## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 6.2

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 $S_{ut}$ 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 $5E8$ cycles.

## Verified Solution

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.

$\begin{array}{ll} \text { bending: } & C_{\text {load }}=1 \\ \text { axial loading: } & C_{\text {load }}=0.70 \end{array}$     (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.

 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 $S_{ut}$ in MPa use For $S_{ut}$ 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 $S_{d} = 0.08 \mu$ Reliability % $C_{reliab}$ 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 $N_{2}$ $z$ 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