Question 6.20: A rotating shaft experiences a steady torque T = 1360LN(1, 0...

Establish the endurance strength. From Eqs. (6–70) to (6–72) and Eq. (6–20), p. 280,

S^{′}{e}= \begin{cases} 0.506 \overline {S}_{ut} LN(1, 0.138)  kpsi   or  MPa & \overline {S}_{ut} ≤ 212  kpsi (1460 MPa) \\ 107LN(1, 0.139)  kpsi & \overline {S}_{ut} > 212  kpsi \\ LN(1, 0.139)  MPa & \overline {S}_{ut} > 1460  MPa \end{cases}             (6-70)

k_{a} = a \overline{S}^{b}_{ut} LN(1,C)                ( \overline{S}_{ut} in kpsi or MPa)                       (6–72)

k_{b}= \begin{cases} (d/0.3)^{−0.107} = 0.879d^{−0.107} & 0.11 ≤ d ≤ 2 in \\ 0.91d^{−0.157} & 2 < d ≤ 10 in \\  (d/7.62)^{−0.107} = 1.24d^{−0.107} & 2.79 ≤ d ≤ 51 mm \\ 1.51d^{−0.107} & 51 < d ≤ 254 mm \end{cases}               (6-20)

S^{′}_{e} = 0.506(86.2)LN(1, 0.138) = 43.6LN(1, 0.138)  kpsi
k_{a} = 2.67(86.2)^{−0.265}LN(1, 0.058) = 0.820LN(1, 0.058)
k_{b} = (1.1/0.30)^{−0.107} = 0.870
k_{c} = k_{d} = k_{f} = LN(1, 0)
S_{e} = 0.820LN(1, 0.058)0.870(43.6)LN(1, 0.138)
\overline{S}_{e} = 0.820(0.870)43.6 = 31.1  kpsi
C_{Se} = (0.058^{2} + 0.138^{2})^{1/2} = 0.150

and so S_{e} = 31.1LN(1, 0.150)  kpsi.

Stress (in kpsi):

σ_{a} =\frac {32K_{f}M_{a}}{πd^{3}} =\frac {32(1.50)LN(1, 0.11)1.26LN(1, 0.05)}{π(1.1)^{3}}

\overline{σ}_{a} =\frac {32(1.50)1.26}{π(1.1)^{3}} = 14.5  kpsi

C_{σ a} = (0.11^{2} + 0.05^{2})^{1/2} = 0.121

τ_{m} =\frac {16K_{f s}T_{m}}{πd^{3}} =\frac {16(1.28)LN(1, 0.11)1.36LN(1, 0.05)}{π(1.1)^{3}}

\overline {τ}_{m} =\frac {16(1.28)1.36}{π(1.1)^{3}} = 6.66  kpsi

C_{τm} = (0.11^{2} + 0.05^{2})^{1/2} = 0.121

\overline{σ}^{′}_{a} =\left(\overline {σ}^{2}_{a} + 3 \overline{τ}^{2}_{a}\right)^{1/2}= [14.5^{2} + 3(0)^{2}]^{1/2} = 14.5  kpsi

\overline {σ}^{′}_{m} =(\overline{σ}^{2}_{m} + 3 \overline {τ}^{ 2}_{m})^{1/2}= [0 + 3(6.66)^{2}]^{1/2} = 11.54  kpsi

r =\frac { \overline{σ}^{′}_{a}}{\overline{σ}^{′}_{m}} =\frac {14.5}{11.54} = 1.26

Strength: From Eqs. (6–80) and (6–81),

\overline {S}_{a} =\frac {r^{2} \overline{S}^{2}_{ut}}{2 \overline {S}_{e}} \left[ -1+\sqrt {1 +\left(\frac {2 \overline{S}_{e}}{r\overline {S}_{ut}}\right)^{2}}\right]                (6-80)

C_{Sa} =\frac {(1 + C_{Sut} )^{2}}{1 + C_{Se}} \frac { \left\{-1+\sqrt {1+\left[ \frac {2 \overline{S}_{e}(1 + C_{Se})}{r \overline{S}_{ut} (1 + C_{Sut} )}\right]^{2}}\right\}} {\left[-1+ \sqrt{1+\left (\frac {2 \overline {S}_{e}}{r \overline {S}_{ut}}\right)^{2} }\right]}-1                      (6-81)

\overline{S}_{a} =\frac {1.26^{2}86.2^{2}}{2(31.1)}  \left\{ -1+ \sqrt{ 1+\left[ \frac {2(31.1)}{1.26(86.2)}\right]^{2} } \right\}= 28.9 kpsi

C_{Sa} =\frac {(1 + 0.045 )^{2}}{1 + 0.150} \frac {-1+\sqrt {1+\left[ \frac {2 (31.1)(1 + 0.15)}{1.26 (86.2) (1 +0.045 )}\right]^{2}}} {-1+ \sqrt{1+\left [\frac {2 (31.1)}{1.26 (86.2)}\right]^{2} }}-1= 0.134

Reliability: Since S_{a} = 28.9LN(1, 0.134) kpsi and σ^{′}_{a} = 14.5LN(1, 0.121) kpsi, Eq. (5–44), p. 242, gives

z =\frac {0 − μ_{y}}{\hat{σ}_{y}} = − \frac{μ_{y}}{σ_{y}} =−\frac {ln μ_{n} − ln \sqrt {1 + C^{2}_{n}}}{ \sqrt {ln(1 + C^{2}_{n})}}\dot{=}  \frac {− ln (μ_{n}/ \sqrt {1 + C^{2}_{n}}} {\sqrt {ln  (1 + C^{2}_{n})}}                 (5-44)

z =−\frac {ln\left(\frac {\overline{S}_{a}}{\overline{σ}_{a}} \sqrt {\frac {1 + C^{2}_{σa}}{1 +C^{2}_{S_{a}}}}\right)}{\sqrt{ln[(1 +C^{2}_{S_{a}})(1 + C^{2}_{σa})]}}=−\frac {ln\left(\frac {28.9}{14.5} \sqrt {\frac {1 + 0.121^{2}}{1 +0.134^{2}}}\right)}{\sqrt{ln[(1 +0.134^{2})(1 + 0.121^{2})]}}= −3.83

From Table A–10 the probability of failure is p_{f} = 0.000 065, and the reliability is, against fatigue,

R = 1 − p_{f} = 1 − 0.000 065 = 0.999 935

Table A–10
Cumulative Distribution Function of Normal (Gaussian) Distribution

The chance of first-cycle yielding is estimated by interfering S_{y} with σ^{′}_{max}. The quantity σ^{′}_{max} is formed from σ^{′}_{a}+ σ^{′}_{m}. The mean of {σ}^{′}_{max} is \overline{σ}^{′}_{a} +\overline{σ}^{′}_{m} = 14.5 + 11.54 = 26.04 kpsi. The coefficient of variation of the sum is 0.121, since both COVs are 0.121, thus C_{σ  max} = 0.121. We interfere S_{y} = 56LN(1, 0.077) kpsi with {σ}^{′}_{max}= 26.04LN (1, 0.121) kpsi. The corresponding z variable is

z =−\frac {ln\left(\frac {56}{26.04} \sqrt {\frac {1 + 0.121^{2}}{1 +0.077^{2}}}\right)}{\sqrt{ln[(1 +0.077^{2})(1 + 0.121^{2})]}}= −5.39

which represents, from Table A–10, a probability of failure of  pproximately 0.0^{7}358 [which represents 3.58( 10^{−8})] of first-cycle yield in the fillet.

The probability of observing a fatigue failure exceeds the probability of a yield failure, something a deterministic analysis does not foresee and in fact could lead one to expect a yield failure should a failure occur. Look at the {σ}^{′}_{a}S_{a} interference and the {σ}^{′}_{max}S_{y} interference and examine the z expressions. These control the relative probabilities. A deterministic analysis is oblivious to this and can mislead. Check your statistics text for events that are not mutually exclusive, but are independent, to quantify the probability of failure:

p_{f} = p(yield) + p(fatigue) − p(yield  and  fatigue)
= p(yield) + p(fatigue) − p(yield)p(fatigue)
= 0.358(10^{−7}) + 0.65(10^{−4}) − 0.358(10^{−7})0.65(10^{−4}) = 0.650(10^{−4})
R = 1 − 0.650(10^{−4}) = 0.999 935

against either or both modes of failure.