## Answers

solution 150KN Hg AC= JAD?+CD² (Ac=lom ☆ Taking moment about A. Hint. (Divide the trapezoidal a Acc"/ a in a reactange a acel and a triangle acc" for ease in calculation of load and censoid of load.

Area of ance!= 20x10 = 200 KN centroid of ance from point A = 10 = 15m Area of aclcll4x10x20 = 100 KCN centroid of ac'c" from point A = 2 x 10 = 20 m centroid of acc" from point c = 1/2 x 10 = 10 m E MA=0 = (200 x 5)+(100X29) + 150X12 – R8 X 18 - Hg X8=0 1888 +8H8 = 3466.67 -0 Taking moment about c. EMC =0 =) (200x5)+(100 x 40) - 150X6-RAX6 +RX12=0 ☆ -6RA+ (2 RB = -433.33 - 40knim GRA-12 RB = 433.33 ☆ from horizontal equilibrium EH-O > 200x60(90-6) tloox cos(90-0) -HB 0 20 km

From A AD g sino = ch - 8 9 HB = 300 COS (90-0) Hp = 300 fino HB = 300 x 80 (Hg = 240 KN) Put H8= 240 kN in eqo 18 R8 + 8X240 = 3466.67 Rg = 85.92660 Put RB=85.926 KN in eq ② GRA-12885.926 = 433.33 TRA= 244.073 kN So support At support Reactions are - A, vertical Reaction (RA= 244.073 KN) At support_B, Vertical Reaction Rp = 85.926 UN Horizontal Reaction HB = 240 KN)

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