For the beam loaded as shown in figure, Calculate the support reactions.[Figure: cantilever AB, 3 m long, fixed at A (left wall). A UDL of 10 kN/m acts over the first 2 m from A. At 2 m from A, a 20 kN force acts downward to the left, inclined at 60°…

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Q.6 (A) For the beam loaded as shown in figure, Calculate the support reactions. [Figure: cantilever AB, 3 m long, fixed at A (left wall). A UDL of 10 kN/m acts over the first 2 m from A. At 2 m from A, a 20 kN force acts downward to the left, inclined at 60° to the beam (angle measured on the right side, between the force and the beam). A vertical load of 15 kN acts downward at the free end B (3 m from A).]

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Loads: UDL resultant = 10 × 2 = 20 kN at 1 m from A. Inclined 20 kN: vertical component 20sin60^° = 17.32 kN (↓), horizontal component 20cos60^° = 10 kN (← towards A). Σ F_x = 0:\ H_A - 10 = 0 ⇒ H_A = 10 kN (→) Σ F_y = 0:\ V_A = 20 + 17.32 + 15 = 52.32 kN (↑) Σ M_A = 0:\ M_A = 20(1) + 17.32(2) + 15(3) = 20 + 34.64 + 45 = 99.64 kN·m (anticlockwise fixing moment at A) Resultant reaction R_A = √(10² + 52.32²) = 53.27 kN at tan⁻¹(52.32/10) = 79.18^° to the horizontal. Answer: H_A = 10 kN →, V_A = 52.32 kN ↑ (R = 53.27 kN at 79.18°), fixing moment M_A = 99.64 kN·m anticlockwise.

For the beam loaded as shown in figure, Calculate the support reactions.[Figure: cantilever AB, 3 m long, fixed at A (left wall). A UDL of 10 kN/m acts over the first 2 m from A. At 2 m from A, a 20 kN force acts downward to the left, inclined at 60°…

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