Problem 11.10. Assembled hyperbolic paraboloid roof

The shell structure given in the Figure is assembled from linear shape hyperbolic paraboloids. The ground plane is a square, the spans are 2l = 8 m in both directions. Height of the structure is f = 2 m.  The thickness of the reinforced concrete shell is h = 60 mm, the weight density is 25 kN/m3. Determine the membrane forces, the loads on the boundary beams and in the tie (dashed line). Draw the free body diagram of the boundary beams. The structure is subjected to

a) a snow load, s = 1.0 kN/m2,

b) self weight.

Solve Problem

Solve

Problem a)

Maximum normal force in the shell, Nx [kN/m]=

Maximum normal force in the shell, Ny [kN/m]=

Maximum shear force in the shell, Nxy [kN/m]=

Normal force in the tie, Ax [kN/m]=

Problem b)

Maximum normal force in the shell, Nx [kN/m]=

Maximum normal force in the shell, Ny [kN/m]=

Maximum shear force in the shell, Nxy [kN/m]=

Do you need help?

Steps

Step by step

Follow steps in Example 11.7.

Problem a)

Step 1. Write the function of the hyperbolic paraboloid.

Check function

Eq(11-72)

z=fl2xy

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

Check partial derivatives

Eq(11-73)

p=zx=fl2y=2.04.02y=0.125y,  q=zy=fl2x=2.04.02x=0.125xzxx=0,  zyy=0,   zxy=fab

Step 3. Determine the projected normal forces.

Check projected normal forces

In the hyperbolic paraboloids only shear force arises:

Eqs.(11-74)-(11-76)

Nxy=l22fpxy=l22fs=4.022×2.0×1.0=4.00 kNmNx=Nxyydx+C1(y)=0Ny=Nxyxdy+C2(x)=0

where constants C1(y) and C2(x) are set equal to zero.

Step 5. Give the membrane forces.

Check membrane forces

The only non zero membrane force is the constant shear force which is equal to the projected shear force:

Eq(11-38)

Nx=Nxy=4.00 kNmNx=Ny=0

Step 6. Determine the loads on the supports, draw the free body diagram.

Check free body diagram

The free body diagrams of the shells and the supporting elements are given in the Figure.

The supporting beams are subjected to the opposite of the tangential reactions of the shell parts. The internal beams support two of the shell parts, thus their load (-2Nxy) is twice the load of the boundary beams (-Nxy). The horizontal reaction of the inclined middle beams are carried by the tie shown in the Figure. The tensile force in the tie is:

Ax=2Nxyf2+l2lf2+l2=2×4.00×4.0=32 kN

Problem b)

Step 1. Calculate the self-weight load.

Check load

pg=γch=25×0.06=1.5 kNm2

Step 2. The function of the parabolic surface and its partial derivatives are given in Problem a).

Step 3. Determine the projected normal forces.

Check projected normal forces

First the shear force in the shell is calculated:

Eq.(11-74) and Footnote e in Section 11.5.

Nxy=l22fpxy=l22fpg1+p2+q2=4.022×2.0×1.51+0.1252y2+0.1252x2=6.001+0.1252y2+0.1252x2

The maximum value of the projected shear force is

Nxyx=l, y=l=6.001+0.1252×4.02+0.1252×4.02=7.348kNm

The projected normal forces are

Eqs.(11-74)-(11-76)

Nx=Nxyydx+C1(y)Ny=Nxyxdy+C2(x)

where the partial derivatives of the projected shear force are

Nxyy=6.00×2×0.1252y21+0.1252y2+0.1252x2=0.09375y1+0.1252y2+0.1252x2Nxyx=6.00×2×0.1252x21+0.1252y2+0.1252x2=0.09375x1+0.1252y2+0.1252x2

Integration is performed numerically. The above integrals (without constants C1(y) and C2(x)) become zero at the coordinate axes (x = 0 and y = 0). We can choose the C1(y) and C2(x) functions arbitrarily as self equilibrated edge forces.

See Figure 11.42.

We set the above functions in such a way, that no normal forces arise at the external supports, as it is shown in the Figure below for one part of the hyper shell.

The maximum values of the projected normal forces are

Nxx=l, y=l=2.601kNmNy(x=l, y=l)=2.601kNm

Step 5. Give the membrane forces.

Check membrane forces

Eq(11-38)

Ny=Ny1+0.1252x21+0.1252y2Nxy=Nxy Nx=Ny1+0.1252y21+0.1252x2

The internal force diagrams are given in the Figure.

The maximum values are:

Nxy(x=l, y=l)=Nxy(x=l, y=l)=7.348 kNmNx(x=l, y=l)=2.6011+0.1252×4.021+0.1252×4.02=2.601kNmNy(x=l, y=l)=2.6011+0.1252×4.021+0.1252×4.02=2.601kNm

Step 6. Determine the loads on the supports, draw the free body diagram.

Check free body diagram

The free body diagrams of one part of the hyper shell and its supports are given in the Figure.

The supporting beams are subjected to the opposite of the reactions of the shell parts, which are given in the internal force diagrams. The external supports can be tangential supports. The vertical components of the shell’s normal force at the internal edges cause the bending of the inclined supports. 

Results

Worked out solution

Follow steps in Example 11.7.

First the function of the hyperbolic paraboloid is written.

Eq(11-72)

z=fl2xy

The partial derivatives of the surface function are:

Eq(11-73)

p=zx=fl2y=2.04.02y=0.125y,  q=zy=fl2x=2.04.02x=0.125xzxx=0,  zyy=0,   zxy=fab

In the hyperbolic paraboloids only shear force arises:

Eqs.(11-74)-(11-76)

Nxy=l22fpxy=l22fs=4.022×2.0×1.0=4.00 kNmNx=Nxyydx+C1(y)=0Ny=Nxyxdy+C2(x)=0

where constants C1(y) and C2(x) are set equal to zero.

The only non zero membrane force is the constant shear force which is equal to the projected shear force:

Eq(11-38)

Nx=Nxy=4.00 kNmNx=Ny=0

The free body diagrams of the shells and the supporting elements are given in the Figure.

The supporting beams are subjected to the opposite of the tangential reactions of the shell parts. The internal beams support two of the shell parts, their load (-2Nxy) is twice the load of the boundary beams (-Nxy). The horizontal reaction of the inclined middle beams are carried by the tie shown in the Figure. The tensile force in the tie is:

Ax=2Nxyf2+l2lf2+l2=2×4.00×4.0=32 kN

Problem b)

The self-weight load of the structure is

pg=γch=25×0.06=1.5 kNm2

The function of the parabolic surface and its partial derivatives are given in Problem a).

First the projected shear force in the shell is calculated:

Eq.(11-74) and Footnote e in Section 11.5.

Nxy=l22fpxy=l22fpg1+p2+q2=4.022×2.0×1.51+0.1252y2+0.1252x2=6.001+0.1252y2+0.1252x2

The maximum value of the projected shear force is

Nxyx=l, y=l=6.001+0.1252×4.02+0.1252×4.02=7.348kNm

The projected normal forces are calculated from the shear force:

Eqs.(11-74)-(11-76)

Nx=Nxyydx+C1(y)Ny=Nxyxdy+C2(x)

where the partial derivatives of the projected shear force are

Nxyy=6.00×2×0.1252y21+0.1252y2+0.1252x2=0.09375y1+0.1252y2+0.1252x2Nxyx=6.00×2×0.1252x21+0.1252y2+0.1252x2=0.09375x1+0.1252y2+0.1252x2

Integration is performed numerically. The above integrals (without constants C1(y) and C2(x)) become zero at the coordinate axes (x = 0 and y = 0). We can choose the C1(y) and C2(x) functions arbitrarily as self equilibrated edge forces.

See Figure 11.42.

We set the above functions in such a way, that no normal forces arise at the external supports, as it is shown in the Figure below for one part of the hyper shell.

The maximum values of the projected normal forces are

Nxx=l, y=l=2.601kNmNy(x=l, y=l)=2.601kNm

The membrane forces are:

Eq(11-38)

Ny=Ny1+0.1252x21+0.1252y2Nxy=Nxy Nx=Ny1+0.1252y21+0.1252x2

The internal force diagrams are given in the Figure.

The maximum values become:

Nxy(x=l, y=l)=Nxy(x=l, y=l)=7.348 kNmNx(x=l, y=l)=2.6011+0.1252×4.021+0.1252×4.02=2.601kNmNy(x=l, y=l)=2.6011+0.1252×4.021+0.1252×4.02=2.601kNm

The free body diagrams of one part of the hyper shell and its supports are given in the Figure.

The supporting beams are subjected to the opposite of the reactions of the shell parts, which are given in the internal force diagrams. The external supports can be tangential supports. The vertical components of the shell’s normal force at the internal edges cause the bending of the inclined supports.