Thin Concrete Barrel Vault

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.)

Thin Concrete Barrel Vault

Wim Kamerling

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

Summary: The paper presents the structural design of a thin barrel vault constructed with Fusée Ceramique infill elements. The load transfer is analyzed and validated. For the structure composed of Fusée Ceramique elements, steel and concrete the stresses are calculated and compared to the stresses given in the codes used from 1950 to the present. The advantages and disadvantages of these low rise barrel vaults are showed. Further the possibilities of a light infill to reduce, for structures of concrete, the need of mortar, the self-weight and environmental load are discussed.

Keywords: Barrel vault, reinforced concrete, Fusée Ceramique infill elements, CO2 emission, environmental load, sustainability, vegetation roof. 1. INTRODUCTION

Just before World War II the French architect Jacques Couëlle invented a system embedding cylindrical tubes of ceramic in concrete walls, roofs and floors. During the fifties of the XX-century building industry was booming. The cost of material and labor were rising and architects and engineers had to find alternatives. Embedding Fusée Ceramique elements in concrete roofs and floors can reduce the self weight and the need of cement and steel substantially. To produce these elements a workshop was erected in Echt, Limburg, The Netherlands, with a capacity of 10 million elements yearly [1]. Nowadays the Fusée Ceramique elements are not used anymore and this system seems almost forgotten. Nevertheless the advantages of an light infill are again interesting. The production of cement is responsible for about 5% of the CO2 emission. Reducing the emission of CO2 can contribute much to

decrease the warming up of the atmosphere. The environmental load of concrete can be decreased by for example using blast furnace cement instead of Portland or by using a light infill. A light infill will reduce the need of mortar, the self weight and the reinforcement too, consequently the environmental burden is decreased too.

Fig. 1. Workshop building Q, constructed in 1955, Woerden, The Netherlands.

The Fusée Ceramique elements were applied mostly for roofs of workshops [2]. Fifty years later most of these buildings are obsolete and pulled down. Nevertheless it is a pity that these buildings are lost forever. At least a few of these buildings must be preserved for the following generations. In 1955 two workshops were constructed on the military complex in Woerden, The Netherlands. Building Q was used as a warehouse for storage and building R was used as a workshop for repairing army tents. When the military complex was closed these buildings were of no use any more and abandoned. The city council bought the site to built domestic houses. Spring 2012 both buildings were pulled down. The demolition revealed an amazing slender roof structure. The thickness of the vaults was 130 mm only. For building Q, with a span of nearly 20 m the ratio thickness/span was 1/154, much lesser than usually recommended in the literature [3]. For example Van Eck and Bish [3] recommended for structures with a span of 15 m at

least two layers of fusées and a thickness of at least 180 mm. It is a pity that the designers did not publish an article about this remarkable thin structure. Calculations of the stresses showed that the vaults in spite of the slenderness of the sections could transfer very well the permanent and live loads to the supporting bearing walls. Due to the minimal need of material the environmental burden of these roofs is pretty small. Especially low rise barrel vaults are structurally very efficient [4] and can transfer easily extra loads for a vegetation and solar systems to improve the sustainability of these buildings and decrease the environmental load. Adding a vegetation will increase the thermal insulation of a roof. During the summer the temperature of a roof surface can be rise up to (273+80)0

K. A vegetation can decrease this surface temperature substantially and drop down the high temperatures of urban areas with circa 20 _{K, so less energy is needed for cooling.}

Further the vegetation on a roof will retain rainwater and discharge the sewage system.

Fig. 2. The demolition of building Q shows the remarkable slenderness of the vault.

This paper describes the structure of building Q. The bearing capacity of the vault to transfer the permanent and variable loads is analyzed. The advantages of an infill to reduce the environmental load of structures of concrete are described. For concrete barrel vaults varying infill elements are discussed to reduce the weight, need of cement, reinforcement and environmental load.

2. ANALYSIS OF THE VAULT OF BUILDING Q

The demolition of this building showed the amazing redundancy of the vault to resist heavy blows. To understand the bearing capacity of these low rise barrel vaults an analysis of the bearing capacity of building Q is made. The calculation is reconstructed conform the theories practiced during the fifties of the twentieth century as described by Van Eck and Bish in Cement [3]. Firstly the geometry and the features of the materials are defined. Next the forces, bending moments and stresses are calculated. Finally the calculation is validated.

and a thickness of 110 mm. For vaults with one layer of fusées a thickness of 110 mm was minimal, see figure 3. For this building a thickness of 130 mm was applied, probably to increase the stiffness needed to create a span of nearly 20 m.

Fig.3. Example of a section of a roof with a minimal thickness of 110 mm, reinforced with rebars center to center 180 mm in a layer at the top

and bottom [5]. 2.1.1. Area and stiffness

The roof structure is composed of concrete, fusées and reinforced with steel. In a section with a width of 1,0 m eleven fusées were placed with a spacing of 10 mm. A fusée is an ceramic cylindrical element with a length of about 350 mm, an outward diameter of Ø80 mm and a thickness of 10 mm. The center to center distance of the ceramic elements is equal to 90 mm, see figure 3. For building Q a thickness of 130 mm was applied. To resist the thrust, steel bars Ø32 were added at a center to center distance of 1,0 m. To decrease the deformations the bars were supported with three ties Ø8 mm hanging down from the vault. The vault was reinforced with bars Ø8 – 180 at the top and bottom. The code of 1950 required for floors with a thickness of 120 mm or more a cover on the reinforcement of 15 mm at minimum. Distribution bars were not used. The roof thickness exceeds a thickness of 120 mm so the minimal cover is equal to 15 mm. For a part of the roof with width of 1,0 m the area and the second moment of the area of the concrete, fusées and the rebars are calculated in table 1 for a section with a width of 1,0 m. The rebars Ø8 – 180 are running parallel to the span and are positioned in a layer at the top and a layer at the bottom of the vault between the fusées with a covering of 15 mm [3].

Table 1: Area and second moment of the area of the fusses, concrete and rebars of the vault with a width of 1 m

Elements Area [mm2_{] and second moment of the area [mm}4_]

Fusees Af = 11 *¼ π ∗ (802-602) = 24,19.103 Concrete Ac = 130 * 1000 - 11 *¼ π.802 = 74,708.103 Steel As = ¼ π ∗ 82 * 1000/180 = 279 Fusees If = 11 .π.(804-604)/64 = 15,12.106 Concrete Ic = 1000*1303/12 - 11 *π.804/64 = 160,97.106 Steel Is = 2 * 279 * (½ * 130 – 15 - ½ * 8)2 = 1,18.106

According to the Theory of Elasticity the stiffness is defined with: EI = Ec * Ic + Ef * If + Es * Is (1)

The Young's modulus of the rebars is equal to Es = 2,1.105 N/mm2. The

stiffness of a concrete structure depends on many variables such as compressive strength, shrinkage, creep and cracks. In the fifties generally the stresses in concrete and reinforcement were calculated with a ratio of the Young's modulus for steel and concrete of n = Es/Ec .

EI = Ec * (Ic + If + n.Is ) (3)

Table 2: Stiffness of the vault for a width of 1,0 m Stiffness

EA= 2,1.104_.(74,71.103_{+ 24,19.10}3 _{+10.2.279) = 2,194.10}9 _N

EI = 2,1.104 _.(160,97.106_+15,12.106_+10*1,18.106_{) = 3,946.1012 Nmm}2

2.1.2. Loads

The roof is subjected to the following dead and live loads as shown in table 3.

Table 3: Permanent and live loads

Equally distributed load

Self weight: pg = 74,704 * 24 + 24,19 * 18 = 2,23 kN/m2

Finishing: pg = 0,17 kN/m2

Permanent load: pg = 2,40 kN/m2

Live load: pe = 1,00 kN/m2

2.2. Forces, bending moments and stresses

The Forces acting at the vault were calculated with the Theory of Elasticity [6]. The vault was schemed as a arch supported with two simple supports. The deformation of the supports was neglected, the ties will lengthen so the supports will move sideward. With a finite element program the effect of the lengthening of the ties will be studied later. The thrust was calculated with the equilibrium of bending moment around the top [6]. Actually the bending moment at the top is not zero, the effect of this assumption is small, this will be validated later with an finite element program.

2.2.1. Full load

The vault is subjected to the dead load qg = 2,4 kN/m2 and a live load qe

= 1,0 kN/m2 _{. The calculation is made for a width of 1,0 m. Due to the}

symmetrical live and dead load the vault is subjected to a full load equal to: q = 2,4 + 1,0 = 3,4 kN/m. The span is equal to 2 * a, with a = 9,9 m.

Fig. 4. Section of the vault subjected to a symmetrical load. The vertical and horizontal reaction force acting on the supports are calculated with respectively:

VA = VB = qg *a + qe * a (4)

φ R

f

a =½ l

q

+q

α

s

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.)

H = (½ qg * a2 + ½ qe * a2)/f (5)

The vault is subjected to compressive normal forces, the normal force acting at the supports is equal to the sum of the vectors V and H:

N = (H2 _{+ V}2₎0,5 ₍₆₎

For x = ½ a = 4,95 m the normal force follows from:

N = [H2 _{+ (q * x)}2_]0,5 ₍₇₎

For the full load the normal forces are shown in table 4. Table 4: Reactions and normal forces due to full load.

Forces [kN] VA = VB = ½ * 2,4 * 19,8 + ½ * 1,0 * 19,8 = 33,66 H = ½ * (2,4 * 9,92 _{+ 1,0 * 9,9}2_{)/2,48 = 67,18} N x = a = [67,182 + 33,662]0,5 = 75,14 N x= a/2 = [67,182 + (3,4 * 9,9/2)2]0,5 = 69,30 2.2.2. Asymmetric load

Due to an asymmetric load the vault is subjected to bending. Assume The vault is subjected to the live load acting at the right side. The permanent load is equal to qg = 3,3 kN/m2. The live load is equal to qe =

1,0 kN/m2_{. The vault is subjected to a minimum load q}

g = 2,4 kN/m2at

the left side and a maximum load equal to qg + e = 3,4 kN/m2 at the right

side.

Fig. 5. Section of the vault subjected to an asymmetrical load. The vertical and horizontal reaction force acting on the supports are respectively:

VA = qg * a + ¼ qe * a (8)

VB = qg * a + ¾ qe * a (9)

H = (½ qg * a2 + ¼ qe *a2/)/f (10)

The resulting normal forces acting at the supports are respectively: NA = (H2 + VA2)0,5 (11)

NB = (H2 + VB2)0,5 (12)

The bending moment is equal to:

M x=a/2 = qe * a2 /16 (13)

For x = ½ a = 4,95 m the normal force follows from:

N x=a/2 = [H2 + (VB - q*x)2]0,5 (14)

Table 5: Normal forces and bending moment due to asymmetrical load

Force: VA = 2,4 * 9,9 + ¼ * 1,0 * 9,9 = 26,24 kN VB = 2,4 * 9,9 + ¾ * 1,0 * 9,9 = 31,19 kN H = ( ½ *2,4 * 9,92 _{+ ¼ *1,0 * 9,9}2 _{)/2,48 = 57,30 kN} NAx = a = (57,32 + 26,242)0,5 = 63,10 kN NBx = a = (57,32 + 31,192)0,5 = 65,20 kN N x=a/2 = [57,32 + (31,19 – 3,4*4,95)2]0,5 = 59,10 kN Moment: M x=a/2 = 1,0 * 9,92 /16 = 6,13 kNm 2.2.3. Buckling

The rise of the parabolic vault is rather small, to define the buckling force the curve is approached by a segment of a circle. For a segment of a circle the critical load qcr can be calculated with the following

expression as defined by Timoshenko et al [7] and Goldenblat et al [8]:

qcr = EI [π2/φ2-1]/ R3 (15)

Thus the ultimate buckling force is equal to:

Ncr = qcr .R = EI [π2/φ2-1]/ R2 (16)

The ratio ncr of the buckling force and the normal force is equal to:

ncr = Ncr/N (17)

Generally it was recommend to design structures with a ratio ncr larger

than 5, thus ncr ≥ 5.

For a circle the angle φ between the radius through of the support and the top is equal to the angle α, see figure 6. For a parabola the angle φ between the tangent and the horizontal line through the supports can be calculated with: tan φ = 2.f/a (18) Thus: tan φ = 2 * 2,48/9,9 = 0,501 → φ = 26,610 _{= 0,465 radials (19)}

φ R

f

a

q

½ α

s

Fig. 6. Circle segment subjected to a radial equally distributed load. The radius of a parabola varies and can be calculated with the following expression:

Rφ = ½ a2 _{* (1+ 4*f}2_/a2₎1/2 _{/ f (20)}

Substituting the rise f and half of the span a = 9,9 m into equation (20) gives:

Rφ = ½ * 9,92 _{(1+ 4 * 2,48}2_/9,92₎0,5 _{/ 2,48 = 22,1 m (21)}

Substitute the radius Rφ = 22,1.103 mm, EI = 3,946.1012 Nmm2 and the angle φ = 0,465 radials into equation (16) to calculate the critical buckling force Ncr : Ncr = 3,946 * 1012 * [π 2 / 0,4652_{-1] / (22,1.10}3₎2 _{= 361,6.10}3 _{N (22)}

φ R

f

½ l

q

q

2.2.4. Calculation of the normal stress, full load.

According to the Theory of Elasticity the normal stress in the concrete, fusées and reinforcement due to a normal force is equal to:

σx = N * Ex /EA (25)

Table 6 shows the stresses, calculated by substituting the Area, Young's Modulus, stiffness and normal load into equation [25] due to normal force N = 69,3 kN caused by the full load for x = ½ a.

Table 6: Stresses acting at the fusées, concrete and steel for the full load Stresses [MPa] σf = -69,3 * 10 3 _{* 2,1 * 10}4 _/2,194.109 = -0,66 σc = - 69,3 * 10 3 _{* 2,1 * 10}4 _/2,194.109 = -0,66 σs = - 69,3* 10 3 _{* 2,1 * 10}5 _/2,194.109 = -6,6

2.2.5. Normal and bending stress, asymmetrical load

Due to the dead load and the asymmetrical live load the structure is subjected to normal stresses and bending stresses.At a quarter of the span the normal force is equal to N = 59,1 kN.

Table 7: Stresses acting at the fusées, concrete and steel due to the normal load N = 59,1 kN caused by the asymmetrical load for x = ½ a

Normal stresses [MPa] σf = -59,1 * 10 3 _{* 2,1 * 10}4 _/2,194.109 = -0,57 σc = -59,1 * 10 3 _{* 2,1 * 10}4 _/2,194.109 = -0,57 σs = -59,1* 10 3 _{* 2,1 * 10}5 _/2,194.109 = -5,7

Due to the asymmetrical loading the vault is subjected to a bending moment equal to: Mo = 6,13 kNm. Conform the Theory of Elasticity the

stress is calculated with:

σc = ncr /( ncr -1) * M * z * Ec/EI (26)

For ncr = 6,1, z = ½ * 130 mm, EI = 3,9458 * 1012 MPaand Ec = 21000

MPa, the bending stress is equal to σc = 2,54 MPa. The stresses due to

compression and bending are calculated by summarizing the normal stress and bending stress, thus:

maximal compressive stress: σc = -0,57 - 2,54 = - 3,11 MPa

maximal tensile stress: σc = -0,57 + 2,54 = 1,97 MPa

The maximum tensile stress due to bending is quite large, probably due to the asymmetrical live load the concrete will be cracked. The vault is reinforced to resist the bending tensile stresses.

3. ULTIMATE STRESS

Langejan [1] gives the following maximum values of the stresses, see table 8. For structures of concrete the maximal compressive stress and shear stress are quite small.

Generally engineers have to design structures with respect to the building codes. Table 9 shows the maximum stresses according to the building code of 1950 [9] for reinforced concrete.

Table 9: Maximum stresses for concrete and steel [9] [MPa] concrete compressive bending stress: fcu = 8

concrete compressive normal stress: fcu = 6

concrete shear stress: τu = 0,5

Steel QR24: fs u = 140

For the fusées the maximum stress was not given in the code of 1950. Experiments showed that the ultimate compressive stress of the fusées is equal to ffu = 100 MPa and the ultimate bending stress is equal to ffu = 13

MPa [1]. For the design the maximum stress is less than the ultimate stress found experimentally. Probably the designers limited for composite structures the ultimate compressive stress of the fusées to a maximum equal to the maximum stress of concrete, then the maximum stress for the fusées is equal to ffu = 8 MPa. Nowadays these maximal

values seem pretty low. At the present an statistic ultimate value is defined by subtracting the statistic uncertainly from the average value, so the ultimate stress f is equal to:

fu = x mean – c * s MPa (27)

with x is the average value, c is a parameter defined with the theory of probability and s is the standard deviation of a certain population. Further the design loads are defined by multiplying the representative loads with a load factor. In the past the stresses due to the representative loads were compared with an allowable stress. The allowable stress is calculated by diving an average value by an safety factor. In the present the classes of concrete are based on the compressive strength found by testing cylinders or cubes. Since 1974 minimal twelve cubes with an edge length of 150 mm were tested according to the code demanded in the Netherlands. Next the characteristic compressive strength is found with:

fck,cube = x12 – c * s MPa. (28)

According to the code of 1950 [9] the compressive strength of concrete was defined by compressing minimal three cubes, the average compressive strength had to be minimal 25 MPa. The edge length of the cubes had to be 200 mm. Between the load and the cube a thin plate of cardboard had to be laid to distribute the load over the area. Later the code of 1962 specified three classification K160, K225 and K 300. The strength of concrete specified in the code of 1950 can be classified as K250 according to the classification of the code of 1962. The classification of the code of 1962 was different from the classification of the code of 1974 and 1984, but the classification of 1974 and 1984 does not differ much from the classification given in the Euro code. According to Van Boom [10] the strength of cubes with a side of 150 mm without patch of cardboard is 16% larger than cubes with sides 200 mm. For the classes K160, K225 and K300, conform the code of 1962, Van Boom gives maximum values for the deviation equal to respectively: s = 4,6 MPa, 6,1 MPa and 7,6 MPa according to the codes of 1974. These maximum values are based on a large number of experiments. For a large population the factor c for a change of 5% is equal to 1,64. Consequently the compressive strength can be calculated with:

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.)

Table 10 shows the classes of the code of 1962 and the classification according to the Euro code 2 based on the strength of cubes.

For K250 (according to the code of 1950) the deviation s will be about 6,6 MPa, then the compressive characteristic strength of the cubes is according to equation (29) equal to:

fck,cube = 1,16 x12 - 1,64 .s = 1,16 * 25 – 1,64 * 6,6 = 18,1 MPa (30)

Table 10: Classes according to the code of 1962 and the class according to the Euro code 2 [11]

Class xmean

[MPa]

Deviation [MPa]

Compressive strength of cubes fck,cube [MPa]

Class Euro code 2 K160 16 4,6 1,16 * 16 –1,64 * 4,6 = 11 C9/11 K225 22,5 6,1 1,16 *22,5–1,64 * 6,1= 16,1 C12/15 K300 30 7,6 1,16 * 30 – 1,64 * 7,6= 22,3 C18/22

The class K250 is approximately equal to the class C15/18 as given in the Euro code [11]. In 1950 for structures of concrete the safety factor concerning the load was 1,8 and the factor concerning the deficiencies and variety of the strength of the material was 1,15 so the maximum allowable stress had to be:

fcu = 18/(1,8*1,15) = 8,7 MPa (31)

So actually the ultimate compressive strength for bending, fcu = 8 MPa,

as described in the code of 1955, was quite reasonable compared with the maximal stress defined in the codes of the present.

3.1. Maximum normal stress, including second order

Conform the Dutch code for structures of concrete of 1950 [9] the ultimate compressive stress in a structure subjected to a normal force was equal to: fcu = 6,0 MPa, see table 7. However for structures

subjected to a normal force the stresses are increased by the second order deformations. To include the second order the Dutch code decreased the maximal stress with a safety factor γ. Thus for columns the maximum stress due to the normal load had to be smaller than the ultimate stress corrected with a factor γ:

σ = N/A < fcu/γ = 6,0/γ MPa (32)

The ratio γ was given in the code [9] art. 32 with respect to the slenderness, see table 11.

Table 11: The factor γ , the slenderness λ and the maximum stress according to the code for structures of reinforced concrete of 1950 [9]

slenderness λ γ σ = f'c/γ [MPa] 60 1,0 6,0 80 1,5 4,0 100 2,0 3,0 120 2,5 2,4 140 3,0 2,0

For a slenderness of λ = 60 the safety factor is equal to γ = 1,0 and the maximum compressive stress is equal to fcu/γ = 6,0 MPa. The safety

factor increases for a slenderness larger than 60. For a slenderness λ = 140 the safety factor is equal to γ = 3,0, then the maximum compressive stress is equal to fcu/γ = 6,0/3,0 = 2,0 MPa.

The slenderness λ is defined with:

λ = lc /i (33)

The factor i is the radius of gyration, this factor is equal to: i = √(I/A)

(34)

A is the area of the section and I is the second moment of area. The slenderness can be calculated also by substituting the radius of gyration (34) into (33):

λ= lc/i = √(A.lc 2

/I) ₍₃₅₎

The slenderness of this barrel vault is quite large. Firstly the radius of gyration is calculated with (34). Substituting EA = 2,294 *109 _{N and EI}

= 3,9458 * 1012 _Nmm2 _{into (35) gives:}

i = (I/A)½ _{= 41,5 mm. (36)}

The buckling length of a column is well defined. For an arch the buckling length is about 1,2 * s, s is the length of the arch from the support to the top. The buckling length can be defined more accurate with the expression of the buckling load according to Timoshenko [7] and Goldenblat [8] and the expression of the buckling load according to Euler. According to Euler a structure subjected to a compressive normal force will collapse in case the load is larger than the buckling load Ncr

with:

Ncr = π2 * E * I / lc2 (37)

With equation (16) and (37) the buckling length of an arch follows from: lc =Rφ.π/(π2/φ2-1)1/2 (38)

Substituting the radius Rφ = 22,1 m, see (21), and the angle φ = 0,465

radials, see (19), into equation (38) gives:

lc =10,39 m. (39)

The length of a low rise arch can approached with:

s = Rφ.φ (40)

For Rφ = 22,1 m, see (21), and the angle φ = 0,465 radials the length of

the arch between support and top is equal to s = 10,3 m. So for this low rise arch the buckling length is 1,01 times the length of the arch. Next the slenderness is calculated with expression (33) and the radius of gyration according to equation (34). For a buckling length of lc = 10,39 m the slenderness of the vault is equal to: λ = 10390/41,5 = 250. For this slenderness table 11 does not give the safety factor and maximum stress. For this slender vault the buckling has to be checked as well as the stress. As mentioned before it is recommended that the buckling load is at least larger than five times the load, or the ratio ncr of

the buckling force and the normal force (17) is larger than 5: ncr = Ncr/N

≥ 5. For the full load the buckling ratio is equal to 5,2 ,see (23), so the buckling ratio of this vault fits to the demand ncr ≥ 5.

The reduction of the maximal stress with a safety factor γ to include the second order was a practical and simple method to design a safe structures, nevertheless especially for this slender vault the question rises in which way the ratio ncr was related to the safety factor γ and the

maximum stress. This can be showed as follows.

Firstly the slenderness (35), buckling force (37) and the normal stress σ = N/A are substituted into the expression for the ratio ncr (17):

ncr = π2.E/(λ2 * σ) (41)

Next the ratio ncr is calculated by substituting the maximal value of the

stress σ = f'c /γ and Young's modulus Ec = 21000 MPa into this

equation (37):

ncr ≥ π2.21000/(λ2 * f'c /γ ) (42)

Table 12 shows the relation of the buckling ratio ncr , the safety factor γ,

and the maximum stress f'c /γ for several values of the slenderness λ,

constructed by extrapolating the values given in table 11, for a slenderness up to 260.

0 0,2

60 80 100 120 140 160 180 200 220 240 slenderness

Fig 7. The ratio ω =1/γ and slenderness λ.

Table 12: Slenderness λ, buckling ratio ncr, maximal stress and the safety

factor γ Slenderness λ γ ω = 1/γ Maximal stress: σ= fcu /γ [MPa] ncr 60 1,0 1,0 6,0 9,6 80 1,5 0,67 4,0 8,1 100 2,0 0,50 3,0 6,9 120 2,5 0,40 2,4 6,0 140 3,0 0,33 2,0 5,3 160 3,5 0,29 1,7 4,7 180 4,0 0,25 1,5 4,3 200 4,5 0,22 1,3 3,9 220 5,0 0,20 1,2 3,6 240 5,5 0,18 1,1 3,3 260 6,0 0,17 1,0 3,0

For a slenderness of λ = 60 the safety factor is equal to γ = 1,0 and ncr =

9,6. the maximum compressive stress following from σ = N/A is equal to σ = f'c/γ = 6,0 MPa. The safety factor increases linear with respect to

the slenderness but the buckling ratio decreases not linear with respect to the slenderness. For a slenderness λ > 140, the buckling ratio ncr is

smaller than 5. The code of 1950 described in table 11 only the stresses for a slenderness up to 140. For slender structures with λ > 140 it is not sufficient to check the stress only, apparently the designers must calculate the buckling ratio ncr as well for these structures.

For a slenderness of 250 the safety factor γ must be equal to: γ = 5,75 so the maximum stress is equal to: fcu/γ = 6,0/5,75 = 1,04 MPa. For the full

load the calculated normal stress, see table 6, is equal to: σc = 0,66

MPa. So the normal stress due to the full load is smaller than the maximum stress showed in table 12: σc = 0,66 < 1,04 MPa.

For the full load the buckling ratio is equal to 5,2 so ncr ≥ 5. The slender

structure fits to the demand of the stress as well as to the demand of the buckling ratio.

4. VALIDATION OF THE CALCULATION

For the parabolic roof the normal forces and bending moments are calculated with a finite element program, Matrix-frame. The line of the system of the vault follows a parabola:

y = f.(x/a)2 ₍₄₃₎

The parabola is partitioned in small parts with ∆x = 1,0 m. The length of the elements is about 5% of the span. The elements Si and nodes Ni are

numbered from the left support to the right support the number of the left support is N1, the number of the right support is N21. The area of the

1 -9.9 -2.48 S1 1 2 2 -9.0 -2.05 S2 2 3 3 -8.0 -1.62 S3 3 4 4 -7.0 -1.24 S4 4 5 5 -6.0 -0.91 S5 5 6 6 -5.0 -0.63 S6 6 7 7 -4.0 -0.41 S7 7 8 8 -3.0 -0.23 S8 8 9 9 -2.0 -0.10 S9 9 10 10 -1.0 -0.03 S10 10 11 11 0 0 S11 11 12 12 1.0 -0.03 S12 12 13 13 2.0 -0.10 S13 13 14 14 3.0 -0.23 S14 14 15 15 4.0 -0.41 S15 15 16 16 5.0 -0.63 S16 16 17 17 6.0 -0.91 S17 17 18 18 7.0 -1.24 S18 18 19 19 8.0 -1.62 S19 19 20 20 9.0 -2.05 S20 20 21 21 9.9 -2.48 S21 1 21

4.1. Permanent and live load

With the finite element program Matrix-frame the bending moments, shear forces and normal forces are calculated for the permanent load q = 2,4 kN/m acting equally distributed on the surface and the equally distributed live load q = 1,0 kN/m, acting asymmetrically at the left and right part of the arch.

Fig 8. Permanent surface load q = 2,4 kN/m.

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.)

Fig 10.Variable load q = 1,0 kN/m, acting at the left part of the arch. Table 14: Results, forces and bending moments acting on the elements due to the permanent and live load

Member Load: Normal

force N [kN] Shear V [kN] Moment M [kNm] S10 Permanent load -48,70 1,46 0,39 Live load -9,97 2,18 0,06 S16 Permanent load -50,78 1,44 0,62 Live load -10,49 0,72 6,09 S20 Permanent load -54,55 1,33 0,41 Live load -12,14 2,43 0 S21 Permanent load 48,65 0 0 Live load 9,90 0 0

Fig. 11. Bending moments due to asymmetrical live load q = 1,0 kN/m, the bending moment is at maximum 6,09 kNm.

Comparing the results of the finite element calculations with the analysis shows a good match of the results. The effect of the deformation of the tie is pretty small. For the analysis the permanent load is equally distributed over the ground face so the parabolic vault is not subjected to bending. For the calculations with the finite element program the permanent load is distributed over the surface, so the sections are subjected to bending moments. The results show that these bending moments are very small. Apparently for this swallow vault the differences between the line of thrust for this load and the line of system are quite small.

4.2. Wind loads

The wind load acting on the roof is calculated according to the Euro code [12]. The city of Woerden is situated at the frontier between South Holland and Utrecht. The height of the structure is equal to 7,555 m. The wind load is calculated with:

p = c.q(z) (44)

For an urban area II, partly without neighboring buildings, the wind load q(z) is given in the code:

For z = 7,0 m: q(z) = 0,75 kN/m2

For z = 8,0 m: q(z) = 0,79 kN/m2

For z = 7,555 the wind load is calculated by linear interpolation: q(z=7,555) = 0,75 + 0,04 * 0,555 = 0,772 kN/m2 ₍₄₅₎

The coefficients for the wind load are showed in table 15, for the windward, top and leeward sides. The sign of a pushing load acting at the surface is positive. The sign of a sucking load acting away from the surface is negative. Three area are defined, at the windward side area A, at the top area B and at the leeward side area C.

Combining internal and external wind pressure two extreme wind loads arise, respectively with wind over pressure and wind under pressure. Table 16 shows the wind loads for the combination with under and over pressure for the area A, B and C.

Table 15: Coefficients of the wind load for area A, B and C

Area Coefficient c

A, windward side Sucking - 1,2

Pressure + 0,2

B, acting at the top Sucking - 0,85

C, leeward side Sucking - 0,4

Internal pressure Over pressure + 0,2 Under pressure - 0,3

Table 16: Wind load combinations with under and over pressure

Area [kN/m2]

With under pressure A p = (+0,2 + 0,3) * 0,772 = + 0,39 B p = (-0,85 + 0,3) * 0,772 = - 0,43 C p = (-0,4 + 0,3) * 0,772 = - 0,08 With over pressure A p = (-1,2 - 0,2) * 0,772 = -1,08 B p = (-0,82 - 0,2) * 0,772 = - 0,81 C p = (-0,4 - 0,2) * 0,772 = - 0,46

Fig. 12. Wind load due to under pressure, pressure at the windward side and sucking at the top and leeward side.

Fig. 13. bending moments due to under pressure, pressure at the windward side and sucking at the top and leeward side.

Fig.14. Wind load due to over pressure and sucking.

Fig.15. Bending moments due to overpressure, pressure at the windward side and sucking at the top and leeward side.

Table 17: Normal forces and bending moments due to the wind load.

Member Wind load: N

[kN] V [kN]

M [kNm] S4 Wind + under pressure + 5,51 0,38 +3,53 S13 Wind + under pressure + 5,26 0,29 -1,62 S4 Wind + over pressure +16,78 0,59 -2,73 S16 Wind + over pressure +17,02 0,36 +2,25 For this structure the bending moments and normal forces, due to the wind loads, are much smaller than the bending moments and forces calculated for an asymmetrical live load equal to q = 1,0 kN/m. The calculations of the full load and asymmetrical live loads are decisive for the calculations of the stresses.

5. CONCLUSIONS

The calculations showed that the forces and bending moments calculated with the finite element program match well with the analysis according to the state of art in 1955. Without computers engineers could design these structures very well, able to transfer the loads during a long time. The vault is designed very slender with a ratio of thickness to the span of 1/154 and a slenderness of λ = 250. The thickness of the vault was much smaller than recommended by van Eck and Bish [3]. Thanks to the efficient load transfer the stresses acting on the sections of the vault are pretty low.

The embodied energy of thin barrel vaults is quite small. For roofs swallow barrel vaults are very suitable to create large spans with a minimal need of material. For roofs of concrete adding a light infill is very profitable. For concrete the emission of CO2 is for the better part

caused by the production of mortar and steel. The Fusée Ceramique elements reduced the self weight and need of cement up to 25% [1]. Of course other another infill can be used too. In stead of Fusée Ceramique elements cardboard tubes, boxes of polystyrene, plastic spherical balls or other light infill elements, by preference made of natural fibers, can be used to reduce the self weight of concrete roofs and floors.

It is a pity that the described two interesting buildings were demolished. Structurally these buildings could be used well for many decennia. Barrel vaults are very efficient for large spans, the need of material is minimal and the construction of a single curved surface is not very

[1] Langejan A., Fusées Ceramiques, een nieuw bouwmateriaal, Bouw (1949) pp. 518-520;

[2] Toeter H.H., Paddestoelen voor de bouw van textielmagazijnen te Goor, Cement No. 11-12 (1955) pp.173-184.

[3] Eck P.J.W. van and Bish J.F., Het Fuseedak, Cement No. 6 (1954) pp. 240-243.

[4] Schodeck D.L., Structures, 4th _{edition, Prentice Hall, Columbus,}

USA (2001).

[5] Vriend J.J., Bouwen, Handboek voor de praktijk van het bouwen, Kosmos, Amsterdam (1955) pp.134-135;

[6] Timoshenko S. and Young D.H, Technische mechanica, Het Spectrum, Utecht Antwerpen, translated from: Engineering Mechanics, Mac GrawHill, New York (1956) pp.127-133. [7] Timoshenko S., Strength of materials, part 2, Advanced Theory

and Problems, 2th edition, D Van Nostrant Company: New York (1952).

[8] Goldenblat L. and Sisow A.M, Die berechnung von Baucon-structionen auf Stabilitat und Schwingungen, VEB verlag Technik, Berlin (1955).

[9] Bish J.F.,Gewapend beton voorschriften 1950, 2e _{druk, L.J.Veen}

uitgevers maatschappij (1950).

[10] Boom G.H. van and Kamerling J.W., Construeren in gewapend beton, deel 1, Delta Press b.v. Oudewater (1977).

[11] NEN EN 1992-1-1 (nl) Eurocode 2, Design of concrete structures-Part1-1: General rules and rules for buildings (2009).

[12] NEN EN 1991-1-4 Eurocode 1, Actions on structures-Part 1-4: General actions, wind actions (2005).