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EXAMPLE 9.2 – Part IIPCI Bridge Design Manual
EXAMPLE 9.2 – Part IIPCI Bridge Design Manual
BULB “T” (BT-72)
THREE SPANS, COMPOSITE DECK
LRFD SPECIFICATIONS
Materials copyrighted by Precast/Prestressed Concrete Institute, 2011. All rights reserved. Unauthorized duplication of the material or presentation
prohibited.
LRFD LIVE LOADS – HL93LRFD LIVE LOADS – HL93
FatigueFatigue
For prestressed beams designed using Service III load combinations,
Fatigue of steel does not need to be considered.
Fatigue does not need to be considered in concrete decks on multi-
beam bridges.
(LRFD 5.5.3.1)
FatigueFatigue
For fully prestressed beams (other than segmental boxes), the
compressive stresses under fatigue loads + ½ of the sum of the
effective prestressing stress (after losses) and permanent load
stresses < 0.4fc’.
(LRFD 5.5.3.1 2009 interim)
DYNAMIC ALLOWANCE FACTORSDYNAMIC ALLOWANCE FACTORS
LRFD 3.6.2:
Condition IM
Deck Joints – All Limit States 75%
Fatigue and Fracture Limit States 15%
All Other Limit States 33%
Multiply the static effect of the TRUCK OR TANDEM live load by
(1+ IM/100)
The lane load is NOT multiplied by (1 + IM/100).
LIVE LOAD SHEARS AND MOMENTS – TRUCK LOAD
LIVE LOAD SHEARS AND MOMENTS – TRUCK LOAD
VLT = (shear force/lane)(DFV)(1+IM/100)
= (shear force/lane)(1.082)(1 + 33/100)
= 1.439 shear force/lane
MLT = (moment/lane)(DFM)(1+IM/100)
= (moment/lane)(0.905)(1 + 33/100)
= 1.204 moment/lane
LIVE LOAD SHEARS AND MOMENTS – LANE LOADLIVE LOAD SHEARS AND MOMENTS – LANE LOAD
VLL = (shear force/lane)(DFV)
= (shear force/lane)(1.082)
= 1.082 shear force/lane
MLL = (moment/lane)(DFM)
= (moment/lane)(0.905)
= 0.905 moment/lane
LIVE LOAD SHEAR AND MOMENT ENVELOPE – TRUCK LOADS INCLUDE IM
LIVE LOAD SHEAR AND MOMENT ENVELOPE – TRUCK LOADS INCLUDE IM
LOAD COMBINATIONS (LRFD 3.4)LOAD COMBINATIONS (LRFD 3.4)
Service I – compression in prestressed concrete (positive moment zones in this example);
compression and tension in reinforced concrete (negative moment zones and the slab in
positive moment zones in this example).
Q = 1.0(DC+DW) + 1.0 (LL+IM)
Service III – longitudinal tension in prestressed concrete.
Q = 1.0(DC+DW) + 0.8 (LL+IM)
LOAD COMBINATIONSLOAD COMBINATIONS
Strength I – ultimate strength of both prestressed and reinforced concrete components.
Minimum:
Q = 0.9DC + 0.65 DW + 1.75 (LL+IM)
Maximum:
Q = 1.25DC + 1.50 DW + 1.75 (LL+IM)
Minimum is used when DL and LL create stresses of opposite signs.
LOAD COMBINATIONSLOAD COMBINATIONS
Sometimes, a permanent load both contributes to and mitigates a critical load effect. For
example, in the three span continuous bridge shown, the DC load in the first and third
spans would mitigate the positive moment in the middle span. However, it would be
incorrect to use a different p for the two end spans. In this case, p would be 1.25 for DC
for all three spans (Commentary C3.4.1 – paragraph 20).
WHY SERVICE III HAS A 0.8 LL FACTOR
WHY SERVICE III HAS A 0.8 LL FACTOR
• Service III is for longitudinal tension in prestressed concrete. It tries to prevent cracking in prestressed members under service load.
• LRFD is statistically calibrated.• Tests show the cracking strength of
prestressed concrete is overestimated.
WHY SERVICE III HAS A 0.8 LL FACTOR
WHY SERVICE III HAS A 0.8 LL FACTOR
• Cracking strength is based on:– Modulus of rupture
• Taken as 7.5(fc’)0.5 , but this is the lower bound.
• Upper bound may be as high as 12(fc’)0.5 .
• Based on design strength, not actual strength.
– Loss of prestressing force• Very hard to estimate exactly.• LRFD overestimates losses.
• Cracking strength is usually overestimated.
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
ASSUME SERVICE III CONTROLS
Find the bending stress due to applied load.
Recall that Mg (beam) and Ms (slab + haunch) are applied to the non-composite beam acting
as a simple span. The remaining moments act on the composite structure as a continuous
span.
bc
ILLwsb
b
sgb S
M8.0MM
S
MMf
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
Here is the first problem with the structure being simple span for some loads and continuous
for others:
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
In the end spans, 0.4L and 0.5L must both be checked for the combination of simple span and
continuous loads.
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
ASSUME SERVICE III CONTROLS
In this example, the center span, interior beam will be designed. Due to symmetry, the
maximum moments for both the simple span load cases and the continuous span load cases
occur at 0.5L.
bc
ILLwsb
b
sgb S
M8.0MM
S
MMf
The non-composite and composite moments cause a large tension on the bottom. The
compression from the prestressing must reduce the total tension to a value below the allowable.
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
ASSUME SERVICE III CONTROLS
bc
ILLwsb
b
sgb S
M8.0MM
S
MMf
M (k-in) S in3 f = M/S ksi
Mg 16688 14915 1.12
Ms 25522 14915 1.71
Mb 876 20545 0.05
Mws 1536 20545 0.09
0.8MLL+IM 0.8(25380) 20545 0.99
sum 3.96 (Tension)
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
Allowable tensile stress in Service III
(LRFD Table 5.9.4.2.2-1)
ft = 0.19(fc’)0.5
= 0.19(7.0 ksi)0.5
= 0.503 ksi
0.19(fc’)0.5
ksi units = 6(fc’)0.5
psi units
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
The applied load cause a TENSION of 3.96 ksi.
The allowable tension is 0.50 ksi.
MINIMUM compressive stress at bottom of the beam due to prestressing AFTER
LOSSES:
3.96 ksi – 0.50 ksi = 3.46 ksi compression = fpb
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
pe pepb
b
P P ef
A S
There are actually two unknowns here.
The eccentricity, e, is not known.
Ppe is the force after all losses and the losses are not known.
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
Assume the centroid of the prestressing tendons will be 5” from the bottom. The eccentricity
(calculated for the non-composite beam) is:
e = yb – 5” = 36.60” – 5” = 31.6”
A = 767 in2
Sb = 14915 in3
fpb = +3.46 ksi (compression)
pe pepb
b
P P ef
A S
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
pe pepb
b
pe pe
2 3
pe
P P ef
A S
P P 31.6in3.46ksi
767in 14915in
1021kips P
Ppe = 1 012 kips is the MINIMUM prestressing force after all losses.
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
Ppe = 1 012 kips MINIMUM after losses
Normally, low relaxation strands are stressed to:
Initial stress = 0.75 fpu = 0.75 (270 ksi) = 202.5 ksi
But this is NOT standard. Either specify it or check with the contractor!
ESTIMATE REQUIRED PRESTRESSING FORCE
ESTIMATE REQUIRED PRESTRESSING FORCE
Losses are usually between 15-30%. Assume 25% loss of prestressing force (just a guess).
Initial stress =202.5 ksi
Effective stress after 25% loss
fpe = 0.75(202.5) = 152 ksi
Ap > Ppe / fpe = 1 012 k / 152 ksi = 6.66 in2
Since a single ½” strand is 0.153 in2
# strands > 6.66/0.153 = 43.6 strands – Use 44
Here is a possible strand
pattern.
Note that each State may use
a different standard pattern.
Check std. drawing for that
state!
STRAND PATTERN
12(2) 12(4) 8(6) 4(8) 2(10 12 14 16)
445.82
36.60 5.82 30.78
bs
bs
b bs
y
y in
e y y in in in
This is close enough
to original assumption
of 5 inches.
STRAND PATTERN
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
As soon as the prestressing forces is applied to the beam,
the strand starts to lose force!
LOSS OF PRESTRESSING FORCE
LOSS OF PRESTRESSING FORCE
• For PRETENSIONED beams, there are 4 sources of loss:– Elastic Shortening– Creep– Shrinkage– Relaxation
• LRFD has a simplified method and more exact method.
ACCURACY OF PRESTRESS LOSSES
ACCURACY OF PRESTRESS LOSSES
• Prestressing losses are APPROXIMATE• One source of loss is elastic shortening
– This uses Eci, modulus of elasticity at release.
– This can vary due to:• Actual strength at release• Material properties
ACCURACY OF PRESTRESS LOSSES
ACCURACY OF PRESTRESS LOSSES
• Creep and Shrinkage– Hard to predict
• C5.4.2.3.1 states the accuracy of creep and shrinkage equations is worse than + 50%
– Creep is affected by when other dead loads are applied.
• Temperature affects losses, so losses vary hour to hour and day to day.
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
2
ppES cgp
ci
i c c g cicgp
g ci cicgp
Ef f
E
Pe e M ePf
A I IM ePeP
fA I I
Elastic shortening:
When the strand is cut, is shortens and compresses the concrete. The CHANGE in the steel
tensile strain must be equal to the compressive strain in the concrete.
LOSS OF PRESTRESSING FORCE
2
ppES cgp
ci
g ci cicgp
Eff
E
M ePePf
A I I
The term, fcgp is the stress in the concrete at the CENTROID of the steel. Note that in Mc/I, c =
e!
Divide fcgp by Eci to get the compressive strain in the concrete. This must be the change in
the steel strain.
Multiply the concrete strain by Ep to get the CHANGE in the steel stress.
LOSS OF PRESTRESSING FORCE
PiPi
• The strand is tensioned to an “initial pull” when the contractor makes the beam.– For low relaxation strand, this is usually
0.75fpu.
• Pi is the force in the steel at release, but at release this is NOT the initial pull.– Loss due to shortening– Loss due to slip– Loss due to relaxation
PiPi
• LRFD C5.9.5.2.3a allows:– Assume the stress in the steel is a % of the initial
pull and iterate until an acceptable accuracy is achieved. Usually 10% is initially assumed. • There are so many unknowns that, often, the
10% loss is simply assumed.– Use Equation C5.9.5.2.3a -1.
2
2
( )
( )
ps pi g m g m g gpES
g g cips g m g
p
A f I e A e M A
A I EA I e A
E
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
Elastic shortening:
In this example, the loss is initially assumed to be 9%. The initial pull stress is:
0.75fpu = 0.75(270ksi) = 202.5 ksi.
Pi = 44 strand(0.153 in.2/strand)(1.00-0.09)(202.5 ksi)
= 1241 k
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
Elastic shortening:
2
2
2 3 3
1241 1241 (30.78 ) 1391 (12 / )(30.78 )
767 545894 5458942.83
g ci cicgp
cgp
cgp
M ePePf
A I I
k k in k ft in ft inf
in in inf ksi
Mg at midspan, based on L=118 ft. (design span). However, many examples use overall length
(assuming the beam will camber up an sit on its ends. The difference is minimal.
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
28500
(2.83 ) 17.94496
ppES cgp
ci
pES
Ef f
E
ksif ksi ksi
ksi
Elastic shortening:
(17.9 ksi)/202.5 ksi = 0.088 or 8.8%
Close enough to initial assumption of 9%
ES USING EQUATIONES USING EQUATION
2
2
22 4 2 2
2 422 4 2
( )
( )
6.73 202.5 545894 30.78 767 30.78 767 16692
767 545894 44966.73 545894 30.78 767
2850017.96
8.
ps pi g m g m g gpES
g g cips g m g
p
pES
pES
pES
A f I e A e M A
A I EA I e A
E
in ksi in in in in in k in
in in ksiin in in in
ksiksi
9%
The equation gives 8.9% without iteration.
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
Creep and Shrinkage of Concrete, Relaxation of Strand:
There are two methods, an approximate method and a more exact method. The approximate
method is used here.
pT pES pLTf f f
10 12pi pspLT h st h st pR
g
f Af f
A
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
10 12pi pspLT h st h st pR
g
f Af f
A (5.9.5.3-1)
(5.9.5.3-2)
(5.9.5.3-3)
1.7 0.01h H
5
1 'stcif
H = average relative humidity in % (so 70 not 0.7)
fci’ = strength of concrete at release, ksi.
fpR = 2.5 ksi for low relaxation strand.
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
1.7 0.01 1.7 0.01 70 1h H
If H = 70%
5 5
0.7691 ' 1 5.5st
cif
Recall that fci’ = 5.5 ksi
2
2
10 12
202.5 6.7310 1 0.769 12 1 0.769 2.5
76725.4
pi pspLT h st h st pR
g
pLT
pLT
f Af f
A
ksi inf ksi
inf ksi
LOSS OF PRESTRESSING FORCELOSS OF PRESTRESSING FORCE
Total Loss:
2
17.9 25.4 43.3
43.30.214 21.5% 25%
202.5202.5 43.3 159.2
44 .153 / 159.2 1072
pT pES pLT
pT
pe
pe
f f f
f ksi ksi ksi
ksiLoss
ksif ksi ksi ksi
P strand in strand ksi k
25% loss was assumed. Since the actual loss is less, the design is probably OK. This will
be verified when service loads are checked .