A slider-crank mechanism is given. The crank length is $r = 0.1$ m, and the connecting rod length is $l = 0.4$ m. The crank angle is $\theta = 30^\circ$. The pressure inside the piston cylinder is $P = 3$ MPa. The crankshaft rotates at a constant angular velocity of $N = 1500$ rpm. The cross-sectional area of the piston is $A = 60$ cm$^2$, and the total mass of the piston is $m = 6$ kg. We need to determine: (i) Velocity of the piston (ii) Acceleration of the piston (iii) Net force acting on the piston (iv) Stress on the connecting rod (v) Side reaction on the piston (vi) Turning moment on the crank shaft

1. Problem Description

A slider-crank mechanism is given. The crank length is $r = 0.1$ m, and the connecting rod length is $l = 0.4$ m. The crank angle is $\theta = 30^\circ$. The pressure inside the piston cylinder is $P = 3$ MPa. The crankshaft rotates at a constant angular velocity of $N = 1500$ rpm. The cross-sectional area of the piston is $A = 60$ cm$^2$, and the total mass of the piston is $m = 6$ kg. We need to determine:

(i) Velocity of the piston

(ii) Acceleration of the piston

(iii) Net force acting on the piston

(iv) Stress on the connecting rod

(v) Side reaction on the piston

(vi) Turning moment on the crank shaft

2. Solution Steps

(i) Velocity of the piston:

First, convert the angular velocity from rpm to rad/s:

$\omega = \frac{2 \pi N}{60} = \frac{2 \pi (1500)}{60} = 50\pi$ rad/s

The piston velocity $v_p$ can be determined using the following formula:

$v_p = r \omega (\sin\theta + \frac{r}{2l} \sin 2\theta)$

$v_p = 0.1 \times 50\pi (\sin(30^\circ) + \frac{0.1}{2 \times 0.4} \sin(2 \times 30^\circ))$

$v_p = 5\pi (\frac{1}{2} + \frac{1}{8} \frac{\sqrt{3}}{2}) = 5\pi (0.5 + 0.108) = 5\pi (0.608) \approx 9.57$ m/s

(ii) Acceleration of the piston:

The piston acceleration $a_p$ can be determined using the following formula:

$a_p = r \omega^2 (\cos\theta + \frac{r}{l} \cos 2\theta)$

$a_p = 0.1 (50\pi)^2 (\cos(30^\circ) + \frac{0.1}{0.4} \cos(60^\circ))$

$a_p = 0.1 (2500 \pi^2) (\frac{\sqrt{3}}{2} + \frac{1}{4} \times \frac{1}{2}) = 250 \pi^2 (\frac{\sqrt{3}}{2} + \frac{1}{8}) = 250 \pi^2 (0.866 + 0.125) = 250 \pi^2 (0.991)$

$a_p \approx 250 (9.87)(0.991) \approx 2443.1$ m/s$^2$

(iii) Net force acting on the piston:

The pressure force acting on the piston is $F_p = P \times A$.

Convert the area from cm$^2$ to m$^2$: $A = 60 \text{ cm}^2 = 60 \times 10^{-4} \text{ m}^2 = 0.006 \text{ m}^2$.

$F_p = 3 \times 10^6 \text{ Pa} \times 0.006 \text{ m}^2 = 18000$ N.

The inertia force is $F_i = m a_p = 6 \text{ kg} \times 2443.1 \text{ m/s}^2 = 14658.6$ N.

The net force acting on the piston is $F_{net} = F_p - F_i = 18000 - 14658.6 = 3341.4$ N.

(iv) Stress on the connecting rod:

We need to calculate the force in the connecting rod.

$F_{connecting\,rod} = \frac{F_p}{\cos \phi}$, where $\phi$ is the angle of the connecting rod with the horizontal.

Using the sine rule: $\frac{\sin\theta}{l} = \frac{\sin\phi}{r}$, therefore, $\sin \phi = \frac{r}{l} \sin\theta = \frac{0.1}{0.4} \sin(30^\circ) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} = 0.125$.

So $\phi = \arcsin(0.125) \approx 7.18^\circ$.

$F_{connecting\,rod} = \frac{18000}{\cos(7.18^\circ)} \approx \frac{18000}{0.992} \approx 18145.16$ N.

To calculate the stress, we need the cross-sectional area of the connecting rod. Since this value is not provided in the problem, let us assume it is $A_{rod}$.

Stress $= \frac{F_{connecting\,rod}}{A_{rod}}$.

Assuming $A_{rod} = 100 mm^2 = 100 \times 10^{-6} m^2= 10^{-4} m^2$:

Stress $= \frac{18145.16}{10^{-4}} = 181.45 \times 10^6 Pa = 181.45 MPa$.

(v) Side reaction on the piston:

$F_{side} = F_{connecting\,rod} \sin \phi = 18145.16 \times \sin(7.18^\circ) \approx 18145.16 \times 0.125 \approx 2268.14$ N.

(vi) Turning moment on the crank shaft:

$T = F_{connecting\,rod} \cos \phi \times r \sin\theta = 18145.16 \times 0.992 \times 0.1 \times 0.5 \approx 900$ Nm. Alternatively, $T = F_p r \sin(\theta + \phi) = 18000 \times 0.1 \times \sin(30^\circ + 7.18^\circ) \approx 1800 \times \sin(37.18) = 1800 \times 0.604 = 1087.2$ Nm

3. Final Answer

(i) Velocity of the piston: 9.57 m/s

(ii) Acceleration of the piston: 2443.1 m/s$^2$

(iii) Net force acting on the piston: 3341.4 N

(iv) Stress on the connecting rod: 181.45 MPa (assuming $A_{rod}=100 mm^2$)

(v) Side reaction on the piston: 2268.14 N

(vi) Turning moment on the crank shaft: 1087.2 Nm