A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the endurance limit and fatigue strength at

Question

A 1015 hot-rolled steel bar has been machined to a diameter of 1 in. It is to be placed in reversed axial loading for 70 000 cycles to failure in an operating environment of 550°F. Using ASTM minimum properties, and a reliability of 99 percent, estimate the endurance limit and fatigue strength at 70 000 cycles.

Accepted Answer

From Table A–20, [latex]S_{u t}=50[/latex] kpsi at 70°F. Since the rotating-beam specimen endurance limit is not known at room temperature, we determine the ultimate strength at the elevated temperature first, using Table 6–4. From Table 6–4,

Table 6–4 Effect of Operating Temperature on the Tensile Strength of Steel.* (ST = tensile strength at operating temperature; SRT = tensile strength at room temperature; 0.099 ≤ ˆσ ≤ 0.110)
Temperature, [latex]{ }^{\circ}{ }^{ C }[/latex]	[latex]S_{T} / S_{R T}[/latex]	Temperature, [latex]{ }^{\circ} F[/latex]	[latex]S_{T} / S_{R T}[/latex]
20	1.000	70	1.000
50	1.010	100	1.008
100	1.020	200	1.02
150	1.025	300	1.024
200	1.020	400	1.018
250	1.000	500	0.995
300	0.975	600	0.963
350	0.943	700	0.927
400	0.900	800	0.872
450	0.843	900	0.797
500	0.768	1000	0.698
550	0.672	1100	0.567
600	0.549

*Data source: Fig. 2–9.

[latex]\left(\frac{S_{T}}{S_{R T}} ight)_{550^{\circ}}=\frac{0.995+0.963}{2}=0.979[/latex]

The ultimate strength at 550°F is then

[latex]\left(S_{u t} ight)_{550^{\circ}}=\left(S_{T} / S_{R T} ight)_{550^{\circ}}\left(S_{u t} ight)_{70^{\circ}}=0.979(50)=49.0 kpsi[/latex]

The rotating-beam specimen endurance limit at 550°F is then estimated from Eq. (6–8) as

[latex]S_{e}^{\prime}=\left\{\begin{array}{ll}0.5 S_{u t} & S_{u t} \leq 200 kpsi (1400 MPa ) \100 kpsi & S_{u t}>200 kpsi \700 MPa & S_{u t}>1400 MPa\end{array} ight.[/latex] (6–8)

[latex]S_{e}^{\prime}=0.5(49)=24.5 kpsi[/latex]

Next, we determine the Marin factors. For the machined surface, Eq (6–19) with Table 6–2 gives

[latex]k_{a}=a S_{u t}^{b}[/latex] (6–19)

[latex]k_{a}=a S_{u t}^{b}=2.70\left(49^{-0.265} ight)=0.963[/latex]

Table 6–2 Parameters for Marin Surface Modification Factor, Eq. (6–19)
Surface Finish	Factor a		Exponent b
Surface Finish	[latex]S_{u t r} ext { kpsi }[/latex]	[latex]S_{ ext {utr }} MPa[/latex]	Exponent b
Ground	1.34	1.58	−0.085
Machined or cold-drawn	2.7	4.51	−0.265
Hot-rolled	14.4	57.7	−0.718
As-forged	39.9	272	−0.995

[latex]k_{e}=1-0.08 z_{a}[/latex] (6–29)

For axial loading, from Eq. (6–21), the size factor [latex] k_{b}=1[/latex], and from Eq. (6–26) the loading factor is [latex] k_{c}=0.85[/latex]. The temperature factor [latex]k_{d}=1[/latex], since we accounted for the temperature in modifying the ultimate strength and consequently the endurance limit. For 99 percent reliability, from Table 6–5, [latex]k_{e}=0.814[/latex]. Finally, since no other conditions were given, the miscellaneous factor is [latex] k_{f}=1[/latex]. The endurance limit for the part is estimated by Eq. (6–18) as

[latex]k_{b}=1[/latex] (6–21)

[latex]k_{c}=\left\{\begin{array}{ll}1 & ext { bending } \0.85 & ext { axial } \0.59 & ext { torsion }^{17}\end{array} ight.[/latex] (6–26)

[latex]S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime}[/latex] (6–18)

[latex]\begin{aligned}S_{e} &=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime} \&=0.963(1)(0.85)(1)(0.814)(1) 24.5=16.3 kpsi\end{aligned}[/latex]

Table 6–5 Reliability Factors ke Corresponding to 8 Percent Standard Deviation of the Endurance Limit
Reliability, %	Transformation Variate [latex]z_{a[/latex]	Reliability Factor [latex]k_{ e }[/latex]
50	0	1.000
90	1.288	0.897
95	1.645	0.868
99	2.326	0.814
99.9	3.091	0.753
99.99	3.719	0.702
99.999	4.265	0.695
99.9999	4.753	0.62

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From p. 285, since [latex]S_{u t}=49<70 kpsi , ext { then } f=0.9[/latex]. From Eq. (6–14)

[latex]a=\frac{\left(f S_{u t} ight)^{2}}{S_{e}}[/latex] (6–14)

[latex]a=\frac{\left(f S_{u t} ight)^{2}}{S_{e}}=\frac{[0.9(49)]^{2}}{16.3}=119.3 kpsi[/latex]

and Eq. (6–15)

[latex]b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}} ight)[/latex] (6–15)

[latex]b=-\frac{1}{3} \log \left(\frac{f S_{u t}}{S_{e}} ight)=-\frac{1}{3} \log \left[\frac{0.9(49)}{16.3} ight]=-0.1441[/latex]

Finally, for the fatigue strength at 70 000 cycles, Eq. (6–13) gives

[latex]S_{f}=a N^{b}[/latex] (6–13)

[latex]S_{f}=a N^{b}=119.3(70000)^{-0.1441}=23.9 kpsi[/latex]

Question 6.8: A 1015 hot-rolled steel bar has been machined to a diameter ...

Related Answered Questions