Four-bar and slider-crank mechanisms and inversions

Grashof's law, the inversions of the four-bar, single and double slider-crank chains, transmission angle, quick-return ratio and piston kinematics with worked numbers.

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Why it matters

The four-bar chain and the slider-crank chain are the two linkages you will meet most often in a vehicle: the slider-crank turns piston force into crankshaft torque, and four-bar linkages drive windscreen wipers, steering (the Ackermann trapezium), suspension arms and bonnet hinges. Knowing which link lengths give a full-rotation crank, and which mechanism you get by fixing each link, lets you pick a linkage for a job before doing any detailed analysis.

Key ideas

Four-bar chain. Four links (frame, input, coupler, output) joined by four revolute pairs; n = 4, j = 4, so F = 1. Name the lengths s (shortest), l (longest), p and q (the other two).

Grashof's law. If s + l ≤ p + q, at least one link can make a complete revolution relative to the others (a Grashof chain). If s + l > p + q, no link can rotate fully (non-Grashof: every inversion is a triple rocker).

Inversions of a Grashof chain (s + l < p + q):

  • Fix a link adjacent to the shortest link: crank-rocker (the shortest link is the crank, the opposite link rocks).
  • Fix the shortest link: double-crank (drag-link); both links pivoted on the frame rotate fully.
  • Fix the link opposite the shortest: double-rocker; only the coupler (the shortest link) rotates fully.

Change-point (s + l = p + q). The chain can rotate, but in some position all four links become collinear, where the output can go either way. Parallelogram and kite (deltoid) linkages are special cases; a locomotive coupling rod is a parallelogram linkage.

Transmission angle μ. The angle between the coupler and the output link. Force transmission is best at μ = 90°; designers keep μ between about 40° and 140°. In a crank-rocker the extreme values occur when the crank is collinear with the frame. When μ reaches 0° or 180° the mechanism is at a toggle (dead-centre) position: a small input force can hold a large output load (toggle clamps, stone crushers).

Slider-crank chain. A four-bar chain in which one revolute pair has been replaced by a sliding pair (the rocker becomes infinitely long). Links: 1 frame (cylinder), 2 crank, 3 connecting rod, 4 slider (piston). Stroke = 2r, independent of the connecting-rod length; the rod length only affects obliquity.

Inversions of the single slider-crank chain.

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

Double slider-crank chain (two sliding and two turning pairs): elliptical trammel, Scotch yoke (gives exact simple harmonic slider motion), Oldham's coupling (joins parallel shafts with a small offset at the same speed).

Quick-return mechanisms. In a crank and slotted-lever mechanism the crank turns at constant speed but sweeps unequal angles during the cutting and return strokes, so the return is faster. The extreme lever positions occur when the crank is perpendicular to the slotted lever.

Formulas

s + l ≤ p + q (Grashof's condition)

  • s, l: shortest and longest link lengths (m); p, q: the other two (m). Applies to four-revolute chains.

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

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

cos(α/2) = r / c and time ratio = (360° − α) / α (crank and slotted lever)

  • α: crank angle turned during the return stroke (°); r: crank radius (m); c: distance between the crank centre and the slotted-lever pivot (m). Ratio is cutting time / return time.

stroke = 2R·r / c

  • R: length of the slotted lever from its pivot to the ram connection (m); valid when the ram line is perpendicular to the line of centres at its extreme positions.

x = r(1 − cos θ) + (r²/2l)·sin²θ (approximate piston displacement)

  • x: piston displacement from the dead centre at θ = 0 (m); r: crank radius (m); l: connecting-rod length (m); θ: crank angle from that dead centre. Good for n = l/r of about 3.5 or more.

v_P = ω·r·(sin θ + sin 2θ / (2n)) and a_P = ω²·r·(cos θ + cos 2θ / n)

  • v_P: piston velocity (m/s); a_P: piston acceleration (m/s²), positive towards the crank centre at θ = 0; ω: crank angular speed (rad/s), ω = 2πN/60 with N in rpm; n = l/r. Approximate forms (series truncated); error is about 1% for n = 4.

Worked examples

Example 1 (standard: Grashof and inversions). A four-bar chain has links of 40, 80, 100 and 120 mm. Classify it and name the mechanism obtained when each link is fixed.

  1. s = 40 mm, l = 120 mm, p + q = 80 + 100 = 180 mm.
  2. s + l = 160 mm < 180 mm, so the chain is Grashof.
  3. Fix the 40 mm link: double-crank. Fix the 80 mm or 120 mm link (both adjacent to the 40 mm link if the order is 40–80–100–120 around the loop): crank-rocker. Fix the 100 mm link (opposite the 40 mm link): double-rocker.

Answer: Grashof chain; double-crank, crank-rocker, crank-rocker, double-rocker.

Example 2 (transmission angle). In the chain above, the 80 mm link is the frame (d = 80 mm), the 40 mm link is the crank (a), the 120 mm link is the coupler (b) and the 100 mm link is the rocker (c). Find the range of transmission angle.

  1. At θ = 0°: cos μ = (120² + 100² − 40² − 80² + 2·40·80·1) / (2·120·100) = (14400 + 10000 − 1600 − 6400 + 6400) / 24000 = 22800 / 24000 = 0.950, so μ_min = 18.2°.
  2. At θ = 180°: cos μ = (14400 + 10000 − 1600 − 6400 − 6400) / 24000 = 10000 / 24000 = 0.4167, so μ_max = 65.4°.

Answer: μ ranges from 18.2° to 65.4°. The minimum is well below 40°, so this linkage would transmit force poorly near θ = 0°.

Example 3 (quick return). In a crank and slotted-lever shaper the distance between fixed centres is 300 mm and the crank radius is 150 mm. The slotted lever is 600 mm long. Find the time ratio and the stroke.

  1. cos(α/2) = r / c = 150 / 300 = 0.5, so α/2 = 60° and α = 120° (return stroke).
  2. Cutting stroke angle = 360° − 120° = 240°.
  3. Time ratio = 240 / 120 = 2.0.
  4. Stroke = 2R·r / c = 2 × 600 × 150 / 300 = 600 mm.

Example 4 (GATE level: piston kinematics). An engine has crank radius 50 mm and connecting rod 200 mm, and the crank turns at ω = 100 rad/s. Find the piston velocity and acceleration at θ = 30° from inner dead centre.

  1. n = l / r = 200 / 50 = 4.
  2. v_P = ω·r·(sin θ + sin 2θ / 2n) = 100 × 0.05 × (0.5 + 0.8660 / 8) = 5 × (0.5 + 0.1083) = 3.04 m/s.
  3. a_P = ω²·r·(cos θ + cos 2θ / n) = 100² × 0.05 × (0.8660 + 0.5 / 4) = 500 × 0.9910 = 495.5 m/s².

An exact calculation gives 3.05 m/s and 497.5 m/s²: the approximate formulas are within 0.5%.

Common mistakes

  • Applying Grashof's condition to the links in the order given (L1 + L2 ≤ L3 + L4) instead of shortest plus longest against the other two.
  • Saying that fixing the shortest link gives a crank-rocker. It gives a double-crank; a crank-rocker comes from fixing a link adjacent to the shortest.
  • Thinking a longer connecting rod gives a longer stroke. Stroke is 2r; a longer rod only reduces obliquity, side thrust and the second-harmonic (cos 2θ / n) part of acceleration.
  • Taking the quick-return ratio as α / (360° − α) with α the cutting angle mixed up with the return angle. The larger angle always belongs to the slow (cutting) stroke.
  • Using v = ωr for the piston at any angle. It is exact only at θ = 90° (where sin 2θ = 0).

For GATE ME

Typical items: classify a four-bar chain by Grashof's law and identify the mechanism for a given fixed link; match slider-crank and double-slider inversions to their applications; compute the quick-return ratio or stroke of a shaper mechanism; find piston velocity or acceleration at a crank angle; find a transmission angle. Practise drawing the extreme positions of a crank-rocker and slotted lever, since most numerical traps come from the geometry of those positions.

Quick check

  1. Links of 30, 50, 60 and 70 mm: is the chain Grashof?
  2. Which link is fixed to get a double-crank from a Grashof chain?
  3. Which inversion of the slider-crank chain is the hand pump?
  4. Does increasing the connecting-rod length change the stroke?
  5. A slotted-lever mechanism has a return angle of 120°. What is the time ratio?

Answers: 1. Yes, 30 + 70 = 100 < 50 + 60 = 110. 2. The shortest link. 3. The one with the slider fixed. 4. No, stroke = 2r. 5. 240° / 120° = 2.

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 links (frame, input crank, coupler, output) joined in a closed loop by four revolute pairs; it has one degree of freedom, so one input gives a definite output. Depending on link lengths (Grashof's law) and which link is fixed, it works as a crank-rocker, double-crank or double-rocker. In vehicles it appears in windscreen wipers, the steering trapezium, double-wishbone suspension and bonnet and boot hinges.

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

    A slider-crank mechanism has a frame (cylinder), a crank, a connecting rod and a slider (piston), with three turning pairs and one sliding pair. As the crank rotates the rod pushes the slider back and forth through a stroke of twice the crank radius; in an engine the gas force on the piston drives the crank instead. The piston motion is close to simple harmonic but not exactly, because of connecting-rod obliquity, which adds a second-harmonic term of relative size r/l.

  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; the relative motion between links stays the same, only the absolute motions change. For a Grashof four-bar chain, fixing a link adjacent to the shortest link gives a crank-rocker, fixing the shortest link gives a double-crank (drag link), and fixing the link opposite the shortest gives a double-rocker. A non-Grashof chain gives a triple rocker whichever link is fixed.

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

    The slider-crank turns the reciprocating gas force on the piston into continuous torque on the crankshaft with only pin and sliding joints, which are simple, compact and can carry very high loads with good lubrication. The piston runs in a cylinder that also forms the combustion chamber, so sealing is straightforward. Its drawbacks, the fluctuating torque and the unbalanced reciprocating inertia forces, are handled with a flywheel, multiple cylinders and balancing.

  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. A longer connecting rod (larger l/r) reduces the obliquity of the rod, so the side thrust of the piston on the cylinder wall and the secondary (cos 2θ / n) part of piston acceleration and inertia force become smaller, and piston motion gets closer to simple harmonic. The cost is a taller, heavier engine, which is why automotive engines use l/r of roughly 3 to 4.

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

    Grashof's law states that for a four-bar linkage, if the sum of the shortest and longest link lengths is less than or equal to the sum of the other two link lengths, at least one link will be able to make a full rotation. This law helps in determining the type of motion (crank-rocker, double-crank, or double-rocker) that the mechanism will exhibit.

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

    Piston velocity is v = ω·r·(sin θ + sin 2θ / 2n) with n = l/r. At θ = 90° sin 2θ = 0, so v = ω·r exactly and ω = v / r = 2 / 0.1 = 20 rad/s. At any other crank angle you cannot use v = ωr; the obliquity term changes the result, which is why the crank angle must be given.

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

    Changing the crank length changes the Grashof sum s + l, so a crank that is made too long may no longer be able to rotate fully and the chain becomes a double- or triple-rocker. Within a crank-rocker, a longer crank increases the swing angle of the rocker but usually narrows the range of the transmission angle, so force transmission gets worse near the extreme positions. Designers therefore choose the crank length to get the required swing while keeping the transmission angle roughly between 40° and 140°.

  9. 9.Explain the difference between a crank-rocker and a double-crank mechanism.Concept

    In a crank-rocker one link pivoted on the frame (the crank, which is the shortest link) rotates fully while the other frame-pivoted link only oscillates; it is obtained by fixing a link adjacent to the shortest one. In a double-crank (drag link) both frame-pivoted links rotate fully; it is obtained by fixing the shortest link. Both require the chain to satisfy Grashof's condition s + l ≤ p + q.

  10. 10.Determine the type of motion in a four-bar linkage with link lengths: 0.2 m, 0.3 m, 0.4 m, and 0.5 m.Numerical

    Shortest plus longest is 0.2 + 0.5 = 0.7 m and the other two sum to 0.3 + 0.4 = 0.7 m, so s + l = p + q: a change-point (borderline Grashof) linkage. Links can rotate fully, but at some position all four links become collinear and the output motion is indeterminate there, so in practice it needs a flywheel or guide to carry it through. With a link adjacent to the 0.2 m link fixed it behaves as a crank-rocker apart from that change point.

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