Answer:
(a) 56 MPa
(b) 39.61 MPa
(c) 142.85 MPa
Explanation:
The stress developed in a member by application of axial load is given by the formula:
Stress = σ = Force/Area
(a)
In this case,
Force = 70 KN = 70,000 N
Area = 25 mm x 50 mm = 0.025 m x 0.05 m = 0.00125 m²
Therefore,
σ = 70,000 N/0.00125 m²
σ = 56 x 10^6 Pa
σ = 56 MPa
(b)
In this case,
Force = 70 KN = 70,000 N
Area = πD²/4 = π(0.15 m)²/4 = 0.001767 m²
Therefore,
σ = 70,000 N/0.001767 m²
σ = 39.61 x 10^6 Pa
σ = 39.61 MPa
(c)
In this case,
Force = 70 KN = 70,000 N
Area = πD²/4 = π(0.025 m)²/4 = 0.00049 m²
Therefore,
σ = 70,000 N/0.00049 m²
σ = 142.85 x 10^6 Pa
σ = 142.85 MPa
