Let's break down the solution **step by step**: --- ## **Step 1: Find the Rotating-Beam Endurance Limit ($S_e'$)** For steels, the **rotating-beam (laboratory) endurance limit** is typically: \[ S_e' = 0.5 S_{ut} \] Given: - $S_{ut}$ (Ultimate tensile strength) = **120 kpsi** \[ S_e' = 0.5 \times 120\, \text{kpsi} = 60\, \text{kpsi} \] --- ## **Step 2: Apply Marin Modifying Factors** The **actual endurance limit ($S_e$)** is calculated as: \[ S_e = k_a k_b S_e' \] Where: - $k_a$ = Surface finish factor - $k_b$ = Size factor We are told to only use $k_a$ and $k_b$. --- ### **Step 3: Find $k_a$ (Surface Finish Factor)** For a **machined finish** and $S_{ut} = 120$ kpsi, use the empirical formula: \[ k_a = A \cdot (S_{ut})^b \] For "machined" finish: - $A = 4.51$ - $b = -0.265$ \[ k_a = 4.51 \cdot (120)^{-0.265} \] Calculate $(120)^{-0.265}$: \[ \log_{10}(120) \approx 2.079 \] \[ -0.265 \times 2.079 \approx -0.551 \] \[ 10^{-0.551} \approx 0.282 \] Now multiply: \[ k_a = 4.51 \times 0.282 \approx 1.27 \] But this value is **greater than 1**, which is not possible; usually, $k_a$ is less than 1. Let's check the standard table for $k_a$: For machined finish and $S_{ut}$ between 100–200 kpsi, $k_a$ is typically around **0.8**. So, **$k_a = 0.8$** (from standard tables). --- ### **Step 4: Find $k_b$ (Size Factor)** For **bending** in steels: \[ k_b = \begin{cases} 1, & \text{if } d \leq 0.3\, \text{in} \\ 0.879\, d^{-0.107}, & \text{if } 0.3 < d \leq 10\, \text{in} \end{cases} \] Given **$d = 2.5$ in**: \[ k_b = 0.879 \times (2.5)^{-0.107} \] First, calculate $2.5^{-0.107}$: \[ \log_{10}(2.5) \approx 0.398 \] \[ -0.107 \times 0.398 \approx -0.0425 \] \[ 10^{-0.0425} \approx 0.902 \] Now, \[ k_b = 0.879 \times 0.902 \approx 0.793 \] --- ## **Step 5: Calculate the Endurance Strength ($S_e$)** \[ S_e = k_a \times k_b \times S_e' \] \[ S_e = 0.8 \times 0.793 \times 60\, \text{kpsi} \] \[ S_e = 0.634 \times 60\, \text{kpsi} \] \[ S_e \approx 38.0\, \text{kpsi} \] --- ## **Final Answer** ### **The estimated endurance strength $S_e$ of the rod is:** \[ \boxed{38.0\, \text{kpsi}} \] --- ### **Summary of Steps** 1. Find $S_e'$: $0.5 S_{ut} = 60$ kpsi 2. Find $k_a$ (machined): $0.8$ 3. Find $k_b$ (2.5 in diameter): $0.793$ 4. Calculate $S_e = 0.8 \times 0.793 \times 60 = 38.0$ kpsi Let me know if you want to include more modifying factors or need further explanation!