Dynamics of Machines
Static Force Analysis
Instructor: Prof. S.K. Dwivedy
Text Books
 J. J Uicker (Jr), G. R Pennock, and J. E Shigley, Theory of Machines and
Mechanisms, 3rd Ed., Oxford International Student Edition, 2009.
 J S Rao and R V Dukkipati, Mechanism and Machine Theory, 2nd Ed., New Age
Intl., 2008
Reference Books
 S. S. Rattan, Theory of Machines, 3rd Ed., Tata McGraw Hill, 2009.
 T. Bevan. Theory of Machines, CBS Publishers and Distributors, 1984.
 L. Meirovitch, Elements of Vibration Analysis, McGraw Hill, 1998.
 W. T. Thomsom and M.D. Dahleh, Theory of Vibration with Applications, 5th Ed.,
Pearson Education, 1999.
Journals
 International Journal of Non-linear Mechanics (ELSEVIER)
 Nonlinear Dynamics (SPRINGER)
 Journal of Sound and Vibration (ELSEVIER)
 Journal of Vibration and Acoustics (ASME)
 Journal of Dynamical Systems, Measurements and Control (ASME)
 Physics D: Nonlinear Phenomena (ELSEVIER)
 Chaos, Solitons and Fractals (ELSEVIER)
 International Journal of Nonlinear Sciences and Numerical Simulation, (Freund
Publishing House)
 Journal of Computational and Nonlinear Dynamics (ASME)
Overview of the Dynamical System
Statics
Static Force Analysis
Mechanics
Kinematics
Dynamics
Kinetics
Dynamic Force Analysis
Theory of
Machine
Rotating
Balancing
Reciprocating
Kinematics
of Machine
Dynamics
of Machine
Governor
Single DoF
Analysis
Synthesis
Gyroscope
Two DoF
Multi DoF
Continuous System
Vibration
Static Force Analysis
 When the inertia forces are neglected in comparison to the externally applied
load, one may go for static force analysis.
 If the body is under equilibrium condition, then this equilibrium is known as
static equilibrium and this condition is also applicable in many machines where
the movement is relatively slow.
Static Force Analysis
Source: https://engineeringlearn.com/types-of-clamps-their-uses-with-pictures/
Static Equilibrium
A body is in static equilibrium if
 the vector sum of the forces acting on the body is zero ∑𝐹 = 0
 the vector sum of all the moments about any arbitrary point is zero ∑ 𝑀 = 0
Hand Clamp
Pipe Clamp
Practice Quiz
 Find out static equilibrium in terms of forces/ moments of the following
figures
Applied and Constrained forces
 When two or more bodies are connected together to form a group or system, the
pair of action and reaction forces between any two of the connecting bodies is
called constrained forces.
 These forces constrain the connected bodies to behave in a specific manner
defined by the nature of the connection.
 Forces acting on this system of bodies from outside the system are called applied
forces.
 Electric, Magnetic and gravitational forces are example of forces that may be
applied without actual physical contact.
 But most of the forces we are concerned in mechanical equipment occur through
direct physical or mechanical contact.
Applied and Constrained forces
3
External force
𝑇
2
𝐹4
4
Constraint force
1
Four bar mechanism showing
external and constraint forces
𝐹34
𝐹23
Equilibrium of a two force member
F
A
B
F
The moment of couple
𝑀 = 𝑅𝐵𝐴 × 𝐹
Equilibrium of three force member
𝐹1
𝐹2
𝐹3
=
=
si n 𝛼 si n 𝛽 si n 𝛾
Example 1: Find the bearing forces and the torque required for static
equilibrium of the four bar mechanism subjected to a force in the
fourth link
C
3
B
Solution: For planar mechanism ∑𝐹𝑥 = 0, ∑𝐹𝑦 = 0,
and ∑ 𝑀𝑧 = 0
𝑇
4
2
𝑅᪄𝐴𝐵 = 𝐴𝐵cos 𝜃2 𝑖ƶ + 𝐴𝐵sin 𝜃2 𝑗ƶ
𝑅᪄ 𝐶𝐵 = 𝐵𝐶cos 𝜃3 𝑖ƶ + 𝐵𝐶sin 𝜃3 𝑗ƶ
𝐹4
A
𝑅᪄ 𝐷𝐶 = 𝐶𝐷cos 𝜃4 𝑖ƶ + 𝐶𝐷sin 𝜃4 𝑗ƶ
C
𝑅᪄ 𝑄𝐷 = 𝐷𝑄cos 𝜃4 𝑖ƶ + 𝐷𝑄sin 𝜃4 𝑗ƶ
𝑃 = 𝑃cos 𝛼𝑖ƶ + 𝑃sin 𝑗ƶ
D
1
𝐹23
𝐹34
B
Here link 3 is a two-force member and at this stage we know only the line of action of
the forces 𝐹23 and 𝐹43 which should be along the line BC.
 Link 4 is a three-force member in which force 𝑃 is
completely known and the line of action of force 𝐹34 which
is equal and opposite to 𝐹43 is known. Only the point of
application of force 𝐹14 , which is at point D, is known. As
link 4 is a three force member, taking moment about D,
𝐹Ԧ34
𝐶
𝑃
𝐹Ԧ14
෍ 𝑀𝑍 = 0 ⇒ 𝑅𝐶𝐷 × 𝐹Ԧ34 + 𝑅𝑄𝐷 × 𝑃 = 0
 As P is completely known one may obtain 𝐹Ԧ34
 As P is completely known one may obtain 𝐹Ԧ34
𝐷
Free body diagram of Link 4



 One may note that link 3 is a two-force member, so F23  F43  F34


 Link 2 which is acted upon by two forcesi.e.,F12 and F32 , and the external applied
torque, will be in equilibrium only if i.e., F12= - F32 these forces are equal and opposite
and the resulting moment of the couple is equal to the applied torque.
 Also one may find the torque by taking moment about point A.
Graphical Method

As link 4 is a three force member, the line of action of F14 should pass through the intersection of the

line of action of P and F34 .


Taking proper scale and by drawing the force polygon one may obtain the magnitude of F34 and F14 .

Then considering equilibrium of link 3, force F23 can be determined.
Then determine the torque taking moment about A.
 When multiple forces act on a mechanism, one may use superposition
theory, which states that in a linear system, the net effect (e.g., bearing
forces or torque) due to all the forces taken simultaneously will be
equal to the summation of the effects due to individual forces taken
one at a time.
 If one wishes to find only the torque acting on the mechanism, the
method of virtual work may be used. It states the work performed
during a virtual displacement from equilibrium is equal to zero.
The virtual displacement is defined as an imaginary infinitesimal
displacement of the system that is consistent with the constraints on the
system. For example, the constraints on the slider-crank mechanism are
that all members including the frame are rigid and all joints maintain
contact
Example 2. Calculate the torque required (assuming no friction in the bearing) for
static equilibrium of an in-line reciprocating engine in the position when crank angle
 = 45 deg (from inner dead center). The dimensions are crank length r =30 mm,
connecting rod length L = 70mm, and the piston force is P = 40 N.
X
Solution
Here OB is link 1, crank OA is the 2nd link, connecting rod AB is the 3rd link and the piston is the 4th
link. Crank radius r =30 mm, Length of connecting rod =70 mm
Letting <ABO = 
r sin   L sin 
 30sin 45 
0
Hence,   sin 1 

17.64

 70 
Taking the positive X axis as shown in the figure
RAO  3045  30cos 45iˆ  30sin 45 ˆj  21.213iˆ  21.213 ˆj
RBA  70342.35  70cos342.35iˆ  70sin 342.35 ˆj  66.70iˆ  21.213 ˆj
It may be observed that link 3 is a two force member and subjected to forces F23
The free-body diagram of link 4, i.e., that of piston is shown below. For the present case, it is a threeforce member subjected to a force P due to gas pressure, vertical reaction force F14 and force of
connecting rod on piston ( F34 ) at the gudgeon pin. Force P is known completely both in magnitude
and direction and the line of action and point of application of force F34 is known. Now drawing the
force polygon as shown in Figure (b) one will be able to find the unknown forces F14 and F34 .
(a) Free body diagram of link-4
(b) Force Polygon
Vector method
As
 F  0, hence, P  F  F  0 ,
14
34
So, (0.953iˆ  0.303 ˆj) F34  Piˆ  F14 ˆj  0
Equating the ith and jth compoment of the forces one may obtain
40
 41.973N
0.953
F14  0.304 F34  12.72N
F34 
Hence F34  41.973342.35 N and F14  12.7290 N .
Using Lami’s formula from the force diagram shown in Figure (b)
F34
F
P
 14 
sin 90 sin  sin(90   )
F34 
40
 41.974 N and
sin(90  17.64)
F23
A
40sin(17.64)
F14 
 12.72 N .
sin(90  17.64)
B
Now considering free-body diagram of link 3 F23   F43
But, F43   F34  41.974342.35
T
So F23   F43  41.974342.35
F12
F43
F32
Considering equilibrium of link 2
Link 2 is subjected to forces F32 and F12 . For equilibrium these two forces must be equal and opposite.
But as they are acting at A and O respectively they will form a couple which will try to rotate the link
OA in anti-clock wise direction. Hence for static equilibrium a torque T must be applied in clockwise
direction whose magnitude should be equal to the couple formed by these forces.
Now F32  F23  41.974342.35  40iˆ  12.7265 ˆj
T  ( RAO  F32 )  (21.213iˆ  21.213 ˆj )  (40iˆ  12.73 ˆj )
= -1118.56kˆ
Negative sign indicate the applied torque should be applied in clock-wise direction.
Static Force Analysis for the system with Friction
rf  r  /( (1   )).
2
𝑟𝑓 𝑊 =
𝑊𝑟𝜇
1 + 𝜇2
= 𝑊𝑟𝜇
Example 4.
3 Calculate the torque required (assuming no friction in the bearing) for the static
equilibrium of an in-line slider crank mechanism in the position when crank angle   450
(from the inner dead center). The dimensions are, Crank length =30 cm, Connecting rod
length=70 cm and the piston force = 40N. Also find the torque required assuming that the coefficient for all bearing is 0.1. The three journal bearings all have radii of 10 mm, and the crank
is rotating in the clockwise.

  45 0

Solutions:
Given data:-  450
Crank length (link OA) =30 cm.
Connecting rod length (link AB)=70 cm.
And the piston force (P)= 40N.
From the figure, and using the sine rule.
We can write,
AB sin   OA sin 
30
 sin   sin 450  0.303
70
   17.640
Case (I):---- Without friction
F34

P
F14
Considering the link4, and using static force analysis,
F34 cos   P  40
& F34 sin   F14
Therefore,
F34  41.97 N ,& F14  12.12 N
Also,
F34   F43 (equal and opposite reaction)
y
F43  41.97 N  F43
x
Considering the link 3.
Since link 3 is a 2-force member,
Therefore,
F23   F43
 F23  41.97 N
And also, we can write, (By equal and opposite reaction)
F32   F23
 F32  41.97 N
Considering link 2.
Now the torque due to reaction force is given by,
T  r  F32
0ˆ
0 ˆ
 T  0.03(cos 45 i  sin 45 j )  41.97( cos  iˆ  sin  ˆj )
ˆ m
 T  1.112kN
Case (II):-- With friction
Considering the link 4.
Radius of friction circle, is given by rf 
r
1  2
Where, r= 10, and =0.1
0.1×10
 𝑟𝑓 =
= 0.99
2
1+(0.1)
the angle ‘’ is given the angle by which the reaction
force lift and is determined by =tan-1() =5.71
Now,
Since the rotation of the crank is clockwise direction,
thus the angle 𝛾 will be decreased and simultaneously,
angle , angle  will increased and deceased. Also the
piston (link 4)
Will move toward the X-axis (to the right). Thus the direction will be towards left & thus 
=+5.71(according to the figure)
Considering the link (3)
Since the link 3 is a 2-force member, then this link can be showed the four possible force
situations.
Same way as above, the forces F43 and F23 will act in the orientation (iii).
Similarly, we can write for the link 2.
Now we have to find  
     
         17.64  tan (
1
    160
rf
0.99
)  17.64  tan (
)  160
DB
35
1
Now considering link .4.
F34 cos    P  F14 sin 
&
F34 sin    F14 cos 
Therefore,
F34  42.84 N
F14  11.87 N
F34  F43  F23  F32  42.84 N as in case (I) considering no friction.