Static force analysis of slider crank

Force members

• Two-force member: a member acted upon by forces at only two points (no couple). For equilibrium the two forces are equal, opposite and collinear (along the line joining the points). The connecting rod (with pin joints at both ends and no load between) is a two-force member.
• Three-force member: a member with forces at three points. For equilibrium the three forces must be coplanar, their lines of action must meet at one point (concurrent) and their vector sum must be zero (a closed force triangle). The slider block (force $P$, guide reaction $F_N$, rod force $F_Q$) is a three-force member.

Geometry

$$\sin\phi = \frac{r\sin\theta}{l} = \frac{\sin45^\circ}{4} = 0.1768 \Rightarrow \phi = 10.18^\circ$$

Free-body of the slider

          F_N (guide reaction, perpendicular)
            ^
            |
   P  <-----[]----- F_Q (rod force)

The rod is a two-force member, so $F_Q$ acts along the rod, at angle $\phi$ to the line of stroke. Equilibrium of the slider:

$$F_Q\cos\phi = P \Rightarrow F_Q = \frac{2000}{\cos 10.18^\circ} = 2032.0\ \text{N}$$ $$F_N = F_Q\sin\phi = P\tan\phi = 2000\tan 10.18^\circ = 359.2\ \text{N}$$

Torque on the crank

The rod force $F_Q$ acts on the crank pin along the rod. The perpendicular distance from the crank centre O to the line of the rod is $r\sin(\theta + \phi)$. So

$$T = F_Q\, r\sin(\theta + \phi) = 2032.0 \times 0.1 \times \sin 55.18^\circ = 166.8\ \text{N m}$$

(The radial component on the crank pin, $F_Q\cos(\theta+\phi) = 1160.2$ N, only loads the bearing.)

Check by virtual work

Slider position from O: $x = r\cos\theta + \sqrt{l^2 - r^2\sin^2\theta}$. For a virtual displacement $\delta\theta$ of the crank, the work of the input torque equals the work done against $P$:

$$T\,\delta\theta = P\,\delta x \Rightarrow T = P\left|\frac{dx}{d\theta}\right| = P r\left[\sin\theta + \frac{\sin2\theta}{2\sqrt{n^2 - \sin^2\theta}}\right]$$ $$T = 2000 \times 0.1\left[0.7071 + \frac{1.0000}{2 \times 3.937}\right] = 166.8\ \text{N m}$$

It agrees with the free-body result.

Answer: $F_Q = 2032.0$ N, $F_N = 359.2$ N, $T = 166.8$ N m.