Let's work through the problem **step by step**: --- ## **Step 1: Problem Understanding** Given: - Beam AB, simply supported at A and B - Uniformly distributed load (UDL) of \( 1.5 \) kN/m over the entire span (6 m) - Two concentrated loads: \( P \) at E (1.2 m from A), \( Q \) at F (4.8 m from A) - Moments at points E and F: - \( M_E = +360 \) Nm at E (1.2 m from A) - \( M_F = +960 \) Nm at F (4.8 m from A) - **Find:** Values of \( P \) and \( Q \), and draw shear and bending moment diagrams. --- ## **Step 2: Sketch and Dimensions** - **A** ( m), **E** (1.2 m), **F** (4.8 m), **B** (6 m) - UDL: \( w = 1.5 \) kN/m from A to B --- ## **Step 3: Reactions at Supports** Let: - \( R_A \): Reaction at A - \( R_B \): Reaction at B **Sum of Vertical Forces:** \[ R_A + R_B = w \cdot L + P + Q \] \[ R_A + R_B = 1.5 \times 6 + P + Q = 9 + P + Q \] --- ## **Step 4: Moment Equation about A** Take moments about A (clockwise positive): \[ \sum M_A = \] \[ R_B \times 6 - w \times 6 \times \frac{6}{2} - P \times 1.2 - Q \times 4.8 = \] \[ R_B \times 6 - 1.5 \times 6 \times 3 - P \times 1.2 - Q \times 4.8 = \] \[ 6R_B - 27 - 1.2P - 4.8Q = \] \[ 6R_B = 27 + 1.2P + 4.8Q \] \[ R_B = \frac{27 + 1.2P + 4.8Q}{6} \] --- ## **Step 5: Moment at E (1.2 m from A)** The bending moment at E (from left): \[ M_E = R_A \times 1.2 - w \times 1.2 \times \frac{1.2}{2} \] Since E is just at the load P, **exclude P** because it acts at the point. \[ M_E = R_A \times 1.2 - 1.5 \times 1.2 \times .6 \] \[ M_E = 1.2R_A - 1.08 \] Given \( M_E = 360 \) Nm \( = .36 \) kNm: \[ 1.2R_A - 1.08 = .36 \] \[ 1.2R_A = 1.44 \] \[ R_A = 1.2 \] --- ## **Step 6: Moment at F (4.8 m from A)** The bending moment at F (from left): \[ M_F = R_A \times 4.8 - w \times 4.8 \times \frac{4.8}{2} - P \times (4.8-1.2) \] \( (4.8 - 1.2) = 3.6 \) m \[ M_F = 4.8R_A - 1.5 \times 4.8 \times 2.4 - P \times 3.6 \] \[ = 4.8R_A - 17.28 - 3.6P \] Given \( M_F = 960 \) Nm \( = .96 \) kNm: \[ 4.8R_A - 17.28 - 3.6P = .96 \] \[ 4.8 \times 1.2 - 17.28 - 3.6P = .96 \] \[ 5.76 - 17.28 - 3.6P = .96 \] \[ -11.52 - 3.6P = .96 \] \[ -3.6P = .96 + 11.52 = 12.48 \] \[ P = -\frac{12.48}{3.6} = -3.47 \text{ kN} \] --- ## **Step 7: Find Q using Equilibrium** Recall: \[ R_A + R_B = 9 + P + Q \] \[ R_A = 1.2 \] \[ R_B = \frac{27 + 1.2P + 4.8Q}{6} \] Plug in \( P = -3.47 \): \[ R_A + R_B = 9 + (-3.47) + Q = 5.53 + Q \] \[ R_B = \frac{27 + 1.2 \times -3.47 + 4.8Q}{6} \] \[ = \frac{27 - 4.164 + 4.8Q}{6} \] \[ = \frac{22.836 + 4.8Q}{6} \] So, \[ 1.2 + \frac{22.836 + 4.8Q}{6} = 5.53 + Q \] Multiply both sides by 6: \[ 7.2 + 22.836 + 4.8Q = 33.18 + 6Q \] \[ 30.036 + 4.8Q = 33.18 + 6Q \] \[ 30.036 - 33.18 = 6Q - 4.8Q \] \[ -3.144 = 1.2Q \] \[ Q = -\frac{3.144}{1.2} = -2.62 \text{ kN} \] --- ## **Step 8: Final Answers** ### **Magnitudes of P and Q** \[ \boxed{P = -3.47\ \text{kN}} \] \[ \boxed{Q = -2.62\ \text{kN}} \] **(Negative sign indicates direction is opposite to the assumed direction in the free body diagram.)** --- ## **Step 9: Shear Force and Bending Moment Diagrams** - **Shear force:** Start from A, add \( R_A \), subtract UDL as you move, adjust for point loads at E and F. - **Bending moment:** Start from A (zero), use areas under shear diagram or calculated at key points. --- ### **Summary Table of Results** | Point | Distance (m) | Shear (kN) | Moment (kNm) | |-------|--------------|------------|--------------| | A | | \( +1.2 \) | | | E | 1.2 | ... | .36 | | F | 4.8 | ... | .96 | | B | 6. | | | --- ### **(Diagrams)** #### **Shear Force Diagram** - Start at \( +1.2 \) at A, drop by \( 1.5 \) kN/m per meter, drop by \( P \) at E, drop by \( Q \) at F, end at zero at B. #### **Bending Moment Diagram** - Start at at A, peak at E (.36 kNm), peak at F (.96 kNm), down to at B. --- ## **Boxed Final Answers** \[ \boxed{P = -3.47\ \text{kN}} \] \[ \boxed{Q = -2.62\ \text{kN}} \] If you need the full shear and moment diagrams drawn, please let me know!