Cable strengthened arches

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Proceedings of the International Association for Shell and Spatial Structures (IASS) Symposium 2013

„BEYOND THE LIMITS OF MAN” 23-27 September, Wroclaw University of Technology, Poland

J.B. Obrębski and R. Tarczewski (eds.)

1

Cable strengthened Arches

Wim Kamerling

1

Lecturer faculty of Architecture, Delft University of Technology, Delft, The Netherlands, e-mail: M.W.Kamerling@tudelft.nl

Summary: The structural efficiency of arches, subjected to several variable loads, can be increased by strengthening these arches with cables. For these structures it can be necessary, especially in case the permanent load is small, to post-tension the cables to avoid any compression acting on the cables. A method to analyse the load transfer is described, to design and optimise these statically indeterminate structures. Consequently the self weight and embodied energy of this structure can be reduced too. The structural analyse is validated with a finite element program.

Keywords: Arches, cable strengthened, post-tensioning, optimising, lightness, sustainability, embodied energy 1. INTRODUCTION

Generally the first priority of most engineers, designing the structure of a building, is to decrease the cost of labor. Reducing the need of material seems not so important nowadays, however in the coming decades saving material will be an important issue again. The warming up of the atmosphere will cause many problems. Consequently the environmental burden of buildings has to be decreased sincerely. Especially the consumption of fossil energy for heating and cooling buildings must be minimized substantially. At the moment the environmental burden caused by the materialization of a building is about 15% of the total environmental load during the life time. This percentage will rise rapidly if fossil energy is not used anymore for heating and cooling buildings, so the reduction of the embodied energy of building materials will be an important issue again. Till the middle of the XX-century engineers had to design structures with a minimal need of material. Studying these designs shows us how to design sophisticated structures consuming less embodied energy. Generally form-active structures, as for example funicular arches, are structurally very efficient [1]. Arches are designed and constructed for at least two thousands years [2]. However the dimensions of the Roman arches were rather substantial, but later, thanks to scientists and researchers, as for example Hook and Poleni, engineers could design slender arches and vaults following the line of thrust. Especially the designs of Gaudi [3] show the beauty of funicular arches. Arches, subjected to normal forces only, can be dimensioned much lighter than section active structures, subjected to bending moments. Nevertheless it is difficult to optimize an arch if this form active structure is subjected to several load combinations and the lines of trust vary too. Arches are subjected to bending if a line of trust, due to one of the load combinations, does not coincide with the line of the system. To optimize such an arch, subjected to varying load combinations, the line of the system has to approach the varying lines of thrust as good as possible.

Fig. 1. V.G.Shukhov, The GUM department store, Moscow At the beginning of the 20th_{century the Russian engineer V.G. Shukhov}

designed extreme light half circular arches to support the glass roof of the GUM department store in Moscow [4], see figure 1. Actually circular arches are not very efficient to transfer vertical loads. A circular arch can resist radial loads very well but due to vertical and horizontal loads these arches will be subjected to bending moments. To decrease the bending moments Shukhov strengthened the half circular arches of

the Gum with cables [4]. Cables can transfer tensile normal forces very well but can not stand compressive normal loads. To avoid buckling Shukhov post-tensioned the cables. Thanks to the post-tensioning the guy rods provided the arch with six additional bearings, capable of working both in compression and tension [4]. Unfortunately the design of an arch strengthened with post-tensioned cables is quite complex and laborious. Nowadays most engineers will use a computer program to design these complex structures [5]. Nevertheless a century ago Shukhov was capable to design these statically indeterminate structures with only a pencil and sliding ruler.

15,0 m

Fig. 2. Scheme of the arches supporting the roof of the GUM department store, strengthened with cables to reduce the bending moments [4]. This paper researches the methods Shukhov could have used to design post-tensioned half circular arches. A simple method, based on the Theory of Elasticity [6], is described to analyze and design these statically indeterminate form active structures. The calculations are validated with a finite element program. The analysis will show the efficiency of an circular arch provided it is strengthened with tensioned cables. Recommendations will be given to tune these post-tensioned arches to minimize the dimensions and reduce the need of material, self-weight and environmental load.

2. HALF CIRCULAR ARCHES

Circular structures can be composed of identical elements, this will simplify the production and assembling, consequently the costs of the whole structure, arches as well as the skin, are reduced. Further an circular arch can resist an equally distributed radial load very well. Due to this load the sections of the arch will be compressed and not loaded by bending so the dimensions can be small. However a circular arch supporting a roof will be subjected to vertical and horizontal loads causing bending moments, consequently the dimensions of the sections must be enlarged to resist these bending moments.

Firstly the load transfer is described for a half circular arch, subjected to an equally distributed vertical load and an anti-metrical radial load. The line of compression of these loads varies from the line of system, so the arch is subjected to bending. To show the effect of strengthening and post-tensioning the ties, the following chapters describe the load transfer

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identical elements. The form of the arch can be described easily with Cartesian as well as polar coordinates. Using polar coordinates (r,φ) the position of the nodes is described with the radius r and the angle φ between the radius and horizontal axis. The length of a chord is equal to: s = 2.r.sin (π/12) = 0,5176.r (1)

4 3 5

2 6

Y

φ

1 X r 7

Fig.3. The facetted arch with Cartesian and polar coordinates. Table 1: Polar coordinates of the nodes of the circular arch.

Nodes Coordinates 1 (r, π) 2 (r, 5.π/6) 3 (r, 2.π/3) 4 (r, π/2) 5 (r, π/3) 6 (r, π/6) 7 (r, 0)

3.1. Half circular arch subjected to vertical loads

The arch is subjected to five concentrated vertical forces Fv acting at the

nodes. The vertical reactions are equal to: V1 = 5/2 Fv en V7 = 5/2 Fv.

Fig. 4. Force diagram and the line of compression (red) due to the vertical concentrated loads acting at the nodes.

M6 = V7.r.[1 – cos (π/6)] – H7. r. sin (π/6) = -0,232.Fv.r (3)

M5 = V7.r.[1 – cos (π/3)] – H7. r. sin (π/3) -Fv .r.[cos (π/6) – cos (π/3)]

M5 = -0,098.Fv.r (4)

Both bending moments are negative and cause tensile stresses at the outer side of the arch. The normal forces acting at the bars are respectively: S45 = H4 . cos (π/12) + ½ Fv * sin (π/12) = 1,225.Fv (5) S56 = H7 . cos (π/4) + (5/2 - 1).Fv . sin (π/4) = 1,863.Fv (6) S67 = H7 . sin (π/12) + V7 . cos (π/12) = 2,708.Fv (7)

H

4

5

6 φ H

7

V

7

Fig. 5. Reactions, forces and loads for the right half of the arch. Nowadays the forces and bending moments can be calculated easily with a finite element program. With a finite element program the forces and bending moments are calculated for a steel arch, HE180 A, with a radius r = 7,5 m, subjected to concentrated loads equal to Fv = 10 kN.

Table 2: Results for the arch, with radius r = 7,5 m, subjected to the vertical loads Fv = 10 kN

Forces and moments:

Vertical reaction: V7 = 25,0 kN 2,5 . Fv Thrust: H7 = 11,33 kN 1,133 . Fv Moments: M5 = - 7,42 kNm -0,091.Fv.r M6 = - 17,5 kNm -0,023.Fv.r Normal forces: S45 = 12,24 kN 1,224 . Fv S56 = 18,62 kN 1,862 . Fv S67 = 27,08 kN 2,708 . Fv

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3

3.2. Anti-metrical radial load

Due to the wind load the arch is subjected to horizontal an vertical loads acting perpendicular to the surface. For the analysis the load is simplified to an anti-metrical radial load, see figure 7.

4

3

5

2

6 1 7

φ r

Fig. 7. Arch subjected to anti-metrical radial load acting at the nodes. The horizontal and vertical components of the loads acting at the nodes are given in the following table.

Table 3: Horizontal and vertical components of the anti-metrical loads Horizontal force Vertical force

2 H2 =+ Fr.cos (π/6) = ½√3.Fr V2 = - Fr.sin (π/6) = -½.Fr

3 H3 = + Fr.cos (π/3) = ½.Fr V3 = - Fr.sin (π/3) = -½√3.Fr

5 H5 = + Fr.cos (π/3) = ½.Fr V5 = + Fr.sin (π/3) = ½√3.Fr

6 H6 = + Fr.cos (π/6) = ½√3.Fr V6 = + Fr.sin (π/6) = ½.Fr

The vectors representing the anti-metrical radial forces act at the center of the arch. The equilibrium of the bending moments around the center is zero, so the vertical reactions must be zero too: V1 = V7 = 0.

The horizontal reactions follow from the equilibrium of the horizontal forces:

H1 = H7 = - [ cos (π/3) + cos (π/6) ] . Fr = -1,366.Fr (8)

The arch is subjected to the following bending moments and normal forces: M6 = H7. r. sin (π/6) = - [½ Fr + ½√3.Fr ]. ½ r = -0,683.Fr.r (9) M5 = - (½ Fr +½√3.Fr) * [2.r.sin (π/12)] ∗ cos (π/12) = -0,683.Fr.r (10) S45 = V4. sin (π/12) = 0,354.Fr (11) (12) S56 = H7 . sin (π/4) - Fr. sin (π/12) = 0,707.Fv (12) S67 = H7 . sin (π/12) = 0,354.Fr (13)

For the arch subjected to the anti-metrical loads Fr = 10 kN, the forces

and bending moments are calculated also with a final element program, see figure 8 and table 4.

Fig.8. Bending moments due to the anti-metrical loads.

Table 4: Results anti-metrical load, Fr = 10 kN, with r = 7,5 m

Forces: Vertical reaction: V7 = 0 kN 0 .Fr Thrust: H7 = 13,66 kN 1,366 .Fr Moment: M5 = -51,2 kNm - 0,683.Fr .r Moment: M6 = -51,2 kNm - 0,683.Fr .r Normal force: S45 = 3,52 kN 0,352 .Fr Normal force: S56 = 7,07 kN 0,707 .Fr Normal force: S67 = 3,52 kN 0,352 .Fr

4. ARCH STRENGTHENED WITH TIES

To reduce the bending moments the arch is strengthened with ties running diagonally from the supports to the nodes. The ties will prevent the arch of bending outward. A circular arch, subjected to vertical loads, is subjected to negative bending moments causing tensile bending stresses at the outer side so the arch is bending outward. The ties prevents these deformations. The ties pulls the arch inward. The bending moments due to the vertical loads are partly compensated by the bending moments caused by the ties. The structure is statically indeterminate. The magnitude of the tensile forces acting into the diagonals depends on the stiffness of the arch and the ties. For the structural analysis the compatibility has to be considered, so the deformation of the arch and ties must be analyzed. Actually the elongations of the ties due to the normal forces are much smaller than the deformations of the arch due to the bending. Neglecting the small deformations of the ties simplifies the analysis much. The load distribution is calculated by replacing the ties by virtual forces acting parallel to the ties. The magnitude of the virtual forces acting on the ties is found by minimizing the bending moments acting at the nodes: Mi = 0. Thus firstly the bending moments are

defined for the arch subjected to the virtual forces. Next the load transfer is defined by compensating the bending moments, due to the loads, by the bending moments, due to the virtual forces.

As calculated before, due to the vertical loads the arch is subjected to bending moments, see (4) and (3): M5 = - 0,098.Fv.r and M6 =

-0,232.Fv.r. Due to the tensile forces acting at the ties these bending

moments are compensated. Next the bending moments are calculated due to the virtual artificial loads acting parallel to the ties, firstly acting at the nodes 3 and 5 next acting at node 2 and 6.

4 3 5

2 6

φ

1 r 7

Fig. 9. Half circular ach subjected to 5 concentrated forces.

4.1.1. Arch subjected to concentrated virtual loads acting

symmetrically at node 3 and 5.

Due to a concentrated forces acting symmetrically at node 3 and node 5 parallel to the diagonals the arch is subjected to bending moments. Thanks to the symmetry the load transfer is defined for a half of the arch.

The vertical reaction force acting at the support follows from:

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equilibrium of horizontal forces: H4 = cos (

π

/6) . P5 – H7 = P5 (17)

Next the bending moments acting into node 5 and 6 are calculated: M5 = [1-sin (π/3)].H4.r = 0,134.P5.r (18)

M6 = [1- cos (π/6)].V7.r + sin (π/6).H7.r = 0 (19)

The normal forces acting at the bars are equal to:

S45 = H4 . cos (π/12) = 0,966.P5 (20) S56 = -H7 . cos (π/4) + V7*sin (π/4)= 0,448.P5 (21) S67 = - H7 . sin (π/12) + V7*cos (π/12) = 0,518.P5 (22)

H

4

5 P

5

6 1 H

7

7 α r V

7

Fig. 10. Arch subjected to artificial load acting at node 5 Table 5 shows the results calculated with a FEM program for an arch with a radius of r = 7,5 m subjected to a concentrated load, acting at node 3 and 5, equal to P3 = P5 = 10 kN.

Table 5: Results arch subjected to virtual loads P3 = P5 = 10 kN

Forces, moments Vertical reaction: V1 = V7 = 5,0 kN 0,5 . P5 Thrust: H1 = H7 = 1,345 kN 0,135 .P5 Bending moment: M2 = M6 = 0 kNm 0 .P5.r Bending moment: M3 = M5 = 10,01 kNm - 0,134.P5.r Normal force: S12 = S67 = - 5,18 kN - 0,518 .P5 Normal force: S23 = S56 = - 4,49 kN - 0,449 .P5 Normal force: S34 = S45 = - 0,967 kN - 0,967 .P5

Fig. 11. Bending moments due to a concentrated Force acting on node 3 and 5

V7 = sin ( π

/6). P5 = 0,259 F6 (23)

At the top of the arch the bending moment is equal to zero: M4 = 0.

According to the equilibrium of bending moments at the top we find: M4 = 0, V7 .r + H7.r – √2.r.sin (π/6). P6 = 0 (24)

With V7 = 0,259.F6:

H7 = -0,259.P6 + ½√2.P6 = 0,448.P6 (25)

The force H4 is calculated with the equilibrium of horizontal forces:

H4 = P6 .cos (π/12) - H7 = 0,518 .P6 (26)

Next the bending moments acting at node 5 and 6 are calculated. M5 = H4.r.[1 - sin (π/3)] = 0,0694.P6.r (27)

M6 = H7.r.sin (π/6) + V7.r.[1- cos (π/6)] = 0,259.P6.r (28)

The normal forces acting at the bars are equal to:

S45 = H4 . cos (π/12) = 0,5.P6 (29) S56 = H4 . cos (π/4) = 0,366.P6 (30) S67 = -H7 . sin ( π /12) + V7.cos ( π /12) = 0,134.P6 (31)

H

4

5 P

6

6 1 H

7

7 α r V

7

Fig. 12. Arch subjected to a concentrated load acting at node 2 and 6 Table 6 shows the results calculated with a FEM program for an arch with a radius of r = 7,5 m subjected to a concentrated load, acting at node 2 and 6, equal to P2 = P6 = 10 kN.

Table 6: Results arch subjected to virtual loads P2 = P6 = 10 kN

Forces, moments Vertical reaction: V1 = V7 = 2,588 kN 0,259.P6 Thrust: H1 = H7 = 4,485 kN 0,449. P6 Bending moment: M2 = M6 = 19,45 kNm 0,259. P6.r Bending moment: M3 = M5 = 5,17 kNm - 0,069. P6.r Normal force: S12 = S67 = - 1,35 kN -0,135.P6 Normal force: S23 = S56 = - 3,66 kN -0,366.P6 Normal force: S34 = S45 = - 5,0 kN -0,5 . P6

Comparing the results shows that the analysis and finite element program give nearly the same forces and bending moments.

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5

Fig. 13. Bending moments due to concentrated forces acting on node 2

and 6.

4.1.3. Calculation of the forces acting on the ties

Next the forces acting on the ties are calculated. In the analysis the diagonals are substituted by virtual forces acting parallel to the ties. The magnitude of the virtual forces is found with the equations of the bending moments caused by the virtual forces and the radial loads. Due to the vertical loads the bending moments acting on node 5 and 6 are: M5 = -0,098.Fv.r and M6 = -0,232.Fv.r. These bending moments are

compensated by the forces P5 en P6 acting at the diagonals.

H4 4

5 P5

P6

6 1 H7 7

α r V7

Fig. 14. Arch subjected to the virtual forces acting parallel to the diagonals

For the two nodes two equations are formulated, describing the equilibrium of bending moments.

ΣM5 = -0,098.Fv.r + 0,134.P5.r + 0,0694.P6.r = 0 (32)

ΣM6 = -0,232.Fv.r + 0.P5 + 0,259.P6.r = 0 (33)

Solving these equations gives:

P5 = 0,268.Fr and P6 = 0,897.Fr (34)

Now the elements of the arch are subjected to normal forces only. The normal forces acting on the members are calculated with the equilibrium of forces. The maximum normal force acts on member S67. The vertical

component of this bar is equal to the loads and forces acting on the diagonals P5 en P6. From this equilibrium of forces follows:

S67 = [5/2 Fv + P5. sin(π/6) + P6.sin(π/12)] / cos(π/12) = 2,967.Fv (35)

The horizontal component of this force is equal to:

H67 = S67. sin (π/12) = 0,768.Fv. (36)

The normal force acting on member S56 is in node 6 in equilibrium with

the load Fv, the force P6 and the normal force Fv acting on S6:

S56 = P6.cos(π/3) + S67.cos (π/6) – Fv. sin(π/4) = 2,310.Fv (37)

The normal force acting on S45 is in node 5 in equilibrium with the

vertical load Fv, the force P5 and de normal force Fv acting on S56:

S45 = P5.cos(π/4) + S56.cos (π/6) – Fv. sin (π/12) = 1,932.Fv (38)

Due to the vertical loads the half circular arch is subjected to negative bending moments giving tensile stresses on the outward side of the arch. Due to the normal forces acting on the diagonals the arch is subjected to

positive bending moments. These bending moments compensate the bending moments due to the vertical load.

The calculations are verified with a finite element program for an arch of steel, radius r = 7,5 m, profile He180A, strengthened with ties ∅25, subjected to vertical loads Fv = 10 kN acting at the nodes, see table 7.

Table 7: Results arch subjected to virtual loads: P3 = P5 = 10 kN

Forces, moments Vertical reaction: V1 = 25,0 kN 2,5 . Fv Thrust: H1 = - 2,42 kN 0,242.Fv Bending moment: M2 = M6 = - 0,9 kNm -0,012.Fv .r Bending moment: M3 = M5 = - 0,7 kNm -0,009.Fv. r Normal force: S12 = S67 = - 29,4 kN 2,94 .Fv Normal force: S23 = S56 = - 22,74 kN -2,27 . Fv Normal force: S34 = S45 = - 18,71 kN -1,87 . Fv Diagonal: S16 = S27 = + 2,27 kN +0,277.Fv Diagonal: S15 = S37 = + 8,55 kN +0,855.Fv

Fig. 15. Bending moments due to the vertical loads.

Comparing the results shows that the analysis match quite well with the output of the finite element program. The analysis neglects the deformations of the ties and arch, nevertheless the differences between the analysis and the output of the finite element program are small. 4.2. Calculation of the normal forces acting at the ties due to the

anti-metrical radial loads

Next the forces acting on the diagonals are calculated. In the analysis the diagonals are substituted by virtual forces acting parallel to the diagonals. The magnitude of the virtual forces is found with the equations of the bending moments caused by the virtual forces and the vertical loads. Due to the anti-metrical load the bending moments acting on node 5 and 6 are: M5 = M6 = 0,683.Fr.r. These bending moments are

compensated by the forces P5 en P6 acting at the diagonals.

Next the bending moments are calculated due to the virtual artificial loads acting parallel to the diagonals.

4.3. Arch subjected to virtual forces acting anti-metrically at node 3 and 5

The virtual forces acts parallel to the diagonals at node 3 and 5. For the right part of the arch the bending moments are defined.

For the anti-metrical load the horizontal reaction at the support is equal to the horizontal component of the load:

H7 = cos (π/6). P5 = ½√3.P5 (39)

At the top the bending moment is zero: M4 = 0, thus:

V7.r + H7.r.sin (π/3) – sin2(π/6).r.P5 = 0 (40)

Substitute H7 = ½ √3.P5 :

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6 1 H

7

7 α r V

7

Fig. 16. Arch subjected to anti-metrical load acting at node 3 and 5. The vertical force acting at the top follows from the equilibrium of forces:

V4 = sin (π/6).P5 - V7 = P5 (42)

Next the bending moments acting in node 5 and 6 are calculated. M5 = r.sin (π/6).V4 = 0,5.r.P5 (43)

M6 = + sin (π/6).r.H7 + [1-cos (π/6)].r.V7 = 0,366.r.P5 (44)

4.4. Arch subjected to virtual forces acting anti-metrically at node 2 and 6

The virtual forces act parallel to the diagonals at node 2 and 6. Using the symmetry the bending moments are defined for the right part of the arch.

V

4

5 P

6

6 1 H

7

7 α r V

7

Fig. 17. Arch subjected to anti-metrical load acting at node 2 and 6. The horizontal reaction acting at the support is equal to the horizontal component of the load:

H7 = cos (π/12). P5 = 0,966.P6 (45)

At the top the bending moment is zero, M4 = 0, Thus:

V7.r + H7.r.sin (π/6)– sin (π/12). cos (π/6).r.P6 = 0 (46)

Substitute H7 = cos ( π /12).P6 : V7 = -sin ( π /12).P6 = -0,259.P6 ↓ (47)

The vertical force acting at the top follows from the equilibrium of forces:

V4 = 2. sin (π/12).P6 = 0,5176381.P6 ↑ (48)

Next the bending moment acting in node 5 and 6 are calculated. M5 = r. sin (π/6).V4 = 0,259.P6 (49)

M6 = + sin (π/6).r.H7 + [1-cos (π/6)].r.V7 = 0,448.r.P6 (50)

ΣM6 = 0,683.Fr.r - 0,366.P5.r - 0,448.P6.r = 0 (52)

Solving these equations gives:

P5 = 1,0.Fr and P6 = 0,708.Fr (53)

The bending moments due to the anti-metrical loads are compensated by the bending moments due to the virtual forces acting at the ties, thus the elements of the arch are subjected to normal forces only. The normal forces acting on the members are calculated with the equilibrium of forces.

The analysis is validated with a finite element program. The arch is subjected to radial concentrated loads equal to: Fr = 10 kN. The results

of the calculation are showed in table 7. Comparing the results shows that the analysis and finite element program give nearly the same forces. The ties are subjected to forces S15 = 9,9 kN = 0,99.Fr and S16 =+ 7,06

kN = 0,706 Fr. These forces are nearly identical to the forces calculated

with the analysis. The bending moments calculated with the finite element program are very small, for node 3: M3 = 0,005.Fr.r.

Fig. 18. bending moments due to the anti-metrical load Table 8: Results arch subjected to radial loads Fr =10 kN

Forces, moments: V1 = 0 kN 0 . Fr H1 = 13,66 kN 1,366 . Fr M2 = 0,3 kNm 0,004.Fr.r M3 = 0,4 kNm 0,005.Fr.r S12 = - 7,0 kN - 0,7 .Fr S23 = - 5,08 kN - 0,508 .Fr S34 = + 0,2 kN + 0,02 .Fr S16 = + 7,06 kN + 0,706.Fr S15 = + 9,9 kN + 0,99 .Fr V7 = 0 kN 0 * Fr H7 = 13,66 kN 1,366 . Fr M5 = - 0,3 kNm - 0,004.Fr.r M6 = - 0,4 kNm - 0,005.Fr.r S67 = + 7,0 kN + 0,7 . Fr S56 = + 5,08 kN + 0,508. Fr S45 = - 0,2 kN - 0,02 . Fr S27 = - 7,06 kN - 0,706 . Fr S37 = - 9,9 kN - 0,99 . Fr

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7

The results show that at the leeward side the ties are tensioned but at the

windward side the ties are loaded by compression. Actually the ties cannot stand any compression, so the compressed ties must be removed or post-tensioned.

5. POST-TENSIONING

To avoid buckling the diagonals have to be post-tensioned in such a way that the compressive stresses are compensated by the tensioning. Due to the post-tensioning of the ties the arch is subjected to bending moments. The wind load is a variable load and can act from all sides, so the pre-tensioning must load the arch symmetrically.

Due to the anti-metrical load the ties at the windward side are compressed.The wind is a variable load and can at from all sides at the structure, consequently the post-tensioning forces must act symmetrically at the structure. The artificial load due to the post-tensioning must compensate the compressive forces acting at the ties so the ties must be loaded by artificial loads as calculated before, see equation (53):

P2 = P6 = 0,71.Fr and P3 = P5 = 1,0.Fr. (54)

Due to this artificial load the arch is subjected to bending moments equal to:

M6 = 0 * P5 + 0,259 * 0,71.Fr.r = 0,184.Fr. r (55)

M5 = + 0,134 * 1,0.Fr .r + 0,069 * 0,71.Fr.r = 0,183. Fr. r (56)

The loads due to the post-tensioning act continuously at the arch, so the bending moments due to the post-tensioning are permanently loading the arch.

The calculations are verified with a finite element program. Table 8 shows the results of the calculation of the arch subjected to the artificial post-tensioning loads equal to P2 = P6 = 7,1 kN and P3 = P5 = 10,0 kN.

Table 9 shows that the bending moments, calculated with the computer for these post-tensioning forces are nearly equal to the bending moments calculated with the analysis, see (55) and (56).

Table 9: Results for the arch, with radius r = 7,5 m, subjected to post-tensioning loads P2 = P6 = 7,1 kN and P3 = P5 =10,0 kN

Forces and moments

Vertical reaction: V1 = V7 = 6,788 kN 0,679 .Fr Thrust: H1 = H1 = 1,852 kN 0,185 .Fr Moments: M2 = M6 = 13,73 kNm 0,183.Fr.r M3 = M5 = 13,58 kNm 0,181.Fr.r Normal forces: S12 = S12 = -6,08 kN - 0,608 .Fr S23 = S12 = -7,04 kN - 0,704 .Fr S34 = S12 = -5,0 kN - 0,50 .Fr

Fig. 19. Arch subjected to the pre-tensioning forces 5.1. Bending moments for the post-tensioned arch

The post-tensioning is acting continuously at the arch so the bending moments caused by the post-tensioning have to be added to the bending moments due to the permanent and variable load. According to the analysis the strengthened arch is not subjected to bending due to the permanent load and the variable load, thus the structure is permanently

subjected to bending moments equal to the bending moments caused by the post-tensioning:

M2 = M6 = 0,184.Fp.r + 0 = 0,184.Fp.r (57)

M3 = M5 = 0,183.Fp.r + 0 = 0,183.Fp.r (58)

The not-strengthened arch was subjected to bending moments due to the vertical loads and the variable loads, see table 10.

Thanks to the post-tensioned ties the bending moments are decreased substantially consequently the dimensions of the arch can be decreased too. For a radial load equal to the permanent load, so Fr = Fv, the

reduction ratio β showing the reduction of the bending moments due to the post- tensioning is equal to:

β = 0,184/(0,232 + 0,683) = 0,2 (59) Thus the height of a rectangular section can be approximately halved. Table 10: Bending moments for the not-strengthened arch according to the analytical approach

Not-strengthened arch permanent load

Not-strengthened arch anti-metrical load

M2 = 0,232.Fv. r 0,683.Fr. r

M3 = 0,098.Fv. r 0,683.Fr. r

M5 = 0,098.Fv. r 0,683.Fr. r

M6 = 0,232.Fv. r 0,683.Fr. r

Table 11: Bending moments for the post-tensioned arch, according to the analytical approach

Post-tensioned arch M2 = +0,184.Fp. r

M3 = +0,183.Fp. r

M5 = +0,183.Fp. r

M6 = +0,184.Fp. r

Due to the deformations of the ties, the arch will be subjected to bending moments. The calculations using a finite element program showed that the effect of the deformations of the ties is rather small, nevertheless the post-tensioned arch is subjected to small bending moments due to the vertical or anti-metrical loads. For example according to the calculation with the finite element program for the permanent load, see table 7, the anti-metrical load, see table 8, and the post-tensioning load, see table 9, the arch is subjected to the following bending moments showed in table 12.

Table 12: Bending moments acting on the post-tensioned arch according to the finite element program calculations

Permanent load Anti-metrical load Post-tensioning M2 = - 0,012.Fv. r + 0,004.Fr. r + 0,183.Fp. r

M3 = - 0,009.Fv. r + 0,005.Fr. r + 0,181.Fp. r

M6 = - 0,012.Fv. r - 0,004.Fr. r + 0,183.Fp. r

M5 = - 0,009.Fv. r - 0,005.Fr. r + 0,181.Fp. r

6. CONCLUSIONS

Due to an equally distributed vertical load a half circular arch is subjected to bending moments, so structurally these arches are not optimal. Otherwise the half circular form can be partitioned easily in identical parts to simplify the construction, production, transport and assembling. Strengthening these arches decreases the bending moments substantially. Especially for half circular arches strengthening is very effective to reduce the bending moments caused by equally distributed vertical loads. Due to the circular form and the vertical load the circular arch is subjected to bending moments causing tensile stresses at the outer side. The arch is curved outward, so the ties are tensioned and

(8)

showed structure the bending moments acting on the post-tensioned arch were about 20% of the bending moments of the not strengthened arch. The post-tensioning decreases the deformations of the arch too, especially in case the structure is subjected to lateral loads.

The analytical approach offers a simple method to analyze the load transfer for arches strengthened with post-tensioned ties and design these arches without using a computer. The validations with the computer show that the deficiencies are quite small. Nevertheless it is amazing that Shukhov could design these elegant structures more than a century ago without the help of a computer, using a pencil and sliding rule only.

To reduce the warming up of the atmosphere and the emissions of CO2

engineers have to design structures with a minimal embodied fossil energy. Studying the slender and elegant structures designed by engineers as Eiffel, Sukhov, Candela, Gaudi [3], Torroga [7] and many other engineers [8] will be most inspiring for designers to create structures fulfilling the needs of the following decades.

7. REFERENCES

[1] Engel H. Structure Systems, Verlag Gerd Hatje OstFildern-Ruit Germany, 2e edition (1999).

[2] Vitruvius, The Ten Books on Architecture, translated by Morris Hicky Morgan, Dover Publications Inc, New York (1960). [3] Molema J., Antoni Gaudi, een weg tot oorspronkelijkheid, Thesis

TUDelft (1987) .

[4] Belenya E., Pre-stressed Load-Bearing Metal Structures, MIR Publishers Moscow (1977).

[5] Haarhuis K. Maximum Transparent Barrel-vaulted Glass roof, proceedings symposium IABSE-IASS, London (2011).

[6] Timoshenko S. and Young D.H, Technische Mechanica, Het Spectrum, Utecht Antwerpen (1976) translated from: Engineering Mechanics, Mac GrawHill, New York (1956).

[7] Torroga E., Logic der Form, Verlag Georg D.W. Callwell. Munchen (1961).

[8] Lux J.A., Architectures de Ingenieurs X1Xe-XXe Siegles, Centre de Creation Industrielle Centre George Pompidou, Expositions itinerantes CCI no_{8, Paris (1978).}