# reinforced concrete serviceability

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Elastic Flexural Analysis for ServiceabilityLecture GoalsServiceabilityCrack widthMoments of inertiaTRANSCRIPT

Elastic Flexural Analysis for Serviceability

Lecture Goals

Serviceability

Crack width

Moments of inertia 1

Introduction

Ultimate Limit States Lead to collapse

Serviceability Limit States Disrupt use of Structures

but do not cause collapse

Recall:

Types of Serviceability Limit States

- Excessive crack width

- Excessive deflection

- Undesirable vibrations

- Fatigue (ULS) 2

Crack Width Control

Cracks are caused by tensile stresses due to loads

moments, shears, etc..

3

Crack Width Control

Cracks are caused by tensile stresses due to loads

moments, shears, etc..

4

Crack Width Control Bar crack development.

5

Crack Width Control

Temperature crack development

6

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?

7

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 )

8

Limits on Crack Width

0.4mm (0.016 in.) for interior exposure

0.33mm (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.

9

Limits on Crack Width

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

10

11

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

1 are generally visible

250l

12

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.)

13

Allowable Deflections

14

Allowable Deflections

15

Deflection Response of RC Beams (Flexure)

A- Ends of Beam Crack

B - Cracking at midspan

C - Instantaneous

deflection under service

load

C - long time deflection

under service load

D and E - yielding of

reinforcement @ ends &

midspan

12

2wlM

12

2wlM

2

16

wlM

16

Moment of Inertia

For normal weight

concrete

psi 33 c1.5

cc fE

For wc = 90 to 155 lb/ft3

psi 57000 cc fE

(ACI 8.5.1)

17

EIM

EI

M

slope

Moment Vs curvature plot

18

Moment Vs Slope Plot

The cracked beam starts to lose strength as the amount of

cracking increases

19

Variation of EI along the length of the beam

20

Moment of Inertia for Deflection Calculation

For (intermediate values of EI) gtecr III

Brandon

derived

Cracking Moment =

Moment of inertia of transformed cross-section

Modulus of rupture =

r g

t

f I

y

c7.5 f

Mcr =

Igt =

fr =

3 3

cr cre g cr

a a

1M M

I I IM M

21

Moment of Inertia for Deflection Calculation

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 3

cr cre g cr

a a

1M M

I I IM M

3

cre cr g cr

a

MI I I I

M

22

Deflection Response of RC Beams (Flexure)

For Continuous beams

ACI 9.5.2.4

ACI Com. 435

Weight Average

e21emideavge 25.050.0 IIII

e21emideavge 15.070.0

:continous ends 2

IIII

1emideavge 15.085.0

:continous end 1

III

2 end @

1 end @

midspan @

ee2

ee1

emide

II

II

II

23

Uncracked Transformed Section

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

Concrete 1 bwh bwh 0.5h 0.5bwh2

As n As (n-1)As d (n-1)Asd

As n As (n-1)As d (n-1)Asd

n A i iy n A

*

i i i

*

i i

y n Ay

n A

Note: (n-1) is to remove area of

concrete

24

Cracked Transformed Section

25

Cracked Transformed Section

s

s

i

ii 2

nAyb

dnAy

yb

A

Ayy

Finding the centroid of singly Reinforced Rectangular

Section

022

0

2

2

ss2

ss

2

ss

2

b

dnAy

b

nAy

dnAynAyb

dnAy

ybynAyb

Solve for the quadratic for y

26

Cracked Transformed Section

022 ss2 b

dnAy

b

nAy

Note:

c

s

E

En

Singly Reinforced Rectangular Section

2s3

cr

3

1ydnAybI

27

Example (Rectangular - Single)

For the following example find centroid and moment

of inertia for an uncracked and cracked section and

compare the results.

Es = 29000 ksi

Ec = 3625 ksi

d = 15.5 in b = 12 in. h = 18 in.

Use 4 # 7 bars for the steel.

28

Example

A #7 bar has an As = 0.6 in2

s

c

E 29000 ksi8

E 3625 ksin

2 2s 4 0.6 in 2.4 inA

The uncracked centroid is

2

s

s

2

2

2

3

2

12

1

12 in 18 in8 1 2.4 in 15.5 in

2 12 in 18 in 8 1 2.4 in

2204.4 in 9.47 in

232.8 in

bhn A d

ybh n A

29

Example The uncracked moment of inertia

232

s

3 22 2

4

112 2

12 in 18 in 18 in9.47 in 12 in 18 in 15.5 in 9.47 in 8 1 2.4 in

12 2

6491 in

bh hI y bh d y n A

The cracked centroid is defined by:

2

s 2.4 in 0.012912 in 15.5 in

A

bd

222 8 0.0129 2 8 0.0129 8 0.0129 0.3627

0.3627 15.5 in 5.62 in

yn n n

d

y

30

Example The cracked moment of inertia is

32

s

3 2 2

4

3

112 in 5.62 in 15.5 in 5.62 in 8 2.4 in

3

2584.2 in

byI d y nA

Notice that the centroid changes from 9.47 in. to 5.62

in. and the moment of inertia decreases from 6491 in4

to 2584 in4 . The cracked section loses more than half

of its strength.

31

Cracked Transformed Section

's s s s2 2 1 2 2 1 2 0n A nA n A d nA d

y yb b

Note:

c

s

E

En

Doubly Reinforced Rectangular Section

2s2

s

3

cr 1

3

1ydnAdyAnybI

32

Uncracked Transformed Section

steel

2

s

2

s

concrete

2

3

gt

11

212

1

dyAndyAn

hybhbhI

Note: 3g

12

1bhI

Moment of inertia (uncracked doubly reinforced beam)

33

Example (Rectangular - Double)

34

35

36

37

38

39

Cracked Transformed Section

Finding the centroid of doubly reinforced T-Section

0

212

2122

w

ss

2

we

w

sswe2

b

dnAAntbb

y

b

nAAnbbty

40

Cracked Transformed Section

Finding the moment of inertia for a doubly reinforced

T-Section

233

cr e e w

beamflange

2 2

s s

steel

1 1

12 2 3

1

tI b y b t y b y t

n A y d nA d y

41

Reinf. 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.

a) Determine Igt, Icr , Mcr(+), Mcr(-)

b) Determine the NA of the beam at ultimate Load.

42

Reinf. Concrete Sections - Example

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

43