For a beam in balance loaded with weights (or other load forces) the reactions forces - R - at the supports equals the load forces - F. The force balance can be expressed as
F1 + F2 + .... + Fn = R1 + R2 (1)
F = force from load (N, lbf)
R = force from support (N, lbf)
In addition for a beam in balance the algebraic sum of moments equals zero. The moment balance can be expressed as
F1 af1 + F2 af2 + .... + Fn afn = R ar1 + R ar2 (2)
a = the distance from the force to a common reference - usually the distance to one of the supports (m, ft)
A 10 m long beam with two supports is loaded with two equal and symmetrical loads F1 and F2 , each 500 kg. The support forces F3 and F4 can be calculated
(500 kg) (9.81 m/s2) + (500 kg) (9.81 m/s2) = R1 + R2
R1 + R2 = 9810 N
= 9.8 kN
Note! Load due to the weight of a mass - m - is mg Newton's - where g = 9.81 m/s2.
With symmetrical and equal loads the support forces also will be symmetrical and equal. Using
R1 = R2
the equation above can be simplified to
R1 = R2 = (9810 N) / 2
= 4905 N
= 4.9 kN
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A 10 m long beam with two supports is loaded with two loads, 500 kg is located 1 m from the end (R1), and the other load of 1000 kg is located 6 m from the same end. The balance of forces can be expressed as
(500 kg) (9.81 m/s2) + (1000 kg) (9.81 m/s2) = R1 + R2
R1 + R2 = 14715 N
= 14.7 kN
The algebraic sum of moments (2) can be expressed as
(500 kg) (9.81 m/s2) (1 m) + (1000 kg) (9.81 m/s2) (6 m) =?R1 (0 m) + R2 (10 m)
R2 = 6377 (N)
= 6.4 kN
F3 can be calculated as:
R1= (14715 N) - (6377 N)
= 8338 N
= 8.3 kN
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