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Introduction
Ultimate Limit States Lead to collapseServiceability Limit States Disrupt use of Structures
but do not cause collapse
Recall:
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Introduction
Types of Serviceability Limit States
- Excessive crack width
- Excessive deflection
- Undesirable vibrations
- Fatigue (ULS)
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Crack Width Control
Cracks are caused by tensile stresses due to loads moments,shears, etc..
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Crack Width Control
Cracks are caused by tensile stresses due to loads moments,shears, etc..
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Crack Width Control
Bar crack development.
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Crack Width Control
Temperature crack development
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Crack Width Control
Appearance (smooth surface > 0.01 to 0.013public concern)
Leakage (Liquid-retaining structures)
Corrosion (cracks can speed up occurrence ofcorrosion)
Reasons for crack width control?
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Crack Width Control
Chlorides ( other corrosive substances) present
Relative Humidity > 60 % High Ambient Temperatures (accelerates
chemical reactions)
Wetting and drying cycles Stray electrical currents occur in the bars.
Corrosion more apt to occur if (steel oxidizes rust )
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L imits on Crack Width
0.016 in. for interior exposure
0.013 in. for exterior exposure
max.. crack width =
ACI Codes Basis
Cracking controlled in ACI code by regulating the
distribution of reinforcement in beams/slabs.
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L imits on Crack Width
Gergely-Lutz Equation
Crack width in units of 0.001 in.Distance from NA to bottom
(tension) fiber, divided by
distance to reinforcement.
=(h-c)/(d-c)
Service load stress in
reinforcement in ksi
= =
fs=
3cs076.0 Adf=
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L imits on Crack Width
Gergely-Lutz Equation
Distance from extreme tension
fiber to center of reinforcement
located closest to it, (in.)
effective tension area of
concrete surrounding tensionbars (w/ same centroid)
divided by # bars. (for 1 layer
of bars A = (2dc
b)/n
dc=
A=
3cs076.0 Adf=
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L imits on Crack Width
ACI Code Eqn 10-5 ( limits magnitude of z term )
Note: = 0.076 z
( =1.2 for beams)
Interior exposure: critical crack width = 0.016 in.
( = 16 ) z = 175k/in
Exterior exposure: critical crack width = 0.013 in.
( = 13) z = 145k/in
3cs Adfz=
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L imits on Crack Width
Tolerable Crack Widths
Tolerable
Crack Width
Dry air or protective membrane - 0.016 in.
Humidity, moist air, soil - 0.012 in.
Deicing chemicals - 0.007 in.
Seawater and seawater spray - 0.006 in.wetting and drying
Water-retaining structures - 0.004 in.
(excluding nonpressure pipes)
Exposure Condition
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L imits on Crack Width
Thin one-way slabs: Use =1.35
z = 155 k/in (Interior Exposure)
z = 130 k/in (Exterior Exposure)
fs= service load stress may be taken as
1.55average load factor
- strength reduction
. factor for flexure
=
0.90
1.5560.0 yys
fff
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Example-Crack
Given: A beam with bw= 14 in. Gr 60 steel 4 #8
with 2 #6 in the second layer with a #4 stirrup.
Determine the crack width limit, z for exteriorand interior limits (145 k/in and 175 k/in.).
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Example-Crack
Compute the center of the steel for the given bars.
) )s
2 2
2
4#8 bars 2#6 bars
4 0.79 in 2 0.44 in
4.04 in
A =
=
=
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Example-Crack
The locations of the center of the bars are
b1 stirrup
b b2
2 b
cover d2
1.0 in.1.5 in. 0.5 in.2
2.5 in.
2.5 in. 2 2
1.0 in. 0.75 in.2.5 in. 1.0 in.
2 2
4.375 in.
dy
d d
y d
=
=
=
=
=
=
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Example-Crack
Compute the center of the steel for the given bars.
) ) ) )
i i
i
2 2
2
2.5 in. 4 0.79 in 4.375 in. 2 0.44 in
4.04 in
2.91 in.
y Ay
A=
=
=
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Example-Crack
Compute number of equivalent bars, n. Use the
largest bar.
Compute the effective tension area
2
i2
bar
4.04 in 5.110.79 in
An
A= = =
) ) 22 2.91 in. 14 in.215.93 in
5.11
ybA
n= = =
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Example-Crack
The effective service load stress is
Compute the effective tension area
)s y0.60 0.6 60 ksi 36 ksif f= = =
) ) )23 3
s c 36 ksi 2.5 in. 15.93 in122.9 k/in.
z f d A= =
=
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Example-Crack
The limits magnitude of z term.
122.9 k/in. < 145 k/in. - Interior exposure
122.9 k/in. < 175 k/in. - Exterior exposure
Crack width is
or = 0.0112 in.
) )
0.076
0.076 1.2 122.9
11.2
w z=
=
=
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Deflection Control
Visual Appearance
( 25 ft. span 1.2 in. )
Damage to Non-structural Elements
- cracking of partitions
- malfunction of doors /windows
(1.)
(2.)
Reasons to Limit Deflection
visiblegenerallyare*
250
1l
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Deflection Control
Disruption of function
- sensitive machinery, equipment
- ponding of rain water on roofsDamage to Structural Elements
- large s than serviceability problem
- (contact w/ other members modify
load paths)
(3.)
(4.)
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Allowable Deflections
ACI Table 9.5(a) = min. thickness unless s are
computed
ACI Table 9.5(b) = max. permissible computeddeflection
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Allowable
Deflections
F lat Roofs( no damageable nonstructural elements
supported)
)
180instLL
l
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Allowable
Deflections
Floors( no damageable nonstructural elements
supported )
)
180instLL
l
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Allowable Deflections
Roof or F loor elements(supported nonstructural elementslikely damaged by large s)
480l
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Allowable Deflections
Roof or F loor elements( supported nonstructural elementsnot likely to be damaged by large
s )
240
l
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Allowable Deflections
Deflection occurring after attachment of
nonstructural elements
Need to consider the specific structures
function and characteristics.
allow
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Moment of I nertia for Deflection Calculation
For (intermediate values of EI)gtecr III
Brandon
derived cr
a
a
cr
gt
a
a
cr
e *1* IM
M
IM
M
I
-
=
Cracking Moment =Moment of inertia of transformed cross-section
Modulus of rupture =
t
gr
y
If
c5.7 f
Mcr=Igt =
fr =
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Moment of I nertia for Deflection Calculation
cr
a
a
crgt
a
a
cre *1* I
M
MI
M
MI
-
=
Distance from centroid to extreme tension fiber
maximum moment in member at loading stage for
which Ie( ) is being computed or at any previous
loading stage
Moment of inertia of concrete section neglect
reinforcement
yt =
Ma =
Ig =
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Moment of I nertia for Deflection Calculation
)
3
a
crcrgcre
cr
3
a
crg
3
a
cre
or
*1*
-=
-
=
M
MIIII
I
M
MI
M
MI
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Moment Vs curvature plot
EIM
EI
M===
slope
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Moment Vs Slope Plot
The cracked beam starts to lose strength as the amountof cracking increases
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Moment of I nertia
For normal weight concrete
)psi33 c1.5
cc fE =
For wc= 90 to 155 lb/ft3
)psi57000 cc fE =
(ACI 8.5.1)
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Deflection Response of RC Beams (F lexure)
A- Ends of Beam CrackB- Cracking at midspan
C- Instantaneous
deflection under serviceload
C- long time deflection
under service loadDand E- yielding of
reinforcement @ ends &
midspan
Note: Stiffness (slope) decreases
as cracking progresses
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Deflection Response of RC Beams (F lexure)
The maximum moments for distributed load actingon an indeterminate beam are given.
=12
2wlM
=
12
2
wlM
=
24
2
wlM
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Uncracked Transformed Section
Part (n) =Ej/Ei Area n*Area yi yi*(n)A
Concrete 1 bw*h bw*h 0.5*h 0.5*bw*h2
As n As (n-1)As d (n-1)*As*d
As n As (n-1)As d (n-1)*As*d
An *ii Any **
=
*ii
*
iii *
An
Anyy
Note:(n-1) is to remove area
of concrete
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Cracked Transformed Section
== s
s
i
ii 2nAyb
dnAy
yb
A
Ayy
Finding the centroid of singly Reinforced RectangularSection
022
0
2
2
ss2
ss
2
ss
2
=-
=-
=
b
dnAy
b
nAy
dnAynAyb
dnAy
ybynAyb
Solve for the quadratic for y
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Cracked Transformed Section
022 ss2 =-
b
dnAy
b
nAy
Note:
c
s
E
En=
Singly Reinforced Rectangular Section
)2s3
cr
3
1ydnAybI -=
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Cracked Transformed Section
) )0
212212 ssss2 =-
--
b
dnAAny
b
nAAny
Note:
c
s
E
En=
Doubly Reinforced Rectangular Section
) ) )2s2
s
3
cr 1
3
1ydnAdyAnybI ---=
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Uncracked Transformed Section
) ) ) )
steel
2
s
2
s
concrete
2
3
gt
11
212
1
dyAndyAn
hybhbhI
----
-=
Note: 3g
12
1 bhI =
Moment of inertia (uncracked doubly reinforced beam)
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Cracked Transformed Section
Finding the centroid of doubly reinforced T-Section
) )
) )0
212
2122
w
ss
2
we
w
sswe2
=
--
-
--
b
dnAAntbb
y
b
nAAnbbty
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Cracked Transformed Section
Finding the moment of inertia fora doubly reinforced T-Section
)
) ) )
steel
2
s
2
s
beam
3
w
flange
2
e
3
ecr
1
3
1
212
1
ydnAdyAn
tybt
ytbybI
---
-
-=
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Reinforced Concrete Sections - Example
Given a doubly reinforced beam with h = 24 in, b = 12 in.,
d = 2.5 in. and d = 21.5 in. with 2# 7 bars in compression
steel and 4 # 7 bars in tension steel. The material
properties are fc= 4 ksi and fy= 60 ksi.
Determine Igt, Icr, Mcr(+), Mcr(-), and compare to the NA of
the beam.
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Reinforced Concrete Sections - Example
The components of the beam
)
)
2 2
s
2 2
s
c c
2 0.6 in 1.2 in
4 0.6 in 2.4 in
1 k57000 57000 4000
1000 lb
3605 ksi
A
A
E f
= =
= =
= =
=
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Reinforced Concrete Sections - Example
The compute the n value and the centroid, I uncracked
n area (in ) n*area (in ) yi(in) yi*n*area (in2) I (in ) d (in) d *n*area(in )
A's 7.04 1.2 8.448 2.5 21.12 - -9.756 804.10
As 7.04 2.4 16.896 21.5 363.26 - 9.244 1443.75
Ac 1 288 288 12 3456.00 13824 -0.256 18.89
313.344 3840.38 13824 2266.74
s
c
29000 ksi8.04
3605 ksi
En
E= = =
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Reinforced Concrete Sections - Example
The compute the centroid and I uncracked
3i i i
2
i i
2 4 4
i i i i
4
3840.38 in12.26 in.313.34 in
I 13824 in 2266.7 in
16090.7 in
y n A
y n A
I d n A
= = =
= =
=
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Reinforced Concrete Sections - Example
The compute the centroid and I for a cracked doublyreinforced beam.
) )0
212212 ssss2 =-
--
b
dnAAny
b
nAAny
) ) ) )
) ) ) ) )
2 2
2
2 2s
2
2 7.04 1.2 in 2 8.04 2.4 in
12 in.
2 7.04 1.2 in 2 8.04 2.4 in 21.5 in.0
12 in.
4.624 72.664 0
y y
y y
- =
- =
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Reinforced Concrete Sections - Example
The compute the centroid for a cracked doubly reinforced
beam.
2
4.624 72.664 0y y - =
) )2
4.624 4.624 4 72.664
26.52 in.
y-
=
=
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Reinforced Concrete Sections - Example
The compute the moment of inertia for a cracked doublyreinforced beam.
) ) )2s2
s
3
cr 1
3
1ydnAdyAnybI ---=
) )
) ) ) ) ) )
3
cr
22
22
4
112 in. 6.52 in.
3
7.04 1.2 in 6.52 in. 2.5 in.
8.04 2.4 in 21.5 in. 6.52 in.
5575.22 in
I =
-
-
=
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Reinforced Concrete Sections - Example
The critical ratio of moment of inertia
4
cr
4g
cr g
5575.22 in
0.34616090.7 in
0.35
I
I
I I
= =
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Reinforced Concrete Sections - Example
Find the components of the beam ) ) )
)
) )
c c
s
s s s
2
s s s c
0.85 0.85 4 ksi 12 in. 0.85 34.68
2.5 in. 0.00750.003 0.003
0.0075 217.529000 0.003 87
217.50.85 1.2 in 87
261
100.32
C f ba c c
cc c
f Ec c
C A f f c
c
= = =
- = = -
= = - = -
= - = -
= -
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Reinforced Concrete Sections - Example
Find the components of the beam
) )
) ) )
)
2
c s
2
2
2.4 in 60 ksi 144 k
261144 k 34.68 100.32 34.68 43.68 261 0
43.68 43.68 4 261 34.68
2 34.68
3.44 in.
T
T C C
c c cc
c
= =
=
= - - - =
=
=
The neutral axis
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Reinforced Concrete Sections - Example
The strain of the steel
Note: At service loads, beams are assumed to act
elastically.
)
)
s
s
3.44 in. 2.5 in.0.003 0.0008 0.00207
3.44 in.
21.5 in. 3.44 in. 0.003 0.0158 0.002073.44 in.
- = =
- = =
3.44 in.
y 6.52 in.
c=
=
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Reinforced Concrete Sections - Example
Using a linearly varying and s= E along the NA is the
centroid of the area for an elastic center
The maximum tension stress in tension is
r c7.5 7.5 4000
474.3 psi 0.4743 ksi
f f= =
=
My
I
s= -
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Reinforced Concrete Sections - Example
The uncracked moments for the beam
) )
)
)
)
4
r
cr
4
rcr
0.4743 ksi 16090.7 in650.2 k-in.
24 in. 12.26 in.
0.4743 ksi 16090.7 in
622.6 k-in.12.26 in.
My IM
I y
f IM
y
f IM y
ss
-
= =
= = =-
= = =
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Homework-12/2/02
Problem 8.7
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