# Answer to Question #119037 in Civil and Environmental Engineering for Hens

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
2. Longitudinal strain produced
3. The total decrease in length
1
2020-06-05T05:37:22-0400

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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