EXAMPLE 9.2 – Part VII PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK...

EXAMPLE 9.2 – Part VIIPCI Bridge Design ManualEXAMPLE 9.2 – Part VIIPCI 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.

CONTINUOUS PRECAST BRIDGESCONTINUOUS 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 BRIDGESCONTINUOUS 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 Mcr . No other calculation is needed.

• 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 > 0.6Mcr. (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 > 0.6Mcr. (5.14.1.4.9a).

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 Mcr . No other calculation is needed.

Here there are two possibilities – bent strand or bent bar connections:


DESIGN UNDER 5.14.1.4.4DESIGN 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 Mcr.

• 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.

Ac = 1868 in2

Ic = 1259300 in4

hc = 80 in.ybc = 59.93 in.(distance to bottom of composite)Sbc = 1259300in4/ 59.93 in. = 21013 in3

Provide 1.2Mcr:

r

cr r bc

cr

f . ksi . ksi

M f S . ksi in k in

. M k in

3

0 24 4 0 480

0 480 21013 10086

1 2 12100

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.

dshpul

pul

f

in inf ksi

( 8)

0.163(26 8 )

1100.163

The 8 in the equation is the 8 inch extension.

DESIGN UNDER 5.14.1.4.4

# Strands Area Aps in2 a in M in-kip > 1.2Mcr ?

8 1.22 0.276 10333 No

10 1.53 0.345 12997 Yes

ps

ps

A ksia

ksi in

d in in in

aM A ksi in

110.4

0.85 4 144

80 3 77

110.4 772

Use 10 extended strand. Use 5 in each of the two bottom rows.



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 > 0.6Mcr. (5.14.1.4.9a)

CREEPCREEP

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:

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

CREEPCREEP

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

CREEPCREEP

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:

1net

eM M

CREEPCREEP

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

0

500

1000

1500

2000

2500

3000

3500

0 10 20 30 40 50 60

Mom

ent

k-ft

Length (ft)

1770 k-ft @ 2.5 ft.

3183 k-ft @ 35.5 ft

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.

k ft ft k ft ftEI k ft ft

EI k ft

2

1770 2.5 3183 1770 3323183 24

2 2 2

160330

CREEP

From basic structural analysis, the rotation at the end of the beam due to beam self weight (w = 0.799k/ft):

k ft ftwLEI

EI k ft

33

2

0.799 / 119

24 24

56100

The prestressing rotation is opposite to the self weight rotation, so:

EI = 160330 – 56100 = 104230 k-ft2

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.

33

2

1.372 / 119

24 24

96340

k ft ftwLEI

EI k ft

So the NET is:EI= 104230 - 96304 = 7926 k-ft2

Again, the (+) sign indicates CCW on left and CW on the right.

CREEP

Based on our theory, the creep angle would be:

EIcreep = EIelastic

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:

r

EIM

L

12

5This 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:

And for a middle span:

Note that the middle span is lower because there is a moment at both ends.

3

2

EIM

L

EIM

L

CREEP

,

,

2,

,

12

512

512

79265 119

160

r cr elastic

r cr elastic

r cr

r cr

EIM

L

M EIL

M k ftft

M k ft

The creep angle is times the elastic angle. Using the multispan formula:

CREEP

Now adjust for rate of creep effect. Assume the creep co-efficient, = 2:

,

2

1

1160 138

cr cr r

cr

eM M

eM k ft k ft

This is a positive moment.

CREEP

ShrinkageShrinkage

• 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.

If the beam were simple span, the differential shrinkage

would deform the beam as though there were end

positive moments:

es = differential shrinkage strain (deck to girder)

Ed = modulus of elasticity of deck concrete=3834 ksi

Ad = area of deck slab = 144*8=1152 in2

y= distance from center of slab to cg of composite section = 80 – (8/2) – 57.2 = 18.8 in

shrinkage s d dM A E y

ShrinkageShrinkage

If a differential shrinkage of 100 x 10-6 is assumed:

,

6 2,

,

100 10 1152 3834 18.8

8300 691

s unres s d d

s unres

s unres

M A E y

M x in ksi in

M k in k ft

Shrinkage

From basic structural analysis, the end rotation for a simple support with a constant moment is:

For the 3 span bridge, we previous used:

So with Ms, unres = 691k-ft, Msr = -829 k-ft; so the NET shrinkage moment is -138 k-ft.

2

ML

EI

1212 12

5 5 2 10s s

sr

M L MEI EIM

L L EI

Shrinkage

This net shrinkage moment is (-), but it will be reduced by creep ( = creep coefficient = 2):

,

2

1

1138

260

s net

eM M

eM k ft

M k ft

Shrinkage

TEMPERATURE GRADIENTTEMPERATURE GRADIENT

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

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

TEMPERATURE GRADIENT

0

i ii

i ii

E T y ydA

E I

The curvature can be calculated from:

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:

0 0

1 1

2 2dL L


, 0

12 6

5 5res temp

EIM EI

L

It the transformed section is used (so that the section can be treated as though it is all made of 7 ksi concrete):

0

,

6

5res temp

EI E T y ydA

M E T y ydA

Again using the restraint moment formula:


0EI E T y ydA 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.


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.


Numerical integration done in a spreadsheet.

(Note, layer 1 can be easily integrated in closed form. If this is done:

a difference of 2%! This is just shown for comparison.)

Area # y' top y' bot T top T bot A y bar Avg T*(y bar)*Ain in Deg F Deg F in^2 in deg F in^3

1 0 4 41 11 436 20.8 235788.82 4 7.5 11 7.791667 381.5 17.05 61115.90263 7.5 8 7.791667 7.333333 16 15.05 1821.054 8 11.5 7.333333 4.125 147 13.05 10990.546885 11.5 13.5 4.125 2.291667 52 10.1 1685.0166676 13.5 15.5 2.291667 0.458333 16 8.22 180.847 15.5 16 0.458333 0 3 7.05 4.846875

Sum 311587.003

y' is measured from top of slaby bar is the distance from the centroid of layer to centroid of girder

3240127OT y ydA F in


,

6 3,

,

6

56

5072 5.5 10 / 311600510430 869

res temp

o ores temp

res temp

M E T y ydA

M ksi x F F in

M k in k ft


NET RESTRAINT MOMENTNET RESTRAINT MOMENT

Moment K-ft

Creep +138

Differential Shrinkage -60

Mrestraint +78

Temperature Gradient +869

0.5 MLL -1000

Mrestraint + 0.5MLL -59Since Mrestraint +MTemp +0.5MLL is slightly negative, assume the joint is fully effective, provide reinforcement for 0.6Mcr. If it had been a large positive value, provide reinforcement for the Mrestraint > 0.6Mcr.

Notes on Restraint MomentsNotes 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 ZONEANCHORAGE ZONE

5.10.10.1 requires reinforcing at the end of the girder to resist splitting forces. The stirrups must resist 4% of Pi:

Asfs > 0.04Pi = 0.04(44 strand)(0.153in2/strand)(202.5ksi)Asfs > 54.5 kips

The term fs cannot exceed 20 ksi, so:

As > 54.5kips / 20 ksi = 2.73 in2 = 5 #5 stirrups.

These must be placed over a distance h/4 = 72/4 = 18 inches at each end of the girder.

CONFINEMENT REINFORCINGCONFINEMENT REINFORCING

5.10.10.2 - For the distance of 1.5d 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 DEFLECTIONCAMBER 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

cip

ci

p

p

e LP e a

E I

in in in ink

ksi in

in

2 2

2 2

2

'

8 6

30.8 1428 11.4 4261241

8 64500 545894

3.79

Standard handbooks give the following formula for a beam with harped strand. The term ec 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.

CAMBER AND DEFLECTION

g

k in inin

ksi in

4

4

5 0.066 / 14281.47

384 4496 545894

Deflection due to self wt (119 ft span) at release:

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 in

ksi in

4

4

5 0.102 / 14161.93

384 5072 545894

Using a structural analysis program, the deflection due to barrier and wearing surface:b+ws = 0.05 in


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


This is the table from the PCI Design Handbook (7th 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 DEFLECTIONLIVE LOAD DEFLECTION

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

EXAMPLE 9.2 – Part VII PCI Bridge Design Manual BULB “T” (BT-72) THREE SPANS, COMPOSITE DECK...

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