This tutorial concerning slider and crank mechanisms covers two aspects directly related to the operation of actual machines.

We look firstly at output characteristics of a crankshaft and flywheel under torque load conditions using crank effort diagrams. We then consider the balancing of inertial forces and moments generated by the mechanism.

Consider the crank mechanism as an engine
operating at constant angular velocity ω against a constant torque load. We use the mechanism
with parameters
from previous tutorials shown below but with the driving force
increased by a factor of 10 to the more practical value of 10 kN. The horizontal driving
force F_{P} at the slider is constant and "double acting",
i.e. the direction of the force reverses at θ = 180°.

The plot below (called ** a crank effort diagram**) shows
the fluctuating torque generated by the engine at the crankshaft
(called the

Hence the area under the engine torque plot is the work done in one crank cycle which we calculate by numerical integration to be 39900 Nm. Thus at constant angular velocity ω = 2π rad/s the power output = 39.9 kW.

We represent the same amount
of work being done by a constant torque T_{av}
called the** mean engine torque** The
calculated value = 6350 Nm is shown on the plot.

T_{av} is also the constant load torque on the
engine running at nominally constant angular velocity ω. Given the
highly fluctuating
engine torque how can constant (or near
constant) angular velocity be maintained? The answer is by the
rotational inertia of a flywheel and other rotating masses with
rotational inertia in the
drive chain.

Referring to the crank effort diagram above, where engine torque is greater than mean engine torque the engine provides energy to the flywheel which increases ω. Conversely, where engine torque is less than mean engine torque the flywheel provides energy to the engine which causes ω.to reduce Thus ω fluctuates between upper and lower values over one cycle.

We now estimate the variation in ω occurring with flywheel having mass moment of inertia I_{FY}.

Let the highest speed = ω_{max} and lowest speed = ω_{min}

Consider the diagram below. From the crank effort diagram we identify which segment of the
cycle has the maximum fluctuation in energy E_{FL} relative
to the mean engine torque. From numerical integration we
establish that segment B has the largest fluctuation E_{FL}
= 5910 Nm . Thus ω_{max} occurs at the
transition point between segments A and B and ω_{min} at the
transition between segments B and C.

For this example a spoked annular ring flywheel shown above which
is twice the diameter of the crank arm and has mass moment of
inertia I_{FY} = 6000 kgm^{2} is attached to the
crankshaft. Using the above expression gives the following
result for E_{FL}
= 5910 Nm

(ω_{max} = ω_{min}) = 0.157 rad/s = 1.50
revs/min (equivalent to ω = ± 0.75 revs/min)

Note the following terms.

In example above the coefficient of fluctuation of speed = 0.025 or 2.5% and the coefficient of fluctuation of energy = 0.148 or 14.8%. These are typical values for double acting, single cylinder steam engines.

However a high degree of control over output speed comes at a cost as the flywheel has a large mass (c.1500 kg) and the high mass moment of inertia makes the engine difficult to start.

To achieve the same output with a smaller flywheel we now
consider three identical engines with the crank arms 120° apart on the same
crankshaft and the driving force F_{P} at each engine = 3.33
kN. The crank effort diagram below shows the effect.

The mean engine torque is unchanged but the maximum fluctuation
of energy E_{FL} by numerical integration = 612 Nm giving
a coefficient of fluctuation of energy = 1.5% as opposed to 14.8% for
a
single crank.

The same coefficient of fluctuation of speed = 2.5% is achieved with a flywheel of identical annular cross section but approximately half the diameter and mass, the consequence of a ten fold decrease in the mass moment of inertia. Note that it is possible to start the engine at almost full load torque.

The example above is relatively simple based on an idealised "engine". Crank effort diagrams for multi-cylinder, two and four stroke internal combustion engines are significantly more complex.

In a previous tutorial we examined inertia forces and moments
generated in a slider and crank mechanism and estimated their effect on crankshaft
torque and reaction forces at bearings. Fluctuating forces
associated with inertial effects can generate vibrations
which are transmitted to the machine supports.
Measures to neutralise these effects are known as *
balancing.*

Firstly, a brief summary of the basic principles of balancing.

In diagram (a) above masses m_{i} are
attached by massless rods of radius r_{i} to a shaft rotating with angular speed ω
. The centripetal acceleration r_{i}ω^{2} of
each mass is associated with a tension force along each rod = m_{i}r_{i}ω^{2}
and corresponding reaction forces on the rotating masses and shaft
bearings*. In this case the sum of these force
vectors is zero thus there is no resultant reaction force at the
shaft bearings arising from rotation of the masses. In this
state we say that the rotating shaft is in ** static balance**. The
general requirement for static balance of rotating masses is Σm

This condition can also be expressed by saying that the vector sum of the centrifugal inertial forces* for the masses is zero.

* The tension force along the rod accounts for the centripetal acceleration of the rotating mass directed towards the centre of rotation The diagrams show the reaction forces of this tension acting on the rotating masses. Centrifugal inertial forces acting on the masses, directed away from the centre of rotation, are "pseudo forces" which allow analysis of accelerating masses using the condition of static equilibrium. ( D'Alembert's principle - see previous tutorial).

In diagram (b) an additional mass m_{4} is attached to the
rotor. Now the vector sum of tension forces in the rods is not zero.
The
rotor is in a state of static unbalance
and a net force arises on the rotor shaft bearings. In
this example the unbalanced force = F4. A
balancing mass m_{B} can be
added to restore static balance such that m_{B}r_{B}ω^{2}
= -m_{4}r_{4}ω^{2 } .

The diagram below illustrates another requirement
known as ** dynamic balancing**.

In the diagram above the rotor with attached masses of equal mass
m at radius r meets the condition for static
balance as Σm_{i}**r**_{i} = 0 for all
angles of rotation.

However, the lower diagrams show a state of *dynamic unbalance*
where centrifugal inertial forces mω^{2}r exert a moment on
the shaft with resultant reaction forces on the bearings. The
directions of the bearing reaction forces
rotate with the shaft; directions of forces at rotation angles of 0°
and 180° are shown on the diagram. The location of
the rotating masses along the axis of the shaft determines the magnitude of the bearing forces,
the further apart the rotational planes the larger the forces.
In the limiting case when the rotating planes converge, reaction
forces are zero.

Dynamic balancing involves placing additional mass on the rotor at positions such that the net moment on the shaft is zero. It is particularly important in the design of multi-cylinder engines. In this tutorial we examine only static unbalance generated in a single crank arrangement.

In a previous tutorial we identified inertial forces generated by the three moving elements of a slider and crank mechanism:

- Rotational inertial force generated by the crank arm
- Reciprocating (translational) inertial force generated by the slider
- Inertial force and moment generated by the connecting rod

We examine balancing these inertial forces and moments from each element in turn using the example mechanism below.

Individual elements are as follows. 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 Length of connecting rod R = 3 m - Mass of crank arm m
_{CA}= 2 kg Length of crank arm L = 1 m

Referring to the diagram below the inertial centrifugal force F_{Ica}
experienced by the
crank arm rotating at constant angular velocity ω produces a
rotating
unbalanced force acting through point O. In this example the C
of G of the crank arm is at position R/2. Thus F_{ICA}
= m_{CA}ω^{2(}R/2) = 2 x (2π^{2}) x 0.5 = 19.7 N

We can balance this force by placing a rotating counterbalance
mass m_{B} at radius R_{B} diametrically opposed to
the crank arm such that m_{B} x R_{B} = m_{CA}
x (R/2) giving for this example **m**_{B}**
= 1/R**_{B }kg

The acceleration a_{B} of
the reciprocating slider of mass m_{S} produced by applied
force F_{P} generates an inertial force F_{IS} such
that F_{IS} = - (m_{S} x a_{B})

In a previous tutorial we derived the following expression for
the acceleration of the slider, a_{B}

In this previous tutorial we derived the above expression for a_{B}
by successive differentiation of the linear displacement of point B as a
function of crank angle θ. When considering balancing it is convenient
to express the second term in brackets as follows*.

* This approximation is obtained by applying the binomial theorem to the expression for displacement of point B.

We now consider the inertial force generated by the slider as two separate parts:

*Primary inertia force*

The diagram above shows the primary inertia force m_{S}ω^{2}Rcosθ generated by
the acceleration of mass m_{S} at crank angle θ. This force
varies in magnitude and direction through
the complete crank cycle as a function of cosθ. In this
example the maximum value of this primary force component (at θ = 0°
and 180° ) = 10 x (2π)2 x 1 = 395 N

This force can be seen as equivalent to the horizontal component
of the centrifugal force generated by an imaginary mass m_{S}
rotating about point O at radius R which can be balanced by the horizontal component of the
centrifugal force generated by a diametrically opposed counterbalance mass m_{B}
at radius R_{B }such that m_{B} x R_{B} = m_{S}
x R.

However, the vertical component of the centrifugal force
generated by the counterbalance mass m_{B} creates an
unbalanced force = m_{B}ω^{2}R_{B}sinθ.
In practical terms it is a question whether suppression of an
oscillating force in in the horizontal plane outweighs the adverse
effect of creating an oscillating force in the vertical plane.

*Secondary inertia force*

which is a function of cos2θ and can be
interpreted as the horizontal component of the centrifugal force
generated by slider mass m_{S} rotating on a secondary
imaginary crank arm with angular velocity 2ω such that

Thus the radius of this secondary crank = R/4n. This
force can be balanced by counterbalance mass m_{B} on a
diametrically opposed secondary crank arm at radius R_{B}
such that m_{S} x (R/4n) = m_{B} x R_{B }as
shown in the diagram below. In common with the counterbalance
for the primary inertia force this method of balancing introduces an
unbalanced vertical force = m_{B}(2ω)^{2}R_{B}sinθ.

The difficulty of constructing a practical arrangement for a secondary crank is such that balancing the secondary inertia force is rarely carried out.

*Connecting rod*

In a previous tutorial we examined the inertial force and torque (or moment) generated by the connecting rod shown in the diagram below for a specific crank angle θ.

We now convert this inertia force and moment io an ** equivalent dynamic
system** representing the total mass of the connecting rod

The general criteria for an equivalent dynamic system with reference to the parameters of right hand diagram above are:

- m = m
_{1}+ m_{2} - m
_{1}x a = m_{2}x b - m
_{1}x a^{2}+ m_{2}x b^{2}= m x.k^{2}where k is the radius of gyration of the component rotating about an axis through its centre of gravity

For the case above where m_{1} = m_{2} = m/2 and a = b = L/2 it is clear that
*No.1* and *No.2* are satisfied.

For *No.3* to be satisfied:

m_{1} x L^{2}/4 + m_{2} x L^{2}/4
= L^{2}/4(m_{1} + m_{2}) = m (L^{2}/4)

For a slender rod rotating on an axis through its C of G,
k = L//√12 giving m x k^{2} = m(L^{2}/12)

Clearly *No.3* is not satisfied*^{1} with this
arrangement of masses*^{2}. However, partial
dynamic equivalence does provide an approximate measure of the
unbalance generated by the inertia of the coupling rod.

*^{1} Complete dynamic
equivalence can be obtained by
applying a "correction couple" = m(a.b = k^{2}).α where α is
the angular acceleration of the connecting rod. Since α varies
continuously the correction has limited value for the purpose of
determining any counterbalancing mass.

*^{2} An equivalent dynamic
system satisfying all three conditions with one mass located at
point A or point B and the other mass located elsewhere on the rod
axis does exist but this arrangement of masses does not suit our
purpose.

Inertia forces generated by masses m_{1} and m_{2}
of the equivalent dynamic system and the
corresponding counterbalancing masses are shown in the diagrams
below. The counterbalance mass for m_{2} can be added
to the counterbalance mass for the rotational inertia force generated by
the crank arm. The counterbalance mass for m_{1} can
be added to the counterbalance mass for the reciprocating primary inertia force
generated by slider mass m_{s}.

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