A prismatic bar has a cross-section of 25 mm x 50 mm and a length of 2000 mm. Under an axial load of 100 kN, the measured elongation of the bar is 2 mm. The tensile stress and % strain in the bar

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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**Problem Statement:**

A prismatic bar has a cross-section of \(25 \, \text{mm} \times 50 \, \text{mm}\) and a length of \(2000 \, \text{mm}\). Under an axial load of \(100 \, \text{kN}\), the measured elongation of the bar is \(2 \, \text{mm}\). The tensile stress and percentage strain in the bar are required to be calculated.

**Solution:**

1. **Calculate the Tensile Stress:**
   - **Formula:** Tensile stress \( \sigma = \frac{F}{A} \)
   - **Given:**
     - Axial load \( F = 100 \, \text{kN} = 100,000 \, \text{N} \)
     - Cross-sectional area \( A = 25 \, \text{mm} \times 50 \, \text{mm} = 1250 \, \text{mm}^2 \)
   - **Convert \( A \) to \( \text{m}^2\) for consistency:** \( 1250 \, \text{mm}^2 = 1250 \times 10^{-6} \, \text{m}^2 \)
   - **Calculation:**
     \[
     \sigma = \frac{100,000 \, \text{N}}{1250 \times 10^{-6} \, \text{m}^2} = 80 \, \text{MPa}
     \]

2. **Calculate the Percentage Strain:**
   - **Formula:** Percentage strain \( \epsilon \% = \left(\frac{\Delta L}{L_0}\right) \times 100 \)
   - **Given:**
     - Elongation \( \Delta L = 2 \, \text{mm} \)
     - Original length \( L_0 = 2000 \, \text{mm} \)
   - **Calculation:**
     \[
     \epsilon \% = \left(\frac{2 \, \text{mm}}{2000 \, \text{mm}}\right) \times 100 = 0.1\%
     \]

**Results:**
- **Tensile Stress:** 80 MPa
- **Percentage Strain:** 0.
Transcribed Image Text:**Problem Statement:** A prismatic bar has a cross-section of \(25 \, \text{mm} \times 50 \, \text{mm}\) and a length of \(2000 \, \text{mm}\). Under an axial load of \(100 \, \text{kN}\), the measured elongation of the bar is \(2 \, \text{mm}\). The tensile stress and percentage strain in the bar are required to be calculated. **Solution:** 1. **Calculate the Tensile Stress:** - **Formula:** Tensile stress \( \sigma = \frac{F}{A} \) - **Given:** - Axial load \( F = 100 \, \text{kN} = 100,000 \, \text{N} \) - Cross-sectional area \( A = 25 \, \text{mm} \times 50 \, \text{mm} = 1250 \, \text{mm}^2 \) - **Convert \( A \) to \( \text{m}^2\) for consistency:** \( 1250 \, \text{mm}^2 = 1250 \times 10^{-6} \, \text{m}^2 \) - **Calculation:** \[ \sigma = \frac{100,000 \, \text{N}}{1250 \times 10^{-6} \, \text{m}^2} = 80 \, \text{MPa} \] 2. **Calculate the Percentage Strain:** - **Formula:** Percentage strain \( \epsilon \% = \left(\frac{\Delta L}{L_0}\right) \times 100 \) - **Given:** - Elongation \( \Delta L = 2 \, \text{mm} \) - Original length \( L_0 = 2000 \, \text{mm} \) - **Calculation:** \[ \epsilon \% = \left(\frac{2 \, \text{mm}}{2000 \, \text{mm}}\right) \times 100 = 0.1\% \] **Results:** - **Tensile Stress:** 80 MPa - **Percentage Strain:** 0.
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