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Support Reactions - Equilibrium

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

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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 = ΣFx = ΣFy = ΣFz = 0                          (1)

ΣM = ΣMx = ΣMy = ΣMz = 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 = ΣFx = ΣFy = 0                          (3)

ΣM = ΣMz = 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

Fx = R1x = R2x = 0                           (5)

Fy - (R1y + R2y) = 0                           (6)

M1 = Fy a - R2y (a + b) = 0                   (7)

Rearranging (7) to express R2y

R2y = F a / (a + b)                         (7b)

Eq. (7b) with values

R2 = (9000 N) (4 m) / ((4 m) + (2 m))

    = 6000 N

    = 6 kN

Rearranging (6) for R1y

R1y = Fy - R2y                        (6b)

Eq. (6b) with values

R1y = (9000 N) - (6000 N)

      = 3000 N

      = 3 kN     

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