# Support Reactions - Equilibrium

## Static equilibrium is achieved when the resultant force and resultant moment equals to zero

Equilibrium of a body requires both a balance of forces to prevent the body from translating or having accelerated motion along a straight or curved path - and a balance of moments to prevent the body from rotating.

Any static force system will be in equilibrium if the resultant force and resultant moment both are equal to zero.

Static equilibrium in a three dimensional system can be expressed as

ΣF =FΣ_{x}=FΣ_{y}=FΣ_{z}= 0 (1)

M =ΣMΣ_{x}=MΣ_{y}=MΣ_{z}= 0 (2)

where

F = force (N, lb)

M = moment (Nm, ft lb)

x, y, z = orthogonal axes

Often the loading of a body can be simplified to a two dimensional system with co-planar forces in the x-y plane. Eq. 1 and 2 can be reduced to

*ΣF = Σ*F

_{x}=

*F*

*Σ*_{y}= 0 (3)

* Σ*M =

*M*

*Σ*_{z}= 0 (4)

The best way to account for all forces acting on a body is to draw the body's free-body diagram. A free-body diagram shows the relative magnitude and direction of all forces acting upon an object in a given situation.

Free-body diagram example - gravity and friction forces acting on a body on an inclined plane

### Example - Support Reactions on a Beam with Eccentric Load

A beam with length *6 m* has an eccentric load of *9000 N* *4 m* from *support 1*. Applying the equations of equilibrium we have

*F _{x} = R_{1x} = R_{2x} = 0 (5)*

*F _{y} - (R_{1y} + R_{2y}) = 0 (6)*

*M _{1} = F_{y} a - R_{2y} (a + b) = 0 (7)*

Rearranging *(7)* to express *R _{2y }*

*R _{2y} = F a / (a + b) (7b)*

Eq. *(7b)* with values* *

*R _{2} = (9000 N) (4 m) / ((4 m) + (2 m)) *

* = 6000 N*

* = 6 kN*

Rearranging *(6)* for *R _{1y} *

*R _{1y} = F_{y} - R_{2y} *(6b)

Eq. *(6b)* with values

*R _{1y} = (9000 N) - (6000 N)*

* = 3000 N*

* = 3 kN *