reasons for torsion
TRANSCRIPT
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Reasons for torsion
Large openings
Skew ends.
Three-line support
Alternating position of columns at both slab ends. Large openings
Shifted support arrangement
Skew supportsLongitudinal support
Cantilevering slab
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Trimmer beam at large opening
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Response to torsion
Rectangular solid section
Rectangular hollow section
Torsion results in shear stresses in outer part of solid section or in actual thin walls of hollow section
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Shear flow – torsional shear stress
tq ⋅= τ
tt W
T
tA
T =⋅
=2
τ
×
tA
1q
2q
3q
4q
1z2z
3z
4z
tb
th
[ ] tttt
tt
tt
tt
t Aqhbqh
bqb
hqh
bqb
hqT 222222 4321 ⋅=⋅==⋅⋅+⋅⋅+⋅⋅+⋅⋅=
Shear flow q (force per unit length) assumed to be constant
Torsional shear stress
tAW tt ⋅= 2
Torsion modulus of section
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Torsional shear stress
• Rectangular solid section• The shear stress is largest at the outside of the section at the
midpoint of the wider side
• Thin walled section• The shear flow is constant and the shear stress is largest in
the thinner wall
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Interaction vertical shear and torsion
Q
e
Eccentric load results in both vertical shear and torsion
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Cracking due to shear and torsion
LivskjuvbrottWeb shear tension crack VridbrottTorsional crack
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Reinforced concrete beam
• Uncracked stage• Torsional stiffness in
uncracked stage
• Torsional cracking
• Cracked stage• Torsional stiffness in
cracked stage
• Torsional resistance
• After inclined cracking the torsional shear is carried by transverse components of inclined compression.
• The inclined compressive struts need to be balanced by transverse and longitudinal steel reinforcement.
• Torsional failure due to crushing of struts or yielding of all steel
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Hollow core slab
• Uncracked stage• Torsional stiffness in
uncracked stage
• Torsional cracking = torsional resistance
• Since transverse and some longitudinal reinforcement is missing, it is not possible to achieve a state of equilibrium in the cracked stage
• Skew cracking in top flange or web means that the torsional resistance is reached
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Torsional deformation and stiffness
• Twist per unit length
• Torsional rigidity (uncracked)
C
T
dx
d =ϕ
(shear modulus)
TGKC =
)1(2 ν+= E
G
TK (cross-sectional factor)
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Cross-sectional factor
• Solid rectangular section b > h
• Thin walled rectangular section
−=b
hbhK 63,01
12
3
T
∑=
i
i
tT
t
uA
K24
b
h
At
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Hollow core section
• Transformation to equivalent thin walled rectangular section
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Cross-sectional factorAt
∑=
i
i
tT
t
uA
K24
( ) ( ))(5,0 ,,, bottomttoptoutwidt hhhbbA +−⋅−=
( )outw
bottomttopt
bottomt
outwid
topt
outwid
i
i
b
hhh
h
bb
h
bb
t
u
,
,,
,
,
,
, )(5,02 +−+
−+
−=∑
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Cracking due to torsion
• Load case:• Normal stress due to prestress and bending moment
(eccentricity of prestress and load)• Shear stress due to vertical shear (load) and torsion
• Different conditions in webs and flanges
• Crack occurs when the principal stress reaches the concrete tensile stress
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Principal tensile stress in flange
Tctct τσσσ +
+=2
I 22
)()()()(
c ccc
t xI
xMexP
A
xP −⋅+⋅−+−=σ
Normal stress due to prestress and bending moment:
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Torsional crack in top flange
Tctct τσσσ +
+=2
I 22
ctdf
• Calculate normal stress in top flange σct
• Assume that the principal stress equals the concrete tensile stress
• Solve the torsional shear stress that creates a skew crack in the top flange
• Torsional moment that results in the crack
ctd
ctctdtopT f
fστ −= 1,
ctd
topctctdtopTtopTtopTtopcr f
fWWT ,,,,, 1
στ −⋅=⋅=
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Torsional modulus of section Wt
tt W
T
tA
T =⋅
=2
τ
For section with constant wall thickness
tAW tt ⋅= 2
For hollow core section – different wall thicknesses
( ) ( ))(5,0 ,,, bottomttoptoutwidt hhhbbA +−⋅−=
)()( thtbhbA ttt −⋅−=⋅=
( )topTtopT tWW =,
( )outwTwebT bWW ,, =
( )bottomTbottomT tWW =,
Different values depending on actual wall thickness
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Principal tensile stress in web
( )VTcc ττσσσ ++
+=2
webI, 22
⋅+
⋅−= )(
1xV
I
S
dx
dP
I
eS
A
A
b c
cp
c
cp
c
cp
wVτ
wc
cpV bI
xVS
⋅⋅
=)(
τ (outside transfer length)
(inside transfer length)
Shear stress due to vertical shear:
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Shear and torsion interaction
• Traditional design approach• Stresses from vertical shear
and torsion are superimposed• The maximum principal stress
creates a crack which causes failure
• One point in web considered• Linear interaction assumed
• Holcotors• Non-linear analysis• Favourable stress
redistribution in cracking concrete and influence of restraint from boundaries
Increase with up to 55 % for 200 mm units and up to 30% for 400 mm units
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Shear and torsion in hollow core slabsHolcotors
• European Commission
• International Prestressed Hollow Core Association
• Bundesverband Spannbeton-Hohlplatten
• Castelo
• Consolis
• Echo
• A. Van Acker
Financiers and collaboration partners
• Strängbetong• VTT• Chalmers
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Holcotors
Karin LundgrenAss. Professor
Chalmers
Helén BrooPh.D. StudentChalmers
Björn EngströmProfessorChalmers
Matti PajariD.Sc. (tech.)VTT
Project started January 1, 2002 and ended December 31, 2004
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Aim of project
• To use the capacity of hollow core slabs better
• To develop methods to design for combined shear and torsion in hollow core slabs
• Single units• Whole floors
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Holcotors, 2002 – 2004
Q
e
VTT, Finland
Chalmers University, Sweden
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Tests on hollow core units
Q
Q
l = 7.0 m
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FE-model of hollow core unit
QBeam element
Strands
Concrete in compression
εc
σc
Concrete in tension
σc
εc
Steel in tensionσs
εs
Bond stress-slip relationτ
δ
26
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Comparison of results
• Maximum load
• Load versus deflection
• Failure mode
• Crack pattern
Maximum load in test [kN]
Maximum load in analysis [kN]
10 %
10 %
0
50
100
150
200
250
300
350
0 50 100 150 200 250 300 350
ST400
ST200
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Comparison of results
0
50
100
150
0.0 1.0 2.0 3.0 4.0 5.0Η [mm]
Q [kN]
FEA
Test b
Test
Q/2 Q/2
0.0
1.0
2.0
3.0
4.0
5.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2
Coordinates [m]
δ [mm]
FEA Q = 86.3 kN
Test Q = 100.0 kN
Test b Q = 98.1 kN
Tests and FEA Q = 50.0 kN
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Comparison of results
0
50
100
150
200
0.0 1.0 2.0 3.0 4.0 5.0€ [mm]
Q [kN]
Q/2 Q/2
Test E1M
FEA E1M Test E1
FEA E1
δ
0
50
100
150
200
250
300
0.0 2.0 4.0 6.0 8.0◊ [mm]
Q [kN]
Test E1M
Test E1
FEA E1
FEA E1M
Q
δ [mm]
Thin plastic sheet
Mortar bed
Neoprene
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Effect of neoprene bearing
1091 870
1037 1093
Strand
Concrete
Neoprene
ST400E1 Web 5
0
50
100
150
200
250
0.00 0.10 0.20 0.30
Microstrain [-]
Load [kN]
El 870
El 1037
El 1091
El 1093
ST400E1-Web 4
0
50
100
150
200
250
0.0 2.0 4.0 6.0Displacement [mm]
Load [kN]
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Critical section for shear tension crack in 400 mm units
Cracking starts here 45°
45°
Critical point
h/2
h/2
z
Analytical model FE-analysis, pure shear
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Conclusions
• FE-analyses are able to capture the overall behaviour in tests
• Failure mode• Maximum load• Crack pattern• Vertical deflection (until first crack)
• Large difference in capacity due to support condition
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FE-analyses for V-T interaction
x
0 1 2 3 4
A’B’ C’
B
A
C
'
'
'
Cy
Cy
By
By
Ay
Ay
δδ
δδ
δδ
=
=
=
x
y z
V T
0
50
100
150
0 100 200 300 400
T [kNm]
V [kN]
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Shear torsion interaction 400 mm
0
50
100
150
200
250
0 100 200 300 400 500 600
neoprene
test
Nominal
FE, nominal
FE anchorage
Anchorage
T [kNm]
V [kN]
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Shear torsion interaction 400 mm
0
50
100
150
0 100 200 300 400
T [kNm]
V [kN]
EN 1168
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Shear tension interaction 200 mm
0
10
20
30
40
50
60
0 50 100 150 200
T [kNm]
V [kN]
EN 1168
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Failure modes
• Torsion dominates• Diagonal crack in top flange
• Vertical shear dominates• Web shear crack in most loaded web and bending crack
• Intermediate situations• Mixed mode
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FE-model of hollow core floor
Tie beam modelled by restraints
Joint modelled by interface element
Hollow core unit modelled by beam element
Reduced torsion
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Integrated model for complete floor
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Design of hollow core slabs
Structural analyses Resistance analyses
FE-model for hollow core unit
V-T capacity
Analytical method in EN1168
V-T capacity
FE-model for floor M, V and T
Distribution diagram (α-factors) in EN 1168
M, V and T
FE-model for floor integrated with FE-model for hollow core unit Load capacity
Tra
ditio
nal
desi
gnIm
prov
ed d
esig
n
IV
III
II
I
<
<
<
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Test on complete floor
Tie beam
Bars
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Design of floor – level III and II
EN 1168
FEA
0
50
100
150
200T [kNm]
0 100 200 300 400V [kN]
1000
6000Q = 357 kN
Q = 449 kN
Q = 521 kN
-200
-100
0
100
200
300
0 2 4 6
V [kN] T [kNm]
z [m]
V
T
Q/2 Q/2
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Simulation of floor test – level I
0
100
200
300
400
500
600
0 1 2 3 4
δ [mm]
Q [kN]
Test
X
Y
Z
FE: Qmax= 480 kN
Test: Qmax= 521 kN
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Floor design, example
Design level
0
100
200
300
400
500
600
III II I
Qmax [kN]
Experiment 521 kN
EN 1168T
V
FEA
T
V X
Y
Z
357 kN
449 kN 480 kN
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Conclusions
• Modelling levels for hollow core slabs were developed• Hollow core floor ⇒ Sectional forces M, V, T
• Reduced torsional moment• Arbitrary geometry and loading
• Hollow core unit ⇒ Shear-torsion capacity• Higher resistance
• The capacity of hollow core units can be used better
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Development of level IV method
Capacity according to EN 1168
Pure shear failure in web (web shear tension)
Mixed mode shear failure
Failure in outer web
Failure in top flange
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Simplified interaction diagram
1) Select section2) Resistance to web shear
tension failure in outer web (pure shear failure) VRdc,st
3) Torsional capacity due to cracking in outer webThe vertical shear is carried by the internal web only (full redistribution)
4) Resistance to web shear tension failure in internal webTo be combined with 3)
5) Check torsional capacity due to cracking in top flange
( ))(
)(
0,
010 zbI
zSVz
innerw
cpRdV ⋅
⋅=τ
( )0,, zWT VwebTwebRd τ⋅=
TtopTtopRd WT τ⋅= ,,
topRdT ,
topRdT ,
topRdT ,
stRdcV ,1RdV
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Transverse distribution of load effects
1 2 3 4 5
• Transfer of vertical
shear
• Lateral restraint
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Forces on the loaded element
The element is subject to a concentrated force and distributed shear along the edges
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Forces on the adjacent element
This element is subject to a distributed shear along one edge downwards, and upwards shear at the other edge – means torsion
50
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Load distribution factors
Note! It is not the load that is distributed, but the load effect.Different factors for bending moment, shear and torsion.
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Distribution of shear, bending moment and torsion
-50
-30
-10
10
30
50
0 2 4 6
Shear force [kN]
Coordinate [m]
1 23
(a) -60
-50
-40
-30
-20
-10
0
0 2 4 6
Bending moment [kNm]
Coordinate [m]1
2
3
(b)
-15
-10
-5
0
5
10
15
0 2 4 6
Torsional moment [kNm]
Coordinate [m]
3
2
1
(c)
1 2 3 4 5
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Distribution of maximum moment and maximum shear
0
10
20
30
40
50
0 5 10 15
Distribution of bending moment at mid span [%]
Span [m](a)
shell
beam
FIP
α3
α2, α4
α1, α5
0
10
20
30
40
50
0 5 10 15
Distribution of shear at the supports [%]
Span [m] (b)
α3, shell
α2, α4, shell
α1, α5, shell
α3, beam
α2, α4, beam
α1, α5, beam