Problem 11.14. Edge disturbance of a spherical dome with skylight

Problem a)

Maximum bending moment, M_max [kNm/m]=

Problem b)

Maximum bending moment, M_max [kNm/m]=

Problem c)

Maximum bending moment, M_max [kNm/m]=

Follow steps in Example 11.12.

Step 1. Give the membrane solution of the dome.

Check membrane solutionHide membrane solution

Membrane forces are derived in Problem 11.3.

Value of the meridian force at the bottom is

$N_{{}_{\mathrm{\alpha }}}^{\mathrm{bottom}}=\frac{–p{a}_1}{a_2\sin 60°}=\frac{–2.0\times 5.0}{10.0\times \sin 60°}=–1.155\frac{\mathrm{kN}}{\mathrm{m}}$

At the top

 ${\alpha }_1={\sin }^{–1}\left(\frac{a_1}{R}\right)={\sin }^{–1}\left(\frac{5.0}{11.55}\right)=25.65°$

and the meridian force becomes

$N_{{}_{\mathrm{\alpha }}}^{\mathrm{top}}=\frac{–p{a}_1}{a_1\sin 25.65°}=–4.619\frac{\mathrm{kN}}{\mathrm{m}}$

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

$$\begin{gathered} N_{{}_{\phi }}^{\mathrm{bottom}}=–N_{{}_{\alpha }}^{\mathrm{bottom}}=1.155 \frac{\mathrm{kN}}{\mathrm{m}} \\ {N}_{{}_{\phi }}^{\mathrm{top}}=–N_{{}_{\mathrm{\alpha }}}^{\mathrm{top}}=–4.619 \frac{\mathrm{kN}}{\mathrm{m}} \end{gathered}$$

Problem a)

Step 2. Determine maximum moment at the top from the edge disturbance assuming hinged edge.

Check maximum momentHide maximum moment

The membrane forces at the top of the dome result in displacements of the edge which are hindered by the top ring. According to Geckeler’s approximation the bending moment at the support is determined by fitting an osculating cylinder to the edge of the dome. Considering hinged support the maximum moment is

Eq.(11-98)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{top}}t=0.093\times 4.619\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1289}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{11.55\times 0.3}=1.117 \mathrm{m}$

Problem b)

Step 2. Determine maximum moment at the bottom from the edge disturbance assuming hinged edge.

Check maximum momentHide maximum moment

The membrane forces at the bottom of the dome result in displacements of the edge which are hindered by the bottom ring. Geckeler’s approximation is applied, an osculating cylinder is fitted to the edge of the dome. In the case of hinged support the maximum moment becomes

Eq.(11-98)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{bottom}}t=0.093\times 1.155\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03222}\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{11.55\times 0.3}=1.117 \mathrm{m}$

Problem c)

Step 2. Determine maximum moment at the bottom from the edge disturbance assuming fixed edge.

Check maximum momentHide maximum moment

If clamped support is assumed the displacement and also the rotation of the edge of the dome is hindered. When the effect of the rotation of the boundary is neglected, the maximum moment is given by

Eq.(11-97)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{\mathrm{bottom}}t=–0.093\times 1155\times 0.3\mathbf{=}\mathbf{–}\mathbf{0}\mathbf{.}\mathbf{1005}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The above maximum negative moment arises at the support.

Follow steps in Example 11.12.

First the membrane solution of the dome is determined.

Membrane forces are derived in Problem 11.3.

Value of the meridian force at the bottom is

$N_{{}_{\mathrm{\alpha }}}^{\mathrm{bottom}}=\frac{–p{a}_1}{a_2\sin 60°}=\frac{–2.0\times 5.0}{10.0\times \sin 60°}=–1.155\frac{\mathrm{kN}}{\mathrm{m}}$

At the top

 ${\alpha }_1={\sin }^{–1}\left(\frac{a_1}{R}\right)={\sin }^{–1}\left(\frac{5.0}{11.55}\right)=25.65°$

and the meridian force becomes

$N_{{}_{\mathrm{\alpha }}}^{\mathrm{top}}=\frac{–p{a}_1}{a_1\sin 25.65°}=–4.619\frac{\mathrm{kN}}{\mathrm{m}}$

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

$$\begin{gathered} N_{{}_{\phi }}^{\mathrm{bottom}}=–N_{{}_{\alpha }}^{\mathrm{bottom}}=1.155 \frac{\mathrm{kN}}{\mathrm{m}} \\ {N}_{{}_{\phi }}^{\mathrm{top}}=–N_{{}_{\mathrm{\alpha }}}^{\mathrm{top}}=–4.619 \frac{\mathrm{kN}}{\mathrm{m}} \end{gathered}$$

Problem a)

The membrane forces at the top of the dome result in displacements of the edge which are hindered by the top ring. According to Geckeler’s approximation the bending moment at the support is determined by fitting an osculating cylinder to the edge of the dome. Considering hinged support the maximum moment is

Eq.(11-98)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{top}}t=0.093\times 4.619\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1289}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{11.55\times 0.3}=1.117 \mathrm{m}$

Problem b)

The membrane forces at the bottom of the dome result in displacements of the edge which are hindered by the bottom ring. Geckeler’s approximation is applied, an osculating cylinder is fitted to the edge of the dome. In the case of hinged support the maximum moment becomes

Eq.(11-98)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{bottom}}t=0.093\times 1.155\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{03222}\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{11.55\times 0.3}=1.117 \mathrm{m}$

Problem c)

If clamped support is assumed the displacement and also the rotation of the edge of the dome is hindered. When the effect of the rotation of the boundary is neglected, the maximum moment is given by

Eq.(11-97)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{\mathrm{bottom}}t=–0.093\times 1155\times 0.3\mathbf{=}\mathbf{–}\mathbf{0}\mathbf{.}\mathbf{1005}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The above maximum negative moment arises at the support.