RC2 Lecture 6.0 - Torsion
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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 =
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
a/b 1.0 1.5 2.0 3.0 5.0
0.208 0.219 0.246 0.267 0.290 1/3
∞
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Cracking StrengthCracking Strength
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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.
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Thin Walled Tube Analogy Thin Walled Tube Analogy
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
Thin Walled Tube Analogy Thin Walled Tube Analogy
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
Reinforced Concrete IIReinforced Concrete II
ττ = T/(2A= T/(2Aoot)t)
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. 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
Reinforced Concrete IIReinforced Concrete II
,,
tensile stress is equal to thetensile stress is equal to the torsionaltorsional shear stress,shear stress,
ττ . Recall that:. Recall that:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
+==
==
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
4enTorsion wh Neglect
33.0 '
cr u
cp
cp
ccr
T T
p f T
φ ≤
⎟ ⎠
⎜⎝
=∴
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einforced Concrete II Hashemite Unive
. 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.
Reinforced Concrete IIReinforced Concrete II
be resisted by the vertical stirrups as well as bybe resisted by the vertical stirrups as well as by
the longitudinal steelthe longitudinal steel
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Reinforcement Requirement forReinforcement Requirement forRC Members in TorsionRC Members in Torsion
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Vertical Stirrup Reinforcement Vertical Stirrup Reinforcement
•• 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
Reinforced Concrete IIReinforced Concrete II
•• Therefore steel area in one leg stirrup is:Therefore steel area in one leg stirrup is:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
s
θ φ cot2 oh yv
ut
A f
sT A =
Vertical Stirrup Reinforcement Vertical Stirrup Reinforcement
•• 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
=
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. 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
Longitudinal Steel ReinforcementLongitudinal Steel Reinforcement
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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 =
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Longitudinal Steel ReinforcementLongitudinal Steel Reinforcement
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=
=
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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 ==
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
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einforced Concrete II Hashemite Unive
. 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
Reinforced Concrete IIReinforced Concrete II
compatibility of deformations). The reduction incompatibility of deformations). The reduction in TTuu is ofis of
the magnitude:the magnitude:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
Reinforced Concrete IIReinforced Concrete II
the structure collapses.the structure collapses.
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Compatibility TorsionCompatibility Torsion A B
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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:
Reinforced Concrete IIReinforced Concrete II
•• NonprestressedNonprestressed membermember withwith an axial force:an axial force:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
It shall be permitted toIt shall be permitted to neglect torsionneglect torsion effects ifeffects if
uu
•• NonprestressedNonprestressed membersmembers withoutwithout axial force:axial force:
Reinforced Concrete IIReinforced Concrete II
•• NonprestressedNonprestressed membersmembers withwith an axial force:an axial force:
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
•• The crossThe cross--sectional dimensions shall be suchsectional dimensions shall be such
τmax
Reinforced Concrete IIReinforced Concrete II
•• If If NOTNOT,, increase section dimensionsincrease section dimensions..Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
•• 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
Reinforced Concrete IIReinforced Concrete II
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
s
A
s
A
s
A t vt v 2
+=+
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
•• 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
Reinforced Concrete IIReinforced Concrete II
prevent excessive loss ofprevent excessive loss of torsionaltorsional stiffness afterstiffness after
cracking, and to control crack widths.cracking, and to control crack widths.
Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
•• 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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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
ACI Requirements for Torsion ACI Requirements for TorsionDesignDesign
•• 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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Example: Equilibrium TorsionExample: Equilibrium Torsion
m
80 kN
180 kN
6 0 0 m
350 mm
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
mmd
mmmm and bh
mml
h
5252/251050600
350600USE
1758
1400
8
9.5(a)Table-9.5.2.1ACItoAccording
min
≈−−−=
==
===
Example: Equilibrium TorsionExample: Equilibrium Torsion
( ) mkN /3.62510
6003501.2weightself beamFactored
6 =×⎟
⎠
⎞⎜⎝
⎛ ××=
kN V
kN
kN
u 82.232
:toleadsAnalysisStructural
2161802.1)(Nload axialFactored
224806.1802.1
..
u
=
=×=
=×+×=
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
kN N
mkN d T
mkN M
kN d V
u
u
u
u
216
.6.33@
.17.286
51.229525.03.682.232@
=
=
=
=×−=
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Example: Equilibrium TorsionExample: Equilibrium Torsion
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 ≥⇒
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
.. .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
Example: Equilibrium TorsionExample: Equilibrium Torsion
mmhb A wc 000,210600350
:Torsionand ShearBothforDesign
2=×==
mm y
mm x
mmhb p
o
o
wcp
490552600
240552350
55mm550stirrupof centercover toAssume
900,16002350222
=×−=
=×−=
=+=
=×+×=+=
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
mm y x p
mm A A
mm y x A
ooh
oho
oooh
460,1)(2
960,9985.0
600,117490240
2
2
=+=
==
=×==
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Example: Equilibrium TorsionExample: Equilibrium Torsion
: 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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
okayissizeBeam
853.2260.2
⇒
≤
Example: Equilibrium TorsionExample: Equilibrium Torsion
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
mmmm f A
T
s
A
yvo
ut /534.04209996075.02
106.33
2
entReinforcemTorsion(a)
26
=×××
×==
φ
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einforced Concrete II Hashemite Unive
. Hazim Dwairi
Example: Equilibrium TorsionExample: Equilibrium Torsion
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
Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite Universit yThe Hashemite Universit y
mmmmd f
V
s
A
yv
sv /691.0525420
103.152 2=×
×==
Example: Equilibrium TorsionExample: Equilibrium Torsion
/759.12
stirrupsselectand entreinforcemshearAdd (c)
2=+=+ mmmms
A
s
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8/13/2019 RC2 Lecture 6.0 - Torsion
http://slidepdf.com/reader/full/rc2-lecture-60-torsion 22/22
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