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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. 

Support reaction forces - equilibrium

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)


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 - forces acting on a body on an inclined plane

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

Support reaction forces - 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     

Example - Reaction Forces on a Structural Frame

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:

∑MA = RBy (230 cm)  - (1000 N) (100 cm) = 0                (1)

∑MB = RAy (230 cm) - (1000 N) (130 cm) = 0                (2)

∑MC-B = RBy (130 cm) - RBx h = 0                (3)

F = RAy + RBy                            (4)

RAx = RBx                           (5)

Eq. 1 can be rearranged to 

RBy = (1000 N) (100 cm) / (230 cm)

      = 435 N

Eq. 2 can be rearranged to 

RAy = (1000 N) (130 cm) / (230 cm)

      = 565 N

The height h in eq. 3 can be calculated as

h = (1532 - 1302)1/2

   = 80.7 cm

Eq. 3 can then be rearranged to

 RBx = (435 N) (130 cm) / (80.7 cm) 

       = 700 N

From eq. 5

RAx  = RBx 

       = 700 N

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  • Engineering ToolBox, (2005). Support Reactions - Equilibrium. [online] Available at: [Accessed Day Mo. Year].

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