontal.

For \( W = 90 \, \text{lb} \) and the stiffness of the spring \( k = 313 \, \text{lb/in} \), determine the static elongation of the spring (\( \delta_{\text{spring}} \)) i.e., \( h = 0 \), [in].

**Diagram Explanation:**

1. **Beam Setup:**
   - A simply-supported beam \( AB \) is shown with supports at points \( A \) and \( B \).
   - The distance between \( A \) and \( B \) is noted as \( 40" \).

2. **Weight Placement:**
   - A weight \( W \) is applied at the mid-span of the beam at point \( C \).
   - The weight is suspended from a height \( h \) directly above point \( C \).

3. **Spring Details:**
   - There is a spring attached vertically at the end of the beam at point \( B \).
   - The spring has a stiffness constant denoted by \( k = 313 \, \text{lb/in} \).

4. **Lengths and Measurements:**
   - The beam \( AB \) measures \( 40" \) from \( A \) to \( B \).
   - The length \( BC \) is \( 1" \).
   - An additional measurement from point \( B \) downwards is labeled \( 2" \).

This scenario examines the static elongation of a spring when a weight is applied at the mid-span of a beam, considering the system reaches equilibrium without any initial height difference (\( h = 0 \)).
Transcribed Image Text:**Transcription of the Problem Statement from the Image:** A weight \( W \) is dropped from a height \( h \) on the mid-span at \( C \) of a simply-supported beam as shown, where the spring always remains vertical and is at its natural length when the beam is horizontal. For \( W = 90 \, \text{lb} \) and the stiffness of the spring \( k = 313 \, \text{lb/in} \), determine the static elongation of the spring (\( \delta_{\text{spring}} \)) i.e., \( h = 0 \), [in]. **Diagram Explanation:** 1. **Beam Setup:** - A simply-supported beam \( AB \) is shown with supports at points \( A \) and \( B \). - The distance between \( A \) and \( B \) is noted as \( 40" \). 2. **Weight Placement:** - A weight \( W \) is applied at the mid-span of the beam at point \( C \). - The weight is suspended from a height \( h \) directly above point \( C \). 3. **Spring Details:** - There is a spring attached vertically at the end of the beam at point \( B \). - The spring has a stiffness constant denoted by \( k = 313 \, \text{lb/in} \). 4. **Lengths and Measurements:** - The beam \( AB \) measures \( 40" \) from \( A \) to \( B \). - The length \( BC \) is \( 1" \). - An additional measurement from point \( B \) downwards is labeled \( 2" \). This scenario examines the static elongation of a spring when a weight is applied at the mid-span of a beam, considering the system reaches equilibrium without any initial height difference (\( h = 0 \)).
Expert Solution
Step 1: Write the given data with suitable variables-

k equals 313 space lb divided by in
W equals 90 space lb

The free-body diagram is-

Mechanical Engineering homework question answer, step 1, image 1



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