Problem 11.6. Parabolic barrel vault subjected to snow load

The parabolic barrel vault shown in the Figure is subjected to snow load, s = 1.5 kN/m². It is supported by arches at both curved ends and by beams at the straight edges. The width of the structure is l = 12 m, the length is b = 10 m and the height is f₀ = 3.5 m. Determine the membrane forces, and the loads on the supporting arches and beams from snow load of the shell, draw the free body diagram.

Solve Problem

SolveHide solve

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]=

Do you need help?

Steps

Step by stepHide steps

Step 1. Write the function of the parabolic surface.

Check functionHide function

Eq(11-52)

$z = \frac{4 f_{0}}{l^{2}} y^{2} = \frac{c}{2} y^{2}, c = \frac{8 f_{0}}{l^{2}}$

Step 2. Give the partial derivatives of the surface function.

Check partial derivativesHide partial derivatives

Eq(11-37)

$p = z_{x} = 0, q = z_{y} = c y z_{x x} = 0, z_{y y} = c = \frac{8 f_{0}}{l^{2}} = 0.1944, z_{x y} = 0$

Step 3. Determine the projected normal forces.

Check projected normal forcesHide projected normal forces

Eqs.(11-53) and (11-51)

$N_{y} = - \frac{p_{y} (y)}{z_{y y} (y)} = - \frac{s}{c} = - \frac{1.5}{0.1944} = 7.417 \frac{kN}{m} N_{x y} = - x \frac{\partial N_{y}}{\partial y} = 0, N_{x} = (\frac{x^{2}}{2} - \frac{b^{2}}{8}) \frac{\partial^{2} N_{y}}{\partial y^{2}} = 0$

Step 4. Calculate the membrane forces at the given points.

Check membrane forcesHide membrane forces

The only non zero membrane force is:

Eq(11-54)

$N_{y} = - \frac{s}{c} \sqrt{1 + c^{2} y^{2}} = - 7.714 \sqrt{1 + 0 . 1944^{2} y^{2}}$

The values of the normal force, N_y at points A and B are

$N_{y, A} (y = 0) = - 7.714 \sqrt{1 + 0 . 1944^{2} \times 0} = - 7.714 \frac{kN}{m} N_{y, B} (y = \frac{l}{2}) = - 7.714 \sqrt{1 + 0 . 1944^{2} \times {(\frac{12.0}{2})}^{2}} = - 11.85 \frac{kN}{m}$

Step 5. Draw the internal force diagrams.

Check internal force diagramsHide internal force diagrams

The barrel vault carries the total load as a cantenary arch.

Step 6. Determine the loads on the supporting arches and beams, draw the free body diagram.

Check free body diagramHide free body diagram

The arches are unloaded, the beams are subjected to the opposite of the normal force, N_y_,A at the vault edges.

Results

Worked out solutionHide solution

The function of the parabolic surface is written in the following form:

Eq(11-52)

$z = \frac{4 f_{0}}{l^{2}} y^{2} = \frac{c}{2} y^{2}, c = \frac{8 f_{0}}{l^{2}}$

The partial derivatives of the surface function are:

Eq(11-37)

$p = z_{x} = 0, q = z_{y} = c y z_{x x} = 0, z_{y y} = c = \frac{8 f_{0}}{l^{2}} = 0.1944, z_{x y} = 0$

The projected normal forces become:

Eqs.(11-53) and (11-51)

Thus the only non zero membrane force is N_y.

Eq(11-54)

$N_{y} = - \frac{s}{c} \sqrt{1 + c^{2} y^{2}} = - 7.714 \sqrt{1 + 0 . 1944^{2} y^{2}}$

The values of the normal force, N_y are calculated at points A and B:

$N_{y, A} (y = 0) = - 7.714 \sqrt{1 + 0 . 1944^{2} \times 0} = - 7.714 \frac{kN}{m} N_{y, B} (y = \frac{l}{2}) = - 7.714 \sqrt{1 + 0 . 1944^{2} \times {(\frac{12.0}{2})}^{2}} = - 11.85 \frac{kN}{m}$

The barrel vault carries the total load as a cantenary arch. The internal force diagrams are given in the Figure.

The arches are unloaded, the beams are subjected to the opposite of the normal force, N_y_,A at the vault edges. The free body diagram is given in the Figure.