Answer:
a) the safety factor according to the maximum normal-stress theory; n = 3
b) the safety factor according to the maximum-shear-stress theory; n = 1.714
c) the safety factor according to the maximum distortion-energy theory is; n = 1.97278
Explanation:
Given the data in the question;
a) What is the safety factor according to the maximum normal-stress theory;
According to the maximum normal-stress theory
n = S / σ
since σ₁ = 20 ksi is greater than σ₂ = -15 ksi
σ = 20 ksi and yield strengths in tension and compression S
= 60 ksi
we substitute
n = 60 ksi / 20 ksi
n = 3
Therefore, the safety factor according to the maximum normal-stress theory; n = 3
b) What is the safety factor according to the maximum-shear-stress theory.
According to maximum-shear-stress theory;
τ = [(σ₁ - σ₂) / 2]
= S / 2n
n = S / 2[(σ₁ - σ₂) / 2]
n = S / (σ₁ - σ₂)
we substitute
n = 60 ksi / (20 ksi - (-15 ksi))
n = 60 ksi / (20 ksi +15 ksi)
n = 60 ksi / 35 ksi
n = 1.714
Therefore, the safety factor according to the maximum-shear-stress theory; n = 1.714
c) the safety factor according to the maximum distortion-energy theory?
By distortion energy theory
σ₁² + σ₂² - σ₁σ₂ = (S/n)²
we substitute
(20)² + (-15)² - ( 20 × -15 ) = ( 60 / n )²
400 + 225 + 300 = 3600 / n²
925 = 3600 / n²
n² = 3600 / 925
n = √( 3600 / 925 )
n = 1.97278
Therefore, the safety factor according to the maximum distortion-energy theory is; n = 1.97278
