Two parallel walls, 8 m apart, are stayed together by a steel rod of 20 mm diameter passing through metal plates and nuts at each end. The nuts are screwed up to the plates while the bar is at a temperature of 400 K. What is the pull exerted by the bar after it has cooled to 300 K, if the total yielding at the two ends is 5 mm? (Take coefficient of thermal expansion for steel as $12 \times 10^{-6}$ per K and Young's modulus of steel as $2 \times 10^5 N/mm^2$)
36.128 kN
The problem involves calculating the pull exerted by a steel rod connecting two parallel walls. We need to account for thermal contraction as the rod cools and the mechanical yielding at the ends. Let's solve this step-by-step.
- Initially, the rod is under no stress at 400 K.
- As the rod cools to 300 K, it contracts. Given the coefficient of thermal expansion for steel is \( \alpha = 12 \times 10^{-6} \) per K, the change in temperature, \( \Delta T \), is \( 400 - 300 = 100 \) K.
- The formula for thermal contraction is: \[ \Delta L_{\text{thermal}} = L \cdot \alpha \cdot \Delta T \] Substituting the values (length \( L = 8 \) m = 8000 mm): \[ \Delta L_{\text{thermal}} = 8000 \cdot 12 \times 10^{-6} \cdot 100 = 9.6 \, \text{mm} \]
- Given that the total yielding at both ends is \( 5 \) mm, the effective contraction for which the rod needs to exert pull is: \[ \Delta L_{\text{effective}} = \Delta L_{\text{thermal}} - \text{Yielding} = 9.6 \, \text{mm} - 5 \, \text{mm} = 4.6 \, \text{mm} \]
- The formula for pull (tensile force) exerted due to this contraction with Young’s modulus \( E = 200,000 \, \text{N/mm}^2 \) and cross-sectional area \( A \) is: \[ F = \frac{E \cdot A \cdot \Delta L_{\text{effective}}}{L} \] First, calculate the cross-sectional area, \( A \), of the rod with a diameter of 20 mm: \[ A = \frac{\pi \cdot d^2}{4} = \frac{\pi \cdot (20)^2}{4} = 314.16 \, \text{mm}^2 \] Using this area in the force formula: \[ F = \frac{200,000 \cdot 314.16 \cdot 4.6}{8000} \approx 36,128 \, \text{N} \] \]
- The pull exerted by the bar is hence \( 36,128 \, \text{N} \), which is equivalent to \( 36.128 \, \text{kN} \).
Thus, the correct answer is 36.128 kN.
Important Questions from Thermal Stresses
Consider the following statements regarding thermal stresses:
1. If the temperature change is uniform throughout the body, the thermal strain is also uniform.
2. If thermal deformation is permitted to occur freely, no internal forces will be induced in the body, and there will be no strain and no stress.
3. If the deformation of a body is restricted, either totally or partially, internal forces will develop that oppose the thermal expansion or contraction. The stresses caused by these internal forces are known as thermal stresses.
Which of the above statements are correct?
