Problem 11.9. Elliptic paraboloid roof with four perfect supports

Consider perfect supports at all the edges of the elliptical paraboloid roof given in the previous problem. The spans are 2a = 2b = 20 m, the height of the parabolas are fa = fb = 1.0 m.  Determine the membrane forces from snow load, s = 1.5 kN/m2 and the loads on the supports.

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. Write the function of the elliptic paraboloid.

Check function

Eq(11-61)

z=faa2x2+fbb2y2

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

Check partial derivatives

Eq(11-62)

p=zx=2faa2x=2×1.010.02x=0.02x,  q=zy=2fbb2y=2×1.010.02x=0.02yzxx=2faa2=0.02,  zyy=2fbb2=0.02,   zxy=0

Step 3. Determine the projected normal forces.

Check projected normal forces

The shell structure carries the load as perpendicular arches along x  and y axes.

Eq.(11-66)

Nx=a22fasxNy=b22fbsyNxy=0 

where

Eq.(11-67)

s=sx+sy

The ratio of the two load components of the undetermined structure can be determined from compatibility condition (assuming identical deflections at the middle). Because of the symmetry of the structure, here

sx=sy=s2=0.75 kNm2

Thus the projected normal forces become:

Nx=10.022×1.0×0.75=37.5 kNmNy=b22fbsy=10.022×1.0×0.75=37.5 kNmNxy=0

Step 4. 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)

Nx=Nx1+p21+q2==37.51+0.022x21+0.022y2Ny=Ny1+q21+p2=37.51+0.022y21+0.022x2Nxy=Nxy=0

The values of the non zero normal forces, Nx and Ny at points A and B areNx,A(x=0, y=0)=37.51+0.022×021+0.022×02=37.5 kNmNy,A(x=0, y=0)=37.51+0.022×021+0.022×02=37.5 kNmNx,B(x=0, y=b)=37.51+0.022×021+0.022×10.02=36.77 kNmNy,B(x=0, y=b)=37.51+0.022×10.021+0.022×02=38.24 kNm

Step 5. Draw the internal force diagrams.

Check internal force diagrams

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

Check free body diagram

All of the perfect supports are subjected to tangential forces.

Results

Worked out solution

First the function of the elliptic paraboloid is written.

Eq(11-61)

z=faa2x2+fbb2y2

The partial derivatives of the surface function are:

Eq(11-62)

p=zx=2faa2x=2×1.010.02x=0.02x,  q=zy=2fbb2y=2×1.010.02x=0.02yzxx=2faa2=0.02,  zyy=2fbb2=0.02,   zxy=0

 The shell structure carries the load as perpendicular arches along x  and y axes. The projected normal forces are:

Eq.(11-66)

Nx=a22fasxNy=b22fbsyNxy=0 

where

Eq.(11-67)

s=sx+sy

The ratio of the two load components of the undetermined structure can be determined from compatibility condition (assuming identical deflections at the middle). Because of the symmetry of the structure, here

sx=sy=s2=0.75 kNm2

Thus the projected normal forces become:

Nx=10.022×1.0×0.75=37.5 kNmNy=b22fbsy=10.022×1.0×0.75=37.5 kNmNxy=0

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

Eq(11-38)

Nx=Nx1+p21+q2==37.51+0.022x21+0.022y2Ny=Ny1+q21+p2=37.51+0.022y21+0.022x2Nxy=Nxy=0

The values of the non zero normal forces, Nx and Ny at points A and B areNx,A(x=0, y=0)=37.51+0.022×021+0.022×02=37.5 kNmNy,A(x=0, y=0)=37.51+0.022×021+0.022×02=37.5 kNmNy,B(x=0, y=b)=37.51+0.022×021+0.022×10.02=36.77 kNmNx,B(x=0, y=b)=37.51+0.022×10.021+0.022×02=38.24 kNm

The internal force diagrams are given in the Figure.

The free body diagram shows that all of the perfect supports are subjected to tangential forces.