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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.