Question 2.14: A horizontal force 1 kN is applied at the end of a cantileve...
A horizontal force 1 kN is applied at the end of a cantilever bent of diameter 40 mm as shown in Fig. 2.22. Determine maximum principal normal stresses 200 mm above the base.
The maximum stress will occur at a point P on the circumference as shown in Fig. 2.23. Consider free body diagram of a small element at point P, Normal stress due to bending,
\sigma_y =\frac{M \times y}{I}=\frac{(1000 \times 300) \times 20}{\pi \times 40^4 / 64} = 47.75 MPa
Shear stress due to torsion,
\tau_{y x}=\frac{T \times y}{J}=\frac{5 \times 10^5 \times 20}{\pi \times 40^4 / 32}=39.79 MPaPrincipal stresses,
\sigma_{x^{\prime},{max}/{min} }=\frac{47.75}{2}+\sqrt{(47.75 / 2)^2+39.79^2} = 70.28 MPa , -22.53 MPa
2 \theta_p=\tan ^{-1} \frac{-\tau_{x y}}{\left(\sigma_x-\sigma_y\right) / 2}=\tan ^{-1} \frac{-39.79}{47.75 / 2}=-59^{\circ}The orientation of principal stresses is shown in Fig. 2.23, which can be verified by Eq. (2.1) or by Mohr’s circle.
\sigma _{x^{\prime}}=\frac{\sigma _x+\sigma _y}{2}+\frac{\sigma _x-\sigma _y}{2}\cos 2\theta -\tau _{xy}\sin 2 \theta (2.1b)
