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## Q. 4.10

A hollow circular column carries a projecting bracket, which supports a load of 25 kN as shown in Fig. 4.29. The distance between the axis of the column and the load is 500 mm. The inner diameter of the column is 0.8 times of the outer diameter. The column is made of steel FeE 200 $\left(S_{y t}=200 N / mm ^{2}\right)$ and the factor of safety is 4. The column is to be designed on the basis of maximum tensile stress and compression is not the criterion of failure. Determine the dimensions of the cross-section of the column. ## Verified Solution

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:

 $\left(d_{0}^{3}+1768.39 d_{0}\right)$ $d_o$ 3 640 258.5 150 4 378 942.4 160 4 223 717.6 158 4 300 853.0 159 4 378 942.4 160

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

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

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