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Problem 9: The 450 kg uniform i-beam supports the load of 220 kg as shown. Determine the reactions at the supports.

Solution

Since the beam is in equilibrium, we apply the two conditions of equilibrium.

∑F = 0 and ∑τ = 0

Now consider the forces on the beam. There are two normal reactions from the two supports, RA and RB, acting upward, weight of the beam W acting downward and weight of the suspended load, say W1, which is also acting down. The first condition of the equilibrium implies that the upward forces must be equal to the downward force.

RA + RB = W + W1

Now mass of the girder = 450 kg ⇒ Weight of the girder, W = 450 × 9.8 = 4410 N

Similarly, mass of the load = 220 kg ⇒ Weight of the load, W1 = 220 × 9.8 = 2156 N

Upward forces, RA and RB are not known and to be determined. However, for equilibrium,

RA + RB = 4410 + 2156 = 6566 N           …          (A)

Now apply the second condition of equilibrium, considering point B to be the pivot. RA is producing clockwise torque = 8 × RA and W and W1 are producing counterclockwise torques 4W and 2.4W1, respectively. Here, 4 m and 2.4 m are the distances of W and W1 from the pivot point A, respectively.

8RA = 4W + 2.4W1. Put the values of W and W1 from the above equations,

8RA = 4 × 4410 + 2.4 × 2156 ⇒ 8RA = 17640 + 5174.4 = 22814.4 N

⇒ RA = 22814.4/8 ⇒ RA = 2851.8 N         …          (X)

Now put this value of RA in equation (A),

2851.8 + RB = 6566 ⇒ RB = 6566 – 2851.8 = 3714.2 N          …           (Y)

Equations (X) and (Y) give the respective reactions of the supports.

 

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