Lecture 21
–
Splices and
Shear
February 5, 2003
CVEN 444
Lecture Goals
Spice
Shear
Shear Design
Bar 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
Tension 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 are
overlapped.
Types of Splices
Class A Spice
(ACI 12.15.2)
When over entire splice
length.
and 1/2 or less of total reinforcement is
spliced win the req’d lay length.
2
d
req'
s
provided
s
A
A
Types of Splices
Class B Spice
(ACI 12.15.2)
All tension lay splices not meeting
requirements of Class A Splices
Tension Lap Splice
(ACI 12.15)
where
A
s (req’d)
= determined for bending
l
d
= development length for bars (not
allowed to use excess reinforcement
modification factor)
l
d
must be greater than or equal to 12 in.
Tension 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)
Compression Lap Splice
(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 psi
Lap, req’d 12 in
For f
c
3000 psi, required lap splice shall be multiply
by (4/3) (ACI 12.16.1)
Compression Lap Splice
(ACI 12.17)
In tied column splices with effective tie area throughout
splice length 0.0015 h
s
factor = 0.83
In spiral column splices,
factor = 0.75 But final splice
length 12 in
.
Example
–
Splice Tension
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 A
s(provided)
/A
s(required)
>2
When 4 bars are spliced and A
s(provided)
/A
s(required)
< 2
When all bars are spliced at the same location.
f
c
= 5 ksi and f
y
= 60 ksi
a.
b.
c.
Example
–
Splice Tension
For #8 bars, db =1.0 in. and
a
= ?
b
=?
g
= ?
l
=?
y
d
tr
b
c
b
3
40
f
l
c K
d
f
d
abgl
Example
–
Splice Tension
The A
s(provided)
/A
s(required)
> 2, class ? splice applies;
The A
s(provided)
/A
s(required)
< 2, class ? splice applies;
Example
–
Splice
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
Example
–
Splice
Compression
For #10 bars, d
b
=? in.
y
d
y
b
c
0.02
0.003
f
l
f
d
f
Check l
s
> 0.005 d
b
f
y
Example
–
Splice
Compression
For #10 bars, d
b
=? in. The l
d
= 2? in.
Check l
s
> (0.0009 f
y
–
24) d
b
So use l
s
= ? in.
Shear Design
Uncracked Elastic Beam
Behavior
Look at the shear and
bending moment
diagrams. The acting
shear stress distribution
on the beam.
Uncracked Elastic Beam
Behavior
The acting stresses distributed across the
cross

section.
The shear stress
acting on the
rectangular beam.
Ib
VQ
Uncracked Elastic Beam
Behavior
The equation of the shear stress for a rectangular beam
is given as:
Note:
The maximum
1st moment occurs at
the neutral axis (NA).
Ib
VQ
ave
max
2
max
3
5
.
1
*
2
3
8
4
*
2
Q
Inertia
of
Moment
12
bh
V
bh
h
bh
bh
I
Uncracked Elastic Beam
Behavior
The ideal shear stress distribution can be described as
:
Ib
VQ
Uncracked Elastic Beam
Behavior
A realistic description of the shear distribution is shown
as:
Uncracked Elastic Beam
Behavior
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.
Inclined Cracking in
Reinforced Concrete Beams
Typical Crack Patterns for a deep beam
Inclined Cracking in
Reinforced Concrete Beams
Flexural

shear crack

Starts out as a flexural
crack and propagates due
to shear stress.
Flexural cracks in beams
are vertical
(perpendicular to the
tension face).
Inclined Cracking in
Reinforced Concrete Beams
For deep beam the cracks are given as:
The shear cracks Inclined (diagonal) intercept crack
with longitudinal bars plus vertical or inclined
reinforcement.
Inclined Cracking in
Reinforced Concrete Beams
For deep beam the cracks
are given as:
The shear cracks fail due
two modes:

shear

tension failure

shear

compression
failure
Shear Strength of RC Beams
without Web Reinforcement
v
cz

shear in compression
zone
v
a

Aggregate Interlock
forces
v
d
= Dowel action from
longitudinal bars
Note
: v
cz
increases from
(V/bd) to (V/by) as crack
forms.
Total Resistance = v
cz
+ v
ay
+ v
d
(
when no stirrups are used
)
Strength of Concrete in Shear
(No Shear Reinforcement)
(1) Tensile Strength of concrete affect inclined
cracking load
Strength of Concrete in Shear
(No Shear Reinforcement)
(2) Longitudinal Reinforcement Ratio,
r
w
d
b
f
V
d
b
A
w
c
c
w
w
s
w
2
:
0025
.
0
0075
.
0
for
cracks
restrains
r
r
Strength of Concrete in Shear
(No Shear Reinforcement)
(3) Shear span to depth ratio, a/d (M/(Vd))
2
d
a
2
d
a
Deep shear spans
more detail design
required
Ratio has little
effect
Strength of Concrete in Shear
(No Shear Reinforcement)
(4) Size of Beam
Increase Depth Reduced shear stress at
inclined cracking
Strength of Concrete in Shear
(No Shear Reinforcement)
(5) Axial Forces

Axial tension Decreases inclined cracking load

Axial Compression
Increases inclined cracking
load (Delays flexural
cracking)
Function and Strength of
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 “clamps” to keep shear cracks from
widening
Function:
Function and Strength of
Web Reinforcement
Uncracked Beam Shear is resisted uncracked
concrete.
Flexural Cracking Shear is resisted by v
cz
, v
ay
,
v
d
bars.
al
longitudin
from
Action
Dowl
force
Interlock
Aggregate
of
component
Vertical
zone
n
compressio
in
Shear
d
ay
cz
V
V
V
Function and Strength of
Web Reinforcement
Flexural Cracking Shear is resisted by
v
cz
, v
ay
, v
d
and v
s
V
s
increases as cracks
widen until yielding of
stirrups then stirrups
provide constant
resistance.
Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
n u
capacity demand
V V
u
n
factored shear force at section
Nominal Shear Strength
0.75 shear strength reduction factor
V
V
Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
n c s
V V V
c
s
V
V
Nominal shear provided by the shear reinforcement
Nominal shear resistance provided by concrete
Shear Strength Provided by
Concrete
Bending only
Simple formula
More detailed
Note:
Eqn [11.5]
Eqn [11.3]
d
b
f
d
b
f
V
w
c
w
c
c
3.5
2
d
b
f
d
b
M
d
V
f
V
w
c
w
u
u
w
c
c
3.5
2500
1.9
r
1
u
u
M
d
V
Shear Strength Provided by
Concrete
Bending and Axial Compression
N
u
is positive for
compression and
N
u
/A
g
are in psi.
Simple formula
Eqn [11.4]
Eqn [11.7]
g
u
w
c
w
c
g
u
c
500
1
3.5
2000
1
2
A
N
d
b
f
d
b
f
A
N
V
Typical Shear Reinforcement
Stirrup

perpendicular to axis of members
(minimum labor

more material)
ACI Eqn 11

15
s
d
f
A
V
a
a
cos
sin
y
v
s
s
d
f
A
V
y
v
s
o
90
a
Typical Shear Reinforcement
Bent Bars (more labor

minimum material) see req’d
in 11.5.6
ACI 11

5.6
s
d
f
A
V
a
a
cos
sin
y
v
s
s
d
f
A
V
y
v
s
o
41
.
1
45
a
Stirrup Anchorage Requirements
V
s
based on assumption stirrups yield
Stirrups must be well anchored.
Stirrup Anchorage Requirements
each bend must enclose a long bar
# 5 and smaller can use standard hooks 90
o
,135
o
, 180
o
#6, #7,#8( f
y
= 40 ksi )
#6, #7,#8 ( f
y
> 40 ksi ) standard hook plus a
minimum embedment
Refer to Sec. 12.13 of ACI 318 for development of web
reinforcement. 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
Design Procedure for Shear
(1) Calculate V
u
(2) Calculate
V
c
Eqn 11

3 or 11

5 (no axial force)
(3) Check
c
u
V
V
2
1
If yes, add web reinforcement (go to 4)
If no, done.
Design Procedure for Shear
(4)
c
u
c
V
V
V
2
1
If
v
w
ys
v
y
w
v
A
b
f
A
s
f
s
b
A
min
for
50
or
50
max
min
Also:
(Done)
11.5.4
"
24
2
max
d
s
Provide minimum
shear reinforcement
Design Procedure for Shear
(5)
c
u
s
c
u
s
s
c
n
u
s
c
u
V
V
V
V
V
V
V
V
V
V
V
V
V
d)
(req'
calulate
,
If
Check:
11.5.4
illegal
otherwise,
8
d
b
f
V
w
c
s
Design Procedure for Shear
Solve for required stirrup spacing(strength)
Assume # 3, #4, or #5 stirrups
s
ys
v
V
d
f
A
s
(6)
from 11

15
Design Procedure for Shear
(7) Check minimum steel requirement (eqn 11

13)
50
max
w
ys
v
b
f
A
s
Design Procedure for Shear
(8) Check maximum spacing requirement (ACI 11.5.4)
illegal
8
If
:
Note
"
12
4
4
If
"
24
2
4
If
c
max
c
max
c
d
b
f
V
d
s
d
b
f
V
d
s
d
b
f
V
w
s
w
s
w
s
Design Procedure for Shear
(9) Use smallest spacing from steps 6,7,8
Note
: A practical limit to minimum stirrup
spacing is 4 inches.
Location of Maximum Shear for
Beam Design
Non

pre

stressed members
:
Sections located less than a distance d from face of
support may be designed for same shear, V
u
, as the
computed at a distance d.
Compression fan
carries load directly
into support
.
Location of Maximum Shear for
Beam 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
:
Location of Maximum Shear for
Beam Design
Compression from support at bottom of
beam tends to close crack at support
Homework
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 180
o
hook and check adequacy.
Homework
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
Homework
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 assume
that the bars are available in 30 ft lengths.
Design a rectangular beam
Determine bar cutoffs.
Locate splices and determine the lap length.
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