Creating the Modified-Goodman Diagram for a Helical Spring Problem Construct the Goodman line for the spring wire of Example 14-1. Given The required cycle life is N = 1E6 cycles. Wire is 0.042-in (1.1-mm) dia. Assumptions The torsional strengths and torsional shear stresses will be used on the

Question

Creating the Modified-Goodman Diagram for a Helical Spring

Problem Construct the Goodman line for the spring wire of Example 14-1.

Given The required cycle life is [latex]N = 1E6[/latex] cycles. Wire is 0.042-in (1.1-mm) dia.

Assumptions The torsional strengths and torsional shear stresses will be used on the Goodman diagram.

Accepted Answer

See Figure 14-16.

1 The material’s ultimate tensile strength from Figure 14-3 or equation 14.3 (p. 792), converted to ultimate torsional strength with equation 14.4 (p. 793) using data from Table 14-4 (p. 792) allows one point on the Goodman line to be determined.

[latex] S_{u t} \cong A d^b [/latex] (14.3)

[latex] S_{u s} \cong 0.67 S_{u t} [/latex] (14.4)

[latex] S_{u t} \cong 184649(0.042)^{-0.1625}=309071 psi [/latex] (a)

[latex] \begin{aligned} S_{u s} & \cong 0.67 S_{u t} \ &=0.67(309071)=207078 psi \end{aligned} [/latex] (b)

This value is plotted as point [latex]A[/latex] on the diagram in Figure 14-16.

2 The [latex]S-N[/latex] diagrams each provide one data point ([latex]S_{fw}[/latex] or [latex]S_{ew}[/latex], depending on whether for finite or infinite life) on the modified-Goodman line for a material/size combination in pure torsional loading. The fatigue strength [latex]S_{fw}[/latex] for that wire material and condition is taken from the S-N line of Figure 14-15 or calculated from the data in Table 14-9 (p. 813) as

[latex] @ N = 1E6 : S_{f w} \cong 0.33(309071)=101993 psi [/latex] (c)

The [latex]x[/latex] and y intercepts are [latex]0.5S_{fw}[/latex] = 50 996 psi. This is plotted as point [latex]B[/latex] on the diagram in Figure 14-16. Note that for infinite life the value of [latex]S_{ew}[/latex] from equation 14.13 would be plotted at [latex]B[/latex] instead of this value of [latex]S_{fw}[/latex] for a finite life.

[latex] \begin{array}{l} S_{e w^{\prime}} \cong 45.0 kpsi (310 MPa ) \quad ext {for unpeened springs} \ S_{e w} \cong 67.5 kpsi (465 MPa ) \quad ext {for peened springs} \end{array} [/latex] (14.13)

3 Note in Figure 14-16 that the wire fatigue strength [latex]S_{fw}[/latex] is plotted at point [latex]B ( au _{a} = au _{m} = 0.5 S_{fw}[/latex]) corresponding to the test conditions of equal mean and alternating stress components (stress ratio [latex]R = au _{min}/ au _{max} = 0[/latex]). Point [latex]B[/latex] is then connected with the ultimate shear strength [latex]S_{us}[/latex] on the mean-stress axis at point [latex]A[/latex] to draw the Goodman line, which is extended to point [latex]C[/latex].

4 We can now find the value of the fully reversed fatigue strength ([latex]R = –1[/latex]), which is point [latex]C[/latex] on the diagram. This value can be found from the equation for the Goodman line, defined in terms of its two known points, [latex]A[/latex] and [latex]B[/latex]:

[latex] \begin{aligned} m &=-\frac{0.5 S_{f w}}{S_{u s}-0.5 S_{f w}} \ S_{f s} &=-m S_{u s} \ S_{f s} &=0.5 \frac{S_{f w} S_{u s}}{S_{u s}-0.5 S_{f w}} \ &=0.5 \frac{101993(207078)}{207078-0.5(101993)}=67658 psi \end{aligned} [/latex] (d)

5 This use of the Goodman line is conservative for stress ratios [latex]R[/latex] ≥ 0, and its use is justified in this case because springs should always be loaded in the same direction. Helical compression springs tend to have stress ratios between 0 and 0.8, which puts their stress coordinates to the right of the 45° line in the figure, where the Goodman line is more conservative than the Gerber line.

6 Any other combination of mean and alternating stress with a stress ratio [latex]R[/latex] ≥ 0 for this material and number of cycles can now be plotted on this diagram to obtain a safety factor.

Table 14-4 Coefficients and Exponents for Equation 14.3 Source: Reference 1
ASTM#	Material	Range		Exponent [latex]b[/latex]	Coefficient [latex]A[/latex]		Correlation Factor
ASTM#	Material	mm	in	Exponent [latex]b[/latex]	MPa	psi	Correlation Factor
A227	Cold drawn	0.5–16	0.020–0.625	–0.182 2	1 753.3	141 040	0.998
A228	Music wire	0.3–6	0.010–0.250	–0.1625	2 153.5	184 649	0.9997
A229	Oil tempered	0.5–16	0.020–0.625	–0.183 3	1 831.2	146 780	0.999
A232	Chrome-v	0.5–12	0.020–0.500	–0.145 3	1 909.9	173 128	0.998
A401	Chrome-s.	0.8–11	0.031–0.437	–0.093 4	2 059.2	220 779	0.991

Table 14-9 Maximum Torsional Fatigue Strength [latex]S_{fw}[/latex]' for Round-Wire Helical Compression Springs in Cyclic Applications (Stress Ratio, R = 0) No Surging, Room Temperature, and Noncorrosive Environment. Source: Ref. 1
	Percent of Ultimate Tensile Strength
Fatigue Life (cycles)	ASTM 228, Austenitic Stainless Steel and Nonferrous		ASTM A230 and A232
Fatigue Life (cycles)	Unpeened	Peened	Unpeened	Peened
[latex]10^5[/latex]	36%	42%	42%	49%
[latex]10^6[/latex]	33	39	40	47
[latex]10^7[/latex]	30	36	38	46

Question 14.2: Creating the Modified-Goodman Diagram for a Helical Spring P...

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