crackwidth (gergely & lutz)

8/10/2019 Crackwidth (Gergely & Lutz)

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Introduction

Ultimate Limit States Lead to collapseServiceability Limit States Disrupt use of Structures

but do not cause collapse

Recall:


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Introduction

Types of Serviceability Limit States

- Excessive crack width

- Excessive deflection

- Undesirable vibrations

- Fatigue (ULS)


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Crack Width Control

Cracks are caused by tensile stresses due to loads moments,shears, etc..


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Crack Width Control

Cracks are caused by tensile stresses due to loads moments,shears, etc..


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Crack Width Control

Bar crack development.


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Crack Width Control

Temperature crack development


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Crack Width Control

Appearance (smooth surface > 0.01 to 0.013public concern)

Leakage (Liquid-retaining structures)

Corrosion (cracks can speed up occurrence ofcorrosion)

Reasons for crack width control?


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Crack Width Control

Chlorides ( other corrosive substances) present

Relative Humidity > 60 % High Ambient Temperatures (accelerates

chemical reactions)

Wetting and drying cycles Stray electrical currents occur in the bars.

Corrosion more apt to occur if (steel oxidizes rust )


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L imits on Crack Width

0.016 in. for interior exposure

0.013 in. for exterior exposure

max.. crack width =

ACI Codes Basis

Cracking controlled in ACI code by regulating the

distribution of reinforcement in beams/slabs.


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Gergely-Lutz Equation

Crack width in units of 0.001 in.Distance from NA to bottom

(tension) fiber, divided by

distance to reinforcement.

=(h-c)/(d-c)

Service load stress in

reinforcement in ksi

= =

fs=

3cs076.0 Adf=


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Gergely-Lutz Equation

Distance from extreme tension

fiber to center of reinforcement

located closest to it, (in.)

effective tension area of

concrete surrounding tensionbars (w/ same centroid)

divided by # bars. (for 1 layer

of bars A = (2dc

b)/n

dc=

A=

3cs076.0 Adf=


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ACI Code Eqn 10-5 ( limits magnitude of z term )

Note: = 0.076 z

( =1.2 for beams)

Interior exposure: critical crack width = 0.016 in.

( = 16 ) z = 175k/in

Exterior exposure: critical crack width = 0.013 in.

( = 13) z = 145k/in

3cs Adfz=


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Tolerable Crack Widths

Tolerable

Crack Width

Dry air or protective membrane - 0.016 in.

Humidity, moist air, soil - 0.012 in.

Deicing chemicals - 0.007 in.

Seawater and seawater spray - 0.006 in.wetting and drying

Water-retaining structures - 0.004 in.

(excluding nonpressure pipes)

Exposure Condition


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Thin one-way slabs: Use =1.35

z = 155 k/in (Interior Exposure)

z = 130 k/in (Exterior Exposure)

fs= service load stress may be taken as

1.55average load factor

- strength reduction

. factor for flexure

=

0.90

1.5560.0 yys

fff


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Example-Crack

Given: A beam with bw= 14 in. Gr 60 steel 4 #8

with 2 #6 in the second layer with a #4 stirrup.

Determine the crack width limit, z for exteriorand interior limits (145 k/in and 175 k/in.).


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Example-Crack

Compute the center of the steel for the given bars.

) )s

2 2

2

4#8 bars 2#6 bars

4 0.79 in 2 0.44 in

4.04 in

A =

=

=


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Example-Crack

The locations of the center of the bars are

b1 stirrup

b b2

2 b

cover d2

1.0 in.1.5 in. 0.5 in.2

2.5 in.

2.5 in. 2 2

1.0 in. 0.75 in.2.5 in. 1.0 in.

2 2

4.375 in.

dy

d d

y d

=

=

=

=

=

=


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Example-Crack

Compute the center of the steel for the given bars.

) ) ) )

i i

i

2 2

2

2.5 in. 4 0.79 in 4.375 in. 2 0.44 in

4.04 in

2.91 in.

y Ay

A=

=

=


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Example-Crack

Compute number of equivalent bars, n. Use the

largest bar.

Compute the effective tension area

2

i2

bar

4.04 in 5.110.79 in

An

A= = =

) ) 22 2.91 in. 14 in.215.93 in

5.11

ybA

n= = =


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Example-Crack

The effective service load stress is

Compute the effective tension area

)s y0.60 0.6 60 ksi 36 ksif f= = =

) ) )23 3

s c 36 ksi 2.5 in. 15.93 in122.9 k/in.

z f d A= =

=


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Example-Crack

The limits magnitude of z term.

122.9 k/in. < 145 k/in. - Interior exposure

122.9 k/in. < 175 k/in. - Exterior exposure

Crack width is

or = 0.0112 in.

) )

0.076

0.076 1.2 122.9

11.2

w z=

=

=


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Deflection Control

Visual Appearance

( 25 ft. span 1.2 in. )

Damage to Non-structural Elements

- cracking of partitions

- malfunction of doors /windows

(1.)

(2.)

Reasons to Limit Deflection

visiblegenerallyare*

250

1l


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Deflection Control

Disruption of function

- sensitive machinery, equipment

- ponding of rain water on roofsDamage to Structural Elements

- large s than serviceability problem

- (contact w/ other members modify

load paths)

(3.)

(4.)


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Allowable Deflections

ACI Table 9.5(a) = min. thickness unless s are

computed

ACI Table 9.5(b) = max. permissible computeddeflection


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Allowable

Deflections

F lat Roofs( no damageable nonstructural elements

supported)

)

180instLL

l


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Allowable

Deflections

Floors( no damageable nonstructural elements

supported )

)

180instLL

l


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Roof or F loor elements(supported nonstructural elementslikely damaged by large s)

480l


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Roof or F loor elements( supported nonstructural elementsnot likely to be damaged by large

s )

240

l


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Deflection occurring after attachment of

nonstructural elements

Need to consider the specific structures

function and characteristics.

allow


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Moment of I nertia for Deflection Calculation

For (intermediate values of EI)gtecr III

Brandon

derived cr

a

a

cr

gt

a

a

cr

e *1* IM

M

IM

M

I

-

=

Cracking Moment =Moment of inertia of transformed cross-section

Modulus of rupture =

t

gr

y

If

c5.7 f

Mcr=Igt =

fr =


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cr

a

a

crgt

a

a

cre *1* I

M

MI

M

MI

-

=

Distance from centroid to extreme tension fiber

maximum moment in member at loading stage for

which Ie( ) is being computed or at any previous

loading stage

Moment of inertia of concrete section neglect

reinforcement

yt =

Ma =

Ig =


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)

3

a

crcrgcre

cr

3

a

crg

3

a

cre

or

*1*

-=

-

=

M

MIIII

I

M

MI

M

MI


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Moment Vs curvature plot

EIM

EI

M===

slope


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Moment Vs Slope Plot

The cracked beam starts to lose strength as the amountof cracking increases


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Moment of I nertia

For normal weight concrete

)psi33 c1.5

cc fE =

For wc= 90 to 155 lb/ft3

)psi57000 cc fE =

(ACI 8.5.1)


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Deflection Response of RC Beams (F lexure)

A- Ends of Beam CrackB- Cracking at midspan

C- Instantaneous

deflection under serviceload

C- long time deflection

under service loadDand E- yielding of

reinforcement @ ends &

midspan

Note: Stiffness (slope) decreases

as cracking progresses


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Deflection Response of RC Beams (F lexure)

The maximum moments for distributed load actingon an indeterminate beam are given.

=12

2wlM

=

12

2

wlM

=

24

2

wlM


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Uncracked Transformed Section

Part (n) =Ej/Ei Area n*Area yi yi*(n)A

Concrete 1 bw*h bw*h 0.5*h 0.5*bw*h2

As n As (n-1)As d (n-1)*As*d

As n As (n-1)As d (n-1)*As*d

An *ii Any **

=

*ii

*

iii *

An

Anyy

Note:(n-1) is to remove area

of concrete


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Cracked Transformed Section

== s

s

i

ii 2nAyb

dnAy

yb

A

Ayy

Finding the centroid of singly Reinforced RectangularSection

022

0

2

2

ss2

ss

2

ss

2

=-

=-

=

b

dnAy

b

nAy

dnAynAyb

dnAy

ybynAyb

Solve for the quadratic for y


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022 ss2 =-

b

dnAy

b

nAy

Note:

c

s

E

En=

Singly Reinforced Rectangular Section

)2s3

cr

3

1ydnAybI -=


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) )0

212212 ssss2 =-

--

b

dnAAny

b

nAAny

Note:

c

s

E

En=

Doubly Reinforced Rectangular Section

) ) )2s2

s

3

cr 1

3

1ydnAdyAnybI ---=


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Uncracked Transformed Section

) ) ) )

steel

2

s

2

s

concrete

2

3

gt

11

212

1

dyAndyAn

hybhbhI

----

-=

Note: 3g

12

1 bhI =

Moment of inertia (uncracked doubly reinforced beam)


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Finding the centroid of doubly reinforced T-Section

) )

) )0

212

2122

w

ss

2

we

w

sswe2

=

--

-

--

b

dnAAntbb

y

b

nAAnbbty


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Finding the moment of inertia fora doubly reinforced T-Section

)

) ) )

steel

2

s

2

s

beam

3

w

flange

2

e

3

ecr

1

3

1

212

1

ydnAdyAn

tybt

ytbybI

---

-

-=


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Reinforced Concrete Sections - Example

Given a doubly reinforced beam with h = 24 in, b = 12 in.,

d = 2.5 in. and d = 21.5 in. with 2# 7 bars in compression

steel and 4 # 7 bars in tension steel. The material

properties are fc= 4 ksi and fy= 60 ksi.

Determine Igt, Icr, Mcr(+), Mcr(-), and compare to the NA of

the beam.


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The components of the beam

)

)

2 2

s

2 2

s

c c

2 0.6 in 1.2 in

4 0.6 in 2.4 in

1 k57000 57000 4000

1000 lb

3605 ksi

A

A

E f

= =

= =

= =

=


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The compute the n value and the centroid, I uncracked

n area (in ) n*area (in ) yi(in) yi*n*area (in2) I (in ) d (in) d *n*area(in )

A's 7.04 1.2 8.448 2.5 21.12 - -9.756 804.10

As 7.04 2.4 16.896 21.5 363.26 - 9.244 1443.75

Ac 1 288 288 12 3456.00 13824 -0.256 18.89

313.344 3840.38 13824 2266.74

s

c

29000 ksi8.04

3605 ksi

En

E= = =


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The compute the centroid and I uncracked

3i i i

2

i i

2 4 4

i i i i

4

3840.38 in12.26 in.313.34 in

I 13824 in 2266.7 in

16090.7 in

y n A

y n A

I d n A

= = =

= =

=


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The compute the centroid and I for a cracked doublyreinforced beam.

) )0

212212 ssss2 =-

--

b

dnAAny

b

nAAny

) ) ) )

) ) ) ) )

2 2

2

2 2s

2

2 7.04 1.2 in 2 8.04 2.4 in

12 in.

2 7.04 1.2 in 2 8.04 2.4 in 21.5 in.0

12 in.

4.624 72.664 0

y y

y y

- =

- =


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The compute the centroid for a cracked doubly reinforced

beam.

2

4.624 72.664 0y y - =

) )2

4.624 4.624 4 72.664

26.52 in.

y-

=

=


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The compute the moment of inertia for a cracked doublyreinforced beam.

) ) )2s2

s

3

cr 1

3

1ydnAdyAnybI ---=

) )

) ) ) ) ) )

3

cr

22

22

4

112 in. 6.52 in.

3

7.04 1.2 in 6.52 in. 2.5 in.

8.04 2.4 in 21.5 in. 6.52 in.

5575.22 in

I =

-

-

=


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The critical ratio of moment of inertia

4

cr

4g

cr g

5575.22 in

0.34616090.7 in

0.35

I

I

I I

= =


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Find the components of the beam ) ) )

)

) )

c c

s

s s s

2

s s s c

0.85 0.85 4 ksi 12 in. 0.85 34.68

2.5 in. 0.00750.003 0.003

0.0075 217.529000 0.003 87

217.50.85 1.2 in 87

261

100.32

C f ba c c

cc c

f Ec c

C A f f c

c

= = =

- = = -

= = - = -

= - = -

= -


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Find the components of the beam

) )

) ) )

)

2

c s

2

2

2.4 in 60 ksi 144 k

261144 k 34.68 100.32 34.68 43.68 261 0

43.68 43.68 4 261 34.68

2 34.68

3.44 in.

T

T C C

c c cc

c

= =

=

= - - - =

=

=

The neutral axis


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The strain of the steel

Note: At service loads, beams are assumed to act

elastically.

)

)

s

s

3.44 in. 2.5 in.0.003 0.0008 0.00207

3.44 in.

21.5 in. 3.44 in. 0.003 0.0158 0.002073.44 in.

- = =

- = =

3.44 in.

y 6.52 in.

c=

=


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Using a linearly varying and s= E along the NA is the

centroid of the area for an elastic center

The maximum tension stress in tension is

r c7.5 7.5 4000

474.3 psi 0.4743 ksi

f f= =

=

My

I

s= -


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The uncracked moments for the beam

) )

)

)

)

4

r

cr

4

rcr

0.4743 ksi 16090.7 in650.2 k-in.

24 in. 12.26 in.

0.4743 ksi 16090.7 in

622.6 k-in.12.26 in.

My IM

I y

f IM

y

f IM y

ss

-

= =

= = =-

= = =


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Homework-12/2/02

Problem 8.7

crackwidth (gergely & lutz)

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