(B) Find the fatigue factor of safety using the Goodman criterion. A shaft has the properties of Se= 35 ksi, Sy = 60 ksi, and Sut = 85 ksi.
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 = ± √(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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