Geotechnical Engineering

DHARM

848 GEOTECHNICAL ENGINEERING

inertial force. Generally speaking, the contribution to the total inertial force from the second-
ary components (due to second and higher harmonics) is considered negligible compared with
that from the primary component.

The inertial force due to rotating masses may be eliminated by what is known as “coun-
ter-balancing”; however, that due to reciprocating mass cannot be avoided.

The above analysis is applicable only for reciprocating machines with a single cylinder.
But most machines have more than one cylinder-that is to say, most machines are multi-
cylinders engines, all cylinders being usually housed in single plane.

The analysis can be extended to a multi-cylinder engine and the unbalanced inertial
forces may be derived as follows:

Pz = (Mrec + Mrot)Rω^2 Σ

i

n

= 1

cos (ωt + αi) ...(Eq. 20.82)

(neglecting secondary inertial forces)

Px = MrotRω^2 Σ

i

n

= 1

sin (ωt + αi) ...(Eq. 20.83)

where αi = the angle between the crank of the i-th cylinder and that of the first cylinder (this
is known as the “crank angle” of the i-th cylinder)

and n = number of cylinders.

(Note:- Moments of these inertial forces about the relevant centroidal axis of the ma-
chine may be determined if the exact relative locations of the engines, and hence the lever
arms, are known).

For a vertical two-cylinder engine, for example, the resultant unbalanced inertial forces
for different crank angles may be obtained as follows:

Crank Angle π/2 (or phase difference is π/2)

This is the most common case.

Pz 1 = (Mrec + Mrot)Rω^2 cos ωt (approx.)

Pz 2 = (Mrec + Mrot)Rω^2 cosFωt+π

HG

I

2 KJ

∴ Pz = PPzz 12 + = (Mrec + Mrot)Rω^2 cosωωcos

tt++F π

HG

I

KJ

L

N

M

O

Q

2 P

or Pz = (Mrec + Mrot)Rω^2 cosω

F t+π

HG

I

4 KJ ...(Eq. 20.84)

Similarly, Px 1 = MrotRω^2 sinωt

Px 2 = MrotRω^2 sinω

F t+π

HG

I

2 KJ

∴ Px = PPxx 12 + = MrotRω^2 sinωωsin

tt++F π

HG

I

KJ

L

N

M

O

Q

2 P

or Px =^2
4

MR^2 t

rot ωω

sinF +π

HG

I

KJ ...(Eq. 20.85)