The work cycle of a mechanical component subjected to completely reversed bending stresses consists of the following three elements:
(i) ± 350 N/mm² for 85% of time
(ii) ± 400 N/mm² for 12% of time
(iii) ± 500 N/mm² for 3% of time
The material for the component is 50C4 $\left(S_{u t}=\right.\left.660 N / mm ^{2}\right)$ and the corrected endurance limit of the component is 280 N/mm². Determine the life of the component.

Question

The work cycle of a mechanical component subjected to completely reversed bending stresses consists of the following three elements:
(i) ± 350 N/mm² for 85% of time
(ii) ± 400 N/mm² for 12% of time
(iii) ± 500 N/mm² for 3% of time
The material for the component is 50C4 [latex] \left(S_{u t}=\right.\left.660 N / mm ^{2}\right) [/latex] and the corrected endurance limit of the component is 280 N/mm². Determine the life of the component.

Accepted Answer

[latex] ext { Given } S_{u t}=660 N / mm ^{2} \quad S_{e}=280 N / mm ^{2} [/latex].

Step I Construction of S–N diagram

[latex] 0.9 ext { Sut }=0.9(660)=594 N / mm ^{2} [/latex].

[latex] \log 10\left(0.9 S_{u t} ight)=\log _{10}(594)=2.7738 [/latex].

[latex] \log _{10}\left(S_{e} ight)=\log _{10}(280)=2.4472 [/latex].

[latex] \log _{10}\left(\sigma_{1} ight)=\log _{10}(350)=2.5441 [/latex].

[latex] \log _{10}\left(\sigma_{2} ight)=\log _{10}(400)=2.6021 [/latex].

[latex] \log _{10}\left(\sigma_{3} ight)=\log _{10}(500)=2.6990 [/latex].

The S–N curve for this problem is shown in Fig. 5.38.

[latex] ext { Step II Calculation of } N_{1}, N_{2} ext { and } N_{3} [/latex].

From Fig. 5.38,

[latex] \overline{E F}=\frac{\overline{D B} imes \overline{A E}}{\overline{A D}}=\frac{(6-3)\left(2.7738-\log _{10} \sigma ight)}{(2.7738-2.4472)} [/latex] (a).

and [latex] \log _{10} N=3+\overline{E F} [/latex] (b).

From (a) and (b),

[latex] \log _{10} N=3+9.1855\left(2.7738-\log _{10} \sigma ight) [/latex].

Therefore

[latex] \log _{10}\left(N_{1} ight)=3+9.1855(2.7738-2.5441) [/latex].

or [latex] N_{1}=128798 [/latex].

[latex] \log _{10}\left(N_{2} ight)=3+9.1855(2.7738-2.6021) [/latex].

or [latex] N_{2}=37770 [/latex].

[latex] \log _{10}\left(N_{3} ight)=3+9.1855(2.7738-2.6990) [/latex].

or [latex] N_{3}=4865 [/latex].

Step III Fatigue life of component
From Eq. 5.34,

[latex] \frac{\alpha_{1}}{N_{1}}+\frac{\alpha_{2}}{N_{2}}+\ldots+\frac{\alpha_{x}}{N_{x}}=\frac{1}{N} [/latex] (5.34).

[latex] \frac{\alpha_{1}}{N_{1}}+\frac{\alpha_{2}}{N_{2}}+\frac{\alpha_{3}}{N_{3}}=\frac{1}{N} [/latex].

[latex] \frac{0.85}{128798}+\frac{0.12}{37770}+\frac{0.03}{4865}=\frac{1}{N} [/latex].

or N = 62 723 cycles.

Question 5.11: The work cycle of a mechanical component subjected to comple...

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