Dynamic force analysis

Unit-III
Dynamic Force Analysis
Dr. Ambuj Saxena
Associate Professor
Department of Mechanical Engineering
G. L. Bajaj Institute of Technology and Management
Plot No. 2, Knowledge Park III,
Greater Noida, Distt. G.B.Nagar (U.P.) India -201306
Theory of Machines
(RME-602)

D-Alembert’s Principle
 Consider a rigid body acted upon by a system of forces.
 The system may be reduced to a single resultant force acting on the body whose
magnitude is given by the product of the mass of the body and the linear
acceleration of the centre of mass of the body.
 The rate of change of momentum is directly proportional to resultant forces.
 According to Newton’s second law of motion, The resultant force acting in the
direction of motion is equal to the product of mass and the acceleration in the
direction of resultant force.
F = m.a ........................................(i)
F = Resultant force acting on the body,
m = Mass of the body, and
a = Linear acceleration of the centre of mass of the body.
Reference : R.S. Khurmi (Theory of Machines)

D-Alembert’s Principle
F – m.a = 0 ………………....(ii)
 A little consideration will show, that if the quantity – m.a be treated as a force, equal, opposite and with the
same line of action as the resultant force F, and include this force with the system of forces of which F is the
resultant, then the complete system of forces will be in equilibrium.
 This principle is known as D-Alembert’s principle. The equal and opposite force – m.a is known as reversed
effective force or the inertia force (briefly written as FI). The equation (ii) may be written as
F + FI = 0………………………….(iii)
Thus, D-Alembert’s principle states that “the resultant force acting on a body together with the
reversed effective force (or inertia force), are in equilibrium.”
D-Alembert’s principle states that a moving body can be brought in equilibrium by adding
inertia forces to the system. In magnitude this inertial force is equal to the product of mass
and acceleration and takes place in a direction opposite to that of acceleration.
This principle is used to reduce a dynamic problem into an equivalent static problem.

Velocity and Acceleration of the Reciprocating Parts in Engines
• Klien’s construction
(a) Klien’s acceleration diagram. (b) Velocity diagram. (c) Acceleration diagram.
Klien’s construction.

Velocity and Acceleration of the Reciprocating Parts in Engines
Klien’s velocity diagram
 First of all, draw OM perpendicular to OP; such that it
intersects the line PC produced at M. The triangle OCM is
known as Klien’s velocity diagram.
 In this triangle OCM, OM may be regarded as a line
perpendicular to PO,CM may be regarded as a line parallel
to PC, and ...(It is the same line.)
 CO may be regarded as a line parallel to CO.
 The velocity diagram for given configuration is a triangle
ocp as shown in Fig.
 If this triangle is revolved through 90°, it will be a triangle
oc1 p1, in which oc1 represents vCO (i.e. velocity of C with
respect to O or velocity of crank pin C) and is parallel to
OC,
 op1 represents vPO (i.e. velocity of P with respect to O or
velocity of cross-head or piston P) and is perpendicular to
OP, andc1p1 represents vPC (i.e. velocity of P with respect to
C) and is parallel to CP. A little consideration will show,
that the triangles oc1p1 and OCM are similar. Therefore,
The velocities of various points may be obtained without
drawing a separate velocity diagram.

Klien’s acceleration diagram
• Draw a circle with C as centre and CM as radius.
• Draw another circle with, PC as diameter. this circle inter-sect the previous circle at
K and L.
• Join KL and produce it to intersect PO at N. Let KL intersect PC at Q. This forms the
quadrilateral CQNO, which is known as Klien’s acceleration diagram.
 c'x represents ar
PC (i.e. radial component of the acceleration of crosshead or piston P
with respect to crank pin C) and is parallel to CP or CQ;
 o'c' represents ar
CO (i.e. radial component of the acceleration of crank pin
with respect to O) and is parallel to CO;
 xp' represents at
PC (i.e. tangential component of the acceleration of P with respect to C)
and is parallel to QN (because QN is perpendicular to CQ); and
 o'p' represents aPO (i.e. acceleration of P with respect to O or the acceleration of
piston P) and is parallel to PO or NO.
 A little consideration will show that the quadrilateral o'c'x p‘ is similar to quadrilateral
CQNO Therefore,

The acceleration of piston P with respect to crank pin C (i.e. aPC) may be
obtained from:
To find the velocity of any point D on the connecting rod PC, divide CM at D1 in the same
ratio as D divides CP. In other words,
To find the acceleration of any point D on the connecting rod PC, draw a line from a point D
parallel to PO which intersects CN at D2 .
If the crank position is such that the point N lies on the right of O instead of to the left as
shown in Fig. then the acceleration of the piston is negative. In other words, the piston is
under going retardation.

The acceleration of the piston P is zero and its velocity is maximum, when N coincides with O. There is no
simple graphical method of finding the corresponding crank position, but it can be shown that for N
and O to coincide, the angle between the crank and the connecting rod must be slightly less than 90°. For
most practical purposes, it is assumed that the acceleration of piston P is zero, when the crank OC and
connecting rod PC are at right angles to each other.

Dynamic force analysis

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Dynamic force analysis