Problem 11.13. Spherical dome with not adequate membrane support

Maximum bending moment, M_max [kNm/m]=

Follow steps in Example 11.11.

Step 1. Calculate the vertical reaction force.

Check vertical reactionHide vertical reaction

The vertical support force, A is calculated from the vertical equilibrium:

See Problem 11.2.

$A=\frac{p_{\mathrm{g}}A}{2a_0\pi }=\frac{{\gamma }_{\mathrm{c}}t2R^2\pi (1–\cos {\alpha }_0)}{2R\sin {\alpha }_0\pi }=\frac{{\gamma }_{\mathrm{c}}tR\left(1–\cos {\alpha }_0\right)}{\sin {\alpha }_0}=\frac{25\times 0.3\times 10.0\left(1–\cos 60°\right)}{\sin 60°}=43.30 \frac{\mathrm{kN}}{\mathrm{m}}$

Step 2. Determine the force component which causes the bending of the edge.

Check perpendicular componentHide perpendicular component

Step 3. Calculate the bending moment from the edge disturbance.

Check momentHide moment

The moment is approximated by the moment of the osculating cylinder subjected to a line load p = A_⊥:

See Figure 10.45a and Eq.(11-82).

${\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=A_{\perp }\frac{0.32}{\lambda }=A_{\perp }\frac{0.32}{1.32}\sqrt{Rt}=21.65\frac{0.32}{1.32}\sqrt{10.0\times 0.3}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{091}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{10.0\times 3.0}=1.039 \mathrm{m}$

from the support.

Note that this bending moment, due to not adequate membrane support is much higher than the bending moment calculated in Problem 11.12.

Follow steps in Example 11.11.

The vertical support force, A is calculated from the vertical equilibrium:

See Problem 11.2.

$A=\frac{p_{\mathrm{g}}A}{2a_0\pi }=\frac{{\gamma }_{\mathrm{c}}t2R^2\pi (1–\cos {\alpha }_0)}{2R\sin {\alpha }_0\pi }=\frac{{\gamma }_{\mathrm{c}}tR\left(1–\cos {\alpha }_0\right)}{\sin {\alpha }_0}=\frac{25\times 0.3\times 10.0\left(1–\cos 60°\right)}{\sin 60°}=43.30 \frac{\mathrm{kN}}{\mathrm{m}}$

A has a component in the direction of the meridian force and one which is perpendicular to it, the latter one, A_⊥ – which is equal to the shear force at the edge – causes the bending of the shell.

$A_{\perp }=A\cos {\alpha }_0=43.30\times \cos 60°=21.65 \frac{\mathrm{kN}}{\mathrm{m}}$

The moment is approximated by the moment of the osculating cylinder subjected to a line load p = A_⊥:

See Figure 10.45a and Eq.(11-82).

${\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=A_{\perp }\frac{0.32}{\lambda }=A_{\perp }\frac{0.32}{1.32}\sqrt{Rt}=21.65\frac{0.32}{1.32}\sqrt{10.0\times 0.3}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{091}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{10.0\times 3.0}=1.039 \mathrm{m}$

from the support.

Note that this bending moment, due to not adequate membrane support is much higher than the bending moment calculated in Problem 11.12.