Problem 11.11. Cylindrical shell

Maximum bending moment, M_max [kNm/m]=

Maximum hoop force, $N_{\phi }^{\max }$ [kN/m]=

Follow steps in Example 11.8.

Step 1. Give the membrane solution for uniform pressure.

Check membrane solutionHide membrane solution

The membrane solution is independent of the axial coordinate:

Eq(11-90)

$$\begin{gathered} w_{\mathrm{inhom}}=R^2\frac{p}{Eh} \\ {N}_{\phi }=pR=10\times 3.0=30.00 \mathrm{kN}/\mathrm{m} \end{gathered}$$

Step 2. Determine the bending moment from the compatibility at the boundaries. Use the analogy of beams resting on elastic foundation.

Check bending momentHide bending moment

The compatibility is ensured by the solution of a strip of plate on elastic foundation.

See Figure 10.45 and Eq(11-95).

$$\begin{gathered} w_{\mathrm{hom}}=–w^B{\eta }_2=–R^2\frac{p}{Eh}{e}^{–\lambda x}\left(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\right) \\ M\left(x\right)=–w^{B}2{\lambda }^{2}D{\eta }_4=–R^2\frac{p}{Eh}2{\lambda }^{2}D{e}^{–\lambda x}(\cos \left(\lambda x\right)–\sin \left(\lambda x\right)) \end{gathered}$$

where D is the bending stiffness, and 

$\lambda =\frac{1.32}{\sqrt{Rh}}=\frac{1.32}{\sqrt{3.0\times 0.2}}=1.704 \frac{1}{\mathrm{m}}$

The maximum bending moment arises at the support. It can be approximated as

Eq(11-96)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{B}h=–0.29\times 30.00\times 0.2\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{74}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

Step 3. Give the maximum hoop force.

Check maximum hoop forceHide maximum hoop force

The hoop force can be calculated from the displacements:

Eq(11-88)

$$\begin{gathered} N_{\phi }=\frac{Eh}{R}\left(w_{\mathrm{inhom}}+w_{\mathrm{hom}}\right)=\frac{Eh}{R}\left(R^2\frac{p}{Eh}–R^2\frac{p}{Eh}{e}^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right))\right)= \\ =Rp(1–e^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\left)\right) \end{gathered}$$

To find the maximum of the hoop force first derivative of N_φ is determined:

$\frac{\partial N_{\phi }}{\partial x}=\frac{\partial }{\partial x}\left[Rp(1–e^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\left)\right)\right]=–Rp\lambda e^{–\lambda x}2\sin \left(\lambda x\right)$

The hoop force reaches its maximum where the above derivative is zero:

$$\begin{gathered} \frac{\partial N_{\phi }}{\partial x}=0=–Rp\lambda e^{–\lambda x}2\sin \left(\lambda x\right) \to \lambda x=0, \mathrm{\pi }, ... \\ {x}_1=0 \\ {x}_2=\frac{\mathrm{\pi }}{\lambda }=\frac{\mathrm{\pi }}{1.704}=1.844 \mathrm{m} \end{gathered}$$

The maximum hoop force is

${\mathit{N}}_{\mathbf{\phi }}^{\mathbf{max}}=Rp(1–e^{–\pi }(\cos \left(\pi \right)+\sin \left(\pi \right)\left)\right)=Rp(1+e^{–\pi })=3.0\times 10.0(1+e^{–\pi })\mathbf{=}\mathbf{31}\mathbf{.}\mathbf{30}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}}$

Step 4. Draw the internal force diagrams. 

Check diagramsHide diagrams

Follow steps in Example 11.8.

First the membrane solution of the cylinder is determined.The membrane solution for uniform pressure is independent of the axial coordinate:

Eq(11-90)

$$\begin{gathered} w_{\mathrm{inhom}}=R^2\frac{p}{Eh} \\ {N}_{\phi }=pR=10\times 3.0=30.00 \mathrm{kN}/\mathrm{m} \end{gathered}$$

Using the analogy of beams resting on elastic foundation the compatibility of the boundary is ensured. The moment becomes:

See Figure 10.45 and Eq(11-95).

$$\begin{gathered} w_{\mathrm{hom}}=–w^B{\eta }_2=–R^2\frac{p}{Eh}{e}^{–\lambda x}\left(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\right) \\ M\left(x\right)=–w^{B}2{\lambda }^{2}D{\eta }_4=–R^2\frac{p}{Eh}2{\lambda }^{2}D{e}^{–\lambda x}(\cos \left(\lambda x\right)–\sin \left(\lambda x\right)) \end{gathered}$$

where D is the bending stiffness, and 

$\lambda =\frac{1.32}{\sqrt{Rh}}=\frac{1.32}{\sqrt{3.0\times 0.2}}=1.704 \frac{1}{\mathrm{m}}$

Note that the origin of the coordinate system is at the top of the cylinder.

The maximum bending moment arises at the support. It can be approximated as

Eq(11-96)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{B}h=–0.29\times 30.00\times 0.2\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{74}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The hoop force can be calculated from the displacements:

Eq(11-88)

$$\begin{gathered} N_{\phi }=\frac{Eh}{R}\left(w_{\mathrm{inhom}}+w_{\mathrm{hom}}\right)=\frac{Eh}{R}\left(R^2\frac{p}{Eh}–R^2\frac{p}{Eh}{e}^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right))\right)= \\ =Rp(1–e^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\left)\right) \end{gathered}$$

To find the maximum of the hoop force first derivative of N_φ is determined:

$\frac{\partial N_{\phi }}{\partial x}=\frac{\partial }{\partial x}\left[Rp(1–e^{–\lambda x}(\cos \left(\lambda x\right)+\sin \left(\lambda x\right)\left)\right)\right]=–Rp\lambda e^{–\lambda x}2\sin \left(\lambda x\right)$

The hoop force reaches its maximum where the above derivative is zero:

$$\begin{gathered} \frac{\partial N_{\phi }}{\partial x}=0=–Rp\lambda e^{–\lambda x}2\sin \left(\lambda x\right) \to \lambda x=0, \mathrm{\pi }, ... \\ {x}_1=0 \\ {x}_2=\frac{\mathrm{\pi }}{\lambda }=\frac{\mathrm{\pi }}{1.704}=1.844 \mathrm{m} \end{gathered}$$

The maximum hoop force is

${\mathit{N}}_{\mathbf{\phi }}^{\mathbf{max}}=Rp(1–e^{–\pi }(\cos \left(\pi \right)+\sin \left(\pi \right)\left)\right)=Rp(1+e^{–\pi })=3.0\times 10.0(1+e^{–\pi })\mathbf{=}\mathbf{31}\mathbf{.}\mathbf{30}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}}$

The internal force diagrams are shown in the Figure.