Question

A beam P Q is 5.0 m long and is supported at its ends in a horizontal position as shown in Figure. Its mass is equivalent to a force of 400 N acting at its centre as shown. Point loads of 12 kN and 20 kN act on the beam in the positions shown. When the beam is in equilibrium, determine (a) the reactions of the supports, RP and RQ, and (b) the position to which the 12 kN load must be moved for the force on the supports to be equal.

Answer 1

(a) At equilibrium,

RP + RQ = 12 + 0.4 + 20

= 32.4 kN (1)

Taking moments about P: clockwise moments = anticlockwise moments i.e.

(b) For the reactions of the supports to be equal,

Let the 12 kN load be at a distance d metres from P (instead of at 1.2 m from P). Taking moments about point P gives:

(12 × d) + (0.4 × 2.5) + (20 × 3.5) = 5.0 RQ

i.e. 12d + 1.0 + 70.0 = 5.0 × 16.2

and 12d = 81.0 − 71.0

from which,d = 10.0/12 = 0.833 m

Hence the 12 kN load needs to be moved to a position 833 mm from P for the reactions of the supports to be equal (i.e. 367 mm to the left of its original position).