EXAMPLE 9.2 – Part II PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK LRFD SPECIFICATIONS Materials copyrighted by Precast/Prestressed

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


• 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.



• 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


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



Here is the first problem with the structure being simple span for some loads and continuous

for others:



In the end spans, 0.4L and 0.5L must both be checked for the combination of simple span and

continuous loads.




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.





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)



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



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



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.



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



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.



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!



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


• 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


• 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



• 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.


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.


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.


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


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


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.


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.


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


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.


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


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 .

Documents

EXAMPLE 9.2 – Part II PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK LRFD SPECIFICATIONS Materials copyrighted by Precast/Prestressed