A 25 -mm-diameter steel rod is 1 m long and is clamped at both ends. If the temperature of the rod is decreased by $$ 60^{\circ} \mathrm{C} $$, calculate the stress induced in the rod if (a) the ends are rigidly fixed; and (b) an allowance of $$ 0,2 \mathrm{~mm} $$ is made for the contraction of the rod. (Let $$ \mathrm{E}=200 \mathrm{GPa} $$ and $$ \alpha=12 imes 10^{-6} /{ }^{\circ} \mathrm{C} $$ )

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A 25 -mm-diameter steel rod is 1 m long and is clamped at both ends. If the temperature of the rod is decreased by $$ 60^{\circ} \mathrm{C} $$, calculate the stress induced in the rod if (a) the ends are rigidly fixed; and (b) an allowance of $$ 0,2 \mathrm{~mm} $$ is made for the contraction of the rod. (Let $$ \mathrm{E}=200 \mathrm{GPa} $$ and $$ \alpha=12 	imes 10^{-6} /{ }^{\circ} \mathrm{C} $$ )

UpStudy AI · Accepted Answer

**Step 1. Compute the free thermal strain**

The free thermal strain is given by  
$$
\epsilon_{\text{thermal}} = \alpha\,\Delta T,
$$  
where  
$$\alpha = 12\times10^{-6}\,/^\circ\text{C}$$ and $$\Delta T = -60^\circ\text{C}$$. Thus,  
$$
\epsilon_{\text{thermal}} = 12\times10^{-6}\times (-60) = -720\times10^{-6} = -0.00072.
$$

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**Part (a): Rigidly Fixed Ends**

With the ends rigidly fixed, the rod cannot contract so that the net strain must be zero. The mechanical strain produced in the rod (which creates stress) must cancel the free thermal strain. Therefore, the induced mechanical strain is  
$$
\epsilon_{\text{mechanical}} = -\epsilon_{\text{thermal}} = 0.00072.
$$

Using Hooke’s law, the induced stress is  
$$
\sigma = E\,\epsilon_{\text{mechanical}},
$$  
with $$E = 200\,\text{GPa} = 200\times10^9\,\text{Pa}$$. Thus,  
$$
\sigma = 200\times10^9 \times 0.00072 = 144\times10^6\,\text{Pa} = 144\,\text{MPa}.
$$

---

**Part (b): Allowance of $$0.2\,\text{mm}$$ Contraction**

The rod’s original length is $$L = 1\,\text{m} = 1000\,\text{mm}$$.

1. **Calculate the free change in length** due to cooling:
   $$
   \Delta L_{\text{free}} = \alpha\,\Delta T\,L = 12\times10^{-6}\times (-60)\times 1000 = -0.72\,\text{mm}.
   $$
   
2. **Determine the constrained change in length.**  
   An allowance of $$0.2\,\text{mm}$$ contraction is made, so the actual change in length is  
   $$
   \Delta L_{\text{actual}} = -0.2\,\text{mm}.
   $$
   
3. **Determine the additional contraction prevented by the constraint:**  
   $$
   \Delta L_{\text{constrained}} = \Delta L_{\text{actual}} - \Delta L_{\text{free}} = (-0.2) - (-0.72) = 0.52\,\text{mm}.
   $$
   
4. **Compute the mechanical strain associated with this constrained change:**
   $$
   \epsilon_{\text{mechanical}} = \frac{\Delta L_{\text{constrained}}}{L} = \frac{0.52\,\text{mm}}{1000\,\text{mm}} = 0.00052.
   $$

5. **Calculate the induced stress:**  
   $$
   \sigma = E\,\epsilon_{\text{mechanical}} = 200\times10^9\,\text{Pa}\times 0.00052 = 104\times10^6\,\text{Pa} = 104\,\text{MPa}.
   $$

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**Final Answers:**

- (a) The induced stress when the ends are rigidly fixed is  
  $$
  144\,\text{MPa}.
  $$
- (b) The induced stress when an allowance of $$0.2\,\text{mm}$$ contraction is made is  
  $$
  104\,\text{MPa}.
  $$
When the rod is rigidly fixed, the induced stress is 144 MPa. When an allowance of 0.2 mm is made for contraction, the induced stress is 104 MPa.

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