The 450-kg uniform I-beam supports a load of 220 kg as shown. Determine the reactions at the supports. (Ans. 2850 N) 





Given:


Mass of I-beam = mb = 450 kg
Weight of I-beam = Wb = Fb = mg = 450 kg ｘ9.8 m s⁻² = 4410 N
Mass of the load = ml = 220 kg
Weight of load = Wl = Fl = mg = 220 kg ｘ9.81 m s⁻² = 2156 N
 

To Find:

The force of Reaction = RA = ?
The force of Reaction = RB = ?

Solution:

First, we will apply the first condition of equilibrium, which is mathematically represented as

Σ F = 0

Σ F (upwards) = Σ F (downwards)

RA + RB = Fb + Fl

RA + RB = 4410 N + 2156 N

RA + RB = 6566 N ------------------(1)


Now considering moments about point A. then
Moment arm for FL = rL = 5.6 cm
Length of I-beam = 2.4 cm + 5.6 cm = 8 cm
Moment arm for RB =rB = 8 cm
Moment arm for Fb = rcg = L/2 = 4/2 = 2 cm

By applying the second condition of equilibrium; which is mathematically written as

Σ T = 0

or

Σ T Anti-clockwise =  Σ T Clockwise 

rb ｘ RB = rb ｘ Fb + rL ｘ RL

8 cm ｘ RB = 4 cm ｘ 4410 N + 5.6 cm ｘ 2156 N c

8 cm ｘ RB = 17,640 N cm ｘ 12,073.6 N

8 cm ｘ RB = 29,713.6 N

RB = 29,713.6 N cm / 8 cm

RB = 3,714.2 N

or

RB = 3,714 N

Putting the value of RB = 3,714 N in equation (1), we get

RA + 3,714 N = 6566 N
RA = 6566 N - 3,714 N

RA = 2,852 N ≅ 2850 N

Thus,
Force of Reaction = RA = 2,850 N ----------Ans.1
Force of Reaction = RB = 3,714 N ----------Ans.2


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