The beam ABCD supports a concentrated force F 36kN and a un

The beam ABCD supports a concentrated force (F = 36kN) and a uniform distributed load (q = 3kN/m) at the locations shown. Point A is a roller, and point D is a fixed support. Assume E = 200GPa and I = 350 times 10^6 mm^4.

Solution

>> Let Reactions at A & D are : Ra & Rd, respectively

>> Now, due to distributed loading, Let F is its resultant

As, F = 3*6 = 18 kN

and it will act at (3 + 3 + 6/2) = 9 m , from A

>> Now, total forces acting on beam are:

1). Ra, Reaction at A, vertically upwards

2). F2 = 36 kN, at 3 m from A, vertically downwards

3). F = 18 kN, at 9 m from A, vertically downwards

4). Rd , Reactio at D, vertically upwards

>> Now, under equilibrium,

Net Forces in Y-Direction = 0

=> Ra + Rd = 36 + 18 = 54 kN ........(1).........

>> Also, under equilibrium, Net Moment = 0

=>Ma = 0

=> 3*36 + 9*18 - 12*Rd = 0

=> Rd = 22.5 kN

So, from (1), Ra = 31.5 kN .....ANSWER.....