Question #296912

reverse bending stress of 300 N/mm2 is acting on 55 mm diameter steel. the ultimate strenth of steel is 590 N/mm2. Consider the factors ka= 0.45,Kb=0.85,Kc=0.89 and remaining factor are 1. Calculate the life of the bar for the reliability of 90%.


1
Expert's answer
2022-02-15T04:22:01-0500

Se=0.5SutS_e'=0.5S_{ut}


Se=0.5×590=295Nmm2S_e'=0.5\times 590 =295 \frac{N}{mm^2}


Se=(ka)(kb)(kc)(Se)=0.45×0.85×0.89×295=100.42Nmm2S_e= (k_a) (k_b) (k_c) (S_e')=0.45\times0.85\times0.89\times295 =100.42 \frac{N}{mm^2}


After plotting SN curve:


log10N=3+1.37=4.37log_{10}^{N}=3+1.37=4.37


N= 23735 Cycles



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