5.1

Chapter 5: Equilibrium of a Rigid Body

Chapter Objectives

• To develop the equations of equilibrium for a rigid body.

• To introduce the concept of a free-body diagram for a rigid body.

• To show how to solve rigid body equilibrium problems using the equations of

equilibrium.

In this chapter, we will describe the various types of supports that are used with rigid

body equilibrium problems.

• We will then show how free-body diagrams and equilibrium equations are used to

determine unknown forces and couples acting on a rigid body.

5.1 Conditions for Rigid-Body Equilibrium

When the force and couple are both equal to zero, the external forces form a

system equivalent to zero and the rigid body is said to be in equilibrium.

• Hence two equations of equilibrium for a rigid body can be summarized as

follows.

∑F = 0 Necessary and sufficient conditions for

∑M = 0 the equilibrium of a rigid body.

• We may express the necessary and sufficient conditions for the equilibrium of

a rigid body in the following scalar form.

∑Fx = 0 ∑Fy = 0 ∑Fz = 0

∑Mx = 0 ∑My = 0 ∑Mz = 0

Equilibrium in Two Dimensions

First, we will consider the case in which the force system acting on a rigid body is in a

single plane.

• This type of force system is referred to as a two-dimensional or coplanar force

system.

• Forces acting on the body are in the same plane.

• Couple moments acting on the body are perpendicular to this plane.

5.2 Free-Body Diagrams

The best way to account for all of the known and unknown external forces acting

on a rigid body is to draw the free-body diagram.

5.2

Support Reactions

“Reactions” are the forces through which the ground and other bodies oppose a

possible motion of the free body.

• Reactions are exerted at points where the free body is supported or connected

to other bodies.

It is important to understand how to replace certain supports with the appropriate

constraining forces.

• In general, if a support prevents translation, then a force is developed on the

body in that direction.

• Likewise, if rotation is prevented, a couple moment is exerted on the rigid body.

Reactions exerted on two-dimensional structure may be divided into three groups,

corresponding to three types of supports or connections.

1. Reaction equivalent to a force with a known line of action (1 unknown).

• This reaction prevents translation of the free body in one direction, but it

cannot prevent the body from rotating about the connection.

• For each of these supports there is only one unknown involved, that is, the

magnitude of the reaction.

• The line of action is known.

The most common examples are as follows.

a. Roller support

A roller support prevents vertical translation.

b. Link

A link prevents translation along the axis of the link.

5.3

c. Cables (weightless)

2. Reaction equivalent to a force with an unknown line of action (2 unknowns).

• This reaction prevents translation of the free body in all directions, but it

cannot prevent the body from rotating about the connection.

• The magnitude and direction of the reactive force is unknown.

• Normally we work with the components, thus fixing the directions but not

knowing the magnitude of the components.

The most common examples are as follows.

a. Pin

A pin prevents vertical and horizontal translation.

b. Ball and socket joint (2-D)

A ball and socket joint prevents vertical and horizontal translation.

3. Reaction equivalent to a force and a couple (3 unknowns).

• This reaction is caused by “fixed supports” which oppose any motion of the

free body and thus constrain it completely, preventing both translation and

rotation.

• Reactions of this group involve three unknowns, consisting usually of the two

components of the force and the moment of the couple.

5.4

The most common example is as follows.

a. Fixed end

The fixed end prevents vertical and horizontal translation, and prevents

rotation.

When the sense of an unknown force or couple is not clearly apparent, no attempt

should be made to determine the correct direction.

• Instead, the sense of the force or couple should be arbitrarily assumed.

• The sign of the answer obtained will indicate whether the assumption is correct

or not.

Internal Forces

If a free-body diagram for the rigid body is drawn, only external forces are shown.

• Internal forces within the rigid body are not represented on a free-body

diagram.

• Internal forces cancel each other, and as a result, do not create an external

effect on the rigid body

Weight and the Center of Gravity

When the weight of a body must be considered, a force resultant representing the

weight is used with its point of application acting through the center of gravity.

Idealized Models

In order to perform a correct force analysis of any object, it is important to

consider a corresponding analytical or idealized model that gives results that

approximate as closely as possible the actual situation.

• Example, consider the underground pump station: Is reinforced concrete slab

acting as simply supported, or fixed-fixed to concrete block walls?

5.5

Procedure for Drawing a Free-Body Diagram

1. Draw the isolated body.

2. Add all the applied forces (including the weight of the body).

3. Show the reactive forces (“constraining forces”).

On a free body diagram, include the dimensions.

• The dimensions may be needed to compute moments.

5.6

Example – Free Body Diagram

Given: The loaded W18 x 40 beam shown.

Find: Draw the free body diagram and

solve for the reactions.

Draw the free body diagram.

Steps:

1. Draw the isolated body.

2. Add the applied forces.

3. Show the reactive forces.

a. Does the support at A prevent . . .

1) Horizontal translation?

2) Vertical translation?

3) Rotation?

b. Does the support at B prevent . . .

1) Horizontal translation?

2) Vertical translation?

3) Rotation?

Solve for the reactions.

∑Fx = 0 = (4/5) 75 + Ax

Ax = - 60 kips Ax = 60 kips ←

∑MA = 0 = 10 By – (3/5) 75 (4)

10 By = 180

By = + 18.0 kips By = 18.0 kips ↑

∑MB = 0 = - 10 Ay + (3/5) 75 (6)

10 Ay = 270

Ay = + 27.0 kips Ay = 27.0 kips ↑

5.7

5.3 Equations of Equilibrium

The conditions of equilibrium in two dimensions are as follows.

∑Fx = 0 ∑Fy = 0 ∑Mz = 0

These three equations (called the “equations of statics”) allow solution for no more

than three unknowns.

Alternative Sets of Equilibrium Equations

These three equations may be replaced with another set of equations.

• An alternate set of equilibrium equations may be as follows.

∑Fx = 0 ∑MA = 0 ∑MB = 0

or ∑MA = 0 ∑MB = 0 ∑MC = 0

5.8

Example

Given: Beam loaded as shown.

Find: Range of values for P for a

safe beam:

RA and RB ≤ 25 kips

(i.e. compression in the

columns at A and B)

∑Fx = 0 = Ax Ax = 0 kips

Let RA = 0 kips: ∑MB = 0 = 6 P - 6 (2) – 6 (4) P = 6.0 kips

Check RB ∑Fy = 0 = - 6 – 6 – 6 + RB

RB = 18.0 kips < 25.0 kips OK

Let RA = 25 kips: ∑MB = 0 = - 25 (9) + 6 P - 6 (2) – 6 (4)

6 P = 225 + 12 + 24 = 261 P = 43.5 kips

Check RB ∑Fy = 0 = 25 - 43.5 – 6 – 6 + RB

RB = 30.5 kips > 25.0 kips NG

Let RB = 25 kips: ∑MA = 0 = - 3 P + 9 (25) – 6 (11) – 6 (13)

3 P = 225 – 66 – 78 = 81 P = 27.0 kips

Check RA ∑Fy = 0 = - 27 + 25 – 6 – 6 + RA

RA = 14.0 kips < 25.0 kips OK

Therefore, 6.0 kips ≤ P ≤ 27.0 kips Answer

5.9

5.4 Two- and Three-Force Members

The solution to some equilibrium problems can be simplified if one is able to

recognize members that are subjected to only two or three forces.

Two-Force Members

If a two-force body is in equilibrium, the two forces must have the same

magnitude, same line of action, and opposite sense.

For the corner plate to be in equilibrium the following equilibrium equations must

be satisfied.

∑MA = 0 = F2r d : either F2r = 0 or d = 0, and

∑MB = 0 = F1r d : either F1r = 0 or d = 0

F1r and F2r must be zero; d = 0 is too restrictive (the rigid body would revert to

a particle).

Characteristics of a two-force member:

1. Coplanar or non-coplanar (any shape).

2. Two forces – same magnitude, same line of

action, opposite direction.

3. Direction of two forces is collinear with line of

action connecting points of application.

4. Points need not be at the end of the member.

5. No couple allowed on member.

A truss member is an example of a two-force member.

5.10

Three-Force Members

If a three-force body is in equilibrium, the lines of action of the three forces must

be either concurrent or parallel.

For the rigid body to be in equilibrium the following equilibrium equation must be

satisfied.

∑MA = 0 = + F3 d: either F3 = 0 or d = 0

F3 ≠ 0 (otherwise the system would revert to a two-force system).

Thus d = 0.

For the rigid body to be in equilibrium, the three forces must be either concurrent

(i.e. d = 0), or parallel.

Concurrent forces Parallel forces

5.11

Equilibrium in Three Dimensions

5.5 Free-Body Diagrams

A simple way of determining the type of reaction corresponding to a given support

or connection and the number of unknowns involved is to find which of the six

fundamental motions (translations in the x, y, and z directions and rotations about

the x, y, and z axes) are allowed and which motions are prevented.

Support Reactions

Reactions exerted on three-dimensional structures may be divided into three

groups, corresponding to three types of supports or connections.

1. Reaction equivalent to a single force and a known line of action (translation

prevented in one direction – one unknown).

Examples: Ball (or roller supports), frictionless surface, cables

2. Reaction equivalent to a single force and an unknown line of action (translation

prevented in 3 directions – 3 unknowns).

Example: Ball and socket joint (3D)

3. Reaction equivalent to a force and a couple with unknown lines of action

(translation and rotation prevented – 6 unknowns)

Example: Fixed support

5.12

5.6 Equations of Equilibrium

Vector Equations of Equilibrium

The two conditions of equilibrium of a rigid body are as follows.

∑F = 0 Necessary and sufficient conditions for

∑M = 0 the equilibrium of a rigid body.

Scalar Equations of Equilibrium

Six scalar equations are required to express the conditions for the equilibrium of a

rigid body in the general three-dimensional case.

∑Fx = 0 ∑Fy = 0 ∑Fz = 0

∑Mx = 0 ∑My = 0 ∑Mz = 0

These equations may be solved for no more than six unknowns.

5.7 Constraints and Statical Determinacy

A problem is said to be “statically determinate” when a rigid body is completely

constrained by its supports and the reactions can be determined using the

equations of statics.

Two-Dimensional Analysis

In a two-dimensional analysis, only three independent equilibrium equations are

available.

• When there are more than three unknown reactions, not all of the unknown

reactions can be determined using the equations of equilibrium.

- This type of problem is said to be “statically indeterminate.”

Three-Dimensional Analysis

In a three-dimensional analysis, only six independent equilibrium equations are

available.

• When there are more than six unknown reactions, not all of the unknown

reactions can be determined using the equations of equilibrium.

- This type of problem is said to be “statically indeterminate.”

Redundant Constraints

Determinate

5.13

Determinate

Determinate

Statically indeterminate (with degree of redundancy)

SI1

SI3

SI1

Mobile structure (improperly constrained)

Improper Constraints

“Partially constrained” – The supports provided are not enough to keep the body

(structure) from moving.

• There are fewer unknowns than equations and one of the equilibrium equations

will not be satisfied.

5.14

• If the reactions involve less than three unknowns (2-D) or less than six

unknowns (3-D), there are more equations than there are unknowns, and some of

the equations of equilibrium cannot be satisfied under a general loading

condition.

• The rigid body is said to be only “partially constrained” and should be avoided

since an unstable condition may result.

“Improperly constrained” – There is a sufficient number of constraints, but these

constraints are not properly arranged and the body (structure) is free to move.

• A rigid body is improperly constrained when the supports, even though they

provide a sufficient number of reactions, are arranged in such a way that at

least one of the equilibrium equations is not satisfied.

- Even with three or more unknowns (2-D), or six or more unknowns (3-D), it is

possible that some of the equations of equilibrium will not be satisfied.

- This may occur when the supports are such that the reactions are forces

that are parallel.

• The rigid body is said to be “improperly constrained” and should be avoided

since an unstable condition may result.