Find the reactions at A and B for the beam loaded as shown below. The weight of

Question

Find the reactions at A and B for the beam loaded as shown below. The weight of the beam may be neglected.

Explanation / Answer

Net moment about point B should be zero for equilibrium: => [520*(12/13) * 1] + [400*(4/5) * 6] = Fay * 4 => Fay = 600 N where, Fay is the force in vertical (y) direction at point A Note: B has roller support hence it can only take vertical force. By equilibrium of force in vertical direction: Fby + Fay -[520*(12/13)] - [400*(4/5)] = 0 => Fby = 800 - 600 = 200 N By equilibrium of force in horizontal direction: => Fax + [520*(5/13)] - [400*(3/5)] = 0 => Fax = -40 N So the reaction at A : sqrt(600^2 + 40^2) = 601.3319 N (at an angle of 86.19 deg. from horizontal) So the reaction at B : 200 N (vertically upwards)

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