concrete cracking check

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Lecture 34 - Elastic Flexural

Analysis for Serviceability

November 20, 2001

CVEN 444



Lecture Goals

• Serviceability

• Crack width

• Moments of inertia



Introduction

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

but do not cause collapse

Recall:



Introduction

Types of Serviceability Limit States

- Excessive crack width

- Excessive deflection

- Undesirable vibrations

- Fatigue (ULS)



Crack Width Control

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



Crack Width Control

Bar crack development.



Crack Width Control

Temperature crack development



Crack Width Control

• Appearance (smooth surface > 0.01 to 0.013

public concern)

• Leakage (Liquid-retaining structures)

• Corrosion (cracks can speed up occurrence of

corrosion)

Reasons for crack width control?



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 )



Limits on Crack Width

0.016 in. for interior exposure

0.013 in. for exterior exposure

max.. crack width =

ACI Code’s Basis

Cracking controlled in ACI code by regulating the

distribution of reinforcement in beams/slabs.




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

= =

f s =

3cs 076.0 Ad f =




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 =

3cs 076.0 Ad f =




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 Ad f z =




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




Thin one-way slabs: Use =1.35

z = 155 k/in (Interior Exposure)

z = 130 k/in (Exterior Exposure)

f s = service load stress may be taken as

1.55 – average load factor

- strength reduction

. factor for flexure

=

0.90

1.5560.0 y

ys

f f f



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.).



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 =

=

=



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.

d y

d d

y d

=

=

=

=

=

=



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 A y

A=

=

=



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.2

15.93 in5.11

yb A

n= = =



Example-Crack

The effective service load stress is

Compute the effective tension area

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

) ) )23 3

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

z f d A= =

=



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 =

=

=



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



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.)



Allowable Deflections

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

computed

ACI Table 9.5(b) = max. permissible computeddeflection




Flat Roofs ( no damageable nonstructural elements

supported)

)

180instLL

l




Floors ( no damageable nonstructural elements

supported )

)

180instLL

l




Roof or Floor elements (supported nonstructural elementslikely damaged by large ’s)

480

l




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

’s )

240

l




Deflection occurring after attachment of

nonstructural elements

Need to consider the specific structures

function and characteristics.

allow



Moment of Inertia for Deflection Calculation

For (intermediate values of EI)gtecr I I I

Brandon

derived cr

a

a

cr

gt

a

a

cr

e *1* I M

M

I M

M

I

-

=

Cracking Moment =Moment of inertia of transformed cross-section

Modulus of rupture =

t

gr

y

I f

c5.7 f

Mcr =Igt =

f r =




cr

a

a

crgt

a

a

cre *1* I

M

M I

M

M I

-

=

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 =




)

3

a

crcrgcre

cr

3

a

crg

3

a

cre

or

*1*

-=

-

=

M

M I I I I

I

M

M I

M

M I



Moment Vs curvature plot

EI M

EI

M ===

slope



“Moment Vs Slope” Plot

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



Moment of Inertia

For normal weight concrete

)psi 33 c

1.5

cc f E =

For wc = 90 to 155 lb/ft3

)psi 57000 cc f E =

(ACI 8.5.1)



Deflection Response of RC Beams (Flexure)

A- Ends of Beam Crack B - Cracking at midspan

C - Instantaneous

deflection under serviceload

C’ - long time deflection

under service loadD and E - yielding of

reinforcement @ ends &

midspan

Note: Stiffness (slope) decreases

as cracking progresses




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

=12

2wl M

=

12

2

wl M

=

24

2

wl M




For Continuous beams

ACI 9.5.2.4

ACI Com. 435

Weight Average

) ) )e21emideavge 25.050.0 I I I I =

) ) )

= e21emideavge 15.070.0

:continousends2

I I I I

) ) )

= 1emideavge 15.085.0

:continousend1

I I I

)

2end@

1end@midspan@

ee2

ee1

emide

I I

I I

I I

=

=

=



Uncracked Transformed Section

Part (n) =E j /Ei Area n*Area yi yi*(n)A

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

A’s n A’s (n-1)A’s d’ (n-1)*A’s*d’

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

An* ii An y **

=

*ii

*

iii *

An

An y y

Note: (n-1) is to remove area

of concrete



Cracked Transformed Section

== s

s

i

ii 2nA yb

d nA y

yb

A

A y y

Finding the centroid of singly Reinforced RectangularSection

022

0

2

2

ss2

ss

2

ss

2

=-

=-

=

b

d nA y

b

nA y

d nA ynA yb

d nA y

yb ynA yb

Solve for the quadratic for y




022 ss2 =-

b

d nA y

b

nA y

Note:

c

s

E

E n =

Singly Reinforced Rectangular Section

)2

s

3

cr

3

1 yd nA yb I -=




) )0

212212 ssss2 =-

--

b

d nA An y

b

nA An y

Note:

c

s

E

E n =

Doubly Reinforced Rectangular Section

) ) )2

s

2

s

3

cr 1

3

1 yd nAd y An yb I ---=



Uncracked Transformed Section

) ) ) )

steel

2

s

2

s

concrete

2

3

gt

11

212

1

d y And y An

h ybhbh I

----

-=

Note: 3

g

12

1 bh I =

Moment of inertia (uncracked doubly reinforced beam)




Finding the centroid of doubly reinforced T-Section

) )

) )0

212

2122

w

ss

2

we

w

sswe2

=

--

-

--

b

d nA Ant bb

y

b

nA Anbbt y




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

yd nAd y An

t ybt

yt b yb I

---

-

-=



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 f c = 4 ksi and f y= 60 ksi.

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

the beam.




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 k 57000 57000 4000

1000 lb

3605 ksi

A

A

E f

= =

= =

= =

=




The compute the n value and the centroid, I uncracked

n area (in2) n*area (in2) yi (in) yi *n*area (in2) I (in4) d (in) d2*n*area(in4)

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

E n

E = = =




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

= = =

= =

=




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

) )0

212212 ssss2 =-

--

b

d nA An y

b

nA An y

) ) ) )

) ) ) ) )

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

- =

- =




The compute the centroid for a cracked doubly reinforced

beam.

2

4.624 72.664 0 y y - =

) )2

4.624 4.624 4 72.664

26.52 in.

y-

=

=




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

) ) )2

s

2

s

3

cr 1

3

1 yd nAd y An yb I ---=

) )

) ) ) ) ) )

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 =

-

-

=




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

= =




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 E c c

C A f f c

c

= = =

- = = -

= = - = -

= - = -

= -




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




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 =

=




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 = -




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 I M

I y

f I M

y

f I M y

s s

-

= =

= = =-

= = =



Homework-12/2/02

Problem 8.7

concrete cracking check

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