BAD 200 Lecture Notes - Lecture 6: Nice Ltd.
18 Dec 2020
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Both end plates are in contact with nearly a full turn of wire. The end plate loads coincide with the spring axis. Analysis: ithe following is only one of many possible solutions. ] The equation with x in meters is: x = 0. 01 + 0. 01 sin (ot at a rotating speed of 3. Fi is: 90 [-0. 01(20 )2sinot -362 sin ot (in newtons) 4. The spring is needed only to oppose the negative acceleration. Both fi and spring force are plotted above. To compensate for small friction and gravity forces plus provide a small safety factor, select minimum and maximum spring forces of 40 n and 400 n: f-400-40-18 n/mm respectively (as shown on graph). Thus, k 20 and the expression for the spring force plotted is: f 40kx 40181010sin ot 220180 sin cot select spring wire having fatigue properties shown in fig. Tma fmax_ 400 tmin min 40 from fig. 12. 16, max 710mpa based on information on fig.
