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Energy Stored In A Rotating Flywheel

A 48-in (121.9-cm) diameter spoked steel flywheel having a 12-in wide X 10-in (30.5-cm X 25.4-cm) deep rim rotates at 200 r/min. How long a cut can be stamped in a 1-in (2.5-cm) thick aluminum plate if the stamping energy is obtained from this flywheel? The ultimate shearing strength of the aluminum is 40,000 lb/in 2 (275,789.9 kPa).

Calculation Procedure:

1. Determine the kinetic energy of the flywheel

In routine design calculations, the weight of a spoked or disk flywheel is assumed to be concentrated in the rim of the flywheel. The weight of the spokes or disk is neglected. In computing the kinetic energy of the flywheel, the weight of a rectangular, square, or circular rim is assumed to be concentrated at the horizontal centerline. Thus, for this rectangular rim, the weight is concentrated at a radius of 48/2 - 10/2 = 19 in (48.3 cm) from the centerline of the shaft to which the flywheel is attached.

Then the kinetic energy K = Wv2/(2g), where K = kinetic energy of the rotating shaft, ft·lb; W = flywheel weight of flywheel rim, lb; v = velocity of flywheel at the horizontal centerline of the rim, ft/s. The velocity of a rotating rim is v = 2p RD/60, where p = 3.1416; R = rotational speed, r/min; D = distance of the rim horizontal centerline from the center of rotation, ft. For this flywheel, v = 2p(200)(19/12)/60 = 33.2 ft/s (10.1 m/s).

The rim of the flywheel has a volume of (rim height, in)(rim width, in)(rim circumference measured at the horizontal centerline, in), or (10)(12)(2)(19) = 14,350 in 3 (235,154.4 cm 3). Since machine steel weighs 0.28 lb/in 3 (7.75 g/cm 3), the weight of the flywheel rim is (14,350)(0.28) = 4010 lb (1818.9 kg). Then K = (4010)(33.2) 2/[2(32.2)] = 68,700 ft·lb (93,144.7 N·m).

2. Compute the dimensions of the hole that can be stamped

A stamping operation is a shearing process. The area sheared is the product of the plate thickness and the length of the cut. Each square inch of the sheared area offers a resistance equal to the ultimate shearing strength of the material punched.

During stamping, the force exerted by the stamp varies from a maximum F lb at the point of contact to 0 lb when the stamp emerges from the metal. Thus, the average force during stamping is ( F + 0)/2 = F/2. The work done is the product of F/2 and the distance through which this force moves, or the plate thickness t in. Therefore, the maximum length that can be stamped is that which occurs when the full kinetic energy of the flywheel is converted to stamping work.

With a 1-in (2.5-cm) thick aluminum plate, the work done is W ft·lb = (force, lb)(distance, ft). The work done when all the flywheel kinetic energy is used is W = K. Substituting the kinetic energy from step I gives W = K = 68,700 ft·lb (93,144.7 N·m) = ( F/2)(1/12); and solving for the force yields F = 1,650,000 lb (7,339,566.3 N).

The force F also equals the product of the plate area sheared and the ultimate shearing strength of the material stamped. Thus, F = ltsu, where l = length of cut, in; t = plate thickness, in; su = ultimate shearing strength of the material. Substituting the known values and solving for l, we get l = 1,650,000/[(1)(40,000)] = 41.25 in (104.8 cm).

Related Calculations: The length of cut computed above can be distributed in any form--square, rectangular, circular, or irregular. This method is suitable for computing the energy stored in a flywheel used for any purpose. Use the general procedure in step 2 for computing the principal dimension in blanking, punching, piercing, trimming, bending, forming, drawing, or coining.

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