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## Q. 2.9

The parabolic arch shown in Figure 2.9 has a second moment of area that varies directly as the secant of the slope of the arch rib. Determine the bending moment at 2, neglecting the effects of axial and shearing forces. ## Verified Solution

The redundant forces consist of the reactions H and V at 3, and the cut-back structure is a curved cantilever.

The equation of the arch axis is:

y = ax(l – x)

The horizontal deflection of 3 due to condition (i) is obtained by considering a virtual-unit load applied horizontally inwards at 3. Then, over the span from x = 0 to x = l/2:

m = y

and

M = Wl/2 – Wx

Thus the horizontal deflection at 3 is:

\begin{aligned}x_{3}^{\prime} &=W \int_{0}^{l / 2}(l / 2-x) y dx / l E I_{o} \\&=Wa \int_{0}^{l / 2}\left(l^{2} x / 2-3 l x^{2} / 2+x^{3}\right) dx / E I_{o} \\&=Wal^{4} / 64 E I_{o}\end{aligned}

The horizontal deflection of 3 due to condition (ii) is:

\begin{aligned}x_{3}^{\prime \prime} &=H \int_{0}^{l} y^{2} dx / E I_{o} \\&=H a^{2} \int_{0}^{l}\left(l^{2} x^{2}-2 l x^{3}+x^{4}\right) dx / E I_{o} \\&=H a^{2} l^{5} / 30 E I_{o}\end{aligned}

The horizontal deflection of 3 due to condition (iii) is:

\begin{aligned}x_{3}^{\prime \prime \prime} &=-V \int_{0}^{l} x y dx / E I_{o} \\&=-V a \int_{0}^{l}\left(l x^{2}-x^{3}\right) dx / E I_{o} \\&=Val^{4} / 12 E I_{o}\end{aligned}

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

$x_{s}=x_{3}^{\prime}+x_{3}^{\prime \prime}+x_{3}^{\prime \prime \prime}=0$

Thus:

W/32 – V/6 + Hal/15 = 0     (1)

The vertical deflection of 3 due to condition (i) is obtained by considering a virtual-unit load applied vertically upwards at 3. Then, over the span from x = 0 to x = l/2:

m = l – x

and

M = Wx – Wl/2

Thus, the vertical deflection of 3 is:

\begin{aligned}y_{3}^{\prime} &=W \int_{0}^{l / 2}(x-l / 2)(l-x) dx / E I_{o} \\&=-W \int_{0}^{l / 2}\left(l^{2} / 2-3 l x / 2+x^{2}\right) dx / E I_{o} \\&=-5 W l^{3} / 48 E I_{o}\end{aligned}

The vertical deflection of 3 due to condition (ii) is:

\begin{aligned}y_{3}^{\prime \prime} &=-H \int_{0}^{l} yx dx / E I_{o} \\&=-H a l^{4} / 12 E I_{o}\end{aligned}

The vertical deflection of 3 due to condition (iii) is:

\begin{aligned}y_{3}^{\prime \prime \prime} &=V \int_{0}^{l} x^{2} dx / E I_{o} \\&=V l^{3} / 3 E I_{o}\end{aligned}

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

$y_{s}=y_{3}^{\prime}+y_{3}^{\prime \prime}+y_{3}^{\prime \prime \prime}=0$

Thus,

-5W/16 + V – Hal/4 = 0     (2)

Solving (1) and (2) simultaneously:

H = 5W/6al

and

$V=25 W / 48$

Thus, the bending moment at 2 is:

\begin{aligned}M_{2} &=V l / 2-H a l^{2} / 4 \\&=5 W l / 96\end{aligned}