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Reasons for torsion

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Page 1: Reasons for torsion
Page 2: Reasons for torsion

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

Page 3: Reasons for torsion

Trimmer beam at large opening

Page 4: Reasons for torsion

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

Page 5: Reasons for torsion

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

Page 6: Reasons for torsion

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

Page 7: Reasons for torsion

Interaction vertical shear and torsion

Q

e

Eccentric load results in both vertical shear and torsion

Page 8: Reasons for torsion

Cracking due to shear and torsion

LivskjuvbrottWeb shear tension crack VridbrottTorsional crack

Page 9: Reasons for torsion

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

Page 10: Reasons for torsion

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

Page 11: Reasons for torsion

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)

Page 12: Reasons for torsion

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

Page 13: Reasons for torsion

Hollow core section

• Transformation to equivalent thin walled rectangular section

Page 14: Reasons for torsion

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 +−+

−+

−=∑

Page 15: Reasons for torsion

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

Page 16: Reasons for torsion

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:

Page 17: Reasons for torsion

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 −⋅=⋅=

Page 18: Reasons for torsion

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

Page 19: Reasons for torsion

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:

Page 20: Reasons for torsion

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

Page 21: Reasons for torsion

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

Page 22: Reasons for torsion

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

Page 23: Reasons for torsion

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

Page 24: Reasons for torsion

Holcotors, 2002 – 2004

Q

e

VTT, Finland

Chalmers University, Sweden

Page 25: Reasons for torsion

Tests on hollow core units

Q

Q

l = 7.0 m

Page 26: Reasons for torsion

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

Page 27: Reasons for torsion

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

Page 28: Reasons for torsion

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

Page 29: Reasons for torsion

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

Page 30: Reasons for torsion

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]

Page 31: Reasons for torsion

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

Page 32: Reasons for torsion

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

Page 33: Reasons for torsion

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]

Page 34: Reasons for torsion

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]

Page 35: Reasons for torsion

Shear torsion interaction 400 mm

0

50

100

150

0 100 200 300 400

T [kNm]

V [kN]

EN 1168

Page 36: Reasons for torsion

Shear tension interaction 200 mm

0

10

20

30

40

50

60

0 50 100 150 200

T [kNm]

V [kN]

EN 1168

Page 37: Reasons for torsion

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

Page 38: Reasons for torsion

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

Page 39: Reasons for torsion

Integrated model for complete floor

Page 40: Reasons for torsion

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

<

<

<

Page 41: Reasons for torsion

Test on complete floor

Tie beam

Bars

Page 42: Reasons for torsion

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

Page 43: Reasons for torsion

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

Page 44: Reasons for torsion

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

Page 45: Reasons for torsion

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

Page 46: Reasons for torsion

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

Page 47: Reasons for torsion

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

Page 48: Reasons for torsion

Transverse distribution of load effects

1 2 3 4 5

• Transfer of vertical

shear

• Lateral restraint

Page 49: Reasons for torsion

Forces on the loaded element

The element is subject to a concentrated force and distributed shear along the edges

Page 50: Reasons for torsion

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

Page 51: Reasons for torsion

Load distribution factors

Note! It is not the load that is distributed, but the load effect.Different factors for bending moment, shear and torsion.

Page 52: Reasons for torsion

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

Page 53: Reasons for torsion

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