DYNAMIC FORCE ANALYSIS OF MECHANISMS
DYNAMIC FORCE ANALYSIS OF MECHANISMS
KINEMATICS
α
ω
DYNAMICS
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DYNAMIC FORCE ANALYSIS OF MECHANISMS
Newton’s Second Law
D’Alembert’s Principle
Dynamic force analysis - examples
NEWTON’S SECOND LAW OF MOTION
Translation:

a
F2
m
F1


∑ F i = ma
F3
Two Scalar Equations
∑F
i.e.
{∑
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= ma x
F iy = ma y
4
NEWTON’S SECOND LAW OF MOTION
Rotation about a fixed point:
∑M
i
= Iα
I - moment of inertia about fixed point
F1
α - angular acceleration
T2
α
F3
4-bar Animation
T3
T1
F2
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Pivot
point
5
NEWTON’S SECOND LAW OF MOTION
GENERAL PLANAR MOTION
Translation of CG


∑ F j = maG
G
F2
Rotation about CG
∑M
jG
= I Gα
F1
AG
α
T1
T2
F3


=
F
m
a
∑ j G
∑M
jG
= I Gα
Subscript G is denoted as the center of mass
IG is the mass moment of inertia about G
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End of topic one
D’ALEMBERT’S PRINCIPLE – FORCE EQUATION
Newton’s law:
F1
AG
Newton’s law:
CG
F2


∑ F j = maG
α
T1
∑ M jG = IGα
T2


∑ F j = maG


⇒ ∑ F j − maG = 0
Define inertial force to be:


FI = −maG
F3
D’Alembert’s Principle:
CI = IGα
We have (mathematically):
F1

∑ Fi = 0
F2
∑M
G
T2
T1
F3
FI = ma
G
i
=0


∑ F j + FI = 0

∑ Fi = 0
Imaginary (mathematic) ‘static’
equilibrium state with an additional

inertial force 
FI = −maG
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D’ALEMBERT’S PRINCIPLE – MOMENT EQUATION
Newton’s law:
F1
AG
Newton’s law:
∑M
CG
F2


∑ F j = maG
α
T1
∑M
T2
jG
= I Gα
CI = IGα
F1

∑ Fi = 0
F2
∑M
G
T2
T1
F3
FI = ma
G
i
=0
= I Gα
⇒ ∑ M jG − I Gα =
0
Define inertial moment to be:
C I = − I Gα
F3
D’Alembert’s Principle:
jG
We have (mathematically):
∑M
jG
∑M
=0
i
+ CI =
0
Imaginary (mathematic) ‘static’
equilibrium state with an additional
inertial moment
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C I = − I Gα
9
D’ALEMBERT’S PRINCIPLE
D’Alembert’s Principle
Newton’s law:
D’Alembert’s Principle:
F1
AG
CI = IGα
CG
F1
F2
F2
α
G
T1
T2
T2
T1
F3
FI
F3


∑ F j = maG

∑ Fi = 0
∑M
∑M
jG
= I Gα
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i

= maG
=0
10
D’ALEMBERT’S PRINCIPLE – LINK EXAMPLE
D’Alembert’s Principle:
Newton’s law:
FI
FI
FA
C.G.
FA


FI = maG
A
A
CI = IGα
TI
TI
AG
B
B
α
FB
C.G.
FB
Dynamic Problem


∑ F j = maG
∑M
jG
Equivalent Static Equilibrium

∑ Fj = 0
∑M
= I Gα
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j
=0
11
D’ALEMBERT’S PRINCIPLE
CONVENTION FOR INERTIAL FORCE & MOMENT
y
G
a
α
x
G
a
x
G
ma
y
G
ma
C.G.
Motion(Kinematics)
IGα
Inertial force & moment
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D’ALEMBERT’S PRINCIPLE
SUMMARY OF THE STEPS FOR DYNAMIC FORCE ANALYSIS
1. Find accel. of CG & angular accel. For each link (or
use given ones)
2. Free individual body(s)
3. Draw inertia force at CG & inertia moment about CG
4. Draw applied force/moments
5. Draw assumed constraint forces
6. Write equilibrium equations for each free body (total
3N)
7. Solve the equations for unknowns
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End of
topic 2
MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY
The mass moment of inertia of a rigid body is
calculated in the following way:
I G = ∫ r 2 ρdV
Where:
ρ isi the density of rigid body
. r
dV
G

r
is the distance from the
centre of mass to an
arbitrary point inside the
body
The mass moment of
inertia is taken about the
centre of mass point G.
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MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY
For a homogenous body of regular geometry, the mass
moment of inertia can be calculated from equation directly.
The following table gives some mass moment of inertia of a
body with simple geometry
L
1
mL2
12
1
I O = mL2
3
IG =
Slender rod
G
O
r
IG =
Disk/Cylinder
1
mr 2
2
a
Rectangular Plate
b
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IG =
1
m(a 2 + b 2 )
12
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MASS CENTRE, MOMENT OF INERTIA FOR RIGID BODY
r2
r1
Ring
r
Thin ring
IG =
1
m ( r12 + r22 )
2
I G = mr 2
IG =
G
Semi-circular plate
r
h
O
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1
mr 2
2
1
I O = mr 2 − mh 2
2
4r
h=
3π
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PARALLEL AXIS THEOREM
For mass moment of inertia being taken about a point A other than the
centre of mass G, we can derive the mass moment of inertia using the
following parallel axis theorem:

I A = I G + m | r AG | 2
Where:
 is the position vector from
rA
IA
IG
G
point A to the centre of mass G

r AG
A
The mass moment of
inertia is taken about an
arbitrary point A.
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EXAMPLE FOR MASS CENTRE, MOMENT OF INERTIA
An assembly of a uniform slender and disk
A uniform slender rod of mass (
and length (
)
) and a uniform disk of
mass (
) and radius (
) are
welded to form an assembly as shown
in the figure.
Determine the mass moment of inertia of the assembly
about its centre of mass G and point C. The moments of
inertia of the disk and the rod about their mass centres A
and B are known to be (
) and (
)
respectively.
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EXAMPLE FOR MASS CENTRE, MOMENT OF INERTIA
An assembly of a uniform slender and disk
Solution:
• Center of Mass:
Using point A as the origin, the centre of mass can located as
Note rAA=0
⇒
(m)
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EXAMPLES FOR MASS CENTRE, MOMENT OF INERTIA
An assembly of a uniform slender and disk
Solution (continued):
• Mass Moment of Inertia about G:
Using the parallel axis theorem, the moment of inertia can be calculated as follows:
the disk:
(kg m2)
the rod:
(kg m2)
and
•
(kg m2)
Mass Moment of Inertia about C:
Similarly, we have
(kg m2)
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End of
topic 3
EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Rotation of a uniform slender rod hinged at point A
A
G
α
g
L
4
Example 1:
The uniform slender rod of mass m and length L is pivoted at A in
the position as shown.
The rod is released from rest. Determine the initial angular
acceleration of the rod and the constraint force at A.
(Given
IG =
1
mL2
12
)
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Acceleration analysis of a uniform
slender rod hinged at point A
Rotation of a uniform slender rod
hinged at point A
G
A
α
y
g
G α
A
x
r AG
L
4
(Given
IG =
1
mL2
12
L
4
)
aG
Solution:
• Kinematics Analysis:
Letting α be the angular acceleration of the rod, we have
a =0
x
G
L
a =− α
4
y
G
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EXAMPLE FOR DYNAMIC FORCE ANALYSIS
• Solution (use D’Alembert’s Principle):
∑M
A
=0
α=
∑F
x
y
FAy
L
L
L
− mg + m α × + I Gα = 0
4
4
4
⇒
FI = maG = m . L α
4
C
=
I
α
I
G
G
12 g
7 L
A
FAx
mg
=0
FAx = 0
∑F
y
G α
A
=0
x
r AG
FAy + m
⇒
y
FAy =
L
α − mg = 0
4
L
4
4
mg
7
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aG
a =0
x
G
L
a =− α
4
y
G
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Example 2:
Analyse the planar in-line slider-crank mechanism for the given dimensions
as shown in diagram. Given the input ω1 and α1, determine the required
torque T1 and the bearing forces at joints.
An in-line slider-crank mechanism
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Solution:
• Kinematics Analysis (velocity):
y
Using
the
relative
velocity,
we
can
find







v C = v B + v C / B = ω 1 × rO1B + ω 2 × rBC


⇒ ω 2 = −2.1k̂ (rad/s)(CW) and vC = −14.85iˆ (in/s)
vB
B
66.4
G1
1"
T1
O1
1.88"
G2
2"
70

vC / B
3.76"
30 
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C G3
vC
27
EXAMPLES FOR DYNAMIC FORCE ANALYSIS
• Kinematics Analysis (acceleration):
Using the relative acceleration, we can find:



n + t + n + t
aC = a B + aC / B = a B
a B aC / B aC / B
⇒ α 2 = 46kˆ (rad/s2)(CCW) and aC = −70iˆ (in/s2).
With the angular velocities and accelerations,
the following quantities can be obtained:


aG1 x = −71.6iˆ (in/s2), aG1 y = −79.5j (in/s2),


aG 2 x = −106iˆ (in/s2), aG 2 y = −80.0j (in/s2).
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Free body diagram of the slider-crank using D’alembert Principle
FBy
FBx
m 2 aG 2 y
B
G2
m 2 aG 2 x
I G 2α 2
FCy
30 
C
FBx
FCy
B
m1aG1 y
FO1 y
FCx
FBy
G1
FCx
m 3 aG 3
m1aG1 x
T1
I G1α1
70
O1
FN
FO1 x
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
Solution (continued):
• Dynamic Analysis of Link 3 (Slider):
From free body diagram, we have
⇒

∑ F =0
⇒
∑F
∑F
FCy
FCx
m 3 aG 3
FN
Free body diagram of the slider using
D’Alembert Principle
⇒
x
= 0 : FCx + m3 aG 3 x = 0
y
= 0 : FN − FCy = 0
FCx = −7.776 ×10 −3 × 70 = −0.543

(lb)
F
=
F
Cy
 N
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
• Dynamic Analysis of Link 2:
From free body diagram, we have

∑MB = 0

rBC × ( − FCx iˆ + FCy ˆj ) − I G 2α 2 + rBG 2 × m 2 (aG 2 x iˆ + aG 2 y ˆj ) = 0
⇒
− rBC y × FCx + rBC x × FCy − I G 2α 2 + rBG 2 y × m 2 aG 2 x + rBG 2 x × m 2 aG 2 y = 0
⇒
⇒
FCy = −0.396
∑F
∑F
x
= 0 : FBx − FCx + m2 aG 2 x = 0
y
= 0 : FBy + FCy + m2 aG 2 y = 0
F = −1.09
⇒  Bx
FBy = −0.018
FBy
FCy
FBx
(lb)
m 2 aG 2 y
B
G2
m 2 aG 2 x
I G 2α 2
FCy
30 
C
FCx
m 3 aG 3
FCx
FN
Free body diagram of the slider-crank using D’Alembert Principle
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EXAMPLES FOR DYNAMIC FORCE ANALYSIS
• Dynamic Analysis of Link 1:
From free body diagram, we have

∑ M O1 = 0
⇒

 
ˆ
ˆ
ˆ
T1 k + rO 1 B × ( − FBx i − FBy j ) − I G 1α 1 + rO 1G 1 × m1 (aG 1 x iˆ + aG 1 y ˆj ) = 0
T1 + rO 1 By × FBx − rO 1 Bx × FBy − I G 1α 1 − rO 1G 1 y × m1aG 1 x + rO 1G 1 x × m1aG 1 y = 0
⇒ T1 = 2.54 (lb-in)
FBy

FBx
m 2 aG 2 y
∑F = 0
⇒
∑F
∑F
x
y
B
= 0 : FO1x − FBx − m1aG1x = 0
= 0 : FO1 y − FBy − m1aG1 y = 0
FO1 x = −1.28
⇒ 
FO1 y = −0.224
G2
FBx
I G 2α 2
m1aG1 y
FO1 y
(lb)
B
FBy
G1
m 2 aG 2 x
FCy
30 
C
FCx
m1aG1 x
T1
I G1α1
70
O1
FO1 x
Free body diagram of the slider-crank using D’Alembert Principle
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End of
topic 4
Example 3 (exam AY00/01 S2): Figure 1 shows a mechanism driven by link 1. At the position shown, link 1 rotates with an
angular acceleration α1. At the output end F, the driven link 5 is pushing a mechanical part, which is idealized as a spring of
stiffness Ks. The mass of link 3 and the mass of the roller link 4 are considered to be negligible. The dimensions of the links are
given in Figure 1. Given that the spring is compressed by a displacement ∆s at this instant, and the following:
Mass
Location
Moment of
of CG inertia about CG
Acceleration
of CG
Angular
acceleration
Link 1
m1
G1 at O1
I1
0
α1 (cw)
Link 2
m2
G2
I2
a2xi+a2yj
α2 (cw)
Link 3 massless




Link 4 massless




G5 at O5
I5
0
α5 (ccw)
Link 5
m5
y
Neglecting the gravitational force and friction forces,
1) Draw the free-body diagram of each link (except the ground
link) for dynamic force analysis;
x
2) Derive the expressions for the constraint force at D between
links 4 and 5, the constraint force at bearing O5, the constraint
force at B (between links 2 and 3), the constraint force at A
(between links 1 and 2), as well as the required input driving
torque T.
D
E
α5
Ks
Link 5
F
Link 4
C
B
O5
G5
massless
G2
G1 O1
A
T
α1
ο
30
α2
Link 3
Link 2
O3
45
ο
Link 1
34
Solution
Free Body Diagrams
35
Solution:
From free body diagrams, we have
• For Free-body link 5:
∑ TO5 = 0
⇒
⇒
r5
2
− I 5α 5 − K s ∆ ⋅ r5 = 0
2 I 5α 5
F45 =
+ 2 K s ∆ = 23 I 5α 5 + 2 K s ∆
r5
F45 ⋅
∑ Fx = 0
⇒
F05x = 0
∑ Fy = 0
⇒
F05y + F45 − K s ∆ = 0
⇒
F05y
2 I 5α 5
= K s ∆ − F45 = −
− Ks∆
r5
36
• For Free-body link 2+4:
∑ TA = 0
⇒ I 2α 2
+ F32 (r2 cos 30o − r3 cos 45o ) sin 45o + m2 a2 x
− m2 a2 y
⇒
r2
2
cos 30o − F45
r2
2
r2
2
sin 30 o
cos 30o = 0
F32 = (3 3m2 a2 y + 2 3I 5α 5 + 6 3K s ∆ − 1.5m2 a2 x − I 2α 2 ) / 2 2
∑ Fx = 0
⇒
⇒
F12x − m2 a2 x + F32 cos 45o = 0
F12x = 85 m2 a2 x − 14 (3 3m2 a2 y + 2 3I 5α 5 + 6 3K s ∆ − I 2α 2 )
∑ Fy = 0
⇒
⇒
F12y − m2 a2 y − F45 + F32 sin 45o = 0
B
4−3 3
4−3 3
m2 a2 y +
I 5α 5
4
6
+ ( 2 − 1.5 3 )K s ∆ + 14 ( 1.5m2 a2 x + I 2α 2 )
F12y =
A
37
•For Free-body link 1:
∑ TO1 = 0
⇒
T + I1α1 + F12y ⋅ 1 = 0
⇒
3 3−4
3 3−4
m2 a2 y +
I 5α 5 + (1.5 3 − 2) K s ∆ − 14 (1.5m2 a2 x + I 2α 2 )
T = − I1α1 +
4
6
38
End of Chapter