Question 16.13: Determine the support reactions in the semicircular two-pinn...

Again we shall choose the horizontal reaction at the support \mathrm{B} as the release so that R_{\mathrm{B}, \mathrm{H}}\left(=R_{1}\right) is given directly by Eq. (16.14)

R_1=\frac{\int_{\text {Profile }}\left(M_0 y / E I\right) \mathrm{d} s}{\int_{\text {Profile }}\left(y^2 / E I\right) \mathrm{d} s}        (16.14)

in which M_{0} and s are functions of x and y. The computation will therefore be simplified if we use an angular coordinate system so that, from the primary structure shown in Fig. 16.27(b)

M_{0}=R_{\mathrm{B}, \mathrm{V}}^{\prime}(5+5 \cos \theta)-\frac{10}{2}(5+5 \cos \theta)^{2}       (i)

in which R_{\mathrm{B}, \mathrm{V}}^{\prime} is the vertical reaction at \mathrm{B} in the primary structure. From Fig. 16.27(b) in which, from symmetry, R_{\mathrm{B}, \mathrm{V}}^{\prime}=R_{\mathrm{A}, \mathrm{V}}^{\prime}, we have R_{\mathrm{B}, \mathrm{V}}^{\prime}=50  \mathrm{kN}. Substituting for R_{\mathrm{B}, \mathrm{V}}^{\prime} in Eq. (i) we obtain

M_{0}=125 \sin ^{2} \theta    (ii)

Also y=5 \sin \theta and \mathrm{d} s=5 \mathrm{~d} \theta, so that from Eq. (16.14) we have

R_{1}=\frac{\int_{0}^{\pi} 125 \sin ^{2} \theta 5 \sin \theta 5 \mathrm{~d} \theta}{\int_{0}^{\pi} 25 \sin ^{2} \theta 5 \mathrm{~d} \theta}

or

R_{1}=\frac{\int_{0}^{\pi} 25 \sin ^{3} \theta \mathrm{d} \theta}{\int_{0}^{\pi} \sin ^{2} \theta \mathrm{d} \theta}     (iii)

which gives

R_{1}=21.2  \mathrm{kN}\left(=R_{\mathrm{B}, \mathrm{H}}\right)

The remaining reactions follow from a consideration of the statical equilibrium of the arch and are

R_{\mathrm{A}, \mathrm{H}}=21.2  \mathrm{kN} \quad R_{\mathrm{A}, \mathrm{V}}=R_{\mathrm{B}, \mathrm{V}}=50  \mathrm{kN}