Start » Exercises » Engineering Mechanics I » Forces on a crank drive

Forces on a crank drive

This exercise is about a statics investigation of the forces on a crank drive with a crosshead guidance for a specified rotation angle. The kinematics of a crank drive is considered here.

Task

The force F acts on the piston of a steam engine. The following parameters have to be calculated:

a) the force in the driving rod,

b) the transverse force in the crosshead guidance (crosshead bearing) and

c) the torque in the crankshaft.

Crank drive for a certain angle of rotation
Crank drive for a certain angle of rotation

Solution

The following video is in german language.

Technische Mechanik 1, Übung 12 - Kräfte am Kurbeltrieb

Sketch for the solution

Forces on the crank drive for a certain angle of rotation
Forces on the crank drive for a certain angle of rotation
\[ \require{cancel} \] \[ \newcommand{\myvec}[1]{{\begin{pmatrix}#1\end{pmatrix}}} \]

The angle α between the force F and the driving rod results from

\[\tag{1} \alpha = arctan \left( \frac{r}{l} \right) \]

Between the force F and the force FS in the driving rod exists the following relationship

\[\tag{2} F = F_S \cdot \cos \alpha \]

\[\tag{3} F_S = \frac{F}{\cos \alpha} \]

The cosine of the arctangent α leads to

\[\tag{4} \cos \left( arctan \left( \frac{r}{l} \right) \right) = \frac{1}{\sqrt{\left( \frac{r}{l}\right)^2 + 1}} \]

So this is the force in the driving rod:

\[\tag{5} F_S = F \cdot \sqrt{\left( \frac{r}{l}\right)^2 + 1} \]

The transversal force in the crosshead guiding results from

\[\tag{6} F_Q = F_S \cdot \sin \alpha \]

The sine of the arctangent α leads to

\[\tag{7} \sin \left( arctan \left( \frac{r}{l} \right) \right) = \frac{\frac{r}{l}}{\sqrt{\left( \frac{r}{l}\right)^2 + 1}} \]

So the transversal force is

\[\tag{8} F_Q = F \cdot \bcancel{\sqrt{\left( \frac{r}{l}\right)^2 + 1}} \cdot \frac{\frac{r}{l}}{\bcancel{\sqrt{\left( \frac{r}{l}\right)^2 + 1}}} \]

\[\tag{9} F_Q = F \cdot \frac{r}{l} \]

The torque at the crankshaft is

\[\tag{10} M_K = F_S \cdot r \]

\[\tag{11} M_K = F \cdot r \cdot \sqrt{\left( \frac{r}{l}\right)^2 + 1} \]

This exercise is part of the collection Engineering Mechanics 1 -- Statics.