Inertia forces in reciprocating parts

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

Cosθ=r2+x2L22rx=>L2=r2+x22rxCosθ

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

L2r2=x22rxCosθ+r2(Cos2θ+Sin2θ1) this expression cna be further written as 

L2r2Sin2θ=(xrCosθ)2 or x=rCosθ±(L2r2Sin2θ)

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

Therefore, dxdt=dxdθ.dθdt=ωrSinθωr2SinθCosθ(L2r2Sin2θ)=>ωr[Sinθ+Sin2θ2((L2r2Sin2θ)]

Generally L / r > 1 and the value of Sin2θ << L / r. Hence we can neglect Sin2θ. Applying this in the equation of velocity we get,

V=ωr[Sinθ+Sin2θ2λ]

Piston acceleration is the rate of change velocity

Therefore, a=dVdt=dVdθ.dθdt, Now dθdt=ω=angularvelocity

Therefore, acceleration,a=ω2r[Cosθ+Cos2θλ] 

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 QcosΦ. This force along the con rod will have two components, one along the crank pin radius and the other perpendicular to it, say Ft. The torque on the crankshaft is given by this force Ft multiplied by the crank radius. So,

Crank effort or Torque, T=Ft.r 

From the figure above we can find, Ft=QcosΦ.sin[90(θ+Φ)]=>QcosΦ.sin(θ+Φ)]

Also, BD=r.sinθ=L.sinΦ or sinΦ=r.sinθL

Expandin expression for Ft, we get , Ft=QcosΦ.[sinθ.cosΦ+cosθ.sinΦ]

or , Ft=Q[sinθ+cosθ.tanΦ]

We also know that sinΦ=r.sinθL, hence cosΦ=(1r2sin2θL2),

therefore tanΦ=rsinθL2r2sin2θ=sinθL2r2sin2θ

since L2r2>>Sin2θ, we assume Sin2θ=0, thus tanΦ=r.sinθL

therefore the force Ft=Q[sinθ+r.cosθ.sinθL]=Q[sinθ+sin2θ2λ], where λ=Lr

Thus the total crank effort, T=Ft.r=Q.r[sinθ+sin2θ2λ]

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.

QcosθP.A=mrec.a, Where mrec = mass of reciprocating parts, We know that acceleration of the piston,

a=ω2.r(cosθ+cos2θλ) , thus Q=P.Amrecω2r(cosθ+cos2θλ)cosθ

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

Also AE = r.sin(θ+Φ)

from the figure above,BD=rsinθ=LsinΦ or sinΦ=sinθλ,whereλ=Lr

We can also express sin(θ+Φ)cosθ=sinθ.cosΦ+cosθ.sinΦcosΦ=sinθ+cosθ.tanΦ

also cosΦ=1sin2θλ2, therefore  tanΦ=sinθλ1sin2θλ2 => tanΦ=sinθλ2sin2θ, since λ2>>sin2θ, hence we can neglect sin2θ,

hence tanΦ=sinθλ, therefore, sin(θ+Φ)cosθ=sinθ+cosθ.sinθλ=sinθ+sin2θλ

Finally the turning moment on the crankshaft is

TurningMoment,T.M=[P.Amrecω2r(cosθ+cos2θλ)].r.(sinθ+sin2θλ)

 

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 m1.a=m2b, also m1+m2=m
    Therefore m1=m.b(a+b),andm2=m.a(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, mc is the mass of the con rod. We can divide mc 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
mc = mass of con rod
L1 = distance of C . G position from C
L = Length of con rod.

Mass of conrod at C=LL1L.mc , Reciprocating mass mrec is also acting at C

Therefore the total mass acting at C = LL1L.mc+mrec 

Therefore inertial force due to equivalent mass at C = FI=[LL1L.mc+mrec].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) 

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