DeflectionConcrete-Nov2010[1]

DEFLECTION CALCULATION EXAMPLE

DEFLECTION CALCULATION EXAMPLE EXAMPLE : METHOD 1 Using closed form formulas determine the deflection at center of panel identified below for combination of DL + LL TNO_123

Slab thickness 8'' (200mm)

(a) Floor plan

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

26'-3'' (8.00m)

VIEW OF TYPICAL LEVEL OF A MULTI-STORY FLOOR SYSTEM SELECTED FOR STUDY 30'-0'' (9.14m)

(b) Panel plan

IDENTIFICATION OF PANEL FOR DEFLECTION CALCULATION

DEFLECTION CALCULATION EXAMPLE

DEFLECTION CALCULATION EXAMPLE w = k (a4*q / E*h3 )



Given: S an len th alon X-X direction Span length along Y-Y direction Slab thickness Ec (modulus of elasticity)

Superimposed dead load

= 30’ 9.14 . m = 26.25’ (8.00 m) = 8 in. (203 mm) = 4.287 * 106 psi , = 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 = deflection normal to slab; = E = Modulus of elasticity; h = slab thickness; and q = 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 = 0.0548 Deflection, Δ = k (a4*q / E*h3 )= 0.0548*8.77 = . For deflection at midpoint of column lines in Xdirection, from Table 5 = . Deflection, Δ = k (a4*q / E*h3 )= 0.0446*8.77 = 0.39 in (10 mm)

DEFLECTION CALCULATIONS

DEFLECTION CALCULATIONS Support

Method 2 :Strip method and assumption of linear elastic response

1



The structure is subdivided into design strips in two orthogonal directions Each strip is extracted and analyzed in isolation  For center of panel, the values obtained from the two solutions are added together

2 d 1 d1 Deflection is 1-2 direction

d2

2

Deflection is 2-2 direction

(a) Total Deflection d = d1 + d2 (a)

b View of the desi n stri alon Y-Y direction extracted from the floor system



EXAMPLE : METHOD 2 Using strip method, linear elastic response and no allowance for crackin , determine the deflection at center of panel identified for load combination (DL + LL)  Design parameters same as previous example _

View of the design strip along Y-Y direction extracted from the floor system Slab thickness 8'' (200mm)



a Floor lan

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

, to deflection to be calculated from the strip in the orthogonal direction. , above value is multiplied by 2, assuming that the deflection from the orthogonal direction is essentially the same

'- '' (8.00m)

'- ''

.

(b) Panel plan

Deflection at center of span 1 from solution of ADAPT-RC is 0.231 Inch



Hence deflection at center of panel: 0.231 * 2 = 0.462 inch


DEFLECTION CALCULATIONS Brief Review of Cracking, Equivalent Moment of Inertia (Ie) and Its Computation Neutral axis

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

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

Reinforcement

(a) Cracked beam elevation

Cracking moment

= = = =

Gross moment of inertia; Moment of inertia of cracked section; Effective moment of inertia; Maximum moment in member at stage deflection is computed; and, = Cracking moment.

Mcr

(b) Applied moment



le

(1)

Ma, is calculated using elastic theory and gross moment of inertia for the uncracked s ection

(c) Effective moment of inertia Ie

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

is assumed to be small, and already accounted for in the empirical formula (1)


DEFLECTION CALCULATIONS Calculation of Cracked Moment of Inertia Icr

Icr depends on:

Cracking Moment Mcr Mcr = fr * Ig /yt Where, fr

(2)

= Modulus of rupture, flexural stress caus ng crac ng. s g ven y:

fr

= 7.5 f’c1/2

yt

= distance of section centroid to farthest tension fiber

(3)



Geometry of section of reinforcement, including prestressing  Location of reinforcement  Amount

For singly reinforced rectangular sections: c = kd Icr = b(kd)3 / 3 + nAs(d-kd)2

For all-li htwei ht concrete, fr is modified as follows: fr

= 0.75 * 7.5 f’c 1/2

c

c

b c c

s

d

(4)

s

(a) Section

(b) Strains

s c Ts

(c) Forces

Post-cracking Stresses and Forces

DEFLECTION CALCULATIONS c

c

b c c

(b) Strains

Simplified Allowance for Cracking through averaging of Equivalent Moment of Inertia (Ie)

c

s

(a) Section

METHOD 3

C s

s

d


Ts

(ACI-318)

(c) Forces

Post-cracking Stresses and Forces Uncracked analysis and design contd… Cracking moment of inertia Icr Break

where, kd = [(2dB + 1) 0.5 – 1] / B

(6)

d = distance of compression fiber to center of tension reinforcement B = b/(nAs) = Es = modulus of elasticity of Steel Ec = modulus of elasticity of concrete

the floor slab into design strips  Select design strips of interest  Using gross moment of inertia obtain elastic uncracked distribution of moment for selected stri s (applied moment Ma)  Design the design strips and determine the amount and location of reinforcement, including any base

contd…

For more com lex eometr and reinforcement the relationship will be more involved

DEFLECTION CALCULATIONS contd…

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


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

,

Cracked analysis and design 

Calculate the cracking moments at face-of-supports and mid-spans (Mcr )  For locations, where Mcr > Ma, calculate the equivalent moment of inertia (Ie)  For each span/cantilever, determine the average Ie,av  For each calculated avera e 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



For continuous members an average equivalent moment 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 is used for the entire cantilever P

DEFLECTION CALCULATIONS EXAMPLE : METHOD 3 Simplified Allowance for Cracking through Averaging of Equivalent Moment of Inertia (Ie) (ACI-318)

DEFLECTION CALCULATIONS contd… EXAMPLE : METHOD 3 (Averaged Ie)

Calculate the cracked deflection at the middle of the panel identified for the load combination (DL + LL). Design arameters same as revious exam les.. f’c = 5000 si TNO_123


(a) Breakdown of slab into design strips (a) Floor plan

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

26'-3'' (8.00m)

30'-0'' (9.14m)

(b) Extraction of design strip for analysis in isolation

(b) Panel plan


DEFLECTION CALCULATIONS Contd…EXAMPLE : METHOD 3 (Averaged Ie) fr = 7.5√ f’c = 7.5 √ 5000 = 530.33 psi g= , n . e mm 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

DISTRIBUTION OF MOMENT (Ma ) DUE TO Consider moment at right support of span 1 Ma = 326.4 k-ft (442.53 kNm) Calculate cracking moment Mcr and a

= b*(kd)3 / 3 + n*As(d - kd)2 = [(2dB + 1)0.5 – 1]/B = . B = b/(nAs) n = Es/Ec = 30000/4287 = 7.0 As = area of steel at face of support =10.12 in2 = . . = . n 0.5 Kd = [(2*6.81*5.08 + 1) – 1] / 5.08 = 1.45 in Icr = (360*1.45 3)/3 + 7.0*10.12* (6.81-1.45) 2 = 2401 in4 kd

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



contd… EXAMPLE : METHOD 3 (Averaged Ie)

contd… EXAMPLE : METHOD 3 (Averaged Ie)

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

Using ACI-318 averaging procedure

Left cantilever : Ie at face of support = Ig

First span Average Ie

= 1.536e+4

Ie at left support centerline Ie at midspan Ie at right support centerline

in 4

= 1.70e+4 in 4 = 1.44e+4 in4 = 5.20e+3 in4

Second Span: Ie at left support centerline Ie at midspan Ie at right support centerline

= 5.63e+3 in4 = 1.536e+4 in 4 = 1.702e+4 in 4

Right cantilever: I at face of su

= 1.536e+4 in 4

ort = I

Left and right cantilevers Ie = Ig

Second span Average Ie


= [(1.70*104+5.20*10 3)/2 +1.44*104]/2 = 12.755e+3 in4

= [(5.63*103+1.70*10 4)/2 +1.536*10 4]/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

Ie = b*he3 /12 ere, strip.

s

ew

Left cantilever: First span Second span Right cantilever

o

e r u ary o

he he he he

e es gn

= 8 in ; no reduction = 7.52 in = 7.63 in = 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 FOR CRACKING, 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 value that is representative of both the midpoint of the panel and midpoint of the line of support  Since the geometry of the panel in two orthogonal directions is somewhat similar, the center of anel deflection is estimated as: 2 * . 0.264 = 0.528 in (13.4 mm)


DEFLECTION CALCULATIONS Contd…

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

Steps to follow:

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

(i) Start by following the same steps as the sim lified rocedure desi n and detail rebar for each design strip (ii) Subdivide each span into a number of segments (10 to 20 division), and determine the equivalent

A

D

V max

t

C

B

t

a L

a

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

e ec e pro e h

A i

i

h

(b) Moment distribution (Ma) A i

le

i

(a) Moment distribution (Ma) (c) Equivalent moment of inertia (Ie) le

(b) Equivalent moment of inertia (Ie)


MEMBER DUE TO CRACKING


contd …

METHOD 5

METHOD 4

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

Equivalent moment of inertia (Ie) combined with numerical integration

ALLOWANCE FOR CRACKING USING NUMERICAL INTEGRATION

a cu a e e ec on y numer ca n egra on 0.235 in. ( 6 mm)  Estimate of deflection at center of panel: 2 * 0.235 = 0.47 in. ( 12.0 mm)

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

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

DEFLECTION CALCULATIONS Contd…

DEFLECTION CALCULATIONS METHOD 6

METHOD 5 Use of Finite Element programs with no cracking

Finite Element Method with allowance for cracking First a brief review of the underlying work Formulation of finite elements with allowance for , post-processor to design and detailing. It requires the detailed information on the following: 

DEFLECTION CONTOUR OF THE FLOOR SYSTEM

The reduction of local stiffness due to cracking and the computation of reduced stiffness depends on:  Presence of local reinforcement,  Amount of local reinforcement,  Orientation and cover of each reinforcing bar  Axial loads due to constraint of supports

(using gross cross-section; no allowance for cracking) -

-

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


 Presence of strands  Orientation of strands  Axial precompression

Structural Modeling for Cracked Deflection CENTROID OF SLAB

SLAB

METHOD 6

BEAM

First a brief review of the underlying work (i)

Since the position of each reinforcing bar and the local stress play a central role in the postcrackin res onse at the local level, and subsequently the overall deflection of a floor system, the authentic modeling of the structure becomes of paramount importance.

(ii)

(iii)

(a) ACTUAL CONDITIONS (b) INCORRECT MODELING

CENTROID OF SLAB

The following are several examples. i

ii

(c) PROPER MODELING beams must be modeled with correct eccentricity with respect to slab

Structural Modeling for Cracked Deflection

Structural Modeling for Cracked Deflection INFLECTION POINT

TENDON Y1

Y2

C

(iv)

W2

C

T

W3

C

(i)

t

T

W1

M

(a) Flanged beam under bending

W4

(a) Traditional schemes substitute each tendon by the force in exerts to concrete TENDON SEGMENT J J'

ero ax a M

Zero axial

I' I

(b) Flanged beam with no interaction Between stem and flange

j i

Proper and Improper Modelling Of Flanged Beam

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

DEFLECTION CALCULATIONS METHOD 6 Finite Element Method with allowance for cracking EXAMPLE  Use ACI318-08 for determine the reinforcement necessary for both the in-service and strength requirements of the code.  Use the calculated minimum reinforcement of the code to determine the cracked deflection. No other reinforcement is added.


METHOD 6 n e

emen

e o w

a owance or crac ng

Steps to follow: 1. Using the geometry, boundary conditions, material properties, and the load combination for which the deflection is sought, obtain a solution based on gross moment of inertia I , and determine a lied moments (Ma) over the entire structure; 2. Design the floor system, if required. Add ,

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

3. Scan the entire structure, to detect and identify all reinforcement. Reinforcement can be in the following orms:  User defined one or more top and bottom reinforcement mesh;  calculated for serviceability or strength  User defined post-tensioning


DEFLECTION CALCULATIONS METHOD 6

METHOD 6 Finite Element Method with allowance for cracking 4. Match the applied moment of each location (Ma) with the cracking moment (Mcr ) at the same location;

Finite Element Method with allowance for cracking 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

5. If Ma > Mcr , using the local geometry, material and reinforcement, calculate the effective moment of inertial (Ie); 6. Re-assembles the system stiffness matrix and solve for deflections

Question: , , 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 cracking pattern.

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

DEFLECTION CALCULATIONS Crack Width


METHOD 6 Finite Element Method with allowance for cracking

Once cracking is initiated, among other factor its width de ends on:  Amount 

and location of reinforcement Size of reinforcement

ACI-318 does not require the computation of crack width. In ACI-318 cracking and its width are controlled roug e a ng an a owa e s resses 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.)

REDUCTION IN EFFECTIVE MOMENT OF INERTIA Ie ABOUT X-X AXIS

DEFLECTION CALCULATIONS Crack Width Example For the same geometry and other conditions, determine the width of probable cracks for the oa com na on +

DEFLECTION CALCULATIONS Crack Width Example For the same geometry and other conditions, determine the width of probable cracks for the load

WIDTH OF PROBABLE CRACKS IN LEFTRIGHT DIRECTION Crack width in left-right direction at the middle of target panel 0.01 in. (0.25 mm)

DIRECTION Crack width in up-down direction at the middle of . . .

COMPARISON OF DEFLECTION CALCULATION METHODS DEFLECTION VALUES AT CENTER OF THE TARGET PANEL FROM DIFFERENT METHODS Calculation Method

e ec on (in)

1

Closed form formulas

0.48 12.2 mm

2

Strip method (uncracked)

0.231

3

Strip method ACI , ,

4

Strip method cracking, Ie and numerical n egra on

0.235 11.9 mm

5

Finite Element Method (FEM) No allowance for cracking

0.540 13.7 mm

6

Finite Element Method (FEM) With allowance for cracking

0.700 17.8 mm

11.7 mm

0.264 .

DEFLECTION OF CONCRETE FLOORS

CALCULATION OF INSTANTANEOUS DEFLECTION

orma ze Deflection

69 %

CONCLUDING REMARKS 

The commonly acceptable methods of deflection ca cu a on y e resu s a can vary up o n 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

66 % 75 %

67 % 

77 %

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



100 %

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

DEFLECTION CALCULATION OF CONCRETE FLOORS Long-Term Deflections Concrete slabs continue to deflect under sustained load.

LONG-TERM DEFLECTIONS CONCRETE FLOORS



Due to shrinkage and creep deflection changes with time.



Shrinkage of concrete is due to loss of moisture.

-

.

 Shrinkage

of concrete

reep un er app e



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



Likewise, under normal conditions, with loss of moisture, the effect of deformation due to shrinkage diminishes.



Restraint of supports to free shortening of a slab due to shrinkage or creep can lead to cracking of slabs and thereb an increase in deflection due to gravity loads.

oa

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 Shrinkage 



LONG-TERM DEFLECTIONS

Shrinkage

It is the long-term shrinkage due to loss of that impacts a slab’s deformation.



Plastic shrinkage that takes place within the first few hours of placing of concrete does not play a significant role in slab’s deflection and its i mpact in long-term deflection is not considered.



On its own, long-term shrinkage does not result in vertical displacement of a floor system. within the depth of a slab that curls the slab (warping) toward the face with none or less reinforcement. N I A R T S E G A K N I R H S

SHRINKAGE

TIME

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 a lied loads, nor its ma nitude.. 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 grav ty oa s. 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

ANSWER

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

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



Slab is on rollers  Slab length 100 ft 

Ultimate shrinkage values are between 400 to 500 micro strain

Slab is on rollers



Ultimate shrinkage values are between 400 to 500 micro strain 

Or

shortening = strain * length * * * . . -6 = 100*12*500*10 = 0.60 in.


QUESTION

Creep 

Creep is stress related.



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

 

Values of creep vary from 1.5 to 4.

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

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



Slab is on rollers  Slab length 100 ft 

Ultimate creep coefficient = 2.5

ELASTIC RECOVERY STRAIN CREEP RECOVERY ELASTIC STRAIN

0

t

0

PERMANENT DEFORMATION

t

TIME

t

8

Cree is due to stress. There are no stresses in the 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 ?

ANSWER

WHAT IS THE TOTAL SHORTENING OF THIS OF 150 psi FROM POST-TENSIONING ? Shortening = P * L / (A*E )



Slab length 100 ft  Creep coefficient 2.5

E = 57000 √4000 = 3,605,000 psi *

Shortening = P * L / (A*E )

*

* ,

,

= 0.05 in short term

E = 57000 √4000 = 3,605,000 psi

Long-term shortening due to creep is between 2 to 3 times more. Use 2.5 times or en ng = .

QUESTION

.

Load

Rebar

The reinforcement in this slab was , . does this impact the deflection of the slab:

Vertical

n ong- erm

ANSWER

Load

How

= .

deflection of the slab Horizontal shortening of the slab.

Rebar

The reinforcement in this slab was , . How



does this impact the deflection of the slab: Shrinkage will cause upward deflection

Horizontal

shortening due to shrinkage will not change



Multiplier Factors for Long-Term Deflections

Restraint of Su

orts

Restraint of supports, such as walls and columns creep can lead a tensile stresses in the slab and early cracking under applied loads. Early crackin will reduce the stiffness of the slab and increase its deflection.

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 . Long-term deflection due to sustained load:

Δl =

Δi

Δl = long-term deflection; Δi = instantaneous deflection; and C = multiplier. ACI-318 suggests the multiplier factor (C ) shown in fi ure of next slide to estimate lon term deflections due to sustained loads





The multiplier can be reduced, if compression reinforcement is present. The factor (λ) for the

2.0 1.5

T

λ = C / (1 + 50ρ’ )

1.0

.5

ρ’ is percentage of compression rebar at

0 01 3 6

12

18

24

30

36

48

Duration of Load, Months Creep and Shrinkage Multiplier for Long-term Deflection (ACI 318-8)

60



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


LOAD COMBINATIONS

Multiplier Factors for Long-Term Deflections i

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

=

Δl

= long-term deflection.

Total Long-Term Displacement From Removal of Forms

Δl = ( 1 + λc + λsh ) * Δi λc

λsh

Where = creep multiplier; = shrinkage multiplier;

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

Source

deflection

λc

λsh

λt

Sbarounis(1984)

1.0

2.8

1.2

5.0

.

.

.

.

Graham and Scanlon (1986b)

1.0

2.0

2.0

5.0

ACI-318

1.0

2.0

= se we g ; = superimposed dead load, (floor cover and partitions); = post-tensioning; = design live load; and = long-term deflection multiplier

SDL

MULTIPLIERS FOR LONG-TERM DEFLECTIONS

λt

PT LL

λt

( λt

= λsh + λc = C )

3.0

LOAD COMBINATIONS

LONG-TERM DEFLECTIONS Change of deflection with time For deflection check in connection with the code

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 remove . e ac or . sugges e or ve 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.

mitigate damage to non-structural elements the following curve is used: D A E G A K G N I N I R N H S T L R A O N I H F S F P O E E E R A T N E C R E P

100%

80

60

40

20

0 1 day

3

7

10

30

50

100

200 1 yr

3 yrs

TIME SCALE

LONG-TERM SHORTENING OF CONCRETE MEMBERS DUE TO CREEP AND

25 yrs

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 calculatin the deflection due to the instantaneous application of live load, the following procedure has to be used:

LONG-TERM DEFLECTION numerical example Check the deflection at the center of the panel identified. The floor supports elements that are

TNO_123


Deflection due to LL = (a) Floor plan Column 18'' x 24'' (460mm x 610mm)

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

26'-3'' (8.00m)

The above accounts for loss of stiffness due to dead load prior to the application of live load. 30'-0'' (9.14m)

(b) Panel plan

Column-supported Two Way Slab

LONG-TERM DEFLECTION numerical example Given: Span length along X-X direction Span length along Y-Y direction Slab thickness Ec (modulus of elasticity)

Superimposed dead load (SDL)

= 30’ ( 9.14 m) = 26.25’ (8.00 m) = 8 in. (203 mm) = 4.287 * 106 psi (29,558 MPa) = 25 psf (1.2 kN/m2)

(includes allowance for partitions)

Live load (LL)

= 40 psf (1.9 kN/m2)

Required Calculation of deflection and code check for the following ncrac e con on  Loads applied at time of removal of forms  SDL applied on day 45; structure placed in service on day 180  Allow

for cracking  SDL applied on day 45  Structure laced in service on da 180

LONG-TERM DEFLECTION numerical example Using ACI-318, since the floor supports non-structural members likely to be damaged from excessive deflection,

For visual impact and comfort For damage mitigation

d/L < 1/240 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. The support spans in the two orthogonal directions are 26’ 3” and 30’ 0”. Hence, the diagonal distance for consideration is: L = ( 26.252 + 302 )0.5 = 39.86 ft

LONG-TERM DEFLECTION numerical example

Using ACI-318, since the floor supports non-structural mem ers e y to e amage rom excess ve e ect on, the following two limits for deflection check apply: For visual impact and comfort For damage mitigation

d/L < 1/240 d/L < 1/480

L = 38.86 ft


UNCRACKED DEFLECTIONS Condition 1: loads applied at time of removal of forms Assumptions: Concrete uncracked under service condition Superimposed 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. , deflections, the principle of superposition applies. A Finite Element (ADAPT-Floor Pro) solution for the slab gives the following results for deflections.

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

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

(39.86 * 12)/480 = 1.0 in.



For Visual and Sense of Comfort

For Damage Mitigation

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 a ona ve oa .



 

Loads are a lied immediatel after removal of forms. Members likely to be damaged installed the same time. Damage is due to long-term effects of initial loads ; and Deflection due to balance of live load (0.7LL)

.Load combination:

Load combination: *



. *

.

= (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

= 2.2 (0.33 + 0.08 + 0.3 * 0.13) + 0.7 * 0.13 = 1.08 in . > L/480 = 1.00 in. ase on eng neer ng u gmen enough to be acceptable.

Note: Sustained portion of live load Balance of design live load -

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

e va ues are c ose

0.30LL 0.70LL . Increase in concrete stiffness with time conservatively not account for



Condition 2: Delayed application of load

contd… Condition 2: Delayed application of load

Loads applied at different times

Loads applied at different times

Assumptions: . 

Uncracked section  Superimposed load 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 Investigate the adequacy of the floor for deflections:

D N A G A K G N I N I R N H E S T L R N I H F S F P E O E E R G C A T E C R E P

100% 48% 80

60

40

20

45

0 1 day

3

7

10

30

180 50

100

200 1 yr

3 yrs

25 yrs

TIME SCALE

LONG-TERM SHORTENING OF CONCRETE Note ,

-

term deflection is 48%

LONG-TERM DEFLECTION numerical example 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 , 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 live load 0.7LL

LONG-TERM DEFLECTION numerical example For Damage Mitigation Damage to non-structural members will be due to deflections that take place subsequent to installation of such members. This is the long-term part of deflection due to selfweight after day 45. When the structure is placed in service. There will be added deflection due to sustained portion of live loads (assumed 30% of its design value) and an additional increase in deflection when the remainder of the live load is likely to apply (assumed 70% of its design value) .

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

Note The delayed application of loads does not impact the e ec on c ec va ues or v sua e ec s..

ong- erm e ec on w

mu p e oa ng mes


LONG-TERM DEFLECTION numerical example contd… Condition 2: Delayed application of load - uncracked

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 deflects . 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 oint C followed b lon -term deflection to D’. When in future, the remainder of the design live load (0.7 LL), is applied, the structure deflects to D The total deflection experienced by the members likely to be damaged is shown by “d” 

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 =

.

. <


=

= .

.


Delayed application of load with allowance for cracking

Assumptions Load-deflection in post-cracking is nonlinear . Consequently, superposition 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 deflection calculation.

 Allow

for cracking of concrete, when local stresses exceed modulus of rupture , location and orientation of local reinforcement, including PT is available

uper mpose oa s app e ays a er e 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

lon -term deflection ma nifier C due to shrinkage and creep remains the same as in the uncracked condition ( C = 2.2)

Load-deflection with intermittent loads and cracking

LONG-TERM DEFLECTION numerical example Delayed application of load with allowance for cracking

LONG-TERM DEFLECTION numerical example contd… Delayed application of load with allowance for cracking 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 . Up to here

The instantaneous displacement at application of 0.7LL , deflection. The increment of deflection due to the application of 0.7LL is the difference between two solutions as follows 0.7LL

The load combination for the cracked deflection of the entire sustained loads is: sw,SDL,0.3LL

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

=

sw,SDL,LL

-

= 0.12 in.

sw,SDL,0.3LL

The observed total deflection for visual effects is: (1 + C) *

sw,SDL,0.3LL

+

0.7LL

= (1 + 2.2) * 0.53 + 0.12

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

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

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

LONG-TERM DEFLECTION numerical example For damage control Damage is caused by deflections subsequent to installation . This includes the long-term portion of deflection due to selfweight after day 45.

LONG-TERM DEFLECTION numerical example Delayed application of load with allowance for cracking The long-term effect of superimposed dead load ( The value of this component from the graph is: 0.48 * C *

SDL,LT

) .

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

The load combination to arrive at

The components of design deflection are illustrated in figure

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

sw

= .

sw

= .

sw

.

.

is:

1.0 * SW

= .

n.

The immediate deflection from SDL does not impact the results, since installation of members subject to damage is , SDL,LT

. =C*

SDL

including and the other excluding the superimposed dead load as given below Long-term deflection with multiple loading times SDL

SDL,LT

=

swSDL ,

. =C*

SDL

sw

= 0.1”

= 2.2 * 0.1 = 0.22 in.



Delayed application of load with allowance for cracking 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 0.3LL of the structure (CC’) contributes to the total displacement (d), likely to damage the structure. Its value is given by subtracting the deflection values sw,SDL,0.3LL from sw,SDL to be obtained from two different solutions.

Delayed application of load with allowance for cracking More than two years, the structure is checked for ( 0.7 LL), w en e ec s o po n . e o a e ec on exper ence by the members likely to be damaged is deflection due: 

Entire live load;  long-term effects of the sustained portion of live. sw,LT

oa com na on or

sw,SDL

= 0.40 in.

s 0.3LL,LT

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

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

sw,SDL,0.3LL

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

=

,

-

, .

= 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> 0.3LL,LT

=C*

0.3LL

= 0.15 in.

= 2.2 * 0.05 = 0.15 in.

For

sw,SDL.LL

(1.0 * SW + 1.0 * SDL + 1.0 * LL) = 0.70 in. For sw,SDL (1.0 * SW + 1.0 * SDL) = 0.48 in. LL


= 0.70 – 0.48 = 0.22 in.

DEFLECTION CALCULATION OF CONCRETE FLOORS

Delayed application of load with allowance for cracking

CONCLUDING REMARKS

Total deflection is the sum of the following: 

Long-term effects of selfweight

 

sw,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.

ns an aneous e ec on ue o = . n. LL ___________________________________________________ Total deflection 0.99 in. < Span/480 = 1.00 in. OK TNO_123

 Reasons for controlling deflections  Deflection control through span/depth ratios  How to measure code-intended values  How to calculate probable deflections  Long-term deflections Ultimate values  Interim values 

Slab thickness 8'' 200mm

(a) Floor plan

 Numerical examples

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

26'-3'' (8.00m )

30'-0'' (9.14m)

 Crack width calculation and control

(b) Panel plan

DeflectionConcrete-Nov2010[1]

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