31-bridgedesign

7/28/2019 31-bridgedesign

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Part 1

Introduction To Bridge Design


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How Do Bridge Engineers DecideOn 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


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Bridge Design ProcessPreliminary Design Process

Bridge Survey Geotechnical Report1. Determine the most

economical type structure andspan arrangement

2. Hydraulic Analysis3. 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


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Types of SuperstructuresBridges are often referred to by their superstructure types.The superstructure system of members carry the roadway over a crossingand 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)


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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


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I - GIRDER

BULB TEE

Prestressed Girders


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Prestressed Concrete I-Girder


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Prestressed Concrete I-Girder Bridge


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Prestressed Concrete Panels


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Prestressed Double Tee Girders


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Steel Plate Girder / Wide Flange Beam / Box Beam


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Steel Plate Girder Bridge


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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 wheredepth of structure is a critical factor.

Typical Slab Bridges : Concrete Box Culverts Solid Slabs Voided Slabs


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14Box Culvert

Triple Box Culvert


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Voided Slab Bridge


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16Solid Slab

Voided Slab Bridge


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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 usedIntermediate 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


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Integral End Abutment


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Semi-Deep End Abutment


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Prestressed I-girder intermediate bent


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Steel girders with open intermediate bent diaphragms


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Footing

Pile Cap Column Footing


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Column Footing


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Preliminary Design

Bridge location Hydraulic design to determine required

bridge length and profile grade Bridge type selection


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Stream Gage Data


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Q = discharge (cfs or m 3/s)k c = constant (1.0 for English units or

0.00278 for metric units)C = Runoff Coefficient

I = Rainfall Intensity (in/hr or mm/hr) A = Drainage Area (acres or hectares)

Rational Method

A I C k Q c


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Drainage Area Delineation


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n1 n2 n3

LeftOverbank

RightOverbank

Channel

Stream Valley Cross-sections


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Mannings Equation

03

2486.1S R A

nQ

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


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n1 n2 n3

LeftOverbank

RightOverbank

Channel

Stream Valley Cross-sections


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Energy Equation

Elevation

1 2

DatumElevation

Pressure

Pressure

Velocity

Velocity

HeadlossEGL

HGL

z 1 z 2

y1

y2

V 12 /2g

V 22 /2g

hl

l h g V

y z g

V y z

22

2

222

2

111


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Constriction of Valley by Bridge

Opening Length

Bridge Deck/Roadway


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Encroachment by Roadway Fill

Flood elevationbefore encroachmenton floodplain

Fill Fill

Bridge Opening Encroachment

Backwater

Encroachment


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Backwater

Normal Water Surface

Water Surface through Structure

Affect of Bridge on Flood

ElevationsDesign High Water Surface (DHW)


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Part 2

Slab Design


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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


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Design Moment

MDL1 = wS 2/10 = 0.106 x 8 2 / 10 = 0.678 MDL2 = wS 2/10 = 0.035 x 8 2 / 10 = 0.224 MLL = 0.8(S+2)P/32 = 0.8(8+2)(16)/32 = 4 M Imp = 30% x M LL = 1.2 M

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

Design For 12.4 k-ft/ft


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Statics, Moment, Shear, Stress?


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Reinforced Concrete Design Basic Equations For Moment Utilize Whitney

Stress Block ConceptDesign Moment = Capacity

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

Compression = Tension0.85f c ba = A s f y

Two Simultaneous Equations, Two Unknowns (a & A s)

d

c

Comp.

Tens.

c = a / b1


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Reinforced Concrete Design (0.85)(4ksi)(12in)(a)=(A s)(60ksi) a=1.47A s 12.4k-ft=(0.9)(A s)(60ksi)(6in-1.47A s/2)/(12in/ft) 12.4=27A s-3.31A s2

ax2+bx+c=0 a=3.31, b=-27, c=12.4, x=A s

As = [-b - (b 2 - 4ac) 1/2]/2a As = [-27 - ((-27) 2-(4)(3.31)(12.4)) 1/2]/[(2)(3.31)] As = 0.49 in 2/ft

5/8 rebar at 7.5 in centers d cComp.

Tens.

c = a / b 1


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Part 3

Steel Beam Design


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Simple Span Beam 50 ft span


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Dead Load = Beam Weight + Deck Weight


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Live Load = HS20 Truck x Distribution Factor

Distribution Factor = S/5.5


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Design Moment = 2358 kip-ft


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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 StressDue to Shear


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Girder Design Moment Of Inertia (I)

1/12bh 3+Ad 2

Parallel Axis Theorem

Section Modulus = S = I/c Stress = Moment/Section Modulus (M/S) For Strength Design Limit Stress to F y Find Shape With S > M/F y S > (2358k-ft)(12in/ft)/50ksi = 566 in 3 A W36x170 Provides 580 in 3


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Part 4

Intermediate Bent Design


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Load Cases

Permanent Loads: DD = Downdrag

DC = Dead LoadComponent 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


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Load Cases (Cont.)

Transient Loads: LL = Live Load

IM = Dynamic Load LS = Live LoadSurcharge

PL = Pedestrian Load

WL = Wind On LiveLoad 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 CombinationLimit State

DC

DDDWEHEV

ESEL

LLIMCEBR

PLLS WA WS WL FR

TU

CR SH TG SE

Use One of These at a Time

EQ IC CT CV

STRENGTH I(unless noted) gp 1.75 1.00 - - -- 1.00 0.50/1.20 gTG gSE -- -- -- --

STRENGTH II gp 1.35 1.00 - - -- 1.00 0.50/1.20 gTG gSE -- -- -- --STRENGTH III gp -- 1.00 1.40 -- 1.00 0.50/1.20 gTG gSE -- -- -- --STRENGTH IV gp -- 1.00 -- -- 1.00 0.50/1.20 -- -- -- -- -- --STRENGTH V gp 1.35 1.00 0.40 1.0 1.00 0.50/1.20 gTG gSE -- -- -- --EXTREME EVENT I gp gEQ 1.00 -- -- 1.00 -- -- -- 1.00 -- -- --EXTREME EVENT II gp 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 gTG gSE -- -- -- --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 gTG gSE -- -- -- --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)(h)= Pbh 2

h

Resultant

P

C o n

t r a c t

i o n S

c o u r

1 0 0 y e a r

P i e r

S c o u r

1 0 0 y e a r

Q 100

b

M


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Water (WA) - Extreme Event

(Cont.)

(b)10000.7VForce

2

C o n

t r a c t

i o n

S c o u r

5 0 0 y e a r

P i e r

S c o u r

5 0 0 y e a r

Q500

b

B

A (B)10000.5VForce

2

A = Of Water Depth 10 B = Sum Of Adjacent Span Length 45

Drift Mat

Pressure = C DV2/1000

CD=0.7

CD=0.5


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

W

W

P (WS)Trans. H H

P (WS)Long.

P Sub.

PVert. = (20psf)(W)(L)PTrans. = (50psf)(H)(L)

PLong. = (12psf)(H)(L T)(%)

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)(L T)(%)

L = Tributary Length

LT = Total Bridge Length% = Long. Distribution %

P (WL)Trans.P (WL)Long.

6


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Int. Bent Analysis


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Cap Beam - Strength Limit State

Basic Equations For Moment Utilize WhitneyStress Block Concept f M

n= f A

s f

y(d-a/2)

f = 0.90

d e c

Comp.

Tens.

c = a / b1


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Cap Beam Service Limit State Crack Control

dc = Concrete Cover To Center Of Closest Bar f s = Service Tensile Stress In Reinforcement h = Overall Section Thickness

ge = 1.00 For Class 1 Exposure (Crack Width = 0.017) = 0.75 For Class 2 Exposure (Crack Width = 0.013)

)d0.7(hd

1c

cs2dc

700s

ss

e

f g


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

x

h d c

f c1

f c2

f s /n

l l CrackSpacing

Primary TensionReinforcement

f c1

f c2

f s /n

f c1

f c2

f s /n

l = =16.03

s s

22c 2sd2

d c


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Simplified Shear Design

LRFD f Vn = f (V c + V s + V p)(kips) f = 0.90

a Set At 90 Set: b=2.0, q =45

Results In:

vvcc db'0.0316Vf

s

)sincot(cotd AV

vyvs

af

LbsToConvertTo1000ByMultiply V c

vvcc db'2.00V f s

d AV vyvs

f

0.0


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Simplifed Shear Design

Section A-A

5 -

# 6 s

( E a c h

F a c e

)

6 - #9s

6 - #9s

#5s @12 or 6 A

A

-400

-200

0

200

400


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Column Design

Column42 Diameter

-1000

3500P (kip)

(P max)

(P min)

1800M (k-ft)

Controlling Point

Axial Load Moment Interaction Diagram

18-#9 Bars

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