Question

1 If the intensity of the distributed load acting on the beam is w = 3 kN/m, determine the reactions at the roller A and pin B. 3 kN/m 30° B 3 m 4 m

Answer 1

A Y 3 kN/m L F_AY OF AX х 60 deg 30° FBX B 3 * sin (30) 3 m 3 * cos (30) F BY 4 m

Given data: w = 3 kN/m. Equilibrium of forces in x-direction = 0 FAX - FBx = 0 (FA * cos (60)) - FBx = 0 0.5FA = Fox Equation-1 Equilibrium of forces in y-direction = 0 FAY - (w* 4) + FBy = 0 (FA * sin (60)) + FBY-(w * 4) = 0 0.866FA + FBy = 12 Equation 2 Equilibrium of moments about point B = 0 (w* 4 * 2) - (FAY* (4 + 3 * cos (30))) - (Fax * 3 * sin (60)) = 0 (24) - (0.866FA * 6.59) - (0.5FA * 1.5) = 0 Reaction at A (FA) = 3.71693 kN.... Answer. Using, Equation-1 we get, Fox = 1.85847 kN. Using, Equation-2 we get, FBy = 8.78114 kN. Reaction at B (RB) = sqrt ((Rex?) + (RBY?)) = sqrt ((1.85847^2) + (8.78114^2)) = 8.975 kN.... Answer.