Clutch Axial Force (Uniform Pressure)
Calculate the axial clamping force of a disc clutch or brake by the uniform-pressure assumption, F = p·(π/4)·(D² − d²), from the contact pressure p (Pa) and the outer D and inner d diameters (m) of the friction annulus. The axial force clamps the discs together (applied by springs in normally-engaged clutches, or by a hydraulic/pneumatic actuator). By the UNIFORM-PRESSURE assumption (valid for new discs, before wear), the force is simply the average contact pressure times the AREA of the friction annulus (the ring between outer and inner diameters). This force is the clutch/brake actuation parameter: it determines the transmissible torque (with friction and mean radius) and must be limited so the contact pressure does not exceed the friction material's allowable (which has a limit, above which it degrades, loses friction by overheating — fading — or wears fast). Design balances: enough axial force for the needed torque, but pressure within the material limit (setting the minimum area and disc count). Enter the contact pressure and the outer and inner diameters.
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Clutch axial force (uniform pressure)
The axial clamping force of a clutch or disc brake, under the uniform pressure assumption, is F = p·(π/4)·(D² − d²), from the contact pressure p, the outer diameter D and the inner diameter d of the friction annulus. The axial force is the force that squeezes the discs against one another (applied by springs in normally engaged clutches, or by a hydraulic or pneumatic actuator). Under the uniform pressure assumption (valid for new discs, before wear sets in), the force is simply the mean contact pressure times the area of the friction annulus (the circular ring between the outer and inner diameters). This force is the actuation parameter of the clutch or brake: it sets the transmissible torque (together with the friction coefficient and the mean radius) and has to be limited so that contact pressure stays below what the friction material allows (there is a ceiling above which the lining degrades, loses grip through overheating — fading — or wears out fast). Sizing balances the two demands: axial force high enough for the required torque, yet pressure within the material limit (which fixes the minimum area and the number of discs). Enter the contact pressure and the outer and inner diameters.
Related Tools
Clutch Max Pressure (Uniform Wear)
Calculate the maximum contact pressure in a disc clutch or brake by the uniform-wear assumption, p_max = 2·F ÷ (π·d·(D − d)), from the axial force F (N) and the outer D and inner d diameters (m); the result is in kPa. Under UNIFORM WEAR (the steady state, after disc run-in), pressure is NOT constant over the friction annulus: it is INVERSELY proportional to radius (highest at the inner radius, lowest at the outer), because wear — proportional to pressure × velocity — only becomes uniform if pressure falls with radius (since velocity grows with radius). So the MAXIMUM pressure occurs at the INNER radius (at diameter d), and that peak limits the design. The maximum pressure must be below the friction material's allowable (linings, pads, ceramic or sintered metallic materials — each with its limit, typically hundreds of kPa to a few MPa). Exceeding the allowable leads to accelerated wear, overheating and friction loss (fading). This calculation checks whether a clutch/brake, under the planned actuation force, operates within its material's pressure limit — an essential durability and safety check. Enter the axial force and the outer and inner diameters.
Mean Friction Radius (Clutch)
Calculate the mean friction radius of a disc clutch or brake by uniform-wear theory, r_m = (D + d) ÷ 4, from the outer D and inner d diameters (m) of the friction annulus. The mean radius is the EFFECTIVE radius at which the resultant friction force is taken to act for torque calculation (T = μ·F·N·r_m). There are two classic assumptions for this radius: UNIFORM WEAR (assuming the disc has 'bedded in' and wears evenly, concentrating pressure at the inner radius; gives r_m = (D+d)/4, the simple mean of radii) and UNIFORM PRESSURE (new disc, constant pressure; gives r_m = (2/3)·(D³−d³)/(D²−d²), slightly larger). Uniform wear is most used in DESIGN, being conservative (slightly lower torque) and representing the run-in steady state. The mean radius shows an interesting design point: discs with a narrow friction annulus (D close to d, thin ring at large radius) have a high mean radius, transmitting more torque per unit force — so high-performance disc brakes use calipers acting near the disc edge (large radius). Enter the outer and inner diameters.
Disc Clutch Torque
Calculate the torque transmissible by a disc clutch (or brake), T = μ·F·N·r_m, from the friction coefficient μ, the axial clamping force F (N), the number of friction surfaces N and the mean friction radius r_m (m). A disc clutch transmits torque between two shafts by FRICTION between surfaces pressed together: an axial force F clamps the discs, and the friction at that interface, acting at the mean radius, generates the torque. The number of friction surfaces N multiplies the capacity — a single disc clutch has N=1 (one face) or N=2 (disc between two faces); MULTI-PLATE clutches (motorcycles, automatic transmissions) stack several discs with high N, transmitting large torque in compact space. The same principle applies to disc and clutch BRAKES: the braking (or transmitting) torque is μ·F·N·r_m. This is central in clutch and brake design: it sets the actuation force (pedal, spring, hydraulic actuator) needed to transmit/brake a given torque, and the area and number of discs. The design torque includes a service factor (1.2-3) over the nominal, to cover peaks and wear. Enter the friction coefficient, axial force, number of surfaces and mean radius.
Hoist Operating Effort
Calculate the effort needed to lift a load with a hoist (block and tackle), F = W ÷ (n·η), from the load weight W (N), the number of supporting rope parts n (parts of the rope supporting the moving block) and the efficiency η (0-1). The hoist (or block and tackle) is a pulley system that MULTIPLIES the applied force, allowing heavy loads to be lifted with little effort — the pulley principle, known since antiquity. A block with n supporting rope parts reduces the needed force to about 1/n of the weight (mechanical advantage n), at the cost of pulling n times more rope length (energy is conserved). But there are friction LOSSES at each pulley (bearings, rope bending): the efficiency η (typically 0.95-0.98 per pulley, accumulating along the system) reduces the real mechanical advantage — so the needed force is slightly more than the ideal W/n. This calculation gives the force the operator (or motor, or winch) must apply at the free rope end to lift the load, accounting for losses. It is essential in sizing manual and electric hoists and choosing the right block: more pulleys (higher n) reduce the force but increase accumulated friction and travel. Enter the weight, the number of supporting rope parts and the efficiency.
Vessel Head Axial Force
Calculate the total axial force the internal pressure exerts on a pressure vessel's cover (or head), F = P · (π·D²/4), from the internal pressure P (MPa) and the internal diameter D (mm); the result is in N. A vessel's internal pressure acts on the ENTIRE internal surface, and on the cover (or closure flange) it generates an axial force tending to PUSH the cover outward — equal to pressure times the cross-sectional area. This force can be ENORMOUS: a modest 1 MPa (10 bar) pressure in a 1-metre-diameter vessel generates a force of nearly 800 kN (80 tonnes!) trying to blow off the cover. This force is what the closure-flange BOLTS (or the head weld) must resist — so flanged pressure vessels have many robust bolts, and computing this force is the starting point of sizing the flange, bolts and gasket. The force also explains why one must NEVER open a still-pressurized vessel: the cover can be hurled with lethal force (serious accidents happen this way, especially with autoclaves and filters). Knowing the cover force is essential for safe closure design and operating procedures. Enter the internal pressure and the diameter.
Belt Centrifugal Tension
Calculate the centrifugal tension in a belt, T_c = m·v², from the mass per unit length m (kg/m) and the belt velocity v (m/s). When the belt wraps a pulley at high speed, its own mass, making the turn, generates a CENTRIFUGAL force tending to 'throw' the belt outward, LIFTING it off the pulley. This creates an additional tension throughout the belt (the centrifugal tension), the same at all points and not contributing to power transmission — it only 'steals' part of the belt's gripping capacity against the pulley. Centrifugal tension grows with the SQUARE of velocity, so it is negligible at low speeds but becomes important in fast belts. The effect is harmful: by lifting the belt off the pulley, centrifugal tension REDUCES the normal contact force and thus the friction available to transmit power — there is an OPTIMAL velocity above which increasing speed reduces transmissible power (the belt starts to 'float'). So belt speed has a practical limit (typically 25-30 m/s for conventional V-belts, more for special belts). Centrifugal tension must be added to the tensions to get the total tight- and slack-side tensions. Enter the mass per unit length and the velocity.
The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.