CE72.52 - Lecture 3b - Section Behavior - Shear and Torsion

1
CE 72.52 Advanced Concrete
Lecture 3b:
Section
Behavior
(Shear-Torsion)
Naveed Anwar
Executive Director, AIT Consulting
Director, ACECOMS
Affiliate Faculty, Structural Engineering, AIT
August - 2014

Shear Behavior of RC
Sections
2

Frame/ Linear Member Sections
3

Shear - Torsion Theories
Major Issues
• No unified Theory Yet
• Several Interactions
• Shear & Torsion, Shear &
Moment, Shear & Load,
Torsion & Moment
• Effect of Confinement,
Flexural Reinforcement,
Hoops
Current Approaches
• Solid Mechanics (principle
stress, Mohr's circle)
• Truss Mechanism, Strut and
Tie, diagonal tension
• Compression Field Theories
• Arch action, dowel action,
Fracture Mechanics,
aggregate interlock and
other considerations
4Advanced Concrete l August-2014

The Origin of Shear Stresses
• Shear stress in beam-column sections is
produced due to:
5
Shear Force Torsion Combination
Advanced Concrete l August-2014

Shear Stress Distribution Due To Shear

Shear Stress Distribution Due to Torsion

Shear Stress Due to Shear + Torsion
8
+
+
-
-
+ =+
+
-
+ =

Concrete Shear Capacity

Shear Stress and Concrete Sections
• Shear stress contributes to principle stresses
• In the absence of direct compressive stress:
• One of the principle stress will be tension
• Concrete does not take much tensions
• So concrete can not resist much shear stress
• Hence concrete alone can not resist much
shear force or torsion
• Shear and torsion capacity of concrete
section is therefore primarily governed by
the tension capacity of concrete

Concrete Shear Capacity

Interaction of Shear Stresses
• Consider Interaction of V + T + M + P for computing Vc
13
M + V (+N) T

Interaction of Shear Stresses

Shear in Cracked Concrete Beam
15
Vc – Intact Concrete
Va – Aggregate Interlock
Vsw – Steel Reinforcement
Vd – Dowel action

Web and Flexure shear cracks

B & D Regions

Average shear stresses between
cracks

Average shear stresses between
cracks
19
and
M M M
T T T
jd jd

   
M
T
jd

 
M V x  
V x
T
jd

 
w
V
v
b jd

w
V
v
b d

w
T
v
b x



   d T d jd
V jd T
dx dx
 
 
d
V Tjd
dx


A Truss and a Beam
20
An RC Beam and “Hidden”
Truss
A Real Truss

Formation of Truss in Beams

Design for Shear
• Design practices
• Failure mechanisms
• Dimensions
• Geometry
• Loading
• Member properties
• Shear strength models
• Stresses in uncracked beams to explain onset
of shear cracking
• Plastic truss models with shear cracks
• ACI code design procedure
• Cracked beam models

Design for Shear
• Compute Cross-sectional properties
• Compute shear capacity of concrete alone
• Determine shear to be resisted by steel
• Compute minimum shear reinforcement
• Compute shear reinforcement separately
• Select rebars and determine bar spacing for
shear
• Check spacing limits on rebars
23

Design for Shear [ACI-318-11]
24
• For Axial compression
• For Axial Tension
• Shear from Stirrups
• If
• Stirrups should be provide as
VsVcVn 
dbwfc
Ag
Nu
Vc '
2000
12 






dbwfc
Ag
Nu
Vc '
500
12 






 
s
d
fyAvVs  cossin 
VnVu 
Vc
Vu
VsorVcVuVn 


Vc= Shear due to concrete
Vs= Shear due to steel stirrups
Vu= Factored shear load
Nu = Axial Load
Ag= Gross area of steel
fc’= concrete strength
d=effective depth
bw= web width
fy= yield strength of steel
α= angle of stirrup
s= spacing of stirrup
Av= area of stirrups

Tensional Stress Distribution
25
A Flanged Section
A Box Section
A Flanged - Box Section

Behavior for Shear and Torsion
26
Pure Torsion produces constant
“Shear Flow” around the cross-
section outer skin.
Shear and
Torsion may
produce Un-
even Shear
Flow

Simplified Behavior for Torsion – RC
Beam
27
Before Cracking
of Concrete
Only this portion
of concrete
section is available
for resisting
Torsion
Therefore “other”
mechanism should
be found to resist
torsion
After Cracking
of Concrete

Beam
Before Cracking of
Concrete
Only this portion
of concrete
section is available
for resisting
Torsion
Therefore “other”
mechanism should
be found to resist
torsion
After Cracking of
Concrete

Beam
• The shear flow in a steel
tube embedded in a
concrete section is
concentrated in the steel
tube. This observation is
relevant to concrete
sections with hoop
reinforcement, where
almost the entire torsion is
resisted by the
reinforcement.
• (green color show high
stress)

Mechanisms of Shear Transfer without
Web Reinforcement
• Shear resistance of the uncracked section
above the flexural crack
• Aggregate interlock
• Longitudinal reinforcing to a friction force
(dowel action)
• Tied-arch type of behavior that exists in
rather deep beams

Shear Strength of Concrete (ACI)
• At locations of large moments, uncracked
area of beam section reduced
Vc < 1.9 √(fc’bwd)
• At locations of small moments, large portion of
section is available to resist shear
Vc ≈ 3.5 √(fc’bwd)
• Considering effects of longitudinal
reinforcements, moments and shear
magnitudes, following equation which is less
conservative but tedious in calculation, is used
(Normally used but conservative)

Shear Design (ACI)
• Determine the factored shear force.
• Determine shear force Vc, resisted by
concrete.
• Determine the reinforcement steel to carry
the balance.
• Check the minimum and maximum shear
reinforcement.
• If the required shear reinforcement is more
than maximum shear reinforcement,
increase the section size.
• Check spacing limits for rebars.

Concrete Shear Strength (Vc) – ACI 318-11
33
• For members subject to shear and flexure only,
• For members subject to axial compression,
• For members subject to significant axial
tension,
• Nu is negative for tension
dbwfc
Ag
Nu
Vc '
500
12 







Shear Reinforcement (Av) – [ACI 318-11]
34
• If Vu ≤ Vc/2,
Av/s = 0
• If Vc/2 < Vu ≤ Vmax,
Av/s = (Vu – Vc) / (fysd)
where
Vmax = Vc +
Minimum shear reinforcement
<

Shear Reinforcement (Av)
35
• Vs is limited by, since a beam
shear strength cannot be increased
indefinitely by adding more and more
shear reinforcing
• The greater the shear transferred by shear
reinforcing to the concrete, the greater will
be the chance of a combination shear
and compression failure.

Shear Reinforcement (Av)

Rebar Spacing Limit
• Maximum Spacing
• If Vs ≤
• d/2 ≤ 24 in
• If
• d/4 ≤ 12 in
• Minimum Spacing
• Approximately 3 or 4 in

Shear Design in Ductile Frames
• Design shear force is based on maximum
probable moment capacities and gravity
shear forces
• Concrete shear strength is not considered in
the hinge region when both of the following
conditions occur.
• The earthquake induced shear force
represents one-half or more of the maximum
required shear strength
• The factored axial compressive force
including earthquake effects is less than 5% of
compressive strength of concrete (Ag fc’/20)

Beams without web reinforcement –
Behavior in Shear

Failures modes of deep beams

Design for Torsion (ACI)
• Determine the factored torsion, Tu.
• Determine special section properties.
• Determine critical torsion capacity.
• Determine the reinforcement steel
required.
• Check the minimum reinforcement

Special Section Properties
• Acp = Area enclosed by outside perimeter
of concrete cross-section
• Aoh = Area enclosed by centerline of the
outermost closed transverse torsional
reinforcement
• Ao = Gross area enclosed by shear flow
path
• pcp = Outside perimeter of concrete cross
section
• pn = Perimeter of centerline of outermost
closed transverse torsional reinforcement

Special Section Properties
• For rectangular beam
• Acp = bh
• Aoh = (b – 2c) (h – 2c)
• Ao = 0.85 Aoh
• pcp = 2b + 2h
• pn = 2 (b – 2c) + 2 (h – 2c)
• For T beam
• Acp = bwh + (bw – bw) ds
• Aoh = (bw – 2c) (h – 2c)
• Ao = 0.85 Aoh
• pcp = 2bf + 2h
• pn = 2 (bw – 2c) + 2 (h – 2c)

Critical Torsion Capacity – [ACI 318-11]
• If the factored torsion is less than critical
torsion, it can be ignored. Tcr is,

Required Reinforcement
46
• If Tu > Tcr,
• The required longitudinal rebar area is
• The required closed stirrup area is

Minimum Reinforcement
47
• Minimum closed stirrup area is
• Minimum longitudinal rebar area is

Maximum Torsional Moment Strength –
[ACI 318-11]
48
• To reduce unsightly cracking and
compressive failure due to shear and
torsion, the size of the cross-section is
limited.
• For solid sections,
• For hollow sections,

Torsional Cracks in Concrete Beams
49

T - ϴ Curve (Torque - Twist Behavior)
50

Torque - Twist Behavior

Shear Deformation
• For short, deep rectangular beam and for
continuous T beams, the deformations
caused by shear my become significant.
• For most relatively slender members,
subjected to low shear, the effect of shear
on deflection is negligible.
• Hence, when service conditions are
examined the designer also needs to be
able to asses the order of expected shear
deflection

Shear Deformation
54
• Uncracked Members
• Before the formation of flexural or diagonal
cracks, the behavior of a beam can be well
predicted by using elasticity principles.
• The shear stiffness of the beam is the amount of
the shear force that applied to a beam of unit
lent, will cause unit shear displacement of one
end relative to other.
psifcEc
Ec
G '57000
)1(2





Shear Deformation
55
• Shear Stiffness of beam
Where, Kv’= Shear Stiffness
bw= width of the web beam
d= effective depth of the beam
f= factor for nonuniform distribution of
shear stresses.
For rectangular section f=1.2 and for T and
I section f=1.0
f
dbG
Kv w
'

Shear Deformation
• Cracked Members
• In beams that are subjected to large shear
forces and are web reinforced accordingly,
diagonal cracks are expected during loading
• These cracks can increase the shear
deformation of beam considerably
• The greater proportion of the load is likely to be
carried by truss mechanism
• Shear distortions, of the beam can be
approximated by using model analogous truss.

Shear Deformation
57
Av ca
a
Av
a
R
S
d
d
R
C
C
V
sV
sV
C

Shear Deformation
58
• Cracked Members
• For simplicity assume vertical stirrups and 45
degree concrete struts
• [Ref. Park and Paulay, 1992]
db
V
vf
bE
V
d
E
f
AE
SV
d
Es
f
Ad
SV
Ajd
SV
f
s
w
s
scd
wc
s
c
cd
c
vs
ss
s
v
s
v
s
s
CRSV
22;
22
2
;
2



Av ca
a
Av
a
R
S
d
d
R
C
C
V
sV
sV
C

Shear Deformation
59
• Shear Distortion
• Shear Stiffness
• Similarly for general, Compression strut
angle =α, Stirrup angle=β
c
s
ws
sv
v
E
E
nn
vdbE
V
d








 ;4
1


dbE
n
K
When
ws
v
v
v
v



41
1
45,







sin
sinsin
)cot(cotsinsin
44
244
w
v
v
ws
v
v
v
sb
A
dbE
n
K





Shear Deformation
• Cracked Members
• Analytical and experimental studies have
verified that the stiffness of the cracked deep
beams, in which shear deformations dominate,
is only about 15 % of the stiffness in the
uncracked state when shear and flexure
distortions are considered.
• [Ref. Park and Paulay, 1992]

Shear Design of Beams in Moment
Resisting Frame
• Assuming beam is
yielding in flexure,
beam end
moments are set
equal to probable
moment strengths.
• Design shear is
based on the
probable moment
to maintain the
moment
equilibrium.
61

Details of Beam in Moment Resisting
Frame
62

Torsion
• Torsion:
• A moment acting about the longitudinal axis of a
member is called a torsional moment. In structure,
torsion results from the eccentric loading of beams
or from deformations resulting from the continuity of
beams or similar members that join at and angle to
each other.
• From the design point of view, torsional moments
are classified into equilibrium moments and
compatibility torsional moments.
• Equilibrium torsional moment:
• It is torsional moment against which a structure
must resist in order to keep the force equilibrium in
the overall structure system. If this torsional moment
is neglected in the calculation of the force
equilibrium of the structure the stability of the
overall structure is destroyed

Torsion
65
• Torsion For Equilibrium and Stability
Cantilever with
eccentrically applied load
Canopy
Section through a
beam supporting
precast floor slab

Torsion
66
• Compatibility Torsional Moment
• It is a moment caused by the compatibility
between the members meeting at a joint(
composing statically indeterminate structure) , and
provides an influence mainly on the elastic
deformation of the structure.
• In general, the torsional rigidity of a concrete
member is greatly reduced after plastic
deformation by torsion. Hence, the torsional
moment action on the member becomes very
small when a concrete member of a statically
indeterminate structure reaches such a state.

Torsion
67
• The mechanical behavior of a concrete
member before torsional crack can be
estimated by using elasticity theory, assuming
that the gross concrete section is effective.
• The angle of twist per unit length is
• Where T = twisting moment,
• Gc= the shear modulus of concrete
• J1= the torsion constant.
• For rectangular section,
• where c and b are the two sides of the rectangle with b<= c.
• The maximum shear stress is at the middle of the longer
side c and its value












 4
4
3
1
12
121.0
3
1
c
b
c
b
cbJ
2max
bc
T

 

Torsion
68
• where μ is a dimensionless coefficient which
varies with the aspect ratio c/b as
• Torsional moment (Mt) acting on the member
section is given by
• Where, Kt is section modulus for torsion and it is
decided by only shape and size of cross-
section
maxtt KM 

Torsion
69
Section Modulus for Torsion (Kt)

Torsion
70
• Torsional Crack (Unreinforced Concrete)
• Pure torsional moment act on a member– Pure
shear stress
• Angle of principle stress is 45 ͦ
• From Mohr’s stress circle, Principe stresses (σ1, σ2 )
• Torsional Crack caused by σ1 when it exceeds the tensile
strength of concrete.
):(max21  tension
K
M
t
t


Torsion
71
• Combined Torsion and Shear
• When a member is subjected to combined shear
and torsion, the two shearing stress components
add on one side face and counteract each other
on other side. As a results, inclined cracking starts
on the face where stresses add (Crack AB) and
extends across the flexural tensile face of the
beam. If the bending moments are sufficiently
large, the cracks will extend almost vertically across
the back face (Crack CD)

Torsion
72
• Combined Torsion and Axial force
• Positive axial force (σn) (Tensile)
• From Mohr’s stress circle
t
n
tttc
t
n
t
t
n
f
fKM
f
f
fCrack
n












11
:;
22
1
2
2
1
=> Torsional moment
capacity reduced.

Torsion
73
• Negative axial force (σn = -σ’n)
(Compression)
• From Mohr’s stress circle
t
n
tttc
t
n
t
t
n
f
fKM
f
f
fCrack
n
'
1
'
1
stress)by tensilecausedisFailure(:;
2
'
2
'
1
2
2
1










 



=> Torsional moment
capacity increased.

Torsion
74
• Space-truss analogy(with reinforcement)
Vx
Vx
VyVy
t
t
xo
yo
Kt=2Amt
Am: area enclosed
by the centerline
of wall thickness
(=xoyo)
t: web thickness xo
yo
s s
M
t
Vx
Vx
Vy
Vy ϴ
Longitudinal Bar
Diagonal compression part
of web concrete
Transverse Bar
o
m
t
o
o
m
t
o
m
t
tt
y
A
M
tyVy
forceshearVx
A
M
txVx
tA
M
KM
2
:
2
2







Torsion Space-truss analogy(with reinforcement)
• Longitudinal Equilibrium force
Vx
Vx
VyVy
t
t
xo
y
o
barsallongitudinofareationalcrosstotalA
A
yx
A
M
yx
A
M
VyVxA
l
l
oo
m
t
oo
m
t
l
sec;
cot)(2
2
cot)(2
2
cot2cot2
1
1










yo
ϴ
σ1
σ1
ϴ
Vy cosϴ
V
y
Longitudinal bar: Al, σ1,fy
Stirrup: Aw, s, σw ,fy
Web Con: σ’c ,f`wc
Shear force: Vx, vy

Torsion Space-truss analogy
(with reinforcement)
76
• Transverse Equilibrium Force
• Diagonal Equilibrium Force
wm
t
w
x
o
wwy
o
ww
A
s
A
M
V
s
x
AV
s
y
A






tan
2
cot
;
cot








sincos
1
2
'
;
sin
cos.';
sin
cos.'
tA
M
V
tx
V
ty
m
t
c
x
oc
y
oc



77
• Torsional Capacity (Mty)
• Longitudinal bar and stirrup yield σ1fy ,
σwfy
• Diagonal Compressive Capacity(Mtcu)



















l
oo
oo
l
ymty
lw
oo
m
ty
y
wm
ty
w
l
oo
m
t
l
oo
m
t
As
Aw
yxand
yxs
AAw
fAM
Then
fyAA
yxs
A
M
f
A
s
A
M
fyA
yx
A
M
A
yx
A
M
)(2tan
)(2
2
)(2
2
tan
2
)(2
2
tan
cot)(2
2
2
1






wcmtuc
m
tuc
wcwcc
ftAM
tA
M
ff
'.sincos2
sincos
1
2
'''






78
• Balance Failure
• Reinforcement yields (Mty)= concrete crushed
(Mtuc)


2
sin
'
)(2tan
'.sincos2
)(2
2
y
wc
w
w
s
w
oo
wcm
oo
sw
ymtucty
f
f
ts
A
sA
A
yx
ftA
yxs
AA
fAMM




Balance Reinforcement ratio

Interaction of Shear,
Flexure and Axial forces

Space Truss Analogy
80
b) Space Truss Model for Torsion
T
Space Truss Model for
Torsion
c) Modified Space Truss for M, V, T
Modified Space Truss for
M, V, T

Rebars for Axial Load - P
81
As : May be needed to resist
compression. Generally not required
if P is small

Rebars for Moment - Mx
82
Ast : To resist main tension due to
moment
Asw : To resist secondary tension due
to moment and prevent web cracks (for
beams more than 90 cm deep)
Asc : To resist compression due to
moment (doubly reinforced beams)

Rebars for Moment - Mx
83
Ast : To resist main
tension due to
moment
Asc : To resist compression due to
moment (may not be neede)
Asw : To resist secondary tension due
to moment and prevent web cracks (for
beams more than 90 cm deep)

Rebars for Shear - V
84
Asv : To resist shear stress due to
shear force exceeding the shear
capacity of concrete

Rebars for Torsion - T
85
Asvt : To resist shear due to Torsion.
Must be closed hoops on sides of the
section
Al : To resist the
longitudinal tension due to
Torsion. Must be distributed
around the perimeter

Rebars for P + Mx + My + V + T
86
Ast + Al/4 : To resist main tension due
to moment and tension due to Torsion
Asw + Al/4 : To resist secondary
tension in deep beams due to
moment and due to Torsion
Asc + Al/4: To resist compression due
to moment Mx (doubly reinforced
beams) and tension due to Torsion
Asvt + Asv/2: To resist shear due
to Torsion. Must be closed hoops on
sides of the section
Ast : To resist
tension due to
My
Asc : To resist compression
due to My (may not be needed)

Design of Beams: Basic Procedure
87
1
2
3
Estimate Cross-section
based on Thumb Rules
Analyze the beam to
obtain design actions
Design for
Bending Moment
Design OK
Design for
Shear and Torsion
Determine the
Layout of Rebars
Design OK
Deflections
Cracking etc
Design
Completed
Y
Y
Y
RevisedSectionMaterial
ReviseSection/Material
ReviseSection/Material

88
Shear-Torsion Design Procedure - ACI
Compute Tc
Section OK Tu > 0
fTc > Tu
Compute Avt for
Ts = Vu
Compute Al
Avt = 0
Y
Y
Y
Vu, Tu, fc, fy
Section
Shear Design
Completed
Av = Avs + 2 Avt
Check
Av (min)
Determine the
Layout of Rebars
Vu > 0 Section OK
Compute Vc
fVc > 0.5Vu
fVc > Vu
Compute Avs for
Vs = Vu - fVc
Minimum
Avs
Avs = 0
Y
Y
Y
Revise Section/Material Revise Section/Material

CE72.52 - Lecture 3b - Section Behavior - Shear and Torsion

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (7)

Similar to CE72.52 - Lecture 3b - Section Behavior - Shear and Torsion

Similar to CE72.52 - Lecture 3b - Section Behavior - Shear and Torsion (20)

More from Fawad Najam

More from Fawad Najam (6)

Recently uploaded

Recently uploaded (20)

CE72.52 - Lecture 3b - Section Behavior - Shear and Torsion