3.4 Four-Bar Linkages, Cam-Follower Systems & Kinematics

Four-Bar Linkages, Cam-Follower Systems & Kinematics

Kinematic mechanisms transform input mechanical motions (continuous rotary, reciprocating linear, or oscillating) into prescribed output trajectories, velocity profiles, and mechanical force advantages. On the NCEES PE Mechanical exam, mechanism problems evaluate five core competencies: system mobility (degrees of freedom), four-bar linkage classifications and Grashof criteria, transmission angle and toggle positions, instantaneous centers of zero velocity (Kennedy's theorem), and cam-follower profile synthesis and pressure angle sizing.

+---------------------------------------------------------------------------------------------------+
|                             PLANAR MECHANISMS TAXONOMY                                            |
|                                                                                                   |
|   [MOBILITY & CRITERIA]    [FOUR-BAR LINKAGES]      [VELOCITY & ACCEL]      [CAM-FOLLOWER]        |
|   - Chebyshev-Grübler      - Grashof: S+L <= P+Q    - Instant Centers (ICs) - Profiles: Cycloidal |
|     M = 3(n-1)-2j1-j2      - Crank-Rocker           - Kennedy's 3-Center    - Simple Harmonic     |
|   - Lower / Higher Pairs   - Drag-Link / Dbl-Crank  - Slider-Crank: v(t)    - Pressure Angle      |
|   - Idle Degrees of Freedom- Transmission Angle mu  - Secondary Accel: 2w   - Base Circle Sizing  |
+---------------------------------------------------------------------------------------------------+

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1. Planar Mechanism Mobility & Degrees of Freedom

The mobility ($M$) or number of degrees of freedom (DOF) of a planar mechanism is the number of independent input parameters (usually actuator joint angles or linear displacements) required to uniquely define the position of all links in space.

+---------------------------------------------------------------------------------------------------+
|                         KINEMATIC PAIR CLASSIFICATIONS (JOINTS)                                   |
|                                                                                                   |
|   LOWER PAIRS (Surface Contact, 1-DOF, j_1):        HIGHER PAIRS (Point/Line, 2-DOF, j_2):        |
|   - Revolute / Pin Joint (R): 1 Rotational DOF      - Cam-Follower (Roll & Slide): 2 DOF          |
|   - Prismatic / Slider Joint (P): 1 Translating DOF - Meshing Gear Teeth: 2 DOF (Roll & Slide)   |
+---------------------------------------------------------------------------------------------------+

The Chebyshev-Grübler-Kutzbach Criterion

For a planar mechanism consisting of $n$ total links (including the fixed ground frame link $n=1$):

$M = 3(n - 1) - 2 j_1 - j_2$

Where:

• $n = \text{Total number of links (including frame link)}$
• $j_1 = \text{Number of single-DOF joints (revolute pins, prismatic sliders)}$
• $j_2 = \text{Number of two-DOF joints (higher pairs, rolling with sliding)}$

Mobility Interpretation Matrix

• $M = 1$: Constrained kinematic mechanism (single motor/actuator drives all links deterministically).
• $M = 2$: Two-degree-of-freedom mechanism (requires two coordinated actuators, e.g., five-bar robot arm or differential).
• $M = 0$: Statically determinate structure (truss; no motion possible).
• $M < 0$: Statically indeterminate, over-constrained structure (pre-stressed / locked).

[!WARNING] Idle / Redundant Degrees of Freedom: In cam mechanisms with a roller follower, the roller can spin freely on its pin without altering the follower's output motion. This free spin is an idle degree of freedom ($M_{\text{idle}} = 1$). To find true working mobility: $M_{\text{effective}} = M_{\text{calculated}} - 1 = 1$.

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2. Four-Bar Linkages & Grashof's Criterion

A planar four-bar linkage consists of four rigid links connected by four pin (revolute) joints: Frame ($L_1$), Input Link ($L_2$), Coupler Link ($L_3$), and Output Link ($L_4$).

+---------------------------------------------------------------------------------------------------+
|                                STANDARD FOUR-BAR LINKAGE                                          |
|                                                                                                   |
|                                     (Joint B)                                                     |
|                                      +-------+                                                    |
|                                     /         \                                                   |
|                  Input Link (L_2)  /           \  Coupler Link (L_3)                              |
|                                   /             \                                                 |
|                                  /               \                                                |
|                        (Joint A)+                 +(Joint C)                                      |
|                                 |                 |                                               |
|                                 |                 | Output Link (L_4)                             |
|                                 |                 |                                               |
|                                 +=================+                                               |
|                              (O_2)  FRAME (L_1)  (O_4)                                            |
+---------------------------------------------------------------------------------------------------+

Grashof's Law

Let $S$ be the length of the shortest link, $L$ be the length of the longest link, and $P, Q$ be the lengths of the two intermediate links.

Case 1: Grashof Linkages ($S + L \le P + Q$)

At least one link can perform a continuous $360^\circ$ revolution relative to the other links. The behavior depends on which link is chosen as the fixed ground frame:

1. Crank-Rocker Mechanism: Ground link is adjacent to the shortest link $S$. The shortest link rotates $360^\circ$ continuously as the input crank, while the opposite grounded link oscillates back and forth as a rocker.
2. Double-Crank / Drag-Link Mechanism: The shortest link $S$ is the fixed ground frame. Both moving links attached to ground perform complete $360^\circ$ continuous revolutions.
3. Double-Rocker Mechanism: The shortest link $S$ is the floating coupler link (opposite the ground). Both grounded links can only oscillate as rockers; the coupler link rotates $360^\circ$.
4. Change-Point Mechanism ($S + L = P + Q$): Parallelogram or deltoid linkage where all links become collinear simultaneously (e.g., locomotive wheel linkages). Requires guide pins to prevent jamming at change points.

Case 2: Non-Grashof Linkages ($S + L > P + Q$)

Triple-Rocker Mechanism: No link can make a complete $360^\circ$ revolution. All three moving links oscillate as rockers regardless of which link is grounded.

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3. Transmission Angle, Mechanical Advantage & Toggle Positions

+---------------------------------------------------------------------------------------------------+
|                         TRANSMISSION ANGLE & MECHANICAL ADVANTAGE                                 |
|                                                                                                   |
|                      Coupler (L_3)                                                                |
|                     ------------------------+                                                     |
|                                              \  <--- Transmission Angle mu                        |
|                                               \      (Ideal mu = 90 deg)                          |
|                                                \                                                  |
|                                                 \ Output Rocker (L_4)                             |
|                                                  \                                                |
|                                                   + (O_4 Ground)                                  |
|                                                                                                   |
|   TOGGLE POSITION: When Crank (L_2) and Coupler (L_3) become collinear:                           |
|   O_2 ---------- A ------------------------ B =======> O_4    (Mechanical Advantage MA -> Infinity)|
+---------------------------------------------------------------------------------------------------+

1. Transmission Angle ($\mu$)

The transmission angle $\mu$ is the acute angle between the floating coupler link ($L_3$) and the output rocker link ($L_4$). It measures the effectiveness with which force is transmitted to produce output torque:

• Ideal: $\mu = 90^\circ$ (100% of coupler force creates useful output torque; zero axial joint push).
• Acceptable Design Standard: $45^\circ \le \mu \le 135^\circ$ throughout the entire motion cycle.
• Extreme Values: Occur when the input crank $L_2$ aligns collinear with the fixed frame ($0^\circ$ or $180^\circ$).

2. Mechanical Advantage ($MA$) and Toggle Position

Neglecting friction, instantaneous power balance gives:

$P_{\text{in}} = P_{\text{out}} \implies T_{\text{in}} \omega_{\text{in}} = T_{\text{out}} \omega_{\text{out}}$

$MA = \frac{F_{\text{out}}}{F_{\text{in}}} \propto \frac{T_{\text{out}}}{T_{\text{in}}} = \frac{\omega_{\text{in}}}{\omega_{\text{out}}}$

When the input crank and coupler become collinear, the output link reaches its extreme oscillation limit where $\omega_{\text{out}} = 0$. Consequently, Mechanical Advantage approaches infinity ($MA \to \infty$). This is called the toggle position, utilized in rock crushers, clamping pliers (Vise-Grip), and punch presses to generate colossal clamping forces with minimal input torque.

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4. Slider-Crank Kinematics & Reciprocating Acceleration

The in-line slider-crank mechanism transforms rotation into reciprocating linear piston motion.

+---------------------------------------------------------------------------------------------------+
|                                 IN-LINE SLIDER-CRANK KINEMATICS                                   |
|                                                                                                   |
|                                (Joint A)                                                          |
|                                   +                                                               |
|                                  / \                                                              |
|                   Crank (r)     /   \  Connecting Rod (L)                                         |
|                                /     \                                                            |
|                               /       \                                                           |
|               (O_2 Ground)   + theta   +-----------------+                                        |
|                              ================|  PISTON (x)     | ===> Line of Stroke              |
|                                              +-----------------+                                  |
|                                                                                                   |
|   Rod-to-Crank Ratio: lambda = r / L                                                              |
+---------------------------------------------------------------------------------------------------+

Exact and Series Expansion Equations

Let $r$ be crank radius, $L$ be connecting rod length, $\theta = \omega t$ be crank angle from Top Dead Center (TDC), and $\lambda = \frac{r}{L}$ (typically $0.20 - 0.30$ in engines):

$\text{Piston Displacement: } x(\theta) \approx r \left[ (1 - \cos\theta) + \frac{\lambda}{2} \sin^2\theta \right]$

$\text{Piston Velocity: } v(\theta) = \dot{x}(\theta) \approx r \omega \left[ \sin\theta + \frac{\lambda}{2} \sin(2\theta) \right]$

$\text{Piston Acceleration: } a(\theta) = \ddot{x}(\theta) \approx r \omega^2 \left[ \cos\theta + \lambda \cos(2\theta) \right]$

[!IMPORTANT] Primary vs. Secondary Reciprocating Forces:

• Primary Acceleration ($r \omega^2 \cos\theta$): Occurs at fundamental rotational engine frequency $\omega$.
• Secondary Acceleration ($r \omega^2 \lambda \cos 2\theta$): Occurs at twice engine frequency ($2\omega$) due to connecting rod angular obliquity, creating high-frequency secondary unbalance in inline-4 engines.

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5. Instantaneous Centers of Velocity (Kennedy's Theorem)

An instantaneous center of zero velocity (IC) is a point common to two bodies in planar motion that has zero relative velocity between them at a given instant.

+---------------------------------------------------------------------------------------------------+
|                         ARONHOLD-KENNEDY THEOREM OF THREE CENTERS                                 |
|                                                                                                   |
|                       I_12 (Joint 1-2)                                                            |
|                             o                                                                     |
|                              \                                                                    |
|                               \                                                                   |
|                                o I_23 (Joint 2-3)                                                 |
|                                 \                                                                 |
|                                  \                                                                |
|                                   o I_13 (Lies on straight line connecting I_12 and I_23!)        |
|                                                                                                   |
|   RULE: Any three bodies (1, 2, 3) have three instant centers (I_12, I_23, I_13)                  |
|         that must lie on ONE SINGLE STRAIGHT LINE.                                                |
+---------------------------------------------------------------------------------------------------+

1. Total Number of Instant Centers

For a mechanism with $n$ links:

$N = \frac{n(n - 1)}{2}$ (For a four-bar linkage, $n=4 \implies N = \frac{4(3)}{2} = 6\text{ instant centers}: I_{12}, I_{23}, I_{34}, I_{14}, I_{13}, I_{24}$).

2. Velocity Ratio Determination Using Instant Centers

The linear velocity of instant center $I_{24}$ can be expressed from link 2 or link 4:

$v_{I_{24}} = \omega_2 \times (I_{24} - I_{12}) = \omega_4 \times (I_{24} - I_{14})$

$\implies \frac{\omega_4}{\omega_2} = \frac{\text{Distance }(I_{12} - I_{24})}{\text{Distance }(I_{14} - I_{24})}$

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6. Cam-Follower Kinematics & Profile Sizing

Cams convert rotary motion into prescribed reciprocating or oscillating follower motion profiles $y(\theta)$.

+---------------------------------------------------------------------------------------------------+
|                              CAM-FOLLOWER MOTION PROFILES                                         |
|                                                                                                   |
|   UNIFORM VELOCITY:            SIMPLE HARMONIC (SHM):       CYCLOIDAL MOTION:                     |
|   - Constant dy/dtheta         - y = (h/2)*(1 - cos(pi*t))  - y = h*(t/beta - sin(2*pi*t)/(2*pi)) |
|   - Infinite accel spikes      - Smooth velocity            - Continuous acceleration             |
|   - Unusable at high speed     - Finite jerk step           - ZERO JERK at boundaries (Best)      |
+---------------------------------------------------------------------------------------------------+

Comparison of Standard Cam Lift Profiles

Cam Pressure Angle ($\phi$) and Base Circle Sizing

The pressure angle $\phi$ is the angle between the follower motion axis and the normal to the cam pitch curve. To prevent excessive side thrust and follower binding in its guides:

• Translating roller follower limit: $\phi_{\max} \le 30^\circ$.
• Oscillating roller follower limit: $\phi_{\max} \le 35^\circ$.

$\tan\phi = \frac{v / \omega}{r_b + y} = \frac{dy/d\theta}{r_b + y}$

Where $r_b$ is the base circle radius, $y$ is follower lift, and $dy/d\theta$ is follower kinematic velocity. To reduce pressure angle below $30^\circ$, increase the base circle radius ($r_b$).

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7. Step-by-Step Worked Problem: Linkage Grashof Check & Cam Sizing

Problem Statement

1. A planar four-bar mechanism has link lengths: Frame $L_1 = 120\text{ mm}$, Input $L_2 = 40\text{ mm}$, Coupler $L_3 = 110\text{ mm}$, and Output $L_4 = 90\text{ mm}$. Classify the mechanism according to Grashof's law when $L_1$ is fixed, and when $L_2$ is fixed.
2. A plate cam with a translating roller follower must provide a lift of $h = 25\text{ mm}$ during a cam rotation angle of $\beta = 90^\circ = \pi/2\text{ rad}$ using simple harmonic motion at $600\text{ rpm}$. Determine the maximum follower velocity and the minimum base circle radius $r_b$ required to limit the maximum pressure angle to $\phi_{\max} = 30^\circ$.

Step-by-Step Solution

Part 1: Four-Bar Linkage Grashof Classification

• Identify lengths: Shortest $S = L_2 = 40\text{ mm}$, Longest $L = L_1 = 120\text{ mm}$, Intermediate $P = L_4 = 90\text{ mm}$, $Q = L_3 = 110\text{ mm}$.
• Test Grashof inequality: $S + L = 40 + 120 = 160\text{ mm}$; $P + Q = 90 + 110 = 200\text{ mm}$.
• Since $S + L (160) \le P + Q (200)$, it is a Grashof Class I mechanism.
• When Frame $L_1$ (adjacent to $S$) is fixed: Crank-Rocker Mechanism ($L_2$ completes $360^\circ$ rotation as a crank, while $L_4$ oscillates as a rocker).
• When Link $L_2$ (the shortest link $S$) is fixed: Double-Crank / Drag-Link Mechanism (both $L_1$ and $L_3$ complete full $360^\circ$ continuous revolutions).

Part 2: Cam Follower Sizing

• Cam angular velocity: $\omega = \frac{2\pi (600)}{60} = 62.83\text{ rad/s}$.
• Cam rotation angle: $\beta = \frac{\pi}{2} = 1.5708\text{ rad}$.
• Maximum follower velocity for SHM (occurs at mid-stroke $\theta = \beta/2$ where $y = h/2 = 12.5\text{ mm}$): $v_{\max} = \frac{\pi}{2} \frac{h \omega}{\beta} = \frac{\pi}{2} \frac{(0.025\text{ m})(62.83\text{ rad/s})}{\pi/2\text{ rad}} = 0.025 \times 62.83 = \mathbf{1.571\text{ m/s}}$
• Maximum kinematic slope: $\left(\frac{dy}{d\theta}\right){\max} = \frac{v{\max}}{\omega} = \frac{1.571}{62.83} = 0.025\text{ m} = 25\text{ mm/rad}$.
• Sizing base circle for $\phi_{\max} \le 30^\circ$ (at mid-stroke $y = 12.5\text{ mm}$): $\tan(30^\circ) = \frac{(dy/d\theta)_{\max}}{r_b + y} \implies 0.57735 = \frac{25\text{ mm}}{r_b + 12.5\text{ mm}}$ $r_b + 12.5 = \frac{25}{0.57735} = 43.30\text{ mm} \implies r_b = 43.30 - 12.5 = \mathbf{30.8\text{ mm}}$ (Choose standard base circle radius $r_b \ge 32\text{ mm}$).

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8. Exam Tips & Common Traps

[!TIP] Rapid Grashof Memory Rules:

• Shortest link grounded = Double-Crank (Drag link).
• Shortest link adjacent to ground = Crank-Rocker.
• Shortest link coupler (opposite ground) = Double-Rocker.
• If $S+L > P+Q$ = Non-Grashof (Triple-Rocker).

[!WARNING] Common Pitfalls:

• Cam Angle in Radians: In cam equations, $\beta$ must always be in radians ($90^\circ = \pi/2\text{ rad}$). Using degrees will produce an error of $\times 57.3$.
• Pressure Angle Location: Maximum cam pressure angle does not occur at the peak lift ($y = h$), but rather near mid-stroke where kinematic velocity $dy/d\theta$ is maximized.