Prefabricated Vaults Strengthened with Ties
Wim Kamerling*Delft University of Technology, Faculty of Architecture Julianalaan 134, 2628 BL, Delft, The Netherlands
M.W.Kamerling@tudelft.nl
Abstract
This paper describes the design of prefabricated low-rise barrel vaults strengthened with ties. Structurally barrel vaults are very efficient, so the need of material is pretty small. Probably in the coming decades the reduction of the environmental load of buildings by reducing the need of material will be most important again. Nowadays concrete barrel vaults are not built often anymore, mostly because of the cost of construction. Prefabrication will reduce the cost of labour, especially if the variation of the elements is restricted and the radius of the curvature of the vault is constant. Generally vaults following a circle segment are subjected to bending moments. Strengthening these vaults with ties is effective to reduce the bending moments. This paper focuses on the design of prefabricated low rise vaults, composed of circular segments and strengthened with slender ties to contribute to the architecturally concept and to reduce the dimensions, self-weight, need of material and environmental load.
Keywords: Barrel vaults, arches, circle segment, optimising, strengthening.
Figure 1: Half circular vault strengthened with ties.
Introduction
Form-Active systems, as defined by Engel [2], are primary subjected to normal forces, consequently the load transfer is very effective. In practice if a structure is subjected to varying load combinations, so the line of the thrust will vary from the line of the system this structure is subjected to bending moments. Strengthening Form-Active systems as barrel vaults and arches with ties will help to reduce the bending moments. Due to an asymmetrical live load the ties can be subjected by compressive normal forces. To prevent buckling the ties have to be post-tensioned or dimensioned large enough to prevent buckling. According to Belenya [1] the Russian engineer V.G.Shukhof strengthened the half
circular arches in the Gum warehouse in Moscow with post-tensioned ties, see figure 1. However post-tensioning these ties will cause bending moments. In practice, especially for buildings with a substantial span, low-rise vaults are preferred above half circular vaults to reduce the surface and volume. This paper focuses on the design of prefabricated low rise barrel vaults, composed of circular segments, strengthened with tensioned ties, to reduce the bending moments, self-weight and need of material. Firstly the transfer of the loads for a not strengthened vault is described. Next alternatives are described to strengthen these vaults, to optimise the dimensions and to reduce the need of material and footprint.
1. The description of the coordinates of circle segments
For a circular segment with the centre of the coordinates at the top, the coordinates of a point P at the curve are described with: P(x,y) = (R.sin φ; R.[1 - cos φ] ). The coordinates of the supports are found for an angle φ = 2.β. The angle β follows from sin (2.β) = a/R and the rise f is equal to f = 2.R.sin2β. The span l is equal to l = 2.a = 2.R.sin (2 β). For a vault with rise f and a span l = 2.a the radius R follows from: R = ½ (a2 + f2)/f. The length of the arch between the top and support is equal to s= 2.β.R .
Figure 2. Segment of a circle described with polar coordinates.
2. Not strengthened vault composed of two circular segments
Firstly the forces and bending moments are defined for a not–strengthened vault composed of two prefabricated segments hinged at the top. The structure is simple supported and statically determinate. The reactions, forces and bending moments are defined with the expressions showing the equilibrium of forces and moments.
Figure 3: Vault subjected to a equally distributed load q.
R 2.β φ ½ l = a q f R 2.β φ a f
2.1 Vault subjected to an equally distributed load
For the three hinged vault subjected to an equally distributed load q the vertical reactions, thrust and bending moment follows from respectively:
VA = VB = q .R.sin (2.β) (1)
Η = q.R.cos2 β (2)
Mφ = q.R2.[cos2β .[1 - cos φ] – ½ .sin2 φ] (3)
The maximum moment is found for an angle φu following from cos φu = cos2 β and is equal to:
Mφ = - ½ q.R2.sin4 β (4)
2.2 Vault subjected to an equally distributed surface load
Due to the dead load the arch will be subjected to a equally distributed surface load q. For this the vertical reactions, thrust and bending moment follows from respectively:
Figure 4: Segment of a vault subjected to an equally distributed surface load.
VA = VB = ∫φ = 0φ =2β q.R.dφ = 2.β.q.R (5)
H = q.R.[2.β / tan β −1] (6)
Mφ = q.R2 [ 2.β × (1- cos φ) - φ.sin φ ] (7) tan β
The maximum moment acting at the arch is found for an angle φu following from: R 2β φ f ½ l = a x y
φu / tan φu = 2.β /tan β - 1 (8)
2.3 Three hinged arch subjected to an asymmetric load
Assume the live load is equally distributed and acting on the right side of the vault. For this loading the vertical reaction forces and thrust follows from respectively:
VA = ¼ q.R × sin (2.β) (9)
VB = ¾ q.R × sin (2.β) (10)
H = ½ q.R × cos2 β (11)
Figure 5: Vault subjected to an asymmetrical equally distributed load.
For the unloaded side the bending moment Mφ is calculated for an angle φ from the top with:
Mφ = ½ q.R2× [cos2β × (1 - cos φ) – sin β × cos β × sin φ ] (12)
The bending moment is maximal for φu = β. For this angle the maximal bending moment is:
Mφ=β = - ½ q.R2 × cos β × (1 − cos β) (13)
For the loaded part of the vault the bending moment Mφ is calculated for a certain angle φ from the top with:
Mφ = ½ q.R2 × [ cos2 β × (1 - cos φ) + sin β × cos β × sin φ – sin2 φ ] (14)
The bending moment is maximal for angle φu, which can be solved numerically from: R
2β φ a
cos2 β × tan φu = 2× sin φu − sin β × cos β (15)
For a low rise vault the angle φu is approximately equal to β. For φu = β the bending moment Mφ is:
Mφ=β = ½ q.R2 × cos (2.β) × [1 - cos β] (16)
This moment is positive and slightly smaller than the bending moment acting at the not loaded part of the vault.
2.4 Example prefabricated vault composed of circular segments.
A prefabricated vault is composed of segments, following a part of a circle. The span is equal to l = 14.4 m, the rise is of the swallow vault is equal to f = l/8 = 1.8 m. The permanent load, including finishing and vegetation is equal to 3.2 kN/m2, the variable load is equal to 5.0 kN/m2. The radius is equal to R = 15.3 m. The coordinates of the supports are found for an angle φ = 2.β. with 2.β = 28.0720 = 0.49 rad. To reduce the weight and environmental load cardboard tubes ∅60 are positioned with a centre-to-centre distance of 90 mm perpendicular to the span. For the section with a width of 1.0 m and a height of 110 mm the area and second moment of the area are respectively:
A = (110 – 60) × 1000 = 50 × 103 mm2
Ic = 1000 × 1103/12 - 1000 × 603/12 = 92.9 × 106 mm4
The forces and bending moments are defined with the expressions (1) to (16). Table 1 and table 2 show the results.
Table 1: Forces and bending moments acting at the vault.
permanent load live load asym. live load
Reaction, loaded side: VB = 24.0 kN 36.0 kN 27.0 kN
Thrust: H = 47.0 kN 72.0 kN 36.0 kN
max. bending moment: Mφ = -0.89 kNm 2.03 kNm 16.95 kNm
Table 2: Maximal normal forces, shear forces, bending moments and stresses.
load H = φ = V = N = σ = N/A M =
σ = M/W
permanent load 47.0 kN 200 17.1 kN 50.0 kN -1.0 MPa 0.89 kNm 0.53 MPa
asym. live load 36.0 kN 14.040 9.0 kN 37.1 kN -0.74 MPa 16.95 kNm 10.04 MPa
To reduce the bending moments it is advisable to strengthen the structure. The following chapter shows a vault strengthened with diagonals.
3. Vault strengthened with diagonals
According to Belenya at the beginning of the 19th century the Russian engineer V.G Shukhof
strengthened the half circular aches of the GUM department store in Moscow with tensioned cables [1], see figure 1. For a low rise vault the effect of the cables is analysed. The vault is strengthened with two cables running form the supports to the vault, see figure 6. The forces and bending moments are calculated with a Finite-Element program, for the loads described in paragraph 2.4. Due to the asymmetrical load one of the diagonals is compressed, F = -19.0 kN. Post-tensioning the diagonals will compensate the compressive load and prevent buckling of these slender ties.
Figure 6: Bending moments due to the a-symmetrical live load
To investigate the effect of the post-tensioning the vault is loaded at the nodes n5 and n13, connecting the diagonals with the vault, with a force Fp = 8.2 kN. Solving this force gives the following horizontal and vertical component: Fp hor = 8.12 kN, Fp vert = 1.0 kN, see figure 7. Due to the post-tensioning the vault is subjected to a bending moment M = 3.65 kNm and a normal force N = 7.8 kN. To compensate the compressive force acting in the tie the post-tensioning must be larger than the compressive force acting at the diagonal: F = 19.0 kN. Due to this post-tensioning force the bending moment is equal to: M = 3.65 × 19.0/8.2 = 8.5 kNm. This moment is quite large, but smaller than the maximum bending moment for the not-strengthened vault Mrep = 16.95 kNm. To avoid the increase of the bending moments due to the post-tensioning the following system is developed.
4 Vault strengthened with ties running from a centre point
To design a Form-Active-System optimal the line of the system has to follow the line of the thrust. For a circular vault the line of the system is identical to the line of the thrust if the structure is subjected to an equally distributed radial load. For any other load the line of the system will vary from the line of thrust. Thus a circular vault is by preference strengthened with radial ties, just like the spokes of a wheel. However in practice it is hard to strengthen a vault with radial ties. By preference the ties are jointed to the supports. The following design shows a vault strengthened with ties running from a centre, positioned at a distance ∆ = a.tan ε, above the horizontal line through the supports. Due to the thrust the ties will be tensioned. The compressive loads will be compensated if the centre is chosen well.
Figure 8: Vault strengthened with ties running from a centre point.
4.1 Parametric design
To analyse the transfer of the loads and tune the structure parametrically the vault is simplified and schemed as a statically determinate truss with hinged members. The span is equal to 2.a with a = R.sin 2.β. The rise of the vault is equal to f = 2.R.sin2 β. The length of the chords is equal to 2.R.sin γ. with γ = ½ β. The length of the vertical tie 1-4 is equal to f ' = f – ∆, with ∆ = a.tan ε.
The structure is subjected by concentrated forces F acting at the nodes. The normal forces acting in the ties and chords are defined with the equilibrium of the bending moments and forces acting horizontal
3
γ 2 4 f ' 6 α 1 3.γ ε β 5 a β βand vertical. Compressed elements are negative and tensioned elements are positive. To define the forces for an asymmetrical load the results of the anti-metrical loads and symmetrical load must be added and divided by 2. With the following table the forces can be defined to find the position of the centre, so the ties are tensioned for the given load combinations.
Table 3: Forces acting at the members of the radial reinforced trussed vault
symmetrical forces F2, F4 symmetrical force F1 anti–metrical F2, - F6
V1 - F - ½ F + ½ F/cos β
V5 - F -½ F - ½ F/cos β
H F.(a – R.sin β)/ f ' ½ F.a/f ' 0
S34 - H/cos γ - ½ H/cos γ 0
S36 2.H.tan γ -F + 2.H.tan γ 0
S23 - H/ cos γ - ½ H/cos γ 0
S56 H.tan α + ½ S36
sin ε + cos ε.tan α H.tan α + ½ Ssin ε + cos ε.tan α 36 + S cos ε 45 cos (3γ) S45 -S56.cos ε
cos (3.γ) -Scos (3.γ) 56.cos ε
- F
sin (3ε) + cos (3.γ).tan α S46 H - S56.cos ε cos α H - S56.cos ε cos α - S45.cos ε cos α S16 H. tan α + ½ S56
sin ε+cos ε.tan α
H. tan α + ½ S56
sin ε+cos ε.tan α - S45 cos (3γ)
cos ε S26 H - S56.cos ε
cos α H - S cos α 56.cos ε + S cos α 56.cos ε S12 - S56.cos ε
cos (3.γ) - S cos (3.γ) 56.cos ε
+ F . sin (3ε) + cos (3.γ).tan α
4.2 Example
The forces are defined for a vault with f = a/4, R = 2.125 × a, tan β = f/a, so β = 14.036 and γ = 7.018. The structure is subjected to concentrated forces acting at the nodes: F = 1.0 kN. For this vault the distance of the centre above the horizontal line between the support is positioned at ∆ = ¼ f, so the lever arm f ' is equal to 0.75 f or f ' = 3/16 a. Table 4 shows the forces due to the symmetrical and asymmetrical loads acting at the nodes, for Fi = 1.0 kN.
The concentrated load acting at the crown causes a compressive force F = -0.343 in the vertical strut S36 but the permanent load will cause a tensile force F = 0.293. The strut will be tensioned if the concentrated load is smaller than 85% of the permanent load.
Due to the asymmetrical load member S46 is compressed, F = -0.514, this force is compensated partly by the permanent load, causing a force F = 0.617.
To prevent the web bars to be subjected by compressive forces the designer can decrease the lever arm or increase the dead load.
Table 4: Forces acting in the members of the strengthened vault
member sym. sym. permanent. anti-metrical asymmetrical.
F2, F4 F3 F2, F3, F4 ½ (-F2, F4) ½(F2,F4)+ ½.(-F2, F4) S34 -2.604 -2.687 -5.291 0 -1.302 S36 +0.636 -0.343 +0.293 0 +0.318 S46 -0.531 +1.148 +0.617 -0.822 -1.088 S56 +3.107 +1.554 +4.661 +0.801 +2.354 S45 -3.323 -1.662 +4.985 -0.856 -2.518 S26 -0.531 +1.148 +0.617 +0.822 -0.556 S23 -2.604 -2.687 -5.291 0 -1.302 S16 +3.107 +1.554 +4.661 -0.801 +0.753 S12 -3.323 -1.662 +4.985 +0.856 -0.805
4.3 Finite element analysis of the truss
To check the analysis the forces, acting at the members of the truss, are calculated with a Finite-Element program. The structure is subjected to concentrated forces acting at the nodes. Comparing the results as given in table 5 and 6 shows that the results of the finite element calculation match with the analyse.
Table 5: Forces acting at the members of the truss
x = y = member top: F3 F2, F4 F4 F2, F3, F4 n1 -7.20 1.800 (S12) M1: -1.66 -3.32 - 0.81 -4.98 n2 -3.71 0.457 (S23) M2: -2.69 -2.60 - 1.30 -5.29 n3 0 0 (S34) M3: -2.69 -2.60 - 1.30 -5.29 n4 3.71 0.457 (S45) M4: -1.66 -3.32 -2.52 -4.98 n5 7.20 1.800 (S56) M5: +1.55 +3.11 +2.36 +4.66 n6 0 1.35 (S16) M6: +1.55 +3.11 +0.75 +4.66 (S26) M7: +1.15 -0.53 +0.56 +0.62 (S36) M8: -0.34 +0.64 +0.32 +0.30 (S46) M9: +1.15 -0.53 - 1.09 +0.62
4.4 Finite-Element analysis of the strengthened vault
Actually the vault is not composed of hinged elements but statically indeterminate. The forces and bending moments are defined with a Finite-Element program for a dead load qg = 3.2 kN/m2 and live load qe = 5.0 kN/m
2
. Due to the a-symmetrical load the tie M21 is compressed, N = -8.0 kN, but due to the permanent load the tie is tensioned, N = 13.3 kN. Thus the tensile force due to the permanent compensates the compressive force due to the live load. The cables can be dimensioned slender and will not disturb the transparency. The structure can be optimised further to reduce the bending moments due to the loads by moving the centre downward.
For the permanent and the asymmetrical live load the bending moment is maximal at node 5. The stresses due to the normal force and bending moment are respectively:
Prep = -71.9 – 54.2 = - 126.1 kNm, σ = N/A = -2.5 N/mm2 Prep = 2.10 + 4.58 = 6.68 kNm, σ = Maze/I = 3.95 N/mm2
Table 6: Output for a surface load PG = 3.2 kNm/m and the asymmetrical load we = 5.0 kNm/m
Node Coord. Member Perm. Perm Live Live Live
x y N M N M V n1 -7.20 0.0 vault M1:n1-n2 - 64.2 0.89 - 27.8 1.23 1.3 n2 -6.360 -0.415 vault M2: n2-n3 - 62.9 1.98 - 27.8 1.42 0.2 n3 -5.497 -0.779 vault M3; n3-n4 - 61.6 1.98 - 27.7 1.42 2.7 n4 -4.612 -1.088 vault M4: n4-n5 - 60.6 2.10 - 27.6 4.58 3.8 n5 -3.711 -1.343 vault M5: n5-n6 - 71.9 2.10 - 53.9 4.58 6.2 n6 -2.795 -1.542 vault M6: n6-n7 - 71.4 1.50 - 54.2 3.91 2.9 n7 -1.869 -1.685 vault M7: n7-n8 - 70.9 1.50 - 54.3 3.91 3.0 n8 -0.937 -1.771 vault M8: n8-n9 - 70.6 1.37 - 54.1 3.51 3.7 n9 0 -1.80 vault M9: n9-n10 - 70.6 1.39 - 54.1 1.55 3.9 n10 0.937 -1.771 vault M10: n10-n11 - 71.0 1.55 - 54.4 2.16 2.6 n11 1.869 -1.685 vault M11: n11-n12 - 71.4 0.41 - 55.2 1.88 3.3 n12 2.795 -1.542 vault M12: n12-n13 - 71.9 2.01 - 56.2 1.33 4.5 n13 3.711 -1.343 vault M13: n13-n14 - 60.6 2.01 - 64.8 0.84 2.7 n14 4.612 -1.088 vault M14: n14-n15 - 61.7 1.62 - 66.1 1.50 2.8 n15 5.479 -0.799 vault M15: n15-n16 - 63.0 1.62 - 67.9 0.92 2.3 n16 6.360 -0.415 vault M16: n16-n17 - 64.2 0.92 - 69.7 0.58 2.3 n17 7.20 0 tie M17: n17-n18 +57.6 +62.1 n18 0 -0.60 tie M18: n1-n18 +57.6 +25.6 tie M19: n9-n18 + 4.3 + 3.2 tie M20: n5-n18 -13.4 +29.1 tie M21: n13-n18 +13.3 - 8.0
Figure 9: Bending moments due to the permanent load
Figure 10: Bending moments due to the a-symmetrical live load
5. Conclusions
Strengthening the prefab low-rise vault is quite efficient. Due to the strengthening the bending moments and shear forces decrease. The structure can be strengthened with cables running from the supports to the vault, in the same way as Shukhof strengthened the GUM department store. For this vault the ties will be subjected to compressive forces in case the structure is loaded asymmetrically. Post-tensioning these ties to compensate the compressive forces will increase the bending moments. Strengthening the vault with ties running from a centre point increases the tensile forces. The ties are tensioned continuously if the centre point is positioned well. For this vault strengthened with ties ramming from the centre the bending moments are much smaller than the bending moments acting at the not-strengthened vault. Consequently the need of material and the embodied energy will be pretty small. Architectonically the slender ties are hardly noticeable and do not spoil the transparency of the interior.
References
[1] Belenya E. Pre-stressed Load-Bearing Structures. MIR publishers, Moscow, 1977. [2] Engel H. Structure systems. Verlag Gerd Hatje (2nd edn), Darmstadt, 1999.
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