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

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

From Table A–20

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

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

and Table 6–2, p. 287, is

is

From Eq. (6–20)

the size factor is

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

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

Table A–16 Approximate Stress- Concentration Factor K_t 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\sigma _0= M/Znet where Z_{net}  is a reduced value of the section modulus and is defined by
Z_{net}=\frac{\pi A }{32D} \left(D^4-d^4\right)
Values of A are listed in the table. Use d = 0 for a solid bar
{d}/{D}
{a}/{D}	A	K_{t}	A	K_{t}	A	K_{t}
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 {a}/{D} ={6}/{4}2 = 0.143 \ and {d}/{D} = {34}/{42} = 0.810,

and using linear interpolation, we obtain A = 0.798 \ and \ K_t= 2.366 \ for \ bending \ ; \ and \ A = 0.89 \ and \ K_{ts}=1.75 for torsion.
Thus, for bending,
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
and for torsion
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

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)

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

The alternating bending stress is now found to be

and the alternating torsional stress is

The midrange von Mises component \sigma '_m is zero. The alternating component \sigma '_m is given by

Since S_e= S_a, the fatigue factor of safety n_f is

The first-cycle yield factor of safety is

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

The corresponding amplitude and steady-stress components are

The steady bending stress component \sigma _{xm}  is

The von Mises components \sigma' _a \ and \sigma' _m are

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

From Table 6–7, p. 307,

the fatigue factor of safety is

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

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

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.