Eleventh Edition
CHAPTER
VECTOR MECHANICS FOR ENGINEERS:
STATICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
David F. Mazurek
Equilibrium of Rigid
Bodies
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Eleventh
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Vector Mechanics for Engineers: Statics
Contents
Introduction
Free-Body Diagram
Rigid Body Equilibrium in Three
Dimensions
Reactions for a Two-Dimensional
Structure
Reactions for a Three-Dimensional
Structure
Rigid Body Equilibrium in Two
Dimensions
Sample Problem 4.8
Sample Problem 4.1
Sample Problem 4.4
Practice
Statically Indeterminate Reactions
and Partial Constraints
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Vector Mechanics for Engineers: Statics
Introduction
• For a rigid body, the condition of static equilibrium means that the
body under study does not translate or rotate under the given loads
that act on the body
• The necessary and sufficient conditions for the static equilibrium of a
body are that the forces sum to zero, and the moment about any point
sum to zero:


 
 F  0  M O   r  F   0
• Equilibrium analysis can be applied to two-dimensional or threedimensional bodies, but the first step in any analysis is the creation of
the free body diagram
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Vector Mechanics for Engineers: Statics
Free-Body Diagram
The first step in the static equilibrium analysis of a
rigid body is identification of all forces acting on
the body with a free body diagram.
• Select the body to be analyzed and detach it
from the ground and all other bodies and/or
supports.
• Indicate point of application, magnitude, and
direction of external forces, including the rigid
body weight.
• Indicate point of application and assumed
direction of unknown forces from reactions of
the ground and/or other bodies, such as the
supports.
• Include the dimensions, which will be needed
to compute the moments of the forces.
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Vector Mechanics for Engineers: Statics
Reactions for a Two-Dimensional Structure
• Reactions equivalent to a
force with known line of
action.
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Vector Mechanics for Engineers: Statics
Reactions at Supports and Connections for a Two-Dimensional Structure
• Reactions equivalent to a
force of unknown direction
and magnitude.
• Reactions equivalent to a
force of unknown
direction and magnitude
and a couple of unknown
magnitude
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Vector Mechanics for Engineers: Statics
Practice
The frame shown supports part of
the roof of a small building. Your
goal is to draw the free body
diagram (FBD) for the frame.
On the following page, you will
choose the most correct FBD for
this problem.
First, you should draw your own
FBD.
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Vector Mechanics for Engineers: Statics
Practice
B
A
150 kN
Choose the most
correct FBD for the
original problem.
B is the most correct, though C is also
correct. A & D are incorrect; why?
C
150 kN
D
150 kN
150 kN
Discuss with a
neighbor why each
choice is correct or
incorrect.
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Vector Mechanics for Engineers: Statics
Rigid Body Equilibrium in Two Dimensions
• For known forces and moments that act on a
two-dimensional structure, the following are
true:
Fz  0 M x  M y  0 M z  M O
• Equations of equilibrium become
 Fx  0  Fy  0  M A  0
where A can be any point in the plane of
the body.
• The 3 equations can be solved for no more
than 3 unknowns.
• The 3 equations cannot be augmented with
additional equations, but they can be replaced
 Fx  0  M A  0  M B  0
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Sample Problem 4.1
STRATEGY:
Draw a free-body diagram to show all of
the forces acting on the crane, then use the
equilibrium equations to calculate the
values of the unknown forces.
A fixed crane has a mass of 1000 kg
and is used to lift a 2400 kg crate. It
is held in place by a pin at A and a
rocker at B. The center of gravity of
the crane is located at G.
MODELING:
Determine the components of the
reactions at A and B.
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Sample Problem 4.1
ANALYSIS:
• Determine B by solving the equation for the
sum of the moments of all forces about A.
 M A  0 :  B1.5m   9.81 kN2m 
 23.5 kN6m   0
B  107.1 kN
• Determine the reactions at A by solving the
equations for the sum of all horizontal forces
and all vertical forces.
 Fx  0 : Ax  B  0
Ax  107.1 kN
 Fy  0 : Ay  9.81 kN  23.5 kN  0
Ay  33.3 kN
• Check the values obtained.
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Vector Mechanics for Engineers: Statics
Sample Problem 4.1
REFLECT and THINK:
You can check the values obtained for
the reactions by recalling that the sum
of the moments of all the external
forces about any point must be zero.
For example, considering point B, you
can show
M B  9.81 kN 2 m   23.5 kN6 m  107.1 kN1.5 m  0
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Vector Mechanics for Engineers: Statics
Sample Problem 4.4
STRATEGY:
- Discuss with a neighbor the steps for
solving this problem
• Create a free-body diagram for the
frame and cable.
The frame supports part of the roof of
a small building. The tension in the
cable is 150 kN.
• Apply the equilibrium equations
for the reaction force components
and couple at E.
Determine the reaction at the fixed
end E.
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Sample Problem 4.4
MODELING:
• Which equation is correct?
4.5
A.  Fx  0 : E x  7.5 150 kN   0
E x  90.0 kN
B.
 Fx  0 : Ex  cos 36.9o 150kN 0
C.
 Fx  0 :
E x  sin 36 .9 o 150 kN   0
E x  90.0 kN
ANALYSIS:
• Apply one of the three
equilibrium equations. Try
using the condition that the
sum of forces in the xdirection must sum to zero.
D.
 Fx  0 : Ex 
E.
 Fx  0 :
6
150kN   0

7.5
E x  sin 36 .9 o 150 kN   0
• What does the negative sign signify?
• Discuss why the others are incorrect.
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Vector Mechanics for Engineers: Statics
Sample Problem 4.4
• Which equation is correct?
A.  Fy  0 : Ey  420kN  sin 36.9 o 150kN   0
B.  Fy  0 : Ey  420kN 
6
150kN  0

7.5
C.  Fy  0 : Ey  420kN  cos 36.9 o 150kN  0
Ey  200 kN
• Now apply the condition
that the sum of forces in
the y-direction must sum
to zero.
6


D.  Fy  0 : E y  4 20 kN  150 kN   0
Ey  200 kN
7.5
6
150kN   0

7.5
• What does the positive sign signify?
E.  Fy  0 : Ey  420kN 
• Discuss why the others are incorrect.
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Vector Mechanics for Engineers: Statics
Sample Problem 4.4
• Three good points are D, E, and F.
Discuss what advantage each point has
over the others, or perhaps why each is
equally good.
• Assume that you choose point E to
apply the sum-of-moments condition.
Write the equation and compare your
answer with a neighbor.
 M E  0 :  20 kN7.2 m   20 kN5.4 m 
• Finally, apply the condition
that the sum of moments about
any point must equal zero.
• Discuss with a neighbor which
point is the best for applying
this equilibrium condition, and
why.
 20 kN3.6 m   20 kN1.8 m 

6
150 kN 4.5 m  M E  0
7.5
M E  180.0 kN  m
• Discuss with a neighbor the origin of
each term in the above equation and
what the positive value of ME means.
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Sample Problem 4.4
REFLECT and THINK:
The cable provides a fourth constraint, making this situation
statically indeterminate. This problem therefore gave us the value
of the cable tension, which would have been determined by
means other than statics. We could then use the three available
independent static equilibrium equations to solve for the
remaining three reactions.
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Vector Mechanics for Engineers: Statics
Practice
A 2100-lb tractor is used to
lift 900 lb of gravel.
Determine the reaction at
each of the two rear wheels
and two front wheels
• First, create a free body diagram.
Discuss with a neighbor
what steps to take to solve
this problem.
• Second, apply the equilibrium
conditions to generate the three
equations, and use these to solve
for the desired quantities.
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Vector Mechanics for Engineers: Statics
Practice
• Draw the free body diagram of the tractor (on your own first).
• From among the choices, choose the best FBD, and discuss the
problem(s) with the other FBDs.
A.
B.
FA
2100 lb
FB
C.
FA
2100 lb
FB
FA
2100 lb
FB
D.
FA
2100 lb
FB
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Vector Mechanics for Engineers: Statics
Practice
Now let’s apply the equilibrium
conditions to this FBD.
• Start with the moment equation:
 M pt  0
FA
2100 lb
FB
Points A or B are equally
good because each results in
an equation with only one
unknown.
Discuss with a neighbor:
• What’s the advantage to
starting with this instead of the
other conditions?
• About what point should we
sum moments, and why?
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Vector Mechanics for Engineers: Statics
Practice
Assume we chose to use point B.
Choose the correct equation for
 M B  0.
FA
2100 lb
FB
A. + FA (60 in.) - 2100lb (40 in.) - 900 lb (50 in.) = 0
B. + FA (20 in.) - 2100lb (40 in.) - 900 lb (50 in.) = 0
C. - FA (60 in.) - 2100lb (40 in.) + 900 lb (50 in.) = 0
D. - FA (60 in.) + 2100lb (40 in.) - 900 lb (50 in.) = 0
FA=650 lb, so the reaction at each wheel is 325 lb
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Vector Mechanics for Engineers: Statics
Practice
Now apply the final equilibrium
condition, Fy = 0.
FA
2100 lb
FB
FA  2100 lb + FB  900 lb = 0
or + 650 lb  2100 lb + FB  900 lb = 0
 FB  2350 lb, or 1175 lb at each front wheel
Why was the third equilibrium
condition, Fx = 0 not used?
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What if…?
W=?
• Now suppose we have a different
problem: How much gravel can this
tractor carry before it tips over?
• Discuss with a neighbor how you
would solve this problem.
2100 lb
FB
W
2100 lb
• Hint: Think about what the free
body diagram would be for this
situation…
FB
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Vector Mechanics for Engineers: Statics
Statically Indeterminate Reactions and Partial Constraints
• More unknowns than
equations
• Fewer unknowns than • Equal number unknowns
and equations but
equations, partially
improperly constrained
constrained
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