Question

2) [40 pts] Extension Spring Design A hard-drawn spring steel extension spring is to be designed to carry a static load F of 18 lbf with an extension of 1/2 in using a design factor of n = 1.5 in bending. Use full-coil end hooks with the fullest bend radius of r = D/2 and r2 = 2d. The free length must be less than 3 in, and the body turns must be fewer than 30. (Note: Integer and half-integer body turns allow end hooks to be placed in the same plane. However, this adds extra cost and is done only when necessary.) Determine the value of the following: The initial tension F, is The spring rate k is Ib/in. Ibf. The number of active coils N, is coils. The number of body coils N, is coils.

Answer 1

Salution to the above problem given data A Hard drawn Spring mone Fstatic = 18 lbf ny extension 1/2 Full coil end hooks = 1/2 nd 22=2d Free length must be less than 3 in body tums must be fewer than 30 To Find - Fi 1 Mbt) initial Tension. k (lblin ) Spring rate Number of actiue Coils Na 1 Namber of body coils NG Salubion and formellos- From Table 10.4 ASTM No A 227 Hand drawn wire 0-190 exponent Relective cost of wire

D = mean diameter

d= wire diameter

А - 140 kpsi in Assume d = 0.067 inch We unow that but 140 A dm 234 kpsi 0-190 0.067) From Table 10.5 HD spring A227 Modulus of Elashüty E - d= 0.064 0.125 28.6 Mpsi Modulus of Rigidity G = 11.5 Mpsi from Table 10.6 Torsjonal Shear bsy = 0.45 Sut 0.45 x 234 Ssy 105.3 kpsi from Table 10.7 Yield Sheng.in Sy= 0.75 sut 0.75 +234 by 175.5 ipsi moumem Tensile shers dice to bending and Aud CA mon Td2 [the 16e + 4 by ny DC ac - spring inden

--> KA - Bending Shess Comreehon Factor for unvature = 40 - 0,-1 46(6-1) => 4C²_col 46(6-1) Đ 4 c²-c-1 46(C-1) (160) + 4 = Id² sy Fmore ny Tx 0.0672x 175.5x 103 18 X 1.5 =91. 6669 =91-669-4 = 87.6669 (462-C-1) 064) 44((-1) 40² – (-1 11 87-6669 4 4c²-(-1 - 21.9167 (-1) 40² – (-1 = 21.91670 – 21.9167 46² - c-21.91670 - 1 +21-9167 =0 40² – 22.91676 +20 9167 =0.

Take 4 out of equahon and Fachrice cone equation 4C (²- 22.9767C + 20.9167! 4. 4 - 5.72910 +5.229175 Flo Ax² + Bx +C =0 -B + B²-4AC) X- 2 A -(-5-7291) (-5-72412 - 14801)* 5-229175) 11 2x1 32.8225 20.9164 5-7291 I - 2 5.729 1 1 3.4505 2 Prefferred "C" C = 4.5898 1-1393 42C512 eq[1018) That why we neglected Take C= 4-5898 1.1393 we know That 4.5898 x 0.067 g=c Daex d = D= 0.3075 in

Initial tension In spring F = Tidati 8D Ti - 7 + [ 33,500 e (0-1050) 1000 14- 6.5 Fi => TTd3 & looo Lao 33500 eco-105xe) (-3 6.51 8D - Ux0.067) 33500 I 2000 (4- 4.5598-31 (0-105x 4.5898) 8x0.3075 6.5 = 3.8409x 33500 60.4819 1000 (4-0.2445) - 3-8409xC74 20689.895 t 3755-5] =73-8409x604 [20689.895 +3755.5] ๆ 3.8409 x 604 [20689.895-3755.5] -> 9-3892, 6-5043 lbf Take Fi = 6-5043 as Prefered one for Simplicity we can take Fi=7 Rescult; Fi= 6.5 or 7

(11) Spring rate K 18-7 Frnan - fi 호 22 lbt ing (111) Number of active coils d4u 8XKXD3 - (0.067) 11.5666 Na= &* 22 x 0.3075) 3 Na = 45.28 Tuns (1) Number of body coils NO Na G E 45.28 11-5 28.6 45.28 – 0.40209 44.88 Tams Resett 6.5 Fi = initial Tension 7 lbf K 22 Abf Spring rate in Number of active coil Na 45.28 colds NG 44.88 Coils Neember of body Cods