Force Analysis of Mechanisms

Inertia force of reciprocating parts

Let $r$ = crank radius, $l$ = connecting rod length, $n = l/r$, $\theta$ = crank angle from the inner dead centre (IDC), $\omega$ = crank speed (constant), $m$ = reciprocating mass.

Displacement of the piston from IDC:

$$x = r(1 - \cos\theta) + l(1 - \cos\phi), \qquad \sin\phi = \frac{\sin\theta}{n}$$

Using $\cos\phi = \sqrt{1 - \sin^2\theta/n^2} \approx 1 - \dfrac{\sin^2\theta}{2n^2}$ and $\sin^2\theta = \dfrac{1 - \cos2\theta}{2}$, the displacement from the crank centre is

$$x_c \approx r\cos\theta + \frac{r}{4n}\cos2\theta + \text{constant}$$

Differentiating twice with $\theta = \omega t$, acceleration of the piston:

$$a = -\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right)$$

The inertia force is equal and opposite to $ma$ (D'Alembert's principle). Taking the working-stroke direction (towards the crank) as positive, its magnitude is

$$F_I = m\omega^2 r\left(\cos\theta + \frac{\cos2\theta}{n}\right)$$

A positive value means the inertia force opposes the gas force (as near IDC); it is negative (aids motion) in the later part of the stroke. The net force on the piston is $F = F_P - F_I$.

Numerical

Data: $D = 100$ mm, $r = 0.1$ m, $l = 0.4$ m so $n = 4$, $N = 600$ rpm, $m = 1.5$ kg, $\theta = 30^\circ$, $p = 0.8$ MPa.

$$\omega = \frac{2\pi \times 600}{60} = 62.832\ \text{rad/s}, \quad \omega^2 = 3947.8$$

Gas force on piston:

$$F_P = p\times\frac{\pi}{4}D^2 = 0.8\times10^6 \times 0.00785 = 6283.2\ \text{N}$$

Inertia force:

$$F_I = 1.5 \times 3947.8 \times 0.1\,(\cos 30^\circ + \tfrac{\cos 60^\circ}{4}) = 1.5 \times 3947.8 \times 0.1\,(0.8660 + 0.1250) = 586.9\ \text{N}$$

(a) Net force on piston (along the line of stroke)

$$F = F_P - F_I = 6283.2 - 586.9 = 5696.3\ \text{N}$$

(b) Thrust on cylinder walls

$$\sin\phi = \frac{\sin30^\circ}{4} = 0.125 \Rightarrow \phi = 7.18^\circ$$ $$F_N = F\tan\phi = 5696.3 \times \tan7.18^\circ = 717.7\ \text{N}$$

(c) Force in connecting rod

$$F_Q = \frac{F}{\cos\phi} = \frac{5696.3}{\cos 7.18^\circ} = 5741.4\ \text{N}$$

(d) Turning moment

Tangential component of the rod force on the crank pin: $F_T = F_Q\sin(\theta + \phi)$, with $\theta + \phi = 37.18^\circ$.

$$F_T = 5741.4 \times \sin 37.18^\circ = 3469.7\ \text{N}$$ $$T = F_T\times r = 3469.7 \times 0.1 = 347.0\ \text{N m}$$

The radial component on the crank pin is $F_R = F_Q\cos(\theta + \phi) = 4574.3$ N, which loads the shaft bearings.

Answer: $F = 5696.3$ N, $F_N = 717.7$ N, $F_Q = 5741.4$ N, $T = 347.0$ N m.

(a) Definitions

• Centrifugal force: the outward radial force on a body in circular motion, $F_c = m\omega^2 r$. It is the reaction to the centripetal force. For analysis it is taken as an inertia force.
• Inertia force: the fictitious force equal to $m a_G$ acting at the centre of mass G, opposite to the acceleration of G: $\vec F_I = -m\vec a_G$.
• Inertia torque (couple): equal to $I_G\alpha$, opposite to the angular acceleration $\alpha$: $C_I = -I_G\alpha$.
• D'Alembert's principle: if the inertia force and inertia torque are added to the actual external forces on a body, the body is in dynamic equilibrium. Then $\sum \vec F + \vec F_I = 0$ and $\sum M + C_I = 0$, so the methods of statics can be used for dynamic problems.

Numerical.

$$\omega = \frac{2\pi \times 300}{60} = 31.416\ \text{rad/s}$$ $$F_c = m\omega^2r = 5 \times 986.96 \times 0.4 = 1973.9\ \text{N}$$

(b) Link

Inertia force at G:

$$F_I = m a_G = 12 \times 25 = 300\ \text{N}$$

directed opposite to the acceleration of G.

Inertia couple:

$$C_I = I_G\alpha = 0.8 \times 40 = 32\ \text{N m}$$

acting clockwise (opposite to $\alpha$).

Single equivalent force. The force $F_I$ and couple $C_I$ can be replaced by the same force $F_I$ acting along a line parallel to its original line, shifted by

$$h = \frac{C_I}{F_I} = \frac{32}{300} = 0.1067\ \text{m} = 106.7\ \text{mm}$$

from G, on the side that gives a moment about G in the same sense as $C_I$ (clockwise).

         h
    ----------|-> F_I (shifted line)
   G (x)  <- C_I clockwise

Answer: (a) $F_c = 1973.9$ N; (b) $F_I = 300$ N, $C_I = 32$ N m (clockwise), shifted force at $h = 106.7$ mm from G.

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.

Force on a spur gear tooth

The tooth force acts along the line of action (common normal), at the pressure angle $\phi$ to the common tangent of the pitch circles. The normal force $W_n$ is resolved into:

• tangential component $W_t$ (transmits the torque),
• radial component $W_r = W_t\tan\phi$ (separates the gears, loads the bearings).

        W_r
         ^   W_n
         |  /
         | /  phi
   ------+------> W_t   (tangent at pitch point)

$$W_t = \frac{T}{d/2}, \qquad W_r = W_t\tan\phi, \qquad W_n = \frac{W_t}{\cos\phi}$$

Calculation

Angular speed: $\omega = \dfrac{2\pi \times 720}{60} = 75.398$ rad/s.

Torque:

$$T = \frac{P}{\omega} = \frac{10000}{75.398} = 132.63\ \text{N m}$$

Pitch circle diameter: $d = mz = 5 \times 20 = 100$ mm.

Pitch line velocity:

$$v = \omega\frac{d}{2} = 75.398 \times 0.050 = 3.770\ \text{m/s}$$

Tangential force:

$$W_t = \frac{T}{d/2} = \frac{132.63}{0.050} = 2653\ \text{N} \quad \left(= \frac{P}{v} = \frac{10000}{3.770}\right)$$

Radial force:

$$W_r = W_t\tan20^\circ = 2653 \times 0.3640 = 965\ \text{N}$$

Normal force:

$$W_n = \frac{W_t}{\cos20^\circ} = \frac{2653}{0.9397} = 2823\ \text{N}$$

The driven gear experiences equal and opposite forces, and each shaft bearing must support the vector sum of $W_t$ and $W_r$ (that is, $W_n$) in addition to the overhung weight.

Answer: $T = 132.63$ N m, $v = 3.770$ m/s, $W_t = 2653$ N, $W_r = 965$ N, $W_n = 2823$ N.

Force components in a bevel gear

The resultant tooth force acts at the mean pitch radius $r_m$ of the face width, normal to the tooth surface. With pitch cone angle $\gamma$ and pressure angle $\phi$:

$$W_t = \frac{T}{r_m}, \qquad W_r = W_t\tan\phi\cos\gamma, \qquad W_a = W_t\tan\phi\sin\gamma$$

The radial force $W_t\tan\phi$ acts normal to the pitch cone; its components are along the radial direction ($\cos\gamma$) and along the axis ($\sin\gamma$). On the pinion the axial force pushes it away from the apex.

Geometry

Pitch cone angle of the pinion:

$$\tan\gamma = \frac{z_1}{z_2} = \frac{18}{54} \Rightarrow \gamma = 18.43^\circ \quad (\sin\gamma = 0.3162,\ \cos\gamma = 0.9487)$$

Large-end pitch diameter: $d = mz_1 = 5 \times 18 = 90$ mm.

Mean pitch diameter:

$$d_m = d - b\sin\gamma = 90 - 40 \times 0.3162 = 77.35\ \text{mm}$$

Torque

$$\omega = \frac{2\pi \times 900}{60} = 94.248\ \text{rad/s}, \qquad T = \frac{P}{\omega} = \frac{7500}{94.248} = 79.58\ \text{N m}$$

Components

$$W_t = \frac{T}{d_m/2} = \frac{79.58}{(77.35/2)\times10^{-3}} = 2058\ \text{N}$$ $$W_r = W_t\tan20^\circ\cos\gamma = 2058 \times 0.3640 \times 0.9487 = 710\ \text{N}$$ $$W_a = W_t\tan20^\circ\sin\gamma = 2058 \times 0.3640 \times 0.3162 = 237\ \text{N}$$

The resultant normal force is $W_n = W_t/\cos20^\circ = 2190$ N. On the gear (shaft at 90 degrees) the pinion's radial force becomes its axial force and vice versa: the gear's axial force $= 710$ N and radial force $= 237$ N.

Answer: $W_t = 2058$ N, $W_r = 710$ N, $W_a = 237$ N on the pinion.

Geometry of the tooth force

In a helical gear the tooth force $W_n$ (normal to the tooth surface) acts in the normal plane, which is inclined to the transverse plane. $\phi_n$ is the normal pressure angle, $\psi$ the helix angle.

 Normal plane:        Pitch plane:
      W_n             axial ^   /
      /|                    |  / helix
     / | W_r                | /  (psi)
    /phi_n                  |/
   +-----               ----+------> tangential
   W_n cos(phi_n)

Step 1: Components in the normal plane

• Radial component: $W_r = W_n\sin\phi_n$.
• Component in the pitch plane, perpendicular to the tooth (helix) direction: $W_n\cos\phi_n$.

Step 2: Resolve the pitch-plane component

The helix line makes angle $\psi$ with the axis. Along the transverse tangent (tangential) and the axis:

• Tangential component: $W_t = W_n\cos\phi_n\cos\psi$
• Axial component: $W_a = W_n\cos\phi_n\sin\psi$

Results

$$W_n = \frac{W_t}{\cos\phi_n\cos\psi}$$ $$W_r = W_n\sin\phi_n = \frac{W_t\tan\phi_n}{\cos\psi} = W_t\tan\phi_t$$ $$W_a = W_n\cos\phi_n\sin\psi = W_t\tan\psi$$

where $W_t = 2T/d$ is found from the torque, and $\tan\phi_t = \tan\phi_n/\cos\psi$.

Axial thrust

$W_a = W_t\tan\psi$ increases with helix angle. For $\psi = 15^\circ$, $W_a = 0.268W_t$; for $\psi = 30^\circ$, $W_a = 0.577W_t$. Thrust bearings (angular contact or taper roller) must take it. The direction of the thrust depends on the hand of the helix and direction of rotation, and the gear and pinion have equal and opposite axial forces.

Double helical (herringbone) gears

A double helical gear is two helical gears of opposite hand cut side by side on one blank. Each half produces an axial thrust of $W_t\tan\psi/2$, and the two thrusts are equal and opposite, so the net axial force on the shaft is zero. This allows larger helix angles (up to 45 degrees), giving smoother and more powerful drives with no thrust bearing needed. Applications: heavy reduction gearboxes, turbines, rolling mills.

Why a spring is needed

A disc cam can push the follower, but it cannot pull it back. During the second half of the rise, the follower decelerates and its inertia tends to carry it forward away from the cam. A spring (or a positive-drive groove) must keep it in contact, otherwise the follower jumps and the motion is lost.

Follower acceleration (SHM rise)

$$a = \frac{\pi^2 h\omega^2}{2\beta^2}\cos\frac{\pi\theta}{\beta}$$

Data: $m = 0.8$ kg, $h = 0.025$ m, $\beta = 90^\circ = \pi/2$, $\omega = \dfrac{2\pi \times 400}{60} = 41.888$ rad/s, $\omega^2 = 1754.6$.

$$a_{max} = \frac{\pi^2 h\omega^2}{2\beta^2} = \frac{\pi^2 (0.025)(1754.6)}{2(\pi/2)^2} = 87.73\ \text{m/s}^2$$

(Here $\pi^2/(2\beta^2) = 2$, so $a_{max} = 2h\omega^2 = 2 \times 0.025 \times 1754.6$; the same value is $\dfrac{h}{2}\left(\dfrac{\pi}{\beta}\right)^2\omega^2$.)

(a) Maximum inertia force

$$F_{I,max} = m\,a_{max} = 0.8 \times 87.73 = 70.18\ \text{N}$$

It acts at the beginning of the rise (positive acceleration, the cam must push) and at the end of rise (negative acceleration, the follower tends to leave the cam).

(b) Minimum spring stiffness

The spring force at the end of the rise (compressed by the lift $h$) must be at least equal to the inertia force at the end of rise:

$$F_0 + k\,h \ge m\,a_{max}$$ $$k \ge \frac{70.18 - 15}{0.025} = 2207\ \text{N/m} \approx 2.21\ \text{N/mm}$$

In practice a margin of 20 to 50 percent is added.

(c) Maximum force from the cam

At the start of the rise, the spring force is the preload $F_0$ and the cam must supply the spring force plus the inertia force:

$$F_{cam} = m a_{max} + F_0 = 70.18 + 15 = 85.2\ \text{N}$$

At any angle the cam force is $N = m a + F_0 + k s$. With the stiffness found it reduces to $N = 42.6(1 + \cos x)$ N where $x = \pi\theta/\beta$, so it falls from the maximum at the start to zero (just touching) at the end of the rise. The spring force at the end of the rise is 70.2 N, equal to the inertia force.

Answer: (a) $F_{I,max} = 70.18$ N; (b) $k_{min} = 2207$ N/m; (c) $F_{cam,max} = 85.2$ N (at start of rise).