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Solution :
Given :
L = 1 in
d = 0.75 in
D = 1 in
Fillet radius, r = 0.063 in
[tex]$K_{t_{bending}=\frac{\sigma_{FEA_{bending }}}{\sigma_{Nominal_{bending }}}$[/tex]
We know that :
[tex]$\sigma_{b} = \frac{32M}{\pi d^3}$[/tex]
[tex]$\sigma_{b} = \frac{32 \times (2998.63 \times 25.4)}{\pi (0.75 \times 25.4)^3}$[/tex]
[tex]$=112.27 \ N/mm^2$[/tex]
[tex]$\sigma_{FEA} = 1.3 \times 10^8 \ N/m^2$[/tex]
[tex]$=1.3 \times 10^8 \times 10^{-6}$[/tex]
[tex]$=1.3 \times 10^2$[/tex] MPa
= 130 MPa
Therefore, the stress concentration factor is :
[tex]$k_t=\frac{130}{112.27}$[/tex]
= 1.157922
