Find the support reactions at A and B for the beam loaded as shown in the given figure.

#### Solution 1

Given : Various forces on beam

To find : Support reactions at A and B

Solution: Draw PQ ⊥to RS

Effective force of uniform load =20 x 6 = 120 kN

2 + `6/2` = 5 m

This load acts at 5m from A

Effective force of uniformly varying load =`1/2` x (80-20) x 6

=180 kN

2+`6/3` x 2 =6m

This load acts at 6m from A

The beam is in equilibrium

Applying the conditions of equilibrium

ΣMA = 0

-120 x 5 -180 x 6 +RBcos30 x 10 -80 x 13 =0

10RBcos 30 = 120 x 5+180 x 6 + 80 x13

RB = 314.0785 N

Reaction at B will be at 60o in second quadrant

ΣFx = 0

R_{A}cosα - RBsin30 = 0

R_{A}cosα - 314.0785 x 0.5 = 0

R_{A}cosα = 157.0393 N ……..(1)

ΣFy = 0

R_{A}sinα - 120 - 180 + RBcos30 - 80 = 0

R_{A}sinα = 12 + 180 -314.0785 x0.866 + 80

R_{A}sinα = 108.008N ……..(2)

Squaring and adding (1) and (2)

R_{A}^{2}(sin^{2}α + cos^{2}α) = 36325.3333

R_{A} = 190.5921 N

Dividing (2) by (1)

`(R_Asinalpha)/(R_Acosalpha)=(108.008)/(157.0393)`

α=tan^{-1}(0.6877)

=34.5173^{o}**Reaction at point A = 190.5921 N at 34.5173o in first quadrantReaction at B = 314.0785 N at 60 ^{o} in second quadrant**

^{}

#### Solution 2

As Shown in FBD,

Taking two components of R_{A} & taking support reaction at pt. B which inclined at angle 60^{∘ }with horizontal. Also converting the trapezoidal load into equivalent point loads we get FBD

Now , applying conditions of equilibrium

`sumF_x = H_A - R_B cos 60 = 0` ------(1)

`sumF_Y = V_A - 120 -180-80 +R_B sin 60 = 0` -----(2)

120(5) + 180(6) - R_{B} sin 60(10) + 80(13) = 0

R_{B} = 314.08 KN at θ = 60°

Putting in eqs. (1) and (2), we get

H_{A} = 157.04KN,V_{A }= 108KN

So, R_{A }= 190.59 KN at θ = 34.52°

Support reaction at A is **190.59 KN at θ = 34.52° & at N is R _{B} = 314.08 KN at θ = 60° **