Question 2.15: A cantilever 50 mm diameter is to carry an axial load 50 kN ...
A cantilever 50 mm diameter is to carry an axial load 50 kN at an eccentricity of 5 mm below horizontal diameter in the vertical plane of symmetry along with a torque of 1.5 x 10^{6} N mm as shown in Fig. 2.24. Calculate the stresses on a plane, normal of which is inclined at 30° clockwise to axis of the cantilever at a point on the lower edge of the vertical plane of symmetry.
Consider an element at a lowermost point P.
Direct axial stress
\sigma_x =\frac{-50000}{\pi \times 50^2 / 4} =-25.46 MPa (\text{Compression})Bending compressive stress at P due to eccentric load,
\sigma_x =\frac{-M \times y}{I}=\frac{-(50000 \times 5) \times 25}{\pi \times 50^4 / 64} =-20.37 MPa (\text{Compression})Total axial load,
\sigma_x =-25.46-20.37 =-45.83 MPa (\text{Compression})Shear stress due to torsion,
\tau _{x y}=\frac{T\times y}{J}=\frac{1.5\times 10^6 \times 25}{\pi \times 50^4 / 30} = 61.11 MPa (Clockwise)
(Complementary shear \tau _{y x}, is not shown in Fig. 2.24 for simplicity)
Plane AB shown in Fig. 2.25 has its normal inclined at 30° clockwise to the axial direction \sigma _x .
Stresses on plane AB,
\sigma_{x^{\prime}}= \frac{-45.83}{2}+\frac{-45.83}{2} \cos -60^{\circ}-61.11\sin -60^{\circ}=18.55 MPa (Tension)
\tau_{x^{\prime}y}=\frac{-45.83}{2} \sin -60^{\circ}+61.11 \cos -60^{\circ} = 50.4 MPa (Clockwise)
