Question

10-26 design by selection.) A compression spring is needed to fit over a 12-mm diameter rod. To allow for er a 12-mm diameter rod. To allow for some clearance. the inside diameter of the spring is to be 15 mm. To ensure a reasonable is to be 15 mm. To ensure a reasonable coil, use a spring index of 10. The spring is to be used in a machine by compressing 125 mm in through a stroke of 75 mm to its solid len ed in a machine by compressing it from a free length of ough a stroke of 75 mm to its solid length. The spring should have squared and ground ends, unpeened, and is to be made from cold-drawn wire. (a) Determine a suitable wire diameter. (b) Determine a suitable total number of coils. (c) Determine the spring constant. (a) Determine the static factor of safety when compressed to solid length. (e) Determine the fatigue factor of safety when repeatedly cycled from free len length. Use the Gerber-Zimmerli fatigue-failure criterion. me clear-

just need answer to part (e)

Answer 1

им ANSWER :- Given data :- * A compression spring is needed to fit over a 12 mm diameter hod. * Inside diameter of the Spring is to be 15mm. * spring index of 10. Length of 125 mm in through a stroke of 75 mm. a) Determine a suitable wire diameter. b) Determine a suitable total number of coils. C) Determine the spring constant . d) Determine the static factor of safety when compressed to Solid length. el petermine the fatigue factor of Safety when repeatedly cycled from free length to solid Tength. use the Gerber - Zimmerli fatique-failure criterion as Suitable wire diameter :- D = Ditd - 0 and spring index Caj -> (2) from 0x -> c= Di+d sed-dadi

=) d = Di cal -) d = 15mm 10-1 = 1.6667 mm -> d= 1.6667 mm Then D=cd = 10x (1.6667 mm ) 10 = 16: 667 mm Suitable total number of coils ;- The solid length Ls = NĚ.d - Nt = les b) = (125mm – 75mm) 1.667mm -> NE = 30 colis c) The Spring Constant : For Squared and ground ends, Na = Nt -2 = 30 - 2 Na = 28 coils/

Note that this spring design exceds the excommended maximum number of coils (15) since Houd drawn means the same thing as cold drawn, then from table 10-5.. d = 1.6667X16?m = 0.06667 in. G = 11.5mpsi 0 = 6.6667 in Then, k=dG - 80? No di = 60.6666780)'.11.58100 K = 3.423 lblin OK - 599.46 Nm static factor of safety when compresed to Solid length : forom table 10-4. for hard - drawn (same as cold drawn) materials. A = 140 kpsi.in", m=0.190 · 5.0 , from equation 10-14, Sut - fo = 140 kpsi.inn 10.0666780 70.190 Sut = 234.2kpsi

Assuming the spring has not been set, then the shear yield strength can be calculated from Parameters in Toble. 10-6. words Before set removed Yeally mean the spring has not been set, and therefore the percent Hensile strength factor should be ous for cold drawn Steer wire. And Bergstyasser factor ik8 should be used on calculating the shear stress Thus, shear yield strength is ; 1 ssy = o.us sut = 0.457 6234-2kpsi) -> Ssy = los.u kpsil . And the Bergstr a sser factor, kg is KB = uct2 4c-3 = (4x10)+2 (4480) - 3 ss188 = 1.135] The maximum force on the spring at the solid fmax = kymax = 3.423 16. Ymax

Ymax = 75mm = 3in Thus, Fmax = 3.423 lb *3.in - TFmax = 10.269 16] maximum shear stress :- Ymar = KB 8 Fmard īd3 = 1.135 x 82 110.26916)x (0.6667in) T1(0.06667 in)3 » Ymex - 66780 psi Then , The yeeld factor of - Safety ny= ssg = 105400 = 105400 Tman 66780 => ny = 1.58 Thus, Static factor of safety =4.58 e) Fatigue factor of safety ;- The Zimmerli data will be used. Ssa - 3 tỷ 3 , SSD = ssles

from cquation 10-30, SSŲ = 0.67Sut -0.67x (234.2 kpsi) - SS4 = 156.9 kpsi | where Fmin zo Fmax = 10269.461 then from cquation 10-31, - Fm = Fa = Fmax = 10269.16 e 2 Fm = Fa=513516 Thos'ta - Ton = Pmax = 66750 = 33.390 psi Then, an. Ssc i sinex = 5$a = 35kp si Sse, Goodman - Ssa - 1-5sm SSU = 35 1 - 55 1569 -> sse , Good man = 53-89 kpsi SSC Gerber = ssa Lissue 2 T-(SAD 35 (1 56.9

-) SSe, Gerber = 39.90 kpsi Sm the fatigue factor of safety are nfi since = sse, sine Ta = 35 33.39 slos nf, Goodman = I'm tla ssa Sse, Goodman 33.39 156.9 33.39 53.89 - 1.2 /27m SSex Gerber] ber LIE ber/ SS.TO ny anerber le more) : + ng Gerber ist ) (1 + Jo pront] - hy. Gerber = 1.13 A. so, sine's model is the most conservative at 1.05 then , Gerbet's model at 1:13, then Goodman at 1:2

comparition of three fatigue failure theorem for springs Goodman Toad line > Desi2 Zimmelesdata A >stie's sol S sercebesty Ssestest of - Zauerber Alternating shear Stress psi so loo is doo Midrange shear stress psi