This problem involves determining the factor of safety for both infinite life and yielding for a round bar subjected to axial and torsional stresses. The given data is:

• Axial stress range: from compressive $\sigma_{\text{min}} = -20 \, \text{ksi}$ to tensile $\sigma_{\text{max}} = 50 \, \text{ksi}$
• Fully reversed torsional shear stress: $\tau_{\text{rev}} = 10 \, \text{ksi}$
• Ultimate strength: $S_u = 80 \, \text{ksi}$
• Yield strength: $S_y = 67 \, \text{ksi}$
• Modified endurance limit: $S_e = 30 \, \text{ksi}$

To solve the problem, we need to calculate the factor of safety for both infinite life (fatigue) and yielding.

(a) Factor of Safety for Infinite Life (Fatigue)

For infinite life, the factor of safety is calculated using the Modified Goodman relation, which accounts for both mean and alternating stresses. The alternating and mean stress components for axial and shear stress need to be calculated.

1. Alternating Stress $\sigma_a$ (axial): $\sigma_a = \frac{\sigma_{\text{max}} - \sigma_{\text{min}}}{2} = \frac{50 - (-20)}{2} = 35 \, \text{ksi}$

2. Mean Stress $\sigma_m$ (axial): $\sigma_m = \frac{\sigma_{\text{max}} + \sigma_{\text{min}}}{2} = \frac{50 + (-20)}{2} = 15 \, \text{ksi}$

3. Shear Stress: Since the torsional shear stress is fully reversed, the alternating shear stress $\tau_a$ is the same as the given shear stress: $\tau_a = \tau_{\text{rev}} = 10 \, \text{ksi}$

Next, using von Mises criteria for combined stress, the effective alternating and mean stress can be calculated as follows:

4. Equivalent Alternating Stress $\sigma_a'$ (von Mises): $\sigma_a' = \sqrt{\sigma_a^2 + 3\tau_a^2} = \sqrt{35^2 + 3 \times 10^2} = \sqrt{1225 + 300} = \sqrt{1525} \approx 39.06 \, \text{ksi}$

5. Equivalent Mean Stress $\sigma_m'$ (von Mises): $\sigma_m' = \sqrt{\sigma_m^2} = 15 \, \text{ksi}$

Now, apply the Modified Goodman equation: $\frac{\sigma_a'}{S_e} + \frac{\sigma_m'}{S_u} = \frac{1}{\text{Factor of Safety (FOS)}}$ $\frac{39.06}{30} + \frac{15}{80} = \frac{1}{\text{FOS}}$ $1.302 + 0.1875 = \frac{1}{\text{FOS}}$ $\text{FOS} = \frac{1}{1.4895} \approx 0.6715$

Thus, the factor of safety for infinite life is approximately 0.6715.

(b) Factor of Safety for Yielding

For yielding, the factor of safety is based on the yield strength, and we use the maximum stress under von Mises criteria:

$\sigma_{\text{equiv}} = \sqrt{\sigma_m^2 + 3 \tau_a^2} = \sqrt{15^2 + 3 \times 10^2} = \sqrt{225 + 300} = \sqrt{525} \approx 22.91 \, \text{ksi}$

The factor of safety for yielding is: $\text{FOS}_{\text{yield}} = \frac{S_y}{\sigma_{\text{equiv}}} = \frac{67}{22.91} \approx 2.93$

Thus, the factor of safety for yielding is approximately 2.93.

Let me know if you would like more details or have any questions!

5 Related Questions:

1. How is von Mises stress used to combine different stress components?
2. What is the significance of the modified Goodman relation in fatigue analysis?
3. How do you determine the alternating and mean stress for fluctuating loading conditions?
4. How does the endurance limit influence the factor of safety for fatigue?
5. What are the differences between yield strength and ultimate strength?

Tip:

When calculating the factor of safety, always ensure that all stress components (axial, shear) are properly combined using appropriate criteria like von Mises for multiaxial states.