Reaction on the beam - mechanics:
Determine algebraically the reaction on the beam loaded as shown in the figure given below. Neglect the thickness and mass of beam.
Resolved all the forces in horizontal and vertical direction.
Let reaction at hinged that is, point A is R_{A}_{H } and R_{AV}, and reaction at the roller support is R_{BV}
Let ∑H & ∑V is the sum of horizontal and vertical component of the forces ,The supported beam is in equilibrium, thus
∑H = ∑H = 0
∑H = R_{A}_{H} - 20cos60° + 30cos45° - 40cos80° = 0
R_{A}_{H} = - 4.26 KN ...(i)
V = R_{A}_{V} - 10 - 20sin60° - 30sin45° - 40sin80° + R_{B}_{V} = 0
R_{A}_{V} + R_{B}_{V} =87.92 KN ...(ii)
Taking moment about A
10 × 2 + 20sin60° × 6 + 30sin45° × 13 - 40sin80° × 17 - R_{B}_{V} × 17= 0
R_{B}_{V} = 62.9 KN ...(iii)
Putting the value of R_{B}_{V} in the equation (ii)
R_{A}_{V} = 25.02 KN
Thus R_{A}_{H} = - 4.26KN, R_{A}_{V} = 25.02KN, R_{B}_{V} = 62.9KN .......ANS