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
Σ F = Σ F _{ x } = Σ F _{ y } = Σ F _{ z } = 0 (1)
Σ M = Σ M _{ x } = Σ M _{ y } = Σ M _{ 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 coplanar forces in the xy plane. Eq. 1 and 2 can be reduced to
Σ F = Σ F _{ x } = Σ F _{ y } = 0 (3)
Σ M = Σ M _{ z } = 0 (4)
The best way to account for all forces acting on a body is to draw the body's freebody diagram . A freebody diagram shows the relative magnitude and direction of all forces acting upon an object in a given situation.
Freebody 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
Example  Reaction Forces on a Structural Frame
A weight F (1000 N) is hanging in a structural frame as shown in the figure above. A structural analyses can be done with the following equations:
∑M _{ A } = R _{ By } (230 cm)  (1000 N) (100 cm) = 0 (1)
∑M _{ B } = R _{ Ay } (230 cm)  (1000 N) (130 cm) = 0 (2)
∑M _{ CB } = R _{ By } (130 cm)  R _{ Bx } h = 0 (3)
F = R _{ Ay } + R _{ By } (4)
R _{ Ax } = R _{ Bx } (5)
Eq. 1 can be rearranged to
R _{ By } = (1000 N) (100 cm) / (230 cm)
= 435 N
Eq. 2 can be rearranged to
R _{ Ay } = (1000 N) (130 cm) / (230 cm)
= 565 N
The height h in eq. 3 can be calculated as
h = (153^{2} 130^{2}) ^{ 1/2 }
= 80.7 cm
Eq. 3 can then be rearranged to
R _{ Bx } = (435 N) (130 cm) / (80.7 cm)
= 700 N
From eq. 5
R _{ Ax } = R _{ Bx }
= 700 N
Related Topics

Mechanics
The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more. 
Statics
Forces acting on bodies at rest under equilibrium conditions  loads, forces and torque, beams and columns.
Related Documents

Acceleration of Gravity and Newton's Second Law
Acceleration of gravity and Newton's Second Law  SI and Imperial units. 
Bodies Moving on Inclined Planes  Acting Forces
Required forces to move bodies up inclined planes. 
Force
Newton's third law  force vs. mass and acceleration. 
Impulse and Impulse Force
Forces acting a very short time are called impulse forces. 
Mass Moment of Inertia
The Mass Moment of Inertia vs. mass of object, it's shape and relative point of rotation  the Radius of Gyration. 
Mass vs. Weight
Mass vs. weight  the Gravity Force. 
Torque or Moment of Force  Online Converter
Torque or moment  the tendency of a force to rotate an object.