Question

A simply supported beam AB of length 5 m is acted upon by a uniformly distributed load of 2 kN/m. The reactions at A and B respectively are
​

Answer 1

Answer:

Answer is 4KN , 6KN

Explanation:

Answer 2

Given,

Beam AB: Simply supported,

Length = 5 m.

Load: Uniformly distributed = 2 kN/m.

To find,

Reactions at A and B.

Solution,

Firstly, let

$R_a,M_a$ = reaction at A, bending moment about A respectively and,

$R_b,M_b$ = reaction and bending moment at support B respectively.

The total reaction should be equal to the total load being applied. That is,

Total load = $R_a+R_b$

Now, the load is acting uniformly at 2 kN/m, over the length of 5 m.

So, the total load will be

= 2 × 5 = 10 kN.

It can be assumed that this load of 10 kN acts at the mid-point of the beam, that is, at 2.5 m from both ends.

From the previous relation,

$R_a+R_b= 10$ kN

Since the total bending moment about any support will be 0.

Taking total bending moment about A = 0,

$\sum M_a=0$

⇒ -10×2.5 + $R_b$×5 = 0

⇒ $R_b=5$ kN.

Since $R_a+R_b= 10$

⇒ $R_a= 5$ kN.

Therefore, for the given beam AB, reactions at A and B will be the same and equal to 5 kN.