A steel rotating-beam test specimen has an ultimate strength of 1100 MPa. Estimate the fatigue strength corresponding to a life of 150000 cycles of stress reversal.

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To estimate the fatigue strength corresponding to a life of 150,000 cycles for a steel specimen with an ultimate strength (${\sigma}_{u}\backslash sigma\_u$σu) of 1100 MPa, we can use the S-N (stress-life) curve approach, particularly focusing on the Basquin’s equation which relates the fatigue strength to the number of cycles to failure.

The general approach involves the following steps:

Determine the endurance limit (${\sigma}_{e}\backslash sigma\_e$σe) for the material:For steels, the endurance limit is often taken as approximately half of the ultimate tensile strength for a high number of cycles (usually greater than $1{0}^{6}10^6$106 cycles). However, since we’re dealing with 150,000 cycles, we need to use a different approach.

${\sigma}_{e}\approx 0.5{\sigma}_{\mathrm{u}}$Given:

${\sigma}_{u}=1100\hspace{0.17em}MPa\backslash sigma\_u\; =\; 1100\; \backslash ,\; \backslash text\{MPa\}$

${\sigma}_{e}\approx 0.5\times 1100=550\hspace{0.17em}MPa\backslash sigma\_e\; \backslash approx\; 0.5\; \backslash times\; 1100\; =\; 550\; \backslash ,\; \backslash text\{MPa\}$

Use the modified Goodman relationship or similar to find the fatigue strength at 150,000 cycles:The Basquin’s equation, which relates the stress amplitude (${\sigma}_{a}\backslash sigma\_a$σa) to the number of cycles to failure (${N}_{f}N\_f$Nf), is given by:

${\sigma}_{a}={\sigma}_{f}^{\prime}({N}_{f}{)}^{b}\backslash sigma\_a\; =\; \backslash sigma\_f\u2019\; (N\_f)^b$

where:

To simplify, many references use empirical relationships derived from testing. For example, for finite life fatigue strength calculations, one common approach is to use a factor based on the number of cycles.

For ${N}_{f}=150,000N\_f\; =\; 150,000$ cycles, we often use the fatigue strength reduction factor (which typically falls between the ultimate strength and the endurance limit).

Using empirical relationships for steels, we can use:

${\sigma}_{a}\approx {\sigma}_{e}{\left(\frac{{N}_{e}}{{N}_{f}}\right)}^{k}$where:

Given:

${N}_{e}=1{0}^{6}\hspace{0.17em}cyclesN\_e\; =\; 10^6\; \backslash ,\; \backslash text\{cycles\}$

${N}_{f}=150,000\hspace{0.17em}cyclesN\_f\; =\; 150,000\; \backslash ,\; \backslash text\{cycles\}$

${\sigma}_{e}=550\hspace{0.17em}MPa\backslash sigma\_e\; =\; 550\; \backslash ,\; \backslash text\{MPa\}$

$k\approx 0.15k\; \backslash approx\; 0.15$

The equation becomes:

${\sigma}_{a}\approx 550{\left(\frac{1{0}^{6}}{150,000}\right)}^{0.15}$Calculate the ratio:

$\frac{1{0}^{6}}{150,000}\approx 6.67$Then raise to the power of 0.15:

$6.6{7}^{0.15}\approx 1.45$So:

${\sigma}_{a}\approx 550\times 1.45\approx 798\hspace{0.17em}MPa\backslash sigma\_a\; \backslash approx\; 550\; \backslash times\; 1.45\; \backslash approx\; 798\; \backslash ,\; \backslash text\{MPa\}$

Thus, the estimated fatigue strength corresponding to a life of 150,000 cycles of stress reversal for a steel specimen with an ultimate strength of 1100 MPa is approximately $798\hspace{0.17em}MPa798\; \backslash ,\; \backslash text\{MPa\}$.