Products

## Holooly Rewards

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

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

## 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. 2.8

The parabolic arch shown in Figure 2.8 has a second moment of area that varies directly as the secant of the slope of the arch rib. The value of the thrust required to restore the arch to its original span is $H_{1}$, and the value of the thrust required to reduce the deflection of 2 to zero is $H_{2}$. Determine the ratio of $H_{2}$ to $H_{1}$, neglecting the effects of axial and shearing forces.

## Verified Solution

The equation of the arch axis, taking the origin of coordinates at 1, is:

y = x(l – x)/l

The second moment of area at any section is given by:

I = $I_{o}$  sec α

= $I_{o}$  ds/dx

where $I_{o}$ is the second moment of area at the crown.

The horizontal deflection at 3 due to W is obtained by considering a virtual unit load applied horizontally inwards at 3. Then from (i) the deflection due to W is obtained by integrating over the length of the arch:

\begin{aligned}1 \times x_{3}^{\prime} &=\int Mm ds/EI \\&=\int Mm dx/EI_{o} \\&=-2 \int_{0}^{l / 2} Wxy dx /2EI_{o} \\&=-W \int_{0}^{l / 2}\left(l x^{2}-x^{3}\right) dx / l E I_{o} \\&=-5 W l^{3} / 192 E I_{o}\end{aligned}

The horizontal deflection at 3 due to $H_{1}$ is:

\begin{aligned}1 \times x_{3}^{\prime \prime} &=2 \int_{0}^{l / 2} H_{1} y^{2} dx /EI_{o} \\&=2 H_{1} \int_{0}^{l / 2}\left(l^{2} x^{2}-2 l x^{3}+x^{4}\right) dx / l^{2} E I_{o} \\&=H_{1} l^{3} / 30 E I_{o}\end{aligned}

The total horizontal deflection of 3 in the original structure is:

\begin{aligned}x_{3} &=x_{3}^{\prime}+x_{3}^{\prime \prime} \\&=0\end{aligned}

Thus,

$H_{l}=25 W / 32$

The vertical deflection of 2 due to W is obtained by considering a virtual unit load applied vertically upwards at 2. Then, from (ii):

\begin{aligned}1 \times y_{2}^{\prime} &=-2 \int_{0}^{l / 2} W x^{2} dx /4EI_{o} \\&=-W \int_{0}^{l / 2} x^{2} dx / 2E I_{o} \\&=-W l^{3} / 48\end{aligned}

The vertical deflection at 2 due to $H_{2}$ is:

\begin{aligned}1 \times y_{2}^{\prime \prime} &=2 \int_{0}^{l / 2} H_{2} xy dx / EI_{o} \\&=H_{2} \int_{0}^{l / 2}\left(l x^{2}-x^{3}\right) dx / l^{2} E I_{o} \\&=5 H_{2} l^{3} / 192 E I_{o}\end{aligned}

The total vertical deflection of 2 in the original structure is:

$y_{2}=y_{2}^{\prime}+y_{2}^{\prime \prime}=0$

Thus,

$H_{2}=4 W / 5$

and

\begin{aligned}H_{2} / H_{1} &=128 / 125 \\ &=1.024 \end{aligned}