Problem 11.5. Truncated cone

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 cone 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 self-weight of the cone part.

Check resultant of the loadHide resultant

The surface of the truncated cone is

$A=\pi \frac{(a^2–a_1^2)}{\cos \alpha }$

Thus the resultant of the gravity load is 

$G=p_{\mathrm{g}}A=p_{\mathrm{g}}\pi \frac{(a^2–a_1^2)}{\cos \alpha }$

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

$N_{\alpha }=\frac{–p_{\mathrm{g}}\pi (a^2–a_1^2)}{2a\pi \sin \alpha \cos \alpha }=\frac{–p_{\mathrm{g}}(a^2–a_1^2)}{2a\sin \alpha \cos \alpha }$

Values of the meridian force at the bottom and at the top are$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{)}=\frac{–12\times \left(20.0^2–10.0^2\right)}{2\times 20.0\times \sin 60°\cos 60°}\mathbf{=}\mathbf{–}\mathbf{207}\mathbf{.}\mathbf{8}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{)}=\frac{–12\times (10.0^2–10.0^2)}{2\times 10.0\times \sin 60°\cos 60°}\mathbf{=}\mathbf{0} \end{gathered}$$

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 meridian of a cone is a straight line, hence one of the principal curvatures is zero, the meridian radius of curvature is infinite, R_α = ∞. Thus, the equilibrium perpendicular to the surface simplifies to the “pressure vessel formula”:

Eq(11-33)

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

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

$p_{\perp }=–p_{\mathrm{s}}\cos \alpha =–12.0\times \cos 60°=–6.0 \frac{\mathrm{kN}}{{\mathrm{m}}^2}$

The hoop radius of curvature, R_φ is:

$R_{\phi }=\frac{a}{\sin \alpha }=\frac{a}{\sin 60°}$

Thus hoop force becomes 

$$\begin{gathered} N_{\phi }=p_{\perp }{R}_{\phi } \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{)}=p_{\perp }\frac{a_1}{\sin \alpha }=–6.0\times \frac{10.0}{\sin 60°}\mathbf{=}\mathbf{–}\mathbf{69}\mathbf{.}\mathbf{28}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{)}=p_{\perp }\frac{a_2}{\sin \alpha }=–6.0\times \frac{20.0}{\sin 60°}\mathbf{=}\mathbf{–}\mathbf{138}\mathbf{.}\mathbf{6}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}} \end{gathered}$$

Step 3.  Draw membrane force diagrams.

Check diagramsHide diagrams

The free body diagram of an arbitrary parallel cut of the cone 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 self-weight of the cone part.

The surface of the truncated cone is

$A=\pi \frac{(a^2–a_1^2)}{\cos \alpha }$

Thus the resultant of the gravity load is 

$G=p_{\mathrm{g}}A=p_{\mathrm{g}}\pi \frac{(a^2–a_1^2)}{\cos \alpha }$

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

$N_{\alpha }=\frac{–p_{\mathrm{g}}\pi (a^2–a_1^2)}{2a\pi \sin \alpha \cos \alpha }$

Values of the meridian force at the bottom and at the top are$$\begin{gathered} {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{)}=\frac{–12\times \left(20.0^2–10.0^2\right)}{2\times 20.0\times \sin 60°\cos 60°}\mathbf{=}\mathbf{–}\mathbf{207}\mathbf{.}\mathbf{8}\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathbf{\alpha }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{)}=\frac{–12\times (10.0^2–10.0^2)}{2\times 10.0\times \sin 60°\cos 60°}\mathbf{=}\mathbf{0} \end{gathered}$$

The meridian of a cone is a straight line, hence one of the principal curvatures is zero, the meridian radius of curvature is infinite, R_α = ∞. Thus, the equilibrium perpendicular to the surface simplifies to the “pressure vessel formula”:

Eq(11-33)

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

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

$p_{\perp }=–p_{\mathrm{s}}\cos \alpha =–12.0\times \cos 60°=–6.0 \frac{\mathrm{kN}}{{\mathrm{m}}^2}$

The hoop radius of curvature, R_φ is:

$R_{\phi }=\frac{a}{\sin \alpha }=\frac{a}{\sin 60°}$

Thus hoop force becomes 

$$\begin{gathered} N_{\phi }=p_{\perp }{R}_{\phi } \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{1}}\mathbf{)}=p_{\perp }\frac{a_1}{\sin \alpha }=–6.0\times \frac{10.0}{\sin 60°}\mathbf{=}\mathbf{–}\mathbf{69}\mathbf{.}\mathbf{28}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}} \\ {\mathit{N}}_{\mathit{\phi }}\mathbf{(}\mathit{a}\mathbf{=}{\mathit{a}}_{\mathbf{2}}\mathbf{)}=p_{\perp }\frac{a_2}{\sin \alpha }=–6.0\times \frac{20.0}{\sin 60°}\mathbf{=}\mathbf{–}\mathbf{138}\mathbf{.}\mathbf{6}\frac{\mathbf{kN}}{{\mathbf{m}}^{\mathbf{2}}} \end{gathered}$$

The membrane force diagrams are given in the Figure below.