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CE6340 Prestressed Beam Example 1/73 Design of a Multi Span Prestressed AASHTO Girder CODE: AASHTO LRFD 4nd edition,
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CE6340
Prestressed Beam Example
1/73
Design of a Multi Span Prestressed AASHTO Girder CODE: AASHTO LRFD 4nd edition, 2007 Unit definitions kN 1000N GPa 109Pa
6
MPa 10 Pa
kip 1000 lbf
ksi 1
kip 2
in
Materials Cast in Place Slab Concrete Strength at 28 Days
fc_s := 5.0 ksi γc_s := 150
lbf ft
Concrete unit weight
3
Precast Beam fci := 5.0 ksi
Concrete Strength at Release
fc := 6 ksi
Concrete Strength at 28 Days
γc := 150
Concrete unit weight
lbf ft
3
Prestressing Strands
0.5 in, low relaxation
d strand := 0.5 in
Strand Diameter 2
Astrand := 0.153 in
Area of one strand
fpu := 270 ksi
Ultimate Strength
Hoop Reinforcement Area: Av := 2
4 in 4 8
π
2
2
Av = 0.39 in
Aditional Dead Loads on Girders p fws := 2 in 140
lbf ft
wbarrier := 0.405
3
kip ft
p fws = 0.162 psi
= 5910.53
kg 2
2 inches of Asphalt
Weight of one New Jersey Barrier
s
p. 1/73
psf 1
lbf ft
2
metro 1m
CE6340
Prestressed Beam Example
p. 2/73
2/73
CE6340
Prestressed Beam Example
Span Information: nspan := 2
Number of Spans in Structure
jspan := 1
Span under Investigation
120.0 L := 120.0 ft 0.0
Lenght of Each Span
Loh := 6 in
Precast Overhang (additional length of beam after CL of Bearing)
Lcc :=
6.0 left support (abutment) in 18.0 right support (pier)
120 L = 120 ft 0
Distance from center line of bearing to centerline of pier or abutment end
Bridge Cross-Section: Number of Beams
Nb := 6
select from 1 to 6 for AASHTO TYPE GIRDERS I to VI :
igirder_type := 5 Wdeck := ( 5 7 ) ft + 8 ft Shoulder_Ratio :=
4 7
= 0.57
Wdeck = 43 ft
width of deck
Ratio S_ext/S
t s_total := 10 in
t s_total = 10 in
Total Slab Thickness
t s := 9 in
t s = 9 in
Structural Slab Thickness Haunch Thickness
t h := 1 in Wbarrier := 19 in
S :=
Wdeck
(Nb - 1 + 2 Shoulder_Ratio)
Sext := Shoulder_Ratio S
Wbarrier = 19 in
Width of Barrier
S = 7 ft
Beam Spacing
Sext = 4 ft
Length of Exterior Shoulder
p. 3/73
3/73
CE6340
Prestressed Beam Example
4/73
Properties of Concrete 1.5
fc_s γc_s Ec_s := 33000 ksi ksi kip ft3
Ec_s = 4286.8 ksi
modulus of elasticity of concrete slab A 5.4.2.4
Eci = 4286.8 ksi
modulus of elasticity at release
A 5.4.2.4
Ec = 4696 ksi
modulus of elasticity at service
A 5.4.2.4
Modulus of Rupture
A5.4.2.6
Yield Strength for low lax
A 5.4.4.1 Table 1
1.5
fci γc Eci := 33000 ksi ksi kip ft3 1.5
fc γc Ec := 33000 ksi ksi kip ft3
fr := 0.37
fc ksi
ksi
fr = 0.91 ksi
Properties of Strands fpy := 0.9 fpu
fpy = 243 ksi
Stress Limits for prestressing strand fpj := 0.75 fpu
fpj = 202.5 ksi
Before Transfer (at jacking)
A 5.9.3 Table 1
fservice := 0.80 fpy
fservice = 194.4 ksi
At Service Limit State (After all losses)
A 5.9.3 Table 1
Ep := 28500 ksi
Modulus of Elasticity
A 5.4.4.2
p. 4/73
CE6340
Prestressed Beam Example
5/73
Reinforcing Bars fy := 60 ksi
Yield Strength
Es := 29000 ksi
Modulus of Elasticity
A 5.4.3.2
Compute Simple Span of precast Beams for dead loads on stage I
C 4.62.2.1 Table 1 - L
ispan := 1 .. nspan
Span of Simple Beam (for dead loads on simple span) Lspan
ispan
Lspan =
:= L
ispan
- Lcc - Lcc 1
2
118 ft 118
Lenght of Span under Investigation
Average Length of Adjacent Spans for dist. factor calculations in Negative Moment Region (for negative moment & interior reaction near int. Support) iave :=
( 1 .. 2 ) if nspan = 1 ( 1 .. nspan + 1 ) otherwise
Lave
iave
:=
L L
iave
if iave = 1 if iave = nspan + 1
iave-1
L
iave- 1
+L
2
iave
120 Lave = 120 ft 120
otherwise
p. 5/73
CE6340
Prestressed Beam Example
6/73
Define Additional Section Data Wroad := Wdeck - 2 Wbarrier Wroad = 39.83 ft
roadway width, face of barrier to face of barrier
Waxle := 6 ft
space between wheels on one axle
Wlane := 12ft
width of design lane
Wroad Wlane
Nlane := floor
Nlane = 3
number of lanes
d top_cover := 2in
Slab: Concrete cover required over bot bars
d bot_cover := 1in
Slab: Concrete cover required over bottom bars
A 5.12.3 Table -1
Cross-Sectional Properties Non Composite Properties from reference: AASHTO GIRDER DATA Reference:F:\Manuel Coll\02 Material de Cursos\002 CE 6340 WI 2012\Worksheets Used in CE6340 WI2010\AASHTO GIRDER DATA.mcd(R)
2
A = 1013 in
St = 16790.59 in
I = 521180 in 3
4
h = 63 in 3
Sb = -16307.26 in
p. 6/73
yb = 31.96 in
CE6340
Prestressed Beam Example
Composite Section Properties
Interior Span
b eff := S 1
b eff :=
b eff
2
b eff =
n :=
2
1
Exterior Span
+ Sext
84 in 90
(Cambio en codigo 2010)
Effective Flange width (Interior, Exterior)
Ec_s Ec
n = 0.91
I I 3 3 n b f_top th n b f_top th 12 12 Inertia := n b eff ts3 n b eff ts3 1 2 12 12
A 4.6.2.6.1
Modular Ratio Between Slab & Beam
Girder Haunch Slab
521180 521180 4 Inertia = 3.2 3.2 in 4658.38 4991.12
A A Area := n b f_top t h n b f_top t h n beff ts n b eff ts 1 2
1013 1013 2 Area = 38.34 38.34 in 690.13 739.43
yb h + th Arm := 2 t h + th + s 2
31.96 Arm = 63.5 in 68.5
p. 7/73
7/73
CE6340
Prestressed Beam Example
j := 1 .. 2
i := 1 .. 3
3
Ac := j
8/73
Ac =
Area
i, j
i=1
1741.47 2 in 1790.77
Area of Composite Section (Interior, Exterior)
3
(Areai, jArmi)
yb_c :=
i =1
j
yb_c =
3
Area
47.13 in 47.72
Distance from centroid to bottom fiber
i, j
i =1
3
Ic := j
3
Inertia
i, j
i=1
+
Area Arm - y 2 b_c i j i, j
(
i=1
)
h c := h + th + t s
h c = 73 in
yt_c := h - yb_c
yt_c =
ys_c := h c - yb_c
ys_c =
Sb_c := j
Sb_c =
Ic
j
yb_c
St_c := j
j
-23006.38 3 in -23188.35
Moment of Inertia Composite Secion
Overall depth of composite section
15.87 in 15.28
Distance from centroid to top of precast girder
25.87 in 25.28
Distance from centroid to bot of slab
Ic
j
yt_c
j
St_c =
1084403.1 4 in 1106618.1
Ic =
Ss_c := j
68351.43 3 in 72436.93
Ic
j
ys_c
Ss_c =
j
41925.32 3 in 43779.67
Section Modulus of Composite Section
Area of bottom half of composite girder section 2
hc - t f_bot 2
Abot := b f_bot t f_bot + Taper + t w
2
Abot = 552 in
p. 8/73
Ac 2
=
870.74 2 in 895.38
CE6340
Prestressed Beam Example
9/73
Define Life Load Shear & Moment Envelopes
Reference:F:\Manuel Coll\02 Material de Cursos\002 CE 6340 WI 2012\Worksheets Used in CE6340 WI2010\Presstress Beam Design - Subrutine 1 - Multi
m := metro
Define ten Points & 0.05*L for Girder of Span being Investigated ten := 10
tentotal := ten + 3
iten := 1 .. tentotal
xten
iten
:=
xten
iten- 1
xten
iten- 1
xspan xspan
jspan
+
+
Lspan
Lspan
jspan
2ten
+ Lcc
jspan + 1
jspan
ten
1
- Lcc
if iten > 2 if ( 1 < iten < 4 ) ( tentotal - 2 < iten < tentotal)
if iten = 1 2
if iten = tentotal
p. 9/73
1 2 3 4 5 6 iten = 7 8 9 10 11 12 13
0.5 6.4 12.3 24.1 35.9 47.7 xten = 59.5 ft 71.3 83.1 94.9 106.7 112.6 118.5
CE6340
Prestressed Beam Example
10/73
Define area outside precompresed tensile zone in span being analized, for tensile stress limits at transfer (Compression zone in final configuration): top_comp_zone
iten
:=
1 if xten
- xspan
1 if xten
- xspan
iten iten
1 if 0.2 L
jspan
jspan jspan
0.7 L 0.3 L
xten
jspan jspan
- xspan
iten
jspan
jspan = 1 jspan = nspan 0.8 L
jspan
1 < jspan < nspan
0 otherwise
bot_comp_zone
iten
:=
1 if xten
- xspan
1 if xten
- xspan
iten iten
1 if 0.25 L
jspan
jspan jspan
(
0.75 L 0.25 L
xten
iten
jspan jspan
- xspan
jspan = 1 jspan = nspan
jspan
) (x
ten iten
- xspan
jspan
) 0.75Ljspan 1 < jspan < nspan
0 otherwise
top_comp_zone
iten
:= 1
bot_comp_zone
iten
:= 0
1 1 1 1 1 1 top_comp_zone = 1 1 1 1 1 1 1
Note: is a simply suported precast beam made continuous by the slab, so top is compression zone.
0 0 0 0 0 0 bot_comp_zone = 0 0 0 0 0 0 0
p. 10/73
CE6340
Prestressed Beam Example
11/73
Live Load Shear & Moment Envelopes 3
4 10
M truck
3
2 10
kip ft M train kip ft
0
M enve kip ft M fatigue kip ft
3
- 2 10
3
- 4 10
0
100
200 xpoi ft
200
Vtruck
100
kip Venve kip
0
Vfatigue kip - 100
- 200
0
100
200 xpoi ft
p. 11/73
CE6340
Prestressed Beam Example
12/73
Define Shear & Moment for unit uniform load on Simple Span girder (NC - Non Composite):
Lspanjspan
Vnc ( x , w) := w
M nc ( x , w) :=
w 2
2
(
x - xspan
(
- x - xspan
jspan
jspan
)
- Lcc
- Lcc Lspan 1
1
jspan
)
Shear at x distance from left support
(
- x - xspan
jspan
- Lcc
)
Moment at x distance from left support
1
Define functions for uniform dead loads on Composite bridge (Stage 2 Loads)
xspannspan+ 1 1 Vec_M := ILMpos( x , k) + ILMneg( x , k) dx k kip 0 ft
xspannspan+1 1 Vec_V := ILVpos( x , k) + ILVneg( x , k) dx k kip 0 ft
M c( x , w) := linterp( xpoi , Vec_M , x) w Vc( x , w) := linterp( xpoi , Vec_V , x) w
Define Negative Moment Regions at span under investigation:
Span_Neg_Zone( x) :=
3 if linterp( xpoi , Vec_M , x) < -0.001 ft 2
L
(1) Positive Moment Region (2) Left Negative Moment Region (3) Right Negative Moment Region
jspan
2
(
x - xspan
2 if linterp( xpoi , Vec_M , x) < -0.001 ft 0 ft x - xspan
Neg
iten
(
:= Span_Neg_Zone xten
)
iten
p. 12/73
) Ljspan
L
2
1 otherwise
jspan
jspan
jspan
2
1 1 1 1 1 1 Neg = 1 1 1 3 3 3 3
CE6340
Prestressed Beam Example
13/73
Calculation of Distribution factors
Figure 1 Sketch of Transverse Section of the Bridge Distribution factors for moment on interior girder (Table 4.6.2.2b-1) Compute transverse to longitudinal stiffness ratio: LDF := L 1
jspan
ts
+ yt + t h
2
Ec
Kg :=
2
Ec_s
jspan
LDF := Lave 3
I + A e 2 g
4
longitudinal stiffness parameter
Kg = 2052545.19 in
0.4
0.3
0.6
0.2
S S Kg gm_int_2 := 0.075 + j 9.5 ft LDFj L t 3 DFj s
(
gm_int := max gm_int_1 , gm_int_2
gm_int_f := j
psi = 1 psi
distance from center of gravity of precast beam to center of gravity of deck
eg = 36.54 in
S S Kg gm_int_1 := 0.06 + j 4300mm LDFj L t 3 DFj s
j
jspan + 1
120 LDF = 120 ft 120
1: for positivmoment 2 & 3 for negative moment regions.
j := 1 .. 3 eg :=
LDF := Lave
gm_int_1 1.20
j
j
j
)
0.1
0.1
t s = 9 in
0.337 gm_int_f = 0.337 0.337
Equation 4.6.2.2.1-1
0.4 gm_int_1 = 0.4 0.4
Distribution Factor for one lane loaded
0.58 gm_int_2 = 0.58 0.58
Distribution Factor for two or more lanes loaded
4
Kg = 2052545.19 in
Fatigue distribution factor for moment on interior girder p. 13/73
0.58 gm_int = 0.58 0.58
120 LDF = 120 ft 120 S = 7 ft
CE6340
Prestressed Beam Example
14/73
Distribution factor for moment on exterior girder (Table 4.6.2.2.2d-1) Lever rule:
Figure 2 Notional Model Applying the Lever Rule c := S + Sext - Wbarrier - 2 ft - Waxle glever_rule :=
Waxle + 2 c 2 S
1.20
c = 1.42 ft
distance from 1st interior girder to inside wheel
glever_rule = 0.757
distribution factor for moment on exterior girder, Lever Rule, including multiple presence factor
Rigid body analysis (C4.6.2.2.2d)
Figure 3 Truck Positions for Rigid Body Analysis
p. 14/73
CE6340
Prestressed Beam Example
15/73
itruck := 1 .. Nlane index for calculation ibeam := 1 .. Nb
xbeam
ibeam
:= ( Nb - ibeam) S -
( Nb - 1) S 2
17.5 10.5 3.5 xbeam = ft -3.5 -10.5 -17.5
Wroad etruck := - 0.6m - 0.9m - ( itruck - 1 ) Wlane itruck 2 Xext := xbeam
Xext = 17.5 ft
1
Wroad 12 ft
= 3.32
distances from center of bridge to each girder
15 etruck = 3 ft -9
distance from centerline of bridge to exterior girder
(x
beam
itruck
Xext R
itruck
:=
itruck Nb
+
ibeam
etruck
j =1
(x
beam
distances from center of bridge to each truck
j
ibeam
)
2
0.47 R = 0.7 0.68
ibeam
)
2
= 857.5 ft
2
Nb = 6
rigid body distribution factors, from Eqn. C4.6.2.2.2d-1
ibeam
1.20 MPF := 1.00 0.85 grigid_body
itruck
:=
multiple presence factor, from Table 3.6.1.1.2-1
(Ritruck MPFitruck) if itruck < 3 (Ritruck MPF3) otherwise
grigid_body := max( grigid_body)
0.57 grigid_body = 0.7 0.58 distribution factor for moment for rigid body analysis
grigid_body = 0.7
p. 15/73
CE6340
Prestressed Beam Example
16/73
Table 4.6.2.2.2d-1 distance from exterior girder to face of parapet d e := Sext - Wbarrier e := 0.77 +
d e = 2.42 ft
de
e = 1.03
2800mm
(
)
gm_ext := max e gm_int_2 , glever_rule , grigid_body j
0.58 gm_int_2 = 0.58 0.58
j
0.757 gm_ext = 0.757 0.757
Table 4.6.2.2.2d-1
0.6 e gm_int_2 = 0.6 j 0.6 for moment on exterior girder distribution factor glever_rule = 0.76
glever_rule gm_ext_f := max ,R 1 j 1.2
gm_ext_f
0.63 = 0.63 0.63
Distribution factors for shear on interior girder gv_int_1 := 0.36 +
gv_int_2 := 0.2 +
S 7600mm S
3600mm
-
S 10700mm
gv_int_1 = 0.64
distribution factor for shear on interior girder, one lane loaded
gv_int_2 = 0.75
distribution factor for shear on interior girder, more than one lane loaded
2
gv_int := max( gv_int_1 , gv_int_2)
gv_int_f :=
gv_int_1
e := 0.64 +
1.20
de 3800mm
gv_ext := max( e gv_int_2 , glever_rule , grigid_body)
glever_rule ,R 1 1.2
gv_ext_f := max
gv_int_f = 0.53
fatigue distribution factor for shear on interior girder
e = 0.83
Table 4.6.2.2.3b-1
gv_ext = 0.76
distribution factor for shear on exterior girder
gv_ext_f = 0.631
fatigue distribution factor for shear in exterior girder
p. 16/73
CE6340
Prestressed Beam Example
17/73
Summary of Distribution Factors For Positive Moment Region Strength & Service moment on interior girder
0.579 gm_int = 0.579 0.579
Fatigue moment on interior girder
0.337 gm_int_f = 0.337 0.337
moment on exterior girder
0.757 gm_ext = 0.757 0.757
moment on exterior girder
gm_ext_f
0.631 = 0.631 0.631
shear on interior girder
gv_int = 0.75
shear on interior girder
gv_int_f = 0.53
shear on exterior girder
gv_ext = 0.757
shear on exterior girder
gv_ext_f = 0.631
Distribution Factors (interior, exterior) j := 1 .. 2
interior, exterior
i := 1 .. 3
positive, negative start, negative end
i gm :=
gv :=
gm_f
gm_inti gm_ext i
gv_int gv_ext
i
gv_f :=
:=
gm_int_f i gm_ext_f i
gv_int_f gv_ext_f
gm =
0.579 0.579 0.579 0.757 0.757 0.757
DF for moment
gv =
0.75 0.76
DF for shear
gm_f =
0.34 0.34 0.34 0.63 0.63 0.63
DF for moment, one lane loaded only & no multiple presence factor
0.53 0.63
DF for shear, one lane loaded only & no multiple presence factor
gv_f =
p. 17/73
CE6340
Prestressed Beam Example
Define functions of moments and shear envelopes for LL+I:
(
)
(
)
(
)
(
)
1 M LLI_max ( j , x) := linterp xpoi , M enve , x gm
2 M LLI_min( j , x) := linterp xpoi , M enve , x gm 1 VLLI_max ( j , x) := linterp xpoi , Venve , x gv
1, j
j , Span_Neg_Zone ( x)
j
2 VLLI_min( j , x) := linterp xpoi , Venve , x gv
j
Define functions of moments and shear envelopes for Fatigue Truck:
(
)
(
)
(
)
(
)
1 M fatigue_max ( j , x) := linterp xpoi , M fatigue , x gm_f 2 M fatigue_min( j , x) := linterp xpoi , M fatigue , x gm_f 1 Vfatigue_max ( j , x) := linterp xpoi , Vfatigue , x gv_f 2 Vfatigue_min( j , x) := linterp xpoi , Vfatigue , x gv_f
1, j
j , Span_Neg_Zone( x)
j
j
Dead Loads placed on simple span:
S wslab := γc_s t s_total S Sext + 2 1 1
wgirder := A γc
wslab =
0.88 kip 0.94 ft
wgirder =
1 1
whaunch := t h b f_top γc
1.06 kip 1.06 ft
whaunch =
0.04 kip 0.04 ft
p. 18/73
18/73
CE6340
Prestressed Beam Example
19/73
Dead Loads Placed on the Continuous Structure: Permanent Loads (curbs & future wearing surface) may be distributed uniformly among all beams if the following conditions are met: -
width of deck is constant beams are parallel & have aproximately the same stiffness curvature in plan less than 4 degrees (changed, check limits in section 4.6.1.2.4) cross section of bridge is consistent with one of the sections given in Table 4.6.2.2.1 Number of beams is not less than 4 road way part of the overhang less than 910 mm (3 ft)
Box_1 :=
"OK!" if Nb 4 d e 910 mm
Box_1 = "OK!"
"NOT GOOD!" otherwise
dead load distribution factor:
S W deck gdead := S Sext + 2 W deck
gdead =
wDC1 := wslab + wgirder + whaunch
wDC2 :=
2 wbarrier Nb
1 1
S wDW := p fws S Sext + 2
A 4.6.2.2.1
0.16 0.17
1.97 kip 2.04 ft
wDC1 =
0.14 kip 0.14 ft
wDC2 =
wDW =
0.16 kip 0.18 ft
wDW := p fws Wdeck gdead
p. 19/73
CE6340
Prestressed Beam Example
20/73
Define function for Factored Forces: ηD := 1.0
Bridge with ductility requirements of the specification and redundant
ηR := 1.0
Typical Bridge
ηI := 1.0
0.95 ηD ηR ηI
η := max
η=1
For Normal superstructure design maximum loads will control
A 1.3.2.1
Define the load factors for the different combinations
1.25 1.00 γDC := 1.00 0.00
0.9
0 0 0
1.50 0.659 1.00 0 γDW := γ := 1.00 0 LL 0.00 0
1.75 1.75 1.00 0 0.80 0 0.75 0
p. 20/73
Strength-I Service-I (Compresion) Service-III (Tension) Fatigue
CE6340
Prestressed Beam Example
21/73
Compute Moment & Shear Envelopes at ten points & at 0.05*L: i := iten
j := 1 .. 2
M slab_haunch M girder
j ,i
M DC1 M DC2 M DW
(
:= M nc xten , wslab + whaunch
(
i
j
:= M nc xten , wgirder i
(
j
:= M nc xten , wDC1
j ,i
i
(
i
(
i
:= M c xten , wDC2
j ,i
j ,i
j ,i
:= M c xten , wDW
j
j
j
)
)
VDC2
)
VDW
(
M f_min
j ,i
j ,i
(
i
:= Vc xten , wDW
j
j
)
)
VLL_min
:= VLLI_min j , xten
(
i
:= M fatigue_min j , xten
M LL_max =
i
)
)
:= M fatigue_max j , xten
(
(
:= Vc xten , wDC2
j ,i
j
:= VLLI_max j , xten
:= M LLI_min j , xten
M f_max
i
VLL_max
M LL_min
i
(
:= Vnc xten , wDC1
j ,i
j ,i
)
i
)
i
(
(
:= Vnc xten , wgirder
j ,i
VDC1
:= M LLI_max j , xten
j ,i
) Vgirder
)
M LL_max
j ,i
j
i
j ,i
)
)
Vf_max Vf_min
j ,i
j ,i
j ,i
(
i
(
i
)
)
(
:= Vfatigue_max j , xten
(
i
:= Vfatigue_min j , xten
i
)
)
30.3 387.3 739 1241.1 1542.6 1674.9 1645.2 1470.3 1157.2 714.7 259.2 128.9 26.1 kip ft 30.3 387.3 739 1241.1 1542.6 1674.9 1645.2 1470.3 1157.2 714.7 259.2 128.9 26.1
T
M enve =
0 372.25 0 1253.54 2137.78 2667.09 2896.8 2837.29 2519.32 1955.1 1160.88 360.79 -0 -165.17 -330.35 -495.52 -660.7 -825.87 -991.05 -1156.22 -1321.4 -1832.58 -2970.77 -1832.33 -
p. 21/73
CE6340
Prestressed Beam Example
22/73
Preliminary Strand Arrangement: Compute Stresses in the bottom beam fibers due to dead load & live loads at midspan
SERVICE III
fbottom
j ,i
:=
s := 3
γDC
k := 1
M DC1
s, k
Sb
j ,i
+
Note: s = Load case; k = max (1), min(0); j = ten point; i = interior (1) exterior (2)
γDC
s,k
M DC2
j ,i
+ γDW
M DW
s, k
Sb_c
j ,i
+ γLL
γLL
M LL_max
s, k
j ,i
j
s, k
= 0.8
max( M DC1) = 3544.46 kip
max( M DC2) = 136.02 kip ft max( M DW ) = 176.32 kip ft
max( M LL_max) = 1674.92 kip fbottom =
-0.02 -0.68 -1.29 -2.26 -2.92 -3.28 -3.36 -3.14 -2.65 -1.88 -0.88 -0.33 0.25 ksi -0.02 -0.7 -1.32 -2.31 -2.99 -3.36 -3.43 -3.22 -2.72 -1.93 -0.9 -0.34 0.26
A 5.9.4.2.2 Table 1
fc
ksi
ften := - 0.19
Sb = -16307.26 Sb_c =
ksi
ften = -0.465 ksi
-23006.38 -23188.35
Service III Concrete Stress Limit for Tension after Losses
Estimate Eccentricity of Centroid of Strands: yb = 31.96 in e := -( yb - 12 in)
e = -19.96 in
Estimated eccentricity between centroid of strands & centroid of girder at point of maximum tension
fb := min( fbottom )
fb = -3.43 ksi
maximum tension in beam
Preq = 1342.8 kip
Estimated total prestress force required:
Preq :=
ften - fb
1 + e A S b
Note: Prestress force acts on noncomposite section
p. 22/73
CE6340
Prestressed Beam Example
23/73
Total Loss in the prestressing steel stress (estimated STANDARD SPECS Table 9.16.2.2)
ΔfpT := 45 ksi fpe := fpj - ΔfpT 2
Astrand = 0.15 in
Pe := Astrand fpe
Nstrand :=
Nstrand :=
fpj = 202.5 ksi
Effective stress in strands after losses
fpe = 157.5 ksi
Effective prestressing force in one strand
Pe = 24.1 kip
Preq
Nstrand = 55.72
Pe
floor( Nstrand + 1) floor( Nstrand + 1 ) floor( Nstrand + 1 ) if = floor 2 2 floor( Nstrand + 2 ) otherwise
Estimated number of strands required
Nstrand = 56
yb = 31.96 in
Nstrand := 56 Nstrand_line := 6
13 13 Nstrand_line = 13 3
para reducir la eccentricidad de los tendones
Nstrand_line := 8
1
select par number of strands
2
Nstrand_line := 8 3
Arrange Strands
nlayer :=
dummy 0 i0 while dummy < Nstrand ii+1 dummy
dummy + Nstrand_line if i < rows( Nstrand_line) i
dummy + Nstrand_line
(
)
rows Nstrand_line
otherwise
i
6 8 Nstrand_line = 8 3
h = 63 in
p. 23/73
CE6340
Prestressed Beam Example
ilayer := 1 .. nlayer
nlayer = 15
Number of Strands per Layer Nstrand_layer
ilayer
:=
Nstrand_line
ilayer
Nstrand_line
if ilayer < rows( Nstrand_line)
(
)
rows Nstrand_line
if ilayer rows( Nstrand_line)
nlayer-1 Nstrand Nstrand_layer if ilayer = nlayer i i =1
0 otherwise h = 63 in
nlayer 2 in = 30 in
Eccentricity of Centroid of Strands:
24/73
6 y = 31.96 in 8 b 82 in nlayer = 30 in 3 3 3 3 Nstrand_layer = 3 3 3 3 3 3 3 1
N
strand_layer
= 56
nlayer
yb_strand :=
ilayer = 1
Nstrand_layerilayer ( ilayer 2 in) Nstrand
yb_strand = 12.79 in yb = 31.96 in
ecc := yb_strand - yb
ecc = -19.17 in
Define Eccentricity at ten points: eten
iten
:= ecc
e = -19.96 in
T
eten = ( -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 ) in
p. 24/73
OK
CE6340
Prestressed Beam Example
25/73
Compute Prestress Loss; Low Relaxation Strand Elastic Shortening
(
T
M g := max M girder j
A 5.9.5.2.3a
)
j
Mg =
1836.59 kip ft 1836.59
Moment due to weight of member
xx es porciento estimado de esfuerzo "at jacking" (fpj) que se pierde por encogimiento Elastico
xx := 1
Pt := ( xx) ( fpj Astrand Nstrand)
% of Total force at transfer assumed
Pt = 1735.02 kip
PP := 1624.5 = 1624.5 Pt := PP kip fcgp :=
Pt A
+
(Pt ecc) ecc I
I = 521180 in +
M g ecc
M g ecc
I
I
1.94 fcgp = ksi 1.94 ΔfpES :=
Ep Eci
fcgp
Pt_revised := ( fpj - ΔfpES) Astrand Nstrand
=
-0.81 ksi -0.81
4
ecc = -19.17 in
2
A = 1013 in
Concrete Stresses at center of gravity of prestressing tendons due to prestressing force at transfer and the self weight of the member at section of maximum moment
12.89 ksi 12.89
ΔfpES =
Losses due to Elastic Shortening
1624.58 kip 1624.58
Pt_revised =
equal to assumed OK!
Programa en mathcad para calcular perdida elastica: (el mismo calculo pero con iteracion programada)
p. 25/73
CE6340
Ptt :=
Prestressed Beam Example
26/73
xx 1.0
j
flag 0
Pt = 1624.5 kip
while flag = 0 pt xx ( fpj Astrand Nstrand) fcgp loss
pt A Ep Eci
+
( pt ecc) ecc I
+
Concrete Stresses at center of gravity of prestressing tendons due to prestressing force at transfer and the self weight of the member at section of maximum moment
M g ecc j
I fcgp :=
fcgp
pt_new ( fpj - loss) Astrand Nstrand flag
0 if
Pt A
+
(Pt ecc) ecc I
+
M g ecc
fcgp =
I
Losses due to Elastic Shortening
pt - pt_new > 0.25 kip
1 if xx < 0.00
ΔfpES :=
1 otherwise
Ep Eci
12.89 ksi 12.89
ΔfpES =
fcgp
xx xx - 0.001 pt
A 5.9.5.4 Refined estimate of Time dependent Losses 2
Aps := Astrand Nstrand
Aps = 8.57 in
Servie life of bridge (days)
t i := 2
Age in days to Transfer
t d := 30
Age in days to concrete deck placement A 5.9.5.4.2 Average annual ambient relative humidity (%)
H := 70 VSratio = 4.56 in
ks := max 1.0 , 1.45 - 0.13 khc := 1.56 - 0.008 H kf :=
max( Ptt ) = -0 kip
Area of prestress strands
t f := 100 365
From Transfer to Deck Placement
5 1+
fci
VSratio in
1.94 ksi 1.94
ks = 1 khc = 1 kf = 0.83
ksi
p. 26/73
CE6340
Prestressed Beam Example
ktd( t ) :=
t fci 64 - 4 + t ksi
ti = 2
ktd( t f - t i) = 1
Ψb( t , t i) := 1.9 ks khc kf ktd( t ) t i
- 0.118
khs := 2.00 - 0.014 H
27/73
t d = 30
Creep Coefficient
ti = 2
A 5.4.2.3.2-1
khs = 1.02 -3
ε bid := ks khs kf ktd( t d - t i) 0.48 10
ε bid = 1.587 10
-4
Shrinkage Strain
A 5.4.2.3.3-1
1
Kid := 1+
2 Ep Aps A ecc ( 1 + 0.7 Ψb( tf , ti) ) 1 + Eci A I
1 1
Ep Eci
t f = 36500
3.78 ksi 3.78
ΔfpSR := ε bid Ep Kid
ΔfpCR :=
Kid = 0.84
ΔfpSR =
fcgp Ψ b( t f , t i) Kid
ΔfpCR =
15.72 ksi 15.72
Losses due to shinkage
Losses Due to Creep
Relaxation of Strands: ΔfpR1 := j
1 30.0
fpj
fpy
- 0.55 max( fpj , 0.55 fpy)
1.91 ksi 1.91
ΔfpR1 =
ΔfpR1 := 1.2 ksi j
Relaxation at transfer
A 5.9.5.4.2c
Losses due to relaxation at transfer use 1.2 ksi acording to AASHTO 5.9.5.4.2c for low relaxation strands
-20.71
(-ΔfpSR - ΔfpCR - ΔfpR1) = -20.71 ksi
Loses from transfer to deck placement
p. 27/73
Ψb( t f , t i) = 1.46
CE6340
Prestressed Beam Example
From deck placement to final time
eccc := ( yb_strand - yb_c)
Kdf := j
eccc =
28/73
A 5.9.5.4.3
-34.35 in -34.94
Distance from centroid of composite section to centroid of strands
1
Ac eccc Ep Aps j j 1+ 1+ Eci Ac Ic j j
2
( ) (1 + 0.7Ψ (t , t ))
b f
i
-3
-4
ε bdf := ks khs kf ktd( t f - t i) 0.48 10 ΔfpSD := ε bdf Ep Kdf
ε bdf = 4.075 10
ΔfpSD =
9.75 ksi 9.75
ΔPt := Aps ( ΔfpSR - ΔfpCR - ΔfpR1)
(
(
7
)
(
7
Δfcd :=
ΔPt A
j
ΔfpCD := j
ΔfpCD =
j
+
1599.08 kip ft 1707.87
Moment due to slab & Haunch
M slab =
M rail =
122.18 kip ft 122.18
Moment due to Railing
)
M fws =
147.83 kip ft 158.39
Moment due to FWS
(ΔP ecc)ecc + M t
-112.55 kip -112.55
)
M rail := M c xten , wDC2
M fws := M c xten , wDW
Losses due to shinkage
ΔPt =
M slab := M nc xten , wslab + whaunch 7
j
I
slab ecc j
I
+
(M
rail j
)(
+ M fws yb_strand - yb_c j
Ic
j
Ep Ep fcgp j ( Ψb( tf , ti) - Ψb( td , ti) ) Kdf j + Δfcd j Ψb( tf , td) Kdf j Eci Ec
3.98 ksi 3.7
0.84 0.84
Kdf =
Losses Due to Creep p. 28/73
j
)
Δfcd =
-1 ksi -1.05
CE6340
Prestressed Beam Example
29/73
Relaxation from deck placement to final time: ΔfpR2 := ΔfpR1 j
j
1.2 ksi 1.2
ΔfpR2 =
Losses due to relaxation after transfer
Deck Shrinkage Gain
VSratio :=
2 S t s_total
VSratio = 8.94 in
2 S + 2 t s_total
ks := max 1.0 , 1.45 - 0.13
VSratio in
khc := 1.56 - 0.008 H
ks = 1
khc = 1
5
kf := 1+
ktd( t ) :=
kf = 1
0.80fc_s ksi
t 0.80fc_s 64 - 4 + t ksi
Ψd( t , t i) := 1.9 ks khc kf ktd( t ) t i
- 0.118
Creep Coefficient
khs := 2.00 - 0.014 H
khs = 1.02 -3
-4
ε ddf := ks khs kf ktd( t f - t d) 0.48 10
(
ed := Arm - yb_c j
Δfcdf := j
3
j
j
ε ddf = 4.89 10
)
ε ddf b eff t s Ec_s 1 + 0.7 Ψ d( t f , t d)
A 5.4.2.3.2-1
( )
eccc ed 1 j j Ac Ic j j
p. 29/73
Shrinkage Strain
A 5.4.2.3.3-1
CE6340
Prestressed Beam Example
30/73
Ψb( t f , t d) = 1.06 ΔfpSS := j
Ep Ec
Δfcdf Kdf ( 1 + 0.7 Ψ b( t f , t d) ) j
ΔfpSS =
j
9.31 ksi 9.68
t s = 9 in Ψd( t f , t d) = 1.27
Total Losses
A 5.9.5.1-1
b eff =
ΔfpT := ΔfpES + ( ΔfpSR + ΔfpCR + ΔfpR1) + ( ΔfpSD + ΔfpCD + ΔfpR2) - ΔfpSS
ΔfpT =
39.22 ksi 38.57
Δfcdf =
Total Losses Refined Estimate
Time Dependen Losses from Lump Sump Estimate: Using Table 1 A5.9.5.3 for T beams
Partial Prestress Ratio
PPR := 1.0
Δfpt := 33.0 ksi 1.0 - 0.15 Δfpt = 33 ksi
A 5.5.4.2.1-2
fc - 6.0 ksi 6.0 ksi
+ 6.0 PPR ksi - 6.0 ksi
1 1
ΔfpT2 =
Total Losses Lump Sum Estimate using Equation 1
A5.9.5.3-1
γh := 1.7 - 0.01 H γh = 1
H = 70
γst :=
5 fci 1 + ksi
A 5.9.5.3 Table 1
Lump sum estimate of time dependent Losses
ΔfpT2 := ΔfpES + Δfpt
45.89 ksi 45.89
γst = 0.83
Δfpr := 2.4 ksi
For Low Relaxation Strand 5.9.5.3
fpj = 202.5 ksi
Stress prior to transfer
p. 30/73
84 in 90
Total Losses Lump Sum Estimate
1.05 ksi 1.09
CE6340
ΔfpLT :=
Prestressed Beam Example
10 fpj Aps A
γh γst + 12.0 ksi ( γh γst) + Δfpr
1 1
ΔfpT2 := ΔfpES + ΔfpLT
ΔfpT2 =
31/73
Total Time Dependent Losses Lump Sum Estimate
39.56 ksi 39.56
A5.9.5.3-1
Total Losses ΔfpLT = 26.67 ksi Lump Sum Estimate
Summary of Loss Calculations: Instantaneous
Tr ansfer to Deck Placement
12.89 ksi 12.89
3.78 ksi 3.78
ΔfpES =
ΔfpSR =
ΔfpCR =
15.72 ksi 15.72 1.2 ksi 1.2
Deck to Final Time ΔfpSD =
9.75 ksi 9.75
ΔfpCD =
3.98 ksi 3.7 1.2 ksi 1.2
ΔfpR1 =
ΔfpR2 =
ΔfpSS =
9.31 ksi 9.68
(GAIN)
ΔfpT := ΔfpT2
ΔfpT =
39.56 ksi 39.56
Total Loss used for calculations
Both methods for calculation of the prestressed losses are valid for the design per AASTHO LRFD 5.9.5. Since we took the time to compute the refined estimates, those will be used for the design. fpt := fpj - ΔfpES j
A 5.9.3.1 Table 1
j
189.61 ksi 189.61
fpt =
(
Effective stress in strands at transfer
)
fpe := min fpj - ΔfpT , 0.8 fpy j
fpe =
162.94 ksi 162.94
j
0.8 fpy = 194.4 ksi
A 5.9.3.1 Table 1
Effective stress in strands after all losses
p. 31/73
CE6340
Prestressed Beam Example
Tr ansfer lenght:
A 5.11.4.1
Ltransfer := 60 d strand
Tr ansfer length of strand
Ltransfer = 30 in
Strand Arrangement: Force reduction factor due to transfer length of strands -Loh 0.0 Ltransfer 1 ,x - x Lfactor := linterp L , + Lcc ten span L i span transfer i jspan 1 jspan 1 0.0 Lspan + Loh jspan
(
fci
ksi
ften_i := -min 0.2 ksi , 0.0948
ftop_i
j ,i
ftop_i =
:=
ksi
)
ften_i = -0.2 ksi
M girder
j ,i
St
0 0.25 0.47 0.84 1.1 1.26 1.31 1.26 1.1 0.84 0.47 0.25 0 ksi 0 0.25 0.47 0.84 1.1 1.26 1.31 1.26 1.1 0.84 0.47 0.25 0
Nstrand_max_i
j ,i
Nstrand_max_i =
32/73
ften_i - ftop_i
1 eteni Astrand Lfactori fptj + St A
:= floor
j ,i
267 100 149 231 290 325 336 325 290 231 149 100 267 267 100 149 231 290 325 336 325 290 231 149 100 267
p. 32/73
0.17 1 1 1 1 1 Lfactor = 1 1 1 1 1 1 0.17
CE6340
ftop
j ,i
ftop =
:=
Prestressed Beam Example
M DC1
j ,i
+
St
γDC
s,k
M DC2
j ,i
+ γDW
M DW
s, k
j ,i
+ γLL
M LL_min
s, k
33/73
j ,i
St_c
j
0 0.47 0.9 1.59 2.07 2.36 2.44 2.31 1.98 1.45 0.69 0.21 -0.32 ksi 0 0.49 0.92 1.63 2.13 2.42 2.5 2.37 2.03 1.48 0.69 0.19 -0.37
Nstrand_max
ften - ftop
j ,i
Nstrand_max =
1 eteni Astrand Lfactori fpej + St A
:= floor
j ,i
724 243 352 531 657 731 751 719 634 497 300 176 227 724 246 358 542 672 747 767 734 646 503 298 169 146
Total Prestressed Force Required at full service loads:
(
Preq := j
1 + ecc A S b
Pe := Astrand fpe
Nreq := j
T
ften - min fbottom
Pe
j
1336.19 kip 1372.72
Preq =
Pe =
( )
max Preq
) j
j
Nreq =
Required Prestress
24.93 kip 24.93
Effective prestressing force in one strand
53.6 55.06
Nstrand = 56
p. 33/73
OK!
CE6340
Prestressed Beam Example
34/73
Design required debonding for Service III loading: (maximum of two different debonding lengths) Ndebond_req
j ,i
(
:= max 0 , Nstrand - Nstrand_max
j ,i
, Nstrand - Nstrand_max_i
Ndebond_req =
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Ndebond :=
dummy 0
i
tentotal 2
j ,i
for k i .. floor
(
dummy max dummy, Ndebond_req
1, k
, Ndebond_req
)
2, k
tentotal +1 2
(
tentotal 2
if i < floor
for k i .. floor
dummy max dummy, Ndebond_req
)
tentotal +1 2
if i > floor 1, k
, Ndebond_req
)
2, k
dummy
T
Ndebond = ( 0 0 0 0 0 0 0 0 0 0 0 0 0 )
Use symetric debonding & only debond up to three differents sequences, debond only 40% of strands in one layer and 25% of total # of strands A 5.11.4.3 floor( 0.25 Nstrand) = 14
Nstrand - floor( 0.25 Nstrand) = 42
floor( 0.125 Nstrand) = 7
Nstrand - floor( 0.25 Nstrand) floor( 0.0 Nstrand) Nstrand := floor( 0.25 Nstrand)
42 Nstrand = 0 14
xspanjspan + Lcc1 - Loh xspanjspan + 1 - Lcc2 + Loh x x ten ten xstrand := 2 tentotal- 1 xten xten 3 tentotal- 2
0 119 xstrand = 6.4 112.6 ft 12.3 106.7
p. 34/73
full length debonded debonded
CE6340
Prestressed Beam Example
35/73
Total Prestress Force: Pt_tot :=
N
Pe_tot :=
N
strand Astrand fpt
strand Astrand fpe
Pt_tot =
1624.58 kip 1624.58
Prestress force at transfer
Pe_tot =
1396.05 kip 1396.05
Effective Prestress force after all losses
Prestress Force at ten points:
xspanjspan + Lcc1 - Loh - 0.001 in 0.0 kip xstrand 0.0 kip k, 1 xstrand + Ltransfer 3 Nstrandk Astrand fptj k, 1 Pt := linterp , , xten j ,i i xstrand - Ltransfer Nstrandk Astrand fptj k, 2 k =1 xstrand 0.0 kip k, 2 0.0 kip x - L + Loh + 0.001 in spanjspan + 1 cc2
Pt =
243.69 1218.44 1218.44 1624.58 1624.58 1624.58 1624.58 1624.58 1624.58 1624.58 1218.44 1218.44 243.69 kip 243.69 1218.44 1218.44 1624.58 1624.58 1624.58 1624.58 1624.58 1624.58 1624.58 1218.44 1218.44 243.69
xspanjspan + Lcc1 - Loh - 0.001 in 0.0 kip xstrand 0.0 kip k, 1 xstrand + Ltransfer Nstrand Astrand fpe 3 k, 1 k j Pe := linterp , , xten j ,i i xstrand - Ltransfer Nstrandk Astrand fpej k, 2 k =1 xstrand 0.0 kip k, 2 0.0 kip x - L + Loh + 0.001 in spanjspan +1 cc2
Pe =
209.41 1047.04 1047.04 1396.05 1396.05 1396.05 1396.05 1396.05 1396.05 1396.05 1047.04 1047.04 209.41 kip 209.41 1047.04 1047.04 1396.05 1396.05 1396.05 1396.05 1396.05 1396.05 1396.05 1047.04 1047.04 209.41
p. 35/73
CE6340
Prestressed Beam Example
36/73
Stress Limits: Temporary before losses
A 5.9.4.1
fci
ksi
ften_i := -min 0.2 ksi , 0.0948
ksi
fcomp_i := 0.60 fci
fc ksi
ften_slab := -0.19
ϕw := 1.0
0.0948
fci ksi
ksi = 0.21 ksi
fcomp_i = 3 ksi
At service limit state after losses ften := -0.19
ften_i = -0.2 ksi
ksi
fc_s ksi
A 5.9.4.2 ften = -0.47 ksi
ksi
ften_slab = -0.42 ksi
AASHTO GIRDER
NOT REQUIRED
A 5.7.4.7.2c-2
A5.9
0.45 fcomp := 0.40 fc 0.60 ϕw
2.7 fcomp = 2.4 ksi 3.6
fcomp_slab := 0.6 fc_s
fcomp_slab = 3 ksi
for permanent loads for LL & 1/2 permanent loads ( ELIMINADO 2010, Fatiga SEC 5.5) for service loads
p. 36/73
CE6340
Prestressed Beam Example
37/73
Compute Stresses, Control Points &/or Lenght of Debonding for Service I Limit State: s := 2
k := 1
a) At release: (Initial prestress & girder dead load) stresses fbot_i
ftop_i
P.t es P de jacking menos perdida elastica
1
j ,i
:= Pt j ,i
A 1
j ,i
:= Pt j ,i
A
+
+
Mgirderj , i + Sb Sb
eten
i
.6 fc = 3.6 ksi
Mgirderj , i + St St
eten
i
fbot_i =
0.53 2.38 2.15 2.65 2.38 2.22 2.16 2.22 2.38 2.65 2.15 2.38 0.53 ksi 0.53 2.38 2.15 2.65 2.38 2.22 2.16 2.22 2.38 2.65 2.15 2.38 0.53
ftop_i =
-0.04 0.06 0.28 0.59 0.85 1.01 1.06 1.01 0.85 0.59 0.28 0.06 -0.04 ksi -0.04 0.06 0.28 0.59 0.85 1.01 1.06 1.01 0.85 0.59 0.28 0.06 -0.04
control points: emax
j ,i
:=
dummy1
Sb Pt
j ,i
fcomp_i -
Pt
j ,i
A
-
M girder
j ,i
Sb
emin
j ,i
:=
dummy1
Sb Pt
j ,i
ften_i -
Pt
j ,i
A
-
M girder
j ,i
Sb
Pt M girder j ,i j ,i dummy2 ften_i Pt A St j ,i
Pt M girder j ,i j ,i dummy2 fcomp_i Pt A St j ,i
dummy max( dummy1 , dummy2)
dummy min( dummy1 , dummy2)
dummy -yb if dummy -yb
dummy yt if dummy yt
dummy
dummy
St
St
emin =
29.48 15.34 12.26 5.75 3.04 1.41 0.86 1.41 3.04 5.75 12.26 15.34 29.48 in 29.48 15.34 12.26 5.75 3.04 1.41 0.86 1.41 3.04 5.75 12.26 15.34 29.48
emax =
-30.36 -22.77 -25.84 -22.7 -25.41 -27.04 -27.58 -27.04 -25.41 -22.7 -25.84 -22.77 -30.36 in -30.36 -22.77 -25.84 -22.7 -25.41 -27.04 -27.58 -27.04 -25.41 -22.7 -25.84 -22.77 -30.36
p. 37/73
CE6340
Prestressed Beam Example
38/73
b) Effective presstress & dead loads:
fbot_permanent
ftop_permanent
1
j ,i
:= Pe j ,i
A
1
j ,i
fbot_permanent =
:= Pe j ,i
A
+
+
γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + + Sb Sb Sb_c j
eten
i
γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + + St St St_c j
eten
i
0.45 1.75 1.28 1.28 0.74 0.44 0.35 0.49 0.85 1.44 1.49 1.98 0.72 ksi 0.45 1.73 1.25 1.22 0.67 0.35 0.27 0.41 0.78 1.39 1.47 1.98 0.72
2.7 fcomp = 2.4 ksi 3.6
fcomp = 2.7 ksi 1
-0.03 0.32 0.75 1.4 1.9 2.19 2.29 2.18 1.86 1.34 0.67 0.24 -0.12 ftop_permanent = ksi -0.03 0.33 0.77 1.45 1.96 2.27 2.36 2.25 1.93 1.39 0.7 0.25 -0.12 control points: emax
j ,i
:=
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + + 1 Pe A Sb Sb_c j ,i j P γ M γ M + γ M ej , i DC DC1 DC DC2 DW DW St s, k j ,i s,k j ,i s, k j , i dummy2 ften - + + Pe A St St_c j ,i j dummy1
Sb
fcomp -
(
dummy max dummy1 , dummy2 , emax
j ,i
)
dummy -yb if dummy -yb dummy
emin
j ,i
:=
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + + Pe A Sb Sb_c j ,i j P γ M γ M + γ M ej , i DC DC1 DC DC2 DW DW St s, k j ,i s, k j ,i s,k j , i dummy2 fcomp - + + 1 Pe A St St_c j ,i j dummy1
Sb
ften -
(
dummy min dummy1 , dummy2 , emin
j ,i
)
dummy yt if dummy yt dummy
p. 38/73
CE6340
Prestressed Beam Example
c) Effective presstress, dead & Live loads
1
MDC1j , i γDCs , k MDC2j , i + γDWs , k MDWj , i + γLLs , k MLL_min j , i + + Sb Sb Sb_c j
eten
i
fbot_service_I
:= Pe
ftop_service_I
:= Pe
fbot_service_I =
0.45 1.77 1.33 1.38 0.89 0.63 0.6 0.79 1.2 1.83 2.03 2.67 1.57 ksi 0.45 1.76 1.32 1.35 0.87 0.61 0.59 0.79 1.23 1.9 2.17 2.86 1.83
j ,i
j ,i
A
1
j ,i
j ,i
A
+
+
MDC1j , i γDCs , k MDC2j , i + γDWs , k MDWj , i + γLLs , k MLL_maxj , i + + St St St_c j
eten
i
fcomp = 3.6 ksi 3
ftop_service_I =
-0.03 0.39 0.88 1.61 2.17 2.49 2.58 2.43 2.06 1.47 0.72 0.26 -0.12 ksi -0.03 0.4 0.9 1.65 2.22 2.54 2.64 2.49 2.12 1.51 0.75 0.27 -0.12
control points: emax
j ,i
:=
dummy1
Sb Pe
j ,i
(
fcomp -
3
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + γLLs , k MLL_minj , i + + A Sb Sb_c j
dummy max dummy1 , e max
j ,i
)
dummy -yb if dummy -yb dummy
emin
j ,i
:=
dummy2
St Pe
j ,i
(
fcomp -
3
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDWs , k MDWj , i + γLLs , k MLL_maxj , i + + A St St_c j
dummy min dummy2 , emin
j ,i
)
dummy yt if dummy yt dummy
p. 39/73
39/73
CE6340
Prestressed Beam Example
d) Live Load & half the sum of effective presstress & dead loads:
fbot_half_I
j ,i
ftop_half_I
j ,i
:=
COTEJO ELIMINADO en 2010 !!!! Añadir Cotejo Fatiga SEC 5.5
1 eteni MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i γLLs , k MLL_min j , i 1 + Pe + + + 2 j ,i A Sb Sb Sb_c Sb_c
:=
40/73
j
j
1 eteni MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i γLLs , k MLL_maxj , i 1 + Pe + + + 2 j ,i A St St St_c St_c
j
fbot_half_I =
0.23 0.9 0.69 0.74 0.52 0.42 0.42 0.54 0.77 1.11 1.28 1.68 1.21 ksi 0.23 0.9 0.69 0.74 0.53 0.43 0.45 0.59 0.84 1.21 1.43 1.88 1.47
ftop_half_I =
-0.01 0.23 0.5 0.92 1.22 1.39 1.43 1.35 1.13 0.8 0.38 0.14 -0.06 ksi -0.01 0.23 0.51 0.93 1.24 1.41 1.45 1.37 1.15 0.81 0.39 0.15 -0.06
j
control points: emax
j ,i
:=
dum1
Sb Pe
j ,i
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i + 2 γLLs , k MLL_min j , i + + A Sb Sb_c j
2 fcomp -
2
(
dummy max dum1 , e max
j ,i
)
dummy -yb if dummy -yb dummy
emin
j ,i
:=
dum2
St Pe
j ,i
2 fcomp -
(
2
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + 2 γLLs , k MLL_maxj , i + + A St St_c j
dummy min dum2 , emin
j ,i
)
dummy yt if dummy yt dummy
p. 40/73
CE6340
Prestressed Beam Example
41/73
d) (Revised) Fatigue I & half the sum of effective presstress & dead loads:
Cotejo Fatiga SEC 5.5
γ de fatiga = 0.75 para fatiga I 2x0.75 = 1.5
fbot_half_I
j ,i
ftop_half_I
j ,i
fbot_half_I =
ftop_half_I =
:=
1 eteni MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i 1.5 M f_minj , i 1 + Pe + + + 2 j ,i A Sb Sb Sb_c Sb_c
:=
j
j
1 eteni MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i 1.5 M f_maxj , i 1 + Pe + + + 2 j ,i A St St St_c St_c
j
0.23 0.89 0.66 0.69 0.44 0.31 0.29 0.38 0.59 0.9 0.95 1.21 0.59 ksi 0.23 0.89 0.67 0.7 0.46 0.35 0.34 0.46 0.69 1.03 1.11 1.39 0.78
j
0.4 fc = 2.4 ksi
-0.01 0.19 0.44 0.8 1.07 1.23 1.27 1.21 1.03 0.73 0.36 0.13 -0.06 ksi -0.01 0.2 0.45 0.82 1.1 1.26 1.31 1.24 1.05 0.75 0.37 0.14 -0.06
control points: emax
j ,i
:=
dum1
Sb Pe
j ,i
(
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDWs , k MDWj , i + 2 1.5 Mf_minj , i + + A Sb Sb_c j
2 fcomp -
2
(
dummy max dum1 , e max
j ,i
)
)
dummy -yb if dummy -yb dummy
emin
j ,i
:=
dum2
St Pe
j ,i
2 fcomp -
(
2
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + 2 1.5 Mf_maxj , i + + A St St_c j
dummy min dum2 , emin
j ,i
)
dummy yt if dummy yt dummy
p. 41/73
CE6340
Prestressed Beam Example
42/73
Compute Stresses & Control PointsI for Service III Limit State: s := 3
k := 1 γLL
e) Effective presstress, dead loads & live loads:
1
fbot_service_III
:= Pe
ftop_service_III
:= Pe
j ,i
j ,i
j ,i
+
A
1
j ,i
+
A
s, k
= 0.8
MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + γLLs , k MLL_maxj , i + + Sb Sb Sb_c j
eten
i
MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + γLLs , k MLL_minj , i + + St St St_c j
eten
i
ften = -0.47 ksi fbot_service_III =
0.44 1.58 0.97 0.76 0.1 -0.26 -0.34 -0.12 0.37 1.14 1.39 1.93 0.71 ksi 0.44 1.57 0.94 0.71 0.03 -0.34 -0.41 -0.2 0.3 1.09 1.36 1.92 0.71
ftop_service_III =
-0.03 0.31 0.73 1.37 1.86 2.14 2.22 2.1 1.77 1.24 0.53 0.05 -0.35 ksi -0.03 0.32 0.76 1.41 1.91 2.2 2.28 2.15 1.81 1.26 0.53 0.03 -0.4
control points:
emax
j ,i
:=
dummy1
St Pe
j ,i
(
ften -
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDWs , k MDW j , i + γLLs , k MLL_minj , i + + A St St_c j
dummy max dummy1 , e max
j ,i
)
dummy -yb if dummy -yb dummy
emin
j ,i
:=
dummy2
Sb Pe
j ,i
(
ften -
Pej , i γDCs , k MDC1j , i γDCs , k MDC2j , i + γDW s , k MDWj , i + γLLs , k MLL_maxj , i + + A Sb Sb_c j
dummy min dummy2 , emin
j ,i
)
dummy yt if dummy yt dummy
p. 42/73
CE6340
Prestressed Beam Example
43/73
Plot Control Points
emin =
29.48 12.74 3.21 -4.86 -12.56 -16.81 -17.66 -15.19 -9.43 -2.85 9.66 15.34 29.48 in 29.48 12.51 2.77 -5.45 -13.34 -17.7 -18.58 -16.07 -10.18 -3.45 9.29 15.34 29.48
emax =
-30.36 -22.77 -25.84 -22.7 -25.41 -27.04 -27.58 -27.04 -25.41 -22.7 -25.84 -22.77 -28.32 in -30.36 -22.77 -25.84 -22.7 -25.41 -27.04 -27.58 -27.04 -25.41 -22.7 -25.84 -22.77 -24.15
T
eten = ( -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 -19.17 ) in xvec
iten
:= xten
iten
- xspan
jspan
- Lcc
1
20 T emin
in T
emax
0
in eten in - 20
0
50
100 xvec ft
ecc = -19.17 in
p. 43/73
CE6340
Prestressed Beam Example
44/73
5. Stresses in bot fibers due to service Loads at top of slab (Service I & III) s := 3
k := 1
fslab_service_III
γDC
:=
j ,i
s, k
M DC2
+ γDW
j ,i
M DW
s, k
S s_c
j ,i
+ γLL
s,k
M LL_min
j ,i
j
n
s := 2
k := 1
fslab_service_I
j ,i
:=
γDC
s,k
M DC2
j ,i
+ γDW
M DW
s, k
Ss_c
j ,i
+ γLL
M LL_max
s, k
j ,i
j
n
0 0.03 0.05 0.07 0.05 -0 -0.09 -0.22 -0.38 -0.58 -0.94 -1.23 -1.56 ksi 0 0.02 0.04 0.04 0 -0.07 -0.18 -0.32 -0.5 -0.71 -1.11 -1.46 -1.84
fslab_service_III =
fslab_service_I =
0.03 0.4 0.76 1.27 1.57 1.7 1.64 1.43 1.07 0.55 -0.01 -0.22 -0.41 ksi 0.03 0.38 0.73 1.22 1.52 1.63 1.58 1.38 1.02 0.53 -0.02 -0.22 -0.41
Check Stresses with maximum allowed: ften_i = -0.2 ksi
1. Stresses at Release Check_Init_Top
j,i
:=
(
"OK!" if ften_i ftop_i
j ,i
)
fcomp_i top_comp_zone = 0 i
fcomp_i = 3 ksi
"BAD!" otherwise ftop_i =
-0.04 0.06 0.28 0.59 0.85 1.01 1.06 1.01 0.85 0.59 0.28 0.06 -0.04 ksi -0.04 0.06 0.28 0.59 0.85 1.01 1.06 1.01 0.85 0.59 0.28 0.06 -0.04
Check_Init_Top =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Init_Bot
:=
j,i
(
"OK!" if ften_i fbot_i
j ,i
)
fcomp_i bot_comp_zone = 0 i
"BAD!" otherwise
Check_Init_Bot =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
p. 44/73
CE6340
Prestressed Beam Example
ften = -0.47 ksi
2. Stresses due to Permanent Loads Check_Permanent_Top
j,i
:=
45/73
(
"OK!" if ften ftop_permanent
j ,i
fcomp
1
)
fcomp = 2.7 ksi 1
"BAD!" otherwise
Check_Permanent_Top =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Permanent_Bot
:=
j,i
(
"OK!" if ften fbot_permanent
j ,i
fcomp
)
1
"BAD!" otherwise Check_Permanent_Bot =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
3. Stresses due to Transient Loads & 1/2 Permanent Loads Check_Transient_Top
j,i
:=
(
"OK!" if ftop_half_I
j ,i
fcomp
2
)
fcomp = 2.4 ksi 2
"BAD!" otherwise
Check_Transient_Top =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Transient_Bot
:=
j,i
(
"OK!" if fbot_half_I
j ,i
fcomp
)
2
"BAD!" otherwise Check_Transient_Bot =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
p. 45/73
CE6340
Prestressed Beam Example
46/73
4. Stresses due to Service Loads Check_Service_Top
j,i
:=
(
"OK!" if ften ftop_service_III
j ,i
ftop_service_I
j ,i
fcomp
3
)
ften = -0.47 ksi fcomp = 3.6 ksi 3
"BAD!" otherwise
Check_Service_Top =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Service_Bot
:=
j,i
(
"OK!" if ften fbot_service_III
j ,i
fbot_service_I
j ,i
fcomp
3
)
"BAD!" otherwise
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Service_Bot =
5. Stresses due to service loads at Top of slab Check_Service_slab
j,i
:=
"OK!" if ften_slab fslab_service_III
j ,i
fslab_service_I
"BAD!" otherwise
Check_Service_slab =
j ,i
fcomp_slab
ften_slab = -0.42 ksi fcomp_slab = 3 ksi
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "BAD!" "BAD!" "BAD!" "BAD!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "BAD!" "BAD!" "BAD!" "BAD!" "BAD!"
p. 46/73
CE6340
Prestressed Beam Example
47/73
Compute Factored Moments & Shears (Stregnth I) at Ten points: s := 1
M u_max
j ,i
:= maxη γDC
(
M DC1
s, 1
j ,i
+ M DC2
j ,i
γDW
)+γ
s, 1
DW
M DW
s, 1
= 1.5
j ,i
γDC
+ γLL
s, 1
s,2
= 1.25
M LL_max
(
γLL
s, 2
)
, η γDCs , 2 MDC1j , i + MDC2j , i + γDW s , 2
j ,i
= 1.75
61 1597 3036 5249 6709 7471 7548 6970 5749 3901 1629 518 -343 kip ft 61 1627 3094 5350 6840 7619 7700 7112 5869 3984 1662 527 -356
M u_max =
(
M u_min =
-3 555 1051 1824 2319 2534 2469 2125 1502 600 -894 -2017 -3569 kip ft -4 549 1037 1790 2255 2431 2319 1919 1231 255 -1419 -2720 -4479
Vu_max
:= maxη γDC
j ,i
s, 1
s, 1
(
VDC1
j ,i
j ,i
+ M DC2
+ VDC2
j ,i
j ,i
)+γ
DW
DW
M DW
s, 1
s,2
VDW
j ,i
j ,i
+ γLL
+ γLL
s,2
s,2
M LL_min
(
:= minη γDC
j ,i
M DC1
)+γ
M u_min
VLL_max
315 287 260 204 150 98 49 10 -27 -62 -93 -32 52 kip 321 293 264 208 153 100 49 9 -29 -64 -96 -35 49
Vu_min
:= minη γDC
Vu_min =
97 85 73 44 4 -39 -83 -135 -188 -240 -291 -309 -325 kip 100 88 76 46 5 -39 -83 -137 -190 -244 -297 -315 -330
s, 1
(
VDC1
j ,i
+ VDC2
j ,i
)+γ
DW
s,2
VDW
j ,i
+ γLL
p. 47/73
s,2
VLL_min
(
)
, η γDCs , 2 VDC1j , i + VDC2j , i + γDWs , 2 VDW
j ,i
Vu_max =
j ,i
)
, η γDCs , 2 MDC1j , i + MDC2j , i + γDWs , 2
j ,i
(
)
, η γDCs , 2 VDC1j , i + VDC2j , i + γDWs , 2 VDW
j ,i
CE6340
Prestressed Beam Example
48/73
3
7.7 10
3
T M u_max
4.655 10
kip ft
3
T M u_min
1.611 10
kip ft
3
- 1.434 10
3
- 4.479 10
0
50
100 xvec ft
3
1.05310
T
518.88
Vu_max kip ft
T
- 15.344
Vu_min kip ft
- 549.567
3
- 1.08410
0
50
100 xvec ft
p. 48/73
CE6340
Prestressed Beam Example
49/73
Nominal Flexural Resistance resitance factor for flexure of prestressed concrete
ϕf := 1.0
Positive Moment capacity Mn_pos 3
Aps := Astrand i
Nstrandk if xstrandk , 1 xteni xstrandk , 2 k =1 0 otherwise
2
As := 0.0 in
2
T
Aps = ( 6.43 6.43 8.57 8.57 8.57 8.57 8.57 8.57 8.57 8.57 8.57 6.43 6.43 ) in
fpy
fpu
k := 2 1.04 -
β1( fc) :=
k = 0.28
A 5.7.3.1.1-2
0.85 if fc 4.0 ksi fc max 0.85 0.05 - 4.0 , 0.65 otherwise ksi
for bonded tendons
fucntion to compute beta factor β1( fc_s ) = 0.8 fc_s = 5 ksi
60.21
ecc = -19.17 in
(
d p := h c - yb + eten i
i
)
60.21 60.21 60.21 60.21 60.21 d p = 60.21 in 60.21 60.21 60.21 60.21 60.21 60.21
Distance from extreme compresion fiber to centroid of bonded tendons
p. 49/73
beta factor for slab
CE6340
Prestressed Beam Example
50/73
A 5.7.3.1.1-3 & 4 c_pos
j,i
Aps fpu + As fy
Aps fpu + As fy
i
:=
0.85 fc_s β1( fc_s ) b eff + k Aps j
i
i
if
fpu
0.85 fc_s β1( fc_s ) b eff + k Aps
dp
i
j
(
)
i
Aps fpu + As fy - 0.85 fc_s β1( fc_s ) b eff - b f_top t s i
j
0.85 fc_s β1( fc_s ) b eff + k Aps j
c_pos =
dp
ts
i
otherwise
fpu
Rectangular Behavior
T section behavior
dp
i
5.91 5.91 7.81 7.81 7.81 7.81 7.81 7.81 7.81 7.81 7.81 5.91 5.91 in 5.52 5.52 7.3 7.3 7.3 7.3 7.3 7.3 7.3 7.3 7.3 5.52 5.52
a := β1( fc_s ) c_pos
a=
i
fpu
t s = 9 in
4.73 4.73 6.24 6.24 6.24 6.24 6.24 6.24 6.24 6.24 6.24 4.73 4.73 in 4.42 4.42 5.84 5.84 5.84 5.84 5.84 5.84 5.84 5.84 5.84 4.42 4.42
fps
j ,i
fps =
:= fpu 1 - k
c_pos
j, i
dp
i
fpy = 243 ksi
A 5.7.3.1.1-1
263 263 260 260 260 260 260 260 260 260 260 263 263 ksi 263 263 261 261 261 261 261 261 261 261 261 263 263
Average stress in the presstresing steel
Compute development length:
fpsj , 1
ksi
l d_strand := max1.6 j
l d_strand =
-
(
)
6.4 fps - fpe d strand 4 fpj d strand j ,1 j d strand , + + 10 in 3 ksi fc fc 2
fpe
j
130.64 in 130.9
p. 50/73
A 5.11.4.2
CE6340
Prestressed Beam Example
51/73
Compute actual Pps including effect of development length:
xspanjspan + Lcc1 - Loh - 0.01 in 0.0 kip xstrand 0.0 kip k, 1 xstrand + Ltransfer Nstrand Astrand fpe k, 1 k j xstrand + l d_strand Nstrand Astrand fps 3 k, 1 j k j ,i Pps := linterp , , xten j ,i i xstrand - l d_strand Nstrandk Astrand fpsj , i k, 2 j k =1 xstrand - Ltransfer N A f k, 2 strandk strand pe j xstrand 0.0 kip k, 2 0.0 kip xspan - Lcc + Loh + 0.01 in jspan + 1 2
209.41 1344.79 1672.04 2229.39 2229.39 2229.39 2229.39 2229.39 2229.39 2229.39 1672.04 1344.79 209.41 kip 209.41 1345.47 1676.1 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 1676.1 1345.47 209.41
Pps =
M n_pos
j ,i
:= Pps d p j ,i
i
a
j,i
a ts j,i + 0.85 fc_s β1( fc_s ) b eff - b f_top ts max 0 in , - j 2 2 2
(
)
Nominal Moment Capacity
1010 6483 7955 10607 10607 10607 10607 10607 10607 10607 7955 6483 1010 A 5.7.3.2.2-1 kip ft 1012 6504 8002 10670 10670 10670 10670 10670 10670 10670 8002 6504 1012
M n_pos =
M r_pos
j ,i
M r_pos =
:= ϕf M n_pos
j ,i
1010 6483 7955 10607 10607 10607 10607 10607 10607 10607 7955 6483 1010 kip ft 1012 6504 8002 10670 10670 10670 10670 10670 10670 10670 8002 6504 1012 61 1597 3036 5249 6709 7471 7548 6970 5749 3901 1629 518 -343 kip ft 61 1627 3094 5350 6840 7619 7700 7112 5869 3984 1662 527 -356
M u_max =
PR_Mn_Pos
j,i
:=
M u_max M r_pos
j ,i
if M u_max
j ,i
j ,i
0 kip ft
0.0 otherwise
p. 51/73
CE6340
Prestressed Beam Example
Negative Moment Capacity resitance factor for flexure of non prestressed concrete
ϕf := 0.9
For negative moment the compression face is the bottom flange of the beam. This section is designed as a non-prestressed R/C section, thus the resistance factor (phi) equals 0.90. The bottom flange of the beam is in compression so f'c of the precast girder at final is used. assume that the deck reinforcement is at mid height of the deck.
ts
d s := h c -
Ru
j ,i
Ru =
:=
d s = 68.5 in
2
(
min M u_min
j ,i
, 0 kip ft
ϕf b f_bot d s
distance from centroid of reinforcement to extreme compresion fiber, assumed at center of slab.
)
2
0 0 0 0 0 0 0 0 0 0 0.09 0.2 0.36 ksi 0 0 0 0 0 0 0 0 0 0 0.14 0.28 0.45
ρ_req
j,i
:=
1 fy
fy 2 Ru j , i 0.85 fc 1 - 1 fy
0.85 fc
ρ_req =
0 0 0 0 0 0 0 0 0 0 0.002 0.003 0.006 0 0 0 0 0 0 0 0 0 0 0.002 0.005 0.008
As_req := ρ_req b f_bot d s
As_req =
0.01 0 0 0 0 0 0 0 0 0 2.93 6.68 12.02 2 in 0.01 0 0 0 0 0 0 0 0 0 4.67 9.08 15.24
p. 52/73
52/73
CE6340
Prestressed Beam Example
Use # 9 bars As bar = 1 si
2
As_bar := 1 in
9
d bar :=
8
53/73
in
As_bar fy 1.25 2 ksi fy in l d_bar := max in , 0.4 d bar 1.0 ksi fc ksi
A 5.11.2.1.1 & 5.11.2.1.2 top reinforcement factor = 1.4 N/A
l d_bar = 30.62 in
Compute provided negative steel reinforcement Number of Bars Required per Girder
max( As_req) As_bar As_bar
2
As := ceil
As = 16 in
As
Nbar_slab_neg := sbar_slab_neg :=
= 16
As_bar S
Nbar_slab_neg
= 5.25 in
Compute minimum cutoff point (tenth point) (colunm 1 Mneg start of span column 2 Mneg end of span) Xcutt
j ,1
:=
dummy 0 in for i 1 .. tentotal
dummy max dummy, xten +
As_req
j ,i
As
i
l d_bar if xten < xspan
i
L jspan
+
jspan
2
2
As_req
> 0 in
As_req
> 0 in
j ,i
dummy Xcutt
j ,2
:=
dummy xspan
jspan + 1
for i 1 .. tentotal
dummy min dummy, xten -
As_req
i
j ,i
As
l d_bar if xten xspan
i
L jspan
+
jspan
2
dummy
0.5 106.23 ft 0.5 105.96
Xcutt =
xspan
jspan
= 0 ft
xspan
p. 53/73
jspan + 1
= 120 ft
j ,i
2
CE6340
Prestressed Beam Example
54/73
Xcutt Revised for Crack Control (See Below)
0 xten9 Xcutt := 0 xten 9
T
xten = ( 0.5 6.4 12.3 24.1 35.9 47.7 59.5 71.3 83.1 94.9 106.7 112.6 118.5 ) ft tentotal = 13
0 83.1 ft 0 83.1
Xcutt =
p. 54/73
CE6340
Prestressed Beam Example
55/73
Now compute provided development length: ld
j ,i
:=
(
max 0 in , Xcutt
j ,1
(
- xten
i
max 0 in , xten - Xcutt
ld =
) )
if xten xspan
j ,2
i
L if xten < xspan i
jspan
+
2 L
i
jspan
jspan
+
jspan
2
0 0 0 0 0 0 0 0 0 141.6 283.2 354 424.8 in 0 0 0 0 0 0 0 0 0 141.6 283.2 354 424.8
As_eff
j ,i
:= min As ,
As l d_bar
ld
j ,i
l d_bar = 30.62 in
As_eff =
0 0 0 0 0 0 0 0 0 16 16 16 16 2 in 0 0 0 0 0 0 0 0 0 16 16 16 16
c_neg
:=
j,i
As_eff
j ,i
fy
0.85 fc β1( fc) b f_bot
As_eff fy if
j ,i
0.85 fc β1( fc) b f_bot
t f_bot
eq A5.7.3.1.1-4
As_eff fy - 0.85 fc β1( fc) ( b f_bot - t w) t s j ,i
c_neg =
0.85 fc_s β1( fc) b f_bot
otherwise
0 0 0 0 0 0 0 0 0 3.04 3.04 3.04 3.04 in 0 0 0 0 0 0 0 0 0 3.04 3.04 3.04 3.04
a := β1( fc) c_neg
a=
0 0 0 0 0 0 0 0 0 2.28 2.28 2.28 2.28 in 0 0 0 0 0 0 0 0 0 2.28 2.28 2.28 2.28
p. 55/73
eq A5.7.3.1.1-3
CE6340
M n_neg
Prestressed Beam Example
j ,i
M n_neg =
:= As_eff fy d s j ,i
a
j,i
a tf_bot + 0.85 fc β1( fc) ( b f_top - tw) ts max0 in , 2 2 2
56/73
eq A5.7.3.2.2-1
0 0 0 0 0 0 0 0 0 5388.74 5388.74 5388.74 5388.74 kip ft 0 0 0 0 0 0 0 0 0 5388.74 5388.74 5388.74 5388.74
M r_neg := ϕf M n_neg
M r_neg =
0 0 0 0 0 0 0 0 0 4849.9 4849.9 4849.9 4849.9 kip ft 0 0 0 0 0 0 0 0 0 4849.9 4849.9 4849.9 4849.9
M u_min =
-2.5 555.4 1050.6 1824.5 2318.8 2533.6 2469.2 2125.4 1502.4 600.1 -894.2 -2017.4 -3569.2 kip ft -4.5 548.5 1036.9 1790 2254.8 2431.1 2319.2 1919.1 1230.9 254.5 -1418.8 -2720.1 -4478.9
PR_Mn_Neg
j,i
:=
M u_min M r_neg
j ,i
(
if M u_min
j ,i
j ,i
)
0 kip ft M r_neg
j ,i
0.0 otherwise
p. 56/73
> 0.001 kip ft
CE6340
Prestressed Beam Example
57/73
Check Limits of Reinforcement Positive Moment Section: Maximum Reinforcement de
j ,i
de =
:=
Pps d p j ,i
A 5.7.3.3.1 SECTION DELETED 2005
i
Pps
j ,i
60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 in 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21 60.21
c_pos Check_Maximum_Reinforcement
j,i
:=
"OK!" if
j,i
de
0.42
j ,i
"BAD!" otherwise Check_Maximum_Reinforcement =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Minimum Reinforcement
A 5.7.3.3.2
At any section, the amount of prestresses and not prestressed tensile reinforcement should be adecuate to develop a factored flexural resistance, Mr, equal to the lesser of 1.2 Cracking Strenght or 1.33 times the factored moment required by the applicable load combination. Cracking Moment:
fce
1
j ,i
:= Pe j ,i
A
+
Sb
eten
i
0.45 2.26 2.26 3.02 3.02 3.02 3.02 3.02 3.02 3.02 2.26 2.26 0.45 fce = ksi 0.45 2.26 2.26 3.02 3.02 3.02 3.02 3.02 3.02 3.02 2.26 2.26 0.45 M cr
j ,i
(
:= maxfr ( -Sb_c) , fr + fce
j
j ,i
)
Sb_cj
( -Sb_c) - M DC1 j
j ,i
Sb
- 1
fr = 0.91 ksi
fr ( -Sb_c ) = 1737.58 kip ft 1
2606 5811.3 5571.4 6623.5 6341.2 6171.9 6115.4 6171.9 6341.2 6623.5 5571.4 5811.3 2606 kip ft 2626.6 5843.4 5589.1 6629.1 6330 6150.5 6090.7 6150.5 6330 6629.1 5589.1 5843.4 2626.6
M cr =
p. 57/73
CE6340
Prestressed Beam Example
Check_Minimum_Reinforcement
:=
j,i
"OK!" if M r_pos
(
min 1.2 M cr
j ,i
j ,i
58/73
, 1.33 M u_max
j ,i
)
M r_pos
"BAD!" otherwise 1.2 M cr
1, 7
= 7338.48 kip ft
1.33 M u_max
1, 7
1, 5
= 10606.67 kip ft
= 10039.08 kip ft
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Minimum_Reinforcement =
Check Limits of Reinforcement, Negative Moment Section: c_neg Check_Maximum_Reinforcement
j,i
:=
"OK!" if
j,i
ds
A 5.7.3.3.1
0.42
"BAD!" otherwise
Check_Maximum_Reinforcement =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
For components conaining non-prestressing steel, the minimum reinforcement provisions, may be considered if: M cr
j ,i
:= ( -fr) Ss_c
Ss_c = j
A 5.7.3.3.2
41925.32 3 in 43779.67
-3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3166.4 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5 -3306.5
M cr =
Check_Minimum_Reinforcement
j,i
:=
"OK!" if M r_neg
j ,i
(
min 1.2 M cr
j ,i
, 1.33 M u_min
j ,i
)
"BAD!" otherwise
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
Check_Minimum_Reinforcement =
p. 58/73
CE6340
Prestressed Beam Example
Check Crack Control:
d c := d top_cover +
βs := 1 +
d bar
Cotejo de Servicio de Tension en la Losa
d c = 2.56 in
2
dc
0.7 (ts - d c)
βs = 1.57
Class 1 Exposure
γe := 1.0
n ratio :=
AASHTO 5.7.3.4
59/73
Es n ratio = 6.76
Ec_s
Calcular esfuerzo del acero bajo cargas de servicio: As_eff
ρ :=
k
j,i
ρ=
b f_bot d s
(ρj , i n ) + (2 ρj , i n ) - ρj , i n
:=
ratio
2
ratio
0 0 0 0 0 0 0 0 0 0.008342 0.008342 0.008342 0.008342 0 0 0 0 0 0 0 0 0 0.008342 0.008342 0.008342 0.008342
ratio
k=
0 0 0 0 0 0 0 0 0 0.28 0.28 0.28 0.28 0 0 0 0 0 0 0 0 0 0.28 0.28 0.28 0.28
It =
0 0 0 0 0 0 0 0 0 329085.77 329085.77 329085.77 329085.77 0 0 0 0 0 0 0 0 0 329085.77 329085.77 329085.77 329085.77
Tr ansformed moment of iner tia of cracked section
It
j ,i
:=
1 3
( j,i )
b f_bot k d s
3
(
+ n ratio As_eff d s - k d s j ,i
j,i
)
2
d s = 68.5 in
Momento negativo de servicio en la losa s := 2
kk := 1
γLL
(
j ,i
M LL_min
=1
)
1.83 23.45 43.98 44.61 3.35 -79.8 -204.47 -370.67 -578.4 -827.66 -1297.18 -1698.82 -2144.47 0.84 10.7 19.44 -5.11 -73.18 -184.78 -339.52 -537.42 -778.47 -1062.68 -1623.63 -2117.26 -2667.15
s , kk
+ γLL
s , kk
M neg_slab_service_I =
j ,i
M DW
γDC
:= γDC
s , kk
+ γDW
=1
M neg_slab_service_I
j ,i
M DC2
s , kk
s , kk
p. 59/73
j ,i
CE6340
Prestressed Beam Example
Service Stress on Steel Rebar:
fs
j ,i
:=
(
)
-n ratio min0 kip ft , M neg_slab_service_I d s - k d s
j ,i
It
j,i
j ,i
if It
4
j ,i
> 0 in
( 0 ksi) otherwise
0 0 0 0 0 0 0 0 0 10.01 15.69 20.55 25.94 ksi 0 0 0 0 0 0 0 0 0 12.85 19.64 25.61 32.26
fs =
scr_min_neg
j ,i
:=
700 γe in - 2 d c if fs > 0 ksi j ,i fsj , i βs ksi ( 1000 in) otherwise
scr_min_neg =
1000 1000 1000 1000 1000 1000 1000 1000 1000 39.45 23.32 16.59 12.08 in 1000 1000 1000 1000 1000 1000 1000 1000 1000 29.59 17.6 12.3 8.71
check_crack_neg
j,i
:=
"Ok!" if sbar_slab_neg scr_min_neg
j ,i
"No Good!" otherwise sbar_slab_neg = 5.25 in check_crack_neg =
"Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!" "Ok!"
Cambiar espaceado de varilla #11 a 6" y extender cut off point hasta punto 106. Verificar de nuevo.
p. 60/73
60/73
CE6340
Prestressed Beam Example
Design for Shear
61/73
Método Simplificado según seccion 5.8.3.4.2 (General Procedure)
ϕv := 0.9
resistance factor for shear in normal weight concrete
b v := t w
minimum thickness of web
b v = 8 in
Assume angle of inclination of principal stressess (theta) Assume angle of crack inclination
θ := 45 deg
1. Compute Effective Shear Depth dv: Mu
j ,i
:=
M u_max M u_min
dv
j ,i
:=
j ,i
j ,i
if
M u_max
> M u_min
j ,i
A5.8.2.9
j ,i
C5.8.3.4.2
otherwise
β1( fc) c_pos
max 0.72 h c , 0.9 d p , d p -
Is conservative to use the maximum moment that occurs at the section
j,i
if Mu > 0 kip ft i i j ,i 2 β1( fc) c_neg max 0.72 h c , 0.9 d s , d s otherwise 2
h c = 73 in
58 57.29 57.29 57.29 57.29 57.29 57.29 57.29 57.29 57.29 68.5 68.5 58 in 58.14 58.14 57.48 57.48 57.48 57.48 57.48 57.48 57.48 57.48 57.48 68.5 68.5
dv =
Critical section is the larger of 0.5*dv*cot(q) or dv from face of support. assume dv Xcritical
:= xspan
Xcritical
:= xspan
j ,1
j ,2
jspan
+ Lcc + d v
jspan + 1
1
max( 1.0 , 1.0)
- Lcc - d v 2
Lspan =
j ,1
j , tentotal
max( 1.0 , 1.0)
118 ft 118 Xcritical =
p. 61/73
5.33 112.79 ft 5.35 112.79
CE6340
Mu
j ,i
Prestressed Beam Example
:=
62/73
j T , Xcritical if xten Xcritical j , 1 i j ,1 j T linterpxten , M u , Xcritical if xten Xcritical j , 2 i j ,2
( )
linterpxten , M u
( )
Mu
j ,i
Mu =
dv
j ,i
otherwise
1319.4 1597.1 3036.4 5248.8 6708.7 7470.8 7548.2 6970.3 5749.4 3901 1628.6 -2017.4 -2067.8 kip ft 1347.3 1627.3 3093.7 5349.7 6839.9 7619 7700 7112.5 5868.6 3983.9 1661.9 -2720.1 -2777.2 j T , Xcritical if xten Xcritical j , 1 i j ,1 j T linterpxten , d v , Xcritical if xten Xcritical j , 2 i j ,2
( )
linterpxten , d v
:=
( )
dv
otherwise
j ,i
Vu
j ,i
j j T T , Xcritical , linterpxten , Vu_max , Xcritical j , 1 j , 1 j j T T max linterpxten , Vu_min , Xcritical , linterpxten , Vu_max , Xcritical j , 2 j , 2
(
max Vu_max
Vu =
(
)
(
)
(
)
(
)
max linterpxten , Vu_min
:=
j ,i
, Vu_min
j ,i
)
otherwise
292.4 287.4 259.6 204.1 150.2 98.2 82.8 135.1 187.5 239.7 291.5 309.2 309.7 kip 297.7 292.7 264.3 207.6 152.6 99.5 83.4 136.8 190.3 243.7 296.5 314.6 315.1
Vp =0 kip for straight strands v
j,i
:=
Vu
j ,i
ϕv b v d v
j ,i
v=
ratio :=
v fc
fpo
j ,i
:= 0.7 fpu
ϕv = 0.9
0.7 0.69 0.63 0.49 0.36 0.24 0.2 0.33 0.45 0.58 0.71 0.63 0.63 ksi 0.71 0.7 0.64 0.5 0.37 0.24 0.2 0.33 0.46 0.59 0.72 0.64 0.64 0.12 0.11 0.1 0.08 0.06 0.04 0.03 0.05 0.08 0.1 0.12 0.1 0.1 0.12 0.12 0.11 0.08 0.06 0.04 0.03 0.06 0.08 0.1 0.12 0.11 0.11
ratio =
fpo =
if xten Xcritical i j ,1 if xten Xcritical i j ,2
189 189 189 189 189 189 189 189 189 189 189 189 189 ksi 189 189 189 189 189 189 189 189 189 189 189 189 189 p. 62/73
CE6340
Prestressed Beam Example
63/73
Compute angle of inclination of crack: Compute longitudinal strain at tendon or bar reinforcement depth:
(
max d v Vu εs
j ,i
dv
:=
j ,i
, Mu
j ,i
j ,i
Ep Aps
(
j ,i
, Mu
j ,i
dv
j ,i
j ,i
(
Es As_req
:=
j ,i
εs j ,i
εs j ,i
i
Es As_req
j ,i
j ,i
j ,i
j ,i
i
0 kip ft
εs
Vu
j ,i
1, 2
= -0.003258021
otherwise
)
( ) ) + E (A
j ,i
c j
)
0 kip ft ε s
j ,i
c
j ,i
i
j ,i
(
Es As_req
- Aps fpo
if M u
if M u
Ep Aps + Ec Abot
( -0.0004) if ε s
εs
j ,i
i
( 0.001) if ε s
j ,i
)+
Ep Aps
(
Vu
j ,i
max d v Vu
εs
)+
- Abot
)
(
if M u
j ,i
j ,i
< 0.0
)
0 kip ft ε s
j ,i
< 0.0
< -0.0004
> 0.006
otherwise
εs =
0 -0.0004 -0.0004 -0.0004 0.00342 0.00193 -0.0004 -0.0004 -0.0004 -0.0004 0 0.00018 0.00018 -0.0004 -0.0004 -0.0004 -0.0004 0 0.00029 0.00029 0.00001 -0.0004 -0.0004 -0.0004 0.00301 0.00181
β
:=
j,i
4.8
(1 + 750ε ) s
j ,i
6.86 6.86 6.86 6.86 4.8 4.23 4.22 4.8 6.86 6.86 6.86 1.35 1.96 6.86 6.86 6.86 6.86 4.8 3.94 3.93 4.76 6.86 6.86 6.86 1.48 2.03
β= θnew
j ,i
:= 29 deg + 3500 ε s deg j ,i
27.6 27.6 27.6 27.6 29 29.63 29.64 29 27.6 27.6 27.6 40.97 35.75 deg 27.6 27.6 27.6 27.6 29 30.02 30.03 29.03 27.6 27.6 27.6 39.52 35.35
θnew =
p. 63/73
CE6340
Prestressed Beam Example
fc b v d vj , i kip ksi 2 in
64/73
Vc
:= min 0.25 fc b v d v , 0.0316 β
Vc =
246.27 246.27 243.25 243.25 170.27 150.08 149.76 170.27 243.25 243.25 243.25 57.11 83.25 kip 246.88 246.88 244.05 244.05 170.84 140.28 140.02 169.57 244.05 244.05 244.05 62.57 86.27
j ,i
j,i
j ,i
AASHTO 5.8.3.3
0.25 fc b v d v
1, 2
Vs
j ,i
Vs =
:=
Vu
j ,i
ϕv
= 695.98 kip
- Vc
j ,i
78.6 73.1 45.2 -16.5 -3.4 -40.9 -57.7 -20.1 -34.9 23.1 80.6 286.5 260.9 kip 83.9 78.3 49.7 -13.3 -1.2 -29.7 -47.4 -17.6 -32.6 26.7 85.4 287 263.9
Required Straigth Stirrup Spacing:
(
Av fy d vj , i cot θnewj , i
smax
j ,i
:= min
)
Vs j ,i ( 1000 in) otherwise
if Vs
j ,i
( min( 0.4 d
> 0 kip , min 0.8 d v
smax =
24 24 24 24 24 24 24 24 24 24 24 6.49 8.6 in 24 24 24 24 24 24 24 24 24 24 24 6.82 8.62
smax
:=
j ,i
j ,i
j ,i
< 0.125 fc b v d v
otherwise
otherwise
j ,i
Vr
j ,i
if Vu
smaxj , i smaxj , i floor in + 0.0 in if 0.001 in < smax - floor in 0.5 in j ,i in in smaxj , i smaxj , i floor in + 0.5 in if smax - floor in > 0.5 in j ,i in in smax
smax =
v
j ,i
) , 12 in) , 24 in
24 24 24 24 24 24 24 24 24 24 24 6 8.5 in 24 24 24 24 24 24 24 24 24 24 24 6.5 8.5
:= ϕv Vc
j ,i
+
(
Av fy d v cot θnew j ,i
smax
j ,i
j ,i
)
319.7 319.7 315.7 315.7 244.6 224.1 223.7 244.6 315.7 315.7 315.7 330.2 312.4 kip 320.5 320.5 316.8 316.8 245.4 214.2 213.9 244.1 316.8 316.8 316.8 327.2 318.6
Vr =
p. 64/73
j ,i
CE6340
Prestressed Beam Example
65/73
Longitudinal Reinforcement:
T_req
j,i
:=
Mu
j ,i
d v 1.0 if M u j ,i
j ,i
> 0 kip ft
+
Vuj , i - 0.5 Vs cot θnew j ,i j ,i ϕv
(
)
0.9 otherwise
T_req =
819.3 871.4 1144.6 1549 1709.4 1792.8 1793.6 1749.1 1636.3 1304.6 883.5 623.4 699.4 kip 830.6 883 1160.2 1571 1735.1 1807.8 1808.9 1774.6 1660.9 1324.1 895.5 779.3 848.2
T_prov
j,i
:=
(A ) f ps i
ps
j ,i
if M u
j ,i
> 0 kip ft
As_eff fy otherwise j ,i
T_prov =
1687.4 1687.4 2229.4 2229.4 2229.4 2229.4 2229.4 2229.4 2229.4 2229.4 2229.4 960 960 kip 1690.4 1690.4 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 2234.8 960 960
Check_long_Reinforcement
j,i
:=
"OK!" if T_req
j,i
< T_prov
j,i
"BAD!" otherwise
Check_long_Reinforcement =
"OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!" "OK!"
p. 65/73
CE6340
Prestressed Beam Example
66/73
Check Horizontal Shear at Slab-Beam Interface: β1( fc_s ) c_pos j,i dp if Mu > 0 kip ft i j ,i 2 β1( fc) c_neg j,i ds otherwise 2
de
:=
de =
57.85 57.85 57.09 57.09 57.09 57.09 57.09 57.09 57.09 57.09 57.09 67.36 67.36 in 58 57.29 57.29 57.29 57.29 57.29 57.29 57.29 57.29 57.29 67.36 67.36 58
j ,i
Vn_req
j ,i
Vuj , i ϕv := de
j ,i
67.4 66.24 60.64 47.66 35.08 22.94 19.34 31.56 43.8 55.99 68.07 61.21 61.31 kip 68.44 67.27 61.52 48.32 35.52 23.17 19.4 31.84 44.29 56.7 69.01 62.27 62.38 ft
Vn_req =
ccohesion := 0.75 ksi
λ := 1.0
μfriction := 0.6 λ
Concrete placed against hardened concrete clean and not free of laitance, but not intentionally roughened.
b v := b f_top Acv
:= smax b v
Acv =
1008 1008 1008 1008 1008 1008 1008 1008 1008 1008 1008 252 357 2 in 1008 1008 1008 1008 1008 1008 1008 1008 1008 1008 1008 273 357
j ,i
j ,i
Vn_prov
j ,i
Vn_prov =
:=
(
min 0.8 ksi Acv , 0.20 fc Acv , ccohesion Acv j ,i
j ,i
j ,i
)
+ μfriction Av fy
smax
A 5.8.4.1
j ,i
385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 403.2 397.96 kip 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 403.2 397.96 ft
ccohesion ( Acv) + μfriction Av fy smax
1, 1
=
385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 101.57 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 385.07 109.44 p. 66/73
CE6340
Prestressed Beam Example
A 5.10.10.1 & C 3.4.3
Compute Anch orage Z one Reinforcement
Pr := 0.04Astrand
Abursting_req :=
N
strand fpj
Pr
2
Abursting_req Av
h 4 = 1.75 in
Nstirrups
Required Bursting Resistance
Pr = 69.4 kip
Abursting_req = 3.47 in
20 ksi
Nstirrups := ceil
67/73
Nstirrups = 9
Stirrup Spacing
number of required closed stirrups whitin h/5 of the end of the beam.
Anchorage Zone changed from h/5 to h/4 in AASHTO fourth Ed
p. 67/73
CE6340
Prestressed Beam Example
Performance Ratios: At Service Limit State: PR_top = 0.88
Top Fiber of Girder
PR_bot = 0.89
Bottom Fiber of Girder
PR_slab = 3.96
Top of Slab
At Ultimate Limit State: PR_MPOS = 0.72
Positive Moment Capacity
PR_MNEG = 0.92 Negative Moment Capacity PR_SHE = 0.99
Shear Capacity
p. 68/73
68/73
CE6340
Prestressed Beam Example
69/73
Compute Deflections Due to Initial Prestress: First Compute Conjugate Moments Applied due to initial prestressing: MCONt
j ,i
:=
Pt
j ,i
(-e ) ten i
Eci I
Prestress Acts on Simple Span Beam, Compute Reaction at Left Support of the Conjugated Beam by taking Moments About the Right End:
(x
ten i+ 1
wtrib :=
- xten
i
2
i
(x
ten i
) )
- xten
i- 1
2
(x
ten i+ 1
i- 1
2
RA :=
ii = 1
xten -
i
(x
ten i
) +x
ten i
- xten
)
i- 1
2
i
if i = 1
if i = tentotal
xten otherwise i
(
)
MCONtj , ii wtribii xtenten - xarmii total
(x
j
- xten
2
i
otherwise
tentotal
-
ten i+ 1
if i = tentotal
)
- xten
(x
xarm :=
if i = 1
ten tentotal
- xten
1
-0.0088 rad -0.0088
RA =
)
Compute Deflection at Ten Points; Equal to the Conjugated Moment at the Section:
δi
j ,i
δi =
(
ten total
:= RA xten - xten j
i
1
) + (MCON ii = 1
t
j , ii
(
wtrib max xten - xarm , 0.0 in ii
i
ii
))
0 -0.62 -1.19 -2.17 -2.87 -3.29 -3.43 -3.29 -2.87 -2.17 -1.19 -0.62 0 in 0 -0.62 -1.19 -2.17 -2.87 -3.29 -3.43 -3.29 -2.87 -2.17 -1.19 -0.62 0
Deflection Due to Losses After Transfer: δloss
j ,i
δloss =
Pej , i - Ptj , i j ,i Pt j ,i
:= δi
0 0.09 0.17 0.31 0.4 0.46 0.48 0.46 0.4 0.31 0.17 0.09 0 in 0 0.09 0.17 0.31 0.4 0.46 0.48 0.46 0.4 0.31 0.17 0.09 0 p. 69/73
CE6340
Prestressed Beam Example
70/73
Deflection due to Dead Weight of the Beam: Using a similar procedure as for Prestress: tentotal -M girder
RA :=
ii = 1
j , ii
Eci I
ii
(x
j
(
wtrib xten
ten tentotal
tentotal
- xten
1
- xarm
ii
)
0.0046 rad 0.0046
RA =
)
Compute Deflection at Ten Points; Equal to the Conjugated Moment at the Section:
(
ten total
-Mgirderj , ii wtrib max xten - xarm , 0.0 in ii i ii Eci I ii = 1
)
)
:= RA xten - xten
δgirder =
0 0.33 0.64 1.23 1.68 1.97 2.07 1.97 1.68 1.23 0.64 0.33 0 in 0 0.33 0.64 1.23 1.68 1.97 2.07 1.97 1.68 1.23 0.64 0.33 0
j ,i
j
i
1
Check at midspan:
+
(
δgirder
d :=
(
5 wgirder Lspan
jspan
)
4
d=
384 Eci I
(
RA :=
ii = 1
Eci I
j , ii
ii
(x
j
(
wtrib xten
ten tentotal
- xten
1
tentotal
- xarm
OK!
M nc xten , wslab + whaunch
Deflection due to Slab dead Load (on simple span): tentotal -M slab_haunch
2.06 in 2.06
ii
)
1
1
) = 0kip ft
1
0.004 rad 0.0043
RA =
)
Compute Deflection at Ten Points; Equal to the Conjugated Moment at the Section:
(
ten total
-Mslab_haunchj , ii wtrib max xten - xarm , 0.0 in ii i ii Eci I ii = 1
)
:= RA xten - xten
δslab =
0 0.28 0.56 1.07 1.46 1.71 1.8 1.71 1.46 1.07 0.56 0.28 0 in 0 0.3 0.6 1.14 1.56 1.83 1.92 1.83 1.56 1.14 0.6 0.3 0
j ,i
j
i
1
+
(
δslab
p. 70/73
)
CE6340
Prestressed Beam Example
71/73
Superimposed Dead Weight Deflection (Curb, Rail and FWS); On the Continuous Beam: (use virtual work method) Define Moment Diagram for Unit Point Load at Tenth Point (use previously defined IL): k := 1 .. n poi
im := 1 .. n unit
(
mten := linterp xunit , unit , xten i
i
unit
im
set counters
:= im
n unit = 81
)
Unit Moment "m" at x due to point load at "i" tenth point:
M m( x , i ) := linterpxpoi , ( ILM)
ceil( mteni)
, x - linterpxpoi , ( ILM) floor( mteni) T + linterpxpoi , ( ILM) , x
xpoi
T
(
T
floor( mteni)
, x mten - floor mten
(
i
(
)) ...
i
)
npoi M ( x , i ) M x , w m c DC2 + wDW 1 j j δsd := dx j ,i Ec Ic 1 kip j xpoi 1
δsd =
0.0018 0.0225 0.043 0.078 0.102 0.1125 0.109 0.0929 0.0678 0.039 0.0137 0.0068 0.0014 in 0.0018 0.0229 0.0438 0.0794 0.1039 0.1145 0.1109 0.0946 0.0691 0.0397 0.014 0.0069 0.0014
δtot
:= δi
j ,i
δtot =
j ,i
+ δloss
j ,i
+ δgirder
j ,i
+ δslab
j ,i
+ δsd
j ,i
0 0.1 0.23 0.51 0.78 0.97 1.03 0.95 0.75 0.47 0.2 0.08 0 in 0 0.12 0.26 0.58 0.88 1.09 1.16 1.07 0.85 0.54 0.23 0.1 0
p. 71/73
CE6340
Prestressed Beam Example
72/73
Compute Long Term Deflection usign PCI's Bridge Design Manual Multipliers: PCIi := 2.88
δi_long := PCIi δi
PCIgirder := 2.88
δgirder_long := PCIgirder δgirder
PCIloss := 2.32
δloss_long := PCIloss δloss
PCIslab := 2.50
δslab_long := PCIslab δslab
PCIsd := 2.50
δsd_long := PCIsd δsd
0 -1.79 -3.43 -6.25 -8.26 -9.47 -9.87 -9.47 -8.26 -6.25 -3.43 -1.79 0 in 0 -1.79 -3.43 -6.25 -8.26 -9.47 -9.87 -9.47 -8.26 -6.25 -3.43 -1.79 0
δi_long =
δgirder_long =
0 0.94 1.86 3.53 4.84 5.67 5.96 5.67 4.84 3.53 1.86 0.94 0 in 0 0.94 1.86 3.53 4.84 5.67 5.96 5.67 4.84 3.53 1.86 0.94 0
δloss_long =
0 0.2 0.39 0.71 0.94 1.07 1.12 1.07 0.94 0.71 0.39 0.2 0 in 0 0.2 0.39 0.71 0.94 1.07 1.12 1.07 0.94 0.71 0.39 0.2 0
δslab_long =
0 0.71 1.4 2.67 3.66 4.29 4.5 4.29 3.66 2.67 1.4 0.71 0 in 0 0.76 1.5 2.85 3.91 4.58 4.81 4.58 3.91 2.85 1.5 0.76 0
δsd_long =
0 0.06 0.11 0.2 0.26 0.28 0.27 0.23 0.17 0.1 0.03 0.02 0 in 0 0.06 0.11 0.2 0.26 0.29 0.28 0.24 0.17 0.1 0.03 0.02 0
δtot_long
:= δi_long
j ,i
j ,i
+ δloss_long
j ,i
+ δgirder_long
j ,i
+ δslab_long
j ,i
+ δsd_long
j ,i
0.0044 0.1183 0.3249 0.8542 1.4299 1.8454 1.9798 1.7966 1.3444 0.7566 0.2516 0.0791 0.0035 in 0.0045 0.1678 0.4223 1.0393 1.6834 2.1422 2.291 2.0924 1.5963 0.9399 0.3477 0.1278 0.0035
δtot_long =
ZETAL := j
δtot_long
j ,2
j ,1
δtot_long ZETAR := j
xten - xten 2
ZETAL =
- δtot_long 1
1.609572752 10- 3 rad - 3 2.306503603 10
p. 72/73
j , tentotal
xten
tentotal
ZETAR =
- δtot_long
j , tentotal- 1
- xten
tentotal- 1
-1.068756965 10- 3 rad - 3 -1.755819829 10
CE6340
Prestressed Beam Example
RL_DL := VDC1 j
j ,1
+ VDC2
j ,1
RL_LL := VLL_max j
RL_DL =
j ,1
122.48 kip 126.17
89.86 kip 90.36
RL_LL =
VDW
j ,1
7.28 kip 7.8
=
p. 73/73
73/73