A rotating cam can close a pivoted shear by lifting its rear arm. The historical arrangement below separates the fixed lower blade from the moving upper blade. Use the calculator to estimate cutting force for a solid round section and compare cycle timing.
Cam-operated Shears Interactive Calculator
Enter solid-wire diameter, shear strength, an allowance factor and stroke rate. The results estimate cutting-edge force and cycle timing. The animation shows an illustrative pivoted shear with a cam that remains tangent to its rear contact face.
Equation Used
- Solid round section, single shear plane and entered uniform shear strength.
- The illustrative fixed blade and moving blade pass in adjacent planes; their silhouettes overlap in the side view.
- One cam revolution per stroke; actual rate is calculated while playback is slowed.
- Tangency is solved for a circular eccentric and straight rear follower face. Return contact is assumed rather than dynamically proved.
- Feed, material fracture, ejection, blade clearance, structural compliance and cam-drive torque are outside the model.
The cam lifts the rear arm of the pivoted shear to close the cutting jaw. Results estimate force at the cutting edge from the entered round-section area and shear strength. The displayed 20% cycle interval is an illustrative timing assumption, not a predicted cutting duration.
A cam closes a pivoted shear
Hiscox figure 1092 shows a cam beneath the long rear arm of a pivoted blade. As the cam lifts that arm, the short upper jaw closes toward a fixed lower jaw. The pivot and lower body remain attached to the frame.
The drawing uses a circular eccentric as an illustrative cam. Its offset centre travels around a fixed shaft, and the rear blade face remains tangent to the circle. That contact determines the blade angle. The cutting edges pass in adjacent planes.
Exploring a mechanically timed cut
The arrangement demonstrates how a rotary input can produce a repeated shear movement. The calculator addresses a solid round workpiece. Hollow tubing, stranded cable, broad sheet and progressive cuts require a model appropriate to their actual section and cutting process.
Cutting force is not cam force
For a solid circular section of diameter d, area A=πd²/4. The simple estimate F=Aτk uses entered shear strength τ and allowance factor k. With millimetres and MPa, F is in newtons. Divide by 1000 for kN, or multiply by 0.2248089431 for lbf.
This is a cutting-edge force estimate. A pivoted blade changes the relationship between that force and the force at the cam. Actual cam load also depends on the moment arms, contact normal, friction, inertia and other loads.
Cycle time is 60000/SPM milliseconds. The timing tile shows exactly 20% of that period as an illustrative interval, not a prediction of cut duration.
A solid round-section example
For d=3 mm, τ=220 MPa and k=1.3, area is approximately 7.07 mm² and estimated cutting force is 2.02 kN, or about 454 lbf. At 800 strokes/min, the period is 75 ms and its 20% interval is 15 ms.
Doubling diameter quadruples the area and this force estimate. Doubling stroke rate halves the period; it does not change the static area-based cutting-force estimate. The animation does not calculate the additional dynamic loading of faster operation.
Interpreting the motion
The cam and blade are constrained geometrically, but their illustrative sizes do not define a production machine. The wire cross-section changes with diameter and the moving edge crosses it. Material failure and removal of the cut end are not simulated.
A complete design would need real lever dimensions, cam contact loading, blade geometry and clearance, return-force analysis, drive torque and structural checks. The displayed force should not be used as an actuator or bearing rating by itself.
Questions about the shear
Why does lifting the rear arm close the blade?
The rear arm and upper blade are one rigid member on opposite sides of the pivot, so their ends move in opposite directions.
Is the displayed force applied at the cam?
No. It estimates force at the cutting edge. Cam force depends on the leverage and contact geometry.
Why can the blade silhouettes overlap?
The cutting blades pass in adjacent planes. A side view projects those planes together.
Is the 20% interval the predicted cutting time?
No. It is an illustrative fraction of the cycle, not a fracture model.
Construction reference
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, page 270, figure 1092. The illustration shows cam-operated pivoted shears. The circular eccentric, numerical cutting-force example and timing assumption here are explicitly stated additions.
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