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3 years ago

A rotating shaft is subjected to a steady torsional stress of 13 ksi and an alternating bending stress of 22 ksi.

Engineering

1 answer:

mixas84 [53]3 years ago

3 0

Answer:

A) б1 = 28 ksi and б2 = -6.02 ksi

B) 1.25

Explanation:

Given data :

Torsional stress = 13 ksi

Alternating bending stress = 22ksi

A) determine yielding factor of safety according to the distortion energy theory

б1,2 = $\frac{22}{2}$ ± √(22/2)² + 13²

= 11 ± 17

therefore б1 = 28 ksi hence б2 = -6.02 ksi

B) determine the fatigue factor of safety

with properties ; Se = 35ksi, Sy = 60 ksi, Sut = 85 ksi

( б1 - б2 )² + ( б2 - б3 )² + ( б3 - б1 )² ≤ 2 ( Sy / FOS ) ²

( 28 + 6.02 ) ² + ( 6.02 - 0 )² + ( 0 - 28 )² ≤ 2 ( 60 / FOS ) ²

solving for FOS = 1.9

Next we can determine FOS with the use of Goodman criterion

бm / Sut + бa / Se = 1 / FOS

= 0 / 85 + 28/35 = 1 / FOS

making FOS the subject of the equation ; hence FOS = 1.25

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