Question #119037

A hollow cast iron cylinder, 4m long 30mm outer diameter and the thickness of 50mm. Is subjected to a central load on the top when standing straight. The stress produced is 750KN/m^2, assume Young's modulus of cast iron = 1.5×10^8KN/m. Find

1. Magnitude of the load

2. Longitudinal strain produced

3. The total decrease in length

1. Magnitude of the load

2. Longitudinal strain produced

3. The total decrease in length

Expert's answer

Outer diameter, D = 30 mm = 0.03 m Thickness, t = 50 mm = 0.05 m Length, L = 4 m

Stress produced, σ = 750 kN/m2

E = 1.5 x 10^8 kN/m2

Here diameter of the cylinder, d = D – 2t = 0.03 – 2 × 0.05 = 0.07 m

(i) Magnitude of the load P: Using the relation,

σ =P/A or P = σ × A = 750 × Π/4 (D2 – d2 ) = 750 × Π/4 (0.03 – 0.07) or P = 23.5619 kN

(ii) Longitudinal strain produced, e : Using the relation, Strain, (e) = stress/E = 750/1.5 x 10^8 = 0.000005

(iii)Total decrease in length, dL: Using the relation, Strain = change in length/original length = dLA/L 50 = dLA/4 dLA = 0.000005 × 4m = 0.00002m=0.02mm

Hence decrease in length = 0.02 mm

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