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## Q. 5.7

A forged steel bar, 50 mm in diameter, is subjected to a reversed bending stress of 250 N/mm². The bar is made of steel 40C8 $\left(S_{u t}\right. \left.=600 N / mm ^{2}\right)$. Calculate the life of the bar for a reliability of 90%.

## Verified Solution

$\text { Given } S_{f}=\sigma_{b}=250 N / mm ^{2}$.

$S_{u t}=600 N / mm ^{2} \quad R=90 \%$.

Step I Construction of S–N diagram

$S_{e}^{\prime}=0.5 S_{u t}=0.5(600)=300 N / mm ^{2}$.

$\text { From Fig. } 5.24,\left(S_{u t}=600 N / mm ^{2}\right. \text { and forged bar), }$.

$K_{a}=0.44$.

$\text { For } 50 mm \text { diameter, } K_{b}=0.85$.

$\text { For } 90 \% \text { reliability, } K_{c}=0.897$.

$S_{e}=K_{a} K_{b} K_{c} S_{e}^{\prime}=0.44(0.85)(0.897)(300)$.

=100.64 N/mm².

$0.9 S_{u t}=0.9(600)=540 N / mm ^{2}$.

$\log _{10}\left(0.9 S_{u t}\right)=\log _{10}(540)=2.7324$.

$\log _{10}\left(S_{e}\right)=\log _{10}(100.64)=2.0028$.

$\log 10\left(S_{f}\right)=\log _{10}(250)=2.3979$.

$\text { Also, } \log _{10}\left(10^{3}\right)=3 \text { and } \log _{10}\left(10^{6}\right)=6$.

The S–N curve for the bar is shown in Fig. 5.31.

Step II Fatigue life of bar
From Fig. 5.31,

$\overline{E F}=\frac{\overline{D B} \times \overline{A E}}{\overline{A D}}=\frac{(6-3)(2.7324-2.3979)}{(2.7324-2.0028)}$.

$=1.3754$.

Therefore,

$\log _{10} N=3+\overline{E F}=3+1.3754$.

$\log _{10} N=4.3754$.

N = 23 736.2 cycles.