Question

A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the minimum possible outside diameter of the tube if the stress is limited to 120 MN/m2. a) 180.85 b) 119.35 c) 210.35 d)102.4s

Answer 1

Answer:

option b) 119.35 mm

Explanation:

Given:

Internal diameter of the hollow steel tube, d = 100 mm = 0.1 m

Tensile load, P = 400 KN = 400000 N

Limiting stress, σ = 120 MN/m² = 120 × 10⁶ N/m²

let the outer diameter of the tube be 'D'

thus,

Cross-sectional area of the tube, A = $\frac{\pi}{4}(D^2-d^2)$

also

stress = Load/Area

therefore,

$120\times10^6=\frac{400000}{\frac{\pi}{4}(D^2-0.1^2)}$

or

$\frac{\pi}{4}(D^2-0.1^2)=\frac{400000}{120\times10^6}$

or

D² - 0.1² = 4.244 × 10⁻³

or

D² = 0.014244 m

or

D = 0.119348 m

or

D = 119.348 ≈ 119.35 mm

hence. option b) is the correct answer