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

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_{ut}[/latex] = 50 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 A–20 Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm ( [latex]\frac {3}{4} to 1\frac {1}{4}[/latex] in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification.] Source: 1986 SAE Handbook, p. 2.15.

8 Brinell Hardness	7 Reduction in Area, %	6 Elongation in 2 in, %	5 Yield Strength, MPa (kpsi)	4 Tensile Strength, MPa (kpsi)	3 Proces-sing	2 SAE and/or AISI No.	1 UNS No.
86	55	30	170 (24)	300 (43)	HR	1006	G10060
95	45	20	280 (41)	330 (48)	CD		G10060
95	50	28	180 (26)	320 (47)	HR	1010	G10100
105	40	20	300 (44)	370 (53)	CD		G10100
101	50	28	190 (27.5)	340 (50)	HR	1015	G10150
111	40	18	320 (47)	390 (56)	CD		G10150
116	50	25	220 (32)	400 (58)	HR	1018	G10180
126	40	15	370 (54)	440 (64)	CD		G10180
111	50	25	210 (30)	380 (55)	HR	1020	G10200
131	40	15	390 (57)	470 (68)	CD		G10200
137	42	20	260 (37.5)	470 (68)	HR	1030	G10300
149	35	12	440 (64)	520 (76)	CD		G10300
143	40	18	270 (39.5)	500 (72)	HR	1035	G10350
163	35	12	460 (67)	550 (80)	CD		G10350
149	40	18	290 (42)	520 (76)	HR	1040	G10400
170	35	12	490 (71)	590 (85)	CD		G10400
163	40	16	310 (45)	570 (82)	HR	1045	G10450
179	35	12	530 (77)	630 (91)	CD		G10450
179	35	15	340 (49.5)	620 (90)	HR	1050	G10500
197	30	10	580 (84)	690 (100)	CD		G10500
201	30	12	370 (54)	680 (98)	HR	1060	G10600
229	25	10	420 (61.5)	770 (112)	HR	1080	G10800
248	25	10	460 (66)	830 (120)	HR	1095	G10950

Table 6–4 Effect of Operating Temperature on the Tensile Strength of Steel.* ([latex]S_{T}[/latex] = tensile strength at operating temperature; [latex]S_{RT}[/latex] = tensile strength at room temperature; 0.099 ≤[latex] \hat {σ}[/latex] ≤ 0.110)

[latex]S_{T}/S_{RT}[/latex]	Temperature, °F	[latex]S_{T}/S_{RT}[/latex]	Temperature, °C
1.000	70	1.000	20
1.008	100	1.010	50
1.020	200	1.020	100
1.024	300	1.025	150
1.018	400	1.020	200
0.995	500	1.000	250
0.963	600	0.975	300
0.927	700	0.943	350
0.872	800	0.900	400
0.797	900	0.843	450
0.698	1000	0.768	500
0.567	1100	0.672	550
		0.549	600

*Data source: Fig. 2–9.

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

The ultimate strength at 550°F is then

[latex](S_{ut} )_{550◦} = (S_{T} /S_{RT} )_{550◦} (S_{ut} )_{70◦} =0.979(50) = 49.0 kpsi[/latex]

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

[latex]S′_{e}=\begin{cases}0.5S_{ut} & S_{ut} ≤ 200 kpsi (1400 MPa) \100 kpsi & S_{ut} > 200 kpsi \ 700 MPa & S_{ut} > 1400 MPa \end{cases} [/latex] (6-8)

[latex]S′_{e} = 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

Table 6–2 Parameters for Marin Surface Modification Factor, Eq.(6–19)

Exponent b	Factor a		Surface Finish
Exponent b	[latex]S_{ut} [/latex],MPa	[latex]S_{ut}[/latex], kpsi	Surface Finish
−0.085	1.58	1.34	Ground
−0.265	4.51	2.70	Machined or cold-drawn
−0.718	57.7	14.4	Hot-rolled
−0.995	272.	39.9	As-forged

[latex]k_{a} = a S^{b}_{ut}[/latex] (6–19)

[latex]k_{a} = aS^{b}_{ut} = 2.70(49^{−0.265}) = 0.963[/latex]

For axial loading, from Eq. (6–21), the size factor [latex]k_{b}[/latex] = 1, and from Eq.(6–26) the loading factor is [latex]k_{c}[/latex]= 0.85. The temperature factor [latex]k_{d}[/latex]= 1, 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}[/latex]= 0.814. Finally, since no other onditions
were given, the miscellaneous factor is [latex]k_{f}[/latex]= 1. The endurance limit for the part is estimated by Eq. (6–18) as

Table 6–5 Reliability Factors [latex]k_{e}[/latex] Corresponding to 8 Percent Standard Deviation of the Endurance Limit

Reliability Factor [latex]k_{e}[/latex]	Transformation Variate [latex]z_{a}[/latex]	Reliability, %
1.000	0	50
0.897	1.288	90
0.868	1.645	95
0.814	2.326	99
0.753	3.091	99.9
0.702	3.719	99.99
0.659	4.265	99.999
0.620	4.753	99.9999

[latex]k_{b} = 1[/latex] (6–21)

[latex]S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}[/latex] (6–18)

[latex]k_{c}=\begin{cases}1 & bending \bending0.85 &axial\axial0.59 &torsion^{17}\end{cases}[/latex]

[latex]S_{e} = k_{a}k_{b}k_{c}k_{d}k_{e}k_{f} S′_{e}[/latex]
= 0.963(1)(0.85)(1)(0.814)(1)24.5 = 16.3 kpsi

For the fatigue strength at 70 000 cycles we need to construct the S-N equation. From p. 277, since [latex]S_{ut}[/latex] = 49 < 70 kpsi, then f 0.9. From Eq. (6–14)

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

[latex]a =\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_{ut}}{S_{e}} ight)[/latex] (6–15)

[latex]b= −\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] = 119.3 (70 000)^{−0.1441 }=23.9 kpsi[/latex]

17 Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such as bending, [latex]k_{c}[/latex] = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 5–5. Note: For pure torsion, the distortion energy predicts that [latex](k_{c})_{torsion} = 0.577[/latex].

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

Related Answered Questions