The values of warping stress coefficient Cx and Cy are 0.87 and 0.72 respectively. The modulus of elasticity of concrete is 3 × 10⁵ kg/cm². Assume thermal coefficient of concrete = 10 × 10⁻⁶ per °C. The maximum temperature difference between the top and bottom of the slab during summer day and night is found to be 15°C. The Poisson’s ratio of concrete is 0.15. Estimate the warping stress at the interior region.
- A22.5 kg/cm²Correct
- B23.0 kg/cm²
- C23.5 kg/cm²
- D24.0 kg/cm²
Solution
Concept
Interior warping stress
Formula
σ = (Eαt/2)(Cx + νCy)/(1 − ν²)
Step-by-step solution
1. Eαt/2 = 3×10⁵ × 10⁻⁵ × 15/2 = 22.5. 2. (0.87 + 0.15 × 0.72)/(1 − 0.0225) = 1.0. 3. σ = 22.5 kg/cm².
Explanation
Bradbury's interior formula.
Why the other options are wrong
Others do not follow.
Correct answer: A) 22.5 kg/cm²