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## Q. 6.1

Determining Estimated S-N Diagrams for Ferrous Materials

Problem    Create an estimated S-N diagram for a steel bar and define its equations. How many cycles of life can be expected if the alternating stress is 100 MPa?

Given    The $S_{ut}$ has been tested at 600 MPa. The bar is 150 mm square and has a hot-rolled finish. The operating temperature is 500°C maximum. The loading will be fully reversed bending.

Assumptions    Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

## Verified Solution

1    Since no endurance-limit or fatigue strength information is given, we will estimate $S_e{ }^{\prime}$ based on the ultimate strength using equation 6.5a (p. 330).

$\text { steels : } \quad\left\{\begin{array}{ll} S_{e^{\prime}} \cong 0.5 S_{u t} & \text { for } S_{u t}<200 kpsi (1400 MPa ) \\ S_{e^{\prime}} \cong 100 kpsi (700 MPa ) & \text { for } S_{u t} \geq 200 kpsi (1400 MPa ) \end{array}\right\}$     (6.5a)

$S_{e^{\prime}} \cong 0.5 S_{u t}=0.5(600)=300 MPa$     (a)

2    The loading is bending so the load factor from equation 6.7a is

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

$C_{load} = 1.0$          (b)

3    The part is larger than the test specimen and is not round, so an equivalent diameter based on its 95% stressed area ($A_{95}$) must be determined and used to find the size factor. For a rectangular section in nonrotating bending loading, the $A_{95 }$ area is defined in Figure 6- 25c (p. 332) and the equivalent diameter is found from equation 6.7d (p. 331):

$d_{\text {equiv }}=\sqrt{\frac{A_{95}}{0.0766}}$      (6.7d)

\begin{aligned} A_{95} &=0.05 b h=0.05(150)(150)=1125 mm ^2 \\ d_{\text {equiv }} &=\sqrt{\frac{A_{95}}{0.0766}}=\sqrt{\frac{1125 mm ^2}{0.0766}}=121.2 mm \end{aligned}       (c)

and the size factor is found for this equivalent diameter from equation 6.7b (p. 331):

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

$C_{\text {size }}=1.189(121.2)^{-0.097}=0.747$      (d)

4    The surface factor is found from equation 6.7e (p. 333) and the data in Table 6-3 for the specified hot-rolled finish.

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

$C_{\text {surf }}=A S_{u t}^b=57.7(600)^{-0.718}=0.584$     (e)

5    The temperature factor is found from equation 6.7f (p. 335):

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

$C_{\text {temp }}=1-0.0058(T-450)=1-0.0058(500-450)=0.71$      (f)

6    The reliability factor is taken from Table 6-4 (p. 335) for the desired 99.9% and is

$C_{\text {reliab }}=0.753$     (g)

7    The corrected endurance limit $S_{e}$ 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_e &=C_{\text {load }} C_{\text {size }} C_{\text {surf }} C_{\text {temp }} C_{\text {reliab }} S_{e^{\prime}} \\ &=1.0(0.747)(0.584)(0.71)(0.753)(300) \\ S_e &=70 MPa \end{aligned}     (h)

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) for bending loading.

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

$S_m=0.90 S_{u t}=0.90(600)=540 MPa$      (i)

9    The estimated $S-N$ diagram is shown in Figure 6-34 with the above values of $S_{m}$ and $S_{e}$. The expressions of the two lines are found from equations 6.10a through 6.10c (p. 338) assuming that $S_{e}$ begins at $10^{6}$ cycles.

$S(N)=a N^b$    (6.10a)

$\begin{array}{c} b=-\frac{1}{3} \log \left\lgroup\frac{S_m}{S_e}\right\rgroup =-\frac{1}{3} \log \left\lgroup\frac{540}{70}\right\rgroup =-0.295765 \\ \log (a)=\log \left(S_m\right)-3 b=\log [540]-3(-0.295765): \quad a=4165.707 \end{array}$       (j)

$\begin{array}{ll} S(N)=a N^b=4165.707 N^{-0.295765} MPa & 10^3 \leq N \leq 10^6 \\ S(N)=S_e=70 MPa & N>10^6 \end{array}$       (k)

10    The number of cycles of life for any alternating stress level can now be found from equations ($k$). For the stated stress level of 100 MPa, we get

\begin{aligned} 100 &=4165.707 N^{-0.295765} \quad 10^3 \leq N \leq 10^6 \\ \log 100 &=\log 4165.707-0.295765 \log N \\ 2 &=3.619689-0.295765 \log N \\ \log N &=\frac{2-3.619689}{-0.295765}=5.476270 \\ N &=10^{5.476270}=3.0 E 5 \text { cycles } \end{aligned}       (l)

Figure 6-34 shows the intersection of the alternating stress line with the failure line at $N = 3E5$ cycles.

11    The files EX06-01 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