Question 10.4: A semi-circular groove of radius 3 mm is machined in a 50 mm...

For the shaft dimensions given, D/d = 50/(50 – 6) = 1.14 and r/d = 3/44 = 0.068

(a) For tensile load

Nominal stress  σ_{nom}=\frac{P}{A} =\frac{50\times 10^{3}}{\pi \times (22\times 10^{-3})^{2}} =32.9  MN/m^{2}

From Fig. 10.23           K_{t}=2.51

Maximum stress  = 2.51 × 32.9 = 82.6  MN/m^{2}

(b) For bending

Nominal stress σ_{nom}=\frac{32M}{\pi d^{3}} =\frac{32\times 150}{\pi \times (44\times 10^{-3})^{3}} =18  MN/m^{2}

and from Fig. 10.24     K_{t}=2.24

Maximum stress Maximum stress  = 2.24 × 18 = 40.3  MN/m^{2}

(c) For torsion

Nominal stress  τ_{nom }=\frac{16 T}{\pi d^{3}} =\frac{16\times 320}{\pi \times (44\times 10^{-3})^{3}} =19.1  MN/m^{2}

and from Fig. 10.25    Kt_{s}=1.65

Maximum stress = 1.65 × 19.1 = 31.5  MN/m^{ 2}

(d) For the combined loading the direct stresses due to bending and tension add to give a total maximum direct stress of 82.6 + 40.3 = 122.9  MN/m^{2} which will then act in conjunction with the shear stress of 3 1.5  MN/m^{2} as shown on the element of Fig. 10.49.

Then either by Mohrs circle or the use of eqn. (13.11)^{†} the maximum principal stress will be
σ_{1} = 130.5  MN/m^{2}.
With a maximum shear stress of τ_{max}= 69  MN/m^{2}.