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Chapter 2

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.

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 H1, and the value of the thrust required to reduce the deflection of 2 to zero is H2.

Step-by-Step

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}