1. **State the problem:** Given a beam with a uniformly distributed load (UDL) of 3 kN/m over 5 m from point A to B, a clockwise moment of 15 kN·m applied at point C, and supports at A and B, determine the reactions at the supports. 2. **Identify loads and distances:** - UDL: $w = 3\ \text{kN/m}$ over length $L_1 = 5\ \text{m}$ starting at $A$. - Moment at $C$: $M_C = 15\ \text{kN\cdot m}$ clockwise. - Distance $AB = 5\ \text{m}$. - Distance $BC = 3\ \text{m}$. - Total beam length $AC = 8\ \text{m}$. 3. **Calculate resultant force of UDL:** $$ F_{UDL} = w \times L_1 = 3 \times 5 = 15\ \text{kN} $$ Acts at the midpoint of the UDL segment, i.e., 2.5 m from $A$. 4. **Define reactions at supports:** Let the vertical reactions be $R_A$ at $A$ and $R_B$ at $B$. 5. **Sum of vertical forces:** $$ R_A + R_B - F_{UDL} = 0 $$ $$ R_A + R_B = 15 $$ 6. **Sum moments about point A:** Positive moments counterclockwise. Taking moments caused by reactions and loads about $A$: - Moment of $R_B$ at $A$: $-R_B \times 5$ (clockwise) - Moment of UDL: $F_{UDL} \times 2.5 = 15 \times 2.5 = 37.5$ (clockwise) - Moment at $C$ applied on beam: $+15$ kN·m (clockwise moment means negative if positive is CCW, so $-15$) Equation: $$ -R_B \times 5 - 37.5 - 15 = 0 $$ Move terms: $$ -5 R_B = 37.5 + 15 $$ $$ -5 R_B = 52.5 $$ $$ R_B = -10.5\ \text{kN} $$ Negative value indicates direction opposite assumed (downward). 7. **Calculate $R_A$:** Using $R_A + R_B = 15$: $$ R_A = 15 - (-10.5) = 25.5\ \text{kN} $$ 8. **Interpretation:** - Reaction at $A$ is $25.5$ kN upward. - Reaction at $B$ is $10.5$ kN downward (unusual for support, indicates moment effect). **Final answer:** $$ R_A = 25.5\ \text{kN} \text{ (upward)}, \quad R_B = 10.5\ \text{kN} \text{ (downward)} $$