reinforced slab

DESIGN OF REINFORCED CONCRETE SLAB

UNIT NO 3: DESIGN OF REINFORCED CONCRETE SLAB

3.1 INTRODUCTION

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

Continuous slab should in principle be designed to withstand the most unfavorable arrangements of loads, in the same manner as beams. Because there are greater opportunities for redistribution of loads in slabs, analysis may however often be simplified by the use of a single load case. Bending moment coefficient based on this simplified method are provided for slabs which span in one direction with approximately equal spans, and also for flat slabs.

The moments in slabs spanning in two directions can also be determined using coefficients tabulated in the code of practice, BS 8110. Slab which are not rectangular in plan or which support an irregular loading arrangement may be analyzed by techniques such as the yield line method or the Helliborg strip method.

Concrete slab behave primarily as flexural members and the design is similar to that 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 the calculations,

2. the shear stress are usually low in a slab except when there are heavy concentrated 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 concrete columns without the use of intermediary beams. The slab may be of constant thickness throughout or in the area of the column it may be thickened as a drop panel. The column may also be of constant section or it 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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http://www.john-doyle.co.uk/Photos/wafer2.jpg%00%E5%A1%B9%EF%92%81%E1%B4%BB%E4%A1%BF%E2%B2%AF%E5%B6%82%E8%97%84%E6%8C%A7%00%00%EA%AE%A5

http://www.civilzone.com/photo/column/column_b/IMG0077.jpg%00%E5%A1%B9%EF%92%81%E1%B4%BB%E4%A1%BF%E2%B2%AF%E5%B6%82%E8%97%84%E6%8C%A7%00%00%EA%AE%A5


Figure 3.2: Drop panels and column head.

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

The flat slab floor has many advantages over the beam and slab floor. The simplified formwork and the reduced storey heights make it more economical. Windows can extend up to the underside of the slab, and there are no beams to obstruct the light and he circulation of air. The absence of sharp corner gives greater fire resistance as there is less danger of the concrete spalling an exposing the reinforcement. Deflection requirements will generally govern slab thickness which should not be less than 125 mm.

Typical ribbed and waffle slab are shown in Figure 3.1[(c), (d)]. Ribbed slabs, which are two-way spanning and are constructed with ribs in both direction of span. Ribbed slab floors are formed using temporary or permanent shuttering system while the hollow block floor is generally constructed with block made of clay tile or with concrete containing a light-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 allowance is made.

The principal advantage of these floors is the reduction in weight achieved by 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 there are long spans, over about 5 m, and light or moderate live loads, such as in hospital wards or apartment buildings. They would not be suitable for structures having a heavy loading, such as warehouses and garages.

Near to the supports the hollow blocks are stopped off and the slab is made solid. This is done to achieve greater shear strength, and if the slab is supported by a monolithic concrete beam the solid section acts as the flange 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 the hollow blocks in adjacent rows so that, as they are stopped off, there is no abrupt change in cross-section extending across the slab. The slabs are usually made solid under partitions and concentrated loads. During construction the hollow tiles should be well soaked in water prior to placing the concrete, otherwise shrinkage cracking of the top concrete flange is liable to occur.

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baypanel


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 provided that the following condition are met;

a) in one-way slab, the area of each bay ≥ 30 m2 b) Live load, Qk ≤ 1.25 Dead load, Gk

c) Live load, Qk ≤ 5 kN/m2 excluding partitions.

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

To determine the value of bending moment coefficient and shear forces coefficient, therefore very important to define the condition of panel type, location and moment considered. Refer to BS 8110: Part 1: 1997, Cl 3.5.3.6 and 3.5.3.7 and also Table 3.14 and Table 3.15 for more information.

Figure 3.3: Slab definition

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 side Iy - length of longer side

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A

C D

B

lx

ly

lx

lx/2

lx

Beam AC and BD

w = n lx / 3

450

C

A

D

B

E F

ly

lx

450

Beam AB and CD

w = n lx / 6 {3- (lx / ly)2}


a) One-way slab

b) Two-way slab

Figure 3.4: Load distribution of slab

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Beam AB and CD w = n lx / 2


3.6 SHEAR IN SLAB

The shear resistance of slab may be calculated by the procedures given in BS 8110, Cl.3.5.5.2. Experimental works has indicated that, compared wit beams, shallow slab fail at slightly higher shear stresses and this is incorporated into the values of design ultimate shear stress vc.(Refer to Table 3.9, BS 8110). The shear stress at a section in a solid slab is given by;

v = V b.d

where V is the shear force due to ultimate load, d is the effective depth of the slab and b is the width of section considered (Refer to Table 3.17 and Cl. 3.5.5.2). Calculation is usually based on strip of slab 1m wide.

The BS 8110 requires that for solid slab;

1. v < 0.8 √ fcu or 5 N/mm2

2. v < vc for a slab thickness less than 200 mm3. if v > vc , shear reinforcement must be provided in slabs more than 200

mm thick.

If shear reinforcement is required, then nominal steel, as for beams, should be provided when v < (vc + o.4) and ‘designed’ reinforcement provided for higher values of v. Since shear stress in slab due to distributed loads are generally small, shear reinforcement will seldom be required for such loads may, however, cause more critical conditions as shown in the following sections. Practical difficulties concerned with bending and fixing of shear reinforcement lead to the recommendation that it should not be used in slabs which are less than 200 mm deep.

3.6.1 PUNCHING SHEAR ANALYSIS

A concentrated load (N) on a slab causes shearing stresses on a 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 the shear 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. No shear reinforcement is required if the punching shear stress, v < vc. The value of vc in Table 3.9, BS 8110, depends on the percentage of reinforcement 100As/bd which should be calculates as an average of a tensile reinforcement in the two directions and should include all the reinforcement crossing the critical section and extending a further distance equal to at least d on either side.

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Check should also be undertaken to ensure that the stress v calculated for the perimeter at the face of the loaded area is less than smaller of 0.8 √ fcu or 5 N/mm2.

Figure 3.5 Punching shear

3.7 SPAN-EFFECTIVE DEPTH RATIOS

Excessive deflections of slab will cause damage to the ceiling, floor finishes and 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 a slab is usually a slender member the restrictions on the span-depth ratio become more important and this can often control the depth of slab required. In terms of the span-effective depth ratio the depth of the slab is given by;

minimum effective depth = span

________ basic ratio x modification factors

The modification factor is based on the area of tension steel in the shorter span when a slab is singly reinforced at mid-span but if a slab has both top 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 plotted in the form of a graph in Figure 3.6.

It can be seen from the figure that a lower service stress gives a higher modification factor and hence a smaller depth of slab would be required. The service stress may be reduced by providing an area of tension reinforcement greater than that required resisting the design moment, or alternatively mild steel reinforcement with its lower service tress may be used.

The span-depth ratios may be checked using the service stress appropriate to the characteristic stress of the reinforcement, as given in Table 1.13, BS 8110. Thus a service stress of 307 N/mm2 would be used when fy is 460 N/mm2. However, if a

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DESIGN OF REINFORCED CONCRETE SLABmore accurate assessment of the limiting span-depth ratio is required the service stress fs, can be calculated from;

fs = 2 x fy x Asreq x 1 3 x Asprov βb

where

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

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 as the minimum area of reinforcement required in a section and limits to the maximum and minimum spacing of bars. Some of these rules are as follows;

a) Minimum areas of reinforcement

Minimum area = 0.13bh / 100 for high yield steel

or = 0.24bh / 100 for mild steel

in both directions.

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b) Maximum Spacing of Reinforcement

The maximum clear spacing given in Table 3.30, and Clause 3.12.11, BS 8110, (apply to bars in beams when a maximum likely crack width of 0.3 mm 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 steel percentage is small, spacing may be increased from those given in Table 3.30, 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 reinforcement in the flange and at right angles to the beam should not be less than 0.15 percent 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 in Figure 3.7.

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 m breadth. The main steel is in the direction of the span and secondary or distribution steel required in the transverse direction. The main steel should from the outer layer of reinforcement to give it the maximum level arm.

The calculations for bending reinforcement follow a similar procedure 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;

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DESIGN OF REINFORCED CONCRETE SLAB As = Mu / 0.87 fy.z

For solid slabs spanning one way the simplified rules for curtailing bars as shown in Figure 3.9 may be used provided the loads are substantially uniformly distributed. With a continuous slab it is also necessary that the spans are approximately equal the simplified single load case analysis has been used.

The % values on the K axis mark the limitfor singly reinforced sections with moment redistribution applied.

Figure 3.8: Lever-arm

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Figure 3.9: Simplified rules for curtailment of bars in slab spanning in one direction

3.9.1.1 Simply Supported Solid Slab

The effective span of the slab is taken as the lesser of:

a) The centre-to-centre distance of the bearings, orb) 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/mm2, plus floor finishes and ceiling load of 1.0 kN/mm2. The characteristic materials strength are fcu = 30 N/mm2, fy = 460 N/mm2. Length of slab is 4.5 m

Solution :

Minimum effective depth, d = span / 20 x modification factor (m.f)= 4500 / 20 m.f = 225 / m.f

For high-yield reinforcement slab;

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Estimating the modification factor to be of the order of 1.3 for a highly reinforcement slab. Try effective depth d = 180 mm. For a mild exposure the cover = 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 103 = 5.0 kN/m2

total dead load, Gk = 1.0 + 5.0 = 6.0 kN/m2

For a 1m width of slab

ultimate load = (1.4Gk + 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

M = 33.4 x 10 6 = 1.03 bd2 1000 x 1802

From Table 3.11 BS 8110, for fs = 307 N/mm2 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 < 0.156 fcubd2 (1000)(1802)(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

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As = M / 0.87fy z = 334 x 106 /(0.87 x 460 x 171) = 447 mm2/m

Provide T10 bars at 150 mm centre, As = 523 mm2/m

3) Shear

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

Shear stress, v = V / bd

= 29.07 x 103/ (1000 x 180)

= 0.16 N/mm2 < 0.8 √ fcu

From Table 3.9, BS 8110,

100As /b d = 100 x 523 / 1000 x 180 = 0.29

vc = 0.51 N/mm2 , v < v c , so no shear reinforcement is required.

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

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DESIGN OF REINFORCED CONCRETE SLAB From Table 3.27 BS 8110, fy = 460 N/mm2

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, As = 314 mm2/m, top layer

6) Cracking check

The bar spacing does not exceed 750 mm or 3d and the minimum reinforcement is less than 0.3%. (Refer Cl. 3.12.11.2.7 and Table 3.30, BS 8110).

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 the span and top reinforcement over the supports. The effective span is the 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 ultimate design load throughout all spans or panels provided that the following condition are met for the single load case analysis, bending moment an shear forces coefficients as shown in Table 3.13, 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/mm2, plus floor finishes and ceiling load of 1.0 kN/mm2. The characteristic materials strength are fcu = 30 N/mm2, fy = 460 N/mm2.

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Figure 3.11 Continuous slab - example

Solution :

From Table 3.10, BS 8110, basic span- effective depth ratio = 26

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

= 173 mm

Try effective depth, d = 170 mm. Assume a mild exposure,

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

h = d + cover + Ø/2

= 170 + 20 + 5 = 195 mm, so taken h = 200 mm

Self-weight of slab = 0.2 x 24 = 4.8 kN/m2

Total dead load, Gk = 1.0 + 4.8 = 5.8 kN / m2

For 1 meter width of slab;

Ultimate load, F = (1.4gk + 1.6qk ) 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 > 30m2, the spans are equal and qk < 1.25 gk

the moment coefficients shown in Table 3.13 Bs 8110 may be used. Thus, assuming that the end support is simply supported, 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 < 0.156 fcubd2 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 / 0.87fy z = 22.5x106 / (0.87 x 460 x 161.5)

= 348 mm2/ m

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2) Span-effective depth ratio

M = 22.5 x 10 6 = 0.778 bd2 1000 x1702



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 give the steel areas shown in Figure 3.12.

3) Distribution / Transverse Steel

From Table 3.27 BS 8110, fy = 460 N/mm2

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, As = 262 mm2/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 103/ (1000 x 170)

= 0.21 N/mm2 < 0.8 √ fcu


100As / bd = 100 x 393 / 1000 x 170 = 0.23

So, v c = 0.47 x (30/25)1/3 = 0.50 N/mm2 ,

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

5) Cracking check

The bar spacing does not exceed 750 mm or 3d and the

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DESIGN OF REINFORCED CONCRETE SLABminimum reinforcement is less than 0.3%. (Refer Cl. 3.12.11.2.7 and Table 3.27 BS 8110).



Figure 3.12: Reinforcement detail in continuous slab

3.9.2 SOLID SLABS SPANNING IN TWO DIRECTIONS

When a slab is supported on all four of it sides it effectively spans in both directions, and it is sometimes more economical to design the slab on this basis. The amount of bending in each direction will depend on the ratio of the two spans and the conditions of restraint at each support.

If the slab is square and the restraints are similar along the four sides then the load will span equally in both directions. If the slab is rectangular then more than one-half of the loads will be carried in the stiffer, shorter direction and less in the longer direction. If one span is much longer than the other, a large proportion of the load will be carried in the short direction and the slab may as well be designed as spanning in only one direction.

Moments in each direction of span are generally calculated using coefficients which are tabulated in the codes of practice, B 8110. Areas of reinforcement to resist the moment’s are determined independently for each direction of span. The slab is reinforced with bars in both directions parallel to the spans with the steel for the shorter span placed furthest from the neutral axis to give it greater effective depth.

The span-effective depth ratios are based on the shorter span and the percentage of reinforcement in that direction.

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With a uniformly distributed load the loads on the supporting beams may 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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DESIGN OF REINFORCED CONCRETE SLAB3.9.2.1 Simply Supported Slab Spanning In Two Directions

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

msx = αsx nlx2 in direction of span lX

and

msy = αsy nlx2 in direction of span ly

where msx and msy are the moments at mid-span on strips of unit width with spans lx and respectively, and

n = (1.4Gk + 1.6Qk), that is, the total ultimate load per unit area ly = the length of the longer sidelx = the length of the shorter side

The area of reinforcement in directions lx and ly respectively are;

Asx = msx / 0.87fyz per metre width

andAsy = msy / 0.87fyz per metre width

The slab should be reinforced uniformly across the full width, in each direction. The effective depth d used in calculating Asy should be less than that for Asx 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 within 0.1lx, or 0.1ly of the appropriate support.

Example 3.3 :

Design the reinforcement for a simply supported slab 200 mm thick and spanning in two directions. The effective span in each direction is 4.5 m and 6.3 m and the slab supports a live load of 10 kN/m2. The characteristic material strengths are fcu = 30 N/mm2 and fy = 460 N/mm2.

Solution :

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

From Table 3.14, αsx = 0.099 and αsy = 0.051.

Self-weight of slab = 0.2 x 24 x 103 = 4.8 kN/m2

Ultimate load, n = 1.4Gk + 1.6Qk

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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 Ø bar = 10mm.

dx = h – c - Ø /2 = 200 – 25 – 5 = 170 mm.

msx = αsx nlx2

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

K = M = 45.5 x 10 6 = 0.052 < 0.156 fcubd2 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 = msx / 0.87fy z = 45.5 x106 / (0.87x 460)(0.94x170)

= 711.5 mm2/ m

Checking Asmin, from Table 3.27 BS 8110, fy = 460 N/mm2

Asmin = 0.13bh / 100 = 0.13(1000 x 200) / 100 = 260 mm2/ mAsx > Asmin → ok


2) Deflection Checking

M = 45.5 x 10 6 = 1.57 bd2 1000 x1702



20 x 1.41 > 4500 / 170

28.2 > 26.5 → ok

3) Shear

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


= 51.12 x 103/ (1000 x 170)

= 0.3 N/mm2 < 0.8 √ fcu


100As / bd = 100 x 786 / 1000 x 170 = 0.46

So, v c = 0.63 x (30/25)1/3 = 0.67 N/mm2 ,


Long Span

1) Bending

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

dy = h – c - Ø/2 = 200 – 25 -10 – 5 = 160 mm.

msy = αsynlx2

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

K = M = 23.5 x 10 6 = 0.031 < 0.156 fcubd2 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 = msy / 0.87fy z = 23.5 x106 / (0.87x 460)(0.95x160)

= 354 mm2/ m


Asmin = 0.13bh / 100 = 0.13(1000 x 200) / 100 = 260 mm2/ mAsx > Asmin → ok


2) Checking for Transverse Steel

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From Table 3.27, fy = 460 N/mm2

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

0.19 > 0.13 (Asmin) → ok

3.9.2.2 Restrained Slab Spanning In Two Directions

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

msX = βsXnlX2 in direction of span lx

and

msy = βsynlX2

in direction of span ly

where βsX and βSy are the moment coefficients given in Table 3.15 of BS 8110 for the specified end conditions, and n = (1.4Gk+ 1.6Qk), the total 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 msx and msy, In the edge strips only nominal reinforcement is necessary, such that 100As/bh = 0.13 for high-yield steel or 0.24 for mild steel.

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

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

2. extend from the edges a minimum distance lx / 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 moment

4. at a corner where the slab is discontinuous in one direction only, have an area of torsion reinforcement only half of that specified in rule 3.

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

Where ly /Ix > 2, the slabs should be designed as spanning in one direction only.

Shear force coefficients are also given in BS 8110 for cases where torsion corner reinforcement is provided, and these are based on a simplified distribution of load to supporting beams which may be used in preference to the distribution shown Figure 3.13.

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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 m and 6 m and the slab supports a live load of 1.5 kN/m2. Given fcu = 30 N/mm2, fy = 250 N/mm2 and slab thickness 150 mm. Design the reinforcement for a continuous slab.

Figure 3.16: Continuous panel spanning in to directions

Solution :

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

Self-weight of slab = 0.15 x 24 x 103 = 3.60 kN/m2

20 mm asphalt = 0.48 kN/m2

50 mm insulting screed = 0.72 kN/m2

Ceiling finishes = 0.24 kN/m 2

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ly = 6m

lx =5m

a b

d c


Total dead load = 5.04 kN/m2

Ultimate load, n = 1.4Gk + 1.6Qk

n = 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

msx = 0.032(9.5)(5)z = 7.6 kNm

msy = 0.024 (9.5)(5) = 5.7 kNm

- ve moment at support (cont)

a long AB & CD, msx = 0.042 (9.5)(5)2 = 10.2 kN.m

a long AD & BC, msx = 0.032 (9.5)(5)2 = 7.6 kN.m

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

dx = h - cover - Ø/2

= 150 - 25 - 10/2 = 120 mm

dy = h - cover – Ø- Ø /2

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

Short Span, l x1) At Mid-Span, msx = 7.6 kNm

K = M = 7.6 x 10 6 = 0.018 < 0.156 fcubd2 30(1000 )(120)2

zx = 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 = msx / 0.87fy z = 7.6 x106 / (0.87x 250)(0.95x120)

= 306.51 mm2/ m width


Asmin = 0.24bh / 100 = 0.24(1000 x 150) / 100 = 360 mm2/ mAsx < Asmin → so use Asmin

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Provide R10 bars at 200 mm centre, As = 393 mm2/m

2) At Support, msx = 10.2 kNm

K = M = 10.2 x 10 6 = 0.024 < 0.156 fcubd2 30(1000 )(120)2

zx = 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 = msx / 0.87fy z = 10.2 x106 / (0.87x 250)(0.95x120)

= 411.37 mm2/ m width


Asmin = 0.24bh / 100 = 0.24(1000 x 150) / 100 = 360 mm2/mAsx > Asmin → ok

Provide R10 bars at 175 mm centre, As = 449 mm2/ m

Long Span, l y

1) At Mid-Span, msy = 5.7 kNm

K = M = 5.7 x 10 6 = 0.016 < 0.156 fcubd2 30(1000 )(110)2

zy = 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 = msy / 0.87fy z = 5.7 x106 / (0.87x 250)(0.95x110)

= 250.78 mm2/ m width


Asmin = 0.24bh / 100 = 0.24(1000 x 150) / 100 = 360 mm2/ mAsy < Asmin → so use Asmin

Provide R10 bars at 200 mm centre, Asprov = 393 mm2/m

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2) At Support, msy = 7.6 kNm

K = M = 7.6 x 10 6 = 0.02 < 0.156 fcubd2 30(1000 )(110)2

zy = 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 = msy / 0.87fy z = 7.6 x106 / (0.87x 250)(0.95x110)

= 334.37 mm2/ m width


Asmin = 0.24bh / 100 = 0.24(1000 x 150) / 100 = 360 mm2/m

Asy < Asmin → so use Asmin

Provide R10 bars at 200 mm centre, Asprov = 393 mm2/ m

→Torsion reinforcement is not necessary because the slab is interior panel.

→Edge strip, provide Asmin (R10 -200mm c/c).

Shear Checking (Critical at Support)

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

From Table 3.16, βvx = 0.39, βvy = 0.33

Vvx = βvx.n.lx

= 0.39(9.5)(5) = 18.5 kN/m width

Vvx = βvy.n.lx

= 0.33(9.5)(5) = 15.7 kN/m width

Shear stress, v = Vmax / bd

= 18.5 x 103 / (1000 x 120)

= 0.15 N/mm2 < 0.8 √ fcu


100As / bd = 100 x 449 / 1000 x 120 = 0.374

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DESIGN OF REINFORCED CONCRETE SLABSo, v c = 0.6 x (30/25)1/3 = 0.64 N/mm2 ,


Deflection Checking (Critical at Mi-span), M sx = 7.6 kN.m M = 7.6 x 10 6 = 0.53

bd2 1000 x1202



26 x 2 > 4500 / 120

52 > 37.5 → ok

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



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

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.

BPLK 93 DCB 3223

Decide concrete grade, concrete cover, fire resistance 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 and 3.16, BS 8110: Part 1: 1985


Figure 3.17: Flowchart for slab design

3.12 REFERENCES

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

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

Design and Construction.

BPLK 94 DCB 3223

Reinforcement deign

Check shear

Check for serviceability limit state

reinforced slab

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continuous slab

ribbed slab floors

situ solid slab

solid slab b flat slab

concrete slab bplk66

types of slabflat slab

ribbed slabs

amonolithic concrete