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
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]=Solve
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Steps
Problem a) Step 2. Give the partial derivatives of the surface function. Step 3. Determine the projected normal forces. In the hyperbolic paraboloids only shear force arises: where constants C1(y) and C2(x) are set equal to zero. Step 5. Give the membrane forces. The only non zero membrane force is the constant shear force which is equal to the projected shear force: Step 6. Determine the loads on the supports, draw the 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: Problem b) Step 1. Calculate the self-weight load. Step 2. The function of the parabolic surface and its partial derivatives are given in Problem a). Step 3. Determine the projected normal forces. First the shear force in the shell is calculated: The maximum value of the projected shear force is The projected normal forces are where the partial derivatives of the projected shear force are 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. 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 Step 5. Give the membrane forces. The internal force diagrams are given in the Figure. The maximum values are: Step 6. Determine the loads on the supports, draw the 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. Step by step
Step 1. Write the function of the hyperbolic paraboloid.
Check function
Check partial derivatives
Check projected normal forces
Check membrane forces
Check free body diagram
Check load
Check projected normal forces
Check membrane forces
Check free body diagram
Results
The partial derivatives of the surface function are: In the hyperbolic paraboloids only shear force arises: 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: 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: Problem b) The self-weight load of the structure is The function of the parabolic surface and its partial derivatives are given in Problem a). The maximum value of the projected shear force is The projected normal forces are calculated from the shear force: where the partial derivatives of the projected shear force are 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. 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 The membrane forces are: The internal force diagrams are given in the Figure. The maximum values become: 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. Worked out solution
First the function of the hyperbolic paraboloid is written.
The free body diagrams of the shells and the supporting elements are given in the Figure.
First the projected shear force in the shell is calculated:
The free body diagrams of one part of the hyper shell and its supports are given in the Figure.