Engineering & Technology · Mechanisms, Machine Elements & Mechanical Design
Mechanisms, linkages and constrained motion
A rotating crank can make a slider move back and forth along a guide. The slider does not follow the crank pin’s circle, because the connecting rod and guide constrain its position. A mechanism model describes those relationships before it asks how hard an actuator must work.
Use Statics, Stress & Deformation; Dynamics, Vibration & Mechanical Response; and Materials Selection, Durability & Degradation. Retain force and moment balances, small elastic models, torque-speed power and units. The slider case supplies elementary trigonometry and a square-root geometry relation. All numbers, records, coefficients and efficiencies are constructed teaching inputs, not hardware selections, permissible loads, maintenance intervals or operating instructions.
Back to the chapterAfter reading, you can
Use joints and geometry to explain constrained motion without treating a position solution as a force calculation.
Name the fixed reference as well as the moving pieces
A link is treated as a rigid body for the selected motion model. A revolute joint permits relative rotation about its axis; a prismatic joint permits sliding along a specified direction. The ground link is the fixed reference connecting supports, even when it is not drawn as a separate bar. Joint types specify permitted motion, not actual bearing friction or clearance.
In a planar slider-crank, the ground, crank, connecting rod and slider form a closed chain. Once a compatible assembly branch and input crank angle are specified, the ideal geometry determines the slider position. One independently specified input describes this regular motion branch. Special alignments, interference or redundant constraints can invalidate a naive freedom count.
Turn the crank angle into two distances
A right triangle has one 90° corner. Pythagoras’ relation says that the two shorter sides a and b obey a²+b²=l², where l is the hypotenuse opposite that corner. It gives the missing side once the other two are known. A projection is the signed distance along an axis: the crank pin’s horizontal and vertical distances from the fixed pivot form such a triangle with the crank radius.
For an acute angle θ between the crank and the positive horizontal guide, cosine is adjacent side divided by hypotenuse and sine is opposite side divided by hypotenuse. Therefore the horizontal projection is r cosθ and the vertical projection is r sinθ. These ratios are dimensionless; multiplying them by the crank length gives distances. At 0° the pin is r to the right and has zero height; at 90° it has zero horizontal projection and height r. Beyond 90° on the upper half turn, the horizontal projection is negative because the pin is left of the pivot.
The 60° values can be built rather than memorised. Split an equilateral triangle of side 20 mm down the middle. The right triangle has hypotenuse 20 mm, horizontal adjacent side 10 mm and height √(20²−10²)=10√3 mm. Thus cos60°=1/2 and sin60°=√3/2≈0.866. A 20 mm crank at 60° puts its pin 10 mm right of the pivot and about 17.32 mm above the guide.
The connecting rod must reach from that elevated pin down to the slider. Its vertical span is the same 17.32 mm. With a 60 mm rod, Pythagoras gives its positive horizontal span √(60²−17.32²), using the exact height 10√3 before rounding. This is √3300≈57.45 mm. Add the crank’s 10 mm projection: the slider lies about 67.45 mm right of the pivot. The next equation packages those same two geometric steps for other compatible angles.
Close the geometry before differentiating motion
Let crank radius be r, rod length l, angle θ measured from the guide’s positive x direction, and slider coordinate x measured from the fixed crank pivot. For the in-line right-hand assembly, the crank pin is at(r cosθ, r sinθ). The rod’s horizontal projection is the positive square root of l²−r²sin²θ. Thus x=r cosθ+√(l²−r²sin²θ).
This expression is a right-triangle result, not a material law. We choose the positive projection because this slider lies to the right of the crank pin. Another assembly branch needs its corresponding sign and feasible geometry. Degrees are used in the worked trigonometric values; software must convert them if its functions expect radians.
Position, velocity and transmitted force need different evidence
The plotted position does not change by equal amounts for equal crank-angle increments. Consequently, a constant crank angular speed does not generally produce constant slider speed. At the two in-line extreme positions, the slider momentarily changes direction. A slider force along that line has no moment arm about the crank pivot exactly at the alignment.
Such dead-point behaviour does not mean a real mechanism supplies unlimited useful force. Compliance, friction, inertia and component limits become important. Kinematic feasibility is only the first check: load paths, joint reactions, deformation, available travel and collisions must also be considered. A paper motion model is useful evidence about geometry rather than approval to operate a machine.
Check a new geometry with the same reasoning
Keep the 20 mm crank but change the rod length to 80 mm in a separate constructed model. At 60°, the pin coordinates are unchanged, but the rod’s horizontal span is now √(80²−(10√3)²)=√6100≈78.10 mm. The slider position is about 88.10 mm. The input angle alone is not enough: the rod length is part of the geometry.
At 0° and 180° this new model gives positions 100 and 60 mm, so its stroke is still 40 mm. At 90° it gives √(80²−20²)≈77.46 mm, not the 80 mm midpoint of those endpoints. The crank radius determines this in-line full stroke, while the rod length changes positions within it. These comparisons concern a compatible rigid geometry only, without inferring its required drive torque or joint load.
Calculate a constructed half stroke
Take r=20 mm and l=60 mm. At θ=0°, x=20+60=80 mm. At 180°, x=−20+60=40 mm, so the full stroke is 40 mm. At 90°, x=√(3600−400)≈56.57 mm, which is not the midpoint 60 mm of the extreme positions.
At 60°, the crank projection is 10 mm and the rod projection√3300≈57.45 mm, giving x≈67.45 mm. The figure’s pose uses the same length scale for both links and the guide. Its position curve covers a half turn only; no speed, force or bearing capacity is inferred from those coordinates.
Try the reasoning
Why must the guide and fixed pivot be included in the model?
Show the answer and reasoning
They establish the constraints and reference geometry that determine the slider motion.
Does constant crank speed imply constant slider speed?
Show the answer and reasoning
No. The position-angle relationship is nonlinear, so its rate changes during the stroke.
Why specify an assembly branch?
Show the answer and reasoning
The geometric closure can have different configurations; the selected branch determines which position solution applies.
Why is the vertical crank-pin height at 60° not 20 mm, and what is it for the stated 20 mm crank?
Show the answer and reasoning
The 20 mm value is the full crank radius, the triangle hypotenuse. Height is its vertical projection: 20 sin60°=20×√3/2≈17.32 mm. It reaches 20 mm at 90°, not at every angle.
A proposed in-line geometry uses crank radius 30 mm and connecting rod length 20 mm. Can the right-hand branch reach the 90° pose?
Show the answer and reasoning
No. The pin would be 30 mm above the guide, farther than the rod’s entire 20 mm length. The expression l²−r²sin²θ becomes 400−900<0, so no real horizontal rod span exists at that angle. This is a geometric incompatibility, not a material-strength result.
What to carry forward
First establish compatible constrained motion, then add the force and deformation models needed for the task.
References
Teaching resources and research supporting this lesson’s explanations and examples.