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.
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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)