Question

PROBLEM 1. A music wire ASTM 228, peened wire will be used in a spring application. The spring must fit in a 1.55 inch hole. The height of the spring Li = 2.4 inch is when a force of 54 lbs. is applied and for a second force of 86 lbs. the height must be 1.7 inch. The spring will be designed for a control valve and are applied repeatedly with a goal life of 10' cycles, and condition of set removed. Your initial decisions based on the suggest values by applicable Code. Determine the specifications for the spring. a. Type of ends b. Wire diameter c. Inner spring diameter. d. Outer spring diameter e. Pitch diameter f. Spring index 8. Sut h. Ssy i. The scale of the spring j. All static factors of safety k. The fatigue factor of safety using Modified-Goodman criteria.

Please answer question and all parts completely. Show all of your work in detailed steps.

Answer 1

Page Given outer dia = 1.55 flength at 54 lbs 2.4 inches 86 lbs = 1.7 inches s Delta wad 86-54 = 32 lbs se - Delta of lengen 2.4-1.7= 0.7 Iload I nich = 32 0-7 Lengen at required for deflecting. I = 45.714 los os findes 45.714 Cress spring rate : 45.714 lbs /inches 1.88 inches compression length by 86 ks = 36 Free length of spring = 1.88 + 17 = 3.58 inches check 54 145.714 - 1.8126 inches compressed length = 3.58 -1.8126=2-3987=2.4 inches Isprung inden old know That. |wjs = Gdy / 8D3, in old) 3 - (s/w) Gd on (old) 3 = (1/45.714 ) Gd/sn young 's Modulus (E) psi q/8n For ASTM 228 Juom databook +E = 30x loto psi & Sut = 360 x 10t3 (d)d )3 - (1145.714 ) 30X 106 old) 3 = 82031 76 d/n D31d4 = 82031.76 In 1.553/d4 = 82031-76/n dun - 4.53955 X 105 idu = mx 4.53955 X 105

classmate Date Page ends For inad ground length must solid length solid length be than llinches solid ht 1.5 us mxd = 1.5 vd = 1.5/n put eq (2) (2) ni u (2) (1.5/n) 4 = mx 4.53955 yios 5.0625 1n4=mx 4.53955 X los ms=5.0625 14.53955 x 105 - 111519.8 m=(111519.8) 15 10.22 046 mou the diameter mxd = 1.5 d= 0.147 inches = 3.73 38 mm n = lo d = 3.658 mm SWG = 9 3.658 us the meareat 3.658 mm = 0.144 inches (quom table 23.2). standard other requerid values :- (a) Type of ground The ends (b) were diameter id= 0.144 inches inner D-20 -1-55 - 2XO-144 ispring diameter 1.262 inches outer spring Pitch diameter - diameter 5.55 = D.d = 1:55/144 tot truces 1.406 inches old = 1.55 1.144 = 10.764 spring inden

(3) Sut psi 360 xlo 3 Suy = 0.67 Sut Say - 0-67 = 36ox to 3 2ч 200 рэс - Чs. Піч Ub» мм, C) Deala , Op rate spring ороля