Products

## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Ads. Manager

Advertise your business, and reach millions of students around the world.

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## 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