Problem 11.17. Truncated cone with not adequate membrane support

Maximum bending moment, M_max [kNm/m]=

Follow steps in Example 11.11.

Step 1. Calculate the vertical reaction force.

Check vertical reactionHide vertical reaction

The vertical support force, A is calculated from the vertical equilibrium:

See Example 11.2.

$A=\frac{p2a_1\pi }{2a_2\pi }=p\frac{a_1}{a_2}=10.0\frac{10.0}{20.0}=5.00 \frac{\mathrm{kN}}{\mathrm{m}}$

Step 2. Determine the force component which causes the bending of the edge.

Check perpendicular componentHide perpendicular component

Step 3. Calculate the bending moment from the edge disturbance.

Check momentHide moment

The moment is approximated by the moment of the osculating cylinder subjected to a line load, A_⊥:

See Figure 10.45a and Eq.(11-82).

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=A_{\perp }\frac{0.32}{\lambda }=A_{\perp }\frac{0.32}{1.32}\sqrt{Rh}=A_{\perp }\frac{0.32}{1.32}\sqrt{\frac{a_2}{\sin \alpha }h}= \\ =2.50\frac{0.32}{1.32}\sqrt{\frac{20.0}{\sin 60°}\times 0.1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9210}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}} \end{gathered}$$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{\frac{20.0}{\sin 60°}\times 0.1}=0.9118 \mathrm{m}$

from the bottom of the cone.

Follow steps in Example 11.11.

The vertical support force, A is calculated from the vertical equilibrium:

See Example 11.2.

$A=\frac{p2a_1\pi }{2a_2\pi }=p\frac{a_1}{a_2}=10.0\frac{10.0}{20.0}=5.00 \frac{\mathrm{kN}}{\mathrm{m}}$

A has a component in the direction of the meridian force and one which is perpendicular to it, the latter one, A_⊥ – which is equal to the shear force at the edge – causes the bending of the shell.

$A_{\perp }=A\cos \alpha =5.00\times \cos 60°=2.50 \frac{\mathrm{kN}}{\mathrm{m}}$

The moment is approximated by the moment of the osculating cylinder subjected to a line load, A_⊥:

See Figure 10.45a and Eq.(11-82).

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{p}{2{\lambda }^{3}D}0.64{\lambda }^{2}D=A_{\perp }\frac{0.32}{\lambda }=A_{\perp }\frac{0.32}{1.32}\sqrt{Rh}=A_{\perp }\frac{0.32}{1.32}\sqrt{\frac{a_2}{\sin \alpha }h}= \\ =2.50\frac{0.32}{1.32}\sqrt{\frac{20.0}{\sin 60°}\times 0.1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9210}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}} \end{gathered}$$

The maximum bending moment occurs at a distance

$0.6\sqrt{Rt}=0.6\sqrt{\frac{20.0}{\sin 60°}\times 0.1}=0.9118 \mathrm{m}$

from the bottom of the cone.