rc2 lecture 6.0 - torsion

8/13/2019 RC2 Lecture 6.0 - Torsion

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einforced Concrete II Hashemite Unive

. Hazim Dwairi

To r si o n i n P l a i n Con c r et eT o r si o n i n P l a i n Con c r et e

M em b er s M em b er s

•• Torsion in rectangular membersTorsion in rectangular members“ ”“ ” – – . .

– – The stress at the corners is zero.The stress at the corners is zero.

– – Stress distribution at any other location is less than that at theStress distribution at any other location is less than that at the

middle andmiddle and

– – greater than zero.greater than zero.

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y

To r si o n i n P l a i n Con c r et eT o r si o n i n P l a i n Con c r et eM em b er s M em b er s

•• Torsion in rectangular membersTorsion in rectangular members

ab

T 2max

α τ =


a/b 1.0 1.5 2.0 3.0 5.0

0.208 0.219 0.246 0.267 0.290 1/3

∞




. Hazim Dwairi

Cracking StrengthCracking Strength


Thin Walled Tube Analogy Thin Walled Tube Analogy

•• The design for torsion is based on a thin walledThe design for torsion is based on a thin walled

, ., .

torsion is idealized as a thintorsion is idealized as a thin--walled tube with thewalled tube with the

core concrete cross section in a solid beam iscore concrete cross section in a solid beam is

neglected.neglected.





. Hazim Dwairi


VV11=V=V33, V, V22=V=V44

According to thin walled theory: According to thin walled theory:

q = Vq = V11/x/xoo=V=V22//yyoo=V=V33/x/xoo=V=V44//yyoo

q = shear force/unit lengthq = shear force/unit length

q = shear flow = constantq = shear flow = constant

Reinforced Concrete IIReinforced Concrete II

q =q = τ.τ.tt

ττ = shear stress= shear stress

Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y


Take moment aboutTake moment about centroidcentroid::

== 11++ 33 yyoo ++ 22++ 44 xxoo

T = 2VT = 2V11yyoo/2 + 2V/2 + 2V22xxoo/2/2

Recall: VRecall: V11 == q.xq.xoo & V& V22 == q.yq.yoo

T = 2VT = 2V11xxooyyoo/2 + 2V/2 + 2V22yyooxxoo/2/2

T = 2 AT = 2 A


ττ = T/(2A= T/(2Aoot)t)





. Hazim Dwairi

Threshold TorsionThreshold Torsion

•• Torques that do not exceed approximately oneTorques that do not exceed approximately one--

cr cr

structurally significant reduction in either thestructurally significant reduction in either the

flexural or shear strength and can be ignored.flexural or shear strength and can be ignored.

•• Cracking is assumed to occur when the principalCracking is assumed to occur when the principal

tensile stress reaches . In atensile stress reaches . In a nonprestressednonprestressed'33.0 c f


,,

tensile stress is equal to thetensile stress is equal to the torsionaltorsional shear stress,shear stress,

ττ . Recall that:. Recall that:


t AT

o2=τ

Threshold TorsionThreshold Torsion

R11.6.1-ACItoAccording

and of valuesngSubstituti

)(2 ; :where

4

3 ;

3

2

2

o

cpcp

cp

cp

cpo

t A

y x p xy A

pt A A

+==

==


4enTorsion wh Neglect

33.0 '

cr u

cp

cp

ccr

T T

p f T

φ ≤

⎟ ⎠

⎜⎝

=∴




. Hazim Dwairi

Torsion inTorsion inReinforcedReinforced

ConcreteConcrete

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit ystirrupof CLtoCL

fromdistance& oo y x

Reinforcement Requirement forReinforcement Requirement forRC Members in TorsionRC Members in Torsion

•• Reinforcement is determined using space trussReinforcement is determined using space truss

..

•• In space truss analogy, the concrete compressionIn space truss analogy, the concrete compression

diagonals (struts), vertical stirrups in tension (ties),diagonals (struts), vertical stirrups in tension (ties),

and longitudinal reinforcement (tension chords)and longitudinal reinforcement (tension chords)

act together as shown in figure on the next slide.act together as shown in figure on the next slide.


be resisted by the vertical stirrups as well as bybe resisted by the vertical stirrups as well as by

the longitudinal steelthe longitudinal steel





. Hazim Dwairi

Reinforcement Requirement forReinforcement Requirement forRC Members in TorsionRC Members in Torsion


Vertical Stirrup Reinforcement Vertical Stirrup Reinforcement

θ

τ

cot

2 :Recall

4 yvt o

yvt oo

o

f A y

f nA yT

qyV

A

T t q

====

==

crossingstirrupsof number=n


θ θ

θ

cot2cot2

2cot

oh

yvt yvt

oon

o

o yvt

o

o

As

f A

s

f A y xT

y

A f A

s

yT

==

=

entreinforcem

shearof strengthYield

stirruplegoneof Area

crack diagonal

=

=

yv

t

f

A




. Hazim Dwairi


•• ForFor No failureNo failure, i.e., torsion capacity greater than, i.e., torsion capacity greater than

•• For torsion capacity equal to or greater thanFor torsion capacity equal to or greater than

torsion demand, we have at the limit state:torsion demand, we have at the limit state:

θ cot2 yvt

A f A

T =

75.0 ; => φ φ un T T


•• Therefore steel area in one leg stirrup is:Therefore steel area in one leg stirrup is:


s

θ φ cot2 oh yv

ut

A f

sT A =


•• ACI 6.3.6 assumes ACI 6.3.6 assumes θθ = 45= 45oo forfor nonprestressednonprestressed

ohoh oo

A Aoo = 0.85A= 0.85Aohoh

•• Therefore:Therefore:

ut

A

sT A

2

=





. Hazim Dwairi

Longitudinal Steel ReinforcementLongitudinal Steel Reinforcement

Diagonal Compression StrutsDiagonal Compression Struts

2

44 cotcot

:TorsiontodueForceAxial

f ys

AV N yvo

t ==Δ θ θ

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y31

2

11

24

cotcot

:Similarly

N N

f xs

AV N

N N

yvot

Δ=Δ

==Δ

Δ=Δ

θ θ


Total axial force is:Total axial force is:

4321total N N N N N Δ+Δ+Δ+Δ=

θ θ

θ θ

22

22

cotcot)(2

cot2cot2

yvht

yvt

oototal

yvot

yvot

total

f ps

A f

s

A y x N

f xs

A f y

s

A N

=+=

+=

Longitudinal Steel Force:Longitudinal Steel Force:

Astirrupof permimeter =

h p


cot yvh yltotal ps

==

entreinforcem

allongitudinof strengthYield

sionresist tor ent toreinforcem

allongitudinof areaTotal

=

=

y

l

f

A

θ 2coth

y

yvt l p

f

f

s

A A =




. Hazim Dwairi


h

yl

yvt l p f

f

s

A A θ cot

2

=

vov

u

lt

o yv

ut

f A f

sT

A A

A f

sT A

φ

φ

φ

2

45forand ,inSubstitute

2

o=

=


h

y

l p f s

A =

yo

hul

f A pT A

φ 2=

Combined Shear and TorsionCombined Shear and Torsion

ot

wv

t AT

d bV

)2/( :StressTorsional

/ :StressShear

=

=

τ

τ

hohoho p ,. :sect oncrac eor ==


27.1 oh

h

w

t v A

Tp

d b

V +=+= τ τ τ 2

2

2 )7.1

()(oh

h

w A

Tp

d b

V +=τ




. Hazim Dwairi

Equilibrium and CompatibilityEquilibrium and CompatibilityTorsionTorsion

•• Equilibr ium TorsionEquilibr ium Torsion:: torsion moment is required fortorsion moment is required for

internal forces redistribution).internal forces redistribution).

•• Compatibility TorsionCompatibility Torsion:: torsionaltorsional moment resultsmoment results

from the compatibility of deformations between membersfrom the compatibility of deformations between members

meeting at a joint (meeting at a joint (torsionaltorsional moment can be reduced bymoment can be reduced by

redistribution of internal forces after cracking if theredistribution of internal forces after cracking if the


compatibility of deformations). The reduction incompatibility of deformations). The reduction in TTuu is ofis of

the magnitude:the magnitude:


Equilibrium TorsionEquilibrium Torsion

The torsion in the beamsThe torsion in the beams

in Fi s a & b must bein Fi s a & b must be

resisted by the structuralresisted by the structural

system if the beam is tosystem if the beam is to

remain in equilibrium. Ifremain in equilibrium. If

the applied torsion is notthe applied torsion is not

resisted, the beam willresisted, the beam will

rotate about its axis untilrotate about its axis until


the structure collapses.the structure collapses.





. Hazim Dwairi

Compatibility TorsionCompatibility Torsion A B


Hinge

ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign

Equilibrium TorsionEquilibrium Torsion:: design for fulldesign for full TTuu

Compatibility TorsionCompatibility Torsion:: reducereduce TTuu to the followingto the following

•• NonprestressedNonprestressed membermember withoutwithout axial force:axial force:


•• NonprestressedNonprestressed membermember withwith an axial force:an axial force:





. Hazim Dwairi


It shall be permitted toIt shall be permitted to neglect torsionneglect torsion effects ifeffects if

uu

•• NonprestressedNonprestressed membersmembers withoutwithout axial force:axial force:


•• NonprestressedNonprestressed membersmembers withwith an axial force:an axial force:



•• The crossThe cross--sectional dimensions shall be suchsectional dimensions shall be such

τmax


•• If If NOTNOT,, increase section dimensionsincrease section dimensions..Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y




. Hazim Dwairi


•• Reinforcement for torsionReinforcement for torsion

eca q.eca q. --

•• Combined shear and torsion reinforcementCombined shear and torsion reinforcement

oo

o yv

ut

A f

T

s

A6003 ;

cot2≤≤= θ

θ φ



s

A

s

A

s

A t vt v 2

+=+


•• Maximum spacing of torsion reinforcementMaximum spacing of torsion reinforcement

•• Spacing is limited to ensure the development ofSpacing is limited to ensure the development of

the ultimatethe ultimate torsionaltorsional strength of the beam, tostrength of the beam, to

⎪⎩

⎪⎨=

mm

p

sh

3008of smallermax


prevent excessive loss ofprevent excessive loss of torsionaltorsional stiffness afterstiffness after

cracking, and to control crack widths.cracking, and to control crack widths.





. Hazim Dwairi


•• Minimum area of closed stirrupsMinimum area of closed stirrups

•• Minimum area of longitudinalMinimum area of longitudinal torsionaltorsional

reinforcementreinforcement

⎪⎪

⎩

⎪⎪

⎨=+

yv

wc

yv

w

t v

f

sb f

f A A

'062.0

.

of larger)2(

1- shall be distributed around the perimeter


yv

wt

y

vt h

t

y

cpc

l

f

b

s

A

f

f p

s

A

f

A f A

175.0:where

42.0 '

min,

≥

⎟ ⎠

⎞⎜⎝

⎛ −=

o e c ose s rrups w a max mum

spacing of 300 mm.

2- The longitudinal bars shall be inside the

stirrups.3- shall have a diameter at least 0.042

times the stirrup spacing, but not less than

φ 10


•• Torsion reinforcement shall be provided for aTorsion reinforcement shall be provided for a

ww

wherewhere TTuu is less thanis less than ΦTΦTcr cr //44..

•• Stirrup DetailingStirrup Detailing





. Hazim Dwairi

Example: Equilibrium TorsionExample: Equilibrium Torsion

m

80 kN

180 kN

6 0 0 m

350 mm


mmd

mmmm and bh

mml

h

5252/251050600

350600USE

1758

1400

8

9.5(a)Table-9.5.2.1ACItoAccording

min

≈−−−=

==

===


( ) mkN /3.62510

6003501.2weightself beamFactored

6 =×⎟

⎠

⎞⎜⎝

⎛ ××=

kN V

kN

kN

u 82.232

:toleadsAnalysisStructural

2161802.1)(Nload axialFactored

224806.1802.1

..

u

=

=×=

=×+×=


kN N

mkN d T

mkN M

kN d V

u

u

u

u

216

.6.33@

.17.286

51.229525.03.682.232@

=

=

=

=×−=




. Hazim Dwairi


441100060.035.0211.01.0

216

1.0if check

'

'

kN A

kN N

A f N

u

gcu

=××××=

=

≥

n.interactio

load axialand bendingfordesigned beshallmemberOtherwise,

design.flexureinneglected becanaffectforceaxialTherefore

1.0 ' A f N gcu ≥⇒


.. .5.342

1960mm254use

1610mmAerrorand By trial

:FlexureforDesign

2

2

s

K O M mkN M un >=

=

=

φ

φ 6

0 0 m m

350 mm

4 25


mmhb A wc 000,210600350

:Torsionand ShearBothforDesign

2=×==

mm y

mm x

mmhb p

o

o

wcp

490552600

240552350

55mm550stirrupof centercover toAssume

900,16002350222

=×−=

=×−=

=+=

=×+×=+=


mm y x p

mm A A

mm y x A

ooh

oho

oooh

460,1)(2

960,9985.0

600,117490240

2

2

=+=

==

=×==




. Hazim Dwairi


: beamof sizeforCheck

( )2166.02117.075.0117600

1460106.33

525350

105.229

66.07.1

2

2

62

3

'

2

2

2

+≤⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ××+⎟⎟

⎠

⎞⎜⎜⎝

⎛

×

×

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +≤⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ c

w

c

oh

hu

w

u f d b

V

A

pT

d b

V φ


okayissizeBeam

853.2260.2

⇒

≤


N Aucp

:TtorsioncriticalforCheck

2

'

c

⎞⎛

φ

T mkN T

T

f A p

uc

c

cgcp

cc

considered bemustTorsion.640.9

2860035033.0

102161

1900

21000021083.075.0

33.0.

32

'

⇒<=

××

×+⎟⎟

⎠

⎞⎜⎜⎝

⎛ ×=

⎠⎝ =

φ

φ


mmmm f A

T

s

A

yvo

ut /534.04209996075.02

106.33

2

entReinforcemTorsion(a)

26

=×××

×==

φ




. Hazim Dwairi


d b f N

V wcu

c 117.0

entReinforcemShear(b)

'⎟ ⎞

⎜⎛ +=φ

kN V V

V

kN V

cu

s

c

g

3.1527.15375.0

51.229

7.15352535021)600350(14

10216117.0

3

=−=−=

=××⎟⎟ ⎠

⎞⎜⎜⎝

⎛

×

×+=

⎠⎝

φ

φ


mmmmd f

V

s

A

yv

sv /691.0525420

103.152 2=×

×==


/759.12

stirrupsselectand entreinforcemshearAdd (c)

2=+=+ mmmms

A

s

A

s

A t vt v

237.0420

35021062.0

292.0420

35035.0

of larger

stirrupsminimumCheck

2

⎪⎪⎩

⎪⎪⎨

⎧

=×

=×

≥+

s

A t v


StirrupsClosed 12@125mmUSE

OK 5.1828/14606.128759.1/226

:s rrupsssume

max

φ ⇒

==⇒===+

mmsmms

mmt v





( )( ) 780420

1460534.0

nfor torsioentreinforcemallongitudinDesign(d)

2==⎟ ⎞

⎜⎛

= mm f

p A

A vt h

t l

( )( ) 3.182420

4201460534.0

420

2100002142.0

42.0

2

min,

'

min,

=−×

=

⎟ ⎠

⎞⎜⎝

⎛ −=

mm A

f

f p

s

A

f

A f A

l

y

vt h

t

y

cpc

l

y


mm10or5.25mm1250.042diameter barmin.

bottomat2and middleat2at top,2

: bars6use300mmspacingMax.

780

=×=

⇒=

=⇒ mm Al


mm328452-780entreinforcemflexuraltoadd

beamtheof half bottominmm452124Use

2

2φ

=

=

2541938mm3281610entReinforcemFlexural 2 φ <=+=⇒

0

0 m m

4 25

2 12

12@125mm


6

350 mm

2 12

rc2 lecture 6.0 - torsion

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