Transversal load distribution - usercontent.one
Transcript of Transversal load distribution - usercontent.one
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Transversal load distribution and
Large openings
Arnold Van Acker
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Transverse load distribution
1 2 3 4 5
Principle
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Historical reviewTest KG Bernander Strängbetong (1970 ?)
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Background guidelines EN 1168
Research programme FEBE BelgiumLoad distribution tests University LLN
o Concentrated load at the floor centre o Concentrated load at the floor edge
Analytical analysis Somers - AVA
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Load distribution tests (1977)Test programme
a) Load test single floor unitFailure at 86,25 kN, deflection 370 mm
b) Load at the floor centreDeflection under static load up to 64 kNDynamic tests : 200.000 load cycles between 32 and 48 kNLoading up to failure
c) Load at the floor edgeDeflection under static load up to 64 kN
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Load distribution testsTest set-up University LLN Belgium
Test floor composed of 6 HC slabs of 200 mm thickness and 6,00 m span
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Load distribution tests
Example of similar test set-up for ribbed floors
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Static test
Step by step increase of load up to 64 kN
0.94
1.12
0.93
1.33 1.36
1.13
Load at the floor center
Deflection in cm
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Loading till failure
Load (kN)
Deflectionmm
3.5092.00128.00156.00191.75225.60255.00280.00300.00310.00337.50
023456789
10failure
Load at the floor center
Load steps and deflection at the floor centre in mm
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Static test up to 48 kNLoad at the floor edge
0.17
0.46
0.26
0.78
1.24
1.93
3.6% 5.5% 9.7% 16.5% 26.1% 38.6%
Deflection in cm
Load distribution in % of total concentrated load
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Calculation programme (1982)Calculation model based on Fourrier analysis
General assumptionsThe distribution is calculated on the basis of the theory of isotropic slabs assuming the compatibility condition of the longitudinal and transversal deflection of the floor elements at their joints. It is also assumed that the longitudinal joints behave in the manner of longitudinal hinges, i.e. they cannot transmit bending moments, but only shear forces.A mathematical calculation based on Fourrier analysis has been carried out for the general case of a floor composed of an unlimited number of floor elements of various widths, one of them carrying a concentrated loading.The theory has been applied on a system of five floor slabs, connected by hinged joints. The loaded element is laterally supported by the adjacent slabs through the joints. Between the two hinges, tensile stresses will occur at the bottom of the loaded slab unit. They are maximum at the place of the linear load, and equal to zero at the joints. At the ULS, the slab is assumed to be cracked right through at the location of the linear load, or at the nearest core. It is further assumed that the crack cannot transmit any bending moment, but only shear forces, just as for the longitudinal joints between the elements.
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Practical calculationsSoftware programme (ERGON)More than 100 calculations on HP pocket calculator
HC 150, 200, 265 and 320Various span lengths20 minutes for each calculation
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Comparison with test results
Floor of 6 units
Test results for P = 1,5 x service load
38,17 26,65 16,84 9,59 5,33 3,41
Calculation results without crack in the loaded unit
36,56 26,10 15,76 9,82 5,68 5,16
Concentrated loading at the edge floor unit
Floor of 5 units
Calculation results without crack in the loaded unit
37,04 26,58 16,56 11,10 8,71
Calculation results with crack in the loaded unit
39,17 25,69 16,00 10,73 8,41
Load distribution in % of linear concentrated load
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Comparison with test results
Floor of 6 units
Test results for P = 2,0 x service load
26,26 24,63 19,73 13,50 9,35 6,53
Calculation results without crack in the loaded unit
26,05 25,75 20,69 12,53 8,40 6,58
Concentrated line load at the second floor unit
Floor of 5 units
Calculation results without crack in the loaded unit
26,58 27,02 21,13 14,16 11,10
Calculation results with crack in the loaded unit
25,89 29,32 20,39 13,67 10,72
Load distribution in % of linear concentrated load
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Comparison with test results
Floor of 6 units
Test results for P = 2,0 x service load
17,34 19,69 20,78 17,97 13,59 10,62
Calculation results without crack in the loaded unit
15,76 20,11 23,02 18,76 12,53 9,82
Concentrated loading at the third or central floor unit
Floor of 5 units
Calculation results without crack in the loaded unit
12,56 21,13 24,62 21,13 12,56
Calculation results with crack in the loaded unit
16,04 20,46 27,00 20,46 16,04
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Comparison with test results
Floor of 6 units
Test results for P = 0,9 x service load
13,01 15,45 20,33 19,92 17,89 13,41
Test results for P = 2,0 x service load
13,98 16,24 19,45 20,58 16,56 13,18
Test results for P = 2,6 x service load
13,64 16,08 19,98 20,34 16,57 13,40
Calculation results without crack in the units
11,97 15,26 22,70 22,70 15,26 11,97
Concentrated loading across the central joint
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Load distribution curves• Curves published in FIP Recommendations 1988
Load distribution factors for linear loads (without topping)
An additional safety margin of 25% has been applied on the directly loaded unit
Linear loading
At the centre
At the edge
Load
dis
tribu
tion
fact
ors
in %
Floor span
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Tests on 400 mm slabs, VTT Finland (1991) Test programme
a) Two tests on 400 mm HC floors without toppingFour HC slabs of 6.00 m spanSix HC slabs of 12.00 m span5 loading cycles with service load located at L/6Loading at floor centre up to failure
b) Load at the floor edge5 loading cycles with service load located at L/6Loading at floor centre up to failure
c) Measurements: deflection + support reactiond) Comparison with FIP curves
(including additional safety of 25% on the most loaded unit)
Study sponsored by IPHA, Partek Concrete, Finnmap and Lohja BetonilaPresented at IECA conference 1991 in Paris
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Tests at VTTTest set-up floor 12 m span
Load cell
400 mm slabs
Peripheral tie beam
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Load distribution factorsComparison with FIP curves
Legend: continuous line: FIP+
VTT
Bending moment distribution factors
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Large floor openingsTests carried out at CBR laboratory, Belgium (1976)
Three test were carried out on floor slabs of 3,60 m x 8,00 m surface with a large opening in the supporting zone.
Test N° 1: Opening 1,80 m x 1,70 m, trimmer beam in reinforced concrete, anchored in the two adjoining units
Test N° 2: Opening 1,80 m x 1,70 m, trimmer beam in reinforced concrete, anchored in the two adjoining units by means of tensile bars ϕ 20
Test N° 3: Opening 1,20 m x 1,20 m, steel trimmer beam, welded to steel plates anchored in the two adjacent slab units
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Test N° 1
Opening 1,80 m x 1,70 m, trimmer beam in reinforced concreter anchored in the two adjacent units
HC 200, 11 strands 3/8”, length 8,00 m
Trimmer beam 200 x 100 mm²
2 reinforcing bars ϕ 12 mm
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Test N° 2
Opening 1,80 m x 1,70 m, trimmer beam in reinforced concreter anchored in the two adjacent units by means of a tensile bar ϕ 20
Reinforcing tie bar ϕ 20 mm
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Test N° 3
Opening 1,20 m x 1,20 m, steel trimmer beam, welded to supports anchored in the two adjacent slab units
Steel trimmer beam fixed by welding to steel plates anchored in filled cores
Trimmer beam composed of vertical plate 180 x 8 x 1206 mm, welded to a horizontal folded plate 100 x 6 mm
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Test set-up
Test N° 1 with opening 1,80 m x 1,70 m and trimmer beam in reinforced concreter anchored in the two adjacent units
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Load (kN)Observations
Lateral units
Central unit
015.515.515.515.515.5
15.5
015.537.044.0
60.00± 80.00
88.00
StartMaximum service loadTheoretical cracking load for the whole floorFirst crack at the corner of slab n° 5Transversal cracks from opening corner to slab edgesLongitudinal cracks in the outer slabs starting from the anchor zones of the trimmer beam in slab n° 6Failure in the anchorage of the trimmer beam in slab N° 6.
Test N° 1Loading steps
P5 P4 P6
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Test N° 1Failure pattern
Test N° 1 with opening 1,80 m x 1,70 m and trimmer beam in reinforced concreter anchored in the two adjacent units
- Failure for 88 kN jack force on central slab. - Maximum allowable imposed floor loading: 5.60 kN/m²- Corresponding load on trimmer beam: 5.60 kN/m² +
3.25 kN/m² self weight = ½ (8.85 kN/m² x 1,80 x 5,33) = 42.5 kN
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Test N° 2Failure pattern
Opening 1,80 m x 1,70 m, trimmer beam in reinforced concreter anchored in the two adjacent units by means of a tensile bar ϕ 20
Bending failure load whole floor = 96 kN on central slab.
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Load (kN)Observations
Lateral units
Central unit
041.0043.0043.0043.0043.0043.0043.0043.0043.00
43.00
044.0037.0050.0053.0060.0080.00110.00120.00127.00
137.00
StartSmall cracks at the corners of the opening e = 4 to 6/100 mmService load edge slabsCrack opening 10/100 mmCracking load edge slabs Several transversal cracks starting from trimmer beamNumerous transversal cracks; no deformation of trimmer beamDeformation of trimmer beamCrushing of exterior void under support of trimmer beamThe anchorages of the trimmer beam in the supporting slabs are gradually being pulled out. The trimmer beam is heavily deformed and carries hardly the middle slab unit. The latter ones still caries the jack load as a cantilevering slab.The central slabs gets broken under negative moment. The anchorages of the trimmer beam are completely extracted, but the central unit is not really collapsing. The failure mode is very slow.
Test N° 6Loading steps
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Test N° 3
Opening 1,20 m x 1,20 m, steel trimmer beam, welded to supports anchored in the two adjacent slab units
Bending failure load whole floor = 13,7 t on central slab. Maximum allowable imposed floor loading: 560 kg/m²Corresponding load on trimmer beam: 560 kg/m² + 325 kg/m² self weight is: ½ (885 kg/m² x 1,20 x 5,33) = 2,83 t
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ConclusionsIn all tests, the joints have transmitted a considerable part of the load. This was demonstrate by the equal deformation of each slab unit near the load.The initial cracks in tests 1 and 2 started from the corner of the opening in the edge slabs, and not at the support of the trimmer beam.At failure, the joints were broken over a length going from 0,8m to 1,5 mThe bearing capacity of the trimmer beam construction is more than the double of the acting load for tests 1 and 2, and about 4,8 times higher in test 3.
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Parametric study on HC floors with large openings Example of design charts
32/20
DIC
ATe
A –
Uni
vers
ity o
f Par
ma
& A
SSA
P R
esea
rch
Prog
ram
“Par
amet
ric n
umer
ical
stu
dy o
n H
C fl
oors
with
larg
e op
enin
gs fo
r the
de
velo
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”
Ap,sup = 279 mm2
Ap,inf = 1302 mm2
Mrd,H32 = 424.79 kNm
H30-H32 untopped floors, H/L = 1/35, imposed service load 10 kN/m2
With
outc
oncr
ete
fillin
gsW
ithco
ncre
te fi
lling
sH32
H30
Ap,sup = 84.84 mm2
Ap,inf = 974 mm2
Mrd,H30 = 382.61 kNm
ftA ftB ftC ftD
ftFftE
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0.00 0.50 1.00 1.50 2.00
M /
M* ,
p/p
*
b / B
H30,endH32, endH30, midspanH32, midspan
ftA-f
mF
ftB-f
mE
ftC
ftD
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0.00 0.50 1.00 1.50 2.00
M /
M* ,
p/p
*
b / B
H30, endH32, endH30, midspanH32, midspan
ftA-f
mF
ftB-f
mE
ftC
ftD
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Parametric study on HC floors with large openings Example of design charts
33/20
DIC
ATeA
–U
nive
rsity
of P
arm
a &
ASS
AP
Res
earc
h Pr
ogra
m“P
aram
etric
num
eric
al s
tudy
on
HC
floo
rs w
ith la
rge
open
ings
for t
he
deve
lopm
ent o
f des
ign
char
ts”
H30-H32 topped floors, H/L = 1/35, imposed service load 10 kN/m2
40m
m c
oncr
ete
topp
ing
80m
m c
oncr
ete
topp
ing
H32
H30
Ap,sup = 84.84 mm2
Ap,inf = 974 mm2
Mrd,H30+4 = 404.67 kNmMrd,H30+8 = 471.38 kNm
ftA ftB ftC ftD
ftFftE
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0.00 0.50 1.00 1.50 2.00
M /
M* ,
p/p
*
b / B
H30, endH32, endH30, midspanH32, midspan
ftA-f
mF
ftB-f
mE
ftC
ftD
0.40
0.50
0.60
0.70
0.80
0.90
1.00
0.00 0.50 1.00 1.50 2.00
M /
M* ,
p/p
*
b / B
H30, endH32, endH30, midspanH32, midspan
ftA-f
mF
ftB-f
mE
ftC
ftD
Ap,sup = 279 mm2
Ap,inf = 1302 mm2
Mrd,H32+4 = 507.88 kNmMrd,H32+8 = 575.63 kNm
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Type solutions
Detailing
Cast in-situ trimmer beamReinforcing bars
Steel trimmer beam Concrete trimmer beam
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DetailingParma Betonila Finland
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DetailingSträngbetong Sweden
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DetailingLarge opening with steel frame
Opening of 3.60 m x 3.60 m in HC floor of 10.00 m span
HC 240 x 600 + 50 mm topping