Increasing the Load Bearing Capacity of Barrel Vaults
Wim KAMERLING
Structural Engineer
Lecturer Technical University
Delft, Faculty Architecture,
Delft, The Netherlands
[email protected]
Wim Kamerling, born 1950, received
his structural engineering degree from
the Technical University of
Eindhoven. He worked for D3BN
Consultants, before becoming lecturer
at the Technical University of Delft.
His main area of research is related to
shells, blobs and floating buildings.
Summary
Just after World War II many barrel vaults and domes were built with a structural system, known as
Fusée Ceramique. This paper analyses the load bearing capacity of these vaults. Schemes, theories,
idealizations and assumptions are analysed, validated and discussed. Methods to increase the load
bearing capacity of barrel vaults are described to preserve the remaining Fusée Ceramique barrel
vaults for the next generations.
Keywords: Barrel vaults; Fusée Ceramique; load bearing capacity; strengthening; stiffening.
Notation
A
cross section;
E
Young's modulus;
F
Force;
H
thrust;
M bending moment;
N
normal force;
V
vertical reaction;
a
half of the span of the barrel vault;
e
eccentricity;
f
rise of the vault;
q
equally distributed load;
qe
live load;
qg
permanent load;
s
length of the curve.
σ
stress;
1.
Fig.1: Barrel vault, workshop former Military
complex, Woerden, The Netherlands.
Introduction
Just after World War II many barrel vaults and domes were built with a structural system, known as
Fusée Ceramique [4]. Thanks to the ceramic infill elements the self-weight of these vaults and
domes was quite small [3]. Technically these structures are very robust. Unfortunately till now
these buildings are architectural not valued highly and most of these buildings are pulled down.
Nevertheless the interest for industrial buildings is rising and creative architects transform old
industrial buildings into interesting and desirable objects. Buildings roofed with a fusée barrel vault
have potentials. Mostly the vaults span from façade to façade. The interiors are not divided by
columns or walls, so these buildings are quite useful for workshops, shopping malls, gymnasia,
cinema's and so on. Of course these buildings must be renovated to meet the standards of the
present. Certainly the insulation of the roof and facades must be increased. Probably the load
capacity of the roofs has to be increased too, especially in case the roofs have to support solar
panels or a vegetation to retain rainwater. Increasing the load bearing capacity of an existing
building can be expensive, but barrel vaults are quite strong and stiff. The form offers potentials to
strengthen these vaults without to much trouble. At this moment only a few fusée ceramique vaults
are remaining. It will be easier to preserve these buildings for the next generations in case the load
bearing capacity of these vaults can be increased cost effective. To strengthen the fusée ceramique
barrel vaults, constructed half a century before, the load bearing capacity of is analysed. Schemes,
theories, idealizations and assumptions, practiced during the fifties of the XX century, are validated
and discussed. Finally methods to strengthen these barrel vaults are described.
2. Design of a barrel vault
Half a century ago the design of a vault of concrete wasn't easy, the methods to analyse the load
transfer were quite restricted and based on mechanics as described by Timoshenko et al [5].
Engineers could not use computers to calculate the load transfer. Numbers were multiplied with a
sliding rule, so the calculations were not very accurate. The analysis is based on an article written
by Van Eck et al in Cement [2]. Van Eck describes the design for a fusée ceramique barrel vault
with a parabolic curvature subjected to an equally distributed dead load and an equally distributed
live load. Actually the dead load acting on the vault is a surface load, so this load is increasing from
the centre to the supports. Due to this load the parabolic vault is subjected to bending moments. The
effect of these bending moments is analysed. Firstly the forces and bending moments are calculated
for the equally distributed load, next the forces and moments are calculated for the surface load.
Finally methods are given to strengthen and stiffen the vault to resist these forces and bending
moments. In the mentioned article [2] the forces, stresses and dimensions were expressed in kg and
cm; for this paper these units are transferred into the units of the present.
2.1
Equally distributed load
The load transfer section of a barrel vault was analysed for a symmetrical equally distributed load
qg and a live load qe acting symmetric or asymmetric on the vault. The curvature of the vault was a
parabola with a rise f and a span ℓ = 2 * a. The structure is calculated for a width equal to 1,0 m.
For a symmetrical equally distributed load q, the thrust H and the vertical reaction V acting at the
support were calculated with respectively expression (1) and (2). The maximum normal force acts
on the arch just above the supports. This force is calculated with expression (3) by adding the
vectors according to the thrust H and the vertical reaction V.
q
s
α
H = q * a2
2f
(1)
.
V=q*a
(2)
.
N = (V2 + H2)0,5
(3)
.
f
a
φ R
Fig. 2: Section of a vault subjected to an
equally distributed load q.
Table 1: Span, rise and load
half span
rise
dead load
depth
live load
a =
f =
qg =
t =
qe =
12,0
3,0
3,3
0,2
1,0
m
m
kN/m
m
kN/m
2.2
Asymmetric load
The vault can be subjected to a live load qe acting at one half of the span. For this load, the thrust H
and the vertical reactions Va and Vb acting at the supports were calculated with respectively
expression (4), (5) and (6). Due to the asymmetric load the vault is subjected to bending. The
bending moment is maximal in case the vault is subjected to a live load acting on the left or right
half of the vault, see figure 3. The eccentricity due to the bending moment and normal force is
calculated with expression (8). The span, rise, dead and live load are given into table 1.
qe
H = qe * a2
4 .f
(4)
.
Va = ¼ qe * a
(5)
.
Vb = ¾ qe * a
(6)
M = qe * a2
16
(7)
.
e = M/N
(8)
.
f
a
φ R
Fig. 3: Section of a vault subjected to an
equally distributed asymmetric live load.
Table 2 shows the forces and bending moments, exclusive second order, acting at the vault
according to the calculations described in the article [2] based on the Theory of Mechanics as
described by for example Timoshenko et al [5].
Table 2. Forces and bending moments acting on vault subjected to an equally distributed load
perm.
live load perm. + asym.
asym. live perm. + asym. live
load
live load
load
load
min. vertical reaction
max. vertical reaction
thrust:
normal force, x = ½ a
bending moment
eccentricity/depth
V
V
H
N
M
e/t
39,6 kN
12 kN
39,6 kN
12 kN
79,2 kN
24 kN
81,6 kN 24,7 kN
51,6 kN
51,6 kN
103,2 kN
106,3 kN
3 kN
9 kN
12 kN
12,4 kN
9 kNm
42,6 kN
48,6 kN
91,2 kN
94,0 kN
9 kNm
0,48
To avoid cracks in a structure of concrete the tensile stresses must be small, the normal force has to
act within the core. Consequently the eccentricity of the normal force must be small. For example, a
rectangular section is compressed in case the eccentricity e = M/N is smaller than 1/6 of the depth.
Due to the asymmetric live load acting at one side the ratio eccentricity/depth is 0,48. The structure
is subjected to tensile bending stresses. To resist bending moments concrete vaults were reinforced
according to the codes [1].
2.3
Surface load
Actually the permanent load is distributed over the surface, see figure 4. Due to the curvature of the
arch the permanent load is increasing from the top to the supports, as showed with expression (8):
qmax = qe* (1+ y'2)½ =
qe* (1+ 4 * f2/a2)½
(9)
For the swallow vault with a span of 2 * a = 24 m and a rise of f = 3 m the reaction force load
acting at the support is increased 11,8%. The
increase of the load due to the curvature affects
the bending moments too. Nowadays the bending
moments due to the curvature can be calculated
easily with a computer. Table 3 shows the results
calculated with a FEM program, based on the
Theory of Elasticity. Comparing table 2 and 3
shows the effect of the surface load due to the
curvature of the structure. Due to the surface load
the maximal bending moment is 0,47 kNm at the
centre and 1,0 kNm between the centre and the
Fig. 4: Surface load qg
support.
The bending moments due to the surface load are
much smaller than the bending moments due to
the asymmetric live load acting at one side.
The effect of the bending moments can be
expressed with the eccentricity. The eccentricity
of the normal force is calculated with expression
(8). For the vault the eccentricity is maximal in
case the live load acts asymmetrically at one side
of the vault. Comparing the bending moments
and eccentricities for this swallow vault shows
that the deficiencies caused by the increase of the
permanent load due to the curvature are quite
Fig.5: Bending moments in the vault subjected
small, but due to the asymmetric live load, acting
to a surface load of qg = 3,3 kN/m.
at one side, the vault is still subjected by
substantial bending moments, causing tensile
stresses, so the vault must be reinforced.
3.37
3.40
3.35
3.31
3.33
3.30
3.30
3.30
3.31
3.33
3.35
3.37
3.40
3.44
3.44
3.48
3.48
3.52
3.52
3.58
3.58
3.63
3.63
1.85
N1
0.00
N2
M1
N3
M2
N4
M3
N7
M6
N6
M5
N5
M4
N8
M7
N9
M8
N10
M9
N11
M10
N12
M11
M12
N13
M13
N14
M14
N15
M15
N16
M16
N17
M17
N18
M1
N198
M1
N209
M20
N21
M21
N22
M22
N23
1.85
M23
N24
M25
-0.47 -0.47
-0.42 -0.42
-0.42 -0.42
-0.35 -0.35
-0.35 -0.35
-0.11 -0.1
-0.15 M11 M12 M13 M14 M15
-0.15 M10
M16 1 M1
M9
7 M1
M8
8 M1
0.04
0.04
9
0.04
0.04
0.33
0.33
0.3 3
0.363
0.56
0.5
0.56
0.56
0.79
0.79
M24
N25
M7
M4
M5
M6
M3
M2
M1
84
8
0.5 84 0.1.01
8
0.
0.5
0.66
1.01
66
0.0.
79
M25
M20
M21
M22
66
0.0.
79
0.66
1.01
M2
3
M2
4
0.
0.
1.0184
0.8 58
0.5
4
8
Table 3. Forces and bending moments acting on vault subjected to the permanent surface load and the
equally distributed live load.
perm.
sym. live
perm. + sym. live asym. live perm. + asym.
surface
load
load
load
live load
load
min. vertical reaction
max. vertical reaction
thrust:
normal force, x = ½ a
max. bending moment:
eccentricity/depth:
V
V
H
N
M
e/t
41,2 kN
41,2 kN
81,0 kN
83,6 kN
1,0 kNm
0,06
12 kN
12 kN
24 kN
24,7 kN
53,2 kN
53,2 kN
105,0 kN
108,3 kN
3 kN
9 kN
12 kN
12,4 kN
9 kNm
44,2 kN
50,2 kN
93,0 kN
96,0 kN
10,0 kNm
0,52
2.4
Surface load approached with linear increasing load
Fifty years ago finite elements programs were not available, engineers had to simplify the
calculations. Actually the surface load could be approached easily with a linear increasing load, see
figure 6. The results calculated for a linear increasing load do not differ much from the results
calculated for a surface load. To show the potentials and the magnitude of the deficiencies this
approach is described. Thus the surface load is simplified into a load increasing linear from the
centre to a maximum at the ends. At the centre the load is equal to q and at the supports the load is
equal to qmax = q + c.q., c is the ratio showing the increase of the load, as showed in expression (10):
c = (1+ 4.f2/a2)½ - 1
(10)
q + c.q
f
a
Fig. 6: Linear increasing load.
The forces and bending moments, due to the
linear increasing load, see figure 6, are defined
by calculating the forces and moments for the
equally distributed load q, see figure 2, and
the triangular load c.q, see figure 7, separately.
Next the forces and bending moments for the
increasing load, according to figure 6, are
calculated by summarizing the results
calculated separately for the equally
distributed load q and the triangular load c.q.
Table 4 shows the forces and bending
moments due to the linear increasing
permanent load and the equally distributed live
load acting symmetrically or asymmetrically at
the vault.
2.5
Triangular load
For the triangular load c.q, increasing linear from the centre to the supports, the reactions and
bending moments can be calculated with the following expressions:
c.q
f
a
vertical reaction:
symmetrical load
V = ½ c.q * a
(11)
thrust:
H = 1/6 c.q * a2/f
(12)
maximal bending
moment:
M=
2
/81 c.q * a2
(13)
Fig.7: Triangular load increasing
from the top to the supports.
2.6
Linear increasing load
The bending moments and forces for the vault subjected to a linear increasing load are calculated by
adding the results for the equally distributed load and triangular load. Table 4 shows the forces and
bending moments for the vault subjected to an increasing permanent load increasing from q at the
centre to q = (1+c).q at the supports, with q = 3,3 kN/m and a ratio c equal to c = 0,118, see figure 6.
Further the vault is subjected to an equally distributed live load equal to qe = 1,0 kN/m.
Table 4 Forces and bending moments acting on vault subjected to a load increasing linearly from the top to
the supports from qg to qg max = q + c.q and an equally distributed live load qe.
perm. load
min. vertical reaction
max. vertical reaction
thrust:
normal force, x = ½ a
max. bending moment
eccentricity/depth
V
V
H
N
M
e
41,9 kN
41,9 kN
82,3 kN
84,8 kN
1,4 kNm
0,08
sym. live perm. + sym. asym. live
perm. + asym.
load
live load
load
live
12,0 kN
53,9 kN
3,0 N
44,9 kN
12,0 kN
53,9 kN
9,0 N
50,9 kN
24,0 kN
106,3 kN
12,0 kN
94,3 kN
24,7 kN
109,5 kN
12,4 kN
97,3 kN
1,4 kNm
9,0 kNm
10,4 kNm
0,06
0,53
The forces and bending moments according to the linear increasing load are slightly larger than the
forces and bending moments calculated with the computer. The deficiencies are small, the approach
of the surface load with a linear increasing load is on the safe side and the method is quite simple,
so fifty years ago the designers could apply this method too. Neglecting the effect of the curvature
underestimates the bending moments acting on the vault. Especially for vaults with a large rise this
effect can not be neglected.
3
Increasing the load-bearing capacity
The load-bearing capacity of an existing vault can be increased by:
• increasing the depth of the vault;
• adding reinforcement;
• adding bars;
The depth can be increased by casting a thin layer of mortar on top of the roof. Due to the added
layer the dead weight of the vault will be increased. The top layer can be reinforced with bars.
Unfortunately the reinforcement at the lower side is not increased so the bending capacity to resist
moments causing tensile stresses at the lower side is not increased.
Reinforcement can be added by gluing thin stripes of steel or carbon on the upper and lower side of
the surface of the vault.
Actually it is cost effective to combine the first and second method. For the upper side it is easier to
add the reinforcement into a thin layer of mortar poured at the top than gluing carbon stripes at the
top. For the lower side the bending capacity is increased by gluing strips to the bottom, so the
reinforcement at the top is embedded in a top layer and the reinforcement at the lower side is glued.
Strengthening the vault with steel bars running through the interior space from the centre to the
support of the vault is not very difficult or labour intensive. The bars can be anchored easily with
bolts, which are glued into holes drilled in the concrete vault. The external bars, running diagonally
from the top to the supports, reduce the bending moments considerately. The following calculations
show the increase of the bearing capacity of these bars for several loads.
3.1
Strengthened vault, equally distributed load
Adding bars to a parabolic vault is very effective. A symmetric or asymmetric equally distributed
load does not cause bending, so the vault is subjected to normal forces only. Especially for vaults of
concrete it is structurally very efficient in case the vault is compressed only and not subjected to
bending moments causing tensile stresses.
For an asymmetric equally distributed load the force
S acting on the diagonal is equal to:
qe
f
a
S = q * a2 (1+ f2/a2)1/2
4f
(18)
.
The thrust and the reactions Va and Vb are not
effected by the added bars and can be calculated
with respectively expression (4), (5) and (6).
Fig. 8: Vault strengthened with ties running
from the top to the support, subjected to
asymmetric load.
Table 5 shows the forces into the vault for the
structure strengthened with ties running from the top
to the supports, subjected to an equally distributed
permanent load, and a live load acting asymmetric or
symmetric on the vault.
The diagonals are subjected to a normal force. Due to the asymmetric live load acting at the right
side of the vault, the bar at the left side is compressed and the right side is tensioned. The normal
force is quite small, a steel tube Ø101,6-5, can resist the normal force, provided the tube is jointed,
halfway between the centre and supports, to the vault to reduce the buckling length.
Table 5: Forces and bending moments acting on vault strengthened with ties running from the top to the
supports, subjected to an equally distributed symmetric or asymmetric load.
max. vertical reaction
thrust:
normal force, x = ½ a
normal force in bar
sym. load
asym. load
perm. load live load perm. + live live load
perm. + live load
V
39,6 kN
12 kN
51,6 kN
9 kN
48,6 kN
H
79,2 kN
24 kN
103,2 kN
12 kN
91,2 kN
N
81,6 kN 24,7 kN
106,3 kN
12,4 kN
94,0 kN
S
12,4 kN
12,4 kN
3.2
Increasing symmetric load
For a symmetrical triangular load the diagonal ties
will reduce the bending moments substantial too.
For the triangular load the force S acting on the
diagonal and the bending moment is equal to:
q
f
a
S = 1/9 q * a2 * (1+ f2/a2)1/2
f
(19).
.
M = 0,013 q * a2
(20)
.
Fig. 9: Vault strengthened with bars running
from the top to the support, subjected to load
increasing from the top to the supports.
3.3
Strengthened vault, subjected to an increasing permanent load.
Next the forces and bending moments are
calculated for the vault subjected to a
permanent surface load of qg = 3,3 kN/m,
increasing linear from the top to the
q
supports , see figure 11, and an equally
distributed live load of qe = 1,0 kN/m
acting symmetric or asymmetric on a part of the
f vault.
Table 6 shows the bending moments and forces
acting on the vault strengthened with
bars running from the top to the supports.
a
Comparing table 6 and table 4 shows the
effect of the strengthening of the vault subjected
to a linear increasing load. Thanks to the
diagonals the moments are reduced considerable.
Fig. 11: Vault strengthened with bars running
To avoid cracks in a structure of concrete the
from the top to the support, subjected to an
tensile stresses must be very small. A section is
asymmetric triangular load.
not subjected to tensile stresses in case the
normal force acts within the core. Consequently
the eccentricity of the normal force must be
small.
For example, a rectangular section is compressed in case the eccentricity is smaller than 1/6 of the
depth. Due to the added diagonals the ratio of eccentricity and depth is 0,04 so the vault is not
subjected to a tensile bending stress.
Table 6. Forces and bending moments acting on vault strengthened with ties running from the top
to the supports, subjected to an increasing load.
sym. load
asym. load
perm. load live load perm. + live live load
perm. + live
Min. vertical reaction V
41,9 kN 12,0 kN
53,9 kN
3,0 N
44,9 kN
Max. vertical reaction V
41,9 kN 12,0 kN
53,9 kN
9,0 N
50,9 kN
Thrust:
H
82,3 kN 24,0 kN
106,3 kN
12,0 kN
94,3 kN
normal force, x = ½ a N
84,8 kN 24,7 kN
109,5 kN
12,4 kN
97,2 kN
max. bending moment M
0,7 kNm
0,7 kNm
0,7 kNm
eccentricity/depth
0,04
0,03
0,04
4
Conclusions
Barrel vaults are structurally and architecturally are very interesting and offer possibilities to design
structures with a minimal self weight. The vaults constructed with the fusée ceramique elements
were capable to resist the loads for decays in spite of the available methods of design, without the
help of computers. If a vault with a large rise was designed for equally distributed loads only then
the forces and bending moments are underestimated. These vaults can be reinforced simply and
cost effective by adding diagonals running from the top to the supports. These bars will reduce the
bending moments considerately, the eccentricity of the compressive normal force will be small, so
the vault is compressed mostly and not cracked. Consequently the stiffness is increased and the
need of maintenance will be reduced too. Especially strengthening the vaults with diagonals will be
helpfully in case the load bearing capacity has to be increased, so the roof can support the loads due
to for example solar panels or a vegetation. Possibly the described technique to strengthen and
stiffen vaults can contribute to preserve these structures for the next generations.
5
[1]
[2]
[3]
[4]
[5]
References
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1955, p.134-135.
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