deflectionconcrete-nov2010[1]

Deflection Calculation of FloorsImmediate; Long-Term; Cracking ; g ; g

D Bij O A l iDr Bijan O AalamiProfessor Emeritus,

San Francisco State University

Principal, ADAPT Corporation

www.adaptsoft.comwww.adaptsoft.com

DEFLECTION CALCULATION OF CONCRETE FLOORS

TOPICS OF PRESENTATION

Reasons for controlling deflectionsDeflection control through span/depth ratiosPermissible deflection valuesPermissible deflection values How to measure code-intended valuesHow to calculate probable deflectionswith or without crackingwith or without cracking

Instantaneous deflectionsLong-term deflections

Ultimate valuesInterim values

Crack width calculation and control

Numerical examples

References


REASONS FOR CONTROLLING DEFLECTION

Aesthetics and Sense of ComfortAesthetics and Sense of ComfortImpression of inadequate in strengthPerception of out-of-level when walking

Limit Damage to Non structural ConstructionLimit Damage to Non-structural ConstructionCracks in plaster or glass

Impairing Design FunctionP di f tPonding of waterMalfunction of doorsbreach of seals

Important note on deflectionsWhether or not an observed deflection is permissible should be viewed in conjunction with the following factors

Consequence of the observed deflectionTime of observation (due to change of deflectionwith times)


REASONS FOR CONTROLLING DEFLECTION

Aesthetics and Sense of ComfortAesthetics and Sense of ComfortImpression of inadequate in strengthPerception of out-of-level when walking

Limit Damage to Non structural ConstructionLimit Damage to Non-structural ConstructionCracks in plaster or glass

Impairing Design FunctionP di f tPonding of waterMalfunction of doorsbreach of seals Crack

Crack in plaster due to excessive C ac p as e due o e cess edeflection


REASONS FOR CONTROLLING DEFLECTIONDamage; impairing Design Function

Crack in glass pane

(a)

BreachBreach of seal at joint

(b)


DEFLECTION CONTROL THROUGHLIMITATIONS ON SPAN/DEPTH RATIOS (L/h)LIMITATIONS ON SPAN/DEPTH RATIOS (L/h)

The major recommendations are for:

One-Way Conventionally Reinforced Slabs andBeams

Two-Way Conventionally Reinforced Slabs andy yBeams

Post-Tensioned Members

DEFLECTION CONTROL THROUGH LIMITATIONS ON SPAN/DEPTH RATIOS

One-Way Conventionally Reinforced Slabs and Beams

MINIMUM THICKNESS OF CONVENTIONALLYREINFORCED BEAMS OR ONE-WAY SLABS

Member Simplysupported

One endcontinuous

Both endscontinuous

Cantilever

Solid one-way slabs L/20 L/24 L/28 L/10

Beams orribbed One-way slabs

L/16 L/18.5 L/21 L/8


Two-Way Conventionally Reinforced Slabs and Beams

MINIMUM THICKNESS OF SLABS WITHOUT INTERIOR BEAMS

Without drop panels*** With drop panels***

fy psi**

Exterior panels Interior panels

Exterior panels Interior panels

Withoutedge

Withedge

#

Withoutedge

Withedge

#beams beams# beams beams#

40,000 Ln/33 Ln/36 Ln/36 Ln/36 Ln/40 Ln/40

60,000 Ln/30 Ln/33 Ln/33 Ln/33 Ln/36 Ln/3675 000 Ln/28 Ln/31 Ln/31 Ln/31 Ln/34 Ln/3475,000 Ln/28 Ln/31 Ln/31 Ln/31 Ln/34 Ln/34

* Ln = clear span in long direction ** For fy between the values interpolate For fy between the values interpolate*** Drop panels extension not less than span/6, less than

slab thickness/4 # Edge beam stiffness, if any, to have minimum specified

stiffness


Post-Tensioned Members

RECOMMENDED SPAN TO DEPTH RATIOS FORPOST-TENSIONED MEMBERS

ContinuousSpans

Simple Spans

Roof Floor Roof FloorOne-way solid slabs 50 45 45 40Two-way solid slabs (supported on columns only)

45-48 40-45

Two-way waffle slabs (1m 40 35 35 30pans)Beams 35 30 30 26One-way joists 42 38 38 35

Note: The above ratios may be increased if calculations verify that deflection, camber, and vibrations are not objectionable.

LIMITS FOR ACCEPTABLE DEFLECTION

Aesthetics and Sense of Comfort

Limit vertical out-of-level to span ratio of 1/240,and for cantilevers in excess of 1/120. This appliesto observation at the time and location, where out-of-level displacement in notedof-level displacement in noted

Example of deflection exceeding sense of comfortp g

LIMITS FOR ACCEPTABLE DEFLECTION

DEFLECTION LIMITS FOR AESTHETICS AND SENSE OF COMFORT

The out-of-level condition of a floor system can be controlled through camber at time of construction, once the long-term deflection of a floor is estimated.

Reduced thickness

Camber Camber

Elevation

Excessive camber can result in unacceptable reduction in thickness, since the top surface of most slabs are made level at construction time.

Camber at Construction

PERMISSIBLE DEFLECTIONS

Important note on permissible deflections

Whether or not an observed deflection is permissible should beWhether or not an observed deflection is permissible should be viewed in conjunction with the following factors

Consequence of the observed deflectionTime of observation (due to change of deflectionTime of observation (due to change of deflectionwith times)

IBC-09 and AC- 318 express the permissible deflection inIBC 09 and AC 318 express the permissible deflection in terms of “span/vertical offset.”

Several other non-USA codes, in addition to a limit on span/vertical offset stipulate a limit on maximum vertical p poffset, irrespective of span.

MAXIMUM PERMISSIBLE COMPUTED DEFLECTIONS (ACI TABLE 9.59(b))

Type of member Deflection to be considered

Deflection limitation

Flat roofs not supporting or attached to nonstructural elements likely to be damaged by large deflection

Immediate deflection due to live load L/180 *

large deflectionFloors not supporting or attached to nonstructural elements likely to be damaged by

Immediate deflection due to live load L/360

likely to be damaged by large deflection

Roof or floor construction supporting

That part of the total deflection occurringconstruction supporting

or attached to nonstructural elements likely to be damaged by large deflection

deflection occurring after attachment of nonstructural elements (sum of the long-time deflection due to all sustained

L/480 **

due to all sustained loads and the immediate deflection due to any additional live load)****

Roof or floor construction supporting or attached to nonstructural elements not likely to be damaged

L/240 ***not likely to be damaged by large deflection


Notes: * Limit not intended to safeguard against ponding.

Ponding should be checked by suitable calculationsf d fl ti i l di dd d d fl ti d tof deflection, including added deflections due to

ponding of water, and considering long-term effects of all sustained loads, camber, constructiontolerances, and reliability of provisions for drainage.

** Limit may be exceeded if adequate measures aretaken to prevent damage to supported or attached elements.

*** But not greater than tolerance provided fornonstructural elements. Limit may be exceeded if camber is provided so that total deflection minuscamber is provided so that total deflection minuscamber does not exceed limit.

**** Long-time deflection shall be determined usingt bli h d d b t b d d bestablished procedures, but may be reduced by

amount of deflection calculated to occur beforeattachment of nonstructural elements. This amountshall be determined on basis of acceptedengineering data relating to time- deflectioncharacteristics of members similar to those beingconsidered.


Exercise engineering judgment when deciding on permissible values

For visual and comfort L/240

Immediate live load , where damage not likelyL/360L/360

Increment of long-term deflection, wheredamage due to increment is likely L/480g y

MEASUREMENT OF DEFLECTION VALUES

MEASUREMENT OF CODE-INTENDED DEFLECTION VALUES

Code-intended values are determined using the notional “deflection/span; d/L” ratio,

Notional span (L) and the vertical offset (d) arenot well defined. A consistent understanding of their values, however, is becoming important, when theapplication of the Finite Element Method is leadingapplication of the Finite Element Method is leading to a greater automation of code check procedures.

The following diagrams help to illustrate the application of the notional values (L) and (d)

Deflection “d” is the maximum offset from the linejoining two adjacent control points on the surfacej g j p

Traditional scheme based on traditional methodsof structural analysisGeneralized scheme, based on automated methods of structural analysis (Finite Elementy (Method (FEM)

MEASUREMENT OF DEFLECTIONVALUES

MEASUREMENT OF DEFLECTION VALUES

GENERALIZED MEASUREMENT OF DEFLECTIONVALUESVALUES

Deflection “d” is the maximum offset from the linejoining two adjacent crests on the surface


GENERALIZED MEASUREMENT OF DEFLECTIONVALUESVALUES

For perimeter regions the notional span (L) isdistance between any interior point of zero slope toany point along the edge of the floor


Example of a cantilever and different notional “L” span optionsp p


EXAMPLE OF A SLAB WITH CANTILEVER

It i i d t d t i th L/d ti f dIt is required to determine the L/d ratio for codecompliance

Plan of column-supported slab showing the cantilever under consideration and distances of

interest in m



D fl t d t f th l bDeflected contours of the slab



It i i d t d t i th L/d ti f dIt is required to determine the L/d ratio for codecompliance

Plan of column-supported slab showing the cantilever under consideration and distances of

interest in m


L/d = 11360/13 = 873

For point along column line: L/d = 3990/19 = 210For point with max deflection: largest L/d = 7259/23 = 315


ESTIMATE OF PROBABLE DEFLECTIONS

Concrete slabs deflect instantaneously due to applicationConcrete slabs deflect instantaneously, due to application of loads, and continue to deflect under sustained load. Hence, estimate of the probable deflection of slabs consists of computation of (i) instantaneous, and (ii) long-term deflectionsterm deflections.

The following presentation is broken down into:

Methods of estimating instantaneous deflections andMethods of estimating instantaneous deflections andexamples: Estimate of long-term deflections and examples

ESTIMATE OF INSTANTANEOUS DEFLECTION

Instantaneous deflection of a concrete slab depends on:The dimensional geometry and support conditions of the slab; Age of the concrete at loading. With time, concretebecomes stifferCracking at higher values of load


Instantaneous DeflectionsThere are several methods commonly used to calculate the instantaneous deflection of a concrete slab Thethe instantaneous deflection of a concrete slab. The most common methods are:

Closed form formulas or tables, available primarilyfor uncracked sections (Method 1)for uncracked sections (Method 1)

Strip method and assumption of linear elastic response with no cracking (Method 2)

Use of equivalent moment of inertia (Ie) for crackingconditions combined with simplified averaging (ACI-318’s simplified procedure) (Method 3);

Use of equivalent moment of inertia (Ie) combinedwith numerical integration - ACI-318’s extendedprocedure, (Method 4)procedure, (Method 4)

Use of Finite Element concrete floor programswithout allowance for cracking (Method 5)

Use of Finite Element Method with allowance forcracking (Method 6)

DEFLECTION CALCULATIONS

Closed Form FormulasSolution are available in literature on plate behavior

DEFLECTION COEFFICIENTS k

a

γ b

DEFLECTION COEFFICIENTS k

xxγ

1 0.0457 0.0143 0.0653 0.0491

1 1 0 0373 0 0116 0 0548 0 0446

1 2 3 4

1.1 0.0373 0.0116 0.0548 0.0446

1.2 0.0306 0.0094 0.0481 0.0422

1.3 0.0251 0.0075 0.0436 0.0403

1.4 0.0206 0.0061 0.0403 0.0387

1.5 0.0171 0.0049 0.0379 0.0369

2.0 0.0071 0.0018 0.0328 0.0326

Poisson’s ratio conservatively assumed 0.25γ = a/b (aspect ratio)

1 = rigid supports; rotationally free;2 = rigid supports; rotationally fixed;


3 = central panel from an array of identical panels supported on columns; deflection at center; and

4 = similar to case 3 but deflection at center of long4 = similar to case 3, but deflection at center of longspan at support line

w = k (a4*q / E*h3 )

Where,w = deflection normal to slab;a = span along X-direction;E = Modulus of elasticity; h = slab thickness; andq = intensity of load.

DEFLECTION CALCULATION EXAMPLE

VIEW OF TYPICAL LEVEL OF A MULTI-STORYFLOOR SYSTEM SELECTED FOR STUDY


EXAMPLE : METHOD 1Using closed form formulas determine the deflection at center of panel identified below for combination of

TNO_123DL + LL

Slab thickness8'' (200mm)

(a) Floor plan Column18'' x 24''(460mm x 610mm)

26'-3''(8.00m)

30'-0'' (9.14m)

(b) Panel plan

IDENTIFICATION OF PANEL FOR DEFLECTION CALCULATION


Given:

Span length along X-X direction = 30’ (9.14 m)Span length along X X direction 30 (9.14 m)Span length along Y-Y direction = 26.25’ (8.00 m)Slab thickness = 8 in. (203 mm)Ec (modulus of elasticity) = 4.287 * 106 psi

(29 558 MPa)(29,558 MPa)

Superimposed dead load = 25 psf (1.2 kN/m2)(includes allowance for partitions)

Live load (LL) = 40 psf (1.9 kN/m2)

Required

Deflection of the panel at midspan for the following load combination

1*DL + 1*LL


w = k (a4*q / E*h3 )

w = deflection normal to slab;a = span along X-direction;a = span along X-direction;E = Modulus of elasticity; h = slab thickness; andq = intensity of load.

Aspect ratio γ = 30/26.25 = 1.14

Total service load q = [(25 + 40) + 150*8/12]/144= 1.146 lb/in2

(a4*q / E*h3 ) = [(30*12)4 *1.146/(4.287*106* 83)= 8.77

For mid-panel deflection, consider case 3 from Table k = 0.0548Deflection, Δ = k (a4*q / E*h3 )= 0.0548*8.77

= 0 48 in (12 mm)= 0.48 in (12 mm)

For deflection at midpoint of column lines in X-direction, from Table 5

k 0 0446k = 0.0446Deflection, Δ = k (a4*q / E*h3 )= 0.0446*8.77

= 0.39 in (10 mm)


Method 2 :Strip method and assumption of linear elastic response

The structure is subdivided into design strips in twoorthogonal directions

Each strip is extracted and analyzed in isolationFor center of panel, the values obtained from the two

l ti dd d t thsolutions are added together

(a)

(b)(b)


Support

1

1d

d1

2

Deflection is 1-2 direction

Deflection is 2-2 direction

d2

2

(a) Total Deflection d = d1 + d2

(b) View of the design strip along Y-Y direction(b) View of the design strip along Y Y directionextracted from the floor system


EXAMPLE : METHOD 2Using strip method, linear elastic response andno allowance for cracking, determine the deflection at g,center of panel identified for load combination (DL + LL)

TNO 123

Design parameters same as previous example

TNO_123


(a) Floor plan Column(a) Floor plan18'' x 24''(460mm x 610mm)

26' 3''26 -3(8.00m)

30' 0'' (9 14m)30 -0 (9.14m)

(b) Panel plan


View of the design strip along Y-Y directionextracted from the floor system

Deflection at center of span 1 from solution of ADAPT-RC is0.231 Inch

For the center of panel the preceding value must be addedFor the center of panel, the preceding value must be addedto deflection to be calculated from the strip in the orthogonaldirection.

Since the panels are roughly square, for expediency theSince the panels are roughly square, for expediency the above value is multiplied by 2, assuming that the deflectionfrom the orthogonal direction is essentially the same

Hence deflection at center of panel:p

0.231 * 2 = 0.462 inch


Brief Review of Cracking, Equivalent Moment of Inertia (Ie) and Its Computation

Neutral axis

Reinforcement

(a) Cracked beam elevation

(b) Applied moment

Cracking moment

(c) Effective moment of inertia Ie

le

Illustration of Effective Second Moment of Area (Ie) ina Partially Cracked Slab

( )


Definition and Evaluation of Equivalent Moment of Inertia (Ie) (ACI-318)

Ie = (Mcr / Ma)3 * Ig + [1-(Mcr / Ma)3] * Icr ≤ Ig (1)

WhWhere, Ig = Gross moment of inertia;Icr = Moment of inertia of cracked section;Ie = Effective moment of inertia;Ma = Maximum moment in member at

stage deflection is computed; and,Mcr = Cracking moment.

Ma, is calculated using elastic theory and grossmoment of inertia for the uncracked section

The change in distribution of moment from crackingThe change in distribution of moment from crackingis assumed to be small, and already accountedfor in the empirical formula (1)


Cracking Moment Mcr

Mcr = fr * Ig /yt (2)Where,

fr = Modulus of rupture, flexural stressi ki It i i bcausing cracking. It is given by:

fr = 7.5 f’c1/2 (3)

yt = distance of section centroid to farthesttension fiber

For all-lightweight concrete, fr is modified as follows:g g ,

fr = 0.75 * 7.5 f’c1/2 (4)


Icr depends on:Calculation of Cracked Moment of Inertia Icr

Geometry of sectionAmount of reinforcement, including prestressingLocation of reinforcement

For singly reinforced rectangular sections:

c = kd

Icr = b(kd)3 / 3 + nAs(d-kd)2

c c

C

b

dc c

s

s

cs

s

C

T

(a) Section (b) Strains (c) Forces

Post-cracking Stresses and Forces


dc c

c

s c

c

sC

b

(a) Section (b) Strains (c) Forces

ssT

contd… Cracking moment of inertia Icr

Post-cracking Stresses and Forces

where,kd = [(2dB + 1)0.5 – 1] / B (6)

d = distance of compression fiber to centerof tension reinforcement

B = b/(nAs) n = Es/Ecn Es/EcEs = modulus of elasticity of SteelEc = modulus of elasticity of concrete

For more complex geometry and reinforcement theFor more complex geometry and reinforcement the relationship will be more involved


METHOD 3Simplified Allowance for Cracking throughaveraging of Equivalent Moment of Inertia (Ie)(ACI-318)

Steps to follow:

Uncracked analysis and design

Steps to follow:

Break the floor slab into design stripsSelect design strips of interestUsing gross moment of inertia obtain elasticuncracked distribution of moment for selected stripsp(applied moment Ma)Design the design strips and determine the amountand location of reinforcement, including any basereinforcement (mesh rebar)reinforcement (mesh rebar)

contd…


METHOD 3

contd…

Simplified Allowance for Cracking throughAveraging of Equivalent Moment of Inertia (Ie) (ACI-318)

Calculate the cracking moments at face-of-supportsand mid-spans (Mcr)

Cracked analysis and design

For locations, where Mcr > Ma, calculate the equivalent moment of inertia (Ie)For each span/cantilever, determine the average Ie,avFor each calculated average Ie.av determine theFor each calculated average Ie.av determine the equivalent uncracked slab thickness (he). Using the calculated equivalent thickness of each span

(he), re-analyze each design strip to determine the resulting deflectionresulting deflection


METHOD 3Simplified Allowance for Cracking through

f f ( )

Ie av = average moment of inertia

Averaging of Equivalent Moment of Inertia (Ie) (ACI-318)

Ie, av = average moment of inertia

For continuous members an average equivalentmoment of inertia is obtained for each span.

Ie, av = 0.5 [ (Ie,left support + Ie,right support )/2 + Ie, midspan ]

For cantilevers, moment of inertia over the support isppused for the entire cantilever

P

ElevationElevation


EXAMPLE : METHOD 3Simplified Allowance for Cracking throughAveraging of Equivalent Moment of Inertia (Ie) (ACI-318)g g q ( ) ( )

Calculate the cracked deflection at the middle of the panel identified for the load combination (DL + LL). Design parameters same as previous examples. f’c = 5000 psiparameters same as previous examples. f c 5000 psi

TNO_123


(a) Floor plan Column18'' x 24''(460mm x 610mm)

26'-3''(8.00m)

30'-0'' (9.14m)

(b) Panel plan


contd…EXAMPLE : METHOD 3 (Averaged Ie)

(a) Breakdown of slab into design strips

(b) Extraction of design strip for analysisin isolation



DISTRIBUTION OF MOMENT (Ma ) DUE TODEAD PLUS LIVE LOAD (ADAPT RC)DEAD PLUS LIVE LOAD (ADAPT-RC)

Consider moment at right support of span 1

Ma = 326.4 k-ft (442.53 kNm)

Calculate cracking moment Mcr and compare with Mcompare with Ma


Contd…EXAMPLE : METHOD 3 (Averaged Ie)

fr = 7.5√ f’c = 7.5 √ 5000 = 530.33 psiIg = 17 019 in4 (6 40e+10 mm4)Ig = 17,019 in4 (6.40e+10 mm4)

Mcr = fr * Ig /yt = 530.33*17,019/(4*12000) = 188.04 k-ft (254.94 kNm) < Ma

Ie = (Mcr / Ma)3 * Ig + [1-(Mcr / Ma)3] * Icr ≤ Ig

where,,Icr = b*(kd)3 / 3 + n*As(d - kd)2

kd = [(2dB + 1)0.5 – 1]/Bd = 6 81 ind 6.81 inB = b/(nAs)n = Es/Ec = 30000/4287 = 7.0As = area of steel at face of support =10.12 in2

B = 360/(7 0* 10 12) = 5 08 /inB = 360/(7.0* 10.12) = 5.08 /inKd = [(2*6.81*5.08 + 1)0.5 – 1] / 5.08 = 1.45 inIcr = (360*1.453)/3 + 7.0*10.12* (6.81-1.45)2

= 2401 in4

Ie = (188.04/326.4)3*17019+[1-(188.04/ 326.4)3] 2401 = 5196 in4 ( = 0.31 Ig )



Using the same procedure , the value of Ie at other locations required by the code formula are calculated and listed below:

Left cantilever:Ie at face of support = Ig = 1.536e+4 in4

First Span:First Span:Ie at left support centerline = 1.70e+4 in4

Ie at midspan = 1.44e+4 in4

Ie at right support centerline = 5.20e+3 in4

Second Span:Ie at left support centerline = 5.63e+3 in4

Ie at midspan = 1.536e+4 in4

Ie at right support centerline = 1.702e+4 in4

Right cantilever:Ie at face of support = Ig = 1.536e+4 in4e pp g



Using ACI-318 averaging procedure

Left and right cantilevers Ie = Ig

First spanAverage Ie = [(1.70*104+5.20*103)/2 +1.44*104]/2

= 12.755e+3 in4

Second spanAverage Ie = [(5.63*103+1.70*104)/2 +1.536*104]/2

= 13.343e+3 in4



The calculated equivalent moments of inertia are used to determine an equivalent thickness (he) for each of the spans. The equivalent thickness is given byby

Ie = b*he3 /12

Wh b i th idth f th t ib t f th d iWhere, b is the width of the tributary of the design strip.

Left cantilever: he = 8 in ; no reductionFirst span he = 7.52 inSecond span he = 7.63 inRight cantilever he = 8 in. ; no reduction

The same computer program used for determination of Ma can calculate the deflections with the reduced slab thickness



DEFLECTED SHAPE WITH ALLOWANCE FORCRACKING, USING SIMPLIFIED METHOD.

Maximum deflection for span 1 is 0.264 in (6.7 mm)allowing for cracking

The strip method, used, provides one deflection valuethat is representative of both the midpoint of the paneland midpoint of the line of supportSince the geometry of the panel in two orthogonal

directions is somewhat similar, the center of panel , pdeflection is estimated as:

2 * . 0.264 = 0.528 in (13.4 mm)


METHOD 4Equivalent moment of inertia (Ie) combined with numerical integration

(i) Start by following the same steps as the simplified procedure, design and detail (rebar) for

Steps to follow:

p p , g ( )each design strip

(ii) Subdivide each span into a number of segments(10 to 20 division), and determine the equivalentmoment of inertia of each of the subdivisionsmoment of inertia of each of the subdivisions

(iii) Use the slope-deflection method of structural analysis, with each segment assigned its own Ieand calculate the deflection

hh

A i i

(a) Moment distribution (Ma)

le

(b) Equivalent moment of inertia (Ie)


METHOD 4Equivalent moment of inertia (Ie) combined with

i l i i

Contd…

numerical integration CBV maxDAA

(a) Deflected profile

L

t

a

t

(a) Deflected profile

hh

(b) Moment distribution (Ma)

A i i

le

VARIABLE MOMENT INERTIA ALONG A

(c) Equivalent moment of inertia (Ie)

VARIABLE MOMENT INERTIA ALONG A MEMBER DUE TO CRACKING


METHOD 4Equivalent moment of inertia (Ie) combined with

contd …

numerical integration

DEFLECTED PROFILE OF THE DESIGN STRIP WITHDEFLECTED PROFILE OF THE DESIGN STRIP WITH ALLOWANCE FOR CRACKING USING NUMERICAL

INTEGRATION

Calculated deflection by numerical integrationCalculated deflection by numerical integration0.235 in. ( 6 mm)Estimate of deflection at center of panel:2 * 0.235 = 0.47 in. ( 12.0 mm)


METHOD 5Use of Finite Element programs with no cracking(linear elastic)

DISCRETIZATION OF THE TYPICAL FLOOR SLAB FOR FINITE ELEMENT ANALYSIS (FLOOR-PRO)

EXAMPLEEXAMPLEUsing the same dimensions and parameters, determine the maximum deflection of panel identified in previous examples


Contd…METHOD 5Use of Finite Element programs with no cracking

DEFLECTION CONTOUR OF THE FLOOR SYSTEM(using gross cross-section; no allowance for cracking)

ADAPT-FLOOR-pro)ADAPT FLOOR pro)

Maximum deflection at the center of the panel under consideration is 0.54 in. (13.7 mm) (ADAPT-FLOOR Pro)


METHOD 6Finite Element Method with allowance for cracking

Formulation of finite elements with allowance forcracking is somewhat involved since it is a

First a brief review of the underlying work

cracking is somewhat involved, since it is apost-processor to design and detailing. It requires the detailed information on the following:

The reduction of local stiffness due to cracking andthe computation of reduced stiffness depends on:

Presence of local reinforcement, Amount of local reinforcement, Orientation and cover of each reinforcing barAxial loads due to constraint of supports

And if post-tensionedAnd if post tensionedPresence of strandsOrientation of strandsAxial precompression


contd…

METHOD 6Finite Element Method with allowance for crackingFinite Element Method with allowance for cracking

First a brief review of the underlying work

Since the position of each reinforcing bar and the local stress play a central role in the post-cracking response at the local level, and g p ,subsequently the overall deflection of a floor system, the authentic modeling of the structure becomes of paramount importance.

The following are several examples.

Structural Modeling for Cracked Deflection

OF SLABCENTROID

SLAB

BEAM

(i) (ii) (iii)

(b) INCORRECT MODELING (a) ACTUAL CONDITIONS

( ) ( ) ( )

CENTROIDOF SLAB

(i) (ii) (iii)

(c) PROPER MODELINGbeams must be modeled with correct eccentricity

with respect to slab

(i) (ii) (iii)


C

CC

MT

T

(i)

(iv)

Ct

(a) Flanged beam under bending

C

Z i l

(b) Flanged beam with no interaction

M Zero axial

Zero axial

Proper and Improper ModellingOf Flanged Beam

Between stem and flange


INFLECTIONTENDON

Y1

X1

POINT

Y2X3

W3

W1

X1

W2

W4

X3

(a) Traditional schemes substitute each tendon by the force in exerts to concrete

SEGMENT

J

J'

TENDON

I' I

j

i

(b) Representation of each post‐tensioning tendon or rebar by a discrete steel element in finite element cells



EXAMPLEUse ACI318-08 for determine the reinforcement necessaryfor both the in-service and strength requirements of thecode. Use the calculated minimum reinforcement of the code todetermine the cracked deflection. No other reinforcement isadded.

DEFLECTION CONTOUR OF SLAB WITH CRACKINGDeflection at center of target panel 0.7 in (17.8 mm) (Floor-Pro)


contd…

METHOD 6Fi it El t M th d ith ll f kiFinite Element Method with allowance for cracking

Steps to follow:

1. Using the geometry, boundary conditions, materialproperties, and the load combination for which thedeflection is sought, obtain a solution based on grossmoment of inertia (Ig), and determine applied moments( g), pp(Ma) over the entire structure;

2. Design the floor system, if required. Addreinforcement if needed;reinforcement, if needed;

3. Scan the entire structure, to detect and identify all reinforcement. Reinforcement can be in the followingforms:forms:

User defined one or more top and bottomreinforcement mesh;calculated for serviceability or strengthUser defined post-tensioning


contd…


4. Match the applied moment of each location (Ma) withth ki t (M ) t th l tithe cracking moment (Mcr) at the same location;

5. If Ma > Mcr, using the local geometry, material andreinforcement, calculate the effective moment of inertial (Ie);

6. Re-assembles the system stiffness matrix and solvefor deflections;for deflections;

Question:Is it adequate from design standpoint to conclude theIs it adequate, from design standpoint, to conclude the computation at this stage, or should one continue the calculation, since the reduced stiffness will result in a different distribution of applied moments, and hence

ki ttcracking pattern.



Maximum deflection at the center of target panel:With allowance for cracking 0.70 in. (ADAPT- FLOOR Pro)Without allowance for cracking 0.54 in.(ADAPT- FLOOR Pro

EXTENT OF CRACKING SHOWN THROUGH REDUCTION IN EFFECTIVE MOMENT OF INERTIA Ie ABOUT Y-Y AXIS



EXTENT OF CRACKING SHOWN THROUGHEXTENT OF CRACKING SHOWN THROUGH REDUCTION IN EFFECTIVE MOMENT OF INERTIA

Ie ABOUT X-X AXIS

DEFLECTION CALCULATIONSCrack Width

Once cracking is initiated, among other factor its width depends on:p

Amount and location of reinforcementSize of reinforcementSpacing of reinforcementSpacing of reinforcement

ACI-318 does not require the computation of crack width. In ACI-318 cracking and its width are controlled through detailing and allowable stressesthrough detailing and allowable stresses

European Code (EC2) does not specify allowable stresses. Rather, it recommends the design to be based on limiting a pre-specified crack width, such as 0.1, 0.2, or 0.3 mm (0.004; 0.008; or 0.01 in.)


ExampleFor the same geometry and other conditions, determine the width of probable cracks for the l d bi ti (DL + LL)load combination (DL + LL)

WIDTH OF PROBABLE CRACKS IN LEFT-RIGHT DIRECTION

Crack width in left-right direction at the middle of target panel 0.01 in. (0.25 mm)


ExampleFor the same geometry and other conditions, determine the width of probable cracks for the load combination (DL + LL)combination (DL + LL)

WIDTH OF PROBABLE CRACKS IN UP-DOWNWIDTH OF PROBABLE CRACKS IN UP-DOWN DIRECTION

Crack width in up-down direction at the middle of target panel 0 009 in (0 23 mm)target panel 0.009 in. (0.23 mm)

COMPARISON OF DEFLECTION CALCULATION METHODS

DEFLECTION VALUES AT CENTER OF THE TARGET PANEL FROM DIFFERENT METHODS

Deflection NormalizedCalculation Method Deflection(in)

Normalized Deflection

1 Closed form formulas

0.4812.2 mm 69 %

2 Strip method(uncracked)

0.23111.7 mm 66 %

3 Strip method ACI -318 (cracked Ie av)

0.26413 4 mm 75 %318 (cracked, Ie,av) 13.4 mm

4

Strip method cracking, Ieand numerical i t ti

0.23511.9 mm 67 %

integration

5

Finite Element Method (FEM)No allowance for

0.54013.7 mm 77 %

cracking

6

Finite Element Method (FEM)With allowance for

0.70017.8 mm 100 %

cracking

DEFLECTION OF CONCRETE FLOORS

CALCULATION OF INSTANTANEOUS DEFLECTION

CONCLUDING REMARKS

The commonly acceptable methods of deflection l l ti i ld lt th t t 30% icalculation yield results that can vary up to 30% in

value. Hence, calculated deflections should be evaluated using engineering judgment

The largest value of calculated deflection is generally given using Finite Element Method with allowance for cracking

The use of strip method combined with equivalent moment of inertia is too cumbersome for hand calculation, but readily available in software based onstrip methodsstrip methods

Closed form formulas, where applicable are simpleand compare well with other approximate methods


Long-Term Deflections

Concrete slabs continue to deflect undersustained load.

Long-Term Deflections are due to:Long Term Deflections are due to:

Shrinkage of concrete

Creep under applied loadCreep under applied load

Values of interest are

Ultimate deflection from initiation of deflection until no change in deflection with time;

Interim deflection over a defined time period

LONG-TERM DEFLECTIONS CONCRETE FLOORS

Due to shrinkage and creep deflection changeswith time.

Shrinkage of concrete is due to loss of moisture.

Creep is increase in displacement under stressCreep is increase in displacement under stress.

Under constant loading, such as self weight, theeffect of creep diminishes with time.

Likewise, under normal conditions, with loss ofmoisture, the effect of deformation due to shrinkagediminishes.

Restraint of supports to free shortening of a slabdue to shrinkage or creep can lead to cracking ofslabs and thereby an increase in deflection due toygravity loads.

LONG-TERM DEFLECTIONS

It is the long-term shrinkage due to loss of moisture through the entire volume of concrete

Shrinkage

moisture through the entire volume of concretethat impacts a slab’s deformation.

Plastic shrinkage that takes place within the firstfew hours of placing of concrete does not play asignificant role in slab’s deflection and its impact inlong-term deflection is not considered.

On its own, long-term shrinkage does not result invertical displacement of a floor system.

It is the presence of non-symmetrical reinforcementIt is the presence of non symmetrical reinforcementwithin the depth of a slab that curls the slab (warping)toward the face with none or less reinforcement.

NK

AG

E ST

RA

IN

TIMESHR

INK SHRINKAGE


Shrinkage

It is important to note that, as such, deflection due to shrinkage alone is independent from the natural deflection of slab. It neither depends on the direction of deflection due to applied loads, nor its magnitude.of deflection due to applied loads, nor its magnitude. The shrinkage deflection depends primarily on the amount and position of reinforcement in slab.

A corollary impact of shrinkage is crack formation due to restraint of the supports. It is the crack formation due to shrinkage that increases deflection under

it l dgravity loads.

Shrinkage values can vary from zero, when concrete is fully immersed in water to 800 micro strain. Typical ultimate shrinkage values are between 400 to 500 micro strain.

QUESTION

WHAT IS THE TOTAL SHORTENING OF THIS SLAB DUE TO SHRINKAGE?

Slab is on rollersSlab length 100 ft

Ultimate shrinkage values arebetween 400 to 500 micro strain

ANSWER

WHAT IS THE TOTAL SHORTENING OF THIS SLAB DUE TO SHRINKAGE?THIS SLAB DUE TO SHRINKAGE?

Slab is on rollersSlab length 100 ftSlab length 100 ft

Ultimate shrinkage values arebetween 400 to 500 micro strain

shortening = strain * length

= 100*12*400*10-6 = 0 48 in= 100*12*400*10 6 = 0.48 in.Or = 100*12*500*10-6 = 0.60 in.


Creep is stress related. Creep

It is a continued magnification of the spontaneousdisplacement of a member with reduced rate with time.

Values of creep vary from 1.5 to 4.

Typical ultimate creep values for commercial andbuilding structures are between 2 to 3.

CREEP

ELASTIC RECOVERY

CREEPSTRAIN

STRAINELASTIC

DEFORMATIONPERMANENT

CREEP RECOVERY

TIME0 t

0t 8t

QUESTION

WHAT IS THE TOTAL SHORTENING OF THIS SLAB DUE TO CREEP?

Slab is on rollersSlab length 100 ft

Ultimate creep coefficient = 2.5

Creep is due to stress. There are no stresses in pthe horizontal direction. Hence there will be no shortening due to creep

QUESTION

WHAT IS THE TOTAL SHORTENING OF THIS SLAB DUE TO A UNIFORM RECOMPRESSION

OF 150 psi FROM POST-TENSIONING ?

Slab is on rollersSlab is on rollersSlab length 100 ft

Creep coefficient 2.5

Shortening = P * L / (A*E )

E = 57000 √4000 = 3,605,000 psi

ANSWER

WHAT IS THE TOTAL SHORTENING OF THIS SLAB DUE TO A UNIFORM PRECOMPRESSIONSLAB DUE TO A UNIFORM PRECOMPRESSION

OF 150 psi FROM POST-TENSIONING ?

Shortening = P * L / (A*E )

E = 57000 √4000 = 3,605,000 psi

Shortening = 150*100*12 / (1 * 3 605 000)Shortening = 150 100 12 / (1 3,605,000)

= 0.05 in short term

Long-term shortening due to creep is between2 to 3 times more. Use 2.5 times

Sh t i 2 5*0 05 0 125 i l tShortening = 2.5*0.05 = 0.125 in long-term

QUESTION

Load

Rebar

The reinforcement in this slab was placed wrongly, top side down.placed wrongly, top side down.

How does this impact the deflection of the slab:

Vertical deflection of the slabHorizontal shortening of the slab.

ANSWER

Load

Rebar

The reinforcement in this slab was placed wrongly, top side down.placed wrongly, top side down.

How does this impact the deflection of the slab:

Shrinkage will cause upward deflection

Horizontal shortening due to shrinkage will not change


Restraint of Supports

Restraint of supports, such as walls and columns to free movement of a slab due to shrinkage and

pp

to free movement of a slab due to shrinkage and creep can lead a tensile stresses in the slab and early cracking under applied loads. Early cracking will reduce the stiffness of the slab and gincrease its deflection.


Multiplier Factors for Long-Term Deflections

For design purposes, the long-term deflection of a floor system due to creep and shrinkage can be expressed as a multiplier to its instantaneous deflectiondeflection.

Long-term deflection due to sustained load:

Δ C * ΔΔl = C * Δi

Δl = long-term deflection;Δi = instantaneous deflection; and i C = multiplier.

ACI-318 suggests the multiplier factor (C ) shown in figure of next slide to estimate long term deflectionsfigure of next slide to estimate long term deflections due to sustained loads



2.0

T 1.0

1.5

.5

Duration of Load, Months0 6 1231 18 24 30 36 48 60

0

Creep and Shrinkage Multiplier for Long-term Deflection (ACI 318-8)



The multiplier can be reduced, if compression reinforcement is present. The factor (λ) for the reduction of the multiplier is given by:reduction of the multiplier is given by:

λ = C / (1 + 50ρ’ )

ρ’ is percentage of compression rebar at

mid-span for simple and continuous members; andat support for cantilevers at support.



Δ = instantaneous deflection; andΔi = instantaneous deflection; andΔl = long-term deflection.

Δl = ( 1 + λc + λsh ) * Δi

Whereλc = creep multiplier;

λsh = shrinkage multiplier;

MULTIPLIERS FOR LONG-TERM DEFLECTIONS

Immediate Creep Shrinkage TotalSource

Immediatedeflection

Creepλc

Shrinkageλsh

Totalλt

Sbarounis(1984) 1.0 2.8 1.2 5.0Branson(1977) 1 0 2 0 1 0 4 0Branson(1977) 1.0 2.0 1.0 4.0Graham and

Scanlon (1986b) 1.0 2.0 2.0 5.0

ACI-318 1.0 3.02.0

LOAD COMBINATIONS

The load combination proposed to be used in evaluating the deflection of a floor system depends on the objective of the floors evaluation.

Total Long-Term Displacement From Removal of Formsof Forms

(1.0*SW + 1.0*SDL + 1.0*PT + 0.3*LL)* λt

SW lf i htSW = selfweight;SDL = superimposed dead load, (floor cover

and partitions); PT = post-tensioning; LL = design live load; andλt = long-term deflection multiplier

( λt = λsh + λc = C )

LOAD COMBINATIONS

Total Long-Term Displacement From Removal of Forms

(1.0*SW + 1.0*SDL + 1.0*PT + 0.3*LL)* λt

The above load combination is conservative since it assumes the application of superimposed loads to be at the time the supports to a floor system are removed The factor 0 3 suggested for liveare removed. The factor 0.3 suggested for live load is for “sustained” load combination. The significance of the above load combination is that it provides a measure for the total deflection from the position of the forms at time concrete is cast. Its application is mostly for aesthetics and drainage of surface water, if applicable.


For deflection check in connection with the code specified maximum values where the purpose is to

Change of deflection with time

specified maximum values where the purpose is to mitigate damage to non-structural elements the following curve is used:

100%ND

100%

80

SHR

INK

AG

E A

NEN

ING

60

40

GE

OF

FIN

AL

SC

REE

P SH

OR

TE

20

0

PER

CEN

TAG C

1 day 3 7 10 30 50 100 200 1 yr 3 yrs 25 yrs

LONG-TERM SHORTENING OF CONCRETE MEMBERS DUE TO CREEP AND

SHRINKAGE WITH TIME (PCI Design Manual)

TIME SCALE

SHRINKAGE WITH TIME (PCI Design Manual)

LIVE LOAD DEFLECTION

Even when linear elastic theory is used for the calculation of a floor system’s deflection, cracking will result in a non-linear response. For the same load, the deflection of a slab depends on the extent of cracking prior to the application of the load. For this reason, when calculating the deflection due to the , ginstantaneous application of live load, the following procedure has to be used:

Deflection due to LL =

(deflection due to DL+LL) – (deflection due to DL)

The above accounts for loss of stiffness due to dead load prior to the application of live load.

LONG-TERM DEFLECTIONnumerical example

Check the deflection at the center of the panel identified. The floor supports elements that are prone to damage due to excessive deflection

TNO_123

prone to damage due to excessive deflection


(a) Floor planColumn

18'' x 24''(460mm x 610mm)

26'-3''(8.00m)

30'-0'' (9.14m)

(b) Panel plan

Column-supported Two Way Slab


Given:Span length along X-X direction = 30’ (9.14 m)Span length along Y-Y direction = 26.25’ (8.00 m)p g g ( )Slab thickness = 8 in. (203 mm)Ec (modulus of elasticity) = 4.287 * 106 psi

(29,558 MPa)

Superimposed dead load (SDL) = 25 psf (1.2 kN/m2)(includes allowance for partitions)Live load (LL) = 40 psf (1.9 kN/m2)Other parameters the same as previous exampleOther parameters the same as previous example

Required Calculation of deflection and code check for the following

U k d ditiUncracked conditionLoads applied at time of removal of formsSDL applied on day 45; structure placed inservice on day 180

Allow for crackingSDL applied on day 45Structure placed in service on day 180S uc u e p aced se ce o day 80


Using ACI-318, since the floor supports non-structuralmembers likely to be damaged from excessive deflection,the following two limits for deflection check apply:the following two limits for deflection check apply:

For visual impact and comfort d/L < 1/240For damage mitigation d/L < 1/480

Where d is deflection and L is its associated span.

The allowance for increase in deflection due to long-term effects of shrinkage and creep is made through a magnification factor (C) to be applied to the immediate deflection. Assume C = 2.2, indicating that the long-term deflections will be 2.2 times the instantaneous values. de ect o s be t es t e sta ta eous a ues

The support spans in the two orthogonal directions are26’ 3” and 30’ 0”. Hence, the diagonal distance fordeflection check at the middle of the panel underdeflection check at the middle of the panel underconsideration is:

L = ( 26.252 + 302 )0.5 = 39.86 ft


Using ACI-318, since the floor supports non-structuralb lik l t b d d f i d fl timembers likely to be damaged from excessive deflection,

the following two limits for deflection check apply:

For visual impact and comfort d/L < 1/240For damage mitigation d/L < 1/480

L = 38.86 ft

For visual impact and comfort (39.86 * 12/240) = 2.0 in.

For damage mitigationFor damage mitigation(39.86 * 12)/480 = 1.0 in.


UNCRACKED DEFLECTIONS

Condition 1: loads applied at time of removal of forms

Assumptions:Concrete uncracked under service conditionS i d l d i li d t ti f l f f dSuperimposed load is applied at time of removal of forms and the slab is placed in service at the same time. Hence, live load is considered effective at day 1.

Since the section is assumed uncracked when calculatingSince the section is assumed uncracked, when calculating deflections, the principle of superposition applies. A Finite Element (ADAPT-Floor Pro) solution for the slab gives the following results for deflections.

Load Deflection (in. )------------------------------------------------------------------------Selfweight (SW) 0.33 Superimposed dead load (SDL) 0.08Superimposed dead load (SDL) 0.08Live load (LL) 0.13


For Visual and Sense of Comfort

On the assumption that the slab is cast with no camber and horizontally, the perceived deflection is the total downward displacement of the slab. The total deflection is due to the instantaneous and long-term values of the loads applied at removal of forms (0.3LL) and the immediate response of the

dditi l li l d (0 7LL)additional live load (0.7LL)

Load combination:

(1 + C) * (SW + SDL + 0 3 * LL) + 0 7LL(1 + C) (SW + SDL + 0.3 LL) + 0.7LL

= (1 + 2.2) * (0.33 + 0.08 + 0.3 * 0.13) + 0.7 * 0.13 = 1.53 in. < d/L = 1/250 = 2.0 in. OK

Note:Sustained portion of live load 0.30LLBalance of design live load 0.70LLC is long-term multiplier assumed 2 2C is long term multiplier assumed 2.2


For Damage Mitigation

Loads are applied immediately after removal of forms. pp yMembers likely to be damaged installed the same time.

Damage is due to long-term effects of initial loads; andDeflection due to balance of live load (0.7LL)

.Load combination:

C * (SW + SDL + 0.3*LL) + 0.7*LL

= 2.2 (0.33 + 0.08 + 0.3 * 0.13) + 0.7 * 0.13 = 1.08 in. > L/480 = 1.00 in.

B d i i j d t th l lBased on engineering judgment the values are close enough to be acceptable.

Note:Note:Increase in concrete stiffness with time conservatively not account for


Condition 2: Delayed application of load

Loads applied at different times

Assumptions:.

Uncracked sectionSuperimposed load applied 45 days after theremoval of forms, when the nonstructural ,members likely to be damaged are also installedStructure is placed in service 180 days after theremoval for forms

Investigate the adequacy of the floor for deflections:


contd…Condition 2: Delayed application of load

Loads applied at different times

100%

E A

ND

80

60

AL

SHR

INK

AG

EO

RTE

NIN

G

52

48%

40

20

NTA

GE

OF

FIN

AC

REE

P SH

O 52

0

PER

CEN

1 day 3 7 10 30 50 100 200 1 yr 3 yrs 25 yrs

TIME SCALE

45 180

LONG-TERM SHORTENING OF CONCRETE

NoteFor a load applied on day 45 the balance of long-For a load applied on day 45, the balance of long-term deflection is 48%


contd…Condition 2: Delayed application of load - uncracked

For Visual and Sense of Comfort

On the assumption that the slab is cast with no camber and horizontally the perceived deflection is the total downwardhorizontally, the perceived deflection is the total downward displacement of the slab. The total deflection is due to the instantaneous and long-term values of the loads applied and the instantaneous response of the additional liveload (0.7LL)load (0.7LL)

Load combination:

(1 + C) * (SW + SDL + 0.3 * LL) + 0.7LL ( ) ( )

= (1 + 2.2) * (0.33 + 0.08 + 0.3 * 0.13) + 0.7 * 0.13 = 1.53 in. < L/240 = 2.0 in. OK

NoteThe delayed application of loads does not impact the d fl ti h k l f i l ff tdeflection check values for visual effects..


For Damage Mitigation

Damage to non-structural members will be due to deflectionsgthat take place subsequent to installation of such members. This is the long-term part of deflection due to selfweight afterday 45. When the structure is placed in service. There will beadded deflection due to sustained portion of live loads(assumed 30% of its design value) and an additional increasein deflection when the remainder of the live load is likely toapply (assumed 70% of its design value) .

L t d fl ti ith lti l l di tiLong-term deflection with multiple loading times


Point B’ is total deflection due to selfweight prior to the application of SDL and installation of members likely to be damaged. Once these loads are applied, the floor deflectsto point Bto point B.The structure is placed in service after six months (point C’). At this time, it is subjected to the sustained portion of live load (0.3 LL). This results in an immediate displacement of the structure to point C, followed by long-term deflection toof the structure to point C, followed by long term deflection toD’. When in future, the remainder of the design live load (0.7 LL),is applied, the structure deflects to DThe total deflection experienced by the members likely to be p y ydamaged is shown by “d”


contd…Condition 2: Delayed application of load - uncracked

Long-term deflection with multiple loading times

Load combination for applicable deflection is:

0.48*C * SW + C * SDL + (1 + C) * 0.30LL + 0.70*LL

= (0.48 * 2.2) * 0.33 + 2.2 * 0.08 + (1 + 2.2) * 0.3 * 0.13 + 0.7 * 0.13

= 0 74 in < d/L = L / 480 = 1 00 in OK 0.74 in. < d/L L / 480 1.00 in. OK


Delayed application of load with allowance for cracking

Load-deflection in post-cracking is nonlinear . Consequently, g ysuperposition does not apply. The increment in deflection at each stage depends on the prior load history of the structure.

In the post-cracking stage, the stiffness of the structure at each point depends on the amount and location of reinforcement available at that point. Hence, the structure must be analyzed, designed and detailed for cracked d fl ti l l tideflection calculation.

Load-deflection with intermittent loadsand cracking


Assumptions

Allow for cracking of concrete, when local stressesexceed modulus of rupture

Reduced stiffness is based on the amountReduced stiffness is based on the amount,location and orientation of local reinforcement, including PT is available

Superimposed load is applied 45 days after theSuperimposed load is applied 45 days after the removal of forms, when the nonstructural members likely to be damaged are also installed.

Structure is placed in service 180 days after the removal for forms

The long-term deflection magnifier (C) due to g g ( )shrinkage and creep remains the same as in the uncracked condition ( C = 2.2)


Visual and sense of comfort



With no camber, and slab cast horizontal, the total downward displacement governs the visual effects.

The observed deflection is the long-term effects of the sustained load on the structure (including 0.3LL) plus instantaneous deflection due to the application of the balance of live load 0.7LL) .balance of live load 0.7LL) .Up to here

The load combination for the cracked deflection of the entire sustained loads is:

Dsw,SDL,0.3LL = (SW + SDL + 0.3LL) = 0.53 in

The above displacement is subject to the long-term effect (C=2.2).


contd…Delayed application of load with allowance for cracking


The instantaneous displacement at application of 0.7LLmust be added to the above in order to give the totalmust be added to the above, in order to give the total deflection.

The increment of deflection due to the application of 0.7LL is the difference between two solutions as follows

D0.7LL = Dsw,SDL,LL - Dsw,SDL,0.3LL = 0.12 in.

The observed total deflection for visual effects is:

(1 + C) * Dsw,SDL,0.3LL + D0.7LL = (1 + 2.2) * 0.53 + 0.12

= 1.82 in < L/240 = 2.0 in. OK

Note that contrary to the uncracked scenarios, in the above the magnification factor C is not included in the load combinationcombination


For damage control

Damage is caused by deflections subsequent to installation. This includes the long-term portion of deflection due tog pselfweight after day 45.

When placed in service, there is instantaneous deflection due to sustained 0.30LL, and additional increase whenremainder live load is applied

The components of design deflection are illustrated in figure

Long-term deflection with multiple loading times



The long-term effect of superimposed dead load (DSDL,LT ) . ,The value of this component from the graph is:

0.48 * C * Dsw

Th l d bi ti t i t D i 1 0 * SWThe load combination to arrive at Dsw is: 1.0 * SW

The long-term effect of displacement due to selfweight is:

D = 0 48 * C * D = 0 48 * 2 2 *0 38 = 0 40 inDsw,LT = 0.48 * C * Dsw = 0.48 * 2.2 *0.38 = 0.40 in.

The immediate deflection from SDL does not impact the results, since installation of members subject to damage is deemed to be in progress but its long term effect doesdeemed to be in progress, but its long-term effect does

DSDL,LT . = C * DSDL

This is obtained from two independent solutions oneThis is obtained from two independent solutions one including and the other excluding the superimposed dead load as given below

DSDL = Dsw SDL - Dsw = 0.1”SDL sw,SDL sw

DSDL,LT . = C * DSDL = 2.2 * 0.1 = 0.22 in.



The structure is placed in service after six months , when (0.3 LL) applies. This results in immediate displacement to point C, followed by long-term deflection to D’. The immediate displacement D0.3LL of the structure (CC’) contributes to the total displacement (d), likely to damage the structure. Its value is given by subtracting the deflection values Dsw,SDL,0.3LL from Dsw,SDL to be obtained from two different solutions.

Load combination for D isLoad combination for Dsw,SDL is

( 1.0*SW + 1.0*SDL) = 0.48 in.

Load combination for D isLoad combination for Dsw,SDL,0.3LL is

(1.0*SW + 1.0*SDL + 0.30*LL) = 0.53 in.

D0 3LL = Dsw SDL 0 3LL - Dsw SDL = 0.53 – 0.48 = 0.05 in.D0.3LL Dsw,SDL,0.3LL Dsw,SDL 0.53 0.48 0.05 in.

The long-term effect of the sustained live load 0.30LL is C times its instantaneous value. It equals>

D0.3LL,LT = C * D0.3LL = 2.2 * 0.05 = 0.15 in.



More than two years, the structure is checked for (0.7 LL), hen it deflects to point D The total deflection e periencedwhen it deflects to point D. The total deflection experienced

by the members likely to be damaged is deflection due:

Entire live load; long term effects of the selfweight after day 45; andlong-term effects of the selfweight after day 45; and long-term effects of the sustained portion of live.

Dsw,LT = 0.40 in.

D0.3LL,LT = 0.15 in.

The instantaneous deflection of live load DLL is given by the difference between two solutions, namely Dsw SDL LL which y sw,SDL.LL includes the entire loads and Dsw,SDL that includes the selfweight and superimposed dead loads. The load combinations for the two solutions are:

For Dsw,SDL.LL (1.0 * SW + 1.0 * SDL + 1.0 * LL) = 0.70 in.

For Dsw,SDL ( * S * S )(1.0 * SW + 1.0 * SDL) = 0.48 in.

DLL = 0.70 – 0.48 = 0.22 in.



Total deflection is the sum of the following:DLong-term effects of selfweight Dsw,LT = 0.40 in.

Long-term effect of SDL DSDL,LT = 0.22 in.Long-term of sustained live load D0.3LL,LT = 0.15 in.Instantaneo s deflection d e to LL D 0 22 inInstantaneous deflection due to LL DLL = 0.22 in.

___________________________________________________Total deflection 0.99 in.

< Span/480 = 1.00 in. OKTNO_123

Slab thickness8'' (200mm)8 (200mm)

(a) Floor plan

Column18'' x 24''(460mm x (610mm)

26'-3''(8.00m

) (b) Panel plan

30'-0'' (9.14m)


CONCLUDING REMARKS

Reasons for controlling deflectionsDeflection control through span/depth ratiosPermissible deflection valuesPermissible deflection values How to measure code-intended valuesHow to calculate probable deflections

Instantaneous deflectionsInstantaneous deflectionsLong-term deflections

Ultimate valuesInterim values

Crack width calculation and control

Numerical examples


References and Further Reading

ACI 318-08, (2008), “Building Code Requirements for Structural Concrete,” American Concrete Institute.

Bares R (1971) “Tables for the Analysis of Plates Slabs andBares, R., (1971), Tables for the Analysis of Plates, Slabs and Diaphragms Based on the Elastic Theory,” Bau verlag GmbH, Wiesbaden und Berlin, 1971, pp. 626

Branson D E (1977) “Deformation of Concrete Structures ”Branson, D. E., (1977), Deformation of Concrete Structures, McGraw-Hill Book Company, New York, pp. 546.

Graham, C. J., and Scanlon, A., (1986), “Long Time Multipliers for Estimating Two-Way Slab Deflections,” ACI Journal,for Estimating Two Way Slab Deflections, ACI Journal, Proceedings V. 83, No.5, pp. 899-908.

PTI (1990). Post-Tensioning Manual, 5th Edition, Post-Tensioning Institute, Phoenix, AZ, pp. 406g pp

Sarbounis, J. A., (1994) “Multistory Flat Plate Buildings: Measured and Computed One-Year Deflections,” Concrete International, Vol. 6, No. 8, August, pp. 31-35.

Thank you for Listening

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