A 1.5-in-diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 16 kip. Because of the ends, and the fillet radius, a fatigue stress-concentration factor Kf is 1.85 for 106or larger life.

Question

A 1.5-in-diameter bar has been machined from an AISI 1050 cold-drawn bar.

This part is to withstand a fluctuating tensile load varying from 0 to 16 kip.

Because of the ends,and the fillet radius, a fatigue stress-concentration factor [latex]K_f \ is \ 1.85 \ for \ 10^6[/latex] or larger life.

Find [latex]S_a \ and \ S_m [/latex] and the factor of safety guarding against fatigue and first-cycle yielding, using (a) the Gerber fatigue line and (b) the ASME-elliptic fatigue line.

Accepted Answer

We begin with some preliminaries. From Table A–20,

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

[latex]S_{ut}= 100 \ kpsi \ and \ S_y= 84 [/latex] kpsi.
Note that [latex]F_a= F_m= 8[/latex] kip.

The Marin factors are, deterministically,
[latex] ka= 2.70(100)^{−0.265}= 0.797[/latex] : Eq. (6–19)

[latex]k_a=as^b_{ut}[/latex]

, Table 6–2,p. 288

Table 6–2 Parameters for Marin Surface Modification Factor, Eq. (6–19)
Surface Finish	Factor a		Exponent b
Surface Finish	[latex]S_{ut}[/latex], kpsi	[latex]S_{ut},[/latex] MPa	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_b= 1[/latex] (axial loading, see[latex] k_c[/latex])

[latex]k_c=0.85[/latex] : Eq. (6–26), p. 290

[latex]k_d=k_e=k_f=1[/latex]

[latex]S_e=0.797\left(1 ight) 0.850\left(1 ight) \left(1 ight) \left(1 ight) 0.5\left(100 ight) =33.9 [/latex] kpsi: Eqs. (6–8), (6–18), p. 282, p. 287
The nominal axial stress components

(6–8)

[latex]S_{e}^{\prime}= \begin{cases}0.5 S_{u t} & S_{u t} \leq 200 \mathrm{kpsi}(1400 \mathrm{MPa}) \ 100 \mathrm{kpsi} & S_{u t}>200 \mathrm{kpsi} \ 700 \mathrm{MPa} & S_{u t}>1400 \mathrm{MPa}\end{cases}[/latex]

(6–18)

[latex]S_{e}=k_{a} k_{b} k_{c} k_{d} k_{c} k_{f} S_{e}^{\prime}[/latex]

[latex]\sigma _{a0} \ and \ \sigma _{m0} are \sigma_{a0}=\frac{4F_a}{\pi d^2} =\frac{4\left(8 ight) }{\pi 1.5^2}=4.53 [/latex] kpsi [latex]\sigma _{m0}=\frac{4F_m}{\pi d^2} =\frac{4\left(8 ight) }{\pi 1.5^2} =4.53[/latex] kpsi.

Applying [latex]K_f[/latex] to both components [latex]\sigma_{a0}[/latex] and [latex]\sigma _{m0}[/latex] constitutes a prescription of no notch
yielding:

[latex]\sigma _a=K_f\sigma_{a0}=1.85\left(4.53 ight) =8.38 \ kpsi \ =\sigma _m[/latex]

(a) Let us calculate the factors of safety first. From the bottom panel from Table 6–7

Table 6–7 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria
Intersecting Equations	Intersection Coordinates
[latex]\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}} ight)^2 =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\frac{r^2S^2_{ut}}{S_e} \left[-1+\sqrt{1+\left(\frac{2S_e}{rS_{ut}} ight)^2 } ight] [/latex] [latex]S_m=\frac{S_a}{r} [/latex]
[latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\frac{rS_y}{1+r} [/latex] [latex]S_m=\frac{rS_y}{1+r} [/latex]
[latex]\frac{S_a}{S_e} +\left(\frac{S_m}{S_{ut}} ight)^2 =1[/latex] [latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex]	[latex]S_m=\frac{2S^2_{ut}}{2S_e} \left[1-\sqrt{1+\left(\frac{2S_e}{S_{ut}} ight)^2 \left(1-\frac{S_y}{S_e} ight) } ight] [/latex] [latex]S_{a}=S_{y}-S_{m}, r_{ ext {crit }}=S_{a} / S_{m}[/latex]
Fatigue factor of safety [latex]n_f=\frac{1}{2} \left(\frac{S_{ut}}{\sigma _m} ight)^2\frac{\sigma _a}{S_e}\left[-1+\sqrt{1+\left(\frac{2\sigma _mS_e}{S_{ut}\sigma _a} ight)^2 } ight] \ \sigma _m\gt 0[/latex]

the factor of safety for fatigue is:

[latex]n_f=\frac{1}{2} \left(\frac{100}{8.38} ight)^2\left(\frac{8.38}{33.9} ight) \left\{-1+\sqrt{1+\left[\frac{2\left(8.38 ight)33.9 }{100\left(8.38 ight) } ight]^2 } ight\} =3.66[/latex]

From Eq. (6–49) the factor of safety guarding against first-cycle yield is

[latex]n_y=\frac{s_y}{\sigma _a+\sigma _m} =\frac{84}{8.38+8.38} =5.01[/latex]

Thus, we see that fatigue will occur first and the factor of safety is 3.68. This can be seen in Fig. 6–28 where the load line intersects the Gerber fatigue curve first at point B.
If the plots are created to true scale it would be seen that

[latex]n_f = {OA}/{OB}[/latex].

From the first panel of Table 6–7, [latex]r =\sigma _a/\sigma _m= 1[/latex],

[latex]S_a=\frac{\left(1 ight)^2100^2 }{2\left(33.9 ight) } \left\{-1+\sqrt{1+\left[\frac{2\left(33.9 ight) }{\left(1 ight)100 } ight]^2 } ight\} =30.7[/latex] kpsi

[latex]S_m=\frac{S_a}{r} =\frac{30.7}{1} =30.7[/latex] kpsi

As a check on the previous result,

[latex] n_f = {OA}/{OB} = {S_a}/{\sigma _a}={S_m}/{\sigma _m} = {30.7}/{8.38}=3.66[/latex]

and we see total agreement.
We could have detected that fatigue failure would occur first without drawing Fig. 6–28 by calculating [latex]r_{crit}[/latex] .

From the third row third column panel of Table 6–7, the intersection point between fatigue and first-cycle yield is

[latex]S_m=\frac{100^2}{2\left(33.9 ight) } \left[1-\sqrt{1+\left(\frac{2\left(33.9 ight) }{100} ight)^2\left(1-\frac{84}{33.9} ight) } ight] =64.0[/latex] kspi

[latex]S_a=S_y-S_m=84-64=20[/latex] kpsi

The critical slope is thus

[latex]r_{crit}=\frac{S_a}{S_m} =\frac{20}{64} =0.312[/latex]

which is less than the actual load line of[latex] r = 1[/latex].

This indicates that fatigue occurs before first-cycle-yield.
(b) Repeating the same procedure for the ASME-elliptic line, for fatigue

[latex]n_f=\sqrt{\frac{1}{\left({8.38}/{33.9} ight)^2+\left({8.38}/{84} ight)^2 } } = 3.75 [/latex]

Again, this is less than [latex]n_y= 5.01 [/latex] and fatigue is predicted to occur first.

From the first row second column panel of Table 6–8

Table 6–8 Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for ASME- Elliptic and Langer Failure Criteria
Intersecting Equations	Intersection Coordinates
[latex]\left(\frac{S_u}{S_e} ight)^2 +\left(\frac{S_m}{S_{y}} ight)^2 =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\sqrt{\frac{r^2S^2_{e}S^2_y}{S^2_e+r^2S^2_y} } [/latex] [latex]S_m=\frac{S_a}{r} [/latex]
[latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex] Load line[latex]r=\frac{S_a}{S_m}[/latex]	[latex]S_a=\frac{rS_y}{1+r} [/latex] [latex]S_m=\frac{S_y}{1+r} [/latex]
[latex]\left(\frac{S_a}{S_e} ight)^2 +\left(\frac{S_m}{S_{y}} ight)^2 =1[/latex] [latex]\frac{S_a}{S_y} +\frac{S_m}{S_y} =1[/latex]	[latex]S_a=0 , \frac{2S_yS^2_e}{S^2_e+S^2_y} [/latex] [latex]S_{m}={S_y} - {S_a}, r_{ctrit}={S_a}/{S_m}[/latex]
Fatigue factor of safety [latex]n_f=\sqrt{\frac{1}{\left({\sigma _a}/{S_{e}} ight)^2+\left({\sigma _m}/{S_{y}} ight)^2 } }[/latex]

, with [latex]r = 1[/latex] , we obtain the coordinates [latex]S_a \ and \ S_m [/latex] of point B in Fig. 6–29 as

[latex]S_a=\sqrt{\frac{\left(1 ight)^233.9^2\left(84 ight)^2 }{33.9^2+\left(1 ight)84^2 } } =31.4[/latex] kpsi

, [latex]S_m=\frac{S_a}{r} =\frac{31.4}{1} =31.4[/latex] kpsi

To verify the fatigue factor of safety, [latex]n_f={S_a}/{\sigma _a}={31.4}/{8.38}=3.75[/latex].

As before, let us calculate [latex]r_{crit}[/latex].

From the third row second column panel of Table 6–8,

[latex]S_a=\frac{2\left(84 ight)33.9^2 }{33.9^2+84^2}=23.5[/latex] , [latex]S_m=S_y-S_a=84-23.5[/latex]=60.5 kpsi

[latex]r_{crit}=\frac{S_a}{S_m} =\frac{23.5}{60.5} =0.388[/latex]

which again is less than [latex]r = 1[/latex], verifying that fatigue occurs first with [latex]n_f = 3.75[/latex].
The Gerber and the ASME-elliptic fatigue failure criteria are very close to each other and are used interchangeably. The ANSI/ASME Standard B106.1M–1985 uses ASME-elliptic for shafting.

Question 2.6.10: A 1.5-in-diameter bar has been machined from an AISI 1050 co...

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