1. **Problem statement:** Given a beam with supports at A and B, a point load $P=20$ at D, a uniformly distributed load $q=5$ over segment AD (4m), and a moment $M=20$ at B. The beam spans are: $AD=4m$, $DB=4m$, $BC=2m$. Find the reaction forces at supports A and B. 2. **Formulas and rules:** - Sum of vertical forces must be zero: $\sum F_y=0$ - Sum of moments about any point must be zero: $\sum M=0$ - Reaction forces at supports are vertical forces $R_A$ and $R_B$ 3. **Calculate total distributed load:** $$q \times AD = 5 \times 4 = 20$$ This load acts at the midpoint of AD, i.e., 2m from A. 4. **Sum of vertical forces:** $$R_A + R_B - P - q \times AD = 0$$ $$R_A + R_B - 20 - 20 = 0$$ $$R_A + R_B = 40$$ 5. **Sum moments about A:** Taking counterclockwise as positive, $$-P \times AD - q \times AD \times 2 + M + R_B \times (AD + DB) = 0$$ Substitute values: $$-20 \times 4 - 20 \times 2 + 20 + R_B \times 8 = 0$$ $$-80 - 40 + 20 + 8 R_B = 0$$ $$8 R_B = 100$$ $$R_B = \frac{100}{8} = 12.5$$ 6. **Find $R_A$:** $$R_A = 40 - R_B = 40 - 12.5 = 27.5$$ **Final answer:** $$R_A = 27.5, \quad R_B = 12.5$$