Дано: P = 12 кН, G = 48 кН, q = 4 кН/м, α=120, M = 24 кНм Q = 3 q = 12 кН ΣFx=0: XA + P cos ( 30) = 0 XA = -P cos ( 30° ) = -12 cos ( 30 ° ) = -10.392 кН ΣMa=0: YB 6 - Q 4.5 - G 3 - M - P cos ( 30° ) 1 + P sin ( 30° ) 3 = 0 Q 4.5 + G 3 + M + P cos ( 30° ) 1 - P sin ( 30° ) 3 6 12 4.5 + 48 3 + 24 + 12 cos ( 30 ° ) - 12 sin ( 30 ° ) 3 YB = = 35.732 кН 6 YB = ΣMb=0: -YA 6 - M + G 3 + Q 1.5 - P cos ( 30° ) 1 - P sin ( 30° ) 3 = 0 -M + G 3 + Q 1.5 - P cos ( 30° ) 1 - P sin ( 30° ) 3 6 -24 + 48 3 + 12 1.5 - 12 cos ( 30 ° ) - 12 sin ( 30 ° ) 3 YA = = 18.268 кН 6 YA = YB = YB = 35.732 ; YA = YA = 18.268 ΣFy=0: YA + YB - G - Q + P sin ( 30° ) = 18.268 + 35.732 - 48 - 12 + 12 sin ( 30 ° ) = 0