# MA2002 Slides-Dynamic Force Analysis-Du Hejun

```DYNAMIC FORCE ANALYSIS OF MECHANISMS
DYNAMIC FORCE ANALYSIS OF MECHANISMS
KINEMATICS
α
ω
DYNAMICS
2
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.
{∑
ix
= ma x
F iy = ma y
4
NEWTON’S SECOND LAW OF MOTION
∑M
i
= Iα
I - moment of inertia about fixed point
F1
α - angular acceleration
T2
α
F3
4-bar Animation
T3
T1
F2
Pivot
point
5
NEWTON’S SECOND LAW OF MOTION
GENERAL PLANAR MOTION
Translation of CG


∑ F j = maG
G
F2
∑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
6
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’

inertial force 
FI = −maG
8
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’
inertial moment
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
i

= maG
=0
10
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
j
=0
11
D’ALEMBERT’S PRINCIPLE
CONVENTION FOR INERTIAL FORCE &amp; MOMENT
y
G
a
α
x
G
a
x
G
ma
y
G
ma
C.G.
Motion(Kinematics)
IGα
Inertial force &amp; moment
12
D’ALEMBERT’S PRINCIPLE
SUMMARY OF THE STEPS FOR DYNAMIC FORCE ANALYSIS
1. Find accel. of CG &amp; angular accel. For each link (or
use given ones)
2. Free individual body(s)
3. Draw inertia force at CG &amp; 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
13
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
centre of mass point G.
15
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
IG =
1
m(a 2 + b 2 )
12
16
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
1
mr 2
2
1
I O = mr 2 − mh 2
2
4r
h=
3π
17
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
arbitrary point A.
18
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 (
) 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.
19
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)
20
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)
21
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
)
23
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
24
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 &times; + 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
aG
a =0
x
G
L
a =− α
4
y
G
25
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
26
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 &times; rO1B + ω 2 &times; rBC


⇒ ω 2 = −2.1k̂ (rad/s)(CW) and vC = −14.85iˆ (in/s)
vB
B
66.4
G1
1&quot;
T1
O1
1.88&quot;
G2
2&quot;
70

vC / B
3.76&quot;
30 
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).
28
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
29
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 &times;10 −3 &times; 70 = −0.543

(lb)
F
=
F
Cy
 N
30
EXAMPLES FOR DYNAMIC FORCE ANALYSIS
• Dynamic Analysis of Link 2:
From free body diagram, we have

∑MB = 0

rBC &times; ( − FCx iˆ + FCy ˆj ) − I G 2α 2 + rBG 2 &times; m 2 (aG 2 x iˆ + aG 2 y ˆj ) = 0
⇒
− rBC y &times; FCx + rBC x &times; FCy − I G 2α 2 + rBG 2 y &times; m 2 aG 2 x + rBG 2 x &times; 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
31
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 &times; ( − FBx i − FBy j ) − I G 1α 1 + rO 1G 1 &times; m1 (aG 1 x iˆ + aG 1 y ˆj ) = 0
T1 + rO 1 By &times; FBx − rO 1 Bx &times; FBy − I G 1α 1 − rO 1G 1 y &times; m1aG 1 x + rO 1G 1 x &times; 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
32
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
Acceleration
of CG
Angular
acceleration
m1
G1 at O1
I1
0
α1 (cw)
m2
G2
I2
a2xi+a2yj
α2 (cw)








G5 at O5
I5
0
α5 (ccw)
m5
y
Neglecting the gravitational force and friction forces,
1) Draw the free-body diagram of each link (except the ground
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
F
C
B
O5
G5
massless
G2
G1 O1
A
T
α1
ο
30
α2
O3
45
ο
34
Solution
Free Body Diagrams
35
Solution:
From free body diagrams, we have
∑ 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
∑ 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
∑ 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
```