Question 4.10: A hollow circular column carries a projecting bracket, which...

Given P = 25 kN e = 500 mm.

S_{y t}=200 N / mm ^{2} \quad(f s)=4 \quad d_{i}=0.8 d_{o} .

Step I Calculation of permissible tensile stress

\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{200}{4}=50 N / mm ^{2} .

Step II Calculation of direct compressive and bending stresses

d_{i}=0.8 d_{0} .

The direct compressive stress is given by,

\frac{P}{A}=\frac{25 \times 10^{3}}{\frac{\pi}{4}\left(d_{o}^{2}-d_{i}^{2}\right)} .

=\frac{25 \times 10^{3}}{\frac{\pi}{4}\left[d_{0}^{2}-\left(0.8 d_{0}\right)^{2}\right]}=\left(\frac{88419.41}{d_{0}^{2}}\right) N / mm ^{2}          (i).

The bending stresses are tensile on the left side and compressive on the right side of the cross-section.
The tensile stress due to bending moment is given by,

\frac{P e y}{I}=\frac{25 \times 10^{3}(500)\left(0.5 d_{0}\right)}{\frac{\pi}{64}\left[d_{0}^{4}-\left(0.8 d_{0}\right)^{4}\right]} .

=\left(\frac{215657104.5}{d_{0}^{3}}\right) N / mm ^{2}          (ii).

Step III Calculation of outer and inner diameters
The resultant tensile stress is obtained by subtracting (i) from (ii). Equating the resultant stress to permissible tensile stress,

\left(\frac{215657104 \cdot 5}{d_{0}^{3}}\right)-\left(\frac{88419.41}{d_{0}^{2}}\right)=50 .

or    \left(d_{0}^{3}+1768.39 d_{0}\right)=4313142 \cdot 09 .

The above expression indicates cubic equation and it is solved by trial and error method. The trial values of d_o and corresponding values of left hand side are tabulated as follows:

From the above table, it is observed that the value of d_o is between 159 and 160 mm

\therefore \quad d_{0}=160 mm .

d_{i}=0.8 d_{o}=0.8(160)=128 mm .