5-flexural analysis service slides 2015.pdf
TRANSCRIPT
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Flexural Analysis of
Prestressed Concrete underService Loads
CEE 453University of Washington
Prof. Marc Eberhard
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Lecture Topics
Beam Behavior
Goals of Service Load Analysis
Prestress Losses
Stresses by Direct Calculation
Stresses by Equivalent Loads
Stresses by Pressure Line (C line)
Kerns
Composite Construction (SKIP FOR NOW)
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Self-weight
Decompression
Balanced
Cracking
Yield
Failure
Concrete becomes non-linear
fr
fpy fps
Curvature
Moment
Typical Beam Behavior
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Tendon
stress
Time
Jacking
Service
Relaxation + thermal
Release PS (initial)
Elastic shortening
Live Load
Shrinkage, creep, relaxation
Typical Beam Behavior
fpj
fpi
fpe
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Goals of Service Load Analysis
Find stresses in the concrete and steel
various times during life of girdervarious x-sections along beam
various (critical) locations within x-section
Compare with (Code) allowable values
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Check stresses at various times:
Jacking (tendon only, fpj)
Initial (directly after release, fpi)
Service (after all PS losses, fpe)
Need to know how prestress is lost between
stages to know how much remains.
Key Times
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Prestress Losses
Calculation Method Pre- Post-tensioned
Lump-Sum Methods
AASHTO 45 ksi 23 - 33 ksi + friction
ASCE/ACI 423 35 ksi 25 ksi + friction
PTI (-) 15 - 35 ksi + friction
Simplified Methods
AASHTO AASHTO LRFD Specifications
PCI Simplified PCI Jo. July 1975
Zia et al. Conc. Int. June 1979
Time-Step Methods
PCI general PCI Jo. July 1975
Ghali et al. PCI Jo. March 1977
NCHRP 446
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Prestress Losses
In this class we do not have time to address
prestress losses in detail. We will:
Use a given value of fpi(will be given in the problem.
Use a given value of h= fpe/fpi(given in the problem).
Typically
fpi 0.9 fpjh 0.85
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Allowable Stresses
ACI and AASHTO specify allowable stresses (as functions offciand fc) to ensure good behavior at service loads.
In our notation:
fci= allowable compression initial (-ve)
fti = allowable tension initial (+ve)
fcs= allowable compression service (-ve)fts = allowable tension service (+ve)
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Allowable Stresses
Design for service loads:Make sure all stresses lie within allowable values:
fci< f1i< fti
Initial conditions: Actual stress
Allowable stress
fci< f2i< fti
Bottom Top
Service conditions: Similar equations
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STRESSES BY DIRECTCALCULATION
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Linear Elastic Material (superposition)
Uncracked
nAp
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PS Alone (M=0)
Prestressing
Moment ppccps FeFeM
Prestressing
Axial Force
c pF F pc
axial
c c
FF
f A A
bMc M
f
I S
FcFp
ec= ep
cgc
cgp
PS alone
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PS Alone (M=0)
FcFp
ec= ep
cgc
cgp
PS alone
S
Ae
A
F
S
Fe
A
Ff
ppppp1
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PS + Applied Moment
FcFp
ec= ep
cgc
cgp
PS alone PS + Mex t
ecFc
ep
Fp
1p p p p pext ext
F e F F e AM Mf
A S S A S S
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Notation and Sign Convention
cgc = origin
y = positive down
Section properties are signed
even powers are always positive (A, I, Cw)
odd powers are signed according to y.
Tension is positive (force, stress, strain deformation)
Beam sign convention Locations numbered from bottom up (but cgc = 0)
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Example: Cross Section30
4
20
cgc
8
6
2
0
1
Ac= 240 in.2
c1= +16.0 in.
c2= -8.0 in.
I= 12,800 in.4
r2
= I/A= +12,800/(+240) = 53.333 in.4
S1= I/c1= +12,800/(+16) = 800 in.3
S2= I/c2=+12,800/(-8) = -1600 in.3
Note the signs on c and S
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Notation and Sign Convention
Stresses due to prestressing alone
S
Ae
A
F
S
Fe
A
Ff
ppppp1
11
1 1
S
Ae
A
F
S
Fe
A
Ff
ppppp
22
2 1
S
Ae
A
F
S
Fe
A
Ff
ppppp
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Example: Prestressing Alone
ksiS
Fe
A
F
f ppp
475.0975.05.01600
120*13
240
120
2
2
ksiS
Fe
A
Ff
ppp450.295.15.0
800
120*13
240
120
1
1 Bot:
Top:
The T-section used in the notation example is prestressed with atendon force of 120 kips, located 3 in. up from the bottom face. Find
the stresses due to prestress alone.
Eccentricity: ep= +16 -3 = +13 in.
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Example: Applied Moment
Positive Applied Moment M = 960 in-kips:
ksi
S
Mf 200.1
800
960
1
1
ksiS
Mf 600.0
1600
960
2
2
Bottom (tension)
Top (compression)
The sign of the stress is automatically correct.
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Example: Superposition
-Fp/A-epFp/S +Mext/S
total
Superposition of stresses
Includes stress due to Msw
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STRESSES BYEQUIVALENT LOADS
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Equivalent LoadsChanges in tendon slope induce transverse loads. The prestressing tendon
can be replaced by the equivalent loads that it causes.
cgc
cgp
Compute the loads caused by all the changes in slope
Apply them to the concrete beam.
Find moments.
Find stresses in the concrete.
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q+dq
Fp+dFp
Fp
q
w
dz
qqq sinsin ppp FddFFwdz
qq sintan dz
deslope
p ''" pppp eFeFw
"''" pppppp eFeFeFw
Equivalent Loads
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Straight tendon: No load
Parabolic, p2(z) : Uniform load
Cubic, p3(z) : Linearly varying load
Sharp angle: Point load.
Equivalent Loads
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Fp= 120 kips
cgc
cgp
12 1216
3
10.8 k 10.8 k
12 16 12
c1= 16
13
144 120 k10.8 k
Tendon is so flat that q= sinq= tanq
Treat horizontal component = diagonal component.
Equivalent Loads
130 ft-k = 1560 k-in.
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cgc
cgp
12 1216
3
10.8 k 10.8 k
12 16 12
130 ft-k = 1560 k-in.
Equivalent Loads
M
2
120 15600.5 0.975 0.475
240 1600f ksi
1
120 1560
0.5 1.95 2.450240 800f ksi
Stresses Between
Harping Points
Same as calculated
directly earlier!. But
now, easier to see
variation over span
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Fp= 120 kipsweq= -8Fp*sag/L2
= -8*120*13/4802
= 0.05517 k/in
= 0.65 k/ft
cgc
cgp
12 1216
13
=
sag
0.60 k/ft
120 ft-k
c1= 16
epmax= 13
ep= 4*sag*(x/L)(1-x/L)
ep= -8*sag/L2
Mpmax= weqL2/8
=0.65*402/8
= 130 ft-kips = 1560 k-in.
Equivalent Loads
Same at midspan,
but varies differently
over span
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STRESSES BY PRESSURELINE (C LINE)
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Effect of Mext on Fc
FcFp
ec= ep
cgc
cgp
PS alone PS + Mex t
ecFc
ep
Fp
Mext= Fp(ep-ec)
We know Mext(x), Fpand ep(x)
We can calculate and plot ec(x)
extc P
P
Me e
F
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Example: Pressure Line
Same T-Beam Fp= 120k
ep_max= 13 in.
W = 0.4 k/ft
Mext_max= 1/8*0.4 * 402
Mext_max= 80 k-ft = 960 k-in.
ec= ep- Mext/ Fp
ec= ep- 8 in. = 5 in.
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Why Should We Care About ec?
1. We want the stress at top and bottom to be within allowable
limits
2. We know Fp, ep
3. We can change (2) to keep location of Fc(ec) within certain
limits
i. Fp too close to bottom: compression on bottom too high
or tension on top too high
ii. Fp too close to top: compression on top too high or
tension on bottom too high
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Kerns
Kern = location of resultant force on cross-section that
causes zero stress on the opposite face
cgc
Kern distance
b
h
k = h/6 for a rectangular section. Middle third rule
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22
2 1S
Ae
A
F
S
Fe
A
Ff c
ppcp
Express the top and bottom stresses in terms of ep/k
2
1 1k
e
A
Ff c
p
1
2 1k
e
A
Ff c
p
ec> k1
cgck1
ec=k1 ec< k1
Kerns
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Keep the resultant force inside the kerns and the
section will never experience tension
cgc
k1
k2OK
zone
Kerns
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Location of compressive resultant if stresses at top and bottomjust equal the allowable stress.
cgc
k1
ec=k1 ec= k1ti
f2= fti
ec
Kern Modified kern
f2= 0.0
Extended Kerns
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If top stress is equal to the initial allowable stress, fti,
1
0
1
2 11
k
ef
k
e
A
Fff ci
cpi
iti
i
ticti
ffkek0
11 1'
Solve for ecto get k1ti
Kern
Modifier
Extended Kerns
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Note:
The (basic ) kern is a property of the section geometry alone.
This is true because, when the (top) stress is zero, the only
relevant characteristic of the compression force is its location.Its magnitude does not affect the (top) concrete stress, which
is zero.
The modif ied kern is a property of the section geometry. The
allowable stresses, and the applied stress, f0i.
Extended Kerns
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Similarly, if the bottom stress is equal to the initial allowablecompressive stress, fci,
2
01 1
k
efff ciici
i
cicci
f
fkek
0
21 1'
Solve for ecto get k1ci
Kern
Modifier
Extended Kerns
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0
21 1'
f
fkek ccc
At any loading stage, four modified kerns exist:
0
11 1'
f
fkek tct
ct kkk 111 ','min'
0
12 1'
f
fkek ccc
0
22 1'
f
fkek tct
ct kkk 222 ','max'
To
prevent
excess
tension
To
prevent
excess
compn.
To avoid
all
trouble
Extended Kerns
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0
21 1'
f
fkek ccc
Note the pattern:
0
11 1'
f
fkek tct
0
12 1'
f
fkek ccc
0
22 1'
f
fkek tct
Note the pattern:
Subscript number stays the same if tension limit.
Subscript number switches if compression limit.
Same subscript:
Different subscript:
Extended Kerns
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At any loading stage, four modified kerns:
k2t
k2c
k1ck1t
k2
k1
cgc
C-linemay be
located
anywhere in this
range
Extended Kerns
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Geometric way of thinking about stress limits:
Stress approach:Keep all stresses within all limiting values.
Geometric approach:Keep stress resultant (C-line) within modified kerns.
Extended Kerns
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Analysis Example
8DT24+2.5
dc,min=3
24 12 24
dc,min=3 dc,min=3
24
5
48 24
8
22
4.5
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Analysis Example
A precast 8DT24+2 1/2 spans 60 ft simply supported.
It is reinforced with ten -in. diameter, low-relaxation
strands stressed to 185 ksi initial and 165 ksi effective.
The strands are harped at 3/8 and 5/8 of the span. The
dcdistance is 10 in. at the ends, 3 in. in the central 25%of the span and varies linearly in between those points.
Under initial conditions, the beam carries only its self-
weight. Under effective (i.e. service) conditions, it carries
10 psf permanent DL plus 50 psf LL.
The material properties are: f'ci= 3500 psi, f'c= 6000 psi
and the concrete density is 150 pcf.
The member is to satisfy ACI 318 allowable stresses
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Analysis Example
The member properties are:
h = 26.5 in. A= 718 in.2 I= 38535 in.4
c2= - 7.191 in. S2= -5359 in.3 k2= -2.779 in.
c1= 19.309 in. S1= 1996 in.3 k1= 7.464 in.
wsw= 748 lbs/ft
Ap= 10*0.153 = 1.530 in.2
dc,min= 3 at midspan, 10 at the ends.
ep= c1dcmin= 16.309 (mid-span), 9.309 (ends).
Check that the stresses satisfy ACI 318 at mid-span
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Analysis Example
Allowable Stresses
From ACI 318, Section 18.4
fci= -0.6 f'ci = -0.6*3.5 = -2.100 ksi
fti= 3.0 f'ci = (3*3500)/1000 = +0.177 ksi
fcs= -0.60 f'c = 0.6*6.0 = -3.600 ksi (for total load)
fts= 7.5 f'c = 7.5*6000)/1000 = +0.581 ksi
A l i E l S iti
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Elastic Analysis Using Stress Superposition
At midspan, initial conditions:
Fpi= 1.530 in.2*185 ksi = 283.1 kips
Mi= Msw= wsw L2/8 = 12*(0.748*602/8) = 4039 in.-kips
bot: f1 = - Fpi/A - epFpi/S1+ Mi/S1
= -(283.1/718) - (16.309*283.1/1996) + (4039/1996)
= -0.684 ksi > - 2.1 ksi = fci OK
top: f2 = - Fpi/A - epFpi/S2+ Mi/S2
= -(283.1/718) - (16.309*283.1/-5359) + (4039/-5359)
= -0.286 ksi < 0.177 ksi = fti OK
Analysis ExampleSuperposition
A l i E l S iti
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At midspan, effective (i.e., service) conditions:
Fpe= 1.530 in.2*165 ksi = 252.45 kips
Ms= Msw + MDL+ MLL = 4039 + 12*(0.080 + 0.400)* 602/8 = 6631 in.-kips
bot: f1= -Fpe/A - epFpe/S1 + Ms/S1
= -(252.45/718) - (16.309*252.45/1996) + (6631/1996)
= 0.908 ksi > 0.581 ksi = fts No Good
top: f1= -Fpe/A - epFpe/S2+ Ms/S2
= -(252.45/718) - (16.309*252.45/-5359) + (6631/-5359)
= -0.821 ksi > -3.600 ksi = fcs OK
The conditions at other locations could be found similarly. Two things are clear:
1. Design is controlled by bottom tension under service conditions. This
condition is very common and can be cured by increasing the prestress force.
2. The calculations are tedious and repetitive and are therefore ideal candidates
for automation.
Analysis ExampleSuperposition
A l i E l E i l t L d
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Analysis ExampleEquivalent Loads
kipsinFM pend 309.9
kipsFFW ppeq2887
12*24"309.9"309.16
Fp
9.309Fp
Fp
7Fp
/288
12 2424
9.309Fp
7Fp
/288
7Fp
/288
7Fp
/288
Elastic Analysis Using Equivalent Loads
The equivalent loads are:
End moment:
Transverse point load:
= 6.881 kips (initial)
= 6.137 kips (effective)
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Initial conditions:-Fpi/A = -283.1/718 in2 = -0.394 ksi
At mid-span:
Mp = -(283.1k*9.309) - (6.881k*288) = -4617 in.-kips
Msw = 0.748 k/ft *602/8*12 = 4039 in.-kipsMi = -4617 + 4039 = -578 in.-kips
f1i = -Fpi/A + Mi/S1 = -0.394 + (-578)/(+1996) =-0.684 ksi
bottom
f2i
= -Fp
i/A + Mi/S2 = -0.394 + (-578)/(-5359) = -0.236 ksi top
Analysis ExampleEquivalent Loads
A l i E l E i l t L d
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Effective conditions:
-Fpi/A = -252.5/718 in2 = -0.352 ksi
At mid-span
Mp = -(252.5k*9.309) - (6.137k*288) = -4118 in.-kips
Msw = 0.748 k/ft *602/8*12 = 4039 in.-kips
MD+L= (0.080+0.400)*602/8*12 = 2592 in.-kips
Me = -4118 + 4039 + 2592 = 2513 in.-kips
f1i = -Fpi/A + Mi/S1 = -0.352 + (2513)/(+1996) = -0.908 ksi
bottomf2i= -Fpi/A + Mi/S2 = -0.352 + (2513)/(-5359) = -0.821 ksi top
These results are the same as those from the stress superposition
calculations.
Analysis ExampleEquivalent Loads
A l i E l P Li
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Elastic Analysis Using Pressure Line and Modified Kerns
Initial conditions:Fpi =283.1 kips
Mext(L/2) = 4039
ep-ec= Mext/Fp= 4039/283.1 = 14.267 in.
ec= ep- Mext/Fp = 16.30914.267 = 2.042 in.
This value must be compared to the range of locations in which the PressureLine may lie, given by the modified kerns. The limits of that range are:
inkki
ti
f
f
ti 027.41779.21' 39 4.017 7.0
220
inkk ici
f
f
ci 319.321464.71' 39 4.010 0.2
120
inkki
ti
f
f
ti 817.101464.71' 39 4.017 7.0
11 0
inkki
ci
f
f
ci 033.121779.21' 39 4.010 0.2
210
ink i 027.4'2
ink i 817.10'1
Analysis ExamplePressure Line
A l i E l P Li
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The C-line lies at y = -2.042 in. This is within the modified kern boundaries (-4.027,
+10.817) under initial conditions, so the results are acceptable.
k1ti
k1ci
k2ci
k2ti
cgc
C-line
OK zone
k2i
k1i
2.042
Analysis ExamplePressure Line
A l i E l P Li
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Under effective conditions:
Fpi=252.5 kipsMext(L/2) = Msw+ MD+ ML= 6631 in.-kips
ep-ec= Mext/Fp= 6631/252.5 = 26.261 in.
ec= ep- Mext/Fp = 16.30926.261 = -9.952 in.
This value must be compared to the range of locations in which the Pressure
Line may lie, given by the modified kerns. The limits of that range are:
inkks
ts
f
f
ts 366.71779.21' 35 2.058 1.0
220
inkks
cs
f
f
cs 872.681464.71' 35 2.060 0.3
120
inkks
ts
f
f
ts 784.191464.71' 35 2.058 1.0
11 0
inkks
cs
f
f
cs 642.251779.21' 352.0600.3
210
ink s 366.7'2
ink s 784.19'1
Analysis ExamplePressure Line
A l i E l P Li
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Under effective conditions,
ec= -9.952 in., but the highest it may be without violating a
stress limit isk2=- 7.366 in.
Thus the C-line lies outside the modified kern boundaries andthe result is unacceptable.
Not surprisingly, k2sis controlled by k2ts, which means that the
bottom tension stress under service conditions controls the
design. This is the same results as we obtained with the othertwo methods.
Analysis ExamplePressure Line
Composite Sections
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Precast beam, cast-in-place slab.
Erect precast without formwork. Use it as form for cast-in-place deck.
CIP deck joins the pieces to work as one structure.
Composite Sections
Behavior of Composite Sections
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Composite action:
Connect CIP slab to PC girderno slip.
Entire section acts as one.
Need transformed section analysis if different Ecvalues. Compute stresses at different load stages, then add.
Behavior of Composite Sections
Behavior of Composite Sections
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Load stages:
1. PC beam, self-weight only, initial prestress.
2. PC beam, self weight + slab weight, reduced prestress.
3. Composite beam, self weight + live load, effective prestress.
Behavior of Composite Sections
For stage 3 check, note that the stresses due to PS and self
weight of beam + slab are already locked in, using the PCbeam section alone. The LL acts on the composite section,
so causes stress in the slab and beam.
Behavior of Composite Sections
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Stage 1: at release
Behavior of Composite Sections
12
1 1S
M
k
e
A
Ff sw
ppi
i
21
2 1S
M
k
e
A
Ff sw
ppi
i
Check these stresses againstfci,fti.
Behavior of Composite Sections
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Stage 2: when slab is cast.Include PC self weight, slab self-weight, construction loading.
Behavior of Composite Sections
Check these stresses againstfcs,fts.
(Depends on how much time-dependent loss has occurred).
112
1 1
S
M
S
M
k
e
A
Ff constrslabsw
ppe
222
2 1S
M
S
M
k
e
A
Ff constrsl absw
ppe
Behavior of Composite Sections
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Stage 3: Service conditions
Include only LL in the composite section analysis.
Add those stresses to the previous locked-in stresses.
Behavior of Composite Sections
Check these stresses againstfcs,fts.
112
1 1S
M
S
M
k
e
A
Ff constrsl absw
ppe
222
2 1S
M
S
M
k
e
A
Ff constrsl absw
ppe
comp
LD
S
Mf
,4
4
comp
LD
S
Mf
,3
3
Behavior of Composite Sections
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Stage 3: Service conditions
Include only LL in the composite section analysis.Add those stresses to the previous locked-in stresses.
Behavior of Composite Sections
PS +
PC sw
Slab
sw constr LL Total
Behavior of Composite Sections
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Stage 3: Service conditions
Behavior of Composite Sections
An 18 x 36 deep rectangular precast beam spans 75 ft simply supported. A 96 x
9 slab is cast on top of it. The composite system carries a LL of 100 psf. Find the
stresses under initial conditions, when the slab has been cast, and under full LL.
Allow for 20 psf construction load when the slab is cast.
Precast concrete: fci= 5.0 ksi, fc= 8.0 ksi, Ec= 5000 ksi, g= 0.150 kcf
c.i.p. concrete: fci= 3.0 ksi, fc= 5.0 ksi, Ec= 4000 ksi, g= 0.150 kcf
Strands: 20 strands, dia., fpi= 185 ksi, h= 0.85, epmax= 15.
Behavior of Composite SectionsThe member properties, all transformed to the precast concrete, are:
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Behavior of Composite Sections
Property pc used comp Property pc used comp pc used comp
h = 36 42 c4 cip = -15.272727b min = 18 18 c3 cip = -9.2727273
b max = 18 96 c2 pc = -18 -9.2727273
A = 648 1108.8 c1 pc = 18 26.7272727 n 1 0.8
Ixx = 6.998E+04 190127.127
rxx^2 = 1.080E+02 171.471074 S4 = -12448.800 k4 =
Iyy = 1.750E+04 371390.4 S3 = -20503.906 k3 =
w = 0.056 S2 -3888 -20503.906 k2 = -6.000
ep.max 15.000 S1 3888 7113.600 k1 = 6.000
Member properties
Behavior of Composite SectionsThe member properties, all transformed to the precast concrete, are:
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Behavior of Composite Sections
Summary of stresses
PS pc beam slab LL total
Initial
2 1.638 -1.465 0.000 0.000 0.173 OK
1 -3.822 1.465 0.000 0.000 -2.357 OK
0 -1.092 0.000 0.000 0.000 -1.092 OK
slab cast
2 1.392 -1.465 -1.302 -0.347 -1.722 OK
1 -3.249 1.465 1.302 0.347 -0.135 OK
0 -0.928 0.000 0.000 0.000 -0.928 OKservice
4 0.000 0.000 0.000 -0.434 -0.434 OK
3 0.000 0.000 0.000 -0.263 -0.263 OK
2 1.392 -1.465 -1.302 -0.329 -1.704 OK
1 -3.249 1.465 1.302 0.949 0.467 OK
0 -0 928 0 000 0 000 0 000 -0 928 OK