Force Analysis
Static Systems
By plane truss we mean a truss that is in one plane, say x  y , and all forces applied to it are in
the same plane. The following equations hold for each link in a static plane truss of rigid links
F
k
0
(1)
k
M
k
0
(2)
k
where Fk is the k th force and M k is its moment about a fixed point of convenience.
Lemma 1. Truss link at static equilibrium with two forces only
If two forces are applied to a link at static equilibrium then they must be equal,
opposite and collinear
Proof
Rigid link
A
From Equation 1 we have F1  F2 so
the forces are equal and opposite.
Suppose that the forces are not collinear
as in Figure 1. Sun of moments about A
gives F2 d  0 , in contradiction to
Equation 2.
B
F2
d
F1
Figure 1
Lemma 2. Truss link at static equilibrium with two forces and a moment
If two forces and a moment are applied to a link at static equilibrium then the
forces must be equal and opposite.
Figure 2 shows such a scenario. It is clear that Equations 1 and 2 are both satisfied provided that
T  F2 d
T
B
A
d
F1
Figure 2
1
F2
Lemma 3. A Three-Force Link
If three forces are applied to a link at static equilibrium then they must close a
triangle and their lines of action must pass through a common point.
Proof
The graphical interpretation of Equation 1 is that the three forces close a triangle, as in Figure 3b.
Suppose that the forces from Figure 3b are applied on the link as in Figure 3a. Note that the lines
of action for these forces do not pass through a common point. Sum of moments about A gives
F2 d  0 , in contradiction to Equation 2.
F1
F1
F2
F3
A
d
F2
F3
Figure 3a
Figure 3b
The link shown in Figure 4a is in static equilibrium. The three forces acting on it close the
triangle shown in Figure 4b, and in addition the three lines of action of the forces pass through
the common point A , shown in Figure 4a.
F1
F2
F2
A
F1
F3
F3
Figure 4a
Figure 4b
Definition: Fij is the force that link i applies on link j
By Newton’s third law
Fij  F ji
(3)
2
Example 1
P
Given force P in the truss of Figure 5. Find the pin
forces F12 , F32 , F23 , F13 .
2
3
Figure 5
Solution
Link 2 is a two-force link. Therefore F13 and F23 are collinear, as shown in Figure 6b. By
Newton’s third law F23 and F32 are equal and opposite. Hence, knowing P and the direction of
F32 determines the direction of F12 , as in Figure 6a, since link 2 is a three-force link the lines of
action must pass through a common point. The force triangle equilibrium shown in Figure 6c
determines all pin forces of the truss which are shown in Figure 7.
F32
F23
P
F32
F12
3
2
P
F12
F13
Figure 6a
Figure 6b
Figure 6c
F32
F23
P
3
2
F12
F13
Figure 7a
Figure 7b
3
Example 2
Given P and Q as shown in the figure below. Find the pin forces.
Solution
The problem is first broken into two sub-problems, ' and ' ' as shown. Each sub-problem is of
the kind of two-force link and three-force link which can be solved as in Example 1. After
solving the two problems, force superposition gives the pin forces F12 , F32 , F23 and F13 . This
process is illustrated below.
Q
P
2
3
The problem
Q
P
2
Problem
superposition
3
2
Problem (‘)
F32
3
Problem (‘’)
F23
F32

F23
P
3
2
Q
3
2
Solution to
Problem (‘)
Solution to
Problem (‘’)
F13
F12
F12
F13
Force superposition
F12
F12
F32
F12
F32
F12  F12  F12
F32  F32  F32
F13
F23
F32
F23

F23
  F23

F23  F23
4
F13
F13
F13  F13  F13
Dynamic systems
For each link in a mechanism in motion the following equations hold:
F
k
 maG
(4)
 I G
(5)
k
M
k
k
Where aG in the acceleration of the center of gravity G , I G is the moment of inertia about G ,
m is the mass of the link, and M k is the moment of the force Fk about G .
D'Alembert's principle
Define an inertia force as
FI  maG
(6)
and inertia moment as
M I   I G .
(7)
Then the dynamics equations (4) and (5) takes the statics equations form
FI   Fk  0
(8)
MI  Mk  0 .
(9)
k
k
The implication of the principle is that if we add the inertia forces and moments to the
mechanism the force analysis may be done by using the statics methodology that was
described above.
5
Stopping a link from accelerating
Suppose that two forces FA and FB
are acting on link AB with mass m
and moment of inertia I G about the
center of gravity G , as shown in the
figure. As a result the link
accelerates. The acceleration of G is
aG and the angular acceleration of
the link is  , as show the figure.
FI  m
maa G
FB
B
FB
FA

G
aG
A
FA
Should the inertia force FI , expressed by equation 6, and shown in the figure, be applied to the
link then the acceleration of G would vanish.
If in addition the inertia force would be placed in a distance e such that
e
I G
FI
as shown it the figure below, then the inertia force FI would also cancel the angular acceleration
 of the link. Note that the moment applied by FI about G should be in an opposite sense to
 , i.e., if  has CCW sense, as in the figure above, then the moment of FI about G should be
with a CW sense, as in the figure below.
FB
B
G
e
I G
FI
FI  ma G
A
FA
6