If we imagine the loaded beam supported at both ends by two giants, it is evident that each giant would have to exert a certain amount of force upwards to keep his end of the beam from tipping. We can therefore imagine in all cases the supports to be resisting or reading with force sufficient to uphold their respective ends. The amount of this reaction for either support is equal to the load multiplied by its distance from the further support, the whole divided by the length, or and q =w.m/l (15)

P=w.n/l (14)

Where p = the amount of the left hand reaction or supporting force.

Where q = the amount of the right hand reaction or supporting force. If there are several loads the same law holds good for each, the reaction being the sum of the products, or p=w1n/l + w11s/l (16) and q = w1m/l + w11r/l (17)

As a check add the two reactions together and their sum must equal the whole load, that is, p+q = w1+w11

Amount of Reaction.

Fig. 7.

Example.

A beam 9' 2" long between bearings carries two loach, one of 200 lbs. 4' 2" from the left-hand support, and the other of 300 lbs. 3' 4" from the right-hand support. What are the right-hand and left-hand reactions?

Referring to Figure 8 we should have w1 = 200 lbs., and w11 = 300 lbs., further I = 110"; m = 50"; n = 60"; s = 40", and r=70", therefore the left-hand reaction would be: p = 200.60 + 300.40 = 218 2 pounds.

110 110 11 and the right-hand reaction would be: q = 200.50 + 300.70 = 281 9 pounds 110 110 11

As a check add p and q together, and they should equal the whole load of 500 lbs., and we have in effect: p + q = 218 2/11+ 281 9/11 = 500 pounds.

If the load on a beam is uniformly distributed, or is concentrated at the centre of the beam, or is concentrated at several points along the beam, each half of beam being loaded similarly, then each support will react just one half of the total load.

Fig. 8.