Calculate the stress developed in the following members that are each subjected to an axial tensile load of 70 kN:

(a) steel bar, 25 mm by 50 mm

(b) 150-mm-diameter wood post (use the given diameter)

(c) 25-mm-diameter steel tie rod

(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