Problem 11.2. Spherical dome

Meridian force at the bottom [kN/m]=

Hoop force at the bottom [kN/m]=

Meridian force at the top [kN/m]=

Hoop force at the top [kN/m]=

Force in the boundary ring [kN]=

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

Eq(11-32)

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

where G is the resultant of the distributed load over the surface of the dome part.

Check resultant of the loadHide resultant

The surface of the sphere part above the cut is

$A=2R^2\pi \left(1–\cos \alpha \right)$

Thus the resultant of the gravity load is 

$G=p_{\mathrm{g}}A={\gamma }_{\mathrm{c}}tA={\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)$

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

$N_{\alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2a\pi \sin \alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2R\pi {\sin }^2\alpha }=\frac{–{\gamma }_{\mathrm{c}}tR}{\left(1+\cos \alpha \right)}$

Values of the meridian force at the bottom and at the top are

$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=\frac{–25\times 0.3\times 10}{\left(1+\cos 60°\right)}\mathbf{=}\mathbf{–}\mathbf{50}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=\frac{–25\times 0.3\times 10}{(1+\cos 0°)}\mathbf{=}\mathbf{–}\mathbf{37}\mathbf{.}\mathbf{5}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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

$p_{\perp }=–p_g\cos \alpha =–{\gamma }_{\mathrm{c}}t\cos \alpha$

The hoop force is expressed as

$N_{\phi }=p_{\perp }R–N_{\alpha }$

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

$$\begin{gathered} {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=–{\gamma }_{\mathrm{c}}tR\cos 60°–N_{\alpha }(\alpha ={\alpha }_0=60°)=–25\times 0.3\times 10\times \cos 60°+50\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=–{\gamma }_{\mathrm{c}}tR\cos 0°–N_{\mathrm{\alpha }}(\mathrm{\alpha }=0°)=–25\times 0.3\times 10+37.5\mathbf{=}\mathbf{–}\mathbf{37}\mathbf{.}\mathbf{5}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

Step 3.  Draw membrane force diagrams.

Check diagramsHide diagrams

Step 4.  Give the force in the boundary ring.

Check ring force at the edgeHide ring force

The ring at the edge is supported vertically and it is unsupported radially. The opposite of the horizontal component of the meridian force loads the ring.

The tensile force of the ring can be calculated from the pressure vessel formula:

Eq(11-35)

$\mathit{H}=a_0{p}_{\mathrm{p}}=a_0{N}_{\alpha }\cos {\alpha }_0=R\sin 60°\times 50\cos 60°\mathbf{=}\mathbf{216}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{kN}$

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

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

Eq(11-32)

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

where G is the resultant of the distributed load over the surface of the dome part.

The surface of the sphere part above the cut is

$A=2R^2\pi \left(1–\cos \alpha \right)$

Thus the resultant of the gravity load is 

$G=p_{\mathrm{g}}A={\gamma }_{\mathrm{c}}tA={\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)$

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

$N_{\alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2a\pi \sin \alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2R\pi {\sin }^2\alpha }=\frac{–{\gamma }_{\mathrm{c}}tR}{\left(1+\cos \alpha \right)}$

Values of the meridian force at the bottom and at the top are

$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=\frac{–25\times 0.3\times 10}{\left(1+\cos 60°\right)}\mathbf{=}\mathbf{–}\mathbf{50}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=\frac{–25\times 0.3\times 10}{(1+\cos 0°)}\mathbf{=}\mathbf{–}\mathbf{37}\mathbf{.}\mathbf{5}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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

$p_{\perp }=–p_g\cos \alpha =–{\gamma }_{\mathrm{c}}t\cos \alpha$

The hoop force is expressed from the equilibrium perpendicular to the surface:

$N_{\phi }=p_{\perp }R–N_{\alpha }$

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

$$\begin{gathered} {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{60}\mathbf{°}\mathbf{)}=–{\gamma }_{\mathrm{c}}tR\cos 60°–N_{\alpha }(\alpha ={\alpha }_0=60°)=–25\times 0.3\times 10\times \cos 60°+50\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\phi }\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=–{\gamma }_{\mathrm{c}}tR\cos 0°–N_{\mathrm{\alpha }}(\mathrm{\alpha }=0°)=–25\times 0.3\times 10+37.5\mathbf{=}\mathbf{–}\mathbf{37}\mathbf{.}\mathbf{5}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

The membrane force diagrams are given in the Figure below.

The ring at the edge is supported vertically and it is unsupported radially. The opposite of the horizontal component of the meridian force loads the ring.

The tensile force of the ring can be calculated from the pressure vessel formula:

Eq(11-35)

$\mathit{H}=a_0{p}_{\mathrm{p}}=a_0{N}_{\alpha }\cos {\alpha }_0=R\sin 60°\times 50\cos 60°\mathbf{=}\mathbf{216}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{kN}$