Question

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

Answer 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 × 6 = 120 kN
2 + `6/2` = 5 m
This load acts at 5m from A
Effective force of uniformly varying load =`1/2` × (80-20) × 6
=180 kN
2 +`6/3` × 2 = 6m
This load acts at 6m from A

The beam is in equilibrium
Applying the conditions of equilibrium
ΣMA = 0
−120 × 5 −180 × 6 + RBcos30 × 10 −80 × 13 = 0
10RBcos 30 = 120 × 5 + 180 × 6 + 80 × 13
RB = 314.0785 N
Reaction at B will be at 60o in second quadrant
ΣFx = 0
R_Acosα − RBsin30 = 0
R_Acosα − 314.0785 × 0.5 = 0
R_Acosα = 157.0393 N ……..(1)
ΣFy = 0
R_Asinα − 120 − 180 + RBcos30 − 80 = 0
R_Asinα = 12 + 180 − 314.0785 × 0.866 + 80
R_Asinα = 108.008N ……..(2)
Squaring and adding (1) and (2)
R_A²(sin²α + cos²α) = 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 quadrant
Reaction at B = 314.0785 N at 60^o in second quadrant

Answer 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°