Problem 11.3. Skylight on spherical dome

Meridian force at the bottom, N_α [kN/m]=

Hoop force at the bottom, N_φ [kN/m]=

Meridian force at the top, N_α [kN/m]=

Hoop force at the top, N_φ [kN/m]=

Step 1.  Determine the meridian force from the vertical equilibrium. Give values at the top and at the bottom of the dome.

Check meridian forcesHide meridian forces

The free body diagram of an arbitrary parallel cut of the dome characterized by the angle, α is given in the Figure below.

The radius of the sphere is

$R=\frac{a_2}{\sin {\alpha }_0}=\frac{10.0}{\sin 60°}=11.55 \mathrm{m}$

The meridian force of the shell of revolution is expressed from the vertical equilibrium.

Eq(11-32)

$N_{\alpha }=\frac{–P}{2a\pi \sin \alpha }$

where P is the resultant of the vertical line load of the skylight.

Check resultant of the loadHide resultant

The resultant of the vertical line load is 

$P=p2a_{\mathit{1}}\pi =2\times 2\times 5.0\times \mathrm{\pi }=62.83 \mathrm{kN}$

Substituting the load resultant into the vertical equilibrium the meridian force becomes

$N_{\alpha }=\frac{–P}{2a\pi \sin \alpha }=\frac{–p2a_1\pi }{2a\pi \sin \alpha }=\frac{–p{a}_1}{a\sin \alpha }=\frac{–p{a}_1}{R{\sin }^2\alpha }$

Value of the meridian force at the bottom is

${\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=\frac{–p{a}_1}{a_2\sin 60°}=\frac{–2.0\times 5.0}{10.0\times \sin 60°}\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{155}\frac{\mathbf{kN}}{\mathbf{m}}$

At the top

 ${\alpha }_1={\sin }^{–1}\left(\frac{a_1}{R}\right)={\sin }^{–1}\left(\frac{5.0}{11.55}\right)=25.65°$

and the meridian force becomes

${\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{65}\mathbf{°}\mathbf{)}=\frac{–p{a}_1}{a_1\sin 25.65°}\mathbf{=}\mathbf{–}\mathbf{4}\mathbf{.}\mathbf{619}\frac{\mathbf{kN}}{\mathbf{m}}$

Step 2.  Determine the hoop force from the equilibrium perpendicular to the surface. Give values at the top and at the bottom of the dome.

Check hoop forcesHide hoop forces

The equilibrium perpedicular to the surface results in the following equation:

Eq(11-33)

$p_{\perp }=\frac{N_{\alpha }}{R_{\alpha }}+\frac{N_{\phi }}{R_{\phi }}=\frac{N_{\alpha }}{R}+\frac{N_{\phi }}{R}$

The curvatures of the sphere is the same in every directions.

The load perpendicular to the surface is zero, thus the hoop force is equal and opposite of the meridian force:

$N_{\phi }=–N_{\alpha }$

The values of the hoop force at the bottom and at the top of the dome are:

$$\begin{gathered} {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=–N_{\alpha }(\alpha ={\alpha }_0=60°)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{155}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{65}\mathbf{°}\mathbf{)}=–N_{\mathrm{\alpha }}(\mathrm{\alpha }=25.65°)\mathbf{=}\mathbf{–}\mathbf{4}\mathbf{.}\mathbf{619}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

Step 3.  Draw membrane force diagrams.

Check diagramsHide diagrams

The free body diagram of an arbitrary parallel cut of the dome characterized by the angle, α is given in the Figure below.

The radius of the sphere is

$R=\frac{a_2}{\sin {\alpha }_0}=\frac{10.0}{\sin 60°}=11.55 \mathrm{m}$

The meridian force of the shell of revolution is expressed from the vertical equilibrium.

Eq(11-32)

$N_{\alpha }=\frac{–P}{2a\pi \sin \alpha }$

where P is the resultant of the vertical line load of the skylight:

$P=p2a_{\mathit{1}}\pi =2\times 2\times 5.0\times \mathrm{\pi }=62.83 \mathrm{kN}$

Substituting the load resultant into the vertical equilibrium the meridian force becomes

$N_{\alpha }=\frac{–P}{2a\pi \sin \alpha }=\frac{–p2a_1\pi }{2a\pi \sin \alpha }=\frac{–p{a}_1}{a\sin \alpha }=\frac{–p{a}_1}{R{\sin }^2\alpha }$

Value of the meridian force at the bottom is

${\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=\frac{–p{a}_1}{a_2\sin 60°}=\frac{–2.0\times 5.0}{10.0\times \sin 60°}\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{155}\frac{\mathbf{kN}}{\mathbf{m}}$

At the top

 ${\alpha }_1={\sin }^{–1}\left(\frac{a_1}{R}\right)={\sin }^{–1}\left(\frac{5.0}{11.55}\right)=25.65°$

and the meridian force becomes

${\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{65}\mathbf{°}\mathbf{)}=\frac{–p{a}_1}{a_1\sin 25.65°}\mathbf{=}\mathbf{–}\mathbf{4}\mathbf{.}\mathbf{619}\frac{\mathbf{kN}}{\mathbf{m}}$

The equilibrium perpedicular to the surface results in the following equation:

Eq(11-33)

$p_{\perp }=\frac{N_{\alpha }}{R_{\alpha }}+\frac{N_{\phi }}{R_{\phi }}=\frac{N_{\alpha }}{R}+\frac{N_{\phi }}{R}$

The curvatures of the sphere is the same in every directions.

The load perpendicular to the surface is zero, thus the hoop force is equal and opposite of the meridian force:

$N_{\phi }=–N_{\alpha }$

The values of the hoop force at the bottom and at the top of the dome are:

$$\begin{gathered} {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=–N_{\alpha }(\alpha ={\alpha }_0=60°)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{155}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{65}\mathbf{°}\mathbf{)}=–N_{\mathrm{\alpha }}(\mathrm{\alpha }=25.65°)\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{619}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

The membrane force diagrams are given in the Figure.