# Beam Loads - Support Force Calculator

## Calculate beam load and supporting forces

### Online Beam Support Force Calculator

The calculator below can be used to calculate the support forces - *R _{1}* and

*R*- for beams with up to 6 asymmetrically loads.

_{2}

Length of beam (m, ft)

Force F1 (N, lb_{f})distance from R_{1}(m, ft)

Force F2 (N, lb_{f})distance from R_{1}(m, ft)

Force F3 (N, lb_{f})distance from R_{1}(m, ft)

Force F4 (N, lb_{f})distance from R_{1}(m, ft)

Force F5 (N, lb_{f})distance from R_{1}(m, ft)

Force F6 (N, lb_{f})distance from R_{1}(m, ft)

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

F_{1}+ F_{2}+ .... + F_{n}= R_{1}+ R_{2}(1)

where

F = force from load (N, lb_{f})

R = force from support (N, lb_{f})

In addition for a beam in balance the algebraic **sum of moments equals zero**. The **moment balance** can be expressed as

F_{1 }a_{f1}+ F_{2 }a_{f2}+ .... + F_{n }a_{fn}= R a_{r1}+ R a_{r2 }(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 *F _{1}* and

*F*, each

_{2}*500 kg*. The support forces

*F*and

_{3}*F*can be calculated

_{4}

(500 kg) (9.81 m/s^{2}) + (500 kg) (9.81 m/s^{2}) = R_{1}+ R_{2}

=>

R_{1}+ R_{2}= 9810 N

= 9.8 kN

**Note!** Load due to the weight of a mass - *m* - is *mg* Newton's - where *g = 9.81 m/s ^{2}*.

With symmetrical and equal loads the support forces also will be symmetrical and equal. Using

R_{1}= R_{2}

the equation above can be simplified to

R_{1}= R_{2}= (9810 N) / 2

= 4905 N

= 4.9 kN

### Related Mobile Apps from The Engineering ToolBox

- free apps for offline use on mobile devices.

### 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 (*R _{1}*), 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/s^{2}) + (1000 kg) (9.81 m/s^{2}) = R_{1}+ R_{2}

=>

R_{1}+ R_{2}= 14715 N

= 14.7 kN

The algebraic sum of moments (2) can be expressed as

(500 kg) (9.81 m/s^{2}) (1 m) + (1000 kg) (9.81 m/s^{2}) (6 m) =?R_{1}(0 m) + R_{2}(10 m)

=>

R_{2}= 6377 (N)

= 6.4 kN

*F _{3} *can be calculated as:

R_{1}= (14715 N) - (6377 N)

= 8338 N

= 8.3 kN