Question 6.2: Determining Estimated S-N Diagrams for Nonferrous Materials ...

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

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

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

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

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

\begin{array}{ll} \text { bending: } & C_{\text {load }}=1 \\ \text { axial loading: } & C_{\text {load }}=0.70 \end{array}      (6.7a)

C_{\text {load }}  = 1.0        (b)

because the applied torsional stress will be converted to an equivalent von Mises normal stress for comparison to the S-N 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:

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

C_{\text {size }}=0.869\left(d_{\text {equiv }}\right)^{-0.097}=0.869(1.5)^{-0.097}=0.835        (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.

C_{\text {surf }} \equiv A\left(S_{\text {ut }}\right)^b \quad \text { if } C_{\text {surf }}>1.0 \text {, set } C_{\text {surf }}=1.0      (6.7e)

C_{\text {surf }}=A S_{u t}^b=39.9(45)^{-0.995}=0.904        (d)

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

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

C_{\text {temp }}=1         (e)

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

C_{\text {reliab }}=0.814             (f)

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

\begin{array}{l} S_e=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^{\prime}} \\ S_f=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_f^{\prime} \end{array}     (6.6)

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

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

\text {bending :} \quad \quad S_m=0.9 S_{u t} \\ \text {axial loading :} \quad \quad S_m=0.75 S_{u t}     (6.9)

S_m=0.90 S_{u t}=0.90(45000)=40500  psi            (h)

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

S(N)=a N^b     (6.10a)

\begin{aligned} b &=\frac{1}{z} \log \left\lgroup\frac{S_m}{S_e}\right\rgroup \quad \text { where } \quad z=\log N_1-\log N_2 \\ \log (a) &=\log \left(S_m\right)-b \log \left(N_1\right)=\log \left(S_m\right)-3 b \end{aligned}        (6.10c)

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

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

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

S(N) is larger than S_{f} 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 S_{f^{\prime}} at some “standard” cycle life (N = 5E8), 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 S-N line so that it passes through the corrected S_{f} at N = 5E8. If we had
created equation 6.10a using the uncorrected S_{f^{\prime}} , solved it for the desired cycle life (N = 2E7), 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.