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Serviceability
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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
Heat of hydration cracking
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Crack Width Control
Bar crack development.
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Crack Width Control
Appearance (smooth surface > 0.25 to 0.33mm= public 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
0.40 mm for interior exposure0.33 mm for exterior exposure
max.. crack width =
ACI Codes Basis Prior to 1999
Now ACI handles crack width
indirectly by limiting the bar spacings and bar cover for beams and
one way slabs ACI 10.6.4.
Bar spacings must also satisfy ACI 7.6.5 (3t or 450mm)
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Example 1 (9-4)
A 20cm thick slab has 12mm diameter bars. The
bars have 420MPa yield stress and a minimum
clear cover of 20mm. Compute the maximumvalue of s.
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Other important issues for crack control
1. Negative moment regions of T-beams.
2. Shrinkage and temperature reinforcement: is intended to
replace the tensile stresses in the concrete
at the time of cracking, using the following
simplified analysis:
For grade 60 steel and 28MPa concrete,
Steel ratio is between 0.004 and 0.005.
This limit is about three times that specified by ACI code
7.12.2.1 which is based on empirical results.
s y g t
s t
g y
A f A f
A f
A f
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3. Web face reinforcement:
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Deflection Control
Visual Appearance
( 7.5m. span 30mm )
Damage to Non-structural Elements
- cracking of partitions
- malfunction of doors /windows
(1.)
(2.)
Reasons to Limit Deflection (Table 9-3)
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 arecomputed
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Allowable Deflections
ACI Table 9.5(b) = max. permissible
computed deflection
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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
2wlM
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Deflection Response of RC Beams (F lexure)
A- Ends of Beam CrackB - Cracking at midspan
C - Instantaneous deflection
under service load
C - long time deflection under
service load
D and E - yielding of
reinforcement @ ends &midspan
Note: Stiffness (slope) decreases as cracking progresses
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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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To avoid complexity in calculations, an overall
average effective moment of inertia
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Moment of I nertia for Deflection Calculation
For (intermediate values of EI)gecr III
Branson
derived cr
3
a
crg
3
a
cre *1* I
M
MI
M
MI
Cracking Moment =
Gross moment of inertia of rc cross-section
Modulus of rupture =
t
gr
y
If
c0.62 f
Mcr=
Ig =
fr =
IfMa /Mcr> 3, the cracking will be extensive,Ie =Icr
IfMa /Mcr< 1, no cracking is likely andIe =Ig
f
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Moment of Inertia
for Deflection
Calculation
Distance from centroid to extreme tension fiber
maximum moment in member at loading stage forwhich Ie ( ) is being computed or at any previousloading stage
yt =
Ma =
3 3
cr cr
e g cra a
3
cre cr g cr
a
* 1 * ,or
, .9.8
M M
I I IM M
MI I I I Eq
M
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Deflection Response of RC Beams (Flexure)
e21emideavge 15.070.0
:continousends2
IIII
e continuouse avg e mid1 end continous:
0.85 0.15I I I
e ei ee mid @ midspan, @ end iI I I I
ACI Com. 435
Weight Average
ACI code
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Definition ofIgACI code: Ig is the moment of inertia of the gross concrete section
neglecting area of tension steel.
Ig might be more accurate if it includes the transformed area of the
reinforcement.
Ig is the moment of inertia of the uncracked transformed section. The
transformed section consists of the concrete area plus the transformed
steel area(=the actual steel area times the modular ratio n =Es /Ec :
Es = 200GPa ,Ec =4700fc ).
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Definition ofIOnce a beam has been cracked by a large moment, it can never
return to its original uncracked state; therefore, the effective
moment of inertiaIe that should be used in deflection
computations must always be equal to the effective moment ofinertia associated with the maximum past moment to which the
beam has been subjected. Often this moment is impossible to
determine for most beams.
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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 fory
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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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Example 2 (9-1)
For the shown beam of 28MPa concrete, Find:
1. Moment of inertia of uncracked section.2. Moment of inertia of cracked section.
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Example 3
(9-2)
For the shown beam of 31.5MPa concrete, Find steel stress at
service loads if the service live-load moment is 70kN.m and
the service dead load moment is 96kN.m
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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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Calculate the Deflections
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
Instantaneous Deflections
due to dead loads( unfactored) , live, etc.
Calculate the
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Calculate the
Deflections
Instantaneous
Deflections
Equations for
calculating Dinst forcommon cases
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Sustained Load Deflections
Creep causes an increase
in concrete strainCurvature
increases
Compression steel
present
Increase in compressive
strains cause increase in
stress in compression
reinforcement (reducescreep strain in concrete)Helps limit thiseffect.
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Sustained Load Deflections
Sustained load deflection = l Di
Instantaneous deflection
l 501 ACI 9.5.2.5
bd
As
at midspan 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
1.41.21.0
2.0
Also see Figure
9.5.2.5 from
ACI code
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The total long time deflection
where
L = immediate live load deflection
D
= immediate dead load deflection
SL = sustained live load deflection (a percentage of the immediate
L determined by expected duration of sustained load)
= time dependant multiplier for infinite duration of sustained
load
t = time dependant multiplier for limited load duration
To calculate L (orSL) due to the live loads, the following procedure
has been found to be generally satisfactory:
LT L D t SL= + + l l
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Calculation of long time deflection
1. Calculate the deflection D+L due to dead and live loads acting
simultaneously. For this calculationIe is found using Eq. 9.8 andthe momentMa is the one produced when both dead and live
loads are acting simultaneously.
2. Calculate the deflection D due to the dead load acting alone. For
this calculationIe is found using Eq. 9.8 and the momentMa isthe one produced when the dead load acts alone.
3. Subtract the deflection D from the deflection D+L to obtain the
desired deflection L.
If the long time deflections exceeds the value permitted, thedesigner may either increase the depth of members, or add
additional compression steel. If the sag produced by the long time
deflections is objectionable from an architectural or functional point
of view, forms may be raised (cambered) a distance equal to that of
the anticipated deflection.
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Example
4 (9-5)
The T-beam shown in Fig. is made of 28MPa concrete and
supports unfactored dead and live loads of 13kN/m and
18kN/m. Compute the immediate midspan deflection. Assume
that the construction loads did not exceed the dead load.
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Example 5 (9-5)If the beam in the previous example is assumed to support
partitions that would be damaged by excessive deflections. If 25%
of the live load is sustained. The partitions are installed at least 3
months after the shoring is removed. Will the computed
deflections exceed the allowable in the end span?
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Problem 1 (9-8 + 9-9)9-8 A simply supported beam with the cross section shown in Figure
next page has a span of 7.5m and supports an unfactored dead load
of 22.5kN/m, including its own self-weight plus an unfactored live
load of 22.5kN/m. The concrete strength is 31.5MPa. Compute
1. the immediate dead load deflection.
2. the immediate dead-plus-live load deflection
3. the deflection occurring after partitions are installed. Assume
that the partitions are installed two months after shoring for the
beam is removed and assume that 20 percent of the live load is
sustained.
9-9 Repeat Problem 9-8 for a beam having the same dimensions and
tension reinforcement, but with two No. 25mm bars as compression
reinforcement.
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Problem 9-10The beam shown in Figure next page is made of 28MPa
concrete and supports unfactored dead and live loads of
15kN/m and 17kN/m respectively. Compute
(a) the immediate dead-load deflection.
(b) the immediate dead-plus-live load deflection.
(c)the deflection occurring after partitions are installed.
Assume that the partitions are installed four months after
the shoring is removed and assumed that 10 percent of thelive load is sustained.
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Problem 9.10