Question 16.14: Determine the support reactions in the parabolic arch shown ...
Determine the support reactions in the parabolic arch shown in Fig. 16.29 assuming that the second moment of area of the cross section of the arch varies in accordance with the secant assumption.
The equation of the arch may be shown to be
y=\frac{4 h}{L^{2}}\left(L x-x^{2}\right) (i)
Again we shall release the arch at B as in Fig. 16.26(b). Then
\begin{aligned} & M_{0}=R_{\mathrm{A}, V^{x}}^{\prime} \quad(0 \leqslant x \leqslant a) \\ & M_{0}=R_{\mathrm{A}, V^{x}}^{\prime}-W(x-a) \quad(a \leqslant x \leqslant L) \end{aligned}
in which R_{\mathrm{A}, \mathrm{V}}^{\prime} is the vertical reaction at \mathrm{A} in the released structure. Now taking moments about \mathrm{B} we have
R_{\mathrm{A}, \mathrm{V}}^{\prime} L-W(L-a)=0
from which
R_{\mathrm{A}, \mathrm{V}}^{\prime}=\frac{W}{L}(L-a)
Substituting in the expressions for M_{0} gives
\begin{array}{ll} M_{0}=\frac{W}{L}(L-a) x \quad (0 \leqslant x \leqslant a) & (\text{ii}) \\ M_{0}=\frac{W}{L}(L-x) \quad(a \leqslant x \leqslant L) & (\text{iii}) \end{array}
The denominator in Eq. (16.15)
R_1=\frac{\int_{\text {Profile }} M_0 y \mathrm{~d} x}{\int_{\text {Profile }} y^2 \mathrm{~d} x} (16.15)
may be evaluated separately. Thus, from Eq. (i)
\int_{\text {Profile }} y^{2} \mathrm{~d} x=\int_{0}^{L}\left(\frac{4 h}{L^{2}}\right)^{2}\left(L x-x^{2}\right)^{2} \mathrm{~d} x=\frac{8 h^{2} L}{15}
Then, from Eq. (16.15) and Eqs (ii) and (iii)
which gives
R_{1}=\frac{5 W a}{8 h L^{3}}\left(L^{3}+a^{3}-2 L a^{2}\right) (iv)
The remaining support reactions follow from a consideration of the statical equilibrium of the arch.
