Question

A uniform steel girder AB is 6.0 m long and has a mass equivalent to 4.0 kN acting at its centre. The girder rests on two supports at C and B as shown in Figure 5.15. A point load of 20.0 kN is attached to the beam as shown. Determine the value of the force F that causes the beam to just lift off the support B.

Answer 1

At equilibrium, R_C + R_B = F + 4.0 + 20.0. When the beam is just lifting off of the support B, then R_B = 0, hence R_C = (F + 24.0) kN. Taking moments about A:

Clockwise moments = anticlockwise moments

i.e. (4.0 × 3.0) + (5.0 × 20.0) = (R_C × 2.5) (RB × 6.0)

i.e. 12.0 + 100.0 = (F + 24.0) × 2.5 + (0)

i.e. 112.0/2.5 = (F + 24.0)

from which, F = 44.8 − 24.0

= 20.8 kN

i.e. the value of force F which causes the beam to just lift off the support B is 20.8 kN.