CE 544_Beam Deflection

7/24/2019 CE 544_Beam Deflection

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KSU. CE 544, (Dr. Asad Esmaeily) 1Related to Chapter 6, Textbook (Nilson)

SERVICEABILITY

Flexural Cracks, Beam Deflection

CE 544


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KSU. CE 544, (Dr. Asad Esmaeily) 2

Goals

Serviceability

Flexural Cracks

Beam deflection


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Control of Flexural Cracks

Types of Cracks:


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ACI Committee 224 Crack Limits:


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ACI Code, sections 10.6.3 and 10.6.4 requires the flexural

tensile reinforcement should be well distributed within the

zone of maximum tension, so that the center-to-center spacing

of reinforcing closest to a tension surface is not greater than:

40000 400001 ACI Equation 10-5 2.5 412cs s

s cf f

=

2Computed tensile stress at working load (or )

3Clear cover of the nearest surface in tension to the

surface of the tensile reinforcement in inch.

ys

c

ff

c

=

=


Nilson Textbook 6.3


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Calculating a flexural crack width (Gergely-Lutz Eqn.):

30.076 s cw f d A=


Calculating a flexural crack width (Frosch Eqn.):

222000

2

sc

s

f sw d

E = +


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estimated crack width in thousandths of inches

ratio of the distance to the neutral axis from the

extreme tension concrete fiber to the distance from

the neutral axis to the centroid of the tensile

w

=

=

steel.

(by working stress method)steel stress in ksi at service load (0.6 is permitted)

the cover of the outermost bar measured from extreme

tension fiber to the center of closest bar or wire

ys

c

ff

d

=

=

.

effective tension area of concrete around the main reinforcing

(with same centroid as the reinforcing) divided by number of bars.

maximum bar spacing (center to center).

A

s

=

=

Control of Flexural Cracks30.076 s cw f d A=

2

220002

sc

s

f sw d

E

= +



permitted by the ACI 224 code

Use the (Gergely-Lutz Equation) for calculating the flexural crack width.



Example (Cont.):

30.076 s cw f d A=



Example (Cont.): 30.076 s cw f d A=

Lets calculate all exactly assuming f 'c=3000psi

We can find that the NA and Icr:

y 9.0206 in= (Nutral Axis Location from Top)

h 27 in= d 24 in=Icr 13669.0831in

4=

dc h d= A2 h d( ) b

3= 2.5

kip

ft= Distributed Load

h y

d y

= 1.2003=

momentLb

2

8= Maximum Moment at the beam center moment 3375kip in=

fsn moment d y( )

Icr= Stress in tensile steel fs 34.3525ksi=

30.076 s cw f d A=A 32in

2= dc 3 in= w 0.000076

in2

kipfs

3dc A=

w 0.0143 in= Compared to 0.015 in. by the inaccurate method.

Ec 57000 fcps i=

Ec 3122.0186 ksi=

nEs

Ec=

n 9.2889=




Replace steel with 5 number 9 bars (f 'c=3000psi), As=5.0 in2

Compared to 0.0127 in. by inaccurate method

y 9.252in= (Nutral Axis Location from Top)

h 27 in= d 24 in=Icr 14325.622in

4=

dc h d= A2 h d( ) b

5= 2.5

kip


h y

d y=

1.2034=

moment Lb

2


fsn moment d y( )


30.076 s cw f d A=A 19.2 in

2

= dc 3 in= w 0.000076 in2

kipfs 3 dc A=

w 0.0114 in=




Replace steel with 6 number 8 bars (f 'c=3000psi), As=4.71 in2

Compared to 0.0119 in.

y 9.0428 in= (Nutral Axis Location from Top)

h 27in= d 24in=Icr 13731.5088in

4=

dc h d= A2 h d( ) b

6= 2.5

kip


h y

d y=

1.2006=

momentLb2


fsn moment d y( )


30.076s c

w f d A=A 16in2= dc 3in= w 0.000076

in2

kipfs

3dc A=

w 0.0113 in=



Example (Cont.):

Calculating flexural crack width using (Frosch Equation)

for the two cases:

2

220002

sc

s

f sw d

E

= +

2

216-2 3 0.6 60 5 in.For 3#11 bars, s= 5 in., 2000 1.2 3 11.632 29000 2 1000

0.0116 in. < 0.012 in. O.K.-but very close to the limit!

w

w

= = + =

=

2

216-2 3 0.6 60 2 in.For 6#8 bars, s= 2 in., 2000 1.2 3 9.425 29000 2 1000

0.0094 in. 0.01< 0.012 in. O.K.

w

w

= = + =

=



Calculating flexural crack width using (Frosch Equation) for the two cases,

with exact calculated values forfs, etc, for 3#11 bars:

Example (Cont.):

Frosch Method:

Ab 1.56 in2

= nb 3=

Calculate Maximum Bar Spacing, s:

2

220002

sc

s

f sw d

E

= +

sb 2cover

nb 1=

fs 34355.1992 psi= 1.2003= s 5 in= dc 3 in=

wf 2000fs

Es dc

2 s

2

2

+1

1000= wf 0.0111in=



Calculating flexural crack width using (Frosch Equation) for the two cases,

with exact calculated values forfs, etc, for 6#8 bars:

Example (Cont.):

Frosch Method:

Ab 0.79 in2

= nb 6=

Calculate Maximum Bar Spacing, s:

2

220002

sc

s

f sw d

E

= +

sb 2cover

nb 1=

fs 33944.0276psi= 1.2009= s 2 in= dc 3 in=

wf 2000fs

Es dc

2 s

2

2

+1

1000= wf 0.0089 in=



SERVICEABILITY

Flexural Cracks, Beam Deflection

Deflection Control



Deflection Control

(1) Visual Appearance( 25 ft. span 1.2 in. )

(2) Damage to Non-Structural (NS) Elements

-cracking of partitions

- malfunction of doors/windows

(3) Disruption of function

-sensitive machinery, equipment

-ponding of rain water on roofs

(4) Damage to Structural Elements

-large deflections than serviceability problem

-(contact w/ other members may lead to modified loadpaths)

Reasons to Limit Deflection

1 are generally visible250

l>


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ACI Table 9.5(b)

Table 6.2 Nilson Textbook


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Allowable Deflections

Flat Roofs (no damageable Non-Structural elements

supported)

( )

180

instLL

l

Floors (no damageable Non-Structural elements

supported )

( )LL inst 360

l


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Allowable Deflections

Roof or Floor elements

(supported Non-Structural elements

likely damaged by large s)

480

l

Roof or Floor elements

(supported Non-Structural NS elements not likely to be

damaged by large s )

240

l

:

allow :

Deflection occurring after attachment ofNon-Structural elements

Need to consider the specific structures

function and characteristics.


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Deflection of Flexural Members, a Review

Load-deflection behavior of a flexural member.


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2

2

So, while linear, the moment curvature response for a RC section

can be written as:

Curvature=d y M

dx EI = =

Relationship between Curvature and Deflection

Where EI is called the flexural rigidity of the section.

Note that the I (Moment of Inertia) is calculated using

the full section, considering the steel (equivalent area of

concrete for steel will b

=M EI

e: [ -1] ) when section is NOTcracked, and after cracking and while in linear behavior

range, the transformed section is used. Then equivalent area

for steel will be:

s

s

n A

nA


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As

(n-1)As

NA

Un-cracked

yd

N.A.

nAs

b

d y

Cracked and linear

( )

( )

( )

2

232

s

32

s

12 2

3

i i iI I y y A

by yby d y nA

byd y nA

= +

= + +

= +

s

c

E

En=

( )

i i

i

2

y A

A

i i i

y

I I y y A

=

= +


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Moment of Inertia for Deflection Calculation

For (intermediate values of EI)gtecr III

Brandon

derived:

a a

cr cr

e gt cr a a

1M M

I I IM M

= +

Cracking Moment =

Moment of inertia of transformed cross-section

Modulus of rupture =

r gt

t

f I

y

c7.5 f

Mcr=

Igt =

fr =

yt =Distance from centroid to extreme tension fiber


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Moment of Inertia for Deflection Calculation

a a

cr cr e g cr

a a

1M M

I I IM M

= +

Maximum moment in member at loading stage for

which Ie ( ) is being computed or at any previous

loading stageMoment of inertia of concrete section neglect

reinforcement

Ma =

Ig =


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ACI Moment of Inertia for Deflection Calculation

r g

cr

t

'r

where: (ACI 9-9)

(ACI 9-1

y

7.5 0)

=

= c

f IM

f f

( )

3 3

cr cr e g cr

a a

3

cr

e cr g cr a

(ACI 9-8)

or:

1M M

I I IM M

MI I I I

M

= +

= +

cr Transformed Moment of Inertia of the cracked sectionI =


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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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Modulus of Elasticity

( )1.5c c c33 psiE f =

For wc = 90 to 155 lb/ft3

( )c c57000 psiE f=

For normal weight concrete


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Deflection Response of RC Beams (Flexure)

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


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A- Ends of Beam CrackB - Cracking at midspan

C - Instantaneous deflection

under service load

C - long time deflection under

service load

D and E - yielding of

reinforcement @ ends &midspan

Note: Stiffness (slope) decreases as cracking progresses


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For Continuous beams ACI 9.5.2.4 permits an average

of the Ie at the center and also ends

( ) ( ) ( )e1 e2e avg e mid ACI 9.5.2.4 0.50 0.25I I I I = + +

( ) ( ) ( )

++= e21emideavge 15.070.0

:continousends2

IIII( ) ( ) ( )

+= 1emideavge 15.085.0

:continousend1

III

( ) e e1 e e2 ee mid @ midspan, @ end 1, @ end 2I I I I I I= = =

ACI Com. 435 Weight Average

e

e

at the center, same for all the beam for simply supported

at the support, same for all the beam for catilever

I

I


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9.5.2.4 For continuous members,Ie shall be

permitted to be taken as the average of values

obtained from Eq. (9-8) for the critical positive andnegative moment sections. For prismatic members,Ie

shall be permitted to be taken as the value obtained

from Eq. (9-8) at midspan for simple and continuous

spans, and at support for cantilevers.

ACI 318 Section 9.5.2.4


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Cracked Transformed Section(used in ACI equation)

+

+

== s

s

i

ii 2nAyb

dnAy

yb

A

Ayy

Finding the centroid of singly Reinforced RectangularSection

2

2

2

2

02

2 20

You can get directly to hereif you write the first moment of area w.r.t assumed N

s s

A

s s

s s

yby nA y by nA d

by nA y nA d

nA nA d y yb b

+ = +

+ =

+ =

Solve for the quadratic for y


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Cracked Transformed Section(used in ACI equation)

022 ss2 =+

b

dnAy

b

nAy

Note:

c

s

E

En=

Singly Reinforced Rectangular Section

( )2s3

cr

3

1ydnAybI +=


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Cracked Transformed Section

( ) ( )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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Finding the neutral axis location of doubly

reinforced T-Section

( ) ( )

( ) ( ) 0212

2122

w

ss

2

we

w

sswe2

=++

+++

b

dnAAntbb

y

b

nAAnbbty


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Finding the moment of inertia for a doubly reinforced T-Section

( )

( ) ( ) ( )

23

beamflange

steel

3

cr e e w

2 2

s s

1 112 2 3

1

tI b y b t y b y t

n A y d nA d y

= + +

+ +


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Calculate the Deflections

(1) Instantaneous (immediate) deflections

(2) Sustained load deflection

Instantaneous Deflections

due to dead loads( unfactored) , live, etc.


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Equations for calculating inst for common cases


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Load Deflections

( ) ( ) ( )D+L (total) D+L inst D inst L inst

Calculated together due to non-linearity

= = +

For instantaneous dead load deflection use the Ma to find Ie and the proper

equation to get the deflection

For dead load and live load use the Ma (under both dead and live loads) to

find Ie and the proper equation to get the deflection for both

For instantaneous live load deflection, deduct the instantaneous dead load

deflection from above

( ) ( )L(inst) D+L inst D inst =

( )D Inst

3 3

cr cr

e g cra a

1

M M

I I IM M

= +

e

e

at the center, same for all the beam for simply supported

at the support, same for all the beam for catilever

I

I


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Sustained Load Deflections

Creep causes an increase

in concrete strainCurvature

increases

Compression steel

present

Increase in compressive

strains cause increase in

stress in compression

reinforcement (reducescreep strain in concrete)Helps limit thiseffect.


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Sustain load deflection =

Instantaneous deflection

(Equation 6.11 in Nilson Textbook)

(Eq. 9-11)1 50

= + ACI 9.5.2.5

bd

As=at midspan for simple and continuous

beams

at support for cantileverbeams

Additional to the instantaneous deflection-

caused by creep and shrinkage.

i


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Sustain load deflection = Instantaneous deflection

(Equation 6.12 in Nilson Textbook)1 50

= +

Where: 1.4 -10,000c

f

=

For concrete strengths 4000 psi and above (called High-

Strength Concrete, HSC), test results have shown that the

following equation is more realistic:.

i

0.4 1.0 (Equation 6.13 in Nilson Textbook)


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= time dependent factor for sustained load

5 years or more

12 months

6 months

3 months

1.4

1.2 1.0

2.0

Figure 9.5.2.5 from

ACI code

S stained Load Deflections


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For dead and live loads (Total Deflection, including instantaneous and long-

term)

( ) ( ) ( ) ( )total D inst L inst D L.T. L L.T.

Calculated together due to non-linearity

(L.T stands for Long Term, sometimes instead

of writing or you may see or

in

D L

some texts or references

DL L

.)

L

= + + +

DL (or D) and LL (or L) may have different factors for LT ( long term )

calculations

( )

Note that this is the total deflection that can happen AFTER we attach

the N/S (Non Structural) parts (excluding the instantanious dead load

after attachment of

N/S component

total D insttotal s

=

deflection happend at the very beginning!)


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For dead and live loads (TOTAL, which means instantaneous and long-term)

( )Total Long Term sustained portion of LLLT( ) D L D SL

D L

t

+

= + + +

( )sustained portion of LL

Long Term without instantaneous dead load

LT L D SLt = + +

For deflection afterthe initial dead load deflection (because that is whatto be considered in 3rdand 4thparts of ACI Table 9.5 (b):


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A note on Sustained Load Deflections

P t l t th d fl ti d t i d


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iw

The general process when a beam is under a

" " and subject to a " " can be

summarized as:

1- Calcul

sustained load

short term heavy load

instantaneo

w P

wate " ":us defl

2- Calc

ection due to

adu dla i nte tio

tw iw

w iw tw

ip i(w+p) iw

w ip

al long-term deflection by

Total deflection

instantaneous deflectionTotal deflection under sustained loa

" ":

3- :

d and sho

4-rt t

's :5- :

w

PPerm oa l d

=

= +

=

= +

Process to evaluate the deflection under sustained

and also short term load(Nilson Textbook)

Example 1:


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Example 1:

c

ss

c

E 57000 3000 3122018 psi=3122 ksi

E 29000E 29000 ksi 9

E 3122

n

= =

= = =

f 7.5 7.5 3000 410.79 psir cf= = =

E l 1(C t )


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Example 1(Cont.):3 3

cr cr e g cr

a a

(1 ACI 9-8)M M

I I IM M

= +

y

( ) ( )

2

2

3 2 4

cr

(12 ) 9 3 (17 ) 6 459 272

(Note: area of each bar=1.0 in ) 6.78 in.

1I 12 6.78 27 10.22 4067 in.

3

yy y y y

y

= =

=

= + =

1 klf, 0.7 klfD L

w w= =

Example 1(Cont ):


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Example 1(Cont.):3 3

cr cr

e g cr

a a

1 (ACI 9-8)M M

I I IM M

= +

ACI Equation 9-8

1.7 klfD Lw + =

Example 1(Cont ):


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Example 1(Cont.):

D SL

+

D SL

+

D SL +


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Example 1(Cont.):

3 years

0.467 2 0.245 1.8 0.07 1.083 in.

Total D L D SL

Total

+ = + +

= + + =

(this is the total deflection that can happen afterthe

initial dead load deflection)

Note that the total deflection, including all deflections is:

Homework


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Homework

Problems: 6.1, 6.2, 6.3 (Nilson Textbook)

Due: See class website

CE 544_Beam Deflection

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