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## Q. 5.6

A rotating bar made of steel 45C8 $\left(S_{u t}=630 N / mm ^{2}\right)$ is subjected to a completely reversed bending stress. The corrected endurance limit of the bar is 315 N/mm². Calculate the fatigue strength of the bar for a life of 90,000 cycles

## Verified Solution

$\text { Given } S_{u t}=630 N / mm ^{2} \quad S_{e}=315 N / mm ^{2}$.

N = 90000 cycles.

Step I Construction of S–N diagram.

$0.9 S_{u t}=0.9(630)=567 N / mm ^{2}$.

$\log _{10}\left(0.9 S_{u t}\right)=\log _{10}(567)=2.7536$.

$\log _{10}\left(S_{e}\right)=\log _{10}(315)=2.4983$.

$\log _{10}(90000)=4.9542$.

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

Figure 5.30 shows the S–N curve for the bar.

Step II Fatigue strength for 90000 cycles
Referring to Fig. 5.30

$\log 10\left(S_{f}^{\prime}\right)=2.7536-\frac{(2.7536-2.4983)}{(6-3)}$

$\times(4.9542-3)=2.5873$.

$S_{f}^{\prime}=386.63 N / mm ^{2}$. 