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Chapter 6

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.

Step-by-Step

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
F6-25
F6-34