lecture31 actual

8/2/2019 Lecture31 Actual

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Lecture 31 - Beam Deflection

April 5, 2001

CVEN 444


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Lecture Goals

Serviceability

Moments and centroids


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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 roofs

Damage 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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AllowableDeflectionsFlat Roofs ( no damageable nonstructural elements

supported)

180

instLL

l


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AllowableDeflectionsFloors ( no damageable nonstructural elements

supported )

180

instLLl


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

480

l


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Roof or Floor 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 Inertia 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 previousloading 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 Inertia

psi33 c1.5

cc fE

For wc = 90 to 155 lb/ft3

psi57000 cc fE

For normal weight concrete

(ACI 8.5.1)


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Deflection Response of RC Beams (Flexure)

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

12

2wlM

12

2

wlM

24

2wlM


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For Continuous beams

ACI 9.5.2.4

ACI Com. 435

Weight Average

e21emideavge 25.050.0 IIII

e21emideavge 15.070.0

:continousends2

IIII

1emideavge 15.085.0

:continousend1

III

2end@

1end@

midspan@

ee2

ee1

emide

II

II

II


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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

AAyy

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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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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Stiffness of Reinforced Concrete Sections

- ExampleGiven 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 materialproperties 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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