Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.
Given,
Length, L = 4 m
Outer diameter, D = 300mm, D= 0.3 m
Thickness, t = 50 mm, t = 0.05 m
Stress produced, σ = 75000 kN/m²
Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²
Calculating the diameter of the cylinder,
Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)
d= 0.2 m
(i) Magnitude of the load P:
Using the relation, σ =P/A
P = σ × A = 75000 × π /4 (D² – d² )
P= 75000 × π/4 (0.3² – 0.2²)
P= 75000 × π/4 (0.09 – 0.04)
P = 2945.2 kN
Hence, Magnitude of the load P is 2945.2 kN.
(ii) Longitudinal strain produced, e :
Using the relation, Strain, (e) = stress/E
e= 75000/(1.5 x 10⁸)= 0.0005
Hence, the Longitudinal strain produced is 0.0005.
(iii)Total decrease in length, dL:
The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.
Strain = change in length/original length
e= dL/L
0.0005 = dL/4
dL = 0.0005 × 4m = 0.002m=2mm
Hence,the decrease in length is 2 mm.
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