Presentation-Moment Capacity of Damaged Beams

7/27/2019 Presentation-Moment Capacity of Damaged Beams

http://slidepdf.com/reader/full/presentation-moment-capacity-of-damaged-beams 1/25

Moment Capacity of Moment Capacity of Damaged GirdersDamaged Girders

Andrea V. Ch Andrea V. Ch áá vezvez , New Mexico State University,, New Mexico State University,NSFNSF --REU 2004REU 2004 August 6, 2004 August 6, 2004

Research Advisor:Research Advisor: Dr. Harry W. ShentonDr. Harry W. Shenton



ObjectiveObjective

•• Analyze bridge girders that have bent flanges Analyze bridge girders that have bent flanges

•• Assume damaged girder is in perfect plastic Assume damaged girder is in perfect plasticregion and calculate plastic moment before andregion and calculate plastic moment before andafter damage occursafter damage occurs

•• Develop plastic moment ratioDevelop plastic moment ratio



Types of Damage to GirdersTypes of Damage to Girders

Severe Case

Our research will focus on aless severe case

Photo courtesy of Connecticut Department of TransportationBureau of Engineering and Highway Operations Division of Research

Photo courtesy of paper on Engineered HeatStraightening



The Focus for this ResearchThe Focus for this Research

DamagedUndamaged

U P M

D P M

( )θ

U P

D P

M

M



Assumptions Assumptions

•• Deck is nonDeck is non --compositecomposite

•• Perfect PlasticityPerfect Plasticity

•• Shear and fatigue not addressed in this studyShear and fatigue not addressed in this study

•• Nonsymmetrical BendingNonsymmetrical Bending



Analysis of Plasticity Analysis of Plasticity

•• Defining the Plastic RegionDefining the Plastic Region

Figure courtesy of Mechanics of Materials (2 nd edition)Gere, Timoshenko

2

)( 21 y y A Z M y

y p

+==

σ σ

y1

A1

A2

y2c2

c1



Analysis of Nonsymmetrical Bending Analysis of Nonsymmetrical Bending

22

)( 21 YAd d YA

y y A Z M T

y y p

==+

==σ

σ

Figure courtesy of Advanced Mechanics of Materials 5 th edition-Boresi, Schmidt,Sidebottom



ConstraintsConstraints

•• Shift in Neutral Axis (which divides theShift in Neutral Axis (which divides the

cross section into two equal areas)cross section into two equal areas)

•• Centroids of areas must be alignedCentroids of areas must be alignedverticallyvertically



Model of Girder Before DamageModel of Girder Before Damage

Load applied

C 1

C 2

N.A.



Model of Girder After DamageModel of Girder After Damage

Load applied

N.A.θ



Choosing a ModelChoosing a Model



Section PropertiesSection Properties

•• Determine the effect of its dimensionsDetermine the effect of its dimensions

•• Calculate the two equal areasCalculate the two equal areas•• Calculate the plastic momentCalculate the plastic moment

•• Locating the neutral axisLocating the neutral axis



Plastic Moment RatioPlastic Moment Ratio

•• RecallRecall

•• Ratio is needed for comparisonRatio is needed for comparison

•• Performance of Calculations basedPerformance of Calculations basedonon

22

)( 21 YAd y y A M y

p=

+=

σ

U

p

D p

M

M

( )( ) U

p

D p

U

D

M

M

y y

y y y y =

+

+

→+

21

2121



Plastic Moment RatioPlastic Moment Ratio

TOP BOT x x=

d

θ

m

Bot y

Bot x

Important:

•m=n

•All thicknesses “t”

are the sameTop x

Top y

n

b

Dm y



Calculating the Plastic Moment RatioCalculating the Plastic Moment Ratio

22

sin28

22

2cos22

22

222

d b

md bd

y

d b

mbmbb

x

BOT

BOT

+

++

=

+

−++−

=

θ

θ

d

θ

b-m

m

d/2

y

x

Bot y

Bot x b




Top x

Top y

d

n

d/2

b-mb

yx

22

8

22

2

2

2

d b

d md y

d b

mbm x

TOP

TOP

+

+

=

+

+−=




2cos32

−

++−=

θ bbm

22

2

22

2cos22

22

22

d b

mbmd

b

mbmbb

x x TOP BOT

+

+−=

+

−++−

=

θ

•Recall

•Solve for m




d b

d db

d b

bd bd

y

yU

m

D

m

+

+

+

++−

=

42

44

cos32sin2

2

22

θ θ

b

d

2=γ

Solve for ratio

γ θ and

Make ratio dependent on only two variables

where



2

2

2+8

cos+34+sin2==

γγ

θ γθ γ

y

y

M

M U

m

Dm

U P

D P




Results of the Plastic Moment RatioResults of the Plastic Moment Ratio

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 5 1 0

1 5

2 0

2 5

3 0

3 5

4 0

4 5

5 0

5 5

6 0

6 5

7 0

7 5

8 0

8 5

9 0

Angle θ (degrees)

r a t i o o

f Y

m D / Y

m U = M

p D / M

p U

0.5

1

23

4

5

10

γ of values



ConclusionsConclusions

••The relative moment capacity decreases for The relative moment capacity decreases for decreasing gammadecreasing gamma

•• As the angle increases the ratio will decrease As the angle increases the ratio will decrease

••The larger the bend of the flange the lower theThe larger the bend of the flange the lower the

capacitycapacity••The shallower the crossThe shallower the cross --section the lower the plasticsection the lower the plasticmoment capacitymoment capacity



Future GoalsFuture Goals

•• Vary the thicknesses for the girder Vary the thicknesses for the girder

•• Calculate the ratio based on varying thicknessesCalculate the ratio based on varying thicknesses

•• Apply new ratio to any girder Apply new ratio to any girder

•• Help DOT with field situations in a timely manner Help DOT with field situations in a timely manner



QuestionsQuestions

Thank You!Thank You!

Contact Info: [email protected]

Presentation-Moment Capacity of Damaged Beams

Documents

Transcript of Presentation-Moment Capacity of Damaged Beams