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• # Crank Torque/Piston Force Question

Discussion in 'Calculations' started by Idiotic, Jun 29, 2012.

1. ### Jonathan JMember

Joined:
Sep 2012
Posts:
5
0
Sounds like all answers are in agreement, but the easiest way to explain where the equations come from is to Sum Forces and Moments about a point.

The following is copied from software output:

Point A is crank, point B is pin joint, point C is piston.

∑M = 0 => −17.160 Ã— RB3[Y] − 5.436 Ã— RB3[X] = −M1
∑F[X] = 0 => RA3[X] + RB3[X] + R38Fp3[X] = 0
∑F[Y] = 0 => RA3[Y] + RB3[Y] + R38Fp3[Y] = 0
∑F[X] = 0 => −RA3[X] = 0
∑F[Y] = 0 => −RA3[Y] = 0
∑F[X] = 0 => −RB3[X] − RB5[X] = 0
∑F[Y] = 0 => −RB3[Y] − RB5[Y] = 0

Resuts
RA3[X] = 0.000 N
RA3[Y] = 0.000 N
RB3[X] = 3939.999 N
RB3[Y] = 267.077 N
RB5[X] = −3939.999 N
RB5[Y] = −267.077 N

And of course,
Shear and Bending
Length CB =3949.04N compression
Length BA = 1444.44N shear & 26000Nmm

(Note the slight variation in answer is because I havent much time and have some rounding errors on inputs. But the answer shows that the preceding post values are certainly correct)

Hope you found that interesting.

2.