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Reinforced Concrete Slabs TYPES OF SLABS In reinforced concrete construction, slabs are used to

provide flat, useful surfaces. A reinforced concrete slab is abroad, flat plate, usually horizontal, with top and bottomsurfaces parallel or nearly so. It may be supported by

reinforced concrete beams (and is usually cast monolithicallywith such beams), by masonry or reinforced concrete walls, bystructural steel members, directly by columns, or continuouslyby the ground.

Slabs may be supported on two opposite sides only, as shownin Fig. 1 a, in which case the structural action of the slab isessentially one-way, the loads being carried by the slab in thedirection perpendicular to the supporting beams.

There may be beams on all four sides, as shown in Fig. 1 b, sothat two-way slab action is obtained. Concrete slabs in somecases may be carried directly by columns, as shown in Fig. 1 d,without the use of beams or girders. Such slabs are describedas flat plates and are commonly used where spans are notlarge and loads not particularly heavy. Flat slab construction,shown in Fig. 1 e, is also beamless but incorporates a thickenedslab region in the vicinity of the column and often employs

flared column tops. Both are devices to reduce stresses due toshear and negative bending around the columns. They arereferred to as drop panels and column capitals, respectively.Closely related to the flat plate slab is the two-way joist, alsoknown as a grid or waffle slab, shown in Fig. 1 f . To reduce thedead load of solid-slab construction, voids ar formed in arectilinear pattern through use of metal or fiberglass forminserts. A two way ribbed construction results. Usually inserts

are omitted near the columns, so a solid slab is formed to resistmoments and shears better in these areas

FIGURE 1

Types of structural slabs

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DESIGN OF ONE-WAY

SLABS The structural action of a one-way slab may be visualized in terms of

the deformed shape of the loaded surface. Figure 2 shows a

rectangular slab, simply supported along its two opposite long edges

and free of any support along the two opposite short edges. If a

uniformly distributed load is applied to the surface, the deflected

shape will be as shown by the solid lines. Curvatures, andconsequently bending moments, are the same in all strips s spanning

in the short direction between supported edges, whereas there is no

curvature, hence no bending moment, in the long strips I parallel to

the supported edges. The surface approximately cylindrical.

For purposes of analysis and design, a unit strip of such a slab cut out

at right angles to the supporting beams, as shown in Fig. 2, may be

considered as a rectangular beam of unit width, with a depth h equalto the thickness of the slab and a span la equal to the distance

between supported edges. This strip can then be analyzed by the

methods that were used for rectangular beams, the bending moment

being computed for the strip of unit width. The load per unit area on

the slab becomes the load per unit length on the slab strip. The loads

recommended by ASCE for different usage of slabs are shown in

table 1. Since all of the load on the slab must be transmitted to the

two supporting beams, it follows that all of the reinforcement should

be placed at right angles to these beams, with the exception of any

bars that may be placed in the other direction to control shrinkage

and temperature cracking. A one-way slab, thus, consists of a set of

rectangular beams side by side.

This simplified analysis, which assumes Poisson's ratio to be zero, is

slightly conservative. Actually, flexural compression in the concreteFig. 2

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in the direction of la will result in lateral expansion in the direction of

lb unless the compressed concrete is restrained. In a one-way slab, this

lateral expansion is resisted by adjacent slab strips, which tend to

expand also. The result is a slight strengthening and stiffening in the

span direction, but this effect is small and can be disregarded.

Factored moments and shears in one-way slabs can be found either by

elastic analysis or through the use ofthe same coefficients as used for

beams . If the slab rests freely on its supports, the span length may be

taken equal to the clear span plus the depth of the slab but need not

exceed the distance between centers of supports, according to ACI

Code 8.9.1. In general, center-to-center distances should be used in

continuous slab analysis, but a reduction is allowed in negative

moments to account for support width as discussed in Chapter 12. For

slabs with clear spans not more than 10 ft that are built integrally withtheir supports, ACI Code 8.9.4 permits analysis as a continuous slab on

knife edge supports with spans equal to the clear spans and the width

of the beams otherwise neglected. If moment and shear coefficients

are used, computations should be based on clear spans.

ACI Code 9.5.2 specifies the minimum thickness in Table 2 for

nonprestressed slabs of normal weight concrete (wc = 145 pcf) using

Grade 60 reinforcement, provided that the slab is not supporting orattached to construction that is likely to be damaged by large

deflections. Lesser thicknesses may be used if calculation of

deflections indicates no adverse effects. For concretes having unit

weight wc in the range from 90 to 115 pcf, the tabulated values should

be multiplied by 1.65 - 0.005wc' but not less than 1.09. For

reinforcement having a yield stress other than 60,000 psi, the

tabulated values should be multiplied by 0.4 +1/100,000. Slab

deflections may be calculated, if required, by the same methods as forbeams . The total slab thickness h is usually rounded to the next higher

Table. 2

Minimum thickness h of

non prestressed one-way slabs

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in. for slabs up to 6 in. thickness, and to the next higher in. for

thicker slabs.

Shear will seldom control the design of one-way slabs, particularly if

low tensile reinforcement ratios are used. It will be found that the

shear capacity of the concrete Vc will, almost without exception, be

well above the required shear strength Vn at factored loads. The concrete protection below the reinforcement should follow the

requirements of ACI Code 7.7.1, calling for in. below the bottom of

the steel . In a typical slab, 1 in. below the center of the steel may be

assumed.

The lateral spacing of the bars, except those used only to control

shrinkage and temperature cracks (see Section 13.3), should not

exceed 3 times the thickness h or 18 in., whichever is less, according

to ACI Code 7.6.5. Generally, bar size should be selected so that the

actual spacing is not less than about 1.5 times the slab thickness, to

avoid excessive cost for bar fabrication and handling. Also, to reduce

cost, straight bars are usually used for slab reinforcement, cut off

where permitted are as described for beams .

Since concrete is weak in tension, these temperature and shrinkage

stresses are likely to result in cracking. Cracks of this nature are not

detrimental, provided their size is limited to what are known ashairline cracks. This can be achieved by placing reinforcement in the

slab to counteract contraction and distribute the cracks uniformly. In

one-way slabs, the reinforcement provided for resisting the bending

moments has the desired effect of reducing shrinkage and

distributing cracks. However, as contraction takes place equally in all

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directions, it is necessary to provide special reinforcement for

shrinkage and temperature contraction in the direction

perpendicular to the main reinforcement. This added steel is

known as temperature or shrinkage reinforcement, or

distribution steel.

Reinforcement for shrinkage and temperature stresses normal

to the principal reinforcement should be provided in a

structural slab in which the principal reinforcement extends in

one direction only. ACI Code 7.12.2 specifies the minimum

ratios of reinforcement area to gross concrete area (i.e., based

on the total depth of the slab) shown in Table 13.2, but in no

case may such reinforcing bars be placed farther apart than 5

times the slab thickness or more than 18 in. In no case is the

reinforcement ratio to be less than 0.0014. The steel required

by the ACI Code for shrinkage and temperature crack control

also represents the minimum permissible reinforcement in the

span direction of oneway slabs; the usual minimums for

flexural steel do not apply.

Table. 3

Minimum ratios of temperature and shrinkage reinforcementin slabs based on gross concrete area

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DESIGN LIMITATIONS ACCORDING TO THE ACI CODE

The following limitations are specified by the ACI Code.

1. A typical imaginary strip 1ft (or 1m) wide is assumed.

2. The minimum thickness of one-way slabs using grade 60

steel according to the ACI Code,

for solid slabs and for beams or ribbed one-way slabs

should be equal to the following:

For simply supported spans: solid slabs, h = Ll20 (ribbed

slabs, h = L/16).

For one-end continuous spans: solid slabs, h = Ll24

(ribbed slabs, h = Ll18.5).

For both-end continuous spans: solid slabs, h = Ll28

(ribbed slabs, h = Ll21).

For cantilever spans: solid slabs, h = LItO (ribbed slabs,

h = Ll8).

Forfy other than 60 ksi, these values shall be

multiplied by 0.4 + 0.01 fy, where fy is in ksi.

This minimum thickness should be used unless

computation of deflection indicates a lesser thickness

can be used without adverse effects.

3. Deflection is to be checked when the slab supports are

attached to construction likely to be damaged by large

deflections. Deflection limits are set by the ACI Code,

Table 9.5b.

4. It is preferable to choose slab depth to the nearest in.

(or to mm).

5. Shear should be checked, although it does not usually

control.

6. Concrete cover in slabs shall not be less than in. (20

mm) at surfaces not exposed to weather or ground. In

this case, d = h - (3/4 in.) - (half-bar diameter).

7. In structural slabs of uniform thickness, the minimum

amount of reinforcement in the direction of the span

shall not be less than that required for shrinkage andtemperature reinforcement (ACI Code, Section 7.12). .

8. The principal reinforcement shall be spaced not farther

apart than three times the slab thickness nor more

than 18 in. (ACI Code, Section 7.6.5).

9. Straight-bar systems may be used in both tops and

bottoms of continuous slabs. An alternative bar

system of straight and bent (trussed) bars placedalternately may also be used.

10. In addition to main reinforcement, steel bars at right

angles to the main must be provided. This additional

steel is called secondary, distribution, shrinkage, or

temperature reinforcement

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One-way slab design. Example.

1. A reinforced concrete slab is built integrally with its supports

and consists of two equal spans, each with a clear span of 15 ft.

The service live load is 100 psf,v and 4000 psi concrete is

specified for use with steel with a yield stress equal to 60,000

psi.Requiremaent

2. Design the slab, following the provisions of the ACI Code.

Solution

3. Thickness of Slab.

h= L/28 Table 3

h= 15x12/28 = 6.43 in Say 6.5 in

4. Factored Loads

Dead Load = 1.2x( 6.5x150/12) = 97 psf

Live Load = 1.6x 100 = 160

Total = 257

5.For this case, factored moments at critical sections may be

found using the ACI moment coefficients(see Table 12.1):

6. Max Steel ratio

Notes

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7. Determine As for max negative BM (6.43 lb ft)

( Assume a=1 and find As)

( Check a for As determined )

As = 0.27 sq in at mid support

As = 0.17 sq in at mid span

As = 0.1 sq in at ext support

8. Temperature and shrinkage Reinforcement

As = 0.0018 X 12 X 6.50 = 0.14 in2

9. Shear Force .

The factored shear force at a distance d from the face of the interiorsupport is

Ok

10. Check spacing of reinforcement etc

Notes

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Two -Way Slab

When the slab is supported on all four sides and the length is

less than twice the width. The slab will deflect in two

directions. and the loads on the slab arc transferred to all four

supports. This slab is referred to as a two-way slab. The

bending moments and deflections in such slabs are less than

those in one-way slabs: thus. the same slab can carry more

load when supported on four sides. The load in this case is

carried in two directions, and the bending moment in each

direction is much less than the bending moment in the slab if

the load were carried in one direction only. Typical two-way

slabs is shown in Fig. 1

To visualize its flexural performance, it is convenient to think of

it as consisting of two sets of parallel strips, in each of the two

directions, intersecting each other. Evidently, part of the load iscarried by one set and transmitted to one pair of edge

supports, and the remainder by the other. Figure 1 shows the

two center strips of a rectangular plate with short span la and

long span lb' If the uniform load is q per square foot of slab,

each of the two strips acts approximately as a simple beam,

uniformly loaded by its share of q. Because these imaginary

strips actually are part of the same monolithic slab, their

deflections at the intersection point must be the same.

Equating the center deflections of the short and long strips

gives

Fig. 1

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where qais the share of the load q carried in the short direction

and qb is the share of the load q carried in the long direction.

Consequently,

One sees that the larger share of the load is carried in the short

direction, the ratio of the two portions of the total load being

inversely proportional to the fourth power of the ratio of the

spans.

This result is approximate because the actual behavior of a slab is

more complex than that of the two intersecting strips. Forinstancefor a simply supportId square slab, qa = qb = q/2 . If only

bending were present, the maximum moment in each strip would

be

The exact theory of bending of elastic plates shows that actually

the maximum moment in such a square slab is only O.048ql2, sothat in this case the twisting moments relieve the bending

moments by about 25 percent. Inelastic redistribution also

modifies these bending moments further. A complicated situation

is therefore obtained. Various design method are suggested to

resolve this situation.

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Choice of Concrete Slabs Various types of floor systems can be used for general

buildings. such as residential. office. And institutional buildings.

The choice of an adequate and economic floor system depends

on the type of building. architectural layout. aesthetic features.

and the span length between columns. In general. the

superimposed live load on buildings varies between 80 and 150

psf. A general guide for the economical use of Boor systems can

be summarized as follows.

Flat plates are most suitable for spans of 20 to 25 ft and live

loads between 60 and 100 psf. The advantages of adopting flat

plates include low-cost formwork, exposed fl al ceilings. and

fast construction. Flat plates have low shear capacity and

relatively low stiffness. which may cause noticeable deflection.

Flat plates are widely used in buildings either as reinforced orprestressed concrete slabs.

Flat slabs are most suitable for spans of 20 to 30 ft and for live

loads of 80 to 150 psf. They need more fonnwork than flat

plates. especially for column capitals. In most cases. only drop

panels without column capitals are used.

Waffle slabs are suitable for spans of 30 to 48 ft and live loads

of 80 to 150 psf They carry heavier loads than flat plates andhave attractive exposed ceilings. Formwork. including the use

of pans. is quite expensive.

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Slabs on beams are suitable for spans between 20 and 30 ft and live

loads of 60 to 120 psf . The beams increase the stiffness of the slabs.

producing relatively low deflection. Additional form wor k for the beams

is needed.

One-way slabs on beams are most suitable for spans of 10 to 20 ft and a

live load of 60 to 100 psf. They can be used for larger spans with

relatively higher cost and higher slab deflection . Additional formworkfor the beams is needed.

A one-way joist floor system is most suitable for spans of 20 to 30 ft and

live loads of 80 to 120 psf. Because of the deep ribs. the concrete and

steel quantities are relatively low, but expensive formwork is expected.

The exposed ceiling of the slabs may look attractive.

Two-way slabs a re extremely complex and statically indeterminate.Many attempts, analytical and empirical, have been made to determine

the division of the moments an d shears between the two spans and the

distribution of these along the principal axes of the slab. Many elastic

analyses of .two-way slabs have been proposed. All have shortcomings.

They neglect Poisson's ratio, torsion, changes in stiffness, ultimate

capacity, edge restraint, variation of moments and shears along span,

and others.

The direct design method. DDM (ACI Code. Section 13.6). is an

approximate procedure for the analysis and design of two-way slabs. It is

limited to slab systems subjected to uniformly distributed loads and

supported on equally or nearly equally spaced columns. The method

uses a set of coefficients to determine the design moments at critical

sections. Two-way slab systems that do not meet the limitations of the

ACI Code. Section 13.6.1, must be analyzed by more accurate

procedures.

Design of Two-way Slabs

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The equivalent frame method. EFM (ACI Code. Section 13.7). isone in which a three dimensional building is divided into a seriesof two-dimensional equivalent frames by cutting the buildingalong lines midway between columns. The resulting frames areconsidered separately in the longitudinal and transverse directionsof the building and treated 1100r by floor. as shown in Fig.

In elastic analysis, a concrete slab may be treated as an elastic

plate. The flexure, shear, and deflection may be calculated by thefourth differential equation relating load to deflection for thinplates with small displacements, as presented by Timoshenko .Finite difference as well as finite element solutions have beenproposed to analyze In the finite element method, the slab isdivided into a mesh of triangles or quadrilaterals. Thedisplacement functions of the nodes (intersecting mesh points)are usually established, and the stiffness matrices are developedfor computer analysis slabs and plates.

For plastic analysis, three methods are available. The yield linemethod was developed by Johansen to determine the limit state ofthe slab by considering the yield lines that occur in the slab as acollapse mechanism. The strip method was developed byHillerborg . The slab is divided into strips, and the load on the slabis distributed in two orthogonal directions. The strips are analyzedas simple beams. The third method is optimal analysis. There hasbeen considerable research into optimal solutions. Rozvany andothers presented methods for minimizing reinforcement based onplastic analysis. Optimal solutions are complex in analysis andproduce complex patterns of reinforcement.

Nonlinear analysis simulates the true load deformationcharacteristics of a reinforced concrete slab when the finiteelement method takes into consideration the nonlinearity of thestress strain-relationship of the individual elements [11,12]. In this

case, the solution becomes complex unless simplified empiricalrelationships are assumed.

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Coefficient Method This method has been used extensively since 1963 for slabs

supported at the edges by walls, steel beams, or monolithic

concrete beams having a total depth not less than about 3 times

the slab thickness. While it was not a part of the 1977 or later ACI

Codes, its continued use is permissible under the ACI 318-08 code

provision (13.5.1) that a slab system may be designed by any

procedure satisfying conditions of equilibrium and geometric

compatibility, if it is shown that the design strength at every

section is at least equal to the requires strength, and that

serviceability requirements are met.

The method makes use of tables of moment coefficients for a

variety of conditions. These coefficients are based on elastic

analysis but also account for inelastic redistribution. In

consequence, the design moment in either direction is smaller byan appropriate amount than the elastic maximum moment in that

direction.

Table 1-4 gives coefficients for different conditions for bending

moments and shear force. These also specify formula to

determine design parameters. The method since falls with in

framework of ACI 318-08; all other specification of code do govern

relevant design parameters. Span length of members not built integrally with supports shall be

considered as the clear span plus the depth of the member, but

need not exceed distance between centers of supports. ACI 8.9.1

Deflection of two-way slab should not exceed the limits specified

in table 9.5(b)

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TABLE 9.5(b)MAXIMUM

PERMISSIBLE COMPUTED

DEFLECTIONS

For slabs with beams spanning between the supports on all sides,the minimum thickness, h, shall be as follows:

For fm greater than 0.2 but not

greater than 2.0, h shall not

be less than but not less than 5 in.

For fm greater than 2.0,

h shall not be less than but

not less than 3.5 in.

. Max spacing of reinforcement should not exceed twice thickness

of slab or that given by temperature and shrinkage .

. The twisting moments are of consequence only at exterior corners

of a two-way slab system, where they tend to crack the slab at thebottom along the panel diagonal, and at the top perpendicular to

the panel diagonal. Special reinforcement should be provided at

exterior corners in both the bottom and top of the slab, for a

distance in each direction from the corner equal to one-fifth the

longer span of the corner panel. The spacing and size should be

same as for positive bending moment.

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Example

1. Design the roof slab, beam and column of house given in figure 1.Concrete compressive strength (fc) = 3 ksi.

Steel yield strength (fy) = 40 ksi.

Load on slab:

4 thick mud. 2 thick brick tile. Live Load = 40 psf.

Requirement

2. Design of slab for room and veranda.Solution

Veranda Slab

3 Ratio of spans

lb/la = 24.75/8 = 3.09 > 2 one way slab

4 Assume slab thickness of slab = 5 in

Clear span = 8 + 5/12 = 8.42 ft ACI 8.9.1

or c/c of support = 8 + 13/24 + .5 = 9.0625 ftTake l = 8.42 ft

Min slab thickness = l/20 x (0.4+fy/100000)x12 ACI 9.5.2.1

= 4.04 in 5 in is ok

5 d = 5 0.75 3/16 = 4 in

6 Factored loads

dead load Slab = 0.15x5/12 = 0.0625 ksf

Mud = 0.12x4/12 = 0.04 ksf

Choka = 0.12x2/12= 0.02 ksf

total =0.1225 ksf

Live load = 0.04 ksf

Wu = 1.2x 0.1225 + 1.6x 0.04

= 0.211 ksf

Mu= (0.211x8.42x8.42/8)x12 = 22.44 in -kip/ft

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7 Asmin = 0.002bhf (for fy 40 ksi, ACI 10.5.4)

= 0.002 12 5 = 0.12 in2

a = Asminfy/ (0.85fcb)

= 0.12 40/ (0.85 3 12) = 0.156 in

Mn(min) = Asminfy (da/2)

= 0.9 0.12 40 (40.156/2)

= 16.94 in-k < Mu

8 Determine As for applied Mu = 22.44 in-kip

Mu/0.9 = As x fy x dsq(Asxfy)/(1.7xfcxb)

As = 0.16 sq in

using #4 bars Ab= 0.2 sq in

Spacing S= 0.2 x 12 /0.16 = 15 in c/cusing #3 bar S=0.11x12/0.16 = 7.5 in or 6 in c/c is ok.

use #3 @ 6 in c/c

9 Temperature and Shrinkage Steel

Ast = 0.002 x b x hf

= 0.002 x 12 x 5 = 0.12 sq in

S =0.11 x 12 /0.12 = 11 in c/c10 Maximum spacing for main steel in one way slab according to

ACI is minimum of:

3hf =3 5 =15 in or 18 in

Therefore 6 spacing is O.K.

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11 Maximum spacing for shrinkage steel in one way slab

according to ACI 7.12.2 is minimum of:

5hf =5 5 =25in or 18in

Therefore 9 in spacing is O.K.

Design of Room Slab

12 Span ratio = 16/12 = 1.33 < 2 two-way slab

Min h = Perimeter/180

= 2 x ( 16 + 12 ) x 12 /180

= 3.73 in not less than 5 in Aci

hf = 5 in is ok.

13 Wu= 0.211 ksf

14 Tables to be consulted for designm = short span/long span

= 12/16 = 0.75

Since slab is continuous on long edge only; case 6 would apply

Coefficient for this slab

Ca,neg = 0.088 Cb,neg = 0

Ca,dl = 0.o48 Cb,dl = 0.012Ca,ll = 0.055 Cb,ll = 0.016

15 Calculate Bending moment

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16 Total design Moments

Max negative BM in Short direction(12 ft) = 32.04 in k

long (16 ft) = 0 in kmax positve BM in short Direction (12 ft ) =18.36 in-k

long (16 ft ) = 8.544 in-k

17 Design of Slab

As min= 0.002x12x5 =0.12 sq in

Mu(min) = 0.9x0.12x40{(40.12x40/(1.7x3x12)}

= 19.94 in-k

Neg ative BM in Short span = # 3 @ 4.5 in c/c

Positive BM in short span = # 3 @ 9.0 in c/c

Negative BM in long span = nil

Positve BM in long span = # 3 @ 11 in c/c not to begreater than 2xhf = 2x5=10 in as per ACI 13.3.2 may reduce to 9in spacing for simplicity.

18 Min Reinforcement is also provided at discont edges againstpartial fixations

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Reinforcement Plan

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Example

1 The floor system shown in Fig. consists of solid slabs and beams intwo directions supported on 20-in. square columns. Design a typicalinterior slab pannel to carry a live load of 100 psf appart from selfweight. Use f`c= 3 ksi and fy = 60 ksi.

2 Solution

Since slab is supported on beams, we determine its thickness. To apply

the equation of ACI we need to find out Ib and Is etc. Effective X-Secof T beam is Shown in fig.

The moment of inertia of the slab in the long direction assuming slabthickness of 7 in

The moment of inertia of the slab in the short direction

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Also, hmin = 3.5 in. Therefore, h = 6.27 in. controls. A slab

thickness of 6.5 in. or 7.0 in. may be adopted. Note that in

most practical cases, Eq. 17.2 controls.

Eq. 9.12

--------

Eq. 9.13

---------

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3. Span ratio = 20/24 = 0.83 two-way slab

thickness of slab = 7 in

4. Factored Load

Wu = 1.2x(7/12x.15) + 1.6x0.1

= 0.265 k/sq ft

5. Coefficient from Tables ( Case 2 )

Ca neg = 0.062 Cb neg =0.029

Ca dl = 0.053 Cb dl = 0.012

Ca ll = 0.039 Cb ll =0.018

6. Bending Moments

Ma neg = 0.062x0.265x20x20x12 = 78.86 in-k

Ma pos = (0.053x0.105+0.039x0.16)x20x20x12 = 56.66 in-kMb neg =0.029x0.265x24x24x12 = 53.12 in-k

Mb pos =(0.012x0.105+0.018x0.16)x24x24 = 28.62 in-k

7. Design of s lab

in short dir for neg moment As= 0.228 sq in #4 @ 10 in c/c

pos As= 0.161 #4 @ 14

In long dir for neg moment As= 0.166 #4 @ 14 pos As= 0.088 #4 @ 14

Max spacing = 2xhf = 14 in or temp & shrinkage rft

Temp= 0.0018xbxhf =0.151 sq in #4 @ 15 in c/c

Max spacing of 14 in c/c governs

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TABLE. 1

COEFFICIENTS FOR NEGATIVE

MOMENTS IN SLABS

*A cross-hatched edge indicates thatthe slab continues across or is fixed at

the support; an unmarked edge

indicates a support at which torsional

resistance is negligible ie simple

support

*

* where w = total uniform Factored

dead plus live load

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TABLE.2

COEFFICIENTS FOR DEAD

LOAD POSITIVE MOMENTS IN

SLABS

*A cross-hatched edge indicates thatthe slab continues across or is fixed at

the support; an unmarked edge

indicates a support at which torsional

resistance is negligible ie simple

support.

*

* where w = total uniform Factored

dead plus live load

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TABLE.3

COEFFICIENTS FOR LIVE LOAD

POSITIVE MOMENTS IN

SLABS

*A cross-hatched edge indicates thatthe slab continues across or is fixed at

the support; an unmarked edge

indicates a support at which torsional

resistance is negligible ie simple

support.

*

* where w = total uniform Factored

dead plus live load

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TABLE.4

RATIO OF LOAD lV INA AND B

DIRECTIONS FOR SHEAR IN SLAB

AND LOAD ON SUPPORTS

*A cross-hatched edge indicates thatthe slab continues across or is fixed at

the support; an unmarked edge

indicates a support at which torsional

resistance is negligible ie simple

support.