Renovating a Fusee Ceramique Barrel Vault

Renovating a Fusee Ceramique Barrel Vault

Delft University of Technology, Faculty of Architecture Julianalaan 134, 2628 BL Delft, The Netherlands

M.W.Kamerling@tudelft.nl.

Abstract

In 1956 two workshops, varying in height and span, were built in Wormerveer, The Netherlands. Both workshops were roofed with a concrete barrel vault with a thickness of 110 mm. The cylindrical vaults were constructed with Fusée Ceramique elements. A Fusee is a small ceramic tube, which was embedded in concrete roofs to save cement and self weight. An analyse of these vaults revealed the following structural problems. Due to the time dependent deformations, as creep and shrinkage, the compressive normal forces acting at the fusee's increase and the normal forces acting on the concrete decrease, consequently the stiffness of the vault and the buckling resistance decrease both. Further the roofs were designed to resist a live load of 0.5 kN/m2. Snow can accumulate at the gutter of the roof of workshop 2, resulting in a much larger load than the live load prescribed in the past. To meet the demands of the present this vault had to be strengthened. To minimize the self-weight and the time needed for the renovation a proposal was made to strengthen this vault with steel diagonals. This document describes the structural design of the strengthening of the vault to increase the bearing capacity.

Keywords: Barrel vault, concrete shell, Fusée Ceramique system, renovation, strengthening.

Figure 1: Section of workshop 1 and 2.

11.15 m 14.65 m

1. Introduction

For the vault of workshop 2 the original structural design is described. The effect of the time dependent effects is analysed. Due to the creep and shrinkage the transfer of the loads changes, the normal forces acting at the infills increase and the normal forces acting at the concrete part of the sections decrease. Consequently the stiffness of the vault and the buckling resistance decrease too. Next the loads according to the Euro-codes of the present are defined. Snow can accumulate at the gutter of the roof of workshop 2. Especially to resist this load the vault of workshop 2 has to be strengthened. A proposal is made to strengthen this roof with steel diagonals running from the top to the supports. For the strengthened structure the forces and bending moments are calculated with a finite element program for the loads prescribed by the present Euro-codes. Finally the ultimate bearing capacity of the strengthened structure is defined.

2. Original design

To reduce the self-weight and need of cement ceramique infill elements were embedded in the concrete vaults. A Fusée is a ceramic tube with a conical top. The outward diameter is Ø80 mm, the inward diameter is Ø60 mm and the length is equal to 350 mm. At the site the elements were joined by pushing the top of the Fusee's into the rear of the next element. The vaults are supported with a frame of beams (section 400 × 400 mm2_{) and columns (section 400 × 400 mm}2_{) with a}

centre-to-centre distance of 3.28 m. The span of the vault of workshop 2, with a thickness of 110 mm, is 14.65 m. The rise of the vault is equal to 1.835 m. The gutter is 5.82 m above the floor. The roofs are reinforced with Ø6 – 180 parallel to the span, at both sides, with a cover of 15 mm. The structure was designed for a permanent load of 2.0 kN/m2 and a live load of 0.5 kN/m2.

Figure 2: Section of the Fusée Ceramique vault according to van Eck and Bish [1]

Features: according to Van Eck and Bish [1] the stiffness and ultimate stresses are: Concrete: σ_c = 6 MPa Ec = 2.1 × 104 MPa

Reinforcement: σ_s = 140 MPa Es = 2.1 × 105 MPa ns = Es/Ec = 10.0

Fusee's: Ef = 1.7 × 104 MPa nf = Ef /Ec = 0.81

Fusée's: Af = 11×π×(802- 602)/4 = 24.2 × 103 mm2 If = 11× π. (804 - 604)/64 = 15.1 × 106 mm4 Concrete: Ac = 110×1000 - 11×¼ π.802 = 54.7 × 103 mm2 Ic = 1000 × 1103/12 – 11 × π.804/64 = 88.8 × 106 mm4 reinforcement: As = 2 × ¼ π × 62 × 1000/180 = 2 × 157 mm 2 Is = 2 × 157 × (110/2-18) 2 = 0.43 × 106 mm4

The reactions, normal forces and bending moments acting at the vault are defined with the following expressions.

symmetrical load asymmetrical load Min. vertical reaction force: VA = ½ q × l VA =

1

/8 qe× l

Max. vertical reaction force: VB = ½ q × l VB = 3/8 qe× l

Thrust: H = 1/8 qg× l2 / f H = 1/16 qe× l2 / f

Normal force: Nx = [H2 + (q.x)2]0,5 Nx = [H2 + (VB – q.x)2]0,5

Bending moment: Mo = qe× l2 /64

Figure 3: Scheme of the vault, subjected to the permanent and asymmetrical live load. Table 1 shows the forces and bending moments due to the permanent and live load.

Table 1. Forces and bending moments due to the permanent and asymmetrical live load Forces, bending moment: permanent load: sym. live load: perm. + sym. live load: asym. live load: perm. + asym. live load: V = 14.7 kN 3.7 kN 18.3 kN 5.5 kN 20.2 kN H = 29.3 kN 7.3 kN 36.6 kN 3.7 kN 33.0 kN M = - - - 1.7 kNm 1.7 kNm x = ¼ l: N = 30.2 kN 7.6 kN 37.8 kN 5.2 kN 34.8 kN φ R f ½ l qe qg

According to the Theory of Elasticity the stress follows from σx = Ex.ε and the specific deformation ε

follows from:

ε0 = N with: mEA, t=0 = 1 + nf × Af /Ac + ns× As /Ac (1)

Ac × Ec× mEA, t=0

Substitution of Af, Ac, As, ns, nf and Ec gives for t = 0:

mEA t=0 = 1 + 24.2 × 103 × 0.81 + 157 × 2 × 10 = 1.42

54.7 × 103 54.7 × 103

For the permanent load the normal force acting in the vault is equal to N = 30.2 kN. The specific deformation ε is equal to:

ε0 = 30200 = 0.0185 ×10 -3

54.7 × 103× 2.1 × 104 × 1.42

For the combination of the permanent and asymmetrical live load, the normal force is equal to N = 34.8 kN. The specific deformation is equal to:

ε0 = 34800 = 0.021 × 10-3

54.7 × 103× 2.1 × 104 × 1.42 Table 2: Normal forces and stresses for t = 0

permanent load permanent and asym. live load

N = σ = N = σ =

Concrete 21.2 kN 0.39 MPa 24.1 kN 0.44 MPa

Fusée's 7.6 kN 0.32 MPa 8.7 kN 0.36 MPa

Reinforcement 1.2 kN 3.9 MPa 1.4 kN 4.4 MPa

The stress due to the bending moment follows from:

σc = M × z × Ec (2)

EI

Substitution of M = 1.7 kNm, EI = 2.2 × 1012 _Nmm2 _{, z = ½ ×110 mm and E}

c = 21000 MPa in (2)

gives the bending stresses:

σc = 1.7 × 106 ×1/2× 110 × 21000 / (2.2 × 1012) = 0.9 MPa

The stresses due to the normal forces and due to the bending moment, are pretty small.

3. The effect of the time dependent deformations

Due to creep and shrinkage the specific deformation increases with ∆ε , thus for t = ∞ the specific deformation of the concrete is: εt=∞ = ε0 + ∆ε. Due to the creep the concrete and fusee's will shorten

with respectively εrc en εrf.. The fusee's and reinforcement are attached firmly with the concrete so the

time dependent deformation of the concrete will be resisted by the fusee's and reinforcement. The sections are subjected to internal forces. The steel rebars and fusee's are subjected to an internal compressive force of respectively Fs and Ff. The concrete is subjected to an internal tensile force Fc.

The sum of the internal forces is equal to zero:

Fc + Ff + Fs = 0 (3)

Figure 4: Forces and deformations due to the time dependent effects.

Due top the internal force Fc the specific deformation of the concrete is reduced with εc = Fc/AcEc .

During a time t the specific deformation due to the internal force Fc is increased by the creep with

εcφ = Fc ×k×φ/AcEc. Actually the internal force is not constant during the time t, consequently the

effect of creep is reduced with a factor k, according to Scherpbier [3] this factor k is equal to ½. Now the specific deformation of the concrete, Fusée's and reinforcement follows respectively from:

εct = εt× ε0 + ε0×φ + εsc - Fc× (1+ ½ φ) (4) Ac Ec εft = ε0 + εsf + Ff (5) Af Ef ε0 ∆ε Nc Nf Ns Ff Fs εsc ε0.ϕ Fc εsf

εst = ε0 + Fs (6)

As Es

The internal force Ff acting at the Fusée's follows from εft = εst and expression (5) and (6):

Ff = Fs× Af Ef - εsf × Af Ef (7)

As Es

The internal force Fc acting at the concrete follows from

εct

=

εst

ε

φ

As Es (1+ ½ ×φ) (1+ ½ ×φ)

The internal force Fs is defined by substituting Ff (7) and Fc (8) into the expression (3):

Fs = As Es× (ε0×φ + εsc )× Ac Ec /(1+ ½×φ) + εsf × Af Ef (9)

Ac Ec /(1+ ½×φ) + Af Ef + As Es

For this vault the shrinkage for concrete and Fusée's is equal to respectively: εsc = 0.4 × 10-3 en εsf =

0.1 ×10-3. Due to the creep the specific deformation of the concrete increases with φ.ε, thus the deformation is equal to ε × (1+φ). The creep factor is equal to φ = 3.8. For the permanent load the specific deformation is: ε0 = 0.0185 ×10-3 . Next the internal forces, acting on the reinforcement,

fusee's and concrete, are defined with respectively (9), (7) and (8). The forces, due to the normal load and internal forces, follow for the concrete, fusee's and reinforcement from respectively: Nc – Fc ,

Nf + Ff and Ns + Fs. Table 3 shows for t = 0 and t = ∞ the normal forces and stresses acting on

the concrete, fusées and reinforcement due to the permanent load. Due to the internal force the concrete section is tensioned. Possibly the structure is cracked, due to the cracks the stiffness and buckling force will decrease.

Table 3: Normal forces, internal forces and stresses in the vault for t = 0 en t = ∞.

N t = 0 Ft = ∞. Nt =0 + Ft = ∞ σ t = ∞.

concrete: -21.2 kN + 81.9 kN +60.7 kN +1.1 MPa

fusée's: -7.6 kN – 73.2 kN -73.2 kN -3.0 MPa

4 Ultimate state

For the ultimate state the resistance of the structure is defined. Successively the design load, stiffness, buckling force and ultimate bending moment are defined. The design load due to permanent and asymmetrical live load follows from: qd = 1.2×2.0 + 1.5×0.5 = 3.15 kN/m.

Vd = 1.2 × 2.0 × 14.65/2 + 3/8 ×1.5 × 0.5 × 14.65 = 21.7 kN

Hd = (1/8 × 1.2 × 2.0 + 1/16 × 1.5 × 0.5) × 14.652/1.835 = 40.6 kN

Nd = [ 40.62 + (21.7 – (1.2 × 2.0 + 1.5 × 0.5)×14.65/4)2]1/2 = 41.9 kN

Md = 1.5 × 1.7 = 2.55 kNm

4.1 Stiffness

The stiffness for the ultimate design of the vault follows from EI = Met/κet . The curvature κet follows

from:

κet = fs/Es (10)

(1 – d/h - kx).h

Figure 5: Forces and deformation for a section with an embedded infill element

With the following procedure the stiffness is defined for varying normal forces: • Define for the normal force N the compressive zone kx.h;

• Define for σs = fs the bending moment Met;

• Define the curvature κet with (10);

• Define the stiffness EI of the vault with: EI = Met/κet;

For the ultimate state: fs = 220/1.15 = 191 MPa; Es = 2×105 MPa;

fc = 12/1.5 = 8 MPa; Ecd = 22500/(1+ 0.8) = 12500 MPa .

According to table 5 the stiffness of the vault is for N = 41.9 kN equal to: EIet = 0.29 × 1012 Nmm2.

Fs1 Fc Fs2 kx.h h εsc εc εst

Table 5: The curvature, bending moment and stiffness for varying normal forces for the ultimate state N = [kN] kx = κ ×106 [1/mm] Met = [kNm] EI×1012 [Nmm2] 0 0.17 13.27 2.54 0.19 20 0.21 14.12 3.42 0.24 40 0.24 14.86 4.27 0.29 60 0.27 15.56 5.09 0.33 80 0.29 16.25 5.87 0.36

4.2.The buckling force

Palkowski [2] gives the following expression for the buckling load:

Ncr = π2 EI (11)

(β.s)2 With: s = the length of the vault, s = 7.5 m

β = reduction coefficient, for the not strengthened vault β = 1.

Substituting s = 7.5 m, β = 1 and EIet = 0.29 × 1012 Nmm2 in (11) gives: Ncr = 50.9 × 103 N.

For the normal force Nd = 41.9 kN the buckling ratio is equal to: n = Ncr/Nd = 50.9/41.9 = 1.21.

The bending moment is equal to: Md = 1.5×1.7× n/(n-1) = 14.7 kNm

4.3 Ultimate bending moment

To define the reinforcement the following graph is made for sections of concrete with embedded infill elements, subjected to an eccentric normal force.

For the normal force Nd = 41.9 kN follows: Nd = 0.048

b.h.fcd

For this ratio, graph 1, gives: Mu = 0.051

b.h2.fcd

Then the ultimate bending moment is equal to Mu = 4.9 kNm. The bending moment acting at the vault

is Md = 14.7 kNm, this moment is significant larger than the ultimate bending moment. The vault

\ Md b.h2.fcd Nd b.h.fcd 0.019 -0.028 0.037 0.012 0.061 0.075 0.086 0.149 0.103 0.200 0.114 0.240 0.120 0.262 0.123 0.278 0.125 0.294 0.127 0.310 0.129 0.326 0.129 0.342 0.129 0.358 0.125 0.384 0.121 0.410 0.116 0.435 0.111 0.458

Graph 1. Ultimate bending moment with respect to the normal force for the Fusee vault.

5. Finite element calculation

With a computer program the forces and bending moments are defined for the structure strengthened with diagonals 2∅76.1-×4, see figure 6. At first the loads are defined according to the codes of the present.

Figure 6: Sketch of the section of the barrel vault strengthened with diagonals

5.1 Wind load

For this building with a height of 10.3 m the wind pressure is qpw = 0.69 kN/m2. The wind load

follows from p = ci ×qp [kN/m 2

]. The roof is partitioned in three zones A, B en C. The coefficients for under and over pressure are respectively 0.3 and 0.2.

Fusee vault, FeB220, C12/15, ω =0.00285, d/h=0.164

-0,1 0 0,1 0,2 0,3 0,4 0,5 0 0,05 0,1 0,15 Md/(b.h.h.fc) N /(b ;h .fc )

Suction and under pressure:

A: p = (-1.2 + 0.3) × 0.69 = -0.62 kN/m2 p = (-1.2 - 0.2) × 0.69 = -0.97 kN/m2 B: p = (-0.83 + 0.3) × 0.69 = -0.37 kN/m2 p = (-0.83- 0.2) × 0.69 = -0.71 kN/m2 C: p = (-0.4 + 0.3) × 0.69 = -0.07 kN/m2 p = (-0.4 - 0.2) × 0.69 = -0.42 kN/m2

Figure 7: Wind suction acting at the cylindrical roof.

5.2 Snowload

The snow load follows from: psn = ui × 0.7 kN/m2.

For an equally distributed snow load ui = 0.8 and psn = 0.8 × 0.7 = 0.56 kN/m2.

For an linear increasing snow load ui = 1.45 and psn max = 1.45 × 0.7 = 1.02 kN/m 2

.

Figure 8: Snow loads acting at the vault.

At the gutter of workshop 2 the snow load can be heaped up. The roof is assumed to be subjected to a linear triangular load at maximum equal to psnmax = 2.8 kN/m2 and acting over a length lo = 2 × ∆h,

∆h is the difference of the height. With ∆h = 2.85 m the length is equal to lo = 2 × 2.85 = 5.7 m. f q sn max ½ q sn max l ∆h l0 q sn max q sn B A C f

5.3 Results of the calculation of the structure with the Finite element method

The forces, deformations and bending moments are calculated with the finite element program. The maximum bending moments and normal forces, due to the ultimate permanent load and the variable loads, are given in table 6.

Figure 9: Bending moments due to the snow load at maximum at the gutter of workshop 2.

Table 6. Maximal normal forces and bending moments included load factors for the ultimate state

Load member Normal force [kN] Bending moment [kNm]

1 permanent + live load S9 1.2 × 30.3 + 1.5 × 7.1 = 47.0 1.2 × 0.6 + 1.5 × 0.1 = 0.9 2 permanent + snow load S2 1.2 × 33.1 + 1.5 × 28.4 = 82.3 1.2 × 0.2 + 1.5 × 1.4 = 2.3 3 permanent + snow load S9 1.2 × 30.3 + 1.5 × 5.3 = 44.3 1.2 × 0.6 + 1.5 × 0.2 = 1.0 4 permanent + wind load S9 0.9 × 30.3 - 1.5 × 8.0 = 15.3 0.9 × 0.6 + 1.5 × 0.2 = 0.8 5 permanent + wind load S9 0.9 × 30.3 - 1.5 × 13.6 = 6.9 0.9 × 0.6 + 1.5 × 0.3 = 1.0

5.4 Buckling

The bending moment and normal force are at maximum for the vault subjected to permanent load and the snow load. For this load combination the normal force, bending moment and stiffness are respectively: Nd = 82.3 kN, Md = 2.3 kNm, EI = 0.36 × 10

-12

Nmm2. Due to the strengthening with diagonals the buckling length is reduced, β = ½. The length of the vault is equal to: s = 7.5 m. Next the buckling force is defined with expression (11):

Ncr = π2.EI = π2 ⋅× 0.36 × 1012 = 252.6 × 103 N

(β.s)2 (0.5 × 7500)2

The buckling ratio is equal to: n = Ncr/N = 252.6/82.3 = 3.1. The maximum bending moment is equal

to: Md = 2.3 × 3.1/(3.1 -1) = 3.4 kNm.

Graph 1 shows for Nd/(b.ht.fd) = 0.094 → Mu/(b.ht2.fd) = 0.067 , thus Mu = 6.5 kNm > Md = 3.4 kNm.

The ultimate bending moment is larger than the maximal bending moment, the vault can transfer the prescribed loads.

5.5 Diagonals

The hangers and ties have a centre to centre distance of 3.28 m. The structure is strengthened with diagonals 2×∅76.1-4 running at both sides of the hangers and ties. The maximum tensile and compressive normal force acting at these diagonals are respectively Nd = ½ × 76.3 kN and Nd = - ½ ×

16.8 kN. The compressive normal force is much smaller than the tensile force, so these elements can be dimensioned quite slender.

Conclusions

Due the strengthening with diagonals the bending moments as well as the buckling length of the vault are decreased substantially. The tensile forces acting at the diagonals ∅76.1-4 are much larger than the compressive forces, so these elements can be dimensioned very slender. The strengthening does not increase the dead weight much. Architecturally the slender elements do not decrease the transparency of the interior. Possible the described system of strengthening, with diagonals running from the top to the supports, can be helpfully to preserve existing arches and vaults for the next generations.

References

[1] Eck P.J.W. van and J.F.Bish. Het Fuseedak. Cement 1954; 6: 240-243.

[2] Palkowski, S., Buckling of parabolic arches with hangers and tie. Engineering Structures 2012;

44: 128-132.

[3] Scherpbier G.: De invloed van het krimpen en kruipen van het beton op samengestelde constructies, PHD-Thesis. Delft, 1965.