Problem 11.16. Edge disturbance of barrel vault

 Determine the bending moment from edge disturbance at the curved edge of the barrel vault of Problem 11.6. (The parabolic barrel vault 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.) Thickness of the structure is 8 cm.

Solve Problem

Solve

Maximum bending moment assuming hinged edges, Mmax [Nm/m]=

Maximum bending moment assuming built-in edge, Mmax [Nm/m]=

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Steps

Step by step

Follow steps in Example 11.12.

Step 1. Give the membrane solution of the barrel vault.

Check membrane solution

Membrane forces are derived in Problem 11.6.

The only non zero membrane force is

Ny=sc1+c2y2=7.7141+0.19442y2

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

Ny,A(y=0)=7.7141+0.19442×0=7.714 kNmNy,B(y=l2)=7.7141+0.19442×12.022=11.85 kNm

Step 2. Determine maximum moment from the edge disturbance.

Check maximum moment assuming hinged edges

The membrane force, Ny of the barrel vault results in displacements of the edges which are hindered by the supporting arches. According to Geckeler’s approximation the bending moment at the support is determined by fitting an osculating cylinder to the edge of the vault. Considering hinged support the maximum moment is

Eq.(11-98)

 

Mmax=0.093NφBt=0.093Nyt

where the hoop force of the osculating cylinder, NφB is equal to the membrane force parallel to the edge, Ny, which varies along the curved edge.

The maximum value of the membrane forces arise along the longitudinal edge y=l2, thus the maximum moment is

Mmax=0.093Ny,Bt=0.093×11.85×0.08=0.08816 kNmm=88.19 Nmm

The location of the positive maximum along x axis is

0.6Rt

from the edge, where the radius, R is the inverse of the curvature of the osculating cylinder:

Eq.(11-16)

Ry=l2=1+zy232zyy=1+c2y232c=1+0.19442(12.02)2320.1944=18.66 m

The location of the maximum positive moment is

y=6.0 mx=b20.6Rt=5.00.618.66×0.08 =4.267 m

Check maximum moment assuming built-in edges

Considering built-in support the maximum negative moment is

Eq.(11-97)

 

Mmax=0.29NφBt=0.29Nyt

The maximum moment arises at the corner x=b2, y=l2:

Mmax=0.29Ny,Bt=0.29×11.85×0.08=0.2749 kNmm=275.0 Nmm

Since in many cases the resistance of a support is hard to calculate, we may design the shell for both bending moments.

Results

Worked out solution

Follow steps in Example 11.12.

First the membrane solution of the barrel vault must be derived.

Membrane forces are determined in Problem 11.6.

The only non zero membrane force is

Ny=sc1+c2y2=7.7141+0.19442y2

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

Ny,A(y=0)=7.7141+0.19442×0=7.714 kNmNy,B(y=l2)=7.7141+0.19442×12.022=11.85 kNm

The membrane force, Ny of the barrel vault results in displacements of the edges which are hindered by the supporting arches. According to Geckeler’s approximation the bending moment at the support is determined by fitting an osculating cylinder to the edge of the vault. Considering hinged support the maximum moment is

Eq.(11-98)

 

Mmax=0.093NφBt=0.093Nyt

where the hoop force of the osculating cylinder, NφB is equal to the membrane force parallel to the edge, Ny, which varies along the curved edge.

The maximum value of the membrane force arise along the longitudinal edge y=l2, thus the maximum moment is

Mmax=0.093Ny,Bt=0.093×11.85×0.08=0.08816 kNmm=88.19 Nmm

The location of the positive maximum along x axis is

0.6Rt

from the edge, where the radius, R is the inverse of the curvature of the osculating cylinder:

Eq.(11-16)

Ry=l2=1+zy232zyy=1+c2y232c=1+0.19442(12.02)2320.1944=18.66 m

The location of the maximum moment is

y=6.0 mx=b20.6Rt=5.00.618.66×0.08 =4.267 m

y=6.0 mx=b20.6Rt=5.00.618.66×0.08 =4.267 m

Considering built-in support the maximum negative moment is

Eq.(11-97)

 

Mmax=0.29NφBt=0.29Nyt

The maximum moment arises at the corner x=b2, y=l2:

Mmax=0.29Ny,Bt=0.29×11.85×0.08=0.2749 kNmm=275.0 Nmm

Since in many cases the resistance of a support is hard to calculate, we may design the shell for both bending moments.