Let's solve the problem step by step. --- ## **Step 1: Cross-Sectional Area Calculation** Given: - \( a = 500\, \text{mm} \) - \( b = 250\, \text{mm} \) The cross-sectional area \( A \) is: \[ A = a \times b = 500\, \text{mm} \times 250\, \text{mm} = 125,000\, \text{mm}^2 \] Convert to \(\text{m}^2\): \[ A = 125,000\, \text{mm}^2 \times \left(\frac{1\, \text{m}}{100\, \text{mm}}\right)^2 = .125\, \text{m}^2 \] --- ## **Step 2: Normal Stress Calculation** Axial force \( P = 227\, \text{kN} = 227,000\, \text{N} \) Normal stress formula: \[ \sigma = \frac{P}{A} \] Plug in the values: \[ \sigma = \frac{227,000\, \text{N}}{125,000\, \text{mm}^2} \] Convert \( \text{mm}^2 \) to \( \text{m}^2 \) (already done above): Or, directly in MPa (since \( 1\, \text{MPa} = 1\, \text{N}/\text{mm}^2 \)): \[ \sigma = \frac{227,000\, \text{N}}{125,000\, \text{mm}^2} = 1.816\, \text{N}/\text{mm}^2 = 1.816\, \text{MPa} \] --- ## **Step 3: Final Answer** **The normal stress at point H is:** \[ \boxed{1.82\, \text{MPa}} \] (Rounded to two decimal places.) --- ### **Summary Table** | Quantity | Value | |-----------------|------------------| | Area (A) | \(125,000\, \text{mm}^2\) | | Axial Force (P) | \(227,000\, \text{N}\) | | Normal Stress | \(1.82\, \text{MPa}\) | --- **Final Answer:** **1.82 MPa**