A machine component is loaded so that stresses at the critical location are σ1 = 20 ksi, σ2 = -15 ksi, and σ3 = 0. The material is ductile, with yield strengths in tension and compression of 60 ksi. What is the safety factor according to (a) the maximum normal-stress theory, (b) the maximum-shear-stress theory, and (c) the maximum distortion-energy theory?

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

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