VIPSolutionsGive step-by-step solution with explanation and final answer: An aluminium spacer used in a machine frame is being assessed for elastic deformation under service conditions. During a proof test, the bar is subjected to significant tensile forces, and the engineer must estimate its dimensional changes to ensure proper fit and maintain clearances with adjacent components. Determine the dimensional change of an aluminium bar that is 1 m long, 9 5 mm wide, and 1 8 0 mm thick when subjected to a unidirectional stretch of 7 5 0 kN along its length and thickness direction. Assume that E = 7 6 0 kN / mm 2 and = 0 . 2 2
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Give step-by-step solution with explanation and final answer: An aluminium spacer used in a machine frame is being assessed for elastic deformation under service conditions. During a proof test, the bar is subjected to significant tensile forces, and the engineer must estimate its dimensional changes to ensure proper fit and maintain clearances with adjacent components. Determine the dimensional change of an aluminium bar that is 1 m long, 9 5 mm wide, and 1 8 0 mm thick when subjected to a unidirectional stretch of 7 5 0 kN along its length and thickness direction. Assume that E = 7 6 0 kN / mm 2 and = 0 . 2 2
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Asked by: manish kumar
Created at: 2025-09-26 14:15:41
Asked by: manish kumar
Created at: 2025-09-26 14:15:41
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# Step-by-Step Solution: Elastic Deformation of Aluminium Bar
Given:
- **Length, \( L \):** \(1\,\text{m} = 100\,\text{mm}\)
- **Width, \( b \):** \(95\,\text{mm}\)
- **Thickness, \( \):** \(180\,\text{mm}\)
- **Applied Force, \( F \):** \(750\,\text{kN} = 750,000\,\text{N}\)
- **Direction:** Along length and thickness (assume force is along length; thickness is affected by Poisson's ratio)
- **Young's Modulus, \( E \):** \(760\,\text{kN}/\text{mm}^2 = 760,000\,\text{N}/\text{mm}^2\)
- **Poisson's Ratio, \( \nu \):** \(.22\)
---
## 1. **Calculate Cross-Sectional Area**
\[
A = b \times t = 95\,\text{mm} \times 180\,\text{mm} = 17,100\,\text{mm}^2
\]
---
## 2. **Calculate Axial Stress**
\[
\sigma = \frac{F}{A} = \frac{750,000\,\text{N}}{17,100\,\text{mm}^2} \approx 43.86\,\text{N}/\text{mm}^2
\]
---
## 3. **Calculate Axial Strain**
\[
\varepsilon = \frac{\sigma}{E} = \frac{43.86\,\text{N}/\text{mm}^2}{760,000\,\text{N}/\text{mm}^2} = 5.77 \times 10^{-5}
\]
---
## 4. **Calculate Change in Length (\( \Delta L \))**
\[
\Delta L = \varepsilon \times L = 5.77 \times 10^{-5} \times 100\,\text{mm} = .0577\,\text{mm}
\]
---
## 5. **Calculate Lateral Strain (for Width and Thickness)**
\[
\text{Lateral Strain} = -\nu \times \varepsilon = -.22 \times 5.77 \times 10^{-5} = -1.27 \times 10^{-5}
\]
---
## 6. **Calculate Change in Width (\( \Delta b \))**
\[
\Delta b = \text{Lateral Strain} \times b = -1.27 \times 10^{-5} \times 95\,\text{mm} \approx -.0012\,\text{mm}
\]
---
## 7. **Calculate Change in Thickness (\( \Delta t \))**
\[
\Delta t = \text{Lateral Strain} \times t = -1.27 \times 10^{-5} \times 180\,\text{mm} \approx -.0023\,\text{mm}
\]
---
# **Summary Table**
| Dimension | Original (mm) | Change (mm) | Final Dimension (mm) |
|--------------|---------------|--------------|----------------------|
| Length (\(L\)) | 100 | +.0577 | 100.0577 |
| Width (\(b\)) | 95 | -.0012 | 94.9988 |
| Thickness (\(t\)) | 180 | -.0023 | 179.9977 |
---
# **Final Answers**
- **Change in length (\( \Delta L \)):** \(\boxed{+.058\,\text{mm}}\) (rounded to 3 decimal places)
- **Change in width (\( \Delta b \)):** \(\boxed{-.0012\,\text{mm}}\)
- **Change in thickness (\( \Delta t \)):** \(\boxed{-.0023\,\text{mm}}\)
These are the dimensional changes in the aluminium spacer under the specified load.
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