1.1 Piston motion equations

Let us first define the crankshaft geometry.

Let the connecting rod length BC = L
Crank Radius AB= r
Crank angle at any instance = θ
Piston pin position = X ----- (based on the direction of motion, origin is at point A).
Piston pin velocity = V
Piston pin acceleration = a
Cranks Angular velocity = ω rad/s

As per the cosine rule

$Cos\theta={r^2+x^2-L^2\over 2rx} => L^2=r^2+x^2-2rxCos\theta$

The position of the piston at any instance can be found as below

$L^2-r^2=x^2-2rxCos\theta +r^2(Cos^2\theta+Sin^2\theta-1)$ this expression cna be further written as 

$L^2-r^2Sin^2\theta = {(x-r Cos\theta)^2}$ or $x=rCos\theta \pm \sqrt(L^2-r^2Sin^2\theta)$

Piston velocity can be found by taking the first derivative of piston position with respect to crank angle.

Therefore, ${dx\over dt}={dx \over d \theta}.{d \theta \over dt}=-\omega rSin\theta - {\omega r^2 Sin\theta Cos\theta \over \sqrt (L^2-r^2Sin^2\theta)} =>-\omega r[Sin\theta+{Sin2\theta \over 2( \sqrt ({L^2 \over r^2}-Sin^2\theta)}]$

Generally L / r > 1 and the value of $Sin^2\theta$ << L / r. Hence we can neglect $Sin^2\theta$. Applying this in the equation of velocity we get,

$V=-\omega r[Sin\theta+{Sin 2\theta \over 2\lambda}]$

Piston acceleration is the rate of change velocity

Therefore, $a={dV \over dt} = {dV \over d\theta}.{d\theta \over dt}$, Now ${d\theta \over dt} = \omega = angular \quad velocity$

Therefore, $acceleration, a=-\omega^2 r[Cos\theta+{Cos 2\theta \over \lambda }]$ 

1.2 Torque on crankshaft neglecting friction and inertia of moving parts

If Q is the thrust force experienced by the piston, then Force along the con rod is $Q \over cos\Phi$. This force along the con rod will have two components, one along the crank pin radius and the other perpendicular to it, say $F_t$. The torque on the crankshaft is given by this force $F_t$ multiplied by the crank radius. So,

Crank effort or Torque, $T=F_t.r$ 

From the figure above we can find, $F_t= {Q\over cos\Phi} . sin [90-(\theta+\Phi)] =>{Q\over cos\Phi} . sin(\theta+\Phi)]$

Also, $BD=r.sin \theta =L. sin\Phi$ or $sin\Phi = {r.sin\theta \over L}$

Expandin expression for $F_t$, we get , $F_t={Q\over cos\Phi}.[sin \theta.cos\Phi +cos\theta.sin\Phi]$

or , $F_t =Q[sin\theta+cos\theta.tan\Phi]$

We also know that $sin\Phi = {r.sin\theta \over L}$, hence $cos\Phi=\sqrt{ (1-{r^2sin^2\theta \over L^2})}$,

therefore $tan\Phi={rsin\theta \over \sqrt {L^2-r^2sin^2\theta}}= {sin\theta \over \sqrt {{L^2 \over r^2}-sin^2\theta}}$

since ${L^2\over r^2}>>Sin^2\theta$, we assume $Sin^2\theta =0$, thus $tan\Phi = r.{sin\theta\over L}$

therefore the force $F_t=Q[sin\theta+r.cos\theta .{sin\theta \over L}] =Q[sin\theta+{sin2\theta\over2\lambda}]$, where $\lambda ={L\over r}$

Thus the total crank effort, $T =F_t.r =Q.r[{sin\theta +{sin2\theta \over 2\lambda}}]$

1.3 ???????Turning moment on crankshaft considering inertial forces

Let AB=r=Crank radius
BC = L = Con rod length
X = Piston motion direction
Q = Force along the con-rod
Gas force = P. A (Cylinder pressure X Piston surface area)
Since the piston is constrained to move in the X direction only.

$Qcos\theta-P.A =m_{rec}.a$, Where $m_{rec}$ = mass of reciprocating parts, We know that acceleration of the piston,

$a=-\omega^2.r(cos\theta+{cos2\theta \over \lambda})$ , thus $Q={P.A - m_{rec}\omega^2r(cos\theta +{cos2\theta\over \lambda}) \over cos\theta}$

Now since, AE is perpendicular to the direction of Q. Moment produced by the force Q, M = Q . AE

Also AE = $r.sin(\theta+\Phi)$

from the figure above,$BD=rsin\theta=Lsin\Phi$ or $sin\Phi={sin\theta \over \lambda}, where \quad \lambda = {L\over r}$

We can also express ${sin(\theta+\Phi)\over cos\theta} ={ sin\theta. cos\Phi+cos\theta.sin\Phi \over cos\Phi} =sin\theta+cos\theta.tan\Phi$

also $cos\Phi = \sqrt {1-{sin^2\theta \over \lambda^2}}$, therefore  $tan \Phi ={{sin\theta \over \lambda}\over \sqrt {1-{sin^2\theta \over \lambda^2}}}$ => $tan\Phi ={{sin\theta}\over \sqrt {\lambda^2-sin^2\theta }}$, since $\lambda^2>> sin^2\theta$, hence we can neglect $sin^2\theta$,

hence $tan\Phi ={{sin\theta} \over \lambda}$, therefore, ${sin(\theta+\Phi)\over cos\theta}=sin\theta+cos\theta.{sin\theta \over \lambda} = sin\theta +{sin2\theta \over \lambda}$

Finally the turning moment on the crankshaft is

$Turning \quad Moment, T.M =[P.A-m_{rec}\omega^2r(cos\theta+{cos2\theta \over \lambda})].r.(sin\theta+{sin2\theta \over \lambda})$

 

1.4 Dynamically equivalent system

A system is said to be dynamically equivalent to another system, when the same set of forces and moments can produce the same linear velocity, angular velocity, linear and angular acceleration in both the system systems. This is possible when the system has

1. Same mass
2. Same centre of mass
3. Same moment of inertia, as that of the first body.

   Dynamically equivalent system for Connecting rod

   

   Mass of con rod = m
   Length of Con rod = a + b = L
   Applying moment about the centre of gravity = 0, we get $m_1.a=m_2b$, also $m_1+m_2 =m$
   Therefore $m_1={m.b \over (a+b)}, \quad and \quad m_2={m.a \over(a+b)}$

1.5 Torque exerted on the crankshaft considering the weight of the connecting rod

G is the centre of gravity of the con rod, $m_c$ is the mass of the con rod. We can divide $m_c$ at point B and C such that the centre of gravity of these two masses lies at G. Point B is in pure rotation and hence the inertia forces due to mass will act radially outwards and will have no effect on the crankshaft torque.
The inertia forces at point C can be obtained as below
$m_c$ = mass of con rod
$L_1$ = distance of C . G position from C
$L$ = Length of con rod.

Mass of conrod at C=${L-L_1\over L}.m_c$ , Reciprocating mass $m_{rec}$ is also acting at C

Therefore the total mass acting at C = ${L-L_1\over L}.m_c$+$m_{rec}$ 

Therefore inertial force due to equivalent mass at C = $F_I= [{L-L_1\over L}.m_c +m_{rec}].a$, where 'a' is the piston acceleration.

FI=(mrec+LLÀL1:mc):a 

Total torque or turning moment acting on the crankshaft is

T:M=FI:AE=>T:MÀFIrSin(Ò+¶) 

Since, BD = r:SinÒ=L:Sin¶=>ÕSinÒ; Where Õ=rL

Therefore Tanφ = ÕSin¶; neglecting Sin2Ò as Õ2>>Sin2Ò

Also, AE = r:Sin(Ò+¶)=r(SinÒ+2ÕSin2ÒÛ 

Finally,
Turning moment on crankshaft,

T:M=Úmrec+LLÀL1:mc)!2r(CosÒ+ÕCos2Ò):r:(SinÒ+2ÕSin2Ò)