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# Kinematics of a crank drive

In this exercise, the kinematics of a crank drive are considered and the functions for the piston speed and piston acceleration are set up.

For a crank drive of a piston machine, the speed and acceleration of the piston have to be calculated. The crank rotates at a constant angular velocity ω. For the ratio of the connecting rod length r and the length of the slider crank l should apply

$l \gg r$

## Solution

The angular velocity is

$\tag{1} \omega = \frac{\varphi}{t}$

The geometric relationship between the angle of rotation φ and the piston position x is:

$\tag{2} r + l = r \cdot cos \varphi + l \cdot cos \psi + x$

It is

$\tag{3} r \cdot sin \varphi = l \cdot sin \psi$

To simplify the calculation, the ratio of the lengths is expressed as follows:

$\tag{4} \frac{r}{l} = \lambda$

This results in

$\tag{5} \frac{x}{r} = \left( 1 - cos \varphi \right) + \frac{1}{\lambda} \left( 1-\sqrt{1-\lambda^2 sin^2 \varphi} \right)$

The velocity is

$\tag{6} \dot{x} = \frac{dx}{d\varphi} \frac{d\varphi}{dt} = \frac{dx}{d\varphi} \omega$

$\tag{7} \frac{\dot{x}}{r \omega} = sin \varphi + \lambda \frac{sin \varphi \cdot cos \varphi}{\sqrt{1-\lambda^2 sin^2 \varphi}}$

It applies

$\tag{8} \lambda \ll 1$

and the term

$\tag{9} \frac{1}{\sqrt{1-\lambda^2 sin^2 \varphi}}$

can be expanded into a power series.

$\tag{10} \frac{1}{\sqrt{1-\lambda^2 sin^2 \varphi}} = 1 + \frac{1}{2} \lambda^2 sin^2 \varphi + \frac{3}{8} \lambda^4 sin^4 \varphi \, + ...$

If one only takes into account the terms up to the 1st power of λ, then from equation (7)

$\tag{11} \frac{\dot{x}}{r \omega} = sin \varphi + \lambda \cdot sin \varphi \cdot cos \varphi = sin \varphi + \frac{\lambda}{2}sin 2 \varphi$

The piston velocity thus is

$\tag{12} \dot{x} = r \omega \left( sin \varphi + \frac{\lambda}{2}sin 2 \varphi \right)$

The acceleration is

$\tag{13} \ddot{x} = \frac{d^2x}{d \varphi^2}\omega^2$

$\tag{14} \frac{\ddot{x}}{r \omega^2} = cos \varphi + \lambda \frac{cos^2 \varphi -sin^2\varphi \left( 1-\lambda^2 sin^2\varphi \right)}{\left( 1 - \lambda^2 sin^2 \varphi \right)^{\frac{3}{2}}}$

For λ much smaller than 1 we get

$\tag{15} \frac{\ddot{x}}{r \omega^2} = cos \varphi + \lambda \cos 2 \varphi$

respectively

$\tag{16} \ddot{x} = r \omega^2 \left( cos \varphi + \lambda \cos 2 \varphi \right)$

So the determination of the speed and the acceleration of the piston in the crank drive is complete.