1. **Stating the problem:** We have a rectangular frame with vertical height $h=4$ m and horizontal length $b=a=6$ m. The frame is subjected to a uniform load $q=2$ t/m on the top horizontal beam, vertical loads $W_1=2$ ton and $W_2=1$ ton on the left vertical beam, and a point load $P=2$ t at the midpoint of the right vertical beam. 2. **Objective:** Calculate the reactions and internal moments in the frame considering the given loads and dimensions. 3. **Key formulas and rules:** - Uniform load total force: $Q = q \times b$ - Sum of vertical forces for equilibrium: $\sum F_y = 0$ - Sum of moments about a point for equilibrium: $\sum M = 0$ - The frame is statically determinate with rigid supports at A and B. 4. **Calculate total uniform load on top beam:** $$Q = 2 \times 6 = 12\text{ ton}$$ 5. **Sum of vertical forces:** Let $R_A$ and $R_B$ be the vertical reactions at supports A and B. $$R_A + R_B = Q + W_1 + W_2 + P = 12 + 2 + 1 + 2 = 17\text{ ton}$$ 6. **Sum moments about A:** Taking moments about point A (counterclockwise positive): - Uniform load $Q$ acts at midpoint of top beam, distance $b/2=3$ m from A. - $W_1=2$ ton acts at left vertical beam, height $h=4$ m from A. - $W_2=1$ ton acts at middle of left vertical beam, height $h/2=2$ m from A. - Point load $P=2$ ton acts at midpoint of right vertical beam, horizontal distance $b=6$ m and vertical height $h/2=2$ m from A. Moment contributions: $$M_A = R_B \times b - Q \times \frac{b}{2} - W_1 \times h - W_2 \times \frac{h}{2} - P \times b$$ Set $M_A=0$ for equilibrium: $$R_B \times 6 = 12 \times 3 + 2 \times 4 + 1 \times 2 + 2 \times 6$$ $$6 R_B = 36 + 8 + 2 + 12 = 58$$ $$R_B = \frac{58}{6} = 9.67\text{ ton}$$ 7. **Calculate $R_A$:** $$R_A = 17 - 9.67 = 7.33\text{ ton}$$ 8. **Summary:** - Reaction at support A: $7.33$ ton upward - Reaction at support B: $9.67$ ton upward These reactions balance the applied loads on the frame. **Final answer:** $$R_A = 7.33\text{ ton}, \quad R_B = 9.67\text{ ton}$$