QUESTION 4 A light, horizontal beam, ABCDE, with \( A \) on the left-hand side is 26 m long. It is supported at two points, \( A \) and \( D \). \( A \) point load of 20 kN is at \( B, 12 \mathrm{~m} \) from \( A \). \( A \) point load of 30 is at \( C, 6 \mathrm{~m} \) from \( B \). A uniform distributed load of \( 8 \mathrm{kN} / \mathrm{m} \) is between \( A \) and \( B \). \( D \) is 4 m from C. \( \begin{array}{ll}\text { 4.1. } \\ 4.2 & \text { Calculate the reactions of the supports at points } A \text { and } D \text { and test your } \\ \text { answers. }\end{array} \) 4.3

**Step 1. Geometry and Load Details** Assign coordinate \( x=0 \) at point \( A \). Then: - \( B \) is \( 12 \, \text{m} \) from \( A \): \( x_B = 12 \, \text{m} \). - \( C \) is \( 6 \, \text{m} \) from \( B \): \( x_C = 12+6 = 18 \, \text{m} \). - \( D \) is \( 4 \, \text{m} \) from \( C \): \( x_D = 18+4 = 22 \, \text{m} \). - The beam ends at \( E \) with \( x_E = 26 \, \text{m} \). Loads applied: - A uniform distributed load (UDL) of \( 8 \, \text{kN/m} \) acts between \( A \) and \( B \). Total load from the UDL: \[ W_{\text{UDL}} = 8 \times 12 = 96 \, \text{kN}. \] Its resultant acts at the centroid of the load distribution, i.e. at \( x = 6 \, \text{m} \). - A point load of \( 20 \, \text{kN} \) at \( B \) (\( x = 12 \, \text{m} \)). - A point load of \( 30 \, \text{kN} \) at \( C \) (\( x = 18 \, \text{m} \)). Supports are at points \( A \) and \( D \) with unknown vertical reactions \( R_A \) and \( R_D \), respectively. --- **Step 2. Vertical Force Equilibrium** The sum of the upward forces must equal the sum of the downward forces: \[ R_A + R_D = 96 + 20 + 30 = 146 \, \text{kN}. \] Thus, \[ (1) \quad R_A + R_D = 146 \, \text{kN}. \] --- **Step 3. Moment Equilibrium About Point \( A \)** Taking moments about \( A \) (with anticlockwise moments as positive): - The UDL of \( 96 \, \text{kN} \) acts at \( x = 6 \, \text{m} \), giving a moment: \[ M_{\text{UDL}} = 96 \times 6 = 576 \, \text{kN}\cdot\text{m} \quad (\text{clockwise, hence negative}). \] - The \( 20 \, \text{kN} \) load at \( x = 12 \, \text{m} \) gives: \[ M_B = 20 \times 12 = 240 \, \text{kN}\cdot\text{m} \quad (\text{clockwise}). \] - The \( 30 \, \text{kN} \) load at \( x = 18 \, \text{m} \) gives: \[ M_C = 30 \times 18 = 540 \, \text{kN}\cdot\text{m} \quad (\text{clockwise}). \] - The reaction \( R_D \) at \( D \) (\( x=22 \, \text{m} \)) produces an anticlockwise moment: \[ M_{R_D} = R_D \times 22. \] Taking anticlockwise as positive, the moment equilibrium equation about \( A \) is: \[ R_D \times 22 - (576 + 240 + 540) = 0. \] Sum the moments caused by the loads: \[ 576 + 240 + 540 = 1356 \, \text{kN}\cdot\text{m}. \] Thus, \[ 22 R_D = 1356. \] Solve for \( R_D \): \[ R_D = \frac{1356}{22} \approx 61.64 \, \text{kN}. \] --- **Step 4. Find \( R_A \)** Substitute \( R_D \) into equation (1): \[ R_A = 146 - R_D = 146 - 61.64 \approx 84.36 \, \text{kN}. \] --- **Step 5. Test of the Answer** 1. Sum of vertical forces: \[ R_A + R_D \approx 84.36 + 61.64 = 146 \, \text{kN}, \] which is consistent with the applied loads. 2. Sum of moments about \( A \): \[ 22 \times 61.64 \approx 1356 \, \text{kN}\cdot\text{m}, \] which matches the calculated total moment from the applied loads. --- **Final Answers:** \[ R_A \approx 84.36 \, \text{kN} \quad \text{(upward)} \] \[ R_D \approx 61.64 \, \text{kN} \quad \text{(upward)} \]