Answer to Question #87069 in Mechanics | Relativity for avi

Question

A 1000 kg truck is parked on a bridge 100 m long at a distance of 25 m from one end of
the bridge. The mass of the bridge is 200 kg m−1
. Calculate the reaction forces at the
supports at the two ends of the bridge. (Take g = 10 ms−2
)

Answer 1

The force with which the truck acts on the bridge

$$F=m\times g=1000\times10=10000 N$$

Distributed load

$$q=m\times g=200\times 10=2000 N$$

Concentrate force

$$Q=q\times l=2000\times100=200000 N$$

The sum of the moments acting about one of the bridge supports (point a)

$$∑M_A =0.25\times l\times F-l\times R_B+0.5\times l\times Q$$

The reaction forces at the supports at the end of the bridge.

$$R_B={(0.5\times l\times Q+0.25\times l\times F)}/{l}={(50\times 200000+25\times10000)}/{100}=102.5 kN$$

The sum of the moments acting about one of the bridge supports (point b)

$$∑M_B =0.75\times l\times F-l\times R_B+0.5\times l\times Q$$

The reaction forces at the supports at the end of the bridge.

$$R_A={(0.5\times l\times Q+0.75\times l\times F)}/{l}={(50\times 200000+75\times 10000)}/{100}=107.5 kN$$