Engine Force Analysis

(a) Conditions for a dynamically equivalent system

A connecting rod of mass $m$ can be replaced by two point masses (one at the small end, one at the big end) if:

1. The sum of the two masses equals the mass of the rod: $m_1 + m_2 = m$.
2. The centre of gravity of the two masses coincides with the centre of gravity of the rod: $m_1 l_1 = m_2 l_2$.
3. The moment of inertia of the two masses about the centre of gravity equals that of the rod: $m_1 l_1^2 + m_2 l_2^2 = m k_G^2$ (so $l_1 l_2 = k_G^2$).

Here $l_1, l_2$ are the distances of the two masses from the centre of gravity and $k_G$ is the radius of gyration of the rod about its centre of gravity. If the third condition is not met, a correcting couple is needed.

(b) Piston velocity and acceleration

        P            x measured from IDC
   O----+----------+------> x
   crank O, crank pin A, piston P
   OA = r, AP = l, angle AOP = theta,
   angle OPA = phi

Let the piston move a distance $x$ from the inner dead centre (IDC). From the geometry,

$$x = r(1-\cos\theta) + l(1-\cos\phi)$$

Also $l\sin\phi = r\sin\theta$, so $\sin\phi = \dfrac{\sin\theta}{n}$ and $\cos\phi = \dfrac{\sqrt{n^2-\sin^2\theta}}{n}$. Hence

$$x = r\left[(1-\cos\theta) + n - \sqrt{n^2-\sin^2\theta}\right]$$

Velocity. With $\theta = \omega t$ and $\omega$ constant, $v = \dfrac{dx}{dt} = \omega\dfrac{dx}{d\theta}$:

$$v = r\omega\left[\sin\theta + \frac{\sin 2\theta}{2\sqrt{n^2-\sin^2\theta}}\right]$$

For $n \ge 4$, $\sin^2\theta$ is small compared with $n^2$, so $\sqrt{n^2-\sin^2\theta}\approx n$:

$$v \approx r\omega\left[\sin\theta + \frac{\sin 2\theta}{2n}\right]$$

Acceleration. Differentiating once more, $a = \omega\dfrac{dv}{d\theta}$:

$$a \approx r\omega^2\left[\cos\theta + \frac{\cos 2\theta}{n}\right]$$

Special positions

The velocity is zero at both dead centres and the acceleration is greatest at IDC. The inertia force of the reciprocating parts is $F_I = m\,a$ acting opposite to the acceleration.

Data: $r = 0.15$ m, $l = 0.6$ m, $n = l/r = 4$, $N = 360$ rpm, $m = 120$ kg, $\theta = 40^\circ$, $p = 1$ MPa, $d = 0.2$ m.

Angular speed and gas force

$$\omega = \frac{2\pi N}{60} = \frac{2\pi \times 360}{60} = 37.70\ \text{rad/s}$$ $$F_P = p\,\frac{\pi}{4}d^2 = 1\times10^{6}\times\frac{\pi}{4}(0.2)^2 = 31416\ \text{N}$$

(i) Inertia force

$$\begin{aligned} F_I &= m r\omega^2\left(\cos\theta + \frac{\cos 2\theta}{n}\right) \\ &= 120 \times 0.15 \times (37.70)^2 \left(\cos 40^\circ + \frac{\cos 80^\circ}{4}\right) \\ &= 25582.0 \times 0.8095 = 20708\ \text{N} \end{aligned}$$

(ii) Net force on piston

During the out stroke the piston is accelerating, so the inertia force opposes the gas force:

$$F = F_P - F_I = 31416 - 20708 = 10708\ \text{N}$$

(iii) Thrust in the connecting rod

$$\sin\phi = \frac{\sin\theta}{n} = \frac{\sin 40^\circ}{4} = 0.1607 \Rightarrow \phi = 9.25^\circ$$ $$F_Q = \frac{F}{\cos\phi} = \frac{10708}{0.9870} = 10849\ \text{N}$$

(iv) Side thrust on the cylinder wall

$$F_N = F\tan\phi = 10708 \times 0.1628 = 1743\ \text{N}$$

(v) Turning moment on the crankshaft

$$T = F_Q \cdot r\sin(\theta+\phi) = 10849 \times 0.15 \times \sin(49.25^\circ) = 1233\ \text{N·m}$$

Answer: Inertia force = 20708 N; piston effort = 10708 N; connecting-rod thrust = 10849 N; side thrust = 1743 N; turning moment = 1233 N·m.