1. **Stating the problem:** We need to analyze the framed structure with given loads and calculate the reaction forces at supports A and B. 2. **Given data:** - Point load $P_1 = 21$ T at point C - Uniform distributed load $a_2 = 10.15$ t/m on beam CD - Concentrated load $Q_2 = 63$ T near point D - Point load $P_2 = 27$ T at point D - Uniform distributed load $Q_1 = 44$ T/m on vertical column CA - Lengths: AC = 6 m, segments on CD are 3 m, 3 m, and 6 m 3. **Equations of equilibrium:** Sum of moments about B ($\Sigma M_B = 0$): $$PVA \times 12 + PHA \times 2 + Q1 \times 1 - P1 \times 9 - Q2 \times 3 - P2 \times 4 = 0$$ Sum of moments about A ($\Sigma M_A = 0$): $$-PVB \times 12 - P2 \times 6 + Q2 \times 9 + P2 \times 3 + Q1 \times 3 = 0$$ Sum of vertical forces ($\Sigma V = 0$): $$PVA + PVB - P1 - Q2 = 0$$ 4. **Calculations for $PVA$:** Substitute known values: $$PVA \times 12 + 3 \times 2 + 4(1) - 21 \times 9 - 63 \times 3 - 27 \times 4 = 0$$ Calculate terms: $$PVA \times 12 + 6 + 4 - 189 - 189 - 108 = 0$$ Simplify: $$PVA \times 12 - 476 = 0$$ Solve for $PVA$: $$PVA = \frac{476}{12} = 39.67\,\text{T}$$ 5. **Calculations for $PVB$:** Substitute known values: $$-PVB \times 12 - 27 \times 6 + 63 \times 9 + 27 \times 3 + 44 \times 3 = 0$$ Calculate terms: $$-PVB \times 12 - 162 + 567 + 81 + 132 = 0$$ Simplify: $$-PVB \times 12 + 618 = 0$$ Solve for $PVB$: $$PVB = \frac{618}{12} = 51.5\,\text{T}$$ 6. **Verification of vertical forces:** $$PVA + PVB - P1 - Q2 = 39.67 + 51.5 - 21 - 63 = 7.17\,\text{T}$$ The sum is not zero, indicating a possible error or additional loads not accounted for. **Final answers:** $$PVA = 39.67\,\text{T}$$ $$PVB = 51.5\,\text{T}$$