Reinforecd Concrete Slab

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DESIGN OF REINFORCED CONCRETE SLAB

UNIT NO 3: DESIGN OF REINFORCED CONCRETESLAB

3.1 INTRODUCTION

Reinforced concrete slabs are used in floors, roofs and walls of buildings and asthe decks of bridges. The floor system of a structure can take many forms suchas in situ solid slab, ribbed slab or pre-cast units. Slabs may span in one directionor in two directions and they may be supported on monolithic concrete beam,steel beams, walls or directly by the structures columns.

Continuous slab should in principle be designed to withstand the mostunfavorable arrangements of loads, in the same manner as beams. Becausethere are greater opportunities for redistribution of loads in slabs, analysis mayhowever often be simplified by the use of a single load case. Bending momentcoefficient based on this simplified method are provided for slabs which span inone direction with approximately equal spans, and also for flat slabs.

The moments in slabs spanning in two directions can also be determined usingcoefficients tabulated in the code of practice, BS 8110. Slab which are notrectangular in plan or which support an irregular loading arrangement may be

analyzed by techniques such as the yield line method or the Helliborg stripmethod.

Concrete slab behave primarily as flexural members and the design is similar tothat for beams, although in general it is somewhat simpler because;

1. the breadth of the slab is already fixed and a unit breadth of 1m is used in thecalculations,

2. the shear stress are usually low in a slab except when there are heavyconcentrated loads, and

3. compression reinforcement is seldom required.

3.2 LEARNING OUTCOMES

After completing the unit, students should be able to :

1. know the requirement for reinforced concrete slab design

2. design reinforced concrete slab

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3.3 TYPES OF SLABS

Type of slab used in construction sectors are:

Solid slab Flat slab Ribbed slab Waffle slab Hollow block floor/slab

(a) Solid slab

(b) Flat slab

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(c) Ribbed slab

(d) Waffle slab

Figure 3.1: Types of slab

Flat slab floor is a reinforced concrete slab supported directly by concretecolumns without the use of intermediary beams. The slab may be of constant thickness throughout or in the area of the column it may bethickened as a drop panel. The column may also be of constant section orit may be flared to form a column head or capital. These various form of construction are illustrated in Figure 3.2.

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Figure 3.2: Drop panels and column head.

The drop panels are effective in reducing the shearing stresses where thecolumn is liable to punch through the slab, and they also provide anincreased moment of resistance where the negative moments aregreatest.

The flat slab floor has many advantages over the beam and slab floor. Thesimplified formwork and the reduced storey heights make it moreeconomical. Windows can extend up to the underside of the slab, andthere are no beams to obstruct the light and he circulation of air. Theabsence of sharp corner gives greater fire resistance as there is lessdanger of the concrete spalling an exposing the reinforcement. Deflectionrequirements will generally govern slab thickness which should not be lessthan 125 mm.

Typical ribbed and waffle slab are shown in Figure 3.1[(c), (d)]. Ribbedslabs, which are two-way spanning and are constructed with ribs in bothdirection of span. Ribbed slab floors are formed using temporary orpermanent shuttering system while the hollow block floor is generallyconstructed with block made of clay tile or with concrete containing alight-weight aggregate. If the block are suitably manufactured and have an

adequate strength they can be considered to contribute to the strength of the slab in the design calculations, but in many designs no such allowanceis made.

The principal advantage of these floors is the reduction in weight achievedby removing part of the concrete below the neutral axis and, in the case of the hollow block floor, replacing it with a lighter form of construction.Ribbed and hollow block floors are economical for buildings where thereare long spans, over about 5 m, and light or moderate live loads, such asin hospital wards or apartment buildings. They would not be suitable forstructures having a heavy loading, such as warehouses and garages.

Near to the supports the hollow blocks are stopped off and the slab ismade solid. This is done to achieve greater shear strength, and if the slab

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bay panel


is supported by a monolithic concrete beam the solid section acts as theflange of a T-section. The ribs should be checked for shear at their

junction with the solid slab. It is good practice to stagger the joints of thehollow blocks in adjacent rows so that, as they are stopped off, there is noabrupt change in cross-section extending across the slab. The slabs areusually made solid under partitions and concentrated loads. Duringconstruction the hollow tiles should be well soaked in water prior toplacing the concrete, otherwise shrinkage cracking of the top concreteflange is liable to occur.

3.4 SIMPLIFIED ANALYSIS

BS 8110 permit the use of simplified load arrangement for all slabs of maximum ultimate design load throughout all spans or panels providedthat the following condition are met;

a) in one-way slab, the area of each bay 30 m 2 b) Live load, Q k 1.25 Dead load, G kc) Live load, Q k 5 kN/m 2 excluding partitions.

If analysis is based on this singled load case, all support moments (exceptat a cantilever) should be reduced by 20 per cent and span momentsincreased accordingly. No further redistribution is then permitted, butspecial attention must be given to cases where a span or panel is adjacentto a cantilever of significant length. In this situation the condition wherethe cantilever is fully loaded and the span unloaded must be examined todetermine possible hogging moments in the span.

To determine the value of bending moment coefficient and shear forcescoefficient, therefore very important to define the condition of panel type,location and moment considered. Refer to BS 8110: Part 1: 1997, Cl3.5.3.6 and 3.5.3.7 and also Table 3.14 and Table 3.15 for moreinformation.

Figure 3.3: Slab definition

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A

C D

B

l x

l y

l x

l x/2

l x

Beam AC and BD

w = n l x

/ 3

450

C

A

D

B

E F

l y

l x

450

Beam AB and CD

w = n l x

/ 6 {3- ( l x

/ l y )2 }


3.5 LOAD DISTRIBUTION FROM SLAB

Define the type of slab either one-way direction or two-way direction, for

determine the shape of load distribution from slab to beam.If Iy / Ix < 2 consider as two-way slab

Iy / Ix 2 consider as one-way slab

where Ix - length of shorter sideIy - length of longer side

a) One-way slab

b) Two-way slab

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Beam AB and CD w = n l x / 2


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section around the load; this effect is referred to a punching shear. The initial critical section for shear is shown in Figure 3.5 and theshear stress is given by;

v = N / (Perimeter of the section x d) = N / (2a + 2b + 12d )d

where a and b are the plan dimensions of the concentrated load. Noshear reinforcement is required if the punching shear stress, v < v c.

The value of v c in Table 3.9, BS 8110, depends on the percentage of reinforcement 100As/bd which should be calculates as an averageof a tensile reinforcement in the two directions and should includeall the reinforcement crossing the critical section and extending afurther distance equal to at least d on either side.

Check should also be undertaken to ensure that the stress vcalculated for the perimeter at the face of the loaded area is lessthan smaller of 0.8 f cu or 5 N/mm 2.

Figure 3.5 Punching shear

3.7 SPAN-EFFECTIVE DEPTH RATIOS

Excessive deflections of slab will cause damage to the ceiling, floor finishesand other architectural details. To avoid this, limits are set on the span-depth ratios. These limits are exactly the same as those for beams. As aslab is usually a slender member the restrictions on the span-depthratio become more important and this can often control the depth of slab required. In terms of the span-effective depth ratio the depth of theslab is given by;

minimum effective depth = span ________

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basic ratio x modification factors

The modification factor is based on the area of tension steel in the shorterspan when a slab is singly reinforced at mid-span but if a slab has bothtop and bottom steel at mid-span the modification factors for the areas of tension and compression steel, as given in Tables 1.13 and 1.14, BS 8110,are used. For convenience, the factors for tension steel have been plottedin the form of a graph in Figure 3.6.

It can be seen from the figure that a lower service stress gives a highermodification factor and hence a smaller depth of slab would be required.

The service stress may be reduced by providing an area of tensionreinforcement greater than that required resisting the design moment, oralternatively mild steel reinforcement with its lower service tress may beused.

The span-depth ratios may be checked using the service stress appropriateto the characteristic stress of the reinforcement, as given in Table 1.13,BS 8110. Thus a service stress of 307 N/mm 2 would be used when f y is 460N/mm 2. However, if a more accurate assessment of the limiting span-depth ratio is required the service stress f s, can be calculated from;

f s = 2 x f y x Asreq x 13 x A sprov b

where

Asreq = the area of reinforcement at mid-spanAsprov = the area of reinforcement provided at mid-span b = the ratio of the mid-span moments after and before any

redistribution.

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Figure 3.6: Modification factors for span-effective depth ratio

3.8 REINFORCEMENT DETAIL

To resist cracking of the concrete, codes of practice specify detail such asthe minimum area of reinforcement required in a section and limits to themaximum and minimum spacing of bars. Some of these rules are asfollows;

a) Minimum areas of reinforcement

Minimum area = 0.13bh / 100 for high yield steel

or= 0.24bh / 100 for mild steel

in both directions.

b) Maximum Spacing of Reinforcement

The maximum clear spacing given in Table 3.30, and Clause 3.12.11, BS8110, (apply to bars in beams when a maximum likely crack width of 0.3mm is acceptable an the cover to reinforcement does not exceed 50 mm),and are similar to beams except that for thin slabs, or if the tensile steelpercentage is small, spacing may be increased from those given in Table3.30, BS 8110 to a maximum of the lesser of 3d or 750 mm.

c) Reinforcement in the flange of a T or L-Beam When the slab from the flange of a T or L beam the area of reinforcementin the flange and at right angles to the beam should not be less than 0.15percent of the longitudinal cross-section of the flange.

d) Curtailment and anchorage of reinforcement

At a simply supported end the bars should be anchored as specified inFigure 3.7.

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Figure 3.7: Anchorage at simple supported for a slab

3.9 SLAB DESIGN

3.9.1 SOLID SLABS SPANNING IN ONE DIRECTION

The slabs are design as if they consist of a series of beams of 1 mbreadth. The main steel is in the direction of the span andsecondary or distribution steel required in the transverse direction.

The main steel should from the outer layer of reinforcement to giveit the maximum level arm.

The calculations for bending reinforcement follow a similarprocedure to that used in beam design. The lever-arm curve of Figure 3.8 is used to determine the lever arm (z) and the area of tension reinforcement is then given by;

As = M u / 0.87 f y.z

For solid slabs spanning one way the simplified rules for curtailingbars as shown in Figure 3.9 may be used provided the loads aresubstantially uniformly distributed. With a continuous slab it is alsonecessary that the spans are approximately equal the simplifiedsingle load case analysis has been used.

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b) The clear distance between supports plus the effective depth of the slab

The basic span-effective depth ratio for this type of slab is 20:1(Refer to Table 3.10 and Cl. 3.4.6.3 in BS 8110).

Example 3.1:

The slab is to be design to carry a live load 3.0 kN/mm 2, plusfloor finishes and ceiling load of 1.0 kN/mm 2. Thecharacteristic materials strength are f cu = 30 N/mm 2, f y = 460N/mm 2. Length of slab is 4.5 m

Solution :

Minimum effective depth, d = span / 20 x modificationfactor (m.f)

= 4500 / 20 m.f = 225 / m.f

For high-yield reinforcement slab ;

Estimating the modification factor to be of the order of 1.3 fora highly reinforcement slab.

Try effective depth d = 180 mm. For a mild exposure thecover = 25 mm.

Allowing, say, 5 mm as half the diameter of the reinforcing bar

overall depth of slab, h = 180 + 25 + 5 = 210 mm

self-weight of slab = 0.21 x 24 x 10 3 = 5.0 kN/m 2

total dead load, G k = 1.0 + 5.0 = 6.0 kN/m 2

For a 1m width of slab

ultimate load = (1.4G k + 1.6Qk) (4.5)

= (1.4 x 6.0 + 1.6 x 3.0)(4.5) = 59.4kN

M = (59.4 x 4.5)/8 = 33.4 kNm

1) Span-effective depth ratio

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M = 33.4 x 10 6 = 1.03bd 2 1000 x 180 2

From Table 3.11 BS 8110, for f s = 307 N/mm 2 the span-effective depth modification factor = 1.29. Therefore;

Allowable span / d > Actual span / d

20 x 1.29 > 4500 / 180

25.8 > 25.0

Thus d = 180 mm is adequate .

2) Bending reinforcement

K = M = 33.4 x 10 6 = 0.034 < 0.156f cu bd 2 (1000)(180 2)(30)

z = d {0.5 + (0.25 K / 0.9)}

= d {0.5 + ( 0.25 0.034 / 0.9)}

= 0.96d > 0.95d, so take z = 0.95d

As = M / 0.87f y z = 334 x 10 6 /(0.87 x 460 x 171) = 447mm 2/m

Provide T10 bars at 150 mm centre, A s = 523 mm 2 /m

3) Shear

Shear, V = W / 2 = 59.4 / 2 = 29.07 kN

Shear stress, v = V / bd

= 29.07 x 10 3/ (1000 x 180)

= 0.16 N/mm 2 < 0.8 f cu

From Table 3.9, BS 8110,

100A s /b d = 100 x 523 / 1000 x 180 = 0.29

v c = 0.51 N/mm 2 , v < v c , so no shear reinforcementis required.

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4) End anchorage (Cl. 3.12.9.4, BS 8110)

v = 0.16 < < v c/2 ok; therefore;

anchorage length > 30 mm or end bearing (support

width)/3

end bearing = 230 mm

Therefore;

anchorage length = 230 / 3 = 77 mm 30 mm

beyond the centre line of the support.

Figure 3.10: End Anchorage

5) Distribution / Transverse Steel

From Table 3.27 BS 8110, f y = 460 N/mm 2

Area of transverse high-yield reinforcement,

As min = 0.13bh/100

= 0.13 x 1000 x 210 /100

= 273 mm 2 /m

Provide T10 at 250 mm centre, A s = 314 mm 2 /m,top layer

6) Cracking check

The bar spacing does not exceed 750 mm or 3d and theminimum reinforcement is less than 0.3%. (Refer Cl.

3.12.11.2.7 and Table 3.30, BS 8110).

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Allowable clear spacing of bars = 3d = 3(180) = 540

mm

Actual clear spacing = 250 10 = 240 mm < 3d

ok

3.9.1.2 Continuous Solid Slab

For a continuous slab, bottom reinforcement is required within thespan and top reinforcement over the supports. The effective span isthe distance between the centre lines of supports. The basic span-effective depth ratio is 26:1 (Refer to Table 3.10 and Cl 3.4.6.3).

If the simplified load arrangement for all slabs of maximum ultimatedesign load throughout all spans or panels provided that thefollowing condition are met for the single load case analysis,bending moment an shear forces coefficients as shown in Table3.13, BS 8110 may be used.

Example 3.2 :

The four-span slab shown in Figure 3.11 support a live load 0f 3.0 kN/mm 2, plus floor finishes and ceiling load of 1.0kN/mm 2. The characteristic materials strength are f cu = 30N/mm 2, f y = 460 N/mm 2.

Figure 3.11 Continuous slab - example

Solution :

From Table 3.10, BS 8110, basic span- effective depth ratio

= 26So depth, d = Span / 26 = 4500 / 26

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= 173 mm

Try ef fec ti ve depth, d = 170 mm . Assume a mild exposure,

cover, c = 20 mm an diameter of bar, = 10 mm

h = d + cover + /2= 17 0 + 20 + 5 = 195 mm, so taken h = 200 mm

Self-weight of slab = 0.2 x 24 = 4.8 kN/m 2

Total dead load, G k = 1.0 + 4.8 = 5.8 kN / m 2

For 1 meter width of slab;

Ultimate load, F = (1.4g k + 1.6q k ) 4.5

= (1.4 x 5.8 + 1.6 x 3.0)(4.5)

F = 58.14 kN per metre width

1) Bending (Refer to CL 3.5.2.3, BS 8110)

Since the bay size > 30m 2, the spans are equal and q k < 1.25g k the moment coefficients shown in Table 3.13 Bs 8110 maybe used. Thus, assuming that the end support is simplysupported, from Table 3.13 for the first span:

M = 0.086FL = (0.086 x 58.14 )(4.5) = 22.5 kNm

K = M = 22.5 x 10 6 = 0.026 < 0.156f cu bd 2 30(1000 )(170) 2

z = d {0.5 + (0.25 K/0.9)}

= d {0.5 + (0.25 0.026 / 0.9)}

= 0.97d > 0.95d, so take z = 0.95d

As = M / 0.87f y z = 22.5x10 6 / (0.87 x 460 x

161.5)

= 348 mm 2/ m

Provide T10 bars at 200 mm centre, A s = 393mm 2 /m

2) Span-effective depth ratio

M = 22.5 x 10 6 = 0.778

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bd 2 1000 x170 2



26 x 1.68 > 4500 / 170

43.68 > 26.5 ok

Thus d = 170 mm is adequate.

Similar calculation for the support and the interior span givethe steel areas shown in Figure 3.12.

3) Distribution / Transverse Steel

From Table 3.27 BS 8110, f y = 460 N/mm 2

Area of transverse high-yield reinforcement,

As min = 0.13bh/100

= 0.13 x 1000 x 200 /100

= 260 mm 2 /m

Provide T10 at 300 mm centre, A s = 262 mm 2 /m,top and bottom layer

4) Shear (Refer Table 3.13 BS 8110)

Shear, V = 0.6 F = 0.6 (58.14) = 34.9 kN


= 34.9 x 10 3/ (1000 x 170)

= 0.21 N/mm 2 < 0.8 f cu

From Table 3.9, BS 8110,100A s / bd = 100 x 393 / 1000 x 170 = 0.23

So, v c = 0.47 x (30/25) 1/3 = 0.50 N/mm 2 ,

v < v c , so no shear reinforcement is required.

5) Cracking check

The bar spacing does not exceed 750 mm or 3d and theminimum reinforcement is less than 0.3%. (Refer Cl.3.12.11.2.7 and Table 3.27 BS 8110).

Allowable clear spacing of bars = 3d = 3(170) = 510

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With a uniformly distributed load the loads on the supporting beamsmay generally be apportioned as shown in Figure 3.13.

Figure 3.13: Loads carried by supporting beams

Figure 3.14: Nine Types of slab panels

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3.9.2.1 Simply Supported Slab Spanning In TwoDirections

A slab simply supported on its four sides will deflect about bothaxes under load and the corners will tend to lift and curl up fromthe supports, causing torsion moments. When no provision hasbeen made to prevent this lifting or to resist the torsion then themoment coefficients of Table 3.14, BS 8110 may be used and themaximum moments are given by equation 14 and 15 in BS 8110;

m sx = sx nlx2 in direction of span l X

and

m sy = sy nlx2 in direction of span l y

where m sx and m sy are the moments at mid-span on strips of unitwidth with spans l x and respectively, and

n = (1.4G k + 1.6Q k), that is, the total ultimate load per unitarea

ly = the length of the longer sidelx = the length of the shorter side

The area of reinforcement in directions l x and l y respectively are;

Asx = m sx / 0.87f yz per metre widthand

Asy = m sy / 0.87f yz per metre width

The slab should be reinforced uniformly across the full width, in eachdirection. The effective depth d used in calculating A sy should be lessthan that for A sx because of the different depths of the two layers of reinforcement.

At least 40 per cent of the mid-span reinforcement should extend to

the supports and the remaining 60 per cent should extend to within0.1l x, or 0.1l y of the appropriate support.

Example 3.3 :

Design the reinforcement for a simply supported slab 200mm thick and spanning in two directions. The effective spanin each direction is 4.5 m and 6.3 m and the slab supports alive load of 10 kN/m 2. The characteristic material strengths aref cu = 30 N/mm 2 and f y = 460 N/mm 2.

Solution :ly / lX = 6.3/4.5 = 1.4 < 2 Two way slab

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From Table 3.14, sx = 0.099 and sy = 0.051.

Self-weight of slab = 0.2 x 24 x 10 3 = 4.8 kN/m 2

Ultimate load, n = 1.4G k + 1.6Q k

n = (1.4 x 4.8) + (1.6 x10) = 22.72kN/m 2 = 22.72 kN/m/m width

Short Span

1) Bending

From Table 3.4, BS 8110, mild exposure conditions,cover, c = 25 mm. Assume bar = 10mm.

d x = h c - /2 = 200 25 5 = 170 mm.

m sx = sx nlx2

= 0.099(22.72)(4.5) 2 = 45.5 kN.m/m

K = M = 45.5 x 10 6 = 0.052 < 0.156f cu bd 2 30(1000 )(170) 2

z = d { 0.5 + (0.25 K/0.9)}

= d { 0.5 + (0.25 0.052/0.9)}

= 0.94d < 0.95d, so take z = 0.94d

Asx = m sx / 0.87f y z = 45.5 x10 6 / (0.87x 460)

(0.94x170)

= 711.5 mm 2/ m

Checking A smin , from Table 3.27 BS 8110, f y = 460 N/mm 2

Asmin = 0.13bh / 100= 0.13(1000 x 200) / 100= 260 mm 2/ m

Asx > A smin ok

Provide T10 bars at 100 mm centre, A s = 786 mm 2 /m

2) Deflection Checking

M = 45.5 x 10 6 = 1.57bd 2 1000 x170 2

From Table 3.11 BS 8110, for f s = 221 N/mm2

the span-effective depth modification factor = 1.41. Therefore;

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20 x 1.41 > 4500 / 170

28.2 > 26.5 ok

3) Shear

Shear, V = WL / 2 = (22.72 x 4.5 ) / 2 = 51.12 kN


= 51.12 x 10 3/ (1000 x 170)

= 0.3 N/mm 2 < 0.8 f cuFrom Table 3.9, BS 8110,

100A s / bd = 100 x 786 / 1000 x 170 = 0.46

So, v c = 0.63 x (30/25) 1/3 = 0.67 N/mm 2 ,

v < v c , so no shear reinforcement is

required.

Long Span

1) Bending

From Table 3.4 BS 8110, mild exposure conditions,cover, c = 25 mm. Assume bar = 10mm.

d y = h c - /2 = 200 25 -10 5 = 160

mm.

m sy = synl x2

= 0.051(22.72)(4.5) 2 = 23.5 kNm/m

K = M = 23.5 x 10 6 = 0.031 < 0.156f cu bd 2 30(1000 )(160) 2

z = d { 0.5 + (0.25 K/0.9)}

= d { 0.5 + (0.25 0.031/0.9)}

= 0.96d > 0.95d, so take z = 0.95d

Asy = m sy / 0.87f y z = 23.5 x10 6 / (0.87x 460)

(0.95x160)

= 354 mm 2/ m

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Asmin = 0.13bh / 100

= 0.13(1000 x 200) / 100= 260 mm 2/ m

Asx > A smin ok

Provide T10 bars at 200 mm centre, A s = 393mm 2 /m

2) Checking for Transverse Steel

From Table 3.27, f y = 460 N/mm2

100As / bh = 100 (393) / 1000 x 200

0.19 > 0.13 (A smin ) ok

3.9.2.2 Restrained Slab Spanning In Two Directions

When the slabs have fixity at the supports and reinforcement isadded to resist torsion and to prevent the corners of the slab fromlifting then the maximum moments per unit width are given by;

m sX = sXnlX2 in direction of span l x

and

m sy = synlX2 in direction of span l y

where sX and Sy are the moment coefficients given in Table 3.15 of BS 8110 for the specified end conditions, and n = (1.4G k+ 1.6Q k), thetotal ultimate load per unit area.

The slab is divided into middle and edge strips as shown in Figure

3.15 and reinforcement is required in the middle strips to resist m sxand m sy , In the edge strips only nominal reinforcement is necessary,such that 100A s/bh = 0.13 for high-yield steel or 0.24 for mild steel.

In addition, torsion reinforcement is provided at discontinuouscorners and it should;

1. consist of top and bottom mats, each having bars in bothdirections of span.

2. extend from the edges a minimum distance l x / 53. at a corner where the slab is discontinuous in both directions

have an area of steel in each of the four layers equal to three-quarters of the area required for the maximum mid-span

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moment4. at a corner where the slab is discontinuous in one direction only,

have an area of torsion reinforcement only half of that specified inrule 3.

Torsion reinforcement is not, however, necessary at any cornerwhere the slab continuous in both directions.

Where l y /Ix > 2, the slabs should be designed as spanning in onedirection only.

Shear force coefficients are also given in BS 8110 for cases wheretorsion corner reinforcement is provided, and these are based on asimplified distribution of load to supporting beams which may beused in preference to the distribution shown Figure 3.13.

Figure 3.15: Division of slabs into middle and edge strips

Example 3.4 :

The panel considered is an interior panel, as shown in

Figure 3.16. The effective span in each direction is 5 mand 6 m and the slab supports a live load of 1.5 kN/m 2.Given f cu = 30 N/mm 2, f y = 250 N/mm 2 and slab thickness 150mm. Design the reinforcement for a continuous slab.

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Figure 3.16: Continuous panel spanning in todirections

Solution :

ly / lX = 6 / 5 = 1.2 < 2 Two way slab

Self-weight of slab = 0.15 x 24 x 10 3 = 3.60 kN/m 2

20 mm asphalt = 0.48 kN/m 2

50 mm insulting screed = 0.72 kN/m 2

Ceiling finishes = 0.24 kN/m 2 Total dead load = 5.04 kN/m 2

Ultimate load, n = 1.4Gk + 1.6Q kn = 1.4x5.04 + 1.6x1.5 = 9.5 kN/m 2

= 9.5 kN/m/m width

From Table 3.15, Case 1 applies;

+ ve moment at mid span

m sx = 0.032(9.5)(5) z = 7.6 kNmm sy = 0.024 (9.5)(5) = 5.7 kNm

- ve moment at support (cont)

a long AB & CD, m sx = 0.042 (9.5)(5)2

= 10.2 kN.ma long AD & BC, m sx = 0.032 (9.5)(5) 2 = 7.6 kN.m

Assume bar = 10 mm, and cover, c = 25 mm

d x = h - cover - /2

= 150 - 25 - 10/2 = 120 mm

d y = h - cover - /2

= 150 - 25 - 10 - 10/2 = 110 mm

Short Span, l x

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l x =5m

d c


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1) At Mid-Span, m sx = 7.6 kNm

K = M = 7.6 x 10 6 = 0.018 < 0.156f cu bd 2 30(1000 )(120) 2

z x = d { 0.5 + (0.25 K/0.9)}

= d { 0.5 + (0.25 0.018/0.9)}

= 0.98d > 0.95d, so take z = 0.95d

Asy = m sx / 0.87f y z = 7.6 x10 6 / (0.87x 250)(0.95x120)

= 306.51 mm 2/ m width


Asmin = 0.24bh / 100= 0.24(1000 x 150) / 100= 360 mm 2/ m

Asx < A smin so use A smin

Provide R10 bars at 200 mm centre, A s = 393 mm 2 /m

2) At Support, m sx = 10.2 kNm

K = M = 10.2 x 10 6 = 0.024 < 0.156f cu bd 2 30(1000 )(120) 2

z x = d { 0.5 + (0.25 K/0.9)}

= d{0.5 + (0.25 0.024/0.9)}

= 0.97d > 0.95d, so take z = 0.95d

Asx = m sx / 0.87f y z = 10.2 x10 6 / (0.87x 250)(0.95x120)

= 411.37 mm2

/ m width


Asmin = 0.24bh / 100= 0.24(1000 x 150) / 100= 360 mm 2/m

Asx > A smin ok

Provide R10 bars at 175 mm centre, A s = 449 mm 2 / m

Long Span, l y

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1) At Mid-Span, m sy = 5.7 kNm

K = M = 5.7 x 10 6 = 0.016 < 0.156f cu bd 2 30(1000 )(110) 2

z y = d{0.5 + (0.25 K/0.9)}

= d{0.5 + (0.25 0.016/0.9)}

= 0.98d > 0.95d, so take z = 0.95d

Asy = m sy / 0.87f y z = 5.7 x10 6 / (0.87x 250)(0.95x110)

= 250.78 mm 2/ m width


Asmin = 0.24bh / 100= 0.24(1000 x 150) / 100= 360 mm 2/ m

Asy < A smin so use A smin

Provide R10 bars at 200 mm centre, A sprov = 393 mm 2 /m

2) At Support, m sy = 7.6 kNm

K = M = 7.6 x 106

= 0.02 < 0.156f cu bd 2 30(1000 )(110) 2

z y = d { 0.5 + (0.25 K/0.9)}

= d { 0.5 + (0.25 0.02/0.9)}

= 0.98d > 0.95d, so take z = 0.95d

Asy = m sy / 0.87f y z = 7.6 x10 6 / (0.87x 250)(0.95x110)

= 334.37 mm 2/ m width


Asmin = 0.24bh / 100= 0.24(1000 x 150) / 100= 360 mm 2/m

Asy < A smin so use A smin

Provide R10 bars at 200 mm centre, A sprov = 393 mm 2 / m

Torsion reinforcement is not necessary because the slab is

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interior panel.

Edge strip, provide A smin (R10 -200mm c/c).

Shear Checking (Critical at Support)

Normally shear reinforcement should not be used in slabs < 200mm deep.

From Table 3.16, vx = 0.39, vy = 0.33

Vvx = vx.n.l x= 0.39(9.5)(5) = 18.5 kN/m width

Vvx = vy.n.l x= 0.33(9.5)(5) = 15.7 kN/m width

Shear stress, v = V max / bd

= 18.5 x 10 3 / (1000 x 120)

= 0.15 N/mm 2 < 0.8 f cuFrom Table 3.9, BS 8110,

100A s / bd = 100 x 449 / 1000 x 120 = 0.374

So, v c = 0.6 x (30/25) 1/3 = 0.64 N/mm 2 ,

v < v c , so no shear reinforcement is required.

Deflection Checking (Critical at Mi-span), M sx = 7.6 kN.m

M = 7.6 x 10 6 = 0.53bd 2 1000 x120 2



26 x 2 > 4500 / 12052 > 37.5 ok

Cracking Checking (Cl 3.12.11.2.7) The bar spacing does not exceed 750 mm or 3d and theminimum reinforcement is less than 0.3%. (Refer Cl.3.12.11.2.7 and Table 3.27 BS 8110).

Allowable clear spacing of bars = 3d = 3(120) = 360 mm

Actual clear spacing = 200 10 = 190 mm < 3d ok

h = 150 mm < 250 mm (for Grade 30) therefore no furtherchecks are required.

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3.11 SUMMARY

In this unit we have studied method for reinforced concrete slab design.Summary of reinforced concrete slab design are shown in Figure 3.17 below.

Figure 3.17: Flowchart for slab design

3.12 REFERENCES

1. W.H.Mosley, J.H. Bungery & R. Husle (1999), Reinforced Concrete Design(5 th Edition) : Palgrave.

2. Reinforced Concrete Modul, (1 st Edition). USM.3. BS 8110, Part 1: 1985, The Structural Use of Concrete. Code of Practice for

Design and Construction.

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Decide concrete grade, concrete cover, fireresistance and durability

Estimate slab thickness for continuous, L/d = 30 or for simply supported, L/d = 24,

where L is shorter span of the slab.

Load calculation and estimationUBBL: 1984 or BS 6339:1984

Structural analysis using Table 3.15 and3.16, BS 8110: Part 1: 1985

Reinforcement deign

Check shear

Check for serviceability limit state


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Documents

Reinforecd Concrete Slab