1. **Stating the problem:** Calculate the reaction forces at supports S and T for the given beam with applied forces: - Vertical downward forces: 0.80 GN and $5 \times 10^6$ MN - Horizontal forces: 1.5 MN leftward at S, 200,000 N rightward at T 2. **Convert all forces to consistent units (Newtons):** - $0.80\ \text{GN} = 0.80 \times 10^9\ \text{N} = 8.0 \times 10^8\ \text{N}$ - $5 \times 10^6\ \text{MN} = 5 \times 10^6 \times 10^6\ \text{N} = 5 \times 10^{12}\ \text{N}$ - $1.5\ \text{MN} = 1.5 \times 10^6\ \text{N}$ - $200 \times 10^3\ \text{N} = 2.0 \times 10^5\ \text{N}$ 3. **Convert all distances to meters:** - Left segment: 4000 mm = 4 m - Middle segment: 450 cm = 4.5 m - Right segment: 25 m 4. **Sum of horizontal forces:** At support S (pinned), horizontal reaction $SRX$ balances horizontal applied forces: $$SRX + 1.5 \times 10^6 = 2.0 \times 10^5$$ $$SRX = 2.0 \times 10^5 - 1.5 \times 10^6 = -1.3 \times 10^6\ \text{N} = -1300\ \text{kN}$$ (Negative means reaction is opposite assumed direction) 5. **Sum of vertical forces:** Let $SRY$ and $TRY$ be vertical reactions at S and T: $$SRY + TRY = 8.0 \times 10^8 + 5 \times 10^{12} = 5.0008 \times 10^{12}\ \text{N}$$ 6. **Sum of moments about S (taking clockwise positive):** Moments due to vertical forces and reactions: - Moment by $TRY$ at distance $4 + 4.5 + 25 = 33.5$ m: $$TRY \times 33.5$$ - Moment by $0.80\ \text{GN}$ at 4 m: $$-8.0 \times 10^8 \times 4 = -3.2 \times 10^9$$ - Moment by $5 \times 10^6\ \text{MN}$ at $4 + 4.5 = 8.5$ m: $$-5 \times 10^{12} \times 8.5 = -4.25 \times 10^{13}$$ Set sum of moments to zero: $$TRY \times 33.5 - 3.2 \times 10^9 - 4.25 \times 10^{13} = 0$$ $$TRY \times 33.5 = 4.25 \times 10^{13} + 3.2 \times 10^9 \approx 4.25 \times 10^{13}$$ $$TRY = \frac{4.25 \times 10^{13}}{33.5} \approx 1.27 \times 10^{12}\ \text{N} = 1.27 \times 10^6\ \text{MN}$$ 7. **Calculate $SRY$ from vertical force equilibrium:** $$SRY = 5.0008 \times 10^{12} - 1.27 \times 10^{12} = 3.73 \times 10^{12}\ \text{N} = 3.73 \times 10^6\ \text{MN}$$ **Final answers:** - $SRX = -1300\ \text{kN}$ (leftward reaction) - $SRY = 3.73 \times 10^{12}\ \text{N}$ - $TRX = 0$ (roller support has no horizontal reaction) - $TRY = 1.27 \times 10^{12}\ \text{N}$ These reactions ensure equilibrium of the beam under the given loads.