A Steel Rotating Beam Test Specimen

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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1. To estimate the fatigue strength corresponding to a life of 150,000 cycles for a steel specimen with an ultimate strength (${\sigma }_{u}\sigma_u$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 (${\sigma }_{e}\sigma_e$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:

${\sigma }_{u}=1100 MPa\sigma_u = 1100 \, \text\left\{MPa\right\}$
${\sigma }_{e}\approx 0.5×1100=550 MPa\sigma_e \approx 0.5 \times 1100 = 550 \, \text\left\{MPa\right\}$

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 (${\sigma }_{a}\sigma_a$σa) to the number of cycles to failure (${N}_{f}N_f$Nf), is given by:

${\sigma }_{a}={\sigma }_{f}^{\prime }\left({N}_{f}{\right)}^{b}\sigma_a = \sigma_f’ \left(N_f\right)^b$
where:

• ${\sigma }_{f}^{\prime }\sigma_f’$σf is the fatigue strength coefficient (often close to ${\sigma }_{u}\sigma_u$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 $1{0}^{6}10^6$106 cycles for endurance limit),
• $kk$k is a material constant (typically around 0.15 for steels).

Given:

${N}_{e}=1{0}^{6} cyclesN_e = 10^6 \, \text\left\{cycles\right\}$
${N}_{f}=150,000 cyclesN_f = 150,000 \, \text\left\{cycles\right\}$
${\sigma }_{e}=550 MPa\sigma_e = 550 \, \text\left\{MPa\right\}$
$k\approx 0.15k \approx 0.15$
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:

${\sigma }_{a}\approx 550×1.45\approx 798 MPa\sigma_a \approx 550 \times 1.45 \approx 798 \, \text\left\{MPa\right\}$

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 $798 MPa798 \, \text\left\{MPa\right\}$.