1

Part 1

Introduction To Bridge Design

2

How Do Bridge Engineers Decide On What Type Of Bridge To Build?

Bridge Survey• flood plain cross sections• inspection reports • existing bridge (scour, etc)• water elevations• photos• existing roadway profile

Geotechnical Report• soil / geological formations• slopes and grading• foundation problems• soil prop.’s - phi angles etc

Factors affecting choice of superstructure• location, city or rural• span length• vertical clearance• maintainability• environmental concerns• transportation to site issues• costFactors affecting choice of substructure• location and geometry• subsoil conditions• height of column

3

Bridge Design ProcessPreliminary Design Process • Bridge Survey• Geotechnical Report1. Determine the most

economical type structure and span arrangement

2. Hydraulic Analysis 3. Preliminary Cost Estimate4. Foundation Borings5. Determine Foundation Type

Final Design Process • Top to Bottom Design (twice)• Design methods per AASHTO and

MoDOT Bridge Manual• Analysis via

•computations•spreadsheets•computer programs

• Detail plans are produced by technicians (Micro-Station)

• Plans are checked• Quantities computed• Special Provisions written• Plans are advertised for bidding• Low Bid Contractor builds the bridge

4

Types of SuperstructuresBridges are often referred to by their superstructure types. The superstructure system of members carry the roadway over a crossing and transfer load to a substructure.

Superstructures are categorized by; • Support type (simply supported or continuous) • Design type (slab on stringer, slab, arch. Rigid frame, etc)• Material type (steel, concrete, timber)

5

Slab on Stringer Bridges

• Most common type of bridge in Missouri.

• Consist of a deck, resting on the girders. The deck distributes the loads transversely to the girders.

• The girders carry the loads longitudinally (down the length of the bridge) to the supports, (abutments and intermediate bents).

• Concrete

• Deck Girder

• Prestressed I Girder

• Prestressed Double Tee

• Prestressed Box

• Steel

• Plate Girder

• Wide Flange

• Steel Box Girder

6

I - GIRDER

BULB TEE

Prestressed Girders

7

Prestressed Concrete I-Girder

8

Prestressed Concrete I-Girder Bridge

9

Prestressed Concrete Panels

10

Prestressed Double Tee Girders

11

Steel Plate Girder / Wide Flange Beam / Box Beam

12

Steel Plate Girder Bridge

13

Slab BridgesIn slab bridges the deck itself is the structural frame or the entire deck is a thin beam acting entirely as one primary member. These types are used where

depth of structure is a critical factor. Typical Slab Bridges : Concrete Box Culverts Solid Slabs Voided Slabs

14Box Culvert

Triple Box Culvert

15

Voided Slab Bridge

16Solid Slab

Voided Slab Bridge

17

SubstructuresThe substructure transfers the superstructure loads to the foundations. End Abutments• Integral Abutment - girders on beam supported by piles, girders “concreted” into the

diaphragm

• Non-Integral Abutment - diaphragms of steel cross-frames, uses expansion devices

• Semi-Deep Abutment - used when spanning divided highways to help shorten span

• Open C.C. Abutment - beam supported by columns and footings, rarely used Intermediate bents• Open Concrete Bent - beams supported by columns and footings (or drilled shafts)

either a concrete diaphragm (Pre-Stressed Girder) or steel diaphragm (Plate Girder) This is the most common type of Pier MoDOT uses.

• Pile Cap Bent - beams supported by piling (HP or C.I.P.) and are used when the column height is less than 15 feet and usually in rural areas.

• Hammer Head Bent - single oval or rectangular column and footing.

• Spread footings - are used when rock or soil can support the structure.

• Pile footings - rectangular c.c. supported by HP or Cast in Place piles

• Drilled Shafts - holes drilled into bedrock filled with concrete

18

Integral End Abutment

19

Semi-Deep End Abutment

20

Prestressed I-girder intermediate bent

21

Steel girders with open intermediate bent diaphragms

22

Footing

Pile Cap Column Footing

23

Column Footing

24

Preliminary Design

• Bridge location

• Hydraulic design to determine required bridge length and profile grade

• Bridge type selection

25

Stream Gage Data

26

Flood-Frequency Rating Curve

0

40000

80000

120000

160000

0 20 40 60 80 100

Return period (years)

Dis

ch

arg

e (

cfs

)

27

Q = discharge (cfs or m3/s)kc = constant (1.0 for English units or

0.00278 for metric units)C = Runoff CoefficientI = Rainfall Intensity (in/hr or mm/hr)A = Drainage Area (acres or hectares)

Rational Method

AICkQ c

28

Drainage Area Delineation

29

n1 n2 n3

LeftOverbank

RightOverbank

Channel

Stream Valley Cross-sections

30

Manning’s Equation

03

2486.1SRA

nQ

n = Roughness CoefficientA = AreaR = Hydraulic Radius = A / PP = Wetted PerimeterS = Hydraulic Gradient (channel slope)

31

n1 n2 n3

LeftOverbank

RightOverbank

Channel

Stream Valley Cross-sections

32

Energy Equation

Elevation

1 2

DatumElevation

Pressure

Pressure

Velocity

Velocity

HeadlossEGL

HGL

z1z2

y1

y2

V12/2g

V22/2g

hl

lhg

Vyz

g

Vyz

22

22

22

21

11

33

Constriction of Valley by Bridge

Opening Length

Bridge Deck/Roadway

34

Encroachment by Roadway Fill

Flood elevation before encroachment on floodplain

Fill Fill

Bridge Opening Encroachment

Backwater

Encroachment

35

Backwater

Normal WaterSurface

Water Surface through Structure

Affect of Bridge on Flood Elevations

Design High WaterSurface (DHW)

36

Part 2

Slab Design

37

Geometry & Loads

16k 16k

Deck Weight = Width x Thickness x Unit Weight

1 ft x (8.5in x12 in/ft) x 150 lb/cf = 106 lb/ft

38

39

40

Design Moment

• MDL1 = wS2/10 = 0.106 x 82 / 10 = 0.678

• MDL2 = wS2/10 = 0.035 x 82 / 10 = 0.224

• MLL = 0.8(S+2)P/32 = 0.8(8+2)(16)/32 = 4

• MImp = 30% x MLL = 1.2

• Mu = 1.3[0.678+0.224+1.67(4+1.2)] = 12.4

Design For 12.4 k-ft/ft

41

Statics, Moment, Shear, Stress?

42

Reinforced Concrete Design• Basic Equations For Moment Utilize Whitney

Stress Block ConceptDesign Moment = Capacity

12.4 k-ft/ft = As fy(d-a/2) = 0.90

Compression = Tension

0.85f’cba = As fy

Two Simultaneous Equations, Two Unknowns (a & As)

d

c

Comp.

Tens.

c = a / 1

43

Reinforced Concrete Design• (0.85)(4ksi)(12in)(a)=(As)(60ksi) a=1.47As

• 12.4k-ft=(0.9)(As)(60ksi)(6in-1.47As/2)/(12in/ft)

• 12.4=27As-3.31As2

• ax2+bx+c=0 a=3.31, b=-27, c=12.4, x=As

• As = [-b - (b2 - 4ac)1/2]/2a

• As = [-27 - ((-27)2-(4)(3.31)(12.4))1/2]/[(2)(3.31)]

• As = 0.49 in2/ft

• 5/8” rebar at 7.5 in centersd

c

Comp.

Tens.

c = a / 1

44

Part 3

Steel Beam Design

45

Simple Span Beam – 50 ft span

46

Dead Load = Beam Weight + Deck Weight

47

Live Load = HS20 Truck x Distribution Factor

Distribution Factor = S/5.5

48

Design Moment = 2358 kip-ft

49

Design Shear = 214 kips

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Steel Girder Design• Design Moment = 2358 k-ft• Design Shear = 214 kips• Limit Bending Stress Due To Moment

• Limit Shear Stress Due to Shear

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52

Girder Design• Moment Of Inertia (I)

– 1/12bh3+Ad2

– Parallel Axis Theorem

• Section Modulus = S = I/c• Stress = Moment/Section Modulus (M/S)• For Strength Design – Limit Stress to Fy

• Find Shape With S > M/Fy

• S > (2358k-ft)(12in/ft)/50ksi = 566 in3

• A W36x170 Provides 580 in3

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Part 4

Intermediate Bent Design

54

Load Cases

• Permanent Loads:– DD = Downdrag– DC = Dead Load

Component– DW = Dead Load

Wearing Surface– EH = Horizontal Earth– ES = Earth Surcharge– EV = Vertical Earth– EL = Locked In Forces

• Transient Loads:– SE = Settlement

– BR = Braking

– CE = Centrifugal Force

– CT = Vehicular Collision

– CV = Vessel Collision

– EQ = Earthquake

– IC = Ice Load

– FR = Friction

55

Load Cases (Cont.)

• Transient Loads:– LL = Live Load

– IM = Dynamic Load

– LS = Live Load Surcharge

– PL = Pedestrian Load

– WL = Wind On Live Load

– WS = Wind On Structure

• Transient Loads:– TG = Temperature

Gradient

– TU = Uniform Temperature

– CR = Creep

– SH = Shrinkage

– WA = Water Load

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Load Combinations

Load Combination

Limit State

DC

DD

DW

EH

EV

ES

EL

LL

IM

CE

BR

PL

LS WA WS WL FR

TU

CR

SH TG SE

Use One of These at a Time

EQ IC CT CV

STRENGTH I(unless noted) p 1.75 1.00 -- -- 1.00 0.50/1.20 TG SE -- -- -- --

STRENGTH II p 1.35 1.00 -- -- 1.00 0.50/1.20 TG SE -- -- -- --

STRENGTH III p -- 1.00 1.40 -- 1.00 0.50/1.20 TG SE -- -- -- --

STRENGTH IV p -- 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --

STRENGTH V p 1.35 1.00 0.40 1.0 1.00 0.50/1.20 TG SE -- -- -- --

EXTREME EVENT I p EQ 1.00 -- -- 1.00 -- -- -- 1.00 -- -- --

EXTREME EVENT II p 0.50 1.00 -- -- 1.00 -- -- -- -- 1.00 1.00 1.00

SERVICE I 1.00 1.00 1.00 0.30 1.0 1.00 0.50/1.20 TG SE -- -- -- --

SERVICE II 1.00 1.30 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --

SERVICE III 1.00 0.80 1.00 -- -- 1.00 0.50/1.20 TG SE -- -- -- --

SERVIE IV 1.00 -- 1.00 0.70 -- 1.00 0.50/1.20 -- 1.0 -- -- -- --FATIGUE – LL, IM & CE ONLY

-- 0.75 -- -- -- -- -- -- -- -- -- -- --

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Water (WA) – Strength

M = (Pbh)(½h)

= ½ Pbh2

½ h

Resultant

PC

ontr

actio

n S

cour

100 y

ear

Pie

r S

cou

r10

0 y

ear

Q100

b

M

58

Water (WA) - Extreme Event (Cont.)

(b)10000.7VForce

2

Con

trac

tion

Sco

ur50

0 y

ear

Pie

r S

cou

r50

0 y

ear

Q500

b

B

A (B)10000.5VForce

2

A = ½ Of Water Depth ≤ 10’

B = ½ Sum Of Adjacent Span Length ≤ 45’

Drift Mat

Pressure = CDV2/1000

CD=0.7

CD=0.5

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Wind on Structure (WS)P(WS)Vert.

W

¼W

P(WS)Trans. H

½H

P(WS)Long.

PSub.

PVert. = (20psf)(W)(L)

PTrans. = (50psf)(H)(L)

PLong. = (12psf)(H)(LT)(%)

PSub. = (40psf)(b)

L = Tributary Length

LT = Total Bridge Length

% = Long. Distribution %

b = Column Or Cap Width

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Wind on Live Load (WL)PTrans. = (100plf)(L)

PLong. = (40plf)(LT)(%)

L = Tributary Length

LT = Total Bridge Length

% = Long. Distribution %

P(WL)Trans.

P(WL)Long.6’

61

Int. Bent Analysis

62

Cap Beam - Strength Limit State

• Basic Equations For Moment Utilize Whitney Stress Block Concept– Mn = As fy(d-a/2)

– = 0.90

d e

c

Comp.

Tens.

c = a / 1

63

Cap Beam – Service Limit State• Crack Control

–

– dc = Concrete Cover To Center Of Closest Bar

– fs = Service Tensile Stress In Reinforcement

– h = Overall Section Thickness

– e = 1.00 For Class 1 Exposure (Crack Width = 0.017”)

= 0.75 For Class 2 Exposure (Crack Width = 0.013”)

)d0.7(h

d1

c

cs

2dc700

sss

e f

64

Cap Beam Service Limit State• Crack Control Is Based On A Physical Model

x

h d c

fc1

fc2

fs/n

l lCrackSpacing

Primary TensionReinforcement

fc1

fc2

fs/n

fc1

fc2

fs/n

l = =16.03”

s s

2 2

c 2sd2

d c

65

Simplified Shear Design

• LRFD– Vn = (Vc + Vs + Vp)(kips) = 0.90

–

– Set At 90°

– Set: =2.0, =45°

–

– Results In:

vvcc d b ' 0.0316V fs

)sincot(cotdAV vyv

s

f

Lbs To Convert To 1000By Multiply Vc

vvcc d b ' 2.00V fs

dAV vyv

s

f

0.0

66

Simplifed Shear Design

Section A-A

5 -

#6’s

(Ea

ch F

ace

)

6 - #9’s

6 - #9’s

#5’s @ 12” or 6”A

A

-400

-200

0

200

400

67

Column Design

Column 42” Diameter

-1000

3500P (kip)

(P max)

(P min)

1800M (k-ft)

Controlling Point

Axial Load – Moment Interaction Diagram

18-#9 Bars