CH 3 - AE Analysis and Design of Two-Way Slabs

8/18/2019 CH 3 - AE Analysis and Design of Two-Way Slabs

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Chapter 3

Analysis and Design of Two-way

Slabs

Abrham E.

Sophonyas A.

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3.0 Introduction

Shear transfer to the column is accomplished by

thickening the slab near the column with drop panels or flaring the top of the column to form a column

capital.

Drop panels commonly extend about one-sixth of the

span in each direction away from the column, giving

extra strength and stiffness in the column region while

minimizing the amount of concrete at mid span.

Slab systems may incorporate beams between someor all of the columns.

The resulting structure is referred to as two-way slabs

with beams.4


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3.2 Distribution of Moments in Slabs Supported

on Stiff Beams and Walls

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3.2 Distribution of Moments in Slabs Supported

on Stiff Beams and Walls

The distribution of the negative moments, m A

, or of the

positive moments, m B, along lines across the slab will be

depicted as shown in Figure (b).

These distributions may be shown as continuous curves,

as shown by the solid lines and shaded areas, or as aseries of steps, as shown by the dashed line.

The height of the curve at any point indicates the

magnitude of the moment at that point.

Occasionally, the distribution of BMs in a strip A-B-C

across the slab will be plotted as shown in Figure (c)

The moments will be expressed in terms of CwL x2, where

L x is the short dimension of the panel. The unit is kNm/m.

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3.3 Analysis of beam/wall supported two way

slabs according to EBCS-2

7

2

x Lwm d ijij


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3.5 Analysis of moments in two-way slabs

Figure (next slide) shows a floor made of simply

supported planks which carries a load of q kN/m2,

supported by simply supported beams.

The moment per meter width in the planks at section

A-A is: m = ql 1 2

/8 kNm/m. The total moment in the entire width of the floor is:

M A-A = (ql 2 )l 1 2 /8 kNm.

This is the familiar equation for maximum moment in

a simply supported floor of width l 2 and span l 1 .

The planks apply a uniform load of ql 1 /2 kN/m on

each beam.

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9

l1

l2


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The moment at section B-B in one beam is thus:

M 1b = (ql 1 /2)l 2 2 /8 kNm/m

The total moment in both beams is:

M B-B = (ql 1 )l 2 2 /8

It is important to note that full load was transferredEast and West by the planks, causing a moment

equivalent to wl 1 2 /8 in the planks where w = ql 2 .

Then the full load was transferred North & South by

beams, causing a similar moment in the beams.

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12

l1

l2


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A similar situation is obtained in the flat plate floor.

Broad strips of the slab centered on the column lines ineach direction serve the same function as beams.

Therefore, for column supported construction (one-

way or two-way), 100% of the applied load must be

carried in each direction, in the case of two-way beam

supported slabs, jointly by the slab and its supporting

beams.

The analysis used to derive the moments in two wayslabs was 1st published by Nichol in 1914, but has not

been fully accepted by ACI until 1971.

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3.6 Distribution of moments in slabs

supported by columns

17

Fig 3.5.1


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3.6 Distribution of moments in slabs supported

by columns

The curvatures and moment diagrams are shown forstrips along lines A-A and B-B.

Both strips have –ve moments adjacent to thecolumns and +ve moments at mid-span.

In Figure 3.5.1(d) the moment diagram from 3.5.1(a) isre-plotted to show the average moments over thewidth of the middle and column strips.

The total static moment, Mo, accounted for here is

(NB: Factor ql n2 gives moment per meter width) M o = ql n

2[(0.122 0.5l 2 ) + (0.041 0.5l 2 ) +(0.053 0.5l 2 ) + (0.034 0.5l 2 )] = 0.125ql n

2

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3.7 Design of Slabs

Two slab design procedures are allowed by the ACI

(EBCS-2). Direct design method

Equivalent frame design method

The two methods differ primarily in the way in whichthe slab moments are computed.

The calculation of the moments in the direct design

method is based on the statical moment Mo.

In this method, the slab is considered panel by panel.

Eq. (A) is used to compute the total moment in each

panel.

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3.7 Design of Slabs

The statical moment is then divided b/n positive and

negative moments, and these are divided b/n middlestrips and column strips.

In the equivalent frame method, the slab is divided

into a series of two-dimensional frames, and the

positive and negative moments are computed via an

elastic frame analysis.

Once the +ve and –ve moments are known, they are

divided b/n middle and column strips in exactly thesame way as in the direct design methods.

Slabs are frequently built with beams from column to

column around the perimeter of the building.21


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3.7.1 Beam-to-slab stiffness ratio, f These beams act to stiffen the edge of the slab and

help to reduce the deflections of the exterior panels ofthe slabs. (Very heavily loaded slabs and long-span

waffle slabs sometimes have beams joining all columns

in the structure).

The effects of beam stiffness on deflections and the

distribution of moments are expressed as a function of

f , defined as the flexural stiffness, 4EI/L, of the beam

divided by the flexural stiffness of a width of slab

bounded by the centerlines of the adjacent panels on

each side of the beam.

f = (4E cb I b /l)/(4E cs I s /l)

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3.7.1 Beam-to-slab stiffness ratio, f Since the length, l , of the beam and the slab are

equal, this quantity is simplified and expressed in the

Code (ACI) as:

f = (E cb I b )/(E cs I s )

If there is no beam, f = 0. (most of the case exceptat the edges where beams are provided for stiffening

edge panels)

The sections considered in computing I b and I s are

shown in Figure (next slide). (NB. Span direction is l 1)

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3.7.1 Beam-to-slab stiffness ratio, f

24

Fig. 3.7.1 Beam and slab sections

for calculations of f

ACI, Section 13.2.4 defines a

beam in monolithic or fullycomposite construction as the

beam stem plus a portion of

the slab on each side of the

beam extending a distanceequal to the projection of the

beam above or below the slab

whichever is greater , but not

greater than four times theslab thickness (see next slide).


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25

Governing

projection


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• Example: Calculation of f for an edge beam:

• A 200 mm-thick slab is provided with an edge beam

that has a total depth of 400 mm and a width of 300

mm as shown in Figure (next slide).• The slab and beam were cast monolithically and have

the same concrete strength and the same E c.

Compute f .

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27

6m

200mm

300mm

400mm

300mm 200mm

200mm

200mm

3150mm

Fig: calculation of f


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• Solution:

• f = I b /I s

1) Compute Ib: The cross section of the beam is as

shown in Figure (slide above). The centroid of the

beam is located 175 mm from the top of the slab. moment of inertia of the beam is: I b =(300 4003 /12) +(300 400) 252+ (200 2003 )/12) +

(200 200) 752 = 2.0333 109 mm4

2) Compute I s: I s = 3150 2003 /12 = 2.1 109 mm4

3) Compute f = 2.0333 109 /2.1 109 = 0.968

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3.7.2 Minimum thicknesses of two-way slabs

ACI code defines minimum thicknesses that are

generally sufficient to limit slab deflections toacceptable values (same as in EBCS-2).

Thinner slabs can be used if it can be shown that thecomputed slab deflections will not be excessive.

Slabs without Beams b/n Interior columns: – 120 mm - Slabs without drop panels,

– 100 mm - Slabs with drop panels.

Slabs with Beams b/n Interior columns:

– 120mm - 0.2 < m < 2.0 :

– 90 mm - m > 2.0 :

– m > 2.0

Where m = average of values for the four sides of the panel. 29


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Slab depth for deflection requirement

According to the EBCS-2, 1995 the effective depth required

for deflection requirement depends on the characteristicstrength of the steel, the position of the panel and the spanratio.

= 0.4 0.6 ∗

400 ∗

f yk - is the characteristic strength of the reinforcement in

Mpa

Le – is the effective length, longer span for flat slabs.

βa – is the appropriate constant value for differentstructural member and support condition given in Table 5.1of EBCS 2, 95, and for slabs carrying partition walls likely tocrack, shall be taken as ≤ 5


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3.7.2 Minimum thicknesses of two-way slabs

31

(MPa)

300

420

520


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3.8 Direct Design Method

The direct-design method could have been called “the

direct-analysis method” because this method essentiallyprescribes values for moments in various parts of the

slab panel without the need for structural analysis.

It has to be noted that this method was introduced in

the era when most engineering calculations were madewith slide rules, and no computer software was

available.

Thus, for continuous-floor slab panels with relatively

uniform lengths and subjected to distributed loading, a

series of moment coefficients were developed that

would lead to safe flexural designs of two-way floor

systems. 32


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3.8 Direct Design Method

• Steps in slab design

1) Choose the layout and type of slab to be used2) Choose the slab thickness (deflection limitation and

shear at both exterior and interior columns)

3) Choose the design method (direct design or equivalent

frame methods)4) Compute +ve and –ve moments in the slab

5) Determine the distribution of the moments across thewidth of the slab

6) If there are beams, a portion of the moments must be

assigned to the beams7) Reinforcement is designed for moments in 5 & 6.

8) The shear strengths at the columns are checked.

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3 8 1 Limitations on the use of the Direct


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3.8.1 Limitations on the use of the Direct

Design Method

1) There must be a minimum of 3 continuous spans in

each direction. Thus a nine-panel structure (3 by 3) is

the smallest that can be divided.

2) Rectangular panels must have a long-span/short-span

ratio not greater than 2. One-way actionpredominates as the span ratio reaches and exceeds 2

3) Successive span lengths in each direction shall not

differ by more than one-third of the longer span

4) Columns should not offset from the basic rectangular

grid of the building more than 0.1 times the span

parallel to the offset.

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3.8.1 Limitations on the use of the Direct

Design Method

5) All loads must be due to gravity only. The directdesign method can not be used for unbraced laterally

loaded frames, foundation mats, or prestressed slabs.

6) The service live load shall not exceed two times the

service dead load (to reduce effects of pattern load).

7) For a panel with beams b/n supports on all sides, the

relative stiffness of the beams in the two directions

given by ( f1l 22 )/( f2l 12 ) shall not be less than 0.2 orgreater than 5. ( is the beam-to-slab stiffness ratiodefined earlier)

35

3 8 2 Di t ib ti f t ithi


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3.8.2 Distribution of moments within

panels-slabs w/o beams b/n all supports

For design, the slab is considered to be a series offrames in the two directions, as shown in Figure

(next slide). These frames extend to the middle of

the panels on each side of the column

In each span of each of the frames, it is necessary to

compute the total statical moment M o:

where wd = factored load per unit area;

l 2 = transverse width of the strip;

l n = clear span between columns 36

8

2

2 nd o

l l w M


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37

Fig: Division of

slab into frames

for design

3 8 2 Di t ib ti f t ithi l


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3.8.2 Distribution of moments within panels

- slabs w/o beams b/n all supports Example: Compute the statical moment, M

o, in the

slab panels in Figure (next 2 slides). The slab is 200

mm thick and supports a live load of 4.54 kN/m2

Sol: (1) Compute the design load:

qd = 1.30.225 + 1.64.54 = 13.76 kN/m2

(2) Consider panel A spanning from column 1 to

column 2. Slab panel A is shown shaded in Figure

(next slide). The moments computed here would be used to

design the reinforcement parallel to lines 1-2 in this

panel .38


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39

600 mm 500 mm

6500 mm

6500 mm

6000 mm

ln

l2

1 2

3 8 2 Di ib i f i hi


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•Now M o = (qd l 2 )l n2 /8; where – l n = clear span of slab panel = 6.5-1/2(0.5)-1/2(0.6) = 5.95m ;

– l 2 = width of panel = 6.5/2 + 6.0/2 = 6.25m

M o = (13.76 6.25 5.95 2 )/8 = 381 kNm• Consider panel B, spanning from column 1 to column 4

(next slide).

• The moments computed here would be used to design

the reinforcement parallel to lines 1-4 in this panel.• For the purpose of computing l n, the circular supports

are replaced by equivalent square columns having a

side length of c1

= 0.886d c

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41

6500 mm5800 mm

6000 mm

1

4

ln

l2

300 mm

600 mm

300 mm

3 8 2 Di t ib ti f t ithi


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l n = 6.0-1/2(0.3)-1/2(0.886 0.6) = 5.59m;

l 2 = 5.8/2 + 6.5/2 = 6.15m;

M o = (13.76 6.15 5.592 )/8 = 331 kNm

Now the total statical moment will be divided b/n thenegative & positive sections of the panel.

In the DDM, the total factored statical moment Mo is

divided into +ve and –ve factored moments accordingto the rules given in ACI Code, Section 13.6.30.

These are illustrated in the Figure (next slide)

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In interior spans, 65% of Mo is assigned to the -ve moment region

35% to the +ve moment region.

The exterior end of an exterior span has considerably

less fixity than the end at the interior support.

The division of Mo in an end span into +ve and –ve

moment regions is given in Table 13.2 below.

In this table, “exterior edge unrestrained” refers to aslab whose edge rests on, but is not attached to, for

example, a masonry wall.

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3.8.3 Positive & Negative Moments in Panels


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44

Assignment of positive- and negative-moment regions

0.35 to 0.63 Mo

0 to 0.65 Mo

0.65 to 0.75 Mo

0.35 Mo

0.65 MoMoMo

Interior span Exterior span


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“Exterior edge fully restrained” refers to a slab whose

exterior edge is supported by, and is continuous with,a concrete wall with a flexural stiffness as large as or

larger than that of the slab.

If the computed –ve moments on two sides of an

interior support are different, the –ve moment section

of the slab is designed for the larger of the two.

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46


L i di l di ib i f S i l


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Longitudinal distribution of Statical moment, Mo

to the critical section (Cont . . . )

Condition of

restraint

The statical moment

at the critical section

(1)

Exterior edgeunrestrained eg. Supported by a masonry wall

(2)

Slabs with

beams b/n

All Supports

Longitudinal distribution of Statical moment M


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Longitudinal distribution of Statical moment, Mo

to the critical section (Cont . . . )

Condition of restraint

Statical moment at the

critical section

Slabs

without

beams

b/n

interior

support

(3)

Without

edge beam eg. Flat Plate

(4)

With edge

beam

(5)

Exterior Edge Fully

restrainedeg. Restrained by monolithic

concrete wall

3 8 4 D fi iti f C l & Middl t i


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3.8.4 Definition of Column & Middle strips

The moments vary continuously across the width of

the slab panels. To aid in steel placement, the design moments are

averaged over the width of column strips over the

columns and middle strips between the column strips.

define column and middle strips Column strips in both directions extend one-fourth of

the smaller span, l min, each way from the column line.

Middle strips are the strips between the columnstrips.

A panel: is bounded by column, beam, or wall

centerlines on all sides.49


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3.8.4 Definition of Column strips & Middle

strips

50

lmin

lmax

3 8 5 Distribution of moments b/n middle strips and


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3.8.5 Distribution of moments b/n middle strips and

column strips-slabs w/o beams b/n interior supports

ACI Section 13.6.4 defines the fraction of the negativeand positive moments assigned to the columns strips.

The remaining amount of negative and positive

moment is assigned to the adjacent half-middle strips.

Table 13.3 gives the percentage of factored negative

moment assigned to the column strip at all interior

supports.

The division is a function of ( f1l 2 /l 1 ) ,which depends onthe aspect ratio of the panel, l 2 /l 1, and the relative

stiffness, f1, of the beams (if any) spanning parallel to

and within the column strip.

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52


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For floor systems w/o interior beams, ( f1l 2 /l 1 ) is takenequal to zero, since f1 = 0.

In this case 75% of the negative moment is distributed

in the column strip, and the remaining 25% is dividedequally b/n the two adjacent half middle strips.

For cases where a beam is present in a column strip

(spanning in the direction of l 1) and ( f1l 2 /l 1) 1.0, the

second row in table 13.3 applies.

For 0 ( f1l 2 /l 1) 1.0 use linear interpolation.

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Table 13.4 gives the percentage of factored positive

moment assigned to the column strip at mid span for

both interior and exterior spans.

For floor systems w/o interior beams, 60% of the +ve

moment is assigned to the column strip and theremaining 40% is divided equally b/n the adjacent half

middle strips.

If a beam is present in the column strip (spanning in

the direction of l1), either the percentages in the 2nd row or a linear interpolation b/n the percentages given

in the 1st or 2nd row in Table 13.4 will apply.

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56


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• At an exterior edge, the division of the exterior-end

factored negative moment distributed to the column

and middle strips spanning to the edge also dependson the torsional stiffness of the edge beam, β

t.

• βt is calculated as the shear modulus, G, times the

torsional constant of the edge beam, C, divided by the

EI of the slab spanning to the edge beam (i.e., EI for

a slab having a width equal to the length of the edgebeam from the center of one span to the center of the

other span) (see Figure)

57

3 8 5 Distribution of moments b/n middle strips and


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Assuming that = 0 G = E/2 so that

The term C is the torsional constant of the edge beamwhich is calculated by subdividing the cross section into

rectangles and carrying out the summation:

where x = shorter side of a rectangle and y = longer side

(NB: Several possible combination of rectangles have tobe tried to get the maximum value of C. To do so, widerectangles should be made as large as possible.

58

363.01

3 y x

y

xC

scs

cb

t I E

C E

2


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59



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Table 13.5 gives percentage of negative factored

moment assigned to column strip at exterior supports.

The set up of this table is similar to the previous ones

(tables 13.3 and 13.4) with the addition of two rows to

account for presence or absence of an edge beam

working in torsion to transfer some of the slab

negative moment into the column.

When there is no edge beam (t = 0), all of the

negative moment is assigned to the column strips. This is reasonable because there is no torsional edge

member to transfer moment from the middle strips all

the way back to the columns.60

/


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61

( )


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• If a stiff beam is present ( t 2.5), the table gives specificpercentages to be assigned to the column strip, depending

on the value of f1 and the l 2 /l 1 ratio, as was done in the

previous tables.

• For values of t between 2.5 and 0.0 and values of ( f1l 2 /l 1 )

b/n 1.0 and 0.0, two or three levels of linear interpolation

may be required to determine the percentage distribution

of negative moment assigned to the column strip.

• If a beam is present in the column strip (spanning in the

direction of l 1), a portion of the column-strip moment is

assigned to the beam (ACI Code, Section 13.6.5).

• If the beam has ( f1l 2 /l 1 ) >1, 85% of the column-strip

moment is assigned to the beam and 15% to the slab.

62

Factored moment in column strip


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p

Aspect ratio of the panel (L2/L1)

0.5 1.0 2.0

Interior negative

moment

1 L2/L1 = 0 75 75 75

1 L2/L1 1.0 90 75 75

Exterior

negative

moment

1 L2/L1 = 0

t = 0 100 100 100

t 2.5 75 75 75

1 L2/L1 1.0

t = 0 100 100 100

t 2.5 90 75 45

Positive Moment1 L2/L1 = 0 60 60 60

1 L2/L1 1.0 90 75 45

3 8 6 Example


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3.8.6 Example

• Calculation of moments in an exterior panel of a flat

plate

• The slab is 200 mm thick and supports a superimposed

service dead load of 0.5 kN/m2 and a service live load

of 3 kN/m2. the beam is 300 mm wide by 400 mm in

overall depth and is cast monolithically with the slab.(1) Compute the factored loads:

wd = 1.3(0.2 25 + 0.5) +1.6 3 = 12 kN/m2

(2) Compute the moments in span BE. (2a) Compute l n and l 2 and divide the slab into middleand column strips.

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3.8.6 Example

65

l1=6.5 m

0.35m

0.40m

6.0m

5.5m

1.5m

1.5m

1.375m

1.375m

l2=5.75 m

3.8.6 Example


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3.8.6 Example

l n = 6.5-1/2(0.35)-1/2(0.4) = 6.125m; l 2 = 5.75m.The column strip extends the smaller of l

2/4 or

l1/4 on each side of the column centerline.

The column strip extends 6/4 = 1.5 m toward ADand 5.5/4 = 1.375 m toward CF from line BE as

shown. The total width of the column strip is 2.875 m.The half middle strip b/n BE and CF has a width of

1.375 m, and the other one is 1.5 m

(2b) Compute M o:

66

kNml l w

M nd o 6.3238

125.685.512

8

22

2

3 8 6 Example


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3.8.6 Example

(2c) Divide M o into positive and negative moments. – The distribution of the total factored moment to the

negative and the positive moment regions is as given

in Table 13-2 (slide 81) under the column “slabs w/o

beams b/n interior supports with edge beam”

– From Table 13-2, the total moment is divided as

follows:

Interior negative: M u = 0.70M o = -226.5 KNm

Positive: M u = 0.50M o = +161.8 KNmExterior negative: M u = 0.30M o = -97.1 KNm

67

3 8 6 E ample


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3.8.6 Example

• (2d) Divide the moments b/n the column and middle

strips – Interior negative moments: This division is a

function of f1l 2 /l 1, which is equal to zero, since

there are no beams to BEInterior column-strip negative moment: 0.75 -

226.5 = -169.9 kNm

i.e. -169.9/2.875= -59.1 kNm/m width of column strip

Interior middle-strip negative moment = -56.6 kNm.Half of this goes to each of the half middle strips

68

3 8 6 Example


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3.8.6 Example

– Positive moments: From ACI Section 13.6.4.4

Column-strip positive moment: 0.60161.8 = 97.1kNm 33.8 kNm/m

Middle-strip positive moment = 64.7 kNm. Half ofthis goes to each half-middle strip.

– Exterior negative moment: From ACI Section13.6.4.2, the exterior negative moment is divided

as a function of f1l 2 /l 1 (again equal to zero, if

there is no beam to l 1

) and t

. See next slide for

attached torsional member for which t will be

calculated

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3.8.6 example

70

300 mm 200 mm

400 mm

200 mm

500 mm

200 mm

300 mm

Fig. 13-30 Slab, column and edge beam

200 mm

3.8.6 example


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3.8.6 example

– For Fig (a): C=[(1-0.63 300/400)3003 400/3+(1-0.63 200/200)2003 200/3] = 2096.3 106 mm4

– For Fig (b): C = 1461.3 106 mm4. The larger of thevalues is used; C = 2096.3 106 mm4

– I s the moment of inertia of the strip of slab beingdesigned, which has b=5.75m and h=200mm.

• I s=5750 2003 /12 = 3833.3 106 mm4 • Since f ck is the same in the slab and beam, E cb=E cs and

t = 2096.3 106 /(2 3833.3 106 ) = 0.273

– Interpolating in Table 13-5, we have:

• For t = 0 100% to column strip

• For t = 2.5 75% to column strip

• for t =0.273 97.3% to column strip and we have:

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3.8.6 example

• Exterior column-strip negative moment:0.973(-97.1)= -94.5 kNm = -32.9 kNm/m

• Exterior middle-strip negative moment: -2.6 kNm

•

Transfer of moments to columns• Shear Strength of Two-way Slabs

• Reinforcement and detailing

72

3 9 f f


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3.9 Transfer of Moments

Exterior Columns: – Moment on an edge column = 0.26 M o - 0.3M o (In

slabs without interior beams)

– The moment is assumed to be about the centroid

of the shear perimeter

– Divided between the column above and below the

slab in proportion to column stiffnesses, 4EI/L.

– The resulting column moments are used in thedesign of the columns.

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3.9 Transfer of Moments

Interior columns:

– Column design moment determined from

unbalanced moment of adjacent spans when longer

span is loaded with factored g d +0.5qd and shorter

with g d only. Unbalanced moment, M

–

Most of the moment is transferred to the columnswhich is used to design the slab-to-column joint

74

88

5.065.0

2

2

2

2 n Dn D D l l g l l q g M

222

25.007.0 n Dn D Dcol l l g l l q g M


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3.10 Shear Strength of Two-way Slab

•Shear Strength (punching shear) of Two-waySlabs without beams

– Design shear resistance for a slab without shear

reinforcement according to EBCS 2, 1995:

75

015.0

2

][0.16.1

0.2501

25.0

2

1

211

eyexe

y x

e

ctd Rd

d d d

metersind d k

k where

d uk k f V

3 10 Sh St th f T Sl b


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Location of the critical perimeter

76


3 10 Sh St th f T Sl b


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Applied load effect

– Centric load or reaction:

–Eccentric load, the equivalent centric load V eq

•

e = eccentricity of the load w.r.t. the centroid of thecritical section; always positive

• Z = section modulus of the critical section correspondingto the direction of the eccentricity

77

1 Rd Sd V V

Sd eq V V

Z d uewhere /1



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3.10 Shear strength of two-way slab

• fraction of the moment transferred by

the eccentricity

• b1 , b2 sides of the rectangle, b1 || to the direction of e

• u = perimeter of the critical section

•

d = effective depth Conservatively, for flat slabs with approx. equal spans

or footings:

– = 1.15 for interior column

– = 1.40 for edge column

– = 1.50 for corner column

78

)/1/(112

bb


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Design shear resistance for a slab with shearreinforcement according to EBCS 2, 1995:

If V sd < V Rd , no shear reinforcement is required. If V sd > V Rd1, punching failure Increase the

capacity by:

Using a drop panel to thicken the slab adjacent to the

column Increasing the column size or a capital around the column

Adding shear reinforcement near the loaded area

79

12 6.1 Rd Rd V V



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Shear Reinforcement

Fig. 13-66 Stirrup-type shear

reinforcement in two-way slabs.

80


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Shear Reinforcement

81

Fig. Headed shear studs.

Critical perimeter for columns with drop panels


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or column capitals

• l H ≤ 1.5 hH – Circular column

– rectangular column

Smaller of

82

c H crit l l d r 5.05.1

ccrit

cccrit

bd r hbd r

64.05.156.05.1

Drop panel Capital



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p p p

or column capitals

•l H > 1.5 hH – Within the drop panel/capital

– Outside the drop panel/capital

83

Drop panel Capital

c H crit l hd r 5.05.1

c H crit l l d r 5.05.1

d ili


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detailing

84

Equivalent Frame Method


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q

It is a more general and more rigorous method than

DDM and is not subject to the limitations of DDM.

The ‘equivalent frame’ concept simplifies the analysis of

a 3D RC building by subdividing it into a series of 2D

(plane) frames (‘equivalent frames’) centered on column

lines in longitudinal as well as transverse directions. The ‘equivalent frame method’ differs from DDM in the

determination of the total ‘negative’ and ‘positive’

design moments in the slab panels — for the condition

of gravity loading.

However, the apportioning of the moments to ‘column

strips’ and ‘middle strips’ (or to beam & slab) across a

panel is common to both methods.



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86

The resultant rigid frames in both directions are analyzed

separately for carrying all the loads in each direction.

q



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87

q

Fig. (b) Equivalent frame for analysis

The bending moments and shear forces are obtained


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The bending moments and shear forces are obtained

in EFM by an elastic analysis

The load transfer in the equivalent frame involvesthree distinct interconnected elements.

the slab-beam members (along span l 1);

the columns (or walls); and

the torsional members, transverse to the frame (along

span l 2) and along the column lines.

To account for the increased flexibility of the slab-to-

column connection, it is recommended to use anequivalent column with stiffness Kec to replace the

actual columns and torsional members.


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Fig. (a) Elements of equivalent frame at a connection


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Fig. (b) Equivalent frame for analysis

For gravity-load analysis, ACI allows analysis of an entire

equivalent frame extending over the height of the

building, or each floor can be considered separately

with the far ends of the columns being fixed.(Comprising horizontal slab-beam members and vertical

‘equivalent columns’)

Properties of Slab –Beams


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p

The horizontal members in the equivalent frame are

referred to as slab-beams. These consist of either only a slab, or a slab and a drop

panel, or a slab with a beam running parallel to the

equivalent frame.

ACI allows determination of the moment of inertia of

slab-beams at any cross section outside of joints or

column capitals using the gross area of concrete.

Variations in the moment of inertia along the lengthshall be taken into account.

Thus for the slab with a drop panel shown in Fig 13


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Thus, for the slab with a drop panel shown in Fig. 13-

32a, the moment of inertia at section A – A is that for a

slab of width l 2 (Fig. 13-32c).At section B –B through the drop panel, the moment of

inertia is for a slab having the cross section shown in

Fig. 13-32d.

Similarly, for a slab system with a beam parallel to l 1 as

shown in Fig. 13-33a, the moment of inertia for

section C –C is that for a slab-and-beam section, as

shown in Fig. 13-33c.

Section D –D is cut through a beam running

perpendicular to the page.

M t f i ti f l b b f t f


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Moment of inertia of slab-beams from center of

column to face of column, bracket, or capital shall be

assumed equal to the moment of inertia of the slab-beam at face of column, bracket, or capital divided by

the quantity (1 – c 2 / l 2)2, where c 2 and l 2 are measured

transverse to the direction of the span for which

moments are being determined.


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Fig. 13-32 EI values for a

slab with a drop panel.


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Fig. 13-33 EI values for a

slab and beam.

Properties of Columns


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In computing the stiffnesses and carryover factors forcolumns, ACI Code Section 13.7.4 states the following:

1. The moment of inertia of columns at any cross sectionoutside of the joints or column capitals may be basedon the gross area of the concrete, allowing forvariations in the actual moment of inertia due to

changes in the column cross section along the length ofthe column.

2. The moment of inertia of columns shall be assumed tobe infinite within the depth of the slab-beam at a joint.

Figure 13-37 illustrates these points for four commoncases. Again, the column analogy can be used to solvefor the moment-distribution constants, or table valuescan be used.


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Fig. 13-37 Sections for the calculations of column stiffness, Kc .

Torsional Members and Equivalent Columns


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In the equivalent frame method of analysis, columnsare considered to be attached to the continuous slabbeam by torsional members transverse to the directionof the span for which moments are being found.

Torsional deformation of these transverse supportingmembers reduces the effective flexural stiffnessprovided by the actual column at the support.

The equivalent column consists of the actual columnsabove and below the slab-beam, plus attached

torsional members on each side of the columnsextending to the centerline of the adjacent panels asshown below.



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In conventional plane frames, the torsional

members are absent, and the skeletal framecomprises only beams and columns.

When conventional frames were used for the

analysis of flat slab, it was found that the test

values of the span moments were more and the

support moments were less than the theoretical

values. This showed that the columns sides werenot as rigid as imagined.

100

In fig 13-38 a & b, the ends ofth b & l th


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Fig. 13-38 Frame action and twisting of

edge member.

the beam & column or theslab & wall undergo equal

rotation. In fig 13-38 c, however, the

rotation at A of strip A –B isgreater than the rotation at

point C , because there is lessrestraint to the rotation ofthe slab at this point.

In effect, the edge of the slab

has twisted, as shown in Fig.13-38d.

The stiffness of the equivalent column, Kec, representsthe combined stiffnesses of the columns and attached


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the combined stiffnesses of the columns and attachedtorsional members:

The stiffness, K ec

, of the equivalent column is obtainedby taking the equivalent (or effective) flexibility(inverse of stiffness) of the connection as equal to thesum of the flexibilities of the actual columns and thetorsional member

Kec = average rotation of the edge beam

=

=

+

Where: Kec = Flexural stiffness of equivalent columnKc = flexural stiffness of actual column

Kt = torsional stiffness of edge beam

If the torsional stiffness of the attached torsional


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If the torsional stiffness of the attached torsional

members is small, will be much smaller than The torsional Stiffness Kt can be calculated by:

= 92 1 − 2 2

Where: Ecs = modulus of elasticity of slab concrete

c2 = size of rectangular column, capital, or bracket

in the direction of l2.

C = cross sectional constant (roughly equivalent topolar moment of inertia)

The torsional constant C can be calculated by:


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y

= 1 − 0.63

3

Where: x is the shorter side of a rectangle and y is

the longer side.

C is calculated by sub-dividing the cross section oftorsional members in to component rectangles and the

sub-division is to maximize the value of C.

Torsional members shall be assumed to have a constant

cross section throughout their length consisting of thelargest of (a), (b), and (c) below:

(a) A portion of slab having a

width equal to that of the


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Fig. 13-40 Torsional

members.

width equal to that of the

column, bracket, or capital in

the direction of the span for

which moments are being

determined;

(b) For monolithic or fully

composite construction, the

portion of slab specified in (a)

plus that part of the transverse

beam above and below the slab;

(c) The transverse beam as

defined in 13.2.4.

Where beams frame into columns in the direction of


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Where beams frame into columns in the direction of

the span for which moments are being determined

(parallel to l1), the torsional stiffness shall be multipliedby the ratio of the moment of inertia of the slab with

such a beam to the moment of inertia of the slab

without such a beam.

In a moment-distribution analysis, the frame analysis iscarried out for a frame with slab-beams having

stiffnesses Ks and with equivalent columns having

stiffnesses Kec

.

Arrangement of Live Loads for Analysis


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If the un-factored live load does not exceed 0.75 times

the un-factored dead load, it is not necessary to considerpattern loadings.

If the un-factored LL exceeds 0.75 times the un-factoredDL the following pattern loadings need to be considered.

For maximum positive moment, factored dead load on allspans and 0.75 times the full factored live load on thepanel in question and on alternate panels.

For maximum negative moment at an interior support,

factored dead load on all panels and 0.75 times the fullfactored live load on the two adjacent panels.

The final design moments shall not be less than for thecase of full factored and live loads on all panels.

Stiffness factors, carry-over factors and fixed-end moments


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108

Moment distribution factor for slabs without drop panel

y



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109

Moment distribution factor for slabs with Drop Panels; with 1.25 thickness ratio



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110

Moment distribution factor for slabs with Drop Panels;

with 1.50 thickness ratio of drop panel and slab



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Stiffness and

Carryover

factors forcolumns

CH 3 - AE Analysis and Design of Two-Way Slabs

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Transcript of CH 3 - AE Analysis and Design of Two-Way Slabs