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.

CONTINUOUS PRECAST BRIDGES
• After final placement, the beams will continue to creep and shrink; cambering up.
• Temperature will also cause camber.
• Positive moments will form causing cracking.

CONTINUOUS PRECAST BRIDGES
• A positive moment connection is required. The requirements for this are given in 5.14.1.4

POSITIVE MOMENT CONNECTIONS -
OPTIONS
• 5.14.1.4.4 – Require the girder to be at least 90 days old when continuity is established. Provide reinforcing for a capacity of1.2 M cr calculation is needed.
. No other
• 5.14.1.4.5 – If the calculated stress at the bottom of the continuity diaphragm for the combination of superimposed permanent loads, settlement, creep, shrinkage, 50 percent live load and temperature gradient, if applicable, is compressive, consider the joint fully effective. Provide reinforcement to resist the restraining moments > 0.6M
cr
. (5.14.1.4.9a)
• 5.14.1.4.5 – Do a more exact analysis to determine the effectiveness of the joints. Provide reinforcing for the restraint moments > 0.6M
cr
. (5.14.1.4.9a).

POSITIVE MOMENT CONNECTIONS -
OPTIONS
5.14.1.4.4 – Require the girder to be at least 90 days old when continuity is established. Provide reinforcing for a capacity of1.2 M cr
. No other calculation is needed.
Here there are two possibilities – bent strand or bent bar connections:

DESIGN UNDER 5.14.1.4.4
• Specify continuity cannot occur until the girders are 90 days old.
• Find the cracking moment of the composite section assuming it is made of diaphragm concrete.
• Provide reinforcement for a capacity of 1.2
M cr
.
• For bent bar, use 5.14.1.4.9b (use hooks and properly embed).
• For bent strand, use the equations of
5.14.1.4.9c.

DESIGN UNDER 5.14.1.4.4
I
A c c
= 1868 in 2
= 1259300 in 4 h c y bc
= 80 in.
= 59.93 in.
(distance to bottom of composite)
S bc
= 1259300in 4 / 59.93 in. = 21013 in 3
Provide 1.2M
cr
: f r

0 24 4 ksi

0 480 ksi
M cr
 f S r bc

.
ksi

21013 in 3


10086 k
 in

12100 k
 in cr

DESIGN UNDER 5.14.1.4.4
Assume bent strand connection 5.14.1.4.9c.
The “tail” will be 18 in long. The extension will be 8 in as required. Assume strands will be extended from the bottom two rows. f pul f pul


(

8) dsh
0.163
(26 in

8 )

110 ksi
0.163
The 8 in the equation is the 8 inch extension.

DESIGN UNDER 5.14.1.4.4
# Strands
8
10
Area A ps in 2 a in
1.22
0.276
1.53
0.345
M in-kip
10333
12997
> 1.2M
cr
?
No
Yes a
 d
A ps


110.4
ksi ksi


144 in


80 in

3 in

77 in
M

A ps

110.4
ksi



77 in
 a
2


Use 10 extended strand. Use 5 in each of the two bottom rows.

DESIGN UNDER 5.14.1.4.5
5.14.1.4.5 – If the calculated stress at the bottom of the continuity diaphragm for the combination of superimposed permanent loads, settlement, creep, shrinkage, 50 percent live load and temperature gradient, if applicable, is compressive, consider the joint fully effective.
Provide reinforcement to resist the restraining moments > 0.6M
cr
. (5.14.1.4.9a)

CREEP
Let us assume a situation the beam is cast, the slab, haunch and barrier are added, but the beam is NOT made continuous. The end of the beam would have a rotation:

CREEP
If the beam is UNRESTRAINED, the beam will camber up and the angle will increase. The
INCREASE,
 creep
, is just the elastic angle multiplied by the CREEP COEFFICIENT.
 
 creep
 elastic

CREEP
However, the beam IS restrained, so as the beam tries to camber up, restraint moments are formed.

CREEP
The restraint moments are POSITIVE, so they cause tension on the bottom, but the creep is caused by compression on the bottom. Thus, the formation of restraint moments starts to mitigate the creep, lessening the formation of further restraint moments. This adjustment can be found from Mattock’s rate of creep equation:
M net

M

1
 e
 



CREEP
First, we need to find the elastic rotations.
Calculate the rotation of the beam end at release:
Here is the arrangement of the harped strand. The force,
P, at release is 1241 kips.
The moment due to prestressing can be found at every point.

CREEP
3500
3000
2500
2000
1500
1000
500
0
0
1770 k-ft @ 2.5 ft.
10 20 30
Length (ft)
3183 k-ft @ 35.5 ft
40 50 60
This diagram accounts for harping and transfer length.

CREEP
The curvature is M/EI. The CHANGE in angle is the area (integral) under M/EI diagram.
The previous diagram is half the beam, so multiply the area by 2 to get the total change in angle.
Because the beam is symmetrical, the angle at each end is ½ the total change in angle, so divide by 2.
EI

EI


2
2



1770 k

2 ft

2.5
ft
 

160330 k
 ft
2
2
 k
 ft

33 ft


3183 k
 ft

24 ft





CREEP
From basic structural analysis, the rotation at the end of the beam due to beam self weight (w =
0.799k/ft):
EI

EI


 wL 3

24
56100 k
 ft
2 k ft

119 ft

3
24
The prestressing rotation is opposite to the self weight rotation, so:
EI

= 160330 – 56100 = 104230 k-ft 2
CCW on left end, CW on right end.

CREEP
The dead load of the slab, haunch and barriers will cause a rotation. As previously shown, the total weight is 1.372 k/ft.
EI

EI


 wL 3

1.372 /
24
96340 k
 ft
2
24

119 ft

3
So the NET is:
EI

= 104230 - 96304 = 7926 k-ft 2
Again, the (+) sign indicates CCW on left and CW on the right.

CREEP
 creep

 elastic

CREEP
Note that there is a problem with this calculation. This calculation ASSUMES the beams are erected and the slab is placed soon after the beams are cast. If the beams are older, they will have a chance to “creep out” and net positive moment will be less.
The previous calculation represents a “worst case”.
This is suggested IF there is no specification requiring the beams to be certain age when they are actually put in the bridge.

CREEP
From basic structural analysis, it can be shown that, if a structure is 3 equal spans, of equal stiffness, the restraint moment is:
M r

12
5
EI

L
This structure does NOT have 3 equal spans, but the difference is only 8%. Also, C5.4.2.3.1 states that creep calculations are accurate to +
50%! Thus, using the given equation will yield reasonable answers.

CREEP
The equation in the previous slide allows for the fact that the adjacent spans do not provide total restraint. The flexibility of the adjacent spans allows some rotation of the joint. In a worst case, the maximum restraint moment would be a “fixed” end.
From basic analysis, the end moments for a end span are:
M

3 EI

L
And for a middle span:
M

2 EI

L
Note that the middle span is lower because there is a moment at both ends.

CREEP
The creep angle is
 times the elastic angle. Using the multispan formula:
M
M
M
M

12
5
EI
 elastic
L
 
12
5 L
EI
 elastic





12
5 119 ft


160 k
 ft


7926 k
 ft
2


CREEP
Now adjust for rate of creep effect. Assume the creep co-efficient,

= 2:
M

M cr
M cr
 

1
 e
 



160 k
 ft


1
 e

2



138 k
 ft
This is a positive moment.

Shrinkage
• This is a difficult effect to determine.
• Conventional wisdom is:
– The deck is younger that beam.
– Deck concrete has a higher shrinkage potential than the beam.
– The net result is that the differential shrinkage of the deck creates a mitigating moment.
• However, the deck could crack and relieve the differential stresses.
• The engineer must determine the most reasonable action.

Shrinkage
If the beam were simple span, the differential shrinkage would deform the beam as though there were end positive moments:
M shrinkage
  
A E s d d
 
 s
= differential shrinkage strain (deck to girder)
E d
= modulus of elasticity of deck concrete=3834 ksi
A d
 area of deck slab = 144*8=1152 in 2
 y= distance from center of slab to cg of composite section = 80 – (8/2) – 57.2 = 18.8 in

Shrinkage
M
M
M
If a differential shrinkage of 100 x 10 -6 is assumed:

  
A E s d d
  x

6

1152 in
2
 
3834 ksi

18.8
in


8300 k
 in

691 k
 ft

Shrinkage
From basic structural analysis, the end rotation for a simple support with a constant moment is:
 
ML
2 EI
For the 3 span bridge, we previous used:
M sr


12 EI

5 L


12 EI
5 L


M L s
2 EI




12 M s
10
So with M s, unres
= 691k-ft, M sr shrinkage moment is -138 k-ft.
= -829 k-ft; so the NET

Shrinkage
This net shrinkage moment is (-), but it will be reduced by creep (

= creep coefficient = 2):
M

M
M
M
1

 e
 
 
138 k
 ft
1
 e

2
2
 
60 k
 ft

TEMPERATURE GRADIENT
From Article 3.12.5, this is the temperature gradient for a girder greater than 16 inches deep, in Zone 2.

TEMPERATURE GRADIENT
This paper shows (in appendix) how to find the curvature for the temperature gradient.

TEMPERATURE GRADIENT
The curvature can be calculated from:

0
 i
 
E i i
 
T y ydA
E I i i
  i

The summation allows for different concrete strengths in the cross section.
If the curvature is constant and the beam is symmetric, the angle at either end is:
 
2
1  
0 dL

1

0
L
2

TEMPERATURE GRADIENT
Again using the restraint moment formula:
M

12
5
EI
L
 
6
5
EI

0
It the transformed section is used (so that the section can be treated as though it is all made of 7 ksi concrete):
EI

0

E
    
M

5
6
E
    

TEMPERATURE GRADIENT
EI

0

E
    

T(y) is the temperature gradient y is the distance from the centroid of the area to the centroid of the cross section.
Given the shape of the section, this is too complex for closed form integration.

TEMPERATURE GRADIENT
Break the section up as shown, and use the average temperature for each section.
Greater accuracy can be achieved by breaking the section into more layers, but given the fact that E and
 are probably only know to + 10%, this integration scheme is probably accurate enough.

TEMPERATURE GRADIENT
Area # y' top in
1
2
3
4
5
6
7
11.5
13.5
15.5
7.5
8
0
4 y' bot T top in Deg F
4
7.5
41
T bot
Deg F
11
11 7.791667
8 7.791667 7.333333
11.5 7.333333
4.125
A in^2
436
381.5
16
147 y bar in
20.8
17.05
Avg T*(y bar)*A deg F in^3
235788.8
61115.9026
15.05
1821.05
13.05
10990.54688
13.5
4.125 2.291667
15.5 2.291667 0.458333
16 0.458333
0
52
16
3
Sum
10.1
8.22
7.05
1685.016667
180.84
4.846875
311587.003
y' is measured from top of slab y bar is the distance from the centroid of layer to centroid of girder
Numerical integration done in a spreadsheet.
(Note, layer 1 can be easily integrated in closed form.
If this is done:     
240127 O F
 in 3 a difference of 2%! This is just shown for comparison.)

TEMPERATURE GRADIENT
M
M
M

5
6
E
    


6 
5072 ksi
 
5
10430 k
 in


6
869 k
 ft o F

311600 o F
 in 3


NET RESTRAINT MOMENT
Moment
Creep
K-ft
+138
Differential Shrinkage -60
M restraint
+78
Temperature Gradient +869
0.5 M
LL
M restraint
+ 0.5M
LL
-1000
-59
Since M restraint
+M
Temp
+0.5M
LL is slightly negative, assume the joint is fully effective, provide reinforcement for 0.6M
cr
. If it had been a large positive value, provide reinforcement for the
M restraint
> 0.6M
cr
.

Notes on Restraint Moments
• The calculation shown assumes that continuity is made at an early age. If this is not true, the beams will undergo creep shrinkage in storage. This will reduce the net, positive creep and shrinkage moments.
• Values of creep co-efficient and differential shrinkage are assumed. The commentary to the AASHTO specifications states that values for creep and shrinkage may vary +50%. Thus, answers can have a wide variation depending on assumptions made.
• If calculated values of the restraint moments are large, the engineer should take steps to reduce these moments. Aging the beams to reduce creep and shrinkage restraint moments seems to work well.

ANCHORAGE ZONE
5.10.10.1 requires reinforcing at the end of the girder to resist splitting forces. The stirrups must resist 4% of P i
:
A s f s
A s f s
> 0.04P
i
= 0.04(44 strand)(0.153in
> 54.5 kips
2 /strand)(202.5ksi)
The term f s cannot exceed 20 ksi, so:
A s
> 54.5kips / 20 ksi = 2.73 in 2 = 5 #5 stirrups.
These must be placed over a distance h/4 = 72/4 = 18 inches at each end of the girder.

CONFINEMENT REINFORCING
5.10.10.2 - For the distance of 1.5
d from the end of the beams other than box beams, reinforcement shall be placed to confine the prestressing steel in the bottom flange. The reinforcement shall not be less than No. 3 deformed bars, with spacing not exceeding 6.0 in. and shaped to enclose the strands.
1.5d = 1.5(72) = 108 inches.

CAMBER AND DEFLECTION
• Camber due to prestressing
• Deflection due to beam self weight
• Deflection due to other permanent loads
• Changes in deflection and camber due to creep and shrinkage
• Live load deflection

CAMBER AND DEFLECTION
Standard handbooks give the following formula for a beam with harped strand. The term e c is eccentricity at the harp point and e’ is the DIFFERENCE in the eccentricity at the end (see figure). The distance to the harp point is a.
P i

 8
2

6
2



4500 ksi
1241 k
 
545894 in 2



30.8
in

1428 in

2
8

11.4
in

426 in

2
6


3.79
in


CAMBER AND DEFLECTION
Deflection due to self wt (119 ft span) at release:
 k in

1428 in
 ksi
 
545894
 in
4
4


1.47
in

Deflection due to slab and haunch (118 ft span).
Beam assumed over 28 days old, load applied to the non-composite section:
  s

 k in

1416 ksi
  in

545894 in
4
4


1.93

Using a structural analysis program, the deflection due to barrier and wearing surface:
 b+ws
= 0.05 in


CAMBER AND DEFLECTION
Camber at transfer:
 transfer
3.79 in – 1.47 in = 2.32 in 
Over time, the beam will camber up due to creep and shrinkage. This is called “growth in storage”.
The PCI Handbook has multipliers to calculate this:
 at erection
= 1.8(3.79) – 1.85(1.42) = 4.20 in 

CAMBER AND DEFLECTION
This is the table from the PCI Design Handbook (7 th ed).
C5.7.3.6.2
allows the use of the these factors for deflection, but these were developed for buildings, not bridges.
Although use is allowed, use with caution.

LIVE LOAD DEFLECTION
Live load deflection is governed by Article 2.5.2.6.2, which is an owner specified OPTIONAL check.