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

F1 af1 + F2 af2 + .... + Fn afn = R ar1 + 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) (9.81 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