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 × 10⁻⁶ per K and Young’s modulus of steel as 2 × 10⁵ N/mm²)
- A75·398 kN
- B240 kN
- C115 kN
- D36·128 kNCorrect
Solution
Concept
Thermal stress when the supports yield
Formula
Restrained contraction = α ΔT L − yield; σ = E × (restrained contraction / L)
Step-by-step solution
1. Free contraction = 12×10⁻⁶ × 100 × 8000 = 9.6 mm. 2. Yielding of supports = 5 mm, so restrained contraction = 4.6 mm. 3. Strain = 4.6 / 8000 = 5.75 × 10⁻⁴; stress = 2×10⁵ × 5.75×10⁻⁴ = 115 N/mm². 4. A = π × 20² / 4 = 314.16 mm²; pull = 115 × 314.16 = 36 128 N = 36.128 kN.
Explanation
When the supports yield, only the part of the thermal contraction that is not absorbed by the yielding produces stress. The pull is that stress times the area of the rod.
Why the other options are wrong
Only 36.128 kN satisfies P = E (αΔT L − δ) A / L. (115 is the stress in N/mm², not the force in kN.) The other values do not follow from the given data.
Shortcut
Net strain = (9.6 − 5)/8000; stress = E × strain; P = σA.
Common mistake
Ignoring the yielding of the supports (that would give 120.6 kN).
Correct answer: D) 36·128 kN