Problem 11.15. Skylight on spherical dome with not adequate membrane support

Maximum bending moment, M_max [kNm/m]=

Follow steps in Example 11.11.

Step 1. Determine the force component which causes the bending of the edge.

Check perpendicular componentHide perpendicular component

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

α₁ is calculated in Problem 11.3.

$p_{\perp }=p\cos {\alpha }_1=2.00\times \cos 25.65°=1.803 \frac{\mathrm{kN}}{\mathrm{m}}$

Step 2. 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_⊥:

See Figure 366a and Eq.(11-82).

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=p_{\perp }\frac{0.32}{\lambda }=p_{\perp }\frac{0.32}{1.32}\sqrt{Rt}=p_{\perp }\frac{0.32}{1.32}\sqrt{\frac{a_2}{\sin {\alpha }_0}t}= \\ =1.803\frac{0.32}{1.32}\sqrt{\frac{10.0}{\sin 60°}\times 0.3}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8135}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}} \end{gathered}$$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{\frac{a_2}{\sin {\alpha }_0}t}=0.6\sqrt{\frac{10.0}{\sin 60°}\times 0.3}=1.117 \mathrm{m}$

from the support.

Follow steps in Example 11.11.

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

α₁ is calculated in Problem 11.3.

$p_{\perp }=p\cos {\alpha }_1=2.00\times \cos 25.65°=1.803 \frac{\mathrm{kN}}{\mathrm{m}}$

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

See Figure 366a and Eq.(11-82).

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=p_{\perp }\frac{0.32}{\lambda }=p_{\perp }\frac{0.32}{1.32}\sqrt{Rt}=p_{\perp }\frac{0.32}{1.32}\sqrt{\frac{a_2}{\sin {\alpha }_0}t}= \\ =1.803\frac{0.32}{1.32}\sqrt{\frac{10.0}{\sin 60°}\times 0.3}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8135}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}} \end{gathered}$$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{\frac{a_2}{\sin {\alpha }_0}t}=0.6\sqrt{\frac{10.0}{\sin 60°}\times 0.3}=1.117 \mathrm{m}$

from the support.