We devoted three tutorials to resolving by various methods the displacement, velocity and acceleration (the kinematics) of the elements in a slider and crank mechanism. We now progress to examining forces and moments in the individual elements of the mechanism using principles of static equilibrium.

In this tutorial we develop simple free body diagrams, neglecting inertia, friction and gravitational forces, for slider and crank mechanisms with loading on power and compression strokes. For preferencel we use the term "piston" (enclosed in a cylinder) rather than "slider".

A free body diagram (shows forces and moments on machine elements as the first step to calculations using static equilibrium conditions (sum of forces = 0, sum of moments = 0). In a succeeding tutorial we explain how inertia forces are included in a static force analysis using D'Alembert's principle.

The figure below shows the outward* power stroke of
a horizontal engine producing counter-clockwise rotation of the
crank arm, shown at crank angle θ = 50° . Dimensions and other parameters are identical to
examples in the kinematics tutorials with piston
force F_{P} added. The free body diagrams and
calculated values of forces and moments are, of course, different
for every crank angle over a complete cycle from 0°to 360°.

* For horizontal engines extremities of
travel of the piston are known as *inner dead centre* at
crank angle = 0° and *outer dead centre *at crank angle 180°.
The corresponding terms for a vertical engine are respectively *
top dead centre* and *bottom dead centre*. The notation
for vertical engines is generally used for internal combustion
engines regardless of their orientation.

*Drawing the diagrams*

We draw free body diagrams for the three separate elements: piston, connecting rod and crank arm. The line of action of forces is taken through the centre of the pin joints. In this case we consider the pins simply as a means of transmitting forces between elements but they could be treated as individual elements with their own diagrams. The length of an arrow has no significance as regards magnitude of the force.

Free body diagrams for each element are shown below. The description which follows explains how the diagrams are drawn.

Starting with pin B linking the piston and connecting rod, we
know the direction and magnitude of F_{P}. The only
other force acting on pin B is from the connecting rod with
direction defined by angle φ. It is convenient to consider
forces as components on x and y co-ordinates thus we designate
forces from the connecting rod acting on the piston as R_{xB}
and R_{yB}. In this instance directions of R_{xB
}and R_{xY} can be assigned intuitively (the connecting
rod exerts a horizontal force in a direction opposed to F_{P
}and a vertical force downward).

It follows that R_{yB} is accompanied by an equal and
opposite force R_{yS} which is the reaction force applied
from the cylinder wall to the piston. (We could continue by showing
action and reaction forces between the cylinder mountings and ground
but this is superfluous for our purpose). It is not always be
possible to draw the direction of a force using intuition, but as
long as the given direction is applied in accordance with the
co-ordinate axes, signs of calculated forces and moments will ultimately be correct.

Because all forces on the piston act through a common point at pin B there are no moments to consider .

Moving to the diagram for the connecting rod, forces acting on
the rod at pin B are equal and opposite to forces R_{xB} and
R_{yB} acting on the piston. At pin A there are forces
paired with the crank arm, designated R_{xA} and R_{yA.}
As there are no other forces acting on the connecting rod, the
directions of R_{xA} and R_{yA} must be
opposed to R_{xB} and R_{yB }from the equilibrium
conditions for forces in x and y directions.

The forces acting on the connecting rod are not directed through a single point hence moments must be considered, either moments of forces at pin A around pin B or vice versa, the choice is arbitrary (see below).

On the crank arm there are forces R_{xA} and R_{yA}
at pin A equal and opposite to the forces on the connecting rod.
At pin O there are forces R_{xO} and R_{yO}
transmitted through crankshaft pin O. In engineering terms pin
O is the crankshaft bearing journal. Because there are no
other forces on the crank arm, directions of R_{xO} and R_{yO
}are opposed to R_{xA} and R_{yA}.

R_{xA} and R_{yA} both generate a
counter-clockwise moment about pin O. For the crank arm to be
in static equilibrium a clockwise moment M_{O} must be
applied.

*Equilibrium equations and calculations*

It is good practice to place all terms initially on the LHS of the equations according to their sign as per the defined x and y axes. This ensures any wrongly assigned direction is highlighted when resolving the equations.

**Piston**

*For Σ horizontal forces = 0 :*

F_{P} - R_{xB} = 0 gives
**R**_{xB}**
= F**_{P}** = 1 kN** --------- (1)

*For Σ vertical forces = 0 :*

R_{yS} - R_{yB} = 0 gives
R_{yB} = R_{yS } (the value of R_{yB}
is derived from equilibrium conditions for the connecting rod)

**Connecting rod**

*For Σ horizontal forces = 0 :*

R_{xB} - R_{xA} = 0 gives
R_{xB} = R_{xA}
gives from (1) ** R**_{xA}** = 1 kN**
-------- (2)

*For Σ vertical forces = 0 :*

R_{yB} - R_{yA} gives
R_{yB} = R_{yA} -------- (3)

*For Σ moments = 0 : *

We choose moments of forces R_{xA} and R_{yA}
around pin B (counter-clockwise moments are +ve). R_{xB}
and R_{yB} have no moment as they act through B.

From the geometry of the main diagram: R_{yA}
x AB.Cosφ - R_{xA} x AB.Sinφ = 0
gives R_{xA} x AB.Sinφ = R_{yA} x
AB.Cosφ

gives R_{yA} = R_{xA }x Tanφ
gives from (2) R_{yA} = 1000 x Tan(14.79°)

gives **R**_{yA }**= 264 N** --------(4) which from
(3) gives **R**_{yB}** = R**_{yS
}**= 264 N **

*Crank arm*

*For Σ horizontal forces = 0 :*

R_{xA} - R_{xO }= 0 gives R_{xA}
= R_{xO} from (2) gives **R**_{xA}** = R**_{xO}**
= 1 kN**

*For Σ vertical forces = 0 :*

R_{yA} - R_{yO} = 0 gives
R_{yA} = R_{yO} from (4)
gives **R**_{yA}** = R**_{yO}** = 264 N
**

*For Σ moments = 0 : *

Moments about O: R_{xA} x
OA.Sinθ + R_{yA} x OA.Cosθ - M_{O} = 0

gives M_{O} = R_{xA} x
OA.Sinθ + R_{yA} x OA.Cosθ = 1000 x 1 x Sin(50°) + 264 x 1 x
Cos(50°)

gives ** M**_{O}**
= 935 Nm **

Calculated values for forces and moment are indicated on the diagrams below.

The following can be deduced from the diagrams.

- At inner dead centre and outer dead centre positions (i.e. crank angles θ = 0° and 180°) all vertical forces at the pin joints are zero and action of all horizontal forces is through crankshaft pin O. Thus there is no moment around pin O. At these crank angles an engine produces zero torque.
- For a double-acting engine where the
inward stroke is also a power stroke (crank angles from 180° to 360°) the direction of
F
_{P}acting on the piston is reversed and consequently all horizontal forces on pin joints reverse. Vertical forces at pin joints do not reverse. To visualise this imagine the connecting rod pushing down on the piston on the outward stroke and pulling the piston down on the inward stroke. On both strokes the piston exerts a downward force on the cylinder wall. - Forces transmitted along the crank arm and connecting rod can be resolved from the components of horizontal and vertical forces along the respective lines as illustrated below. Forces are considered at pin joint B for the connecting rod and pin joint A for the crank arm. Pin joints A and O respectively could equally well be used. In this instance forces have been drawn to scale.

The figure below shows the configuration for the inward compression stroke which could apply to a reciprocating gas compressor, the compression stroke of an internal combustion engine or a positive displacement pump. The crank angle in this example is 310°. This angle is the mirror image of the crank angle of the engine considered above on its outward stroke.

The free body diagrams are shown below. Intuitively we visualise the connecting rod pushing upwards on the piston with the driving torque applied from the crankshaft, resulting in an upward force on the cylinder wall. Note that the "balancing" torque at crankshaft pin O required for static equilibrium is counter-clockwise.

Reversing the direction of rotation* of the crankshaft does not change the direction of forces in the two sets of diagrams above. The power and compression strokes would now occur when the crank arm is respectively below and above the horizontal centre line between the piston and the crankshaft pin.

* In practice the direction of rotation of internal combustion engines is not reversible being determined by the fixed arrangement of fuel intake and exhaust valves. Steam engines can be reversed by adjusting the timing of steam inlet and exhaust valves by independent movement of the valve gear.

In the next tutorial on the kinetics of crank mechanisms we examine the effects of inertia forces and moments and the calculation of crankshaft torque.

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