A Steel Rotating Beam Test Specimen

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:

1. 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 $10^{6}10^6$106 cycles). However, since we’re dealing with 150,000 cycles, we need to use a different approach.

   ${\sigma }_e\approx 0.5{\sigma }_{\mathrm{u<span></span>}}$Given:

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

2. 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:

   ${\sigma }_a={\sigma }_f^{\prime }(N_f{)}^b\backslash sigma\_a\; =\; \backslash sigma\_f’\; (N\_f)^b$
   where:

   • ${\sigma }_f^{\prime }\backslash sigma\_f’$σf′​ is the fatigue strength coefficient (often close to ${\sigma }_u\backslash sigma\_u$σu​),
   • $bb$b is the fatigue strength exponent (typically around -0.12 to -0.15 for steels),
   • $N_{f}N\_f$Nf​ is the number of cycles to failure.

   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:

   • $N_{e}N\_e$Ne​ is the reference number of cycles (typically $10^{6}10^6$106 cycles for endurance limit),
   • $kk$k is a material constant (typically around 0.15 for steels).

   Given:

   $N_e=10^6\hspace{0.17em}cyclesN\_e\; =\; 10^6\; \backslash ,\; \backslash text\{cycles\}$
   $N_f=150,000\hspace{0.17em}cyclesN\_f\; =\; 150,000\; \backslash ,\; \backslash text\{cycles\}$
   ${\sigma }_e=550\hspace{0.17em}MPa\backslash sigma\_e\; =\; 550\; \backslash ,\; \backslash text\{MPa\}$
   $k\approx 0.15k\; \backslash approx\; 0.15$
   The equation becomes:

   ${\sigma }_a\approx 550{\left(\frac{10^6}{150,000}\right)}^{0.15}$Calculate the ratio:

   $\frac{10^6}{150,000}\approx 6.67$Then raise to the power of 0.15:

   $6.67^{0.15}\approx 1.45$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 $798\hspace{0.17em}MPa798\; \backslash ,\; \backslash text\{MPa\}$.