CANDIDATE SOLUTIONS
Candidate   (dimensions mm, forces N)		A	B	C	D	E
	trial wire diameter, d	4.5	5	5.6	6.3	7.1
Compute corresponding spring index to satisfy chosen fatigue safety factor of 1.1, from ( a)
	Fut from Table 3	22000	26600	32800	40600	50600
	initial trial C	7.5	7.5	7.5	7.5	7.5
	updating C from (a) :	4.69	5.67	6.99	8.65	10.8
	C = Fut /1.1 ( 1430 Ks + 2310 Kh )	4.33	5.43	6.92	8.81	11.2
	where K's based on last C-update	4.25	5.39	6.91	8.83	11.3
	final converged maximum C	4.22	5.38	6.91	8.83	11.3
	D   = Cd	19.0	26.9	38.7	55.6	80.0
	Di = D - d = ( C - 1)d > 15 ?	too small	21.9	33.1	49.3	73.1
	Do = D + d = ( C + 1)d < 0.96 * 65 = 62.4 ?	-	31.9	44.3	61.9	too big
Calculate corresponding number of turns to give required stiffness of 12 N/mm, from ( 2)
	na = Gd/8kC3 = 79E3*d/8*12*C3	-	26.4	14.0	7.5	-
Finalise solid and free lengths, and verify close-coiled assumption
	total turns   nt = na + 2     (Table 1)	-	28.4	16.0	9.5	-
	solid length   Ls = ntd     (Table 1)	-	142	90	60	-
	free length   Lo = Ls + δs	-	198	146	116	-
	pitch   p = ( Lo - 2d )/na     (Table 1)	-	7.1	9.6	13.8	-
	helix angle   α = arctan( p/π D)	-	4.8o	4.5o	4.5o	-
Check buckling at maximum deflexion of 50 mm, assuming guided ends ( λ = 0.5 )
	λLo / c2 D     (absolutely stable if ≤ 1 )	-	1.40	0.72	0.40	-
	δcrit     from ( 3a) (conditionally stable if > 50 mm)	-	48	-	-	-
	conclusion re stability	-	unstable	OK	OK	-
Check for probability of resonance, from ( 6)
	fund'l natural frequency,   fn = 358E3/naCD     Hz	-	-	96	97	-
	ratio natural/running frequencies = fn*60/400	-	-	14.4	14.6	-
Check for yielding when solidified
	tensile ultimate,   Sut   (Table 2)   MPa	-	-	1330	1305	-
	shear yield,   Sys = 0.48 Sut   (Table 2)   MPa	-	-	640	625	-
	solid stress,   τs = Ks 8 Fs C / π d2     MPa	-	-	403	404	-

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    Copyright 1999-2005 Douglas Wright,   doug@mech.uwa.edu.au
      last updated May 2005