In a previous tutorial we examined action and reaction forces generated in the elements of a crank mechanism by using simple free body diagrams for an engine and a compressor. In this tutorial we expand free body diagrams to include inertia forces and inertia torques generated in the individual elements (slider, connecting rod and crank arm). We also consider how to calculate the torque produced at the crankshaft of an engine delivering power through a crank mechanism.

The following free body diagrams illustrate the concepts of inertia forces and torques.

(note: the terms *inertia torque* and
*inertia moment *are synonymous)

In diagram (i) * F*_{P} is
the force which must be applied to accelerate mass * m* at

By contrast, in diagram
(ii) force *F*_{P}
must be applied to decelerate mass ** m** against
the same constant load

note: arrows** v **and
**a** in the above diagrams illustrate directions of
velocity and acceleration vectors.

Diagrams (iii) and (iv) show the inertia torques *T*_{I}
when torque
*T*_{P} is applied a rotating mass
with mass moment of inertia
*I*_{G} and clockwise rotation against a
constant load torque *T*_{L}
with angular acceleration
** α ** radians/sec

We use of inertia forces and moments to take account of forces in the crank mechanism associated with acceleration of masses.

The basis of this analysis is that inertia forces and moments act through or about the centre of gravity of the moving machine element. Thus we begin by determining the mass and centre of gravity of each element which for this example are shown on the diagram below.

For simplicity we assume the crank arm and connecting rod have uniform cross section along their length, thus the centres of gravity are at the mid-points.

Firstly we will identify the inertia forces and moments generated by acceleration of the individual elements and then illustrate with a numerical example.

In this tutorial we restrict our analysis to a crank arm
rotating with constant angular velocity ** ω**,
thus the crank arm has no angular acceleration and correspondingly
no inertia torque. We take account of the centripetal acceleration of point A
directed radially towards centre of rotation O when considering connecting rod
AB.

We established in an earlier tutorial that the slider experiences
linear acceleration which varies constantly throughout the crank cycle.
There are four distinct phases in the cycle when the absolute
slider velocity is increasing or decreasing. In the diagram
below we show the direction of inertia forces *F*_{Is}
for each phase of the crank cycle. For a horizontal slider the
first phase starts at the inner dead centre position (crank angle
** ω **
= 0°). The transition between second and third phases occurs
at outer dead centre (crank angle

In all cases *F*_{Is}
= *m*_{s} x** a**_{B}

To determine the inertia forces generated by the mass of the
connecting rod we require: (i) the linear acceleration of the centre
of gravity (point g_{AB}) relative to point O which we
designate *a*_{gAB},
(ii) the inertia moment of AB about the C of G which we designate
*T*_{IAB}
=* I*_{g}*.α*_{AB}
where *I*_{g} is the mass moment of
inertia about the C of G and *α*_{AB}
is the angular acceleration of AB.

To obtain *a*_{gAB}
we use the acceleration diagram developed in
a previous tutorial shown below.
Note this diagram is specifically for crank angle ω = 50° .

To recap, accelerations shown on this diagram are:

*a*_{rA/O}the acceleration of point A relative to point O in a radial direction along crank arm AO. Because angular velocity ω of the crank arm is constant in this analysis there is no tangential component of acceleration of A relative to O.*a*_{rB/A}the acceleration of point B relative to point A in a radial direction along connecting rod AB.*a*_{tB/A}the acceleration of point B relative to point A tangential to connecting rod AB.*a*_{B/O}the acceleration of point B relative to point O, which is the linear acceleration of the slider mass considered above.

To develop this acceleration diagram to find *a*_{gAB} (see below) firstly show the acceleration of point B relative to point A, *a*_{B/A}
which is the sum of radial and tangential components. In our
example the acceleration of point g_{AB}
is at the mid-point of this line thus line Og_{AB}
represents *a*** _{gAB}.
**This acceleration is the absolute value
relative to ground point O and is opposed by inertia force

To find the inertia moment about point g_{AB} viz. *
T*

*α*_{AB}
= AB x *a*** _{tB/A} **
is obtained from the acceleration diagram (see the numerical
example below).

We illustrate accelerations, inertia force and inertia torque acting on the connecting rod in the diagrams below. Note again that this diagram illustrates the direction of acceleration for crank angle θ = 50° .

The inertia force and torque result in reaction forces at pin joints A and B. The object of the following numerical example is to find these reaction forces.

The physical parameters of the crank mechanism in the example,
shown below, are the same as the examples in the kinematics
tutorials. In this idealised arrangement the load torque on
the crank arm varies such that the angular velocity ω remains
constant through a complete cycle while maintaining a constant force
**F**_{P} on the slider. We also ignore
the effects of friction.

Individual elements have the following mass. The C of G is located at the mid-point of each element.

- Mass of slider (m
_{S}) = 10 kg - Mass of connecting rod (m
_{CR}) = 5 kg - Mass of crank arm (m
_{CA}) = 2 kg

Force *F*_{P} = 1.0 kN is
applied to the slider in the direction shown (for crank angles >
180° the direction of *F*_{P}
would reverse)

A free body diagram of connecting rod AB is shown below for crank angle = 50°. We use sum of forces in x direction = 0, sum of forces in y direction = 0 and moments about C of G =0 to find the unknown reaction forces.

*R*_{xB}
is the horizontal reaction force exerted on B by the slider and acts
in a +ve direction on the x axis. We reasonably deduce that the
horizontal reaction force at point A, *R*_{xA}
, acts in the -ve direction. We also reasonably deduce that
the vertical reaction forces *R*_{yB}
at point B and *R*_{yA}
at point A will be +ve and -ve respectively on account of the
direction of thrust along the connecting rod.

*M*_{Ig}
is the inertia moment about the C of G.

As a first step we obtain the value of *R*_{xB} from the free body diagram for the slider (below) where *R*_{xB}
is the equal and opposite reaction force on the slider.

The inertia force on slider mass m_{s}, *F*_{Is} =
m_{s} x a_{B} where a_{B} is the linear
acceleration of point B. From a previous tutorial
we know for this mechanism and crank angle θ = 50° that a_{B} = 23.4 m/s^{2}.
Note in passing that reaction force *R*_{yB}
on the slider is opposed by reaction force *R*_{ys}
from the cylinder wall. This vertical reaction force also
opposes gravity force m_{s}.g.

For sum of horizontal forces on the slider = 0:

F_{P} - F_{Is} - R_{xB} = 0
Hence R_{xB} = F_{P} - F_{Is}

Inserting known values gives: *
R*_{xB} = 1000 - (10 x 23.4) N =
7**66 N**

We now calculate inertia force *F*_{IgAB}
acting on the connecting rod through its C of G. From the
acceleration diagram below derived in a previous
tutorial for this mechanism we obtain by geometry the value of
*a*_{gAB} and its
direction relative to the horizontal axis.

Referring to the free body diagram for the connecting rod above the inertia forces through the Cof G are:

*F*_{xI} = m_{cr} x a_{gAB} x cos(31.79°) =
5 x 28.68 x cos(31.79°) =** 121.9 N**

*F*_{yI} = m_{cr} x a_{gAB} x sin(31.79°) =
5 x 28.68 x sin(31.79°) = **75.5 N**

For sum of horizontal forces on connecting rod = 0:

R_{xB} - F_{xI} - R_{xA} = 0
gives: R_{xA} = R_{xB} - F_{xI}
gives: *R*_{xA}
= 766 - 121.9 = **644.1 N**

For sum of vertical forces on connecting rod = 0:

F_{yI} - R_{yA} - m_{cr}.g + R_{yB}
= 0 gives: R_{yA}
- R_{yB} = F_{yI} - m_{cr}.g
gives: R_{yA} - R_{yB} = 75.5 - (5 x 9.81)
gives: **R**_{yA} -*
R*_{yB} = **
26.5 N **

For sum of moments about Cof G of connecting rod = 0 (counterclockwise moments +ve):

R_{yA} x 1.5 x cos(14.79°) - R_{xA} x 1.5 x
sin(14.79°) + R_{yB} x 1.5 x cos(14.79°) - R_{xB} x
1.5 x sin(14.79°) - M_{Ig} = 0

M_{Ig} = Ig x α_{AB} and from a
previous tutorial
for this mechanism we know α_{AB} = 9.91 rad/sec^{2}

Substitute R_{yA} = 26.5 + R_{yB}

gives: (26.5 + R_{yB} ) x 1.5 x cos(14.79°) - 644.1
x 1.5 x sin(14.79°) + R_{yB} x 1.5 x cos(14.79°)
- 766 x 1.5 x sin(14.79°) - (3.8 x 9.91) = 0

gives: 38.4 + 1.45 x R_{yB} - 246.6 + 1.45 x R_{yB}
- 293.3 - 37.7 = 0

gives: 2.9 x R_{yB} = 593.2
gives: *R*_{yB}
= **185.9 N ** and
* R*_{yA}
= 185.9 + 26..5 = **212.4 N**

From the free body diagram for the slider:

vertical reaction force R_{ys} = R_{yB} + m_{s}.g
gives: *R*_{ys}
= 185.9 + (10 x 9.81) = **284 N**

The completed free body diagram for the connecting rod is shown below. Values are rounded to whole numbers.

It should be emphasised again that inertia moments and forces vary continuously throughout the cycle of the crank mechanism. This example is a snapshot at crank angle = 50° where inertia forces oppose the driving force and consequently reduce the torque available at the crankshaft. At other points in the cycle inertia forces aid the driving force and increase the crankshaft torque. In the next section we look at crankshaft torque in more detail.

It is important to know the torque *T*_{ca} produced by an
engine at the crankshaft. We can obtain crankshaft torque
directly from the
reaction forces on the crank arm OA at crank pin A which are equal
and opposite to the reaction forces *R*_{xA}
and *R*_{yA}
on the connecting rod derived in the previous example. The free body diagram for
the crank arm is shown below. We have ignored the gravity force ( = 10
N) which is insignificant in this instance.

Crank arm AO = 1.0 m

Hence crankshaft torque *T*_{ca}
= R_{xA} x 1.0 x sin(50°) + R_{yA} x 1.0 x
cos(50°) = 644 x sin(50°) + 212 x cos(50°) =** 626 Nm**

Note: the complete free body diagram
representing static equilibrium for the crank arm would include a
clockwise moment about point O equal in magnitude to *T*_{ca}.

Consider the diagram below derived in a previous tutorial showing
velocities relative to ground point O of slider B (*v*_{B})
and crank pin A (*v*_{A})
and construction of the velocity pole for connecting rod AB.

We add to this diagram the external horizontal force *
F*_{P} applied to the slider and the resultant
force *F*_{caT}
at crank pin A perpendicular to crank arm OA which delivers torque
*T*_{ca}
to the crankshaft such that *T*_{ca}
= *F*_{caT}
x OA

Using the principle that the rate of work done at point B must equal the rate of work done at point A it follows that:

From the diagram above, it follows from velocity pole P
that: *v*_{B} = ** PB**
x

Now on the velocity pole diagram extend line BA and take a perpendicular line from point O to intersect extended line BA at point M forming triangle AOM. Triangles APB and AOM are similar such that:

In this example OM is calculated from the extended velocity pole diagram above by simple geometry giving the value OM = 0.94 m.

We defined *F*_{P} as the
external force on the slider which in the previous example = 1 kN,
giving the crankshaft torque by this method *T*_{ca} =
1000 x 0.94 = **940 Nm **compared to **626 Nm**
from the previous calculation.

Thus this second method gives a first approximation of crankshaft
torque **excluding** inertia forces and torques for
the slider and connecting rod and gravity forces.

A better approximation of crankshaft torque takes account of the inertia force generated by the slider mass.
From the previous example the net horizontal reaction force at B taking account of
this inertia force, *R*_{xB}
= 766 N, giving *T*_{ca}
= 766 x 0.94 = **720 Nm.** The remaining
discrepancy is accounted for by the inertia force and inertia torque
generated by the connecting rod and the gravity force of the rod.

For convenience we used a crank angle = 50° for the above examples but crankshaft torque varies continuously throughout a full crank cycle. In previous tutorials we developed general expressions for velocities and accelerations of elements in the mechanism. For crankshaft torque there is a very convenient general expression derived from the extended acceleration diagram (see above) where we noted that:

The diagram below plots crankshaft torque *T*_{ca}
against crank angle using the above formula for the parameters applied in the
previous examples. The diagram shows a full cycle for crank
angle θ from 0° to 360° with the direction of the external
force **F**_{p} = 1 kN on the
slider reversed at θ = 180°.

The effect of the inertia force generated by acceleration of the slider is illustrated (the inertia force and torque generated by acceleration of the connecting rod is not included). This clearly shows where the slider inertia force has a positive or negative effect on crankshaft torque.

In a real engine the torque delivered to the load is smoothed by the rotational energy of a flywheel which cyclically transfers rotational energy to and from the crank mechanism resulting in a relatively smooth "mean torque " in continuous operation. We look at this aspect in another tutorial.

I welcome feedback at: