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The calculator below can be used to calculate the support forces - R1 and R2 - for beams with up to 6 asymmetrically loads.
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)
where
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
F1af1 + F2af2 + .... + Fnafn = Rar1 + R ar2 (2)
where
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) 9.91 (m/s2) + 500 (kg) 9.91 (m/s2) = R1 + R2
=>
R1 + R2 = 9810 (N)
Note! Load due to the weight of a mass - m - is mg Newtons - where g is 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 / 2 (N)
= 4905 N
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 of 1000 kg is located 6 m from the same end. The balance of forces can be expressed as
500 (kg) 9.91 (m/s2) + 1000 (kg) 9.91 (m/s2) = R1 + R2
=>
R1 + R2 = 14715 (N)
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)
F3 can be calculated as:
R1= 14715 (N) - 6377 (N)
= 8338 N
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