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## Q. 12.4

Determine the forces in columns A and B of the vessel shown in Fig. 12.9. ## Verified Solution

Axial force in columns A and B due to W is

$F=\frac{W}{N}=\frac{240}{8}=30^{k}$

Axial force in column A due to M is

\begin{aligned}F &=\frac{2 M}{N R} \\&=\frac{2 \times 2000}{8 \times 5}=100^{k}\end{aligned}

total axial load in column A = -50 -100 = -150 kip

total axial load in column B = -50 kip

The shearing stresses transferred to column A are zero. Those transferred to column B are determined from

$H=\frac{V Q}{I t}$

The moment of inertia I of the whole cross section in Fig. 12.9 c is given by $\pi r^{3} t$. The quantity Q of the crosshatched area in this figure is given by

\begin{aligned}Q &=(\pi r t)\left(\frac{2}{\pi} r\right) \\&=2 r^{2} t\end{aligned}

The force H is then given by

$H=\frac{V\left(2 r^{2} t\right)}{\left(\pi r^{3} t\right)(2 t)}=\frac{V}{\pi r t}=\frac{2 V}{A}$

$H=\frac{2 \times 50}{2 \pi(60)}=0.2653 lb / in.$

Horizontal force in column B is

\begin{aligned}H &=(0.2653)\left(\frac{2 \pi r}{N}\right) \\&=12.50 kips\end{aligned}

This force H is normally resolved into two components as shown in Fig. 12.9 d.

Force U is a radial force on the shell and force X is a horizontal force in the plane of the cross-bracing.

$U=H \cot \alpha=12.5 \times 0.414=5.18 \text { kips }$

$X=\frac{H}{\sin \alpha}=\frac{12.5}{0.924}=13.53 kips$

The force X introduces additional compressive force in column B as shown in Fig. 12.9 e. The distance between columns is

$l=\frac{2 \pi r}{N}=3.93 ft$

The approximate height of the columns is 20 ft. Hence, angle β is about 11 degrees and the axial force F in column B is

$F=\frac{X}{\tan \beta}=69.61 kips$

$\text { force } E=\frac{X}{\sin \beta}=70.91 kips$

total force in column A = -150 kip

total force in column B = -50 -69.61 = -119.61 kip

total force in bracing = 70.91 kips