1. EXAMPLE 9.2 – Part II PCI Bridge Design Manual
2011/12 Edition
EXAMPLE 9.2 – Part II PCI Bridge Design Manual
BULB “T” (BT-72)
THREE SPANS, COMPOSITE DECK
LRFD SPECIFICATIONS
Materials copyrighted by Precast/Prestressed Concrete Institute, All rights reserved. Unauthorized duplication of the material or presentation prohibited.

2. LRFD LIVE LOADS – HL93

3. Fatigue
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 )

4. Fatigue
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 interim)

5. DYNAMIC 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).

6. LIVE LOAD SHEARS AND MOMENTS – TRUCK LOAD
VLT = (shear force/lane)(DFV)(1+IM/100)
= (shear force/lane)(1.082)(1 + 33/100)
= shear force/lane
MLT = (moment/lane)(DFM)(1+IM/100)
= (moment/lane)(0.905)(1 + 33/100)
= moment/lane

7. LIVE LOAD SHEARS AND MOMENTS – LANE LOAD
VLL = (shear force/lane)(DFV)
= (shear force/lane)(1.082)
= shear force/lane
MLL = (moment/lane)(DFM)
= (moment/lane)(0.905)
= moment/lane

8. LIVE LOAD SHEAR AND MOMENT ENVELOPE – TRUCK LOADS INCLUDE IM

9. 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) (LL+IM)
Service III – longitudinal tension in prestressed concrete.
Q = 1.0(DC+DW) (LL+IM)

10. LOAD COMBINATIONS
Strength I – ultimate strength of both prestressed and reinforced concrete components.
Minimum:
Q = 0.9DC DW (LL+IM)
Maximum:
Q = 1.25DC DW (LL+IM)
Minimum is used when DL and LL create stresses of opposite signs.

11. LOAD COMBINATIONS
2011/12 Edition
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).

12. 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.

13. 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.

14. 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.

15. ESTIMATE REQUIRED PRESTRESSING FORCE
Here is the first problem with the structure being simple span for some loads and continuous for others:

16. 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.

17. 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.

18. ESTIMATE REQUIRED PRESTRESSING FORCE
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.

19. ESTIMATE REQUIRED PRESTRESSING FORCE
ASSUME SERVICE III CONTROLS
M (k-in)
S in3
f = M/S ksi Mg
16688
14915
1.12 Ms
25522
1.71 Mb
876
20545
0.05
Mws
1536
0.09
0.8MLL+IM
0.8(25380)
0.99
sum
3.96 (Tension)

20. ESTIMATE REQUIRED PRESTRESSING FORCE
Allowable tensile stress in Service III
(LRFD Table )
ft = 0.19(fc’)0.5 = 0.19(7.0 ksi)0.5 = ksi
0.19(fc’)0.5 ksi units = 6(fc’)0.5 psi units

21. 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

22. ESTIMATE REQUIRED PRESTRESSING FORCE
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.

23. 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 = in3
fpb = ksi (compression)

24. ESTIMATE REQUIRED PRESTRESSING FORCE
Ppe = kips is the MINIMUM prestressing force after all losses.

25. ESTIMATE REQUIRED PRESTRESSING FORCE
Ppe = kips MINIMUM after losses
Normally, low relaxation strands are stressed to:
Initial stress = 0.75 fpu = 0.75 (270 ksi) = ksi
But this is NOT standard. Either specify it or check with the contractor!

26. 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 = k / 152 ksi = 6.66 in2
Since a single ½” strand is in2
# strands > 6.66/0.153 = 43.6 strands – Use 44

27. STRAND PATTERN Here is a possible strand pattern.
Note that each State may use a different standard pattern.
Check std. drawing for that state!

28. STRAND PATTERN This is close enough to original assumption
of 5 inches.

29. LOSS OF PRESTRESSING FORCE
As soon as the prestressing forces is applied to the beam, the strand starts to lose force!

30. 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.

31. 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

32. ACCURACY OF PRESTRESS LOSSES
Creep and Shrinkage
Hard to predict
C 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.

33. LOSS OF PRESTRESSING FORCE
Elastic shortening:

34. LOSS OF PRESTRESSING FORCE
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.

35. LOSS OF PRESTRESSING FORCE
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.

36. Pi
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

37. Pi
LRFD C a 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 C a -1.

38. LOSS 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) = ksi.
Pi = 44 strand(0.153 in.2/strand)( )(202.5 ksi)
= 1241 k

39. LOSS OF PRESTRESSING FORCE
Elastic shortening:
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.

40. LOSS OF PRESTRESSING FORCE
Elastic shortening:
(17.9 ksi)/202.5 ksi = or 8.8%
Close enough to initial assumption of 9%

41. ES USING EQUATION
The equation gives 8.9% without iteration.

42. LOSS 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.

43. LOSS OF PRESTRESSING FORCE
( )
( )
( )
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.

44. LOSS OF PRESTRESSING FORCE
If H = 70%
Recall that fci’ = 5.5 ksi

45. LOSS OF PRESTRESSING FORCE
Total Loss:
25% loss was assumed. Since the actual loss is less, the design is probably OK. This will be verified when service loads are checked .