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Solve the question as stated with a graph and solve the question in 2 steps with examples and the final answer.10.5-12 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P = 175 kN, acting as shown in the figure. The reaction ¢ of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b =300 mmand 4 = 250 mm. Calculate the maximum bending stress or, in the tie due to the loads P, assuming the distance L = 1500 mmand the overhang length ¢ = 500 mm . JZ JZ i L—> “— [— > BE q

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Solve the question as stated with a graph and solve the question in 2 steps with examples and the final answer.Uploaded Image10.5-12 A railroad tie (or sleeper) is subjected to two rail loads, each of magnitude P = 175 kN, acting as shown in the figure. The reaction ¢ of the ballast is assumed to be uniformly distributed over the length of the tie, which has cross-sectional dimensions b =300 mmand 4 = 250 mm. Calculate the maximum bending stress or, in the tie due to the loads P, assuming the distance L = 1500 mmand the overhang length ¢ = 500 mm . JZ JZ i L—> “— [— > BE q

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Asked by: Rahul
Created at: 2026-01-02 19:14:37

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# Solution to Maximum Bending Stress in Railroad Tie Let's break the solution into **2 steps** and include a diagram for clarity. --- ## **Given Data** - Load \( P = 175\, \text{kN} = 175,000\, \text{N} \) - Overhang \( a = 500\, \text{mm} \) - Distance between loads \( L = 150\, \text{mm} \) - Width \( b = 300\, \text{mm} \) - Height \( h = 250\, \text{mm} \) --- ## **Step 1: Find the Maximum Bending Moment** The tie is simply supported by the ballast (uniformly distributed support), with two concentrated loads. - **Total length supported by ballast:** \( L + 2a \) - **Maximum moment occurs under each load.** The **reaction per unit length** \( q \) is: \[ q = \frac{2P}{L + 2a} \] The **maximum bending moment** at the point under the load (distance \( a \) from the end) is: \[ M_{\text{max}} = P \cdot a - q \cdot \frac{a^2}{2} \] ### **Calculations:** \[ L + 2a = 150 + 2 \times 500 = 250\, \text{mm} \] \[ q = \frac{2 \times 175,000}{250} = 140\, \text{N/mm} \] \[ M_{\text{max}} = 175,000 \times 500 - 140 \times \frac{500^2}{2} \] \[ M_{\text{max}} = 87,500,000 - 140 \times 125,000 \] \[ M_{\text{max}} = 87,500,000 - 17,500,000 = 70,000,000\, \text{N}\cdot\text{mm} \] --- ## **Step 2: Find the Maximum Bending Stress (\( \sigma_{\text{max}} \))** The maximum bending stress is: \[ \sigma_{\text{max}} = \frac{M_{\text{max}} \cdot c}{I} \] Where: - \( c = \frac{h}{2} = 125\, \text{mm} \) - \( I = \frac{b h^3}{12} = \frac{300 \times 250^3}{12} \) ### **Calculations:** \[ I = \frac{300 \times 15,625,000}{12} = \frac{4,687,500,000}{12} = 390,625,000\, \text{mm}^4 \] \[ \sigma_{\text{max}} = \frac{70,000,000 \times 125}{390,625,000} \] \[ \sigma_{\text{max}} = \frac{8,750,000,000}{390,625,000} \] \[ \sigma_{\text{max}} \approx 22.4\, \text{MPa} \] --- ## **Graph Illustration** ```plaintext |<---a-->|<------L------>|<---a---> |-------|==============|-------| ^ ^ | | P P Uniform distributed reaction q over entire length b (width) = 300 mm, h (height) = 250 mm (cross-section) ``` --- ## **Final Answer** **The maximum bending stress in the tie is:** \[ \boxed{22.4\, \text{MPa}} \]

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