Ground Bearing Slab

African Concrete Code symposium- 2005

156

DESIGN OF GROUND BEARING CONCRETE SLABS

JOHN KNAPTON

John Knapton Consulting Engineers Ltd. 85 Monkseaton Drive

Whitley Bay NE26 3DQ UK

www.john-knapton.com

SYNOPSIS: A series of ground bearing slab design methods is presented in this Paper. The methods are all similar in that they model the loading regime numerically, they assign elastic properties to the ground and they use the resulting information to produce data which allows a slab to be understood and installed. They differ in the way in which they transform the loading regime and ground properties into design output data. The appropriate design procedure depends upon the nature of the loading, upon the engineering properties of the materials from which the slab is to be constructed and upon the end use of the slab. For each design method, a worked example is presented. The paper also includes information on loads likely to be applied to industrial floors or hard standings and gives guidance on the provision of movement joints.


157

INTRODUCTION The design of ground bearing concrete slabs poses several problems for the structural engineer. Design methods for ground bearing slabs are not usually included in codes of practice because they differ radically from those design methods adopted for conventional reinforced or prestressed concrete elements. Yet there are usually far greater savings to be made in accurate design of ground bearing concrete than there are in any other structural element. The floor to a distribution warehouse is often of area 12,000m2 and thickness 200mm. Bringing this thickness down by 25mm can save 300m3 of reinforced concrete. The design methods presented in this paper ensure that ground bearing floor slabs are designed optimally. Also, the load induced tensile flexural stresses in ground bearing slabs are accommodated by the concrete, not by reinforcement. Usually, ground bearing slabs are reinforced, but the reinforcement is selected on the basis of controlling cracking resulting from restraint to shrinkage. Shrinkage occurs as the concrete dries and results in linear shrinkage and curling. The friction between the ground and the shrinking concrete develops tensile stress within the concrete and the self-weight of the concrete develops flexural stress as it attempts to reposition the curled slab on the ground. Additionally, the load regime applied to ground bearing slabs is more complex than is often the case for conventional structural elements. Whereas loads can frequently be represented as distributed or point loads, in the case of ground bearing slabs, the effect of either static or rolling patch loads needs to be considered. Furthermore, the effect of the interaction of nearby patch loads has to be assessed. Stresses resulting from restraint to shrinkage or usually secondary but in the case of ground bearing concrete slabs, they are of primary importance. The design methods presented in this paper address these issues. Data is included which can form the basis for the development of a


158

design code for ground bearing slabs. The failure of a ground bearing floor is one of the most feared structural problems. The disruption to the user is often total, involving decanting the operation to a temporary industrial unit while the floor is replaced. But perhaps the crucial matter is the fact that as Africa begins to prosper, there will be an overwhelming demand for new industrial buildings of all types and limited funding will create an imperative for a cost effective engineered ground bearing floor slab. The design methods described in this Paper are: 1. Using Hetenyi's equations for a slab subjected to a uniformly

distributed load over part of its area. 2. Using Westergaard's equations for patch loads. 3. Using Meyerhof's equations for patch loads 4. Using the Eurocode 2 method for punching shear 5. Using design charts developed from Westergaard equations for a

slab subjected to patch loads Table 1 shows the situations where each of the 5 design methods is appropriate. Additionally, this paper provides design guidance on the selection of an arrangement of joints so as to eliminate cracking.

Table 1: Design methods reviewed in this Paper Design Method

Type of slab Type of load Output

Hetenyi

Floor slabs Distributed Slab thickness

Modified Westergaard

All ground bearing slabs

Patch load Slab thickness & elastic deflexion

Meyerhof

Floor slab where cracking is acceptable

Patch load/Point load Floor Slab thickness

Eurocode 2


Patch load Confirmation that slab will not suffer shear failure

Westergaard design charts


Patch load/Point load Slab thickness


159

In all of the design methods, concrete characteristic flexural strength is used as the basis of stress assessment and load induced stresses are computed using classical methods. See Table 2 for concrete strength values.

The following factors have to be taken into account in ground floor design. a) Loading Regime b) Strength of Concrete c) Strength of existing ground and effect of the sub-base

STRENGTH OF CONCRETE Concrete slabs are frequently constructed from C30, C35 or C40 concrete with a minimum cement content of 300kg/m3 with a slump of 50mm or less. Design is based upon comparing concrete characteristic flexural strength with calculated flexural stresses whereas specification is by characteristic compressive strength. Table 2 shows flexural strength values for a range of commonly used concretes.

LOADING Loading on ground bearing floors comprises distributed loads and patch or point loads, typically applied by vehicles and materials handling equipment. Industrial floors are commonly required to accommodate distributed loads of 37.5kN/m2. This is a Table 2: Mean and characteristic 28-day flexural strength values for various concrete mixes. See Table 1.1 for other properties of concrete.


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Flexural Strength (N/mm2) Mean Characteristic ________________________________________________________________ Plain C30 concrete 2.0 1.8 Microsilica C30 concrete 2.4 1.9 C30 concrete 20kg/m3 ZC 60/1.00 steel fibre 2.8 2.0 C30 concrete 30kg/m3 ZC 60/1.00 steel fibre 3.2 2.2 C30 concrete 40kg/m3 ZC 60/1.00 steel fibre 3.8 2.7 ________________________________________________________________ Plain C35 concrete 2.2 1.95 Microsilica C35 concrete 2.6 2.0 C35 concrete 20kg/m3 ZC 60/1.00 steel fibre 3.0 2.1 C35 concrete 30kg/m3 ZC 60/1.00 steel fibre 3.4 2.3 C35 concrete 40kg/m3 ZC 60/1.00 steel fibre 4.0` 2.9 __________________________________________________________ Plain C40 Concrete 2.4 2.1 Microsilica C40 Concrete 2.8 2.15 C40 concrete20kg/m3 ZC 60/1.00 steel fibre 3.2 2.2 C40 concrete 30kg/m3 ZC 60/1.00 steel fibre 3.6 2.5 C40 concrete 40kg/m3 ZC 60/1.00 steel fibre 4.2 3.2 Plain C45 Concrete 2.7 2.3 Microsilica C45 Concrete 3.1 2.4 C45 concrete20kg/m3 ZC 60/1.00 steel fibre 3.5 2.5 C45 concrete 30kg/m3 ZC 60/1.00 steel fibre 4.0 2.8 C45 concrete 40kg/m3 ZC 60/1.00 steel fibre 4.8 3.6 Plain C50 Concrete 3.0 2.5 Microsilica C50 Concrete 3.4 2.6 C50 concrete20kg/m3 ZC 60/1.00 steel fibre 3.8 2.7 C50 concrete 30kg/m3 ZC 60/1.00 steel fibre 4.2 3.0 C50 concrete 40kg/m3 ZC 60/1.00 steel fibre 5.0 3.8 Plain C55 Concrete 3.4 2.7


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Microsilica C55 Concrete 3.8 2.8 C55 concrete20kg/m3 ZC 60/1.00 steel fibre 4.2 2.9 C55 concrete 30kg/m3 ZC 60/1.00 steel fibre 4.6 3.2 C55 concrete 40kg/m3 ZC 60/1.00 steel fibre 5.4 4.1 ZC 60/1.00 refers to a commonly used anchored bright wire fibre of total length 60mm and wire diameter 1.00mm. particularly high load which is rarely applied. It is specified in order that an industrial floor can be guaranteed to accommodate any future loading which a floor might need to sustain. It is in fact rare for a floor/hardstanding to be loaded by a distributed load. However, it has become a simple form of common currency which is used by the developers of industrial premises to assess the suitability of industrial premises for different end users. Even when a floor/hardstanding is specified in terms of distributed load, the designer should nonetheless consider the likely point or patch loads. This is because they often produce much higher stress levels than do distributed loads. Floors subjected to loads applied by storage racking are often designed to accommodate 10 tonne or 12 tonne end frames. End frames each have two legs so the patch load is half of these values. However, it is common for storage arrangements to cause two such frames to be positioned side by side. Therefore, the floor is commonly designed for point or patch loads of 10 tonne or 12 tonne. External hard standings may need to accommodate highway vehicles which will typically have patch loads of up to 6tonne. In container handling areas, such as ports, Reach Stackers or Front Lift Trucks handling 40ft containers have front axle loads of up to 110 tonne so the half-axle patch load can be up to 55 tonne. Dynamics may need to be applied to account for braking, turning and general dynamics. These factors may add 25% to the load so the total design patch load may be 70t.


162

Container stacking areas need to be designed to accommodate the patch loads applied by the corner castings of container feet. A typical loaded 40ft container weighs 26t so the corner casting load is 6.5t. Often containers are stored in blocks with four corner castings in close proximity. This applies an effective patch load of 26t to the slab. Containers are often stacked three high so the load applied to the slab is 78 tonne, although a reduction is sometimes made of say 20% to account for the fact that it is unlikely that all stacked containers will be loaded to their capacity. Sometimes containers are stacked five high which increases the loading further. Sometimes, the repetition of loads needs to be taken into account. A simple expedient is to increase the load by 50% and assume that it can be applied an infinite number of times. Alternatively, a full fatigue analysis will need to be undertaken which can prove mathematically difficult if the loads vary in intensity and duration of application. Load and strength partial safety factors of 1 are usually applied in the case of industrial floors or hard standings. This is because their failure is unlikely to lead to severe consequences such as death or injury. It is accepted that if the slab is overloaded or is under strength then cracking will develop. The alternative would be to provide an uneconomic solution.

STRENGTH OF EXISTING GROUND AND EFFECT ON SUB-BASE

The design methods require a value for the modulus of subgrade reaction (K) which defines the stiffness of the material beneath the slab. The following four values of K are used in the design procedures: K = 0.013 N/mm3 - very poor ground


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K = 0.027 N/mm3 - poor ground K = 0.054 N/mm3 - good ground K = 0.082 N/mm3 - very good ground (no sub-base needed) A subgrade with a K value of 0.027 N/mm3 will deflect vertically by 1mm when a vertical pressure of 0.027 N/mm2 or 27kN/m2 is applied through a standard disk. The beneficial effect of a granular sub-base may be taken into account by increasing K according to the thickness and strength of the sub-base. However, the above figures can be used conservatively.

STRESS IN CONCRETE The stress in a floor slab depends upon: a) The properties of the subgrade b) The loading regime: i) Uniformly Distributed Load (UDL) ii) Patch/Point loads c) The thickness of the floor slab d) The strength of the sub-base

HETENYI DESIGN METHOD FOR UNIFORMLY DISTRIBUTED LOADING

The common loading system comprises alternate unloaded aisles and loaded storage areas. The maximum negative bending moment (hogging) occurs within the centre of the aisles and is given by:

( )''22 bahog BBqM λλλ−−=

where a = width of the aisle b = width of the loaded area


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a' = a/2 b' = a/2 + b q = UDL(characteristic load x load factor)(N/mm2) � = (3K/Eh)1/4 K = modulus of subgrade reaction E = concrete modulus (10,000N/mm2 for sustained load) h = slab thickness The maximum positive bending moment (sagging) occurs beneath the centre of the loaded area (loading from adjacent blocks is ignored ): Msag = q/2�2.B(1/2)�b where Bx = e-x .sin x The two moment equations can be simplified into a single conservative equation for any combination of aisle width and stacking zone width: Mmax = -0.168q/�2 The corresponding maximum flexural stress is given by: �max = 6.Mmax /h2

�max=1008 q/(�2 h2) (N/mm2)

where the maximum flexural strength cannot exceed the relevant characteristic value from Table 2. To ease calculation, Table 3 shows values of �2h2 for common combinations of modulus of subgrade reaction (K) and slab thickness.


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Table 3 Values of λ2h2 for combinations of slab thickness and modulus of sub grade reaction. ________________________________________________________________ �2h2 for slab thickness (mm): 150 175 200 225 250 Modulus of subgrade reaction K (N/mm3) 0.082 0.061 0.066 0.070 0.074 0.078 0.054 0.049 0.053 0.057 0.060 0.063 0.027 0.035 0.038 0.040 0.043 0.045 0.013 0.024 0.026 0.028 0.029 0.031 Uniformly Distributed Load Example Consider a 150mm thick floor slab carrying a uniformly distributed load of 50kN/m2 between aisles on poor ground.

222max /

.008.1 mmN

hq

λσ =

22 /10.02/05.0 mmNxmmNq ==

(In the above equation, 2 is the fatigue factor. A value of 2 means a load of 0.05N/mm2 can be repeated indefinitely)

For poor ground K = 0.027N/mm3 Therefore:

2max /88.2

035.010.0008.1 mmN=

=σ


166

From the characteristic strength values in Table 2, this stress can be withstood by a C40 concrete incorporating 40kg/m2 of 60mm long 1mm diameter anchored steel fibre. This design solution can be used for any combination of aisle and stacking zone width.

MODIFIED WESTERGAARD DESIGN METHOD FOR PATCH LOADS

Highway vehicles and handling equipment apply loads to the surface of a floor as patch loads. In most cases, sufficient accuracy is gained by assuming the patch to be circular and to apply uniform stress throughout the patch. These assumption lead to minor errors - the contact patch shape in the case of a commercial vehicle is nearer to a rectangle than to a circle but to assume accurate shaped patch loads would preclude the use of Westergaard equations. The error in assuming constant stress circular loading is very small and is conservative i.e. true analysis would lead to a slightly thinner floor. The maximum flexural tensile stress occurs at the bottom (or top, in the case of corner loading) of the slab under the heaviest wheel load. The maximum stress under a patch load can be calculated by the elastic Westergaard4.1 and Timoshenko equations, sometimes referred to as the modified Westergaard equations: a) patch load in mid-slab (i.e. more than 0.5m from slab edge)

1.........................36.0log.)1(275.04

3

2max

+=

KbEhP

hµσ

b) patch load at edge of slab

( ) 2................................20.0log54.01529.0 4

3

2max

+=

KbEh

hPµσ


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c) patch load at slab corner

( )3........................................

112

41.113

6.0

25.0

2

32max

−

−=

KEh

bhP

µ

σ

where �max = flexural stress (N/mm2) P = point load (N) i.e. characteristic wheel load x fatigue factor � = Poisson's ratio, usually 0.15 h = slab thickness (mm) E = elastic modulus, usually 20,000N/mm2 K = modulus of sub grade reaction (N/mm3) b = radius of tyre contact zone (mm) = (P/�.p)1/2 p = contact stress between wheel and floor (N/mm2) Twin wheels bolted side by side are assumed to be one wheel transmitting half of the axle load to the floor. In certain cases wheel loads at one end of an axle magnify the stress beneath wheels at the opposite end of the axle, distance S away (S is measured between load patch centres). To calculate the stress magnification, the characteristic length (radius of relative stiffness, l) has to be found from Equation 4.

( ) 4........................112

25.0

2

3

−

=K

Ehlµ

Values of l are shown in Table 4. Once Equation 4 has been evaluated, the ratio S/l can be determined so that Fig. 1 can be used


168

to find Mt/P. Where Mt is the tangential moment. The stress under the heaviest wheel is to be increased to account for the other wheel. The stress to be added is calculated from the following equation.

5.............................622 P

hPMt

add =σ

where: P = greater patch load, P2 = other patch load. Sum the stresses and verify that the characteristic flexural strength in Table 4.1 has not been exceeded for the appropriate concrete mix.

Figure 1: Relationship between ratio S/l and ratio Mt/P used in assessing the influence of load proximity, Highway Vehicle Example using Westergaard Equations


169

Consider a highway vehicle with a rear axle load of 8000kg and assume a slab thickness of 200mm on good ground (K=0.054N/mm3). The fatigue factor is 2.0 so the slab can withstand an infinite number of such repetitions. The design axle load is 160,000N (i.e.80,000N x fatigue factor(2.0)) so the design wheel load is 80,000N p = contact stress between wheel and floor = 0.7N/mm2 b = radius of contact area = (W/�p)1/2 = (80,000/0.7)1/2 =191mm Substituting known values into Equation 2:

( )

+= 4

3

2max 191054.0200200002.0log

2008000015.054.01529.0

xxxσ

2max /03.3 mmN=σ

beneath one wheel.

Assume that the wheel at the other end of the axle is 2.7m away (S = 2700mm) The radius of relative stiffness (l), using Equation 4 is:

( )25.0

2

3

054.015.011220020000

−

=xl

mml 709=

8.37092700

==lS

From Figure 1

005.0=P

Mt


170

Therefore, using Equation 5

22

6 PhP

Mtadd =σ

80000.200

6005.0 2

=

2/06.0 mmN= Total Stress = 3.03 + 0.06N/mm2

= 3.09N/mm2 By comparing this stress with the characteristic strength of C40 concrete reinforced with 40kg/m3 steel fibres(3.2N/mm2), it can be seen that the proposed mix is satisfactory. A C40 concrete incorporating 40kg/m2 60mm long 1mm diameter anchored steel fibres is inadequate for this design. However, the inclusion of a 250mm thick granular sub-base would enhance K from 0.054 to 0.073 which would reduce the stress to below the characteristic value.

MEYERHOF DESIGN METHOD FOR POINT & PATCH LOADS

The Meyerhof design method is based upon yield line analysis and provides a relatively simple way of selecting the thickness of a concrete slab for a single patch load and for some combinations of similar patch loads. Meyerhof developed his equations by considering how a concrete slab would fail. His initial work related to suspended slabs but he later applied the method to ground bearing slabs. He considered the way in which a slab would ultimately fail and equated the potential energy lost when the load moved downwards to the strain energy absorbed along the cracks which form at the time of failure.


171

Consider a patch load applied to a ground bearing slab gradually increasing in magnitude. The slab directly beneath the load sags and develops tension towards the underside of the slab. Further away from the patch load, the tension develops in the upper part of the slab as the bending transforms from sagging to hogging. For failure to occur, the concrete has to crack in these tension zones. The yield line method assumes that at failure, the slab breaks into a series of plates and that sufficient cracks (or "plastic hinges") need to form to generate a collection of hinged plates. Figure 2 shows typical patterns of hinges generated by one, two and four patch loads.

SINGLE PATCH LOAD YIELD LINE PATTERN

Twin point load yield line pattern

FOUR POINT LOAD YIELD LINE PATTERN

Figure 2: Yield lines or plastic hinges resulting from single, twin and four patch loads.


172

These are the Meyerhof equations for point loads: i) Single point load applied within the body of the slab:

Pu = 2π Mp + Mn[ ] ii) Single point load applied along a free edge of a slab:

Pu = π Mp + Mn[ ]/ 2( )+ 2Mn iii) Single point load applied at a corner of a slab:

nu MP 2= iv) Two similar point loads spaced x apart and away from edges or corners of the slab:

Pu = 2π +1.8x

l

Mp + Mn[ ]

v) Four similar point loads applied at corners of a rectangle of side lengths x and y, all four loads away from edges or corners of the slab.

( ) [ ]npu MMl

yxP +

+

+=8.12π

The corresponding equations for patch loading are:

i) Single patch load applied within the body of the slab:

Pu = 4π Mp + Mn[ ]/ 1 −a3l

ii) Single patch load applied along a free edge of a slab:


173

[ ]( )

−++=

laMMMP nnpu 3

21/4π

iii) Single patch load applied at a corner of a slab:

nu MlaP

+=

412

iv) Two similar patch loads of radius a spaced x apart and away from edges or corners of the slab:

[ ]npu MMal

x

la

P +

−

+

−

=

2

8.1

31

4π

v) Four similar patch loads applied at corners of a rectangle of side lengths x and y, all four loads away from edges or corners of the slab.

( ) [ ]npu MMal

yx

la

P +

−

++

−

=

2

8.1

31

4π

Where: Pu = Ultimate load Mp = Ultimate positive (sagging) moment of resistance of the slab Mn = Ultimate negative (hogging) moment of resistance of the slab The above two values can be obtained from the equation:

=

6

2

,hfM np


174

where: f = characteristic flexural strength of concrete - see Table 2 h = slab thickness a = radius of patch load x = spacing of two point of patch loads (centre to centre of load) y = spacing of four point loads forming corners of a rectangle - y measured at right angles to x. l = radius of relative stiffness:

( )25.0

2

3

112

−

=K

Ehlµ

(See Table 4.4 for values of l. Table 4.4 reproduced here for convenience)

Table 4: Radius of relative stiffness values for different slab thicknesses and support conditions

Modulus of subgrade reaction (K) (N/mm3) Slab thickness (mm) 0.013 0.027 0.054 0.082 150 816 679 571 515 175 916 763 641 578 200 1012 843 709 639 225 1106 921 774 698 250 1196 997 838 755 275 1285 1071 900 811 300 1372 1143 961 E = Elastic modulus = 20000 N/mm2 � = Poissons Ratio = 0.15


175

Note that the Meyerhof patch load equations apply when :

al( )≥ 0.2

For values of a/l between 0.2 & zero, interpolate between the patch load and the point load Mu values. Highway Vehicle Example using Meyerhof Equations Consider a same highway vehicle with a rear axle load of 8000kg. Assume a slab thickness of 200mm on good ground (K = 0.054N/mm3). The load is dynamic so a factor of 1.6 can be applied. In the case of a highway vehicle, this will take care of both dynamics and fatigue. The design axle load is 128,000N (i.e.80,000N x dynamic/fatigue factor(1.6)) so the design wheel load is 64,000N (64kN). Consider a slab of thickness 200mm and try C40 concrete containing 40kg/m3 of anchored steel fibres. Radius of patch load, a = (64,000/π0.7)1/2 = 170mm From Table 2, characteristic strength of concrete = 3.2N/mm2. Apply the Material Partial Safety Factor of 1.3 to obtain design strength: 3.2/1.3=2.5N/mm2.

Now use:

=

6

2

,hfM np to obtain the ultimate positive and

negative moments of resistance of the proposed slab: Mp = Mn = 2.5.2002/6 = 16,667N-mm per mm width of slab. From Table 4, the Radius of Relative Stiffness is 709mm. Note that this is the distance from the centre of the load to where the bending moment in the slab changes from sagging to hogging.


176

Consider the situation where this single patch load is applied away from the slab edges or corners.

[ ]

−+=

laMMP npu 3

1/4π

Pu = 455,312N = 455kN (64kN required) Consider the slab edge condition:

[ ]( )

−++=

laMMMP nnpu 3

21/4π

Pu = 204kN (64kN required) Consider the load applied at the corner of the slab:

nu MlaP

+=

412

Pu= 65,304N = 65kN (64kN required) Consider also both wheels separated by 2.7m. Introduce all of the now known parameters into the equation:

[ ]npu MMal

x

la

P +

−

+

−

=

2

8.1

31

4π

(x is the spacing of the rear wheels, which in this case is 2700mm.)

[ ]667,16667,16

21702127

4860

21271701

4+

−

+

−

=π

uP


177

334,3320424860

92.056.12

+=uP

= 534.344N = 534kN

i.e. each wheel load = 534/2 = 267kN (64kN required) This result indicates that a 200mm thick slab comprising C40 steel fibre reinforced concrete can sustain between 455kN and 65kN depending upon the position of the load in relation to the slab. The Westergaard solution suggested that such a slab could sustain 80kN. Both methods used similar levels of safety factor, although they were applied in different ways. The reason for the difference between Westergaard (or other elastic methods set out in this Paper) and Meyerhof solutions is the redundancy of a concrete slab. The Westergaard solution considers the critical stress point and ensures that the actual stress remains below the strength of the concrete by the safety factor at that point. Meyerhof recognises that the slab can crack only when sufficient plastic hinges have formed to develop one of the mechanisms shown in Figure 2. This means that there is a significant reserve of strength beyond the condition when stress reaches strength at the critical point i.e. there is a significant reserve of strength beyond the Westergaard elastic failure criterion. Loading beyond this condition will not lead to failure, rather the stress pattern is redistributed until stress has reached concrete strength along all of the plastic hinges which form the collapse mechanism. As can be appreciated from the Meyerhof example, the load can grow to several times the Westergaard limiting value. However, there remains the possibility that cracks will develop in the critical stress location once the Westergaard elastic condition has been surpassed. In the case of an unreinforced slab, this could lead to the slab's breaking into two which would be unacceptable in most circumstances. However, in the case of steel mesh or steel fibre reinforced concrete, the crack would be bridged by the reinforcement. The Author's tests have shown that steel fibre


178

reinforced concrete has a residual flexural strength after the development of the first crack. Therefore, it is likely that steel fibre reinforced concrete slabs will crack before the Meyerhof condition is reached but that the cracks will be controlled by the steel fibres. Also, it should be recognised that the benefit of redundancy is less marked in the situation where a single patch load is applied at the corner of a slab. For this reason, Meyerhof is most beneficial in jointless slabs where this condition can be omitted. Which should be used, Westergaard (or the other elastic methods set out in this Paper) or Meyerhof? The Author has been involved in full scale tests in which a steel fibre reinforced slab (C40 concrete, 30kg/m3 steel fibre content) of thickness 140mm withstood a patch load of 290kN, suggesting that the Meyerhof method is not unconservative. For several years, floor construction contractors have reported that slabs much thinner than those which Westergaard's equations would require appear to function satisfactorily. Therefore, it would seem sensible to use the Meyerhof equations for ground bearing floors subjected to point or patch loads in situations where cracks are acceptable. However, in situations where any degree of cracking would be unacceptable, then Westergaard equations or the other elastic methods presented in this Paper should be used. This will be the case generally for external slabs subjected to the effects of weather. The Meyerhof equations should be used for steel mesh or steel fibre reinforced concrete floor slabs where some controlled cracking is acceptable. However, it should be recognised that stresses induced by patch or point loads may represent a small part of the stress regime. Restraint to thermal and moisture induced movement may be the principal structural action. In the past, it is likely that the conservative Westergaard equations provided the requisite factor of ignorance which allowed those additional stresses to be accommodated. If Meyerhof equations are to be used in design, then a full understanding of the behaviour of the slab will be required. In particular, the possibility of punching shear failure must be investigated and the stresses arising from restraint to shrinkage and curling must be evaluated. The slab must then be


179

proportioned on the basis of the most adverse combination of all of the stresses which are likely to occur, not just those resulting from patch or point loads.

EUROCODE 2 DESIGN METHOD FOR PUNCHING SHEAR CHECK OF PATCH LOADS

Even when elastic and plastic design methods confirm that a slab is adequate, there is the possibility that the load can punch directly through the slab as a result of shear stresses exceeding the shear strength of the slab around the perimeter of the patch load. A check should be made as follows. Firstly, calculate the area over which the shear stress acts. This is the depth of the slab multiplied by the perimeter of the load patch. Divide the factored load by this area to obtain the shear stress. This stress should not exceed the shear strength of the concrete vm which is calculated from:

cdm vfv 5.0= where fcd= design cylinder strength (see Table 1.1 - divide the characteristic cylinder strength by 1.3)

and

−= 25016.0 ckfv

(fck = characteristic cylinder strength - see Table 1.1) Consider a 175mm thick C35 slab loaded by a circular 90kN patch load of diameter 250mm. The area over which the vertical shear force acts is: π x load patch diameter x slab thickness = π.250.175 = 137,444mm2. The shear stress is 90,000/137,444 = 0.65N/mm2. Now calculate the allowable shear stress as follows:

−= 25016.0 ckfv


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[ ]2505.2816.0 −=v

(see Table 1 for the fck value of 28.5) Therefore, 53.0=v Substitute in: cdm vfv 5.0= where the cylinder design strength fcd is 28.5/1.3 = 22N/mm2. (1.3 is the Material Partial Safety Factor for concrete) Therefore, the allowable shear stress vm = 0.5 x 0.53 x 22 = 5.83N/mm2. By comparing this figure with the applied shear stress of 0.65N/mm2, it can be concluded that there is no likelihood of punching shear failure in this instance. Note that for suspended slabs, a further shear check is made at a distance of twice the slab depth from the perimeter of the patch load. Because of the support offered to a ground bearing slab, this is rarely necessary.

DESIGN METHOD FOR PATCH LOADS USING DESIGN CHARTS DERIVED FROM WESTERGAARD EQUATIONS The above example illustrates the complexity of undertaking calculations in the case of patch loads. A simplified version of the elastic Westergaard method has been developed by the Author specifically for designing industrial floors/hardstandings. In order to eliminate much of the effort, a series of 10 design charts has been developed using Westergaard calculations. These charts are reproduced at the end of this Paper. The procedure is as follows: 1. Assess the existing conditions: 2. Determine the Actual Point Load (APL) and Modulus of

Subgrade Reaction (K) values to confirm the category of subgrade.

3. Calculate the additional stress generated by point or patch loads or wheels in close proximity.


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• If the distance between loads (S) is greater than 3m, the APL can be used directly (depending on the radius of contact zone) to calculate the thickness of the slab using the relevant Design Chart. In this case, go directly to Stage 6.

• If the distance between loads is less than 3m, the radius of relative stiffness (l) has to be determined. Table 4.4 shows values of radius of relative stiffness for different K values and slab thicknesses.

4. From Figure 1, determine Mt/P, the ratio of the tangential

moment to the greater point load by calculating S/l. Then use the following equation to determine the stress to add.

5........................622 P

hPMt

add =σ

where P = greater patch load (N) P2 = other patch load (N) 5. From Table 2, select a proposed concrete mix, hence

characteristic strength, �char. 6. When two patch loads are acting in close proximity (i.e. less

than 3m apart), the greater patch load (P) produces a flexural strength �max directly beneath it's point of application. The nearby smaller patch load (P2) produces additional stress �add beneath the larger load. Calculate �max ,

6...........................max addchar σσσ −=

7. Calculate the Single Point Load (SPL) which, acting alone

would generate the same flexural stress as the actual loading configuration.

7.........................max

=

σσ charAPLSPL


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where APL = Actual Point Load 8. Prior to using the Design Charts, it is necessary to modify the

SPL (Single Point Load) to account for contact area as well as wheel proximity to obtain the ESPL (Equivalent Single Point Load). Design Charts 1 to 10 apply directly when patch loads have a radius of contact between 150mm and 250mm. Some racking systems and pallet transporters have a contact radius of less than 150mm and some vehicles have a contact radius greater than 250mm. In these cases multiply the patch load by a factor in Table 5 prior to use in the Design Chart.

9. Use the corresponding Design Chart for the mix selected in 3) to determine slab thickness and return to 3) if an alternative concrete mix is required.

Table 5: Patch load multiplication factors for loads with a radius of

contact outside the range 150mm to 250mm Modulus of Subgrade Reaction (K) N/mm3 Radius of contact (mm) 0.013 0.027 0.054 0.082 50 1.5 1.6 1.7 1.7 100 1.2 1.2 1.3 1.3 150 1.0 1.0 1.0 1.0 200 1.0 1.0 1.0 1.0 250 1.0 1.0 1.0 1.0 300 0.9 0.9 0.9 0.9 Design Example for multiple patch loading using design chart A concrete floor is subjected to two patch loads. A 60kN patch load is applied 1m away from a 50kN patch load. The 60kN patch load has a contact zone radius of 100mm and the 50kN patch load has a


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300mm radius. The existing ground conditions are poor (K=0.027N/mm3). Assume thickness of slab = 225mm The radius of relative stiffness, l is given by Table 4 From Table 4 l = 921mm distance apart = 1m (i.e. S = 1000mm)

086.1921

1000==

lS

From Figure 1

053.0=P

Mt

so from Equation 5,

50000200

6053.0 2

=addσ

�add = 0.4N/mm2 Try steel fibre reinforced C30 concrete with a characteristic strength of 2.2N/mm2 (30kg/m3 steel fibre dosage - see Table 2) i.e. �char = 2.2N/mm2

addchar σσσ −=max = 2.2 - 0.4 = 1.8N/mm2 This is the maximum flexural stress which the 60kN load can be allowed to develop. Calculate the Single Point Load using Equation7:


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=

maxσσ charAPLSPL

kN738.12.260 =

=

From Table 5, the modified factor to be applied to the SPL to obtain the ESPL is 1.2:

kNx 882.173 = From Design Chart No.4, thickness of slab required = 225mm. A 225mm thick C30 concrete slab incorporating 30kg/m3 steel fibre is adequate for this design.

SPACING OF JOINTS Both construction and movement joints are required in ground bearing industrial slabs. Construction joint spacings will depend upon the type of equipment used to install the floor. Recent developments in large pour technology allow floors/external hardstandings of up to 3,000m2 to be constructed without construction joints in one day. If no movement joints were included in such a floor/hardstanding, it would crack. This is because the friction between the ground and the shrinking concrete develops tensile stress within the concrete and the self-weight of the concrete develops flexural stress as it attempts to reposition the curled slab on the ground. This cracking is controlled by including movement joints at spacings which experience indicates cracking would otherwise occur. The crack spacings are rarely calculated because it is difficult at design stage to predict the factors which will govern those spacings. Table 6 shows joint spacings for various types of concrete. The joint spacings in the Table have been used successfully for many


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years in the UK. Some consider that spacings can be greater than 12m for steel fibre reinforced floors/hardstandings. Whilst 14m joint spacings will probably be acceptable for most applications, the additional movement which would occur at joints might lead to loss of aggregate interlock and joint degradation. Joints spacings are frequently designed to coincide with column spacings or other features such as recesses or floor width changes. In any project, it is necessary to develop a joint layout identifying all practical issues prior to the placing of concrete. Table 6: Common Concretes and suggested joint spacings.

Concrete type Joint spacing (m) Plain C30 concrete 6 Microsilica C30 concrete 6 20kg/m3 ZC 60/1.00 steel fibre reinforcement C30 concrete 6 30kg/m3 ZC 60/1.00 steel fibre reinforcement C30 concrete 8 40kg/m3 ZC 60/1.00 steel fibre reinforcement C30 concrete 10 Plain C40 concrete 6 Microsilica C40 concrete 6 20kg/m3 ZC 60/1.00 steel fibre reinforcement C40 concrete 6 30kg/m3 ZC 60/1.00 steel fibre reinforcement C40 concrete 10 40kg/m3 ZC 60/1.00 steel fibre reinforcement C40 concrete 12 Joint spacings can be increased beyond the figures shown in Table 6 by the use of steel mesh reinforcement. Table 7 shows commonly available steel mesh sizes. Steel mesh is assumed to carry the tensile forces developed in the concrete resulting from restraint to shrinkage caused by loss of moisture or temperature. Consequently mesh allows greater joint spacings. Steel mesh is often used in long strip floor construction as it can be placed conveniently without the need for cutting. Where necessary, mesh sheets should be lapped at their edges and ends by 450mm. Overlapping can result in an unacceptable build up of thickness of reinforcement. There is often debate regarding the positioning of the mesh in relation to the depth of the concrete. From an installation point of view, it is simpler to position the mesh near the bottom of the concrete. The benefit of positioning the mesh near the upper


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surface is that it can control any surface cracking which might develop as a result of installation defects. Table 7: Commonly available steel mesh sizes.

Longitudinal wires Cross wires Mesh

reference Nominal size (mm)

Pitch (mm)

Area (mm2/m)

Nominal wire size

(mm)

Pitch (mm)

Area (mm2/m)

Mass (kg/m2)

Square meshA393 A252 A193 A142 A98

10 8 7 6 5

200 200 200 200 200

393 252 193 142 98

10 8 7 6 5

200 200 200 200 200

393 252 193 142 98

6.16 3.95 3.02 2.22 1.54

Structural mesh

B1131 B785 B503 B385 B283 B196

12 10 8 7 6 5

100 100 100 100 100 100

1131 785 503 385 283 196

8 8 8 7 7 7

200 200 200 200 200 200

252 252 252 193 193 193

10.9 8.14 5.93 4.53 3.73 3.05

Long mesh C785 C636 C503 C385 C283

10 9 8 7 6

100 100 100 100 100

785 636 503 385 283

6 6 5 5 5

400 400 400 400 400

70.8 70.8 49 49 49

6.72 5.55 4.34 3.41 2.61

Wrapping mesh D98 D49

5

2.5

200 100

98 49

5

2.5

200 100

98 49

1.54 0.77

Longitudinal wires Cross wires Sheet area Stock sheet size Length 4.8m Width 2.4m 11.52m2

Pitch : The centre-to-centre spacing of wires in a sheet of mesh. Width : When specifying mesh the width is the overall dimension measured in the direction of the cross wires. Wrapping mesh: Wire usually of Grade 250 for use in wrapping mesh. Sheet size: Mesh types "A" and "B" is delivered in standard sheets of 4.8 x 2.4m, or in scheduled size sheets. Mesh type "C" is available in sheets or rolls.


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It is common for long mesh to be used to enhance joint spacings. Table 8 shows joint spacings for four commonly used structural mesh sizes.

Table 8: Enhanced joint spacing with Long Mesh reinforcement

Structural Mesh Name

Weight of Reinforcement

(kg/m2)

Wire Sizes (Diameter in mm)

Joint Spacing (m)

C283 2.61 6mm main wires 5mm cross wires 16




CONCLUSIONS The paper presents a series of simple design tools which can be used to proportion the thickness of a concrete ground bearing industrial floor or external hard standing and which can be used to select an arrangement of movement joints. The methods have proved successful in many parts of the world and the ground bearing floors and hard standings which they produce can be expected to have a design life in excess of 40 years. The designer has a choice between no reinforcement, using polypropylene fibres which provide a small level of reinforcement, steel fibres which provide a significant level of reinforcement or steel mesh which provides an enhanced level of reinforcement. In the case of ground bearing industrial concrete floors and hard standings, the reinforcement is not provided to accommodate flexural and/or tensile stresses as is the case with conventional reinforced concrete. Rather, it controls the cracking resulting from restraint to shrinkage. Therefore, the provision of progressively higher levels of reinforcement permits wider joint spacing without the risk of cracking, or rather with a series of narrow cracks which are of no consequence.


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The information included in the Paper has been selected to include that which would be of relevance to an African Concrete Code. It is a self-contained document and would allow designers to proportion ground bearing concrete slabs with confidence.

Ground Bearing Slab

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