Question

A stock helical-coil compression spring has a total of 14 coils, both ends squared and ground, a free length of 1.25 in, and an outside coil diameter of 7/16 (0.438) in giving the number of active coils, Ng=12. The material is hard- drawn wire 0.042-inch in diameter and S=259,000 psi. The operating forces are Fmin=1.5 lb and Fmax=3.5 lb. F VUONI 0.438 in N=14 coils 0.042-in dia 1.25-in free length Note, choose one or more than one of the following MC answers and then type the letter (s) corresponding to the question number in the essay section of this problem with supporting calculations contained on the scratch sheet. If the answer requires "the correct answer is "type that in the essay section as well.

Answer 1

1.5 lb. Fmax. 3.5lb # Total number of coils (Nt) = 14 .. Free length (24) = 1.25 in. outside Coil diameter (Do) = 0.438 in Number of active Coils (N) = 12 wire diameter (d) = 0.042 in. Su=259000 psi we know, Mean diameter of coil (1) = (Do-d) 6.438-0.042) 0.396 in. spring inden (c) = 2 0-396 0.042 or/ 9.43 For squared & Ground ends. -), Solid length (Ls) d*Ne (14% 0.042) =0.588 in Shear stress Correction factor (ks) (1+ za ) 26 t 2x 9.4. 92 or Ks 1.053

If F=3db then, y = 8 Fmax Maximum shear stress (Tmax ) = * D *ks. nd 3 8*3.5*0.396*1.053 T * (0.0423 = 50, 163.04 psi Minimum shear stress (Tmin) = 8 * Fimin *D*Ks 703 8x1.5 x 0.396*1.053 a(0.042) = 21, 498.44 psi For hand drawn spring wire, G= 11500000 psi we know, spring Constant (K) 4 OND3 11500000 Ⓡ (0.04291 8*12* 0.39673 30 6 Ablin we know, F = kay Loy=deflection of spring] 10.5 in

Deflection at minimum force Femin min ( 1.5 6 Ymin 6.25 in. Deflection at Maximum force max "max K mange 3.5 6 Imax = 0.583 in Deflection at Solid length = Y = (hp - Ls) = (1.25 -0.588) y = 0.662 in force acting at selid length (Fs) = (k*%) - (6 x 0.662) or ifs - 4 lb.

1. a Fab) 3.5 K= 6eblin 1.5 0.25 0.583 (in) 11.6 9.43 spring index, ca Ks = tos option (a), and (c) are correct 1.C. F = 4 lb ; Solid length 0.588 in. option (f) is Correct Tod K = 6. lblin. would design the helical spring against Shear we Stress. options (a) and (a) are Correct.
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