Problem 11.4. Paraboloid of revolution

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 is given in the Figure below.

The radii of curvature are derived in the literature:

$R_{\alpha }=\frac{2a^2}{f{\cos }^3\alpha }, R_{\phi }=\frac{2a^2}{f\cos \alpha }$

Geometry of the cut is characterized by the angle, α

$$\begin{gathered} r=R_{\phi }\sin \alpha =\frac{2a^2\tan \alpha }{f} \\ \alpha ={\tan }^{–1}\frac{fr}{2a^2} \end{gathered}$$

From the inital data:

${\alpha }_0\left(r=a\right)={\tan }^{–1}\frac{f}{2a}={\tan }^{–1}\frac{4.0}{2\times 10.0}=11.31°$

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

Eq(11-32)

$N_{\alpha }=\frac{–P}{2r\pi \sin \alpha }=\frac{–p_{\mathrm{s}}{r}^2\mathrm{\pi }}{2r\pi \sin \alpha }=\frac{–p_{\mathrm{s}}r}{2\sin \alpha }=\frac{–p_{\mathrm{s}}}{2\sin \alpha }\frac{2a^2\tan \alpha }{f}=\frac{–p_{\mathrm{s}}}{\cos \alpha }\frac{a^2}{f}$

where P is the resultant of the distributed snow load over the dome part.

Values of the meridian force at the bottom and at the top are$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{31}\mathbf{°}\mathbf{)}=\frac{–1.00\times 10.0^2}{4\times \cos 11.31°}\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{50}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=\frac{–1.00\times 10.0^2}{4\times \cos 0°}\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{00}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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)

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

The normal component of the snow load referred to the unit area of tangent plane is

$p_{\perp }=–p_{\mathrm{s}}{\cos }^2\alpha$

The hoop force is expressed as

$N_{\phi }=\left(p_{\perp }–\frac{N_{\alpha }}{R_{\alpha }}\right)R_{\phi }$

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

$$\begin{gathered} R_{\alpha }\left(\mathrm{\alpha }={\mathrm{\alpha }}_0=11.31°\right)=\frac{2\times 10.0^2}{4.0\times {\cos }^{3}11.31°}=50.03 \mathrm{m}, R_{\phi }=\frac{2\times 10.0^2}{4.0\times \cos 11.31°}=50.99 \mathrm{m} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{31}\mathbf{°}\mathbf{)}=\left(–1.00\times {\cos }^{2}11.31°+\frac{25.50}{50.03}\right)50.99\mathbf{=}\mathbf{–}\mathbf{24}\mathbf{.}\mathbf{51}\frac{\mathbf{kN}}{\mathbf{m}} \\ {R}_{\alpha }(\mathrm{\alpha }={\mathrm{\alpha }}_0=0°)=\frac{2\times 10.0^2}{4.0\times {\cos }^{3}0°}=50.00 \mathrm{m}, R_{\phi }=\frac{2\times 10.0^2}{4.0\times \cos 0°}=50.00 \mathrm{m} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=(–1.00\times {\cos }^{2}0°+\frac{25.00}{50.00})50.00\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{00}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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)

$\mathit{H}=a{p}_{\mathrm{p}}=a{N}_{\alpha }\cos {\alpha }_0=10.0\times 25.50\times \cos 11.31°\mathbf{=}\mathbf{250}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kN}$

The free body diagram of an arbitrary parallel cut of the dome is given in the Figure below.

The radii of curvature are derived in the literature:

$R_{\alpha }=\frac{2a^2}{f{\cos }^3\alpha }, R_{\phi }=\frac{2a^2}{f\cos \alpha }$

Geometry of the cut is characterized by the angle, α

$$\begin{gathered} r=R_{\phi }\sin \alpha =\frac{2a^2\tan \alpha }{f} \\ \alpha ={\tan }^{–1}\frac{fr}{2a^2} \end{gathered}$$

From the inital data:

${\alpha }_0\left(r=a\right)={\tan }^{–1}\frac{f}{2a}={\tan }^{–1}\frac{4.0}{2\times 10.0}=11.31°$

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

Eq(11-32)

$N_{\alpha }=\frac{–P}{2r\pi \sin \alpha }=\frac{–p_{\mathrm{s}}{r}^2\mathrm{\pi }}{2r\pi \sin \alpha }=\frac{–p_{\mathrm{s}}r}{2\sin \alpha }=\frac{–p_{\mathrm{s}}}{2\sin \alpha }\frac{2a^2\tan \alpha }{f}=\frac{–p_{\mathrm{s}}}{\cos \alpha }\frac{a^2}{f}$

where P is the resultant of the distributed snow load over the dome part.

Values of the meridian force at the bottom and at the top are$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{31}\mathbf{°}\mathbf{)}=\frac{–1.00\times 10.0^2}{4\times \cos 11.31°}\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{50}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=\frac{–1.00\times 10.0^2}{4\times \cos 0°}\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{00}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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

Eq(11-33)

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

The normal component of the snow load referred to the unit area of tangent plane is

$p_{\perp }=–p_{\mathrm{s}}{\cos }^2\alpha$

The hoop force is expressed as

$N_{\phi }=\left(p_{\perp }–\frac{N_{\alpha }}{R_{\alpha }}\right)R_{\phi }$

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

$$\begin{gathered} R_{\alpha }\left(\mathrm{\alpha }={\mathrm{\alpha }}_0=11.31°\right)=\frac{2\times 10.0^2}{4.0\times {\cos }^{3}11.31°}=50.03 \mathrm{m}, R_{\phi }=\frac{2\times 10.0^2}{4.0\times \cos 11.31°}=50.99 \mathrm{m} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathbf{\alpha }\mathbf{=}{\mathbf{\alpha }}_{\mathbf{0}}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{31}\mathbf{°}\mathbf{)}=\left(–1.00\times {\cos }^{2}11.31°+\frac{25.50}{50.03}\right)50.99\mathbf{=}\mathbf{–}\mathbf{24}\mathbf{.}\mathbf{51}\frac{\mathbf{kN}}{\mathbf{m}} \\ {R}_{\alpha }(\mathrm{\alpha }={\mathrm{\alpha }}_0=0°)=\frac{2\times 10.0^2}{4.0\times {\cos }^{3}0°}=50.00 \mathrm{m}, R_{\phi }=\frac{2\times 10.0^2}{4.0\times \cos 0°}=50.00 \mathrm{m} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{°}\mathbf{)}=(–1.00\times {\cos }^{2}0°+\frac{25.00}{50.00})50.00\mathbf{=}\mathbf{–}\mathbf{25}\mathbf{.}\mathbf{00}\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

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)

$\mathit{H}=a{p}_{\mathrm{p}}=a{N}_{\alpha }\cos {\alpha }_0=10.0\times 25.50\times \cos 11.31°\mathbf{=}\mathbf{250}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kN}$