Question 10.3: A rectangular bar with shoulder fillet is subjected to a uni...

For applied moment
From simple bending theory, nominal stress (related to smaller part of the bar) is:

 \sigma _{nom} =\frac{My}{I} =M\times \frac{d}{2}\times \frac{12}{hd^{3}}=\frac{6M}{hd^{2}}
=\frac{6\times 100}{10\times 10^{-3}\times (25\times 10^{-3})^{2}}=96  MN/m^{2}

Now from Fig. 10.22 the elastic stress concentration factor for D/d = 2 and r/d = 0.1 is:

                K_{t} = 1.85
Maximum stress = 1.85 × 96 = 177.6  MN/m^{2}.

(a) For tensile load
Again for smallest part of the bar

 \sigma _{nom} =\frac{P}{hd} =\frac{20\times 10^{3}}{10\times 10^{-3}\times 25\times 10^{-3}}=80  MN/m^{2}

and from Fig. 10.2,  K_{t} = 2.44

Maximum stress = 2.44 × 80 = 195.2  MN/m^{2}

(b) For combined bending and tensile load
Since the maximum stresses arising from both the above conditions will be direct stresses in the fillet radius then the effects may be added directly, i.e. the most adverse stress condition will arise in the bending tensile fillet when the maximum stress due to combined tension and bending will be:

\sigma _{max} =K_{t}\sigma _{b_{nom}} +K^{′}_{t} \sigma _{d_{nom}}

= 177.6 + 195.2 = 372.8  MN/m^{2}