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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.

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

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)


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

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 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

F x = R 1x = R 2x = 0                           (5)

F y - (R 1y + R 2y ) = 0                           (6)

M1 = 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

R2= (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

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 C-B = 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 = (1532- 1302) 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.

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