Question

3. For the simply supported beam below, determine the reactions at the supports, A and B, so that the beam is in static equilibrium. (20 pts.) 25 kN 10 kN Зm 330 2m → 25 kN

the total distance from A - B is 20m

Answer 1

SOLUTION :- At A there is hinge or pin support which will have two reactions a horizontal reaction to the rightwards direction ( Ax) and a vertical reaction to the upward direction (Ay)

And at B there is a roller support which will have only vertical reaction to the upward direction (By)

It is mentioned that distance between A to B is 20 m.

Now we can write

AC = 2m

CD = 5m

DE = 3 m

EB = 10 m

Now load of 10 KN at C will have two components 10cos33° to leftwards direction and 10sin33° to the downwards

Now from statically equilibrium equations we can write

(a) Sum of all forces is x-direction is equal to zero

Ax = 10cos33°

Ax = 8.386 KN

(b) Sum of all forces in y - direction is equal to zero

Ay + By + 25 = 10 sin 33° + 25

Ay + By = 5.446..... (1)

(c) Sum of moments about any point is equal to zero

Taking all moments about A equal to zero

By*20 + 25*7 = 25*10 + 10sin33*2

By = 4.2946 KN

Now from equation (1) putting the value of By we will get

Ay = 1.151 KN

Hence reactions at A, Ax= 8.386 KN

Ay = 1.151 KN

At B, By = 4.2946 KN