Problem 11.7. Parabolic barrel vault subjected to self-weigth

A parabolic barrel vault with the same geometry given in the previous problem is subjected to self-weigth. The supports are different: one arch is free, the other arch and the straight edges are perfect supports. The thickness of the reinforced concrete shell is h = 80 mm. The weight density is 25 kN/m3. Determine the membrane forces, and the loads on the supports, draw the free body diagram.

Solve Problem

Solve

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

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

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

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

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

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

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Steps

Step by step

Step 1. Calculate the self-weight load.

Check load

pg=γch=25×0.08=2.00 kNm2

Step 2. Write the function of the parabolic surface.

Check function

Eq(11-52)

z=4f0l2y2=c2y2,    c=8f0l2

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

Check partial derivatives

Eq(11-37)

p=zx=0,  q=zy=cyzxx=0,  zyy=c=0.1944,   zxy=0

Step 4. Determine the projected normal forces.

Check projected normal forces

Eqs.(11-56) and (11-60)
Ny=pzzyy=pgc1+c2y2Nxy=xNyy=pgy1+c2y2cx Nx=x222Nyy2=pg(1+c2y2)32cx22

 

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

Check membrane forces

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

Eq(11-38)

Ny=Ny1+c2y2=pgc1+c2y2=2.000.19441+0.19442y2Nxy=Nxy=pgy1+c2y2cx=2.00×0.1944×xv1+0.19442y2 Nx=Nx1+c2y2=pgc(1+c2y2)2x22=2.00×0.1944(1+0.19442y2)2x22

The values of the meridian forces at points A and B areNy,A(x=b, y=0)=2.000.19441+0.19442×02=10.29 kNmNxy,A(x=b, y=0)=2.00×0.1944×0×10.01+0.19442×02=0 Nx,A(x=b, y=0)=2.00×0.1944(1+0.19442×02)210.022=19.44 kNm

Ny,B(x=b, y=l2)=2.000.19441+0.19442×6.02=24.29 kNmNxy,B(x=b, y=l2)=2.00×0.1944×6.0×10.01+0.19442×6.02=15.19 kNmNx,B(x=b, y=l2)=2.00×0.1944(1+0.19442×6.02)210.022=3.488 kNm

Step 6. Draw the internal force diagrams.

Check internal force diagrams

The barrel vault carries a part of the load as a cantenary arch, while the remaining part analogously to a cantilever beam.

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

Check free body diagram

Results

Worked out solution

First the self-weight of the structure is calculated:

pg=γch=25×0.08=2.00 kNm2

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

Eq(11-52)

z=4f0l2y2=c2y2,    c=8f0l2

The partial derivatives of the surface function are:

Eq(11-37)

p=zx=0,  q=zy=cyzxx=0,  zyy=c=0.1944,   zxy=0

Then the projected normal forces are determined.

Eqs.(11-56) and (11-60)
Ny=pzzyy=pgc1+c2y2Nxy=xNyy=pgy1+c2y2cx Nx=x222Nyy2=pg(1+c2y2)32cx22

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

Eq(11-38)

Ny=Ny1+c2y2=pgc1+c2y2=2.000.19441+0.19442y2Nxy=Nxy=pgy1+c2y2cx=2.00×0.1944×xv1+0.19442y2 Nx=Nx1+c2y2=pgc(1+c2y2)2x22=2.00×0.1944(1+0.19442y2)2x22

The values of the meridian forces at points A and B areNy,A(x=b, y=0)=2.000.19441+0.19442×02=10.29 kNmNxy,A(x=b, y=0)=2.00×0.1944×0×10.01+0.19442×02=0Nx,A(x=b, y=0)=2.00×0.1944(1+0.19442×02)210.022=19.44 kNm

Ny,B(x=b, y=l2)=2.000.19441+0.19442×6.02=24.29 kNmNxy,B(x=b, y=l2)=2.00×0.1944×6.0×10.01+0.19442×6.02=15.19 kNmNx,B(x=b, y=l2)=2.00×0.1944(1+0.19442×6.02)210.022=3.488 kNm

The barrel vault carries a part of the load as a cantenary arch, while the remaining part analogously to a cantilever beam. The internal force diagrams are given in the Figure.

The free body diagram and the loads on the supports are shown below.