introduction ultimate limit states lead to collapse serviceability limit states disrupt use of...

Introduction

Ultimate Limit States Lead to collapse

Serviceability Limit States Disrupt use of Structures but do not cause collapse

Recall:

IntroductionIntroduction

Types of Serviceability Limit States

- Excessive crack width

- Excessive deflection

- Undesirable vibrations

- Fatigue (ULS)

Crack Width ControlCrack Width Control

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


• Heat of hydration cracking

Crack Width ControlCrack Width ControlBar crack development.


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

• Leakage (Liquid-retaining structures)

• Corrosion (cracks can speed up occurrence of corrosion)

Reasons for 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 Limits on Crack WidthWidth

0.40 mm for interior exposure

0.33 mm for exterior exposure

max.. crack width =

ACI Code’s 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)

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 maximum value of s.

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

3. Web face reinforcement:

Deflection ControlDeflection 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)

visiblegenerally are *250

1l

Deflection ControlDeflection 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.)

Allowable Allowable DeflectionsDeflections

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

Allowable DeflectionsAllowable Deflections

• ACI Table 9.5(b) = max. permissible computed deflection

Deflection Response of RC Beams (Flexure)Deflection Response of RC Beams (Flexure)

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

12

2wlM

12

2wlM

24

2wlM

Deflection Response of RC Beams (Flexure)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

Note: Stiffness (slope) decreases as cracking progresses

Moment Vs curvature plotMoment Vs curvature plot

EIM

EI

M

slope

““Moment Vs Slope” PlotMoment Vs Slope” Plot

The cracked beam starts to lose strength as the amount of cracking increases

• To avoid complexity in calculations, an overall average effective moment of inertia

Moment of Inertia for Deflection CalculationMoment of Inertia 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 =

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

If Ma / Mcr < 1, no cracking is likely and Ie =Ig

Moment of Inertia Moment of Inertia for Deflection for Deflection CalculationCalculation

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

yt =

Ma =

3 3

cr cre g cr

a a

3

cre cr g cr

a

* 1 * ,or

, .9.8

M MI I I

M M

MI I I I Eq

M

Deflection Response of RC Beams (Flexure)

e21emideavge 15.070.0

:continous ends 2

IIII

e continuouse avg e mid

1 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

Definition of Ig

ACI 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 ).

Definition of I

Once a beam has been cracked by a large moment, it can never return to its original uncracked state; therefore, the effective moment of inertia Ie that should be used in deflection computations must always be equal to the effective moment of inertia associated with the maximum past moment to which the beam has been subjected. Often this moment is impossible to determine for most beams.

Uncracked Transformed SectionUncracked 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*h*h

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 Any **

*ii

*iii *

An

Anyy

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

Cracked Transformed SectionCracked Transformed Section

s

s

i

ii 2

nAyb

dnAy

yb

A

Ayy

Finding the centroid of singly Reinforced Rectangular Section

022

02

2

ss2

ss2

ss2

b

dnAy

b

nAy

dnAynAyb

dnAy

ybynAyb

Solve for the quadratic for y


022 ss2

b

dnAy

b

nAy

Note:

c

s

E

En

Singly Reinforced Rectangular Section

2s

3cr

3

1ydnAybI


'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

2s

2s

3cr 1

3

1ydnAdyAnybI

Uncracked Transformed SectionUncracked Transformed Section

steel

2s

2s

concrete

2

3gt

11

212

1

dyAndyAn

hybhbhI

Note:3

g

12

1bhI

Moment of inertia (uncracked doubly reinforced beam)

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.

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


Finding the centroid of doubly reinforced T-Section

w s s2

w

2 'w s s

w

2 2 1 2

2 1 2 0

t b b n A nAy y

b

b b t n A d nA d

b


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

e w

tI b y b t y b y t

n A y d nA d y

b b b

Calculate the DeflectionsCalculate the Deflections

(1) Instantaneous (immediate) deflections

(2) Sustained load deflection

Instantaneous Deflections

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

Calculate the Calculate the DeflectionsDeflectionsInstantaneous Deflections

Equations for calculating inst for common cases

Sustained Load DeflectionsSustained Load Deflections

Creep causes an increase in concrete strain

Curvature increases

Compression steel present

Increase in compressive strains cause increase in stress in compression reinforcement (reduces creep strain in concrete)

Helps limit this effect.


Sustained load deflection = i

Instantaneous deflection

501 ACI 9.5.2.5

bd

As at midspan for simple and continuous beams

at support for cantilever beams


= time dependent factor for sustained load

5 years or more 12 months 6 months 3 months

1.4 1.2 1.0

2.0

Also see Figure 9.5.2.5 from ACI code

The total long time deflection

where

δL = immediate live load deflection

δD = immediate dead load deflection

δSL = sustained live load deflection (a percentage of the immediate δL determined by expected duration of sustained load)

λ = time dependant multiplier for infinite duration of sustained load (assumed to occur after partitions are installed)

λt0, = time dependant multiplier for infinite duration minus that at the time t0 when partitions are installed

To calculate δL (or δSL) due to the live loads, the following procedure has been found to be generally satisfactory:

0,LT L t D SL = + +

Calculation of long time deflection

1. Calculate the deflection δD+L due to dead and live loads acting simultaneously. For this calculation Ie is found using Eq. 9.8 and the moment Ma is the one produced when both dead and live loads are acting simultaneously.

2. Calculate the deflection δD due to the dead load acting alone. For this calculation Ie is found using Eq. 9.8 and the moment Ma is the one produced when the dead load acts alone.

3. Subtract the deflection δD from the deflection δD+L to obtain the desired deflection δL.

Handling long term deflections

Surveys of partition damage have shown that damage to brittle partitions can occur with deflections as small as L/1000. A frequent limit on deflections that cause damage is L/480 after attachment of nonstructural elements. The value L/480 shall be compared with the value of the total long time deflection.

If the long time deflections exceeds the value

permitted, the designer 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.

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.•

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?

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.

Problem 9-10

The 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 the live load is sustained.

•

Problem 9.10

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