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Lecture 21Lecture 21 Splices and Splices and Shear Shear
February 5, 2003
CVEN 444
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Lecture Goals Lecture Goals
Spice
Shear
Shear Design
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B ar Splices B ar Splices
Why do we need bar splices? -- for long spans
Types of Splices
1. Butted &Welded
2. Mechanical Connectors
3. Lay Splices
Must develop 125%
of yield strength
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T ension Lap Splices T ension Lap Splices
Why do we need bar splices? -- for long spans
Types of Splices
1. Contact Splice
2. Non Contact Spice (distance 6´ and 1/5
splice length)
Splice length is the distance the two bars areoverlapped.
e e
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Ty pes of Splices Ty pes of Splices Class A Spice (ACI 12.15.2)
When over entire splicelength.
and 1/2 or less of total reinforcement is
spliced win the req¶d lay length.
2dreq's
provideds
u A
A
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Ty pes of Splices Ty pes of Splices
Class B Spice (ACI 12.15.2)
All tension lay splices not meetingrequirements of Class A Splices
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T ension Lap SpliceT ension Lap Splice(ACI 12.15)(ACI 12.15)
whereAs (req¶d) = determined for bending
ld = development length for bars (not
allowed to use excess reinforcementmodification factor)
ld must be greater than or equal to 12 in.
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T ension Lap Splice (ACI 12.15)T ension Lap Splice (ACI 12.15)
Lap Spices shall not be used for bars larger than No. 11.
(ACI 12.14.2)
Lap Spices should be placed in away from regions of
high tensile stresses -locate near points of inflection(ACI 12.15.1)
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Compression Lap SpliceCompression Lap Splice
(ACI 12.16.1)(ACI 12.16.1)
Lap, req¶d = 0.0005f y d b for f y < 60000 psi
Lap, req¶d = (0.0009f y -24) d b for f y > 60000 psiLap, req¶d 12 in
For f c 3000 psi, required lap splice shall be multiply
by (4/3) (ACI 12.16.1)
u
e
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Compression Lap SpliceCompression Lap Splice(ACI 12.17)(ACI 12.17)
In tied column splices with effective tie area throughout
splice length 0.0015 hs factor = 0.83
In spiral column splices, factor = 0.75 But final splice
length 12 in.
u
u
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Ex ampleEx ample Splice T ension Splice T ension
Calculate the lap-splice length for 6 #8 tension
bottom bars in two rows with clear spacing 2.5 in.
and a clear cover, 1.5 in., for the following cases
When 3 bars are spliced and As(provided) /As(required) >2
When 4 bars are spliced and As(provided) /As(required) < 2
When all bars are spliced at the same location.
f c= 5 ksi and f y = 60 ksi
a.
b.
c.
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Ex ampleEx ample Splice T ension Splice T ension
For #8 bars, db =1.0 in. and E = ? F =? K = ?P =?
yd
tr b c
b
3
40
f l
c K d f
d
EFKP¨ ¸© ¹© ¹!
© ¹© ¹ª º
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Ex ampleEx ample Splice T ension Splice T ension
The As(provided) /As(required) > 2, class ? splice applies;
The As(provided) /As(required) < 2, class ? splice applies;
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Ex ampleEx ample SpliceSpliceCompression Compression
Calculate the lap splice length for a # 10
compression bar in tied column when f c= 5 ksi and
when a) f y = 60 ksi and b) f y = 80 ksi
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Ex ampleEx ample SpliceSpliceCompression Compression
For #10 bars, d b =? in.
ydy
b c
0.020.003
f l f
d f ! u
Check ls > 0.005 d b f y
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Ex ampleEx ample SpliceSpliceCompression Compression
For #10 bars, d b =? in. The ld = 2? in.
Check ls > (0.0009 f y ±24) d b
So use ls= ? in.
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Shear Design Shear Design
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
Look at the shear and
bending moment
diagrams. The actingshear stress distribution
on the beam.
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
The acting stresses distributed across the
cross-section.
The shear stressacting on the
rectangular beam.
IbVQ!X
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
The equation of the shear stress for a rectangular beam
is given as:
Note: The maximum1st moment occurs at
the neutral axis (NA).
IbVQ!X
avemax
2
max
3
5.1*2
3
84*
2Q
Inertiaooment12
X X !¹ º ¸©
ª¨!
!¹ º
¸©ª
¨¹ º
¸©ª
¨!
!
bh
V
bhhbh
bh I
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
The ideal shear stress distribution can be described as:
Ib
VQ!X
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
A realistic description of the shear distribution is shown
as:
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U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior
The shear stress acting along the beam can be described
with a stress block:
Using Mohr¶s circle, the stress block can be
manipulated to find the maximum shear and the crack
formation.
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Inclined Cracking in Inclined Cracking in Reinforced Concrete B eams Reinforced Concrete B eams
Typical Crack Patterns for a deep beam
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Inclined Cracking in Inclined Cracking in Reinforced Concrete B eams Reinforced Concrete B eams
Flexural-shear crack -
Starts out as a flexural
crack and propagates due
to shear stress.
Flexural cracks in beams
are vertical
(perpendicular to thetension face).
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Inclined Cracking in Inclined Cracking in Reinforced Concrete B eams Reinforced Concrete B eams
For deep beam the cracks are given as:
The shear cracks Inclined (diagonal) intercept crack
with longitudinal bars plus vertical or inclined
reinforcement.
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Inclined Cracking in Inclined Cracking in Reinforced Concrete B eams Reinforced Concrete B eams
For deep beam the cracks
are given as:
The shear cracks fail due
two modes:
- shear-tension failure- shear-compression
failure
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Shear Strength of RC B eams Shear Strength of RC B eams without Web Reinforcement without Web Reinforcement
vcz - shear in compression
zone
va - Aggregate Interlock
forces
vd= Dowel action from
longitudinal bars
Note: vcz increases from
(V/bd) to (V/by) as crack
forms.
Total Resistance = vcz + vay + vd (when no stirrups are used)
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Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)
(1) Tensile Strength of concrete affect inclined
cracking load
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Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)
(2) Longitudinal Reinforcement Ratio, Vw
d b
f V
d b
A
wccw
w
sw
2:0025.00075.0or
cracksrestrains
d$
ee
!
V
V
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Strength of Concrete in Shear Strength of Concrete in Shear
(No Shear Reinforcement)(No Shear Reinforcement)(3) Shear span to depth ratio, a/d (M/(Vd))
2d
a
2
"
e
d
a Deep shear spansmore detail design
required
Ratio has littleeffect
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Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)
(4) Size of Beam
Increase Depth Reduced shear stress at
inclined cracking
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Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)
(5) Axial Forces
- Axial tension Decreases inclined cracking load
- Axial Compression Increases inclined crackingload (Delays flexural
cracking)
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F unction and Strength of F unction and Strength of Web Reinforcement Web Reinforcement
Web Reinforcement is provided to ensure that
the full flexural capacity can be developed.
(desired a flexural failure mode - shear failure
is brittle)
- Acts as ³clamps´ to keep shear cracks from
widening
Function:
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F unction and Strength of F unction and Strength of Web Reinforcement Web Reinforcement
Uncracked Beam Shear is resisted uncracked
concrete.
Flexural Cracking Shear is resisted by vcz, vay, vd
bars.allongitudinromctiono l orcenterlock ggregateocom onentertical
zonencom ressioinShear
d
ay
cz
V
V
V
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Designing to Resist Shear Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
n u
capacity demand
V V J u
u
u
n
factored shear force at section
Nominal Shear Strength
0.75 shear strength reduction factor
V
V
J
!
!
!
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Designing to Resist Shear Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
n c sV V V !
c
s
V
V
!
! Nominal shear provided by the shear reinforcement
!
Nominal shear resistance provided by concrete
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Shear Strength Provided by Shear Strength Provided by ConcreteConcrete
Bending only Bending only
Simple formula
More detailed
Note:
Eqn [11.5]
Eqn [11.3]
d b f
d b f V
wc
wcc
3.5
2
e
!
d b f
d bM
d V
f V
wc
w
u
u
wcc
3.5
25001.9
e
¹¹
¹
º
¸
©©
©
ª
¨
¹
¹
º
¸
©
©
ª
¨
! V
1
u
u e¹¹
º
¸
©©
ª
¨
M
d V
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Shear Strength Provided by Shear Strength Provided by ConcreteConcrete
Bending and Axial Compression Bending and Axial Compression
Nu is positive for
compression and
Nu
/Ag
are in psi.
Simple formula
Eqn [11.4]
Eqn [11.7]
g
uwc
wc
g
uc
50013.5
200012
A N d b f
d b f A
N V
e
¹¹¹ º
¸
©©©ª
¨
!
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Ty pical Shear Reinforcement Ty pical Shear Reinforcement
Stirrup - perpendicular to axis of members
(minimum labor - more material)
ACI Eqn 11-15
s
d f V
EE cossinyv
s
!
s
d f V
yv
s
o90 !!E
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Ty pical Shear Reinforcement Ty pical Shear Reinforcement
Bent Bars (more labor - minimum material) see req¶d
in 11.5.6
ACI 11-5.6
s
d f
V
EE cossinyv
s
!
s
d f V yv
s
o 41.145 !!E
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Stirrup Anchorage Requirements Stirrup Anchorage Requirements
Vs based on assumption stirrups yield
Stirrups must be well anchored.@
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Stirrup Anchorage Requirements Stirrup Anchorage Requirements
each bend must enclose a long bar
# 5 and smaller can use standard hooks 90o,135o, 180o
#6, #7,#8( f y = 40 ksi )
#6, #7,#8 ( f y > 40 ksi ) standard hook plus aminimum embedment
Refer to Sec. 12.13 of ACI 318 for development of web
reinforcement. Requirements:
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Stirrup Anchorage Requirements Stirrup Anchorage Requirements
Also sec. 7.10 requirement for minimum stirrups
in beams with compression reinforcement, beams subject to stress reversals, or beams
subject to torsion
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Design Procedure for Shear Design Procedure for Shear
(1) Calculate Vu
(2) Calculate JVc Eqn 11-3 or 11-5 (no axial force)
(3) Check
°¯®
pu cu V V J
2
1
If yes, add web reinforcement (go to 4)
If no, done.
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Design Procedure for Shear Design Procedure for Shear
(4) pee cuc V V V J J
2
1 I
¹¹ º
¸©©ª
¨!! v
w
y sv
y
wv A
b
f A s
f
sb A minfor
50or 50 maxmin
Also:
(Done) 11.5.4 "24
2max
d s
Provide minimum
shear reinforcement
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Design Procedure for Shear Design Procedure for Shear
(5)
cu
scu s
scnu
scu
V V
V V V V
V V V V
V V V
!!
!e
pu
J J J
J J J
J d)(req'calulate, I
Check:
11.5.4 illegalother wise, 8 d b f V wc sde
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Design Procedure for Shear Design Procedure for Shear
Solve for required stirrup spacing(strength)
Assume # 3, #4, or #5 stirrups
s
y sv
V
d f A s e
(6)
from 11-15
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Design Procedure for Shear Design Procedure for Shear
(7) Check minimum steel requirement (eqn 11-13)
50
max
w
y sv
b f A s !
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Design Procedure for Shear Design Procedure for Shear
(8) Check maximum spacing requirement (ACI 11.5.4)
illegal 8 I:ote
"124
4 I
"242 4 I
c
maxc
maxc
d b f V
d sd b f V
d s
d b f V
w s
w s
w s
du
eepdu
eepd
e
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Design Procedure for Shear Design Procedure for Shear
(9) Use smallest spacing from steps 6,7,8
Note: A practical limit to minimum stirrup
spacing is 4 inches.
L ti f M i Sh fL ti f M i Sh f
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Location of Max imum Shear for Location of Max imum Shear for B eam Design B eam Design
Non-pre-stressed members:
Sections located less than a distance d from face of
support may be designed for same shear, Vu, as the
computed at a distance d.
Compression fan
carries load directly
into support.
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Location of Max imum Shear for B eam Design
The support reaction introduces compression
into the end regions of the member.
No concentrated load occurs with in d from
face of support .
1.
2.
When:
L ti f M i Sh f
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Location of Max imum Shear for B eam Design
Compression from support at bottom of
beam tends to close crack at support
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H omework H omework Determine the development length required for the bars
shown . f c =4-ksi and f y = 60-ksi. Check the anchorage
in the column. If it is not satisfactory, design an
anchorage using a 180o
hook and check adequacy.
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H omework H omework Considering the anchorage of the beam bars into a
column, determine the largest bar that can be used with
out a hook. f c = 3-ksi and f y= 40ksi
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H omework H omework A simple supported uniformly loaded beam carries a
total factored design load of 4.8 k/ft (including self-
weight) on a clear span of 34 ft. f c =3 ksi and f y=40 ksi.
Assume that the supports are 12 in wide and assumethat the bars are available in 30 ft lengths.
Design a rectangular beam
Determine bar cutoffs.Locate splices and determine the lap length.