Question 2.6.18: Estimate Kf and q for the steel shaft given in Ex. 6–6, p. 2...

Estimate [lstex] K_{f}[/latex] and q for the steel shaft given in Ex. 6–6, p. 296.

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From Ex. 6–6, a steel shaft with S_{ut}= 690 MPa and a shoulder with a fillet of 3 mm was found to have a theoretical stress-concentration factor of K_t.= 1.65. From
Table 6–15,

Table 6–15 Heywood’s Parameter √a and coefficients ofvariation C_{Kf} for steels
\sqrt{a} \left(\sqrt{in} \right) \sqrt{a} \left(\sqrt{mm} \right)
Notch Type S_{ut} \ in \ kpsi S_{ut} \ in   MPa Coefficient of
Variation CKf
Transverse hole {5}/{S_{ut}} {174}/{S_{ut}} 0.10
Shoulder {4}/{S_{ut}} {139}/{S_{ut}} 0.11
Groove {3}/{S_{ut}} {104}/{S_{ut}} 0.15
\sqrt{a} =\frac{139}{S_{ut}} =\frac{139}{690} =0.2014\sqrt{mm}

From Eq. (6–78),

K_f=\frac{K_t}{1+\frac{2\left(K_t-1\right) }{K_t} \frac{\sqrt{a} }{\sqrt{r} } } =\frac{1.65}{1+\frac{2\left(1.65-1\right) }{1.65} \frac{0.2014 }{\sqrt{3} } } =1.51

which is 2.5 percent lower than what was found in Ex. 6–6.

From Table 6–15, CK f = 0.11. Thus from Eq. (6–79),

K_f=\bar{K} _fLN\left(1,C_{K_f}\right)

 

K_f=1.51 LN\left(1,0.11\right)

From Eq. (6–77), with K_t= 1.65

 

\bar{q} =\frac{\bar{K}_f-1 }{K_t-1} \\\hat{\sigma } _q=\frac{C\bar{K}_f}{K_t-1} \\C_q=\frac{C\bar{K}_f }{\bar{K}_f-1 }

 

\bar{q} =\frac{1.51-1 }{1.65-1}=0.785 \\C_q=\frac{C\bar{K}_f }{\bar{K}_f-1 }=\frac{0.11\left(1.51\right) }{1.51-1 }= 0.326\\\hat{\sigma } _q=C_q\bar{q}=0.326\left(0.785\right) =0.256

 

So,

q = LN\left(0.785,0.256\right)

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