Problem 11.12. Edge disturbance of a spherical dome

Problem a)

Maximum bending moment, M_max [kNm/m]=

Problem b)

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 11.2.

The meridian force is:

$N_{\alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2a\pi \sin \alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2R\pi {\sin }^2\alpha }=\frac{–{\gamma }_{\mathrm{c}}tR}{\left(1+\cos \alpha \right)}$

The hoop force is:

$N_{\phi }=p_{\perp }R–N_{\alpha }$

For examining the edge disturbance meridian forces at the bottom of the dome must be calculated:

$N_{\mathrm{\alpha }}(\mathrm{\alpha }={\mathrm{\alpha }}_0=60°)=\frac{–25\times 0.3\times 10}{\left(1+\cos 60°\right)}=–50\frac{\mathrm{kN}}{\mathrm{m}}$

$N_{\phi }(\mathrm{\alpha }={\mathrm{\alpha }}_0=60°)=–{\gamma }_{\mathrm{c}}tR\cos 60°–N_{\alpha }(\alpha ={\alpha }_0=60°)=–25\times 0.3\times 10\times \cos 60°+50=12.5 \frac{\mathrm{kN}}{\mathrm{m}}$

Problem a)

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

Check maximum momentHide maximum moment

The membrane forces of the dome result in displacements of the edge which are hindered by the support. 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)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{B}}t=0.093\times 12.5\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3488}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{10.0\times 0.3}=1.039 \mathrm{m}$

Problem b)

Step 2. Determine maximum moment 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)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{\mathrm{B}}t=–0.29\times 12.5\times 0.3\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{088}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

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 11.2.

The meridian force is:

$N_{\alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2a\pi \sin \alpha }=\frac{–{\gamma }_{\mathrm{c}}t2R^2\pi \left(1–\cos \alpha \right)}{2R\pi {\sin }^2\alpha }=\frac{–{\gamma }_{\mathrm{c}}tR}{\left(1+\cos \alpha \right)}$

The hoop force is:

$N_{\phi }=p_{\perp }R–N_{\alpha }$

For examining the edge disturbance meridian forces at the bottom of the dome must be calculated:

$N_{\mathrm{\alpha }}(\mathrm{\alpha }={\mathrm{\alpha }}_0=60°)=\frac{–25\times 0.3\times 10}{\left(1+\cos 60°\right)}=–50\frac{\mathrm{kN}}{\mathrm{m}}$

$N_{\phi }(\mathrm{\alpha }={\mathrm{\alpha }}_0=60°)=–{\gamma }_{\mathrm{c}}tR\cos 60°–N_{\alpha }(\alpha ={\alpha }_0=60°)=–25\times 0.3\times 10\times \cos 60°+50=12.5 \frac{\mathrm{kN}}{\mathrm{m}}$

Problem a)

The membrane forces of the dome result in displacements of the edge which are hindered by the support. 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)

${\mathit{M}}_{\mathbf{max}}=0.093N_{\phi }^{\mathrm{B}}t=0.093\times 12.5\times 0.3\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3488}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The location of the positive maximum is

$0.6\sqrt{Rt}=0.6\sqrt{10.0\times 0.3}=1.039 \mathrm{m}$

Problem b)

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)

${\mathit{M}}_{\mathbf{max}}=–0.29N_{\phi }^{\mathrm{B}}t=–0.29\times 12.5\times 0.3\mathbf{=}\mathbf{–}\mathbf{1}\mathbf{.}\mathbf{088}\mathbf{ }\frac{\mathbf{kNm}}{\mathbf{m}}$

The above maximum negative moment arises at the support.