Problem 11.8. Elliptic paraboloid roof with two perfect supports

Normal force at point A, N_x [kN/m]=

Normal force at point A, N_y [kN/m]=

Shear force at point A, N_xy [kN/m]=

Normal force at point B, N_x [kN/m]=

Normal force at point B, N_y [kN/m]=

Shear force at point B, N_xy [kN/m]=

Step 1. Write the function of the elliptic paraboloid.

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Eq(11-61)

$z=\frac{f_a}{a^2}{x}^2+\frac{f_b}{b^2}{y}^2$

Step 3. Give the partial derivatives of the surface function.

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Eq(11-62)

$$\begin{gathered} p=z_x=\frac{2f_a}{a^2}x=\frac{2\times 1.0}{10.0^2}x=0.02x, q=z_y=\frac{2f_b}{b^2}y=\frac{2\times 1.0}{10.0^2}x=0.02y \\ {z}_{xx}=\frac{2f_a}{a^2}=0.02, z_{yy}=\frac{2f_b}{b^2}=0.02, z_{xy}=0 \end{gathered}$$

Step 4. Determine the projected normal forces.

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The shell structure carries the load as an arch along x axis.

Eq.(11-65)

$$\begin{gathered} {\overline{N}}_x=–\frac{a^2}{2f_a}s=–\frac{10.0^2}{2\times 1.0}1.5=–75.00 \frac{\mathrm{kN}}{\mathrm{m}} \\ {\overline{N}}_y=0 \\ {\overline{N}}_{xy}=0 \end{gathered}$$

Step 5. Calculate the membrane forces at the given points.

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The membrane forces can be calculated from the projected forces as:

Eq(11-38)

$$\begin{gathered} N_x={\overline{N}}_x\frac{\sqrt{1+p^2}}{\sqrt{1+q^2}}=–75.00\frac{\sqrt{1+0.{02}^2{x}^2}}{\sqrt{1+0.{02}^2{y}^2}} \\ {N}_y={\overline{N}}_y\frac{\sqrt{1+q^2}}{\sqrt{1+p^2}}=0 \\ {N}_{xy}={\overline{N}}_{xy}=0 \end{gathered}$$

The values of the non zero normal force, N_x at points A and B are$$\begin{gathered} {\mathit{N}}_{\mathit{x}\mathbf{,}\mathbf{A}}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{)}=–75.00\frac{\sqrt{1+0.{02}^2\times 0^2}}{\sqrt{1+0.{02}^2\times 0^2}}\mathbf{=}\mathbf{–}\mathbf{75}\mathbf{.}\mathbf{00}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathit{x}\mathbf{,}\mathit{B}}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{=}\mathit{b}\mathbf{)}=–75.00\frac{\sqrt{1+0.{02}^2\times 0^2}}{\sqrt{1+0.{02}^2\times 10.0^2}}\mathbf{=}\mathbf{–}\mathbf{73}\mathbf{.}\mathbf{54}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

Step 6. Draw the internal force diagrams.

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Step 7. Determine the loads on the supports, draw the free body diagram.

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First the function of the elliptic paraboloid is written.

Eq(11-61)

$z=\frac{f_a}{a^2}{x}^2+\frac{f_b}{b^2}{y}^2$

The partial derivatives of the surface function are:

Eq(11-62)

$$\begin{gathered} p=z_x=\frac{2f_a}{a^2}x=\frac{2\times 1.0}{10.0^2}x=0.02x, q=z_y=\frac{2f_b}{b^2}y=\frac{2\times 1.0}{10.0^2}x=0.02y \\ {z}_{xx}=\frac{2f_a}{a^2}=0.02, z_{yy}=\frac{2f_b}{b^2}=0.02, z_{xy}=0 \end{gathered}$$

The shell structure carries the load as an arch along x axis. The projected normal forces are:

Eq.(11-65)

$$\begin{gathered} {\overline{N}}_x=–\frac{a^2}{2f_a}s=–\frac{10.0^2}{2\times 1.0}1.5=–75.00 \frac{\mathrm{kN}}{\mathrm{m}} \\ {\overline{N}}_y=0 \\ {\overline{N}}_{xy}=0 \end{gathered}$$

The membrane forces can be calculated from the projected forces as:

Eq(11-38)

$$\begin{gathered} N_x={\overline{N}}_x\frac{\sqrt{1+p^2}}{\sqrt{1+q^2}}=–75.00\frac{\sqrt{1+0.{02}^2{x}^2}}{\sqrt{1+0.{02}^2{y}^2}} \\ {N}_y={\overline{N}}_y\frac{\sqrt{1+q^2}}{\sqrt{1+p^2}}=0 \\ {N}_{xy}={\overline{N}}_{xy}=0 \end{gathered}$$

The values of the non zero normal force, N_x at points A and B are$$\begin{gathered} {\mathit{N}}_{\mathit{x}\mathbf{,}\mathbf{A}}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{=}\mathbf{0}\mathbf{)}=–75.00\frac{\sqrt{1+0.{02}^2\times 0^2}}{\sqrt{1+0.{02}^2\times 0^2}}\mathbf{=}\mathbf{–}\mathbf{75}\mathbf{.}\mathbf{00}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \\ {\mathit{N}}_{\mathit{x}\mathbf{,}\mathit{B}}\mathbf{(}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{y}\mathbf{=}\mathit{b}\mathbf{)}=–75.00\frac{\sqrt{1+0.{02}^2\times 0^2}}{\sqrt{1+0.{02}^2\times 10.0^2}}\mathbf{=}\mathbf{–}\mathbf{73}\mathbf{.}\mathbf{54}\mathbf{ }\frac{\mathbf{kN}}{\mathbf{m}} \end{gathered}$$

The internal force diagrams are shown in the Figure.

The free body diagram is given below. The perfect supports are subjected to tangential forces.