Lesson 6: Force Analysis and Mechanism Synthesis

Identify the members. The main link carries a force along its own line (two-force member). The clamp arm then has three forces: the main-link push at one joint, the pad reaction at the workpiece, and the pivot reaction at . ✅

Draw the triangle. Choose a force scale and mark it (for example 1 cm = 20 N). Lay the known main-link force tip to tail with the pad-reaction direction; the pivot reaction closes the triangle. Measuring the sides gives the pad force and the pivot force. ✅

The over-centre amplification. Geometrically the ideal force ratio of the toggle is , where is the angle of the links from the collinear (dead-centre) line. The closer to centre, the larger the amplification. ✅

Mechanical advantage at the lock margin. Ideal geometry first, then the efficiency, or both in one step:

taking it in two stages, and . ✅

Clamping force:

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The closer the rest position is set to top-dead-centre, the higher this rises, the over-centre design from Lessons 1 and 3 seen as force.

Pin shear at a representative link/pin force of N (the internal forces exceed the pad force near the joints). Taking the conservative single-shear case:

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Pin safety factor, against the shear yield, not the tensile yield:

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Had the pin been in double shear the stress would halve to MPa and would double to . ✅

Link bending stress, with N·mm:

The same result through : mm⁴ and mm, so MPa. ✅

Link safety factor and the verdict:

Both parts clear the required , so the design is acceptable. The link in bending governs at , against for the pin, so the clamp would fail by the link bending before the pin ever sheared. Any weight saving should come off the pin, and any increase in clamping force is limited by the link. ✅

Open the simulator (siwit.co/TCM) and set the hand force, lock margin, efficiency, and the pin and link sizes. ✅

Confirm the clamping force rises sharply as the lock margin shrinks toward centre, and read the pin-shear and link-bending stresses with their pass/fail verdict against the allowable. They match the hand calculation. ✅

Measure as the angle between the coupler and follower at joint , taken as the value between and . At it is , an excellent transmission. ✅

Sweep the crank. Across a full turn rises to about and falls to about . The dips below the guide are the poor-force zones, where a large coupler force produces only a small useful drive on the follower. ✅

Design response. If those zones fall inside the working stroke, change the link lengths (Application 4 synthesis) or re-time the load so the heavy work happens where is large. ✅

Split the coupler force at the joint. With coupler force arriving at at transmission angle to the follower: ✅

The output torque is , so all of the angle’s effect lands on . ✅

Price the good instant and the bad one. At , and : almost the whole coupler force turns the follower. At the worst instant, and . For the same output torque you therefore need ✅

Rather more than twice the coupler force, and with it rather more than twice the pin and link stresses of Application 1, for no extra useful work. ✅

Note where the rest of it goes. At , , so nearly 90% of the coupler force is driven straight along the follower into its bearing. It does no work at all; it only wears the pin and deflects the link. That is what a poor transmission angle actually costs. ✅

Read the guide as a force statement. The rule is just : a promise that no more than about a third of the coupler force is wasted. This linkage spends about a quarter of every turn below that line, which is acceptable only if the working stroke avoids those zones. ✅

Set it up. Open the simulator (siwit.co/FBL) and set , , , . Transmission angle is already in degrees, so unlike the velocity and acceleration charts there is no conversion to do. ✅

Match the two readouts. The panel reports Min µ and Max µ directly: ✅

Locate the poor-force zones on the axis. The trace crosses the guide at and again at , so the linkage is below the guide from round through zero to : about 26% of every revolution, shaded in the figure above. If your working stroke lives in that band, the design is wrong regardless of how good the motion looks. ✅

Cross-check against mechanical advantage. The Minimum MA should bottom out at exactly the crank angles where dips, because both measure the same loss of leverage from opposite ends: from the geometry, MA from the force ratio. ✅

Draw the lift to scale at the angle of interest, say , choosing and marking a length scale (1 cm = 50 mm works for mm). ✅

Mark the two lengths that change. The platform height and the base spread the actuator works across: ✅

Note that they move in opposite senses. Raising the platform ( increasing) makes grow and shrink, so the actuator pulls the base pins together as the load goes up. That opposition is the whole reason a appears rather than a . ✅

Differentiate both lengths with respect to the scissor angle, giving the virtual displacements for a small rotation : ✅

Equate the work. Over a frictionless virtual displacement the actuator’s work equals the work done lifting the load. The actuator shortens while the load rises, so both are positive: ✅

Cancel and read the result. Both and drop out, which is why the answer depends on the angle alone and not on the size of the lift: ✅

This is for an actuator mounted horizontally across the base. A diagonal or pantograph placement changes the geometry factor, and the simulator reports the value for each type. ✅

Read the runaway. At , N; at , N; at , N. As the platform nears the floor, the mechanical advantage works against the actuator and the force spikes. ✅

Design response. This is why scissor lifts work over a limited low-angle band, use a diagonal or pantograph actuator placement to improve the low-angle advantage, and never start fully flat. ✅

Set it up and read three points off the trace. Open the simulator (siwit.co/SLM), set the platform load to N and the actuator type to Horizontal Base. Forces are already in newtons, so these are read directly with no conversion: ✅

The row is the one to anchor on: it is the angle where the actuator force exactly equals the load, so the mechanical advantage is one and the curve crosses its own reference. Everything below that angle costs you, everything above it pays. ✅

Watch Max Actuator Force follow the lowest angle. The panel’s Max Actuator Force is not a property of the lift; it is whatever the force reaches at the bottom of the stroke you allow. Raise the minimum height and it falls immediately, which is the cheapest design fix available here. ✅

Change the actuator placement. Switch the actuator type to Diagonal Base and watch the low-angle end of the curve drop: the geometry factor has changed, so no longer describes it, and the trade-off between the two ends of the stroke becomes visible. This is the mounting choice that makes low-angle work practical. ✅

Check the link stress. With the actuator force known, the arm stress follows as in Application 1. The lowest angles are the worst case for both force and stress at once, which is why a lift is rated from its lowest working height. ✅

Read the family off the task. A continuous input turning an oscillating output points to a crank-rocker four-bar; a straight-line output would point to a slider-crank; a clamping action would point to a toggle. ✅

Confirm one input drives it. The mobility check gives one degree of freedom for a four-bar, so a single motor on the crank fully determines the motion. ✅

Set out the frame. Choose and mark a length scale, say 1 cm = 20 mm. Draw the ground mm. ✅

Draw the rocker at both extremes. Swing the rocker mm about into its two limit positions and , separated by the required . Where that swing sits relative to the ground is your choice, not the specification’s; here the bisector is placed at to the ground line. ✅

Measure the two distances from the crank centre to each rocker extreme: ✅

Solve the pair. Adding and subtracting the two measurements separates the unknowns: ✅

Grashof, so the crank can actually turn all the way round. Shortest plus longest against the other two: . It passes, and since the shortest link is the crank and it is adjacent to the ground, the linkage is a crank-rocker, which is the family asked for. ✅

Swing, to confirm the construction did its job. Running the position analysis over a full turn gives a rocker oscillation of , exactly the target. The motion requirement is met. ✅

Transmission angle, and here it fails. These are the very link lengths analysed in Application 2, where was found to dip to and to spend about a quarter of every turn below the guide. By the force reading of Step 2 there, that means stretches of the cycle where nearly of the coupler force is wasted into the bearings. ❌

This is the ordinary outcome of a first synthesis, not a mistake in the construction. The limit-position method controls the motion and says nothing whatever about force quality. ✅

Identify what is actually free. The ground is fixed by the mounting and the swing is the requirement. That leaves the rocker length and the placement of the swing, and both feed back into and through the same construction. ✅

Shorten the rocker and re-place the swing. Repeating Step 2 with mm and the bisector moved to gives ✅

Re-run every check. Grashof still passes (), still a crank-rocker. The rocker swing comes out at , within a fifth of a degree of the target. And the transmission angle now has a minimum of : ✅

The linkage no longer drops below the guide at any point in the cycle, against of the cycle before. Same motion, far better force. ✅

Recognise the loop. Nothing here was solved in one pass. Synthesis is propose, analyse, refine, repeated until every check clears, and the analysis tools of the previous five lessons are precisely the tests being run. ✅

Check the swing without trusting your drawing. Enter the refined lengths in the simulator (siwit.co/FBL) and read the Oscillation result: it reports the rocker’s total swing directly, and should show about . That is the synthesis target verified by an independent route. ✅

Compare both candidates side by side. Put in , , , and note the minimum transmission angle, then the refined , , , . The reported minimum should rise from about to about , and the transmission-angle trace should stop crossing the guide. ✅

Run the full sequence on the survivor. Mobility, position, velocity and mechanical advantage, acceleration and inertia loads, then joint forces and pin and link stresses as in Application 1. A design is finished only when all of them pass. ✅

Take it to CAD. The surviving link lengths and limit positions become the sketch constraints of a real part in the Parametric Mechanical CAD with FreeCAD course. ✅