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 Sut has been tested at 600 MPa. The bar is 150 mm square and has a

Question

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 [latex]S_{ut}[/latex] 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.

Accepted Answer

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

[latex] ext { steels : } \quad\left\{\begin{array}{ll} S_{e^{\prime}} \cong 0.5 S_{u t} & ext { for } S_{u t}<200 kpsi (1400 MPa ) \ S_{e^{\prime}} \cong 100 kpsi (700 MPa ) & ext { for } S_{u t} \geq 200 kpsi (1400 MPa ) \end{array} ight\} [/latex] (6.5a)

[latex] S_{e^{\prime}} \cong 0.5 S_{u t}=0.5(600)=300 MPa [/latex] (a)

2 The loading is bending so the load factor from equation 6.7a 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_{load} = 1.0 [/latex] (b)

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

[latex] d_{ ext {equiv }}=\sqrt{\frac{A_{95}}{0.0766}} [/latex] (6.7d)

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

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

[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 }}=1.189(121.2)^{-0.097}=0.747 [/latex] (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.

[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=57.7(600)^{-0.718}=0.584 [/latex] (e)

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

[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 }}=1-0.0058(T-450)=1-0.0058(500-450)=0.71 [/latex] (f)

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

[latex] C_{ ext {reliab }}=0.753 [/latex] (g)

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

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

[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(600)=540 MPa [/latex] (i)

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

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

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

[latex] \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} [/latex] (k)

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

[latex] \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 ext { cycles } \end{aligned} [/latex] (l)

Figure 6-34 shows the intersection of the alternating stress line with the failure line at [latex]N = 3E5[/latex] 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 [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

Question 6.1: Determining Estimated S-N Diagrams for Ferrous Materials Pro...

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