Determining Estimated S-N Diagrams for Nonferrous Materials Problem Create an estimated S-N diagram for an aluminum bar and define its equations. What is the corrected fatigue strength at 2E7 cycles? Given The Sut for this 6061-T6 aluminum has been tested at 45 000 psi. The forged bar is 1.5 in

Question

Determining Estimated S-N Diagrams for Nonferrous Materials

Problem Create an estimated [latex]S-N[/latex] diagram for an aluminum bar and define its equations. What is the corrected fatigue strength at [latex]2E7[/latex] cycles?

Given The [latex]S_{ut}[/latex] for this 6061-T6 aluminum has been tested at 45 000 psi. The forged bar is 1.5 in round. The maximum operating temperature is 300°F. The loading is fully reversed torsion.

Assumptions A reliability factor of 99.0% will be used. The uncorrected fatigue strength will be taken at [latex]5E8[/latex] cycles.

Accepted Answer

1 Since no fatigue-strength information is given, we will estimate [latex] S_{f^{\prime}} [/latex] based on the ultimate strength using equation 6.5c (p. 330).

[latex] ext {aluminums :} \left\{\begin{array}{lll} S_{f_{@ 5 E 8}} \cong 0.4 S_{u t} & ext { for } S_{u t}<48 kpsi (330 MPa ) \ S_{f_{@ 5 E 8}} \cong 19 kpsi (130 MPa ) & ext { for } S_{u t} \geq 48 kpsi (330 MPa ) \end{array} ight\} [/latex] (6.5c)

[latex] \begin{array}{l} S_{f^{\prime}} \cong 0.4 S_{u t} \quad ext { for } S_{u t}<48 ksi \ S_{f^{\prime}} \cong 0.4(45000)=18000 psi \end{array} [/latex] (a)

This value is at [latex]N = 5E8[/latex] cycles. There is no knee in an aluminum [latex]S-N[/latex] curve.

2 The loading is pure torsion, so the load factor from equation 6.7a (p. 330) is

[latex] \begin{array}{ll} ext { bending: } & C_{ ext {load }}=1 \ ext { axial loading: } & C_{ ext {load }}=0.70 \end{array} [/latex] (6.7a)

[latex]C_{ ext {load }} = 1.0 [/latex] (b)

because the applied torsional stress will be converted to an equivalent von Mises normal stress for comparison to the [latex]S-N[/latex] strength.

3 The part size is greater than the test specimen and it is round, so the size factor can be estimated with equation 6.7b (p. 331), noting that this relationship is based on steel data:

[latex] ext {for} d \leq 0.3 ext {in} (8 mm) : \quad C_{ ext {size }}=1 \ ext {for} 0.3 ext {in} \lt d \leq 10 ext {in} : \quad C_{ ext {size }}=0.869 d^{-0.097} \ ext {for} 8 mm \lt d \leq 250 mm : \quad C_{ ext {size }}=1.189 d^{-0.097}[/latex] (6.7b)

[latex] C_{ ext {size }}=0.869\left(d_{ ext {equiv }} ight)^{-0.097}=0.869(1.5)^{-0.097}=0.835 [/latex] (c)

4 The surface factor is found from equation 6.7e (p. 333) using the data in Table 6-3 (p. 333) for the specified forged finish, again with the caveat that these relationships were developed for steels and may be less accurate for aluminum.

[latex] C_{ ext {surf }} \equiv A\left(S_{ ext {ut }} ight)^b \quad ext { if } C_{ ext {surf }}>1.0 ext {, set } C_{ ext {surf }}=1.0 [/latex] (6.7e)

[latex] C_{ ext {surf }}=A S_{u t}^b=39.9(45)^{-0.995}=0.904 [/latex] (d)

5 Equation 6.7f (p. 335) is only for steel so we will assume:

[latex] ext {for} T \leq 450^\circ C (840^\circ F) : \quad C_{ ext {temp }}=1\ ext {for} 450^\circ C \lt T \leq 550^\circ C : \quad C_{ ext {temp }}=1-0.0058(T-450)\ ext {for} 840^\circ F \lt T\leq 1020^\circ F : C_{ ext {temp }}=1-0.0032(T-840) [/latex] (6.7f)

[latex] C_{ ext {temp }}=1 [/latex] (e)

6 The reliability factor is taken from Table 6-4 (p. 335) for the desired 99.0% and is

[latex] C_{ ext {reliab }}=0.814 [/latex] (f)

7 The corrected fatigue strength [latex]S_{f}[/latex] at [latex]N = 5E8[/latex] can now be calculated from equation 6.6 (p. 330):

[latex] \begin{array}{l} S_e=C_{ ext {load }} C_{ ext {size }} C_{ ext {surf }} C_{ ext {temp }} C_{ ext {reliab }} S_{e^{\prime}} \ S_f=C_{ ext {load }} C_{ ext {size }} C_{ ext {surf }} C_{ ext {temp }} C_{ ext {reliab }} S_f^{\prime} \end{array} [/latex] (6.6)

[latex] \begin{aligned} S_f &=C_{ ext {load }} C_{ ext {size }} C_{ ext {surf }} C_{ ext {temp }} C_{ ext {reliab }} S_{f^{\prime}} \ &=1.0(0.835)(0.904)(1.0)(0.814)(18000)=11063 psi \end{aligned} [/latex] (g)

8 To create the [latex]S-N[/latex] diagram, we also need a number for the estimated strength [latex]S_{m}[/latex] at 10³ cycles based on equation 6.9 (p. 337). Note that the bending value is used for torsion.

[latex] ext {bending :} \quad \quad S_m=0.9 S_{u t} \ ext {axial loading :} \quad \quad S_m=0.75 S_{u t} [/latex] (6.9)

[latex] S_m=0.90 S_{u t}=0.90(45000)=40500 psi [/latex] (h)

9 The coefficient and exponent of the corrected [latex]S-N[/latex] line and its equation are found using equations 6.10a through 6.10c (p. 338). The value of [latex]z[/latex] is taken from Table 6-5 (p. 338) for [latex]S_{f}[/latex] at [latex]5E8[/latex] cycles.

[latex] S(N)=a N^b [/latex] (6.10a)

[latex] \begin{aligned} b &=\frac{1}{z} \log \left\lgroup\frac{S_m}{S_e} ight group \quad ext { where } \quad z=\log N_1-\log N_2 \ \log (a) &=\log \left(S_m ight)-b \log \left(N_1 ight)=\log \left(S_m ight)-3 b \end{aligned} [/latex] (6.10c)

[latex] \begin{array}{c} b=-\frac{1}{5.699} \log \left\lgroup\frac{S_m}{S_f} ight group=-\frac{1}{5.699} \log \left\lgroup\frac{40500}{11063} ight group=-0.0989 \ \log (a)=\log \left(S_m ight)-3 b=\log [40500]-3(-0.0989): \quad a=80193 \end{array} [/latex] (i)

10 The fatigue strength at the desired life of [latex]N = 2E7[/latex] cycles can now be found from the equation for the corrected [latex]S-N[/latex] line:

[latex] S(N)=a N^b=80193 N^{-0.0989}=80193(2 e 7)^{-0.0989}=15209 psi [/latex] (j)

[latex]S(N)[/latex] is larger than [latex]S_{f}[/latex] because it is at a shorter life than the published fatigue strength.

11 Note the order of operations. We first found an uncorrected fatigue strength [latex] S_{f^{\prime}} [/latex] at some “standard” cycle life ([latex]N = 5E8[/latex]), then corrected it for the appropriate factors from equations 6.7 (pp. 330–335). Only then did we create equation 6.10a (p. 338) for the [latex]S-N[/latex] line so that it passes through the corrected [latex]S_{f}[/latex] at [latex]N = 5E8[/latex]. If we had
created equation 6.10a using the uncorrected [latex] S_{f^{\prime}} [/latex], solved it for the desired cycle life ([latex]N = 2E7[/latex]), and then applied the correction factors, we would get a different and incorrect result. Because these are exponential functions, superposition does not hold.

12 The files EX06-02 are on the CD-ROM.

Table 6-3 Coefficients for Surface-Factor Equation 6.7e Source: Shigley and Mischke, Mechanical Engineering Design, 5th ed., McGrawHill, New York, 1989, p. 283 with permission
	For [latex]S_{ut}[/latex] in MPa use		For [latex]S_{ut}[/latex] in kpsi (not psi) use
Surface Finish	A	b	A	b
Ground	1.58	–0.085	1.34	–0.085
Machined or cold-rolled	4.51	–0.265	2.7	–0.265
Hot-rolled	57.7	–0.718	14.4	–0.718
As-forged	272	–0.995	39.9	–0.995

Table 6-4 Reliability Factors for [latex]S_{d} = 0.08 \mu[/latex]
Reliability %	[latex]C_{reliab}[/latex]
50	1.000
90	0.897
95	0.868
99	0.814
99.9	0.753
99.99	0.702
99.999	0.659
99.9999	0.620

Table 6-5 z-factors for Eq. 6-10c
[latex]N_{2}[/latex]	[latex]z[/latex]
1.0E6	–3.000
5.0E6	–3.699
1.0E7	–4.000
5.0E7	–4.699
1.0E8	–5.000
5.0E8	–5.699
1.0E9	–6.000
5.0E9	–6.699

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