Four-bar and slider-crank mechanisms and inversions

Four-bar and slider-crank chains: Grashof's law, inversions, transmission angle, toggle, slider displacement and quick-return ratio, with worked numericals.

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Why it matters

The four-bar chain and the slider-crank chain are the two linkages behind most single-input machines: engine crank trains, pumps, compressors, shaper quick-return drives, wiper drives, presses and many grippers. Knowing which link to fix, whether the input can rotate fully, and how well force is transmitted lets you choose a linkage before you size a motor or actuator.

Key ideas

Four-bar chain. Four binary links joined by four revolute pairs: frame (d), input or crank (a), coupler (b) and output or follower (c). Mobility F = 3(4 − 1) − 2(4) = 1.

Grashof's law. Let s = shortest, l = longest, p and q = the other two links.

  • If s + l ≤ p + q (Grashof chain), at least one link can make a full revolution relative to the others. The inversion depends on which link is fixed:
    • fix a link adjacent to the shortest → crank-rocker (shortest link is the crank);
    • fix the shortest link → double crank (drag link); both side links rotate fully;
    • fix the link opposite the shortest → double rocker; only the coupler rotates fully.
  • If s + l > p + q (non-Grashof), no link rotates fully; every inversion is a triple rocker (double rocker).
  • s + l = p + q is the change-point case: the linkage can pass through a flattened position where all links are collinear and the output can switch branch. A parallelogram linkage is the best-known example.

Inversions of the four-bar chain (examples). Beam engine and windscreen-wiper drive (crank-rocker); locomotive coupling rod (parallelogram double crank); Watt's indicator mechanism and Ackermann steering (double rocker).

Transmission angle (μ). The angle between the coupler and the output link. Only the component of coupler force perpendicular to the output link produces torque, so force transmission is good when μ is near 90° and poor when it falls below about 40°. Designers check μ_min, which occurs when the crank is collinear with the frame.

Toggle and mechanical advantage. Mechanical advantage of a four-bar = output torque / input torque = ω_input / ω_output (ideal, no friction). At a toggle position the input link and coupler are collinear, ω_output → 0 and the mechanical advantage → ∞. Stone crushers, clamps and toggle presses exploit this. When the coupler and output link are collinear (μ = 0°) the linkage is at a dead centre and cannot be driven from the output side.

Slider-crank chain (single slider-crank). A four-bar chain in which one revolute pair is replaced by a sliding pair. Links: frame and guide (1), crank (2), connecting rod (3), slider (4). Inversions:

  • fix 1 → reciprocating engine and compressor;
  • fix 2 (crank) → Whitworth quick-return mechanism, rotary (Gnome) engine;
  • fix 3 (connecting rod) → oscillating-cylinder engine, crank and slotted-lever quick-return mechanism;
  • fix 4 (slider) → hand pump (pendulum pump).

Double slider-crank chain. Two sliding and two turning pairs. Inversions: elliptical trammel, Scotch yoke (pure simple-harmonic output) and Oldham's coupling (connects parallel shafts with a small offset).

Quick-return action. In a shaper the cutting stroke should be slow and the idle return fast. In the crank and slotted-lever mechanism the crank turns through a larger angle for the cutting stroke than for the return, at constant crank speed.

Formulas

s + l ≤ p + q (Grashof condition for full rotation of at least one link)

cos μ = (b² + c² − a² − d² + 2ad·cos θ) / (2bc)

  • a = crank, b = coupler, c = output link, d = frame (m); θ = crank angle measured from the frame line at the crank pivot; μ = transmission angle. Extremes at θ = 0° and 180°.

x = r(1 − cos θ) + l − √(l² − r²·sin² θ) (exact slider displacement, in-line slider-crank) x ≈ r[(1 − cos θ) + sin² θ / (2n)] (approximate), with n = l / r

  • r = crank radius (m), l = connecting rod length (m), θ = crank angle from the dead centre where the slider is farthest from the crank axis. Stroke = 2r, independent of l.

cos(α/2) = r / d (crank and slotted lever; α = crank angle for the return stroke) Time ratio = (360° − α) / α Stroke = 2·L·(r / d)

  • r = crank radius, d = distance between the crank centre and the slotted-lever pivot, L = slotted-lever length (all in m).

Worked examples

Example 1 (standard): classify a four-bar and find its worst transmission angle. Given: crank a = 40 mm, coupler b = 120 mm, rocker c = 80 mm, fixed link d = 100 mm.

  1. s = 40, l = 120, p + q = 80 + 100 = 180 mm. s + l = 160 mm ≤ 180 mm → Grashof.
  2. The shortest link (40 mm) is adjacent to the fixed link → crank-rocker.
  3. At θ = 0°: cos μ = (b² + c² − (d − a)²) / (2bc) = (14 400 + 6 400 − 3 600) / 19 200 = 0.8958 → μ = 26.4°.
  4. At θ = 180°: cos μ = (b² + c² − (d + a)²) / (2bc) = (20 800 − 19 600) / 19 200 = 0.0625 → μ = 86.4°. Answer: crank-rocker, μ_min ≈ 26.4° (below the usual 40° guide, so lengthen the rocker or shorten the frame before using it for heavy loads).

Example 2 (GATE level): crank and slotted-lever quick return. Given: distance between crank centre and lever pivot d = 300 mm, crank radius r = 150 mm, slotted lever length L = 600 mm.

  1. cos(α/2) = r / d = 150 / 300 = 0.5 → α/2 = 60°, α = 120° (return stroke).
  2. Cutting stroke angle = 360° − 120° = 240°.
  3. Time ratio = 240 / 120 = 2
  4. Stroke = 2·L·(r / d) = 2 × 600 × 0.5 = 600 mm Answer: time ratio = 2.0 (cutting takes twice as long as return), stroke = 600 mm.

Example 3 (check of the slider formula). r = 0.1 m, l = 0.4 m, θ = 60°. Exact: x = 0.1(1 − 0.5) + 0.4 − √(0.16 − 0.01 × 0.75) = 0.05 + 0.4 − 0.39051 = 0.05949 m. Approximate (n = 4): x = 0.1(0.5 + 0.75/8) = 0.05938 m. Answer: x ≈ 59.5 mm; the approximation is within 0.2%.

Common mistakes

  • Concluding "crank-rocker" from s + l ≤ p + q alone. The inversion depends on which link is fixed.
  • Writing stroke = 2l or saying a longer connecting rod gives a longer stroke. Stroke is always 2r; a longer rod only reduces obliquity and side thrust.
  • Using arc length r·θ for slider displacement. The slider moves along a straight line; use the displacement equation.
  • Taking the cutting-stroke angle as the smaller one in a quick-return mechanism. The slow (cutting) stroke takes the larger crank angle.
  • Mixing up toggle (input and coupler collinear, infinite mechanical advantage) with dead centre (coupler and output collinear, cannot drive from the output).
  • Calling the Scotch yoke an inversion of the single slider-crank; it belongs to the double slider-crank chain.

For GATE ME

Typical questions: classify a four-bar from given lengths and the fixed link; match mechanisms to their parent chain and inverted link; compute the quick-return ratio or stroke of a crank and slotted-lever or Whitworth mechanism; find slider displacement or the transmission angle at a given crank angle. Practise Grashof checks with all four fixed-link choices and the quick-return geometry.

Quick check

  1. Lengths 30, 70, 90, 100 mm with the 30 mm link fixed. What is the mechanism?
  2. Which slider-crank inversion is the hand pump?
  3. What is the stroke of a slider-crank with r = 60 mm and l = 250 mm?
  4. A crank and slotted-lever mechanism has a return angle of 150°. What is the time ratio?
  5. At what transmission angle is force transmission best?

Answers: 1. 30 + 100 = 130 ≤ 160, shortest fixed → double crank. 2. Slider fixed. 3. 120 mm. 4. 210/150 = 1.4. 5. 90°.

Interview questions

All Theory of Machines and Vibrations interview questions

Try answering each one aloud before you open it.

  1. 1.What is a four-bar mechanism, and where is it commonly used?Concept

    A four-bar mechanism is four rigid links (frame, input crank, coupler, output link) joined in a closed loop by four revolute pairs, giving one degree of freedom. Depending on the Grashof condition and which link is fixed, it converts continuous rotation into oscillation (crank-rocker), rotation into rotation (double crank) or oscillation into oscillation (double rocker). Typical uses are windscreen-wiper drives, beam engines, locomotive coupling rods, Ackermann steering, toggle clamps and parallel-jaw grippers, and coupler-point curves are used for path generation.

  2. 2.Explain the working principle of a slider-crank mechanism.Concept

    A slider-crank mechanism is a mechanical linkage that converts rotational motion into linear motion or vice versa. It consists of a crank, a connecting rod, and a slider. As the crank rotates, it moves the connecting rod, which in turn moves the slider back and forth in a linear path. This mechanism is commonly used in engines and pumps.

  3. 3.What are inversions in the context of four-bar mechanisms?Concept

    An inversion is the mechanism obtained by fixing a different link of the same kinematic chain; relative motions between links do not change, but the absolute motions, and so the application, do. For a Grashof four-bar (s + l ≤ p + q), fixing a link adjacent to the shortest gives a crank-rocker, fixing the shortest link gives a double crank, and fixing the link opposite the shortest gives a double rocker. A slider-crank chain has four inversions too: reciprocating engine, Whitworth quick return or rotary engine, oscillating-cylinder engine or slotted-lever quick return, and the hand pump.

  4. 4.Why is a slider-crank mechanism preferred in internal combustion engines?Application

    A slider-crank mechanism is preferred in internal combustion engines because it efficiently converts the reciprocating motion of the pistons into rotational motion of the crankshaft. This conversion is essential for transferring the energy generated by combustion into usable mechanical work to drive the vehicle.

  5. 5.What happens if the length of the connecting rod in a slider-crank mechanism is increased?Application

    The stroke does not change: it is always twice the crank radius, 2r. A longer connecting rod (larger n = l/r) reduces the obliquity of the rod, so the side thrust of the piston on the cylinder wall and the secondary (2ω) component of piston acceleration and inertia force both fall, and the piston motion is closer to simple harmonic. The cost is a taller engine and a heavier rod.

  6. 6.How does the Grashof's law apply to four-bar mechanisms?Concept

    Grashof's law states that if s + l ≤ p + q, where s and l are the shortest and longest link lengths and p and q are the other two, then at least one link can make a full revolution relative to the others. Fixing a link adjacent to the shortest gives a crank-rocker, fixing the shortest gives a double crank, and fixing the link opposite the shortest gives a double rocker. If s + l > p + q no link can rotate fully and every inversion is a double rocker; s + l = p + q is the change-point case.

  7. 7.In a slider-crank mechanism the crank is 0.1 m, the connecting rod 0.5 m, and the slider velocity is 2 m/s when the crank is at 90° to the line of stroke. Find the crank's angular velocity.Numerical

    Slider velocity is v = ω·r·(sin θ + sin 2θ / (2n)) with n = l/r = 5. At θ = 90°, sin 2θ = 0, so v = ω·r exactly. Then ω = v / r = 2 / 0.1 = 20 rad/s (about 191 rpm). At any other crank angle the connecting rod length enters the answer, so v = ω·r is not a general relation.

  8. 8.What is the effect of increasing the crank length in a four-bar mechanism?Application

    Changing the crank length changes s + l relative to p + q, so it can change whether the crank can rotate fully at all (the Grashof condition). Within a crank-rocker, a longer crank increases the rocker's swing angle but lowers the minimum transmission angle, so force transmission near the extreme positions worsens and the joint loads rise. Designers usually keep the transmission angle above about 40° and adjust the crank length to get the swing they need within that limit.

  9. 9.Explain the concept of a double-crank mechanism and its applications.Concept

    A double-crank (drag-link) mechanism is a four-bar in which both links pivoted to the frame make full revolutions. It requires the Grashof condition s + l ≤ p + q with the shortest link fixed. With equal opposite links it becomes a parallelogram linkage that transmits uniform rotation, as in locomotive coupling rods; the general drag link gives a non-uniform output speed and is used to drive slow-advance, quick-return feeds, for example in shaping and packaging machines.

  10. 10.How is the mechanical advantage of a four-bar linkage defined, and when does it become very large?Numerical

    For an ideal four-bar, power in equals power out, so mechanical advantage = output torque / input torque = ω_input / ω_output; it changes with position and cannot be found from link lengths alone. When the input link and the coupler become collinear (toggle position) the output angular velocity becomes zero and the mechanical advantage tends to infinity. Toggle clamps, stone crushers and riveting presses use this to get a large force from a small input.

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