A rotating shaft is made of 42- × 4-mm AISI 1018 cold-drawn steel tubing and has an 6-mm-diameter hole drilled transversely through it. Estimate the factor of safety guarding against fatigue and static failures using the Gerber and Langer failure criteria for the ollowing loading condition

Question

A rotating shaft is made of [latex]42- imes 4-[/latex]mm AISI 1018 cold-drawn steel tubing and has a 6-mm-diameter hole drilled transversely through it.

Estimate the factor of safety guarding against fatigue and static failures using the Gerber and Langer failure criteria for the following loading conditions:
(a) The shaft is subjected to a completely reversed torque of 120 N · m in phase with a completely reversed bending moment of 150 N · m.
(b) The shaft is subjected to a pulsating torque fluctuating from 20 to 160 N · m and a steady bending moment of 150 N · m.

Accepted Answer

Here we follow the procedure of estimating the strengths and then the stresses, followed by relating the two.

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](\frac{3}{4} \text { to } 1 \frac{1}{4} \text { in ). }[/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 MPa (kpsi) Strength,	Yield Strength, MPa (kpsi)	Elongation in 2 in, %
UNS No.	SAE and/or AISI No.	Processing	Tensile MPa (kpsi) Strength,	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

we find the minimum strengths to be [latex]S_{ut}= 440 \ MPa \ and \ S_y=370 \ MPa[/latex].

The endurance limit of the rotating-beam specimen is [latex]0.5\left(440\right) = 220 \ MPa[/latex].
The surface factor, obtained from Eq. (6–19)

[latex]K_a=aS^B_{ut}[/latex]

and Table 6–2, p. 287, is

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

is

[latex]K_a=1.51S^{-0.265}_{ut}=4.51\left(440\right) ^{-0.265}=0.899[/latex]

From Eq. (6–20)

[latex] k_{b}=\left\{\begin{array}{ll}(d / 0.3)^{-0.107}=0.879 d^{-0.107} & 0.11 \leq d \leq 2 \text { in } \0.91 d^{-0.157} & 2

the size factor is

[latex]k_b=\left(\frac{d}{7.62} \right) ^{-0.107}=\left(\frac{42}{7.62} \right)^{-0.107} =0.833[/latex]

The remaining Marin factors are all unity, so the modified endurance strength [latex]S_e[/latex] is
[latex]S_e=0.899\left(0.833\right) 220=165[/latex] MPa

(a) Theoretical stress-concentration factors are found from Table A–16.

Table A–16 Approximate Stress- Concentration Factor[latex] K_t [/latex] for Bending of a Round Bar or Tube with a Transverse Round Hole Source: R. E. Peterson, Stress-Concentration Factors, Wiley,New York, 1974, pp. 146, 235
The nominal bending stress is[latex]\sigma _0= M/Znet[/latex] where[latex] Z_{net} [/latex] is a reduced value of the section modulus and is defined by
[latex]Z_{net}=\frac{\pi A }{32D} \left(D^4-d^4\right) [/latex]

Values of A are listed in the table. Use d = 0 for a solid bar
[latex]{d}/{D}[/latex]
[latex]{a}/{D}[/latex]	A	[latex]K_{t}[/latex]	A	[latex]K_{t}[/latex]	A	[latex]K_{t}[/latex]
0.050	0.92	2.63	0.91	2.55	0.88	2.42
0.075	0.89	2.55	0.88	2.43	0.86	2.35
0.100	0.86	2.49	0.85	2.36	0.83	2.27
0.125	0.82	2.41	0.82	2.32	0.80	2.20
0.15	0.79	2.39	0.79	2.29	0.76	2.15
0.175	0.76	2.38	0.75	2.26	0.72	2.10
0.20	0.73	2.39	0.72	2.23	0.68	2.07
0.225	0.69	2.40	0.68	2.21	0.65	2.04
0.25	0.67	2.42	0.64	2.18	0.61	2.00
0.275	0.66	2.48	0.61	2.16	0.58	1.97
0.30	0.64	2.52	0.58	2.14	0.54	1.94

Using [latex]{a}/{D} ={6}/{4}2 = 0.143 \ and {d}/{D} = {34}/{42} = 0.810[/latex],

and using linear interpolation, we obtain [latex] A = 0.798 \ and \ K_t= 2.366 \ for \ bending \ ; \ and \ A = 0.89 \ and \ K_{ts}=1.75 [/latex] for torsion.
Thus, for bending,
[latex]Z_{net}=\frac{\pi A }{32D} \left(D^4-d^4\right)=\frac{\pi \left(0.798\right) }{32\left(42\right) } \left[\left(42\right)^4-\left(34\right)^4 \right] =3.31\left(10^3\right) \ mm^3[/latex]
and for torsion
[latex]J_{net}=\frac{\pi A }{32} \left(D^4-d^4\right)=\frac{\pi \left(0.8\right) }{32\left(42\right) } \left[\left(42\right)^4-\left(34\right)^4 \right] =155\left(10^3\right) \ mm^4[/latex]

Next, using Figs. 6–20 and 6–21, pp. 295–296, with a notch radius of 3 mm we find the notch sensitivities to be 0.78 for bending and 0.81 for torsion.
The two corresponding fatigue stress-concentration factors are obtained from Eq. (6–32)

[latex]K_f=1+q\left(K_t-1\right) [/latex] or [latex]K_{fs}=1+q_{shear}\left(K_{ts}-1\right)[/latex]
as

[latex]K_f=1+q\left(K_t-1\right)=K_f=1+0.78\left(2.366-1\right)=2.07[/latex]

[latex]K_{fs}=1+0.81\left(1.75-1\right)=1.61[/latex]

The alternating bending stress is now found to be

[latex]\sigma _{xa}=K_f\frac{M}{Z_{net}} =2.07\frac{150}{3.31\left(10^{-6}\right) } =93.8\left(10^6\right) \ Pa=93.8 \ MPa[/latex]

and the alternating torsional stress is

[latex]\tau _{xya}= K_{fs}\frac{TD}{2J_{net}} =1.61\frac{120\left(42\right)\left(10^{-3}\right) }{2\left(155\right)\left(10^{-9}\right) } =26.2\left(10^6\right) \ Pa= 26.2 \ MPa[/latex]

The midrange von Mises component [latex]\sigma '_m[/latex] is zero. The alternating component [latex]\sigma '_m[/latex] is given by

[latex]\sigma '_a=\left(\sigma ^2_{xa}+3\tau ^2_{xya}\right)^{{1}/{2}}=\left[93.8^2+3\left(26.2^2\right) \right]^{{1}/{2}}=104.2 \ MPa[/latex]

Since [latex]S_e= S_a[/latex], the fatigue factor of safety [latex]n_f[/latex] is

[latex]n_f=\frac{S_a}{\sigma '_a} =\frac{165}{104.2} =1.58[/latex]

The first-cycle yield factor of safety is

[latex]n_y=\frac{S_y}{\sigma '_a} =\frac{370}{105.6} =3.50[/latex]

There is no localized yielding; the threat is from fatigue. See Fig. 6–32.
(b) This part asks us to find the factors of safety when the alternating component is due to pulsating torsion, and a steady component is due to both torsion and bending.

We have [latex]T_a={\left(160-20\right) }/{2}=70 \ N.m \ and \ T_m={\left(160+20\right) }/{2}=90 \ N.m[/latex].

The corresponding amplitude and steady-stress components are

[latex]\tau _{xya}=K_{fs}\frac{T_aD}{2J_{net}} =1.61\frac{70\left(42\right)\left(10^{-3}\right) }{2\left(155\right)\left(10^{-9}\right) } =15.3 (10^6) pa =15.3 \ MPa[/latex]

[latex]\tau _{xym}=K_{fs}\frac{T_mD}{2J_{net}} =1.61\frac{90\left(42\right)\left(10^{-3}\right) }{2\left(155\right)\left(10^{-9}\right) } =19.7 (10^6) pa = 19.7 \ MPa[/latex]

The steady bending stress component [latex]\sigma _{xm}[/latex] is

[latex]\sigma _{xm}=K_f=\frac{M_m}{Z_{net}} =2.07\frac{150}{3.31\left(10^{-6}\right)} =93.8\left(10^6\right) \ Pa=93.8 \ MPa[/latex]

The von Mises components [latex]\sigma' _a \ and \sigma' _m[/latex] are

[latex]\sigma' _a=\left[3\left(15.3\right)^2 \right] ^{{1}/{2}}=26.5 \ MPa[/latex].

[latex]\sigma' _m=\left[93.8^2+3\left(19.7\right)^2 \right] ^{{1}/{2}}=99.8 \ MPa[/latex]

From Table 6–7, p. 307,

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}}\right)^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}} \right)^2 } \right] [/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}}\right)^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}} \right)^2 \left(1-\frac{S_y}{S_e} \right) } \right][/latex] [latex]S_{a} = S_{y} - S _{m} , r_{crit}={S_a}/{S_m}[/latex]
Fatigue factor of safety [latex]n_f=\frac{1}{2} \left(\frac{S_{ut}}{\sigma _m} \right)^2\frac{\sigma _a}{S_e}\left[-1+\sqrt{1+\left(\frac{2\sigma _mS_e}{S_{ut}\sigma _a} \right)^2 } \right] \ \sigma _m\gt 0[/latex]

the fatigue factor of safety is

[latex]n_f=\frac{1}{2}\left(\frac{440}{99.8} \right) ^2\frac{26.5}{165} \left\{-1+\sqrt{1+\left[\frac{2\left(99.8\right)165 }{440\left(26.5\right) } \right]^2 } \right\} =3.12 [/latex]

From the same table, with [latex]r={\sigma '_a}/{\sigma'_m}={26.5}/{99.8}=0.28[/latex] , the strengths can be shownto be [latex]S_a= 85.5 \ MPa \ and \ S_m= 305 \ MPa[/latex].

See the plot in Fig. 6–32.
The first-cycle yield factor of safety [latex]n_y[/latex] is

[latex]n_y=\frac{S_y}{\sigma '_a+\sigma '_m} =\frac{370}{26.5+99.8} =2.93[/latex]

There is no notch yielding. The likelihood of failure may first come from first-cycle yielding at the notch. See the plot in Fig. 6–32.

Question 2.6.14: A rotating shaft is made of 42- × 4-mm AISI 1018 cold-drawn ...

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