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Beam Loads - Support Force Calculator

Calculate beam load and supporting forces

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Online Beam Support Force Calculator

The calculator below can be used to calculate the support forces - R1 and R2 - for beams with up to 6 asymmetrically loads.

Length of beam (m, ft)

Force F1 (N, lbf) distance from R1 (m, ft)

Force F2 (N, lbf) distance from R1 (m, ft)

Force F3 (N, lbf) distance from R1 (m, ft)

Force F4 (N, lbf) distance from R1 (m, ft)

Force F5 (N, lbf) distance from R1 (m, ft)

Force F6 (N, lbf) distance from R1 (m, ft)


Load Calculator!

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)

Example - A beam with two symmetrical loads

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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Example - A beam with two not symmetrical loads

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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