A rotating steel shaft at 2,400 rpm. The finite life portion of the S-N diagram

Question

A rotating steel shaft at 2,400 rpm. The finite life portion of the S-N diagram for the shaft obeys the relation Sf=229.73N^(-0.14)ksi (Sf is the fatigue strength in ksi, N is in revolutions). The shaft operates in an overload condition for 30 seconds during which time the stress amplitude is 72 ksi. The life at this stress level is 3,974 revolutions. The load is decreased to 60 ksi and th eshaft continues to operate at this stress level for another 60 seconds. The part life at this stress level is 14,614 revolutions. The load is then reduced to 50 ksi. The life of an undamaged part at this stress level is 53,745 revolutions. Given the previous overload operating conditions, how many cycles would you expect the part to survive at 50 ksi? a.) 18,111 b.) 28,688
c.) 53,745 d.) an infinite number of revolutions

Explanation / Answer

The answer is b) 28688.

Use Miner's rule:

at 72 ksi, N1 = 3974 sycles, n1 = 1200 cycles (30 second at 2400 rpm = 1200 cycles total)

at 60 ksi, N2 = 14614 sycles, n2 = 2400 cycles (60 second at 2400 rpm = 2400 cycles total)

at 50 ksi N3 = 53745 sycles, n3 = ?

From Miner's rule:

n1/N1 + n2/N2 + n3/N3 =1   which gives n3 = 28688

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