The Slider Crank - Velocities, Accelerations, Forces and Moments, graphical and hand calculations

This note shows the approach needed (if computer software is not available) to determine the torque applied to a crank taking into account the gas pressure above the piston, the mass of the piston and for the final part, the mass of the connecting rod also (but ignoring friction losses in all cases) for one crank position.

The first diagram is the layout diagram:

The next two diagrams are the velocity and the acceleration diagrams:


Assuming that the connecting rod has a uniformly distributed mass, then the centre of mass (CoM) is at the mid - point of the rod. The acceleration of the CoM is at g', half way between the points showing the accelerations of the 2 ends of the connecting rod, a' and b'.

If in the above example the connecting rod had a uniformly distributed mass of 0.1 kg, determine the torque on the crankshaft and the normal force on the cylinder bore.

This is done by determining the linear motion of the CoM of the connecting rod, resolved in vertical and horizontal directions, then the rotation of the connecting rod about it's CoM.

The diagram below shows the forces acting on the connecting rod and the accelerations of the connecting rod:

On the above diagram, the value of RT, allowing for the mass of the piston, is: [2800000 x 3.14159 X (0.068)2/4] - 0.3 x 4344 = 8865.5 N (downwards).

The acceleration of the CoM of the connecting rod, the vertical component of c'g', is:

4207 cos(15.06) = 4062.5 m/s2, downwards, -ve. Using: force = mass x acceleration in the vertical direction: RV - RT - (mass connecting rod)g = (mass connecting rod) x acceleration .. upwards +ve.

RV - 8865.5 - 0.1 x 9.8 = 0.1 x (- 4062.5)

RV = 8460.2 N

Forces and acceleration of the connecting rod CoM in the horizontal direction, +ve to right:

Acceleration of connecting rod CoM to right is: 4207 sin(15.06) = 1093.1 m/s2

RW + RH = 0.1 x 1093.1 = 109.31 N Taking moments of the forces acting on the connecting rod about CoM, ccw +ve:

The moment of inertia of the connecting rod about the CoM (Icrg) = mass x (length)2/12 = 0.1(0.1)2/12 = 0.000083333kgm2

0.05 x RH x cos(7.18) - 0.05 x RW x cos(7.18) - 0.05 x RT x sin(7.18) - 0.05 x RV x sin(7.18) = Icrg x ang. acc. Con. rod

RH x cos(7.18) - RW x cos(7.18) - RT x sin(7.18) - RV x sin(7.18) = (Icrg x ang. acc. Con. rod)/0.05

Substituting RH = 109.31 - RW and other known forces: 109.31 - 2 x RW x cos(7.18) - (8865.5 + 8460.23) sin(7.18) = 0.000083333 x 20922/0.05

109.31 - 1.984317 x RW - 2165.49 = 34.86986

RW = (109.31 - 2165.49 - 34.86986)/1.984317

RW = - 1053.79 N, and so RH = 1163.1 N

RW, the force on the connecting rod at the piston pin, A, is to the left. The equal and opposite reaction on the cylinder wall is therefore to the right.

The value of crank torque about C calculated, 130.9 Nm, is slightly lower than that on the acceleration diagram as some of the force is required to accelerate the connecting rod, 406.25 N (there is a very small positive contribution from the weight of the connecting rod, 0.98 N) which was not considered in the calculations on the acceleration diagram.

David Grieve, revised: 29th April 2013, original: 27th November 2006.