Download - Reinforced concrete Serviceability
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Elastic Flexural Analysis for Serviceability
Lecture Goals
Serviceability
Crack width
Moments of inertia 1
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Introduction
Ultimate Limit States Lead to collapse
Serviceability Limit States Disrupt use of Structures
but do not cause collapse
Recall:
Types of Serviceability Limit States
- Excessive crack width
- Excessive deflection
- Undesirable vibrations
- Fatigue (ULS) 2
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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 0.013 public concern)
Leakage (Liquid-retaining structures)
Corrosion (cracks can speed up occurrence of corrosion)
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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Limits on Crack Width
0.4mm (0.016 in.) for interior exposure
0.33mm (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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Limits 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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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
1 are generally visible
250l
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Deflection Control
Disruption of function
- sensitive machinery, equipment
- ponding of rain water on roofs
Damage to Structural Elements
- large s than serviceability problem
- (contact w/ other members modify
load paths)
(3.)
(4.)
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Allowable Deflections
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Allowable Deflections
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Deflection Response of RC Beams (Flexure)
A- Ends of Beam Crack
B - Cracking at midspan
C - Instantaneous
deflection under service
load
C - long time deflection
under service load
D and E - yielding of
reinforcement @ ends &
midspan
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2wlM
12
2wlM
2
16
wlM
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Moment of Inertia
For normal weight
concrete
psi 33 c1.5
cc fE
For wc = 90 to 155 lb/ft3
psi 57000 cc fE
(ACI 8.5.1)
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EIM
EI
M
slope
Moment Vs curvature plot
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Moment Vs Slope Plot
The cracked beam starts to lose strength as the amount of
cracking increases
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Variation of EI along the length of the beam
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Moment of Inertia for Deflection Calculation
For (intermediate values of EI) gtecr III
Brandon
derived
Cracking Moment =
Moment of inertia of transformed cross-section
Modulus of rupture =
r g
t
f I
y
c7.5 f
Mcr =
Igt =
fr =
3 3
cr cre g cr
a a
1M M
I I IM M
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Moment of Inertia for Deflection Calculation
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 =
3 3
cr cre g cr
a a
1M M
I I IM M
3
cre cr g cr
a
MI I I I
M
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Deflection Response of RC Beams (Flexure)
For Continuous beams
ACI 9.5.2.4
ACI Com. 435
Weight Average
e21emideavge 25.050.0 IIII
e21emideavge 15.070.0
:continous ends 2
IIII
1emideavge 15.085.0
:continous end 1
III
2 end @
1 end @
midspan @
ee2
ee1
emide
II
II
II
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Uncracked Transformed Section
Part (n) =Ej /Ei Area nArea yi yi(n)A
Concrete 1 bwh bwh 0.5h 0.5bwh2
As n As (n-1)As d (n-1)Asd
As n As (n-1)As d (n-1)Asd
n A i iy n A
*
i i i
*
i i
y n Ay
n A
Note: (n-1) is to remove area of
concrete
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Cracked Transformed Section
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Cracked Transformed Section
s
s
i
ii 2
nAyb
dnAy
yb
A
Ayy
Finding the centroid of singly Reinforced Rectangular
Section
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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Example (Rectangular - Single)
For the following example find centroid and moment
of inertia for an uncracked and cracked section and
compare the results.
Es = 29000 ksi
Ec = 3625 ksi
d = 15.5 in b = 12 in. h = 18 in.
Use 4 # 7 bars for the steel.
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Example
A #7 bar has an As = 0.6 in2
s
c
E 29000 ksi8
E 3625 ksin
2 2s 4 0.6 in 2.4 inA
The uncracked centroid is
2
s
s
2
2
2
3
2
12
1
12 in 18 in8 1 2.4 in 15.5 in
2 12 in 18 in 8 1 2.4 in
2204.4 in 9.47 in
232.8 in
bhn A d
ybh n A
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Example The uncracked moment of inertia
232
s
3 22 2
4
112 2
12 in 18 in 18 in9.47 in 12 in 18 in 15.5 in 9.47 in 8 1 2.4 in
12 2
6491 in
bh hI y bh d y n A
The cracked centroid is defined by:
2
s 2.4 in 0.012912 in 15.5 in
A
bd
222 8 0.0129 2 8 0.0129 8 0.0129 0.3627
0.3627 15.5 in 5.62 in
yn n n
d
y
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Example The cracked moment of inertia is
32
s
3 2 2
4
3
112 in 5.62 in 15.5 in 5.62 in 8 2.4 in
3
2584.2 in
byI d y nA
Notice that the centroid changes from 9.47 in. to 5.62
in. and the moment of inertia decreases from 6491 in4
to 2584 in4 . The cracked section loses more than half
of its strength.
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Cracked Transformed Section
's s s s2 2 1 2 2 1 2 0n A nA n A d nA d
y yb b
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
1bhI
Moment of inertia (uncracked doubly reinforced beam)
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Example (Rectangular - Double)
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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 for a doubly reinforced
T-Section
233
cr e e w
beamflange
2 2
s s
steel
1 1
12 2 3
1
tI b y b t y b y t
n A y d nA d y
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Reinf. 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.
a) Determine Igt, Icr , Mcr(+), Mcr(-)
b) Determine the NA of the beam at ultimate Load.
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Reinf. 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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Reinf. Concrete Sections - Example
Compute the n value and the centroid, I uncracked
n area (in2) narea (in2) yi (in) yi n area (in
2) I (in4) d (in) d
2 n area(in4)
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
The compute the centroid and I uncracked 3
i i i
2
i i
2 4 4 4
i i i i
3840.38 in12.26 in.
313.34 in
I 13824 in 2266.7 in 16090.7 in
y n Ay
n A
I d n A
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Reinf .Concrete Sections - Example
Compute the centroid and I for a cracked doubly reinforced
beam.
's s s s2 2 1 2 2 1 2 0n A nA n A d nA d
y yb b
2 2
2
2 2
2
2 7.04 1.2 in 2 8.04 2.4 in
12 in.
2 7.04 1.2 in (2.5) 2 8.04 2.4 in 21.5 in.0
12 in.
4.624 72.664 0
y y
y y
2
4.624 4.624 4 72.6646.52 in.
2y
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Reinf. Concrete Sections - Example
Compute the moment of inertia for a cracked doubly
reinforced beam.
2s2
s
3
cr 1
3
1ydnAdyAnybI
3 22
cr
22 4
112 in. 6.52 in. 7.04 1.2 in 6.52 in. 2.5 in.
3
8.04 2.4 in 21.5 in. 6.52 in. 5575.22 in
I
The critical ratio of moment of inertia
4
crcr g4
g
5575.22 in0.346 0.35
16090.7 in
II I
I
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Reinf. Concrete Sections - Example
The maximum tension stress in tension is
r c7.5 7.5 4000 474.3 psi 0.4743 ksif f
The uncracked moments for the beam
4
r
cr
4
r
cr
0.4743 ksi 16090.7 in650.2 k-in.
24 in. 12.26 in.
0.4743 ksi 16090.7 in622.6 k-in.
12.26 in.
My IM
I y
f IM
y
f IM
y
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Reinf. Concrete Sections - Example
b) Find the components of the beam (at ultimate)
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
217.5 2610.85 1.2 in 87 0.85(4) 100.32
C f ba c c
c
c c
f Ec c
C A f fc c
Find the components of the beam
2
2
c s
2.4 in 60 ksi 144 k
261 144 k 34.68 100.32 34.68 43.68 261 0
T
T C C c c cc
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Reinf. Concrete Sections - Example
The neutral axis
243.68 43.68 4 261 34.68
3.44 in.2 34.68
c
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 0.00207
3.44 in.
3.44 in.
y 6.52 in.
c
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Calculate the Deflections
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
Instantaneous Deflections
due to dead loads (unfactored) , live, etc.
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Instantaneous (Immediate)
deflections
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Instantaneous (Immediate)
deflections
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Deflection Coefficients
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Sustained Load Deflections
Creep causes an increase in
concrete strain
Curvature
increases
Compression steel
present
Increase in compressive strains
cause increase in stress in
compression reinf. (reduces
creep strain in concrete) Helps limit this effect.
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Sustained Load Deflections
Sustained load deflection = D Di
Instantaneous deflection 1 50
D
ACI 9.5.2.5
bd
As at mid-span for simple and continuous beams
at support for cantilever beams
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Sustained Load Deflections
= time dependent factor for sustained load
5 years or more
12 months
6 months
3 months
1.4 1.2
1.0
2.0
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Sustained Load Deflections
For dead and live loads
total DL inst LL inst
DL L.T. LL L.T.
D D D
D D
DL and LL may have different factors for LT ( long
term) D calculations
instDLtotal
components N/S
of attachmentafter total
DDD
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Example 9-5:
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Example 9-6:
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