Example Three Spans

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EXAMPLE 9.2 – Part I

PCI Bridge Design Manual

EXAMPLE 9.2 – Part I

PCI 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.

BRIDGE LAYOUT - LongitudinalBRIDGE LAYOUT - Longitudinal

Continuous for Live Load


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BRIDGE LAYOUT – Cross SectionBRIDGE LAYOUT – Cross Section

DESIGN SPECIFICATIONSDESIGN SPECIFICATIONS

• LRFD – 5th Edition (2010)

• HL-93 Truck Loading

• No Skew

• Composite Deck


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• Concrete:

– f c’ = 7.0 ksi @ 28 days

– f ci’ = 5.5 ksi @ release

– wc = 0.150 kcf

– Ecb = 33000w1.5 (f c’)

0.5 (LRFD 5.4.2.4)

= 33000(0.150)1.5(7.0)0.5 = 5072 ksi

• Prestressing Steel:

– GR 270 (f pu = 270 ksi; f py = 243 ksi) – ½” strand (Ap = 0.153 in

2 / strand)

– Ep = 28500 ksi


• Mild Steel

– GR 60 (f y = 60 ksi)

– Es = 29000 ksi

• Future Wearing Surface

– 2” thick

– wws =0.150 kcf • Barriers

– New Jersey type

– 0.300 k/ft


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• Deck

– 7.5” Structural thickness

– 0.5” wearing surface

– Total thickness = 8”

– f c’ = 4.0 ksi @ 28 days

– wc = 0.150 kcf

– Ecs = 33000w1.5 (f c’)

0.5 (LRFD 5.4.2.4)

= 33000(0.150)1.5(4)0.5 = 3 834 ksi

• Note – LRFD uses kip, inch, foot units in

formulae

CONTINUOUS FOR LL PRECAST BRIDGESCONTINUOUS FOR LL PRECAST BRIDGES

• Precast beams are made in a factory

and shipped to site. The beam is set on

simple supports – beam carries self

weight and prestressing force as a

simple beam.


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The deck is formed and poured. Since the

beams are NOT shored, the beams carry the

deck load as simple beams.


The deck is cast continuous over the piers. When the

deck hardens, a continuous structure is formed. The

negative moment connection is usually made with

non-prestressed steel over the piers. Thus, the

negative moment region is conventionally

reinforced.


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• Once the deck hardens and continuity isestablished, any superimposed dead load

(asphalt surfaces, barriers, utilities) is carried by

the beams as a continuous structure.

• All live load is carried as a continuous structure.


• After the slab is poured, the beams will

continue to creep and shrink; cambering

up.

• Temperature will also cause camber.

• Positive moments will form causing

cracking.


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• A positive moment connection is

required. The requirements for this will

be discussed later. (LRFD 5.14.1.4)


• It is thought that creep and shrinkage

will redistribute dead load, so some

states design using simple spans for all

dead load and assuming a continuous

bridge for live load only.

• Some states completely ignore the

continuity and design as simple span forall loads.


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DESIGN SPANSDESIGN SPANS

• Beams:

– Overall Length• 110 ft. end spans

• 119 ft center span

– Design Spans –Simple Span beam• 109 ft. end spans

• 118 ft. center span

– Design Spans – Continuous Beam• 110 ft. end spans

• 120 ft. center span

PROPERTIES OF BT-72PROPERTIES OF BT-72

A = 767 in.2

h = 72 in.

I = 545 894 in.4

yb = 36.60 in.

yt = 35.40 in.

Sb = 14 915 in.3

St = 15 421 in.3

w = 0.799 k/ft


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PROPERTIES OF COMPOSITE

BT-72


BT-72

Ecs = 3834 ksi Ecb = 5072 ksi (prev. defined)

Modular ratio: n = Ecs /Ecb = 3834/5072 = 0.7559

LRFD 4.6.2.6.1 (NEW IN 2009):

The effect ive flange width is now the TRIBUTARY

AREA:

bf = 144 inches


BT-72


BT-72

Note: ½ inch haunchassumed.

Shaded area is

transformed.


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PROPERTIES OF COMPOSITE BT-72PROPERTIES OF COMPOSITE BT-72

Transformed Flange Width =

(Effect ive Flange Width)*n = 144(0.756)= 108.9 in.

Transformed Flange Area = 108.9” (7.5” ) = 816.8 in2

Note: only 7.5” of deck thickness is structural.


BT-72


BT-72Haunch – assumed ½” over BT-72 flange width to account

for di fferential camber in the beams.

Transformed Haunch Width = 0.756(42” ) = 31.75 in.

Transformed Haunch Area = 31.75” (0.5”) = 15.87 in2


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Atr in2

ybin.

Aybin.3

A(ybc-yb)2

in.4

Iin.4

I+A(ybc-yb)in.4

Beam 767.00 36.60 28072 325484 545894 871378

Haunch 15.87 72.25 1147 3601 0 3601

Deck 816.8 76.25 62280 296420 3829 300249

Sum 1599.7 91500 1175230

ybc = 91500/1599.7 = 57.20 in.(distance to bottom of composite)


Ac = 1599 in2

Ic = 1175230 in4

hc = 80 in.

ybc = 91477/1599.4 = 57.20 in.

(distance to bottom of composite)

ytc = 80 – 57.20 = 22.80 in.

(distance to top of composite)ytg = 72 – 57.20 = 14.80 in.

(distance from composite neutral axis to top of

beam)


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BT-72


BT-72

Composite Section Modulus to Bottom:Sbc = Ic / ybc = 1175230 /57.20 = 20545 in

.3

Composite Section Modulus to Top of

Composite:

Stc = Ic /nytc = 1175230/(0.756*22.8) = 68180 in.3

Note: 1/n converts stress in transformed

concrete to stress in actual concrete.

Composite Section Modulus to Top of Beam:

Stg = Ic / ytg = 1175230 /14.8 = 79400 in.3

DEAD LOADS - DCDEAD LOADS - DC

DC – Applied to precast only.

Beam self weight wg = 0.799 kip/ft.

Slab weight – include ½” integral wearing

surface.

ws = (8” /12” /ft)(12 ft.)(0.150 kcf) = 1.20 kip/ft

Haunch

wh = (0.5” /12)(42” /12)(0.150 kcf) = 0.022 kip/ft


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DC – Applied to composite section. To

determine if the barrier weight and the

future wearing surface can be equally

distributed, the following must be met

(LRFD 4.6.2.2.1):

1) Width of deck constant OK

2) Number of beams > 4 OK

3) Curvature < specified in 4.6.2.1.4 OK

straight4) Cross section matches one given in LRFD

Spec. table 4.6.2.2.1-1 OK type “ k”

k


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DEAD LOADS - DCDEAD LOADS - DC5) The overhang of the roadway, from the

outside of the web, de < 3.0 ft.

de = 3 ft OK

Def. of de changed in2008 interim (LRFD 4.6.2.2.1).


DC – Applied to composite section

Barrier weight – 0.30 kip/ft

wb = 2 barriers (0.3 k/ft) / (4 beams) =

0.150 k/ft /beam

Diaphragm weight – assumed steel X

braces. Weight ignored in this example.

Typically, they weigh a few hundred

pounds.


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DEAD LOADS - DWDEAD LOADS - DW

DW – Future wearing surface and uti lities.

Future wearing surface 2” @ 0.150 kcf

(2” /12)(0.150 kcf) = 0.025 ksf

0.025 ksf (42’ roadway width) / 4 beams

= 0.263 k/ft /beam

UNFACTORED DEAD LOADSUNFACTORED DEAD LOADS

Al l loads are uniform. DL moments and

shears on the precast can be found from:

0.5

0.5

x

x

V w L x

M wx L x

Use overall length at ini tial (release)condition.

Center to center of bearing at deck

placement.


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UNFACTORED DEAD LOADSUNFACTORED DEAD LOADS

The shears and moments due to the future

wearing surface and the barrier weight are

computed by considering the bridge as a

continuous, three span structure.

The span lengths after continuity is

established are center of support to center of

pier for end spans and center of pier to

center of pier for the middle span.

Shears and moments can be found us ing any

analysis program or by a hand calculation.

Unfactored DL MomentsUnfactored DL Moments

End

Spans

MiddleSpan


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LIVE LOAD DISTRIBUTION FACTORSLIVE LOAD DISTRIBUTION FACTORS

To use distr ibution factors, the following must bemet:

1) Width of deck constant OK

2) Number of beams > 4 OK

3) Curvature < specified in 4.6.2.1.4 OK straight

4) Cross section matches one given in LRFD Spec.

table 4.6.2.2.1-1 OK type “ k”

5) de < 3 f t. OK 3 ft.

6) Beams parallel and approximately same

stiffness. OK

LIVE LOAD DISTRIBUTION

FACTORS - MOMENT


FACTORS - MOMENTNumber of design lanes = integer part of

42 ft. / (12 ft./lane) = 3 lanes

42 ft. is clear roadway width.

Interior Beams (Table 4.6.2.2.2b-1):

0.10.6 0.2

3

0.10.4 0.3

3

0.075

9.5 12

0.0614 12

g

s

g

s

Two Lanes

K S SDFM

L Lt

One Lane

K S SDFM

L Lt


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FACTORS - MOMENT


FACTORS - MOMENT

To use these factors:

1) 3.5’ < S < 16’ S = 12 ft. OK

2) 4.5” < ts < 12” ts = 7.5 in. OK

3) 20’ < L < 240’ L = 120 ft. OK

4) Nb > 4 beams Nb = 4 beams OK

Note: Although this is a 3 lane bridge, there is

NO reduction to the LL for mul tiplepresence. The distribution factors already

account for multiple presence.


FACTORS - MOMENT


FACTORS - MOMENT

2g gc

cs

K n I Ae

E n

E

n = 5072/3834 = 1.3229eg = (7.5/2)+0.5+35.4 = 39.65

= distance between centroids of beam and slab

A = area of non-composi te beam

I = moment of inertia of non-composi te beam


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FACTORS - MOMENT


FACTORS - MOMENT

2g gK n I Ae

Kg = (1.323)[545894 + 767(39.65)2]

= 2 317 340 in4


FACTORS - MOMENT


FACTORS - MOMENT

0.10.6 0.2

30.075

9.5 12

g

s

Two Lanes

K S SDFM

L Lt

S = 12 ft.

L = 120 ft.

Kg = 2 317 340 in

4

ts = 7.5”

DFM = 0.905 lanes/beam


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FACTORS - MOMENT


FACTORS - MOMENT

0.10.4 0.3

30.06

14 12

g

s

One Lane

K S SDFM

L Lt

S = 12 ft.

L = 120 ft.

Kg = 2 317 340 in4

ts = 7.5”

DFM = 0.596 lanes/beam

DFM = 0.905 lanes/beam –two lanes CONTROLS


FACTORS - SHEAR


FACTORS - SHEAR

2

0.212 35

0.3625

Two Lanes

S SDFV

One Lane

SDFV

Interior Beams:


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FACTORS - SHEAR


FACTORS - SHEAR

To use these factors:

1) 3.5’ < S < 16’ S = 12 ft. OK

2) 4.5” < ts < 12” ts = 7.5 in. OK

3) 20’ < L < 240’ L = 120 ft. OK

4) Nb > 4 beams Nb = 4 beams OK

5) 10 000 < Kg < 7 000 000

Kg = 2 317 340 OK


FACTORS - SHEAR


FACTORS - SHEAR

2

0.212 35

0.3625

Two Lanes

S SDFV

One Lane

SDFV

S = 12 ft.

DFV = 1.082 lanes/beam two lane CONTROLS

DFV = 0.840 lanes/beam one lane

Example Three Spans

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