## Step-by-step solution ### Step 1: Understand the problem - The bar is a rectangular steel strip. - Width \( W = 3.000 \) inches - Thickness \( t = \frac{2}{8} = .25 \) inches - Length \( L = 600 \) inches - The bar is loaded in push-pull fatigue with axial force \( F_a \). - The goal: Find the maximum axial force \( F_a \) that can be applied without causing failure, considering the design factor \( n_d = 2 \). ### Step 2: Determine the cross-sectional area \[ A = W \times t = 3. \times .25 = .75 \text{ in}^2 \] ### Step 3: Find the yield or ultimate strength of steel Assuming standard steel properties: - Ultimate tensile strength \( S_{ut} \) ≈ 75,000 psi (common value for structural steel) - Since the problem involves fatigue, the allowable stress is often taken as the endurance limit or a safe fraction of the ultimate strength. ### Step 4: Calculate the allowable axial stress The design factor \( n_d = 2 \) reduces the maximum allowable stress: \[ \sigma_{allow} = \frac{S_{ut}}{n_d} = \frac{75,000}{2} = 37,500 \text{ psi} \] ### Step 5: Calculate the maximum force Using the allowable stress: \[ F_{max} = \sigma_{allow} \times A = 37,500 \times .75 = 28,125 \text{ lbs} \] ### Final answer: Round to three decimal places: \[ \boxed{28,125.000 \text{ kips}} \] **Note:** 1 kip = 100 lbs, so converting lbs to kips: \[ F_{max} = \frac{28,125}{100} = 28.125 \text{ kips} \] ## **Final answer:** **The largest force that can be applied is \(\boxed{28.125}\) kips.**