servicability limits

7/27/2019 servicability limits

1/51

Serviceability


2/51

Introduction

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

but do not cause collapse

Recall:


3/51

Introduction

Types of Serviceability Limit States

- Excessive crack width

- Excessive deflection

- Undesirable vibrations

- Fatigue (ULS)


4/51


5/51

Crack Width Control

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


6/51

Crack Width Control

Heat of hydration cracking


7/51

Crack Width Control

Bar crack development.


8/51

Crack Width Control

Appearance (smooth surface > 0.25 to 0.33mm= public concern)

Leakage (Liquid-retaining structures)

Corrosion (cracks can speed up occurrence ofcorrosion)

Reasons for crack width control?


9/51

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 )


10/51

L imits on Crack

Width

0.40 mm for interior exposure0.33 mm for exterior exposure

max.. crack width =

ACI Codes Basis Prior to 1999

Now ACI handles crack width

indirectly by limiting the bar spacings and bar cover for beams and

one way slabs ACI 10.6.4.

Bar spacings must also satisfy ACI 7.6.5 (3t or 450mm)


11/51

Example 1 (9-4)

A 20cm thick slab has 12mm diameter bars. The

bars have 420MPa yield stress and a minimum

clear cover of 20mm. Compute the maximumvalue of s.


12/51

Other important issues for crack control

1. Negative moment regions of T-beams.

2. Shrinkage and temperature reinforcement: is intended to

replace the tensile stresses in the concrete

at the time of cracking, using the following

simplified analysis:

For grade 60 steel and 28MPa concrete,

Steel ratio is between 0.004 and 0.005.

This limit is about three times that specified by ACI code

7.12.2.1 which is based on empirical results.

s y g t

s t

g y

A f A f

A f

A f


13/51

3. Web face reinforcement:


14/51

Deflection Control

Visual Appearance

( 7.5m. span 30mm )

Damage to Non-structural Elements

- cracking of partitions

- malfunction of doors /windows

(1.)

(2.)

Reasons to Limit Deflection (Table 9-3)

visiblegenerallyare*

250

1l


15/51

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.)


16/51

Allowable

Deflections

ACI Table 9.5(a) = min.

thickness unless s arecomputed


17/51

Allowable Deflections

ACI Table 9.5(b) = max. permissible

computed deflection


18/51

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

2wlM


19/51

Deflection Response of RC Beams (F lexure)

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


20/51

Moment Vs curvature plot

EIM

EI

M

slope


21/51

Moment Vs Slope Plot

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


22/51

To avoid complexity in calculations, an overall

average effective moment of inertia


23/51

Moment of I nertia for Deflection Calculation

For (intermediate values of EI)gecr III

Branson

derived cr

3

a

crg

3

a

cre *1* I

M

MI

M

MI

Cracking Moment =

Gross moment of inertia of rc cross-section

Modulus of rupture =

t

gr

y

If

c0.62 f

Mcr=

Ig =

fr =

IfMa /Mcr> 3, the cracking will be extensive,Ie =Icr

IfMa /Mcr< 1, no cracking is likely andIe =Ig

f


24/51

Moment of Inertia

for Deflection

Calculation

Distance from centroid to extreme tension fiber

maximum moment in member at loading stage forwhich Ie ( ) is being computed or at any previousloading stage

yt =

Ma =

3 3

cr cr

e g cra a

3

cre cr g cr

a

* 1 * ,or

, .9.8

M M

I I IM M

MI I I I Eq

M


25/51

Deflection Response of RC Beams (Flexure)

e21emideavge 15.070.0

:continousends2

IIII

e continuouse avg e mid1 end continous:

0.85 0.15I I I

e ei ee mid @ midspan, @ end iI I I I

ACI Com. 435

Weight Average

ACI code


26/51

Definition ofIgACI code: Ig is the moment of inertia of the gross concrete section

neglecting area of tension steel.

Ig might be more accurate if it includes the transformed area of the

reinforcement.

Ig is the moment of inertia of the uncracked transformed section. The

transformed section consists of the concrete area plus the transformed

steel area(=the actual steel area times the modular ratio n =Es /Ec :

Es = 200GPa ,Ec =4700fc ).


27/51

Definition ofIOnce a beam has been cracked by a large moment, it can never

return to its original uncracked state; therefore, the effective

moment of inertiaIe that should be used in deflection

computations must always be equal to the effective moment ofinertia associated with the maximum past moment to which the

beam has been subjected. Often this moment is impossible to

determine for most beams.


28/51


29/51

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 fory


30/51


022 ss2

b

dnAy

b

nAy

Note:

c

s

E

En

Singly Reinforced Rectangular Section

2s3

cr

3

1ydnAybI


31/51


0

212212 ssss2

b

dnAAny

b

nAAny

Note:

c

s

E

En

Doubly Reinforced Rectangular Section

2s2

s

3

cr 1

3

1ydnAdyAnybI


32/51

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)


33/51

Example 2 (9-1)

For the shown beam of 28MPa concrete, Find:

1. Moment of inertia of uncracked section.2. Moment of inertia of cracked section.


34/51

Example 3

(9-2)

For the shown beam of 31.5MPa concrete, Find steel stress at

service loads if the service live-load moment is 70kN.m and

the service dead load moment is 96kN.m


35/51


Finding the centroid of doubly reinforced T-Section

0

212

2122

w

ss

2

we

w

sswe2

b

dnAAntbb

y

b

nAAnbbty


36/51


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


37/51

Calculate the Deflections

(1) Instantaneous (immediate) deflections

(2) Sustained load deflection

Instantaneous Deflections

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

Calculate the


38/51

Calculate the

Deflections

Instantaneous

Deflections

Equations for

calculating Dinst forcommon cases


39/51


40/51

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.


41/51


Sustained load deflection = l Di

Instantaneous deflection

l 501 ACI 9.5.2.5

bd

As

at midspan for simple and continuous beams

at support for cantilever beams


42/51


= time dependent factor for sustained load

5 years or more

12 months

6 months

3 months

1.41.21.0

2.0

Also see Figure

9.5.2.5 from

ACI code


43/51


44/51

The total long time deflection

where

L = immediate live load deflection

D

= immediate dead load deflection

SL = sustained live load deflection (a percentage of the immediate

L determined by expected duration of sustained load)

= time dependant multiplier for infinite duration of sustained

load

t = time dependant multiplier for limited load duration

To calculate L (orSL) due to the live loads, the following procedure

has been found to be generally satisfactory:

LT L D t SL= + + l l


45/51

Calculation of long time deflection

1. Calculate the deflection D+L due to dead and live loads acting

simultaneously. For this calculationIe is found using Eq. 9.8 andthe momentMa is the one produced when both dead and live

loads are acting simultaneously.

2. Calculate the deflection D due to the dead load acting alone. For

this calculationIe is found using Eq. 9.8 and the momentMa isthe one produced when the dead load acts alone.

3. Subtract the deflection D from the deflection D+L to obtain the

desired deflection L.

If the long time deflections exceeds the value permitted, thedesigner may either increase the depth of members, or add

additional compression steel. If the sag produced by the long time

deflections is objectionable from an architectural or functional point

of view, forms may be raised (cambered) a distance equal to that of

the anticipated deflection.


46/51

Example

4 (9-5)

The T-beam shown in Fig. is made of 28MPa concrete and

supports unfactored dead and live loads of 13kN/m and

18kN/m. Compute the immediate midspan deflection. Assume

that the construction loads did not exceed the dead load.


47/51

Example 5 (9-5)If the beam in the previous example is assumed to support

partitions that would be damaged by excessive deflections. If 25%

of the live load is sustained. The partitions are installed at least 3

months after the shoring is removed. Will the computed

deflections exceed the allowable in the end span?


48/51

Problem 1 (9-8 + 9-9)9-8 A simply supported beam with the cross section shown in Figure

next page has a span of 7.5m and supports an unfactored dead load

of 22.5kN/m, including its own self-weight plus an unfactored live

load of 22.5kN/m. The concrete strength is 31.5MPa. Compute

1. the immediate dead load deflection.

2. the immediate dead-plus-live load deflection

3. the deflection occurring after partitions are installed. Assume

that the partitions are installed two months after shoring for the

beam is removed and assume that 20 percent of the live load is

sustained.

9-9 Repeat Problem 9-8 for a beam having the same dimensions and

tension reinforcement, but with two No. 25mm bars as compression

reinforcement.


49/51


50/51

Problem 9-10The beam shown in Figure next page is made of 28MPa

concrete and supports unfactored dead and live loads of

15kN/m and 17kN/m respectively. Compute

(a) the immediate dead-load deflection.

(b) the immediate dead-plus-live load deflection.

(c)the deflection occurring after partitions are installed.

Assume that the partitions are installed four months after

the shoring is removed and assumed that 10 percent of thelive load is sustained.


51/51

Problem 9.10

servicability limits

Documents

Transcript of servicability limits