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Design of Slabs-on-Grade

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1 Design of Slabs-on-Grade
CE A433 – RC Design T. Bart Quimby, P.E., Ph.D. Spring 2007

2 Introduction Slabs on grade are PAVEMENTS not generally structural elements Pavements pass loads through compression to the supporting soil As long as the soils deformations are low, there is negligible bending in the slab Slabs on grade are deemed to be successful if there is little or no cracking

3 Pavement Apply load to top of slab
Since the slab is stiffer than the soil the load is distributed over a larger area of soil A thicker slab is stiffer and distributes the load over a larger area of soil

4 Types of Cracks Structural Shrinkage
Structural cracks are the result of subgrade settlement and/or stiffness discontinuity Often occur when a floor is over loaded Shrinkage Shrinkage cracks occur soon after a floor slab DRIES and will not increase in length, width or number after the drying process is completed.

5 Causes of Structural Cracking
Virtually all structural cracks are the result of subgrade failure The failure may result from one or more of the following conditions The subgrade is improperly designed or prepared The slab thickness is too thin for applied loads and the stiffness of the subgrade The concrete does not have sufficient strength It is necessary to determine the stiffness of the subgrade and the magnitude of the expected loads so that the proper slab thickness can be determined

6 Structural Cracks Cracks form when the Moment exceeds the Cracking Moment

7 Thickness Design of Slabs on Grade
Slabs on grade are, to a limited extent, beams on elastic foundations. The softer the supporting soil and/or the larger the load, the stronger and stiffer the slab must be to spread the load over more of the supporting soil Slab stiffness is a function of slab thickness Slab cracking strength is a function of concrete strength and slab thickness

8 Thickness Design Procedures
Portland Cement Association Wire Reinforcing Institute Corp. of Engineers

9 PCI Method A series of charts for various loading conditions (wheels, racks, posts, etc) Example of slab thickness determination for a wheeled vehicle: Data for lift truck Axle load = 25 k Wheel spacing = 37 in Number of wheels = 2 Tire inflation pressure = 110 psi

10 PCI Example Continued Contact area = wheel load/inflation pressure
Contact area = (25,000 lb / 2 wheels) / 110 psi = 114 in2 Subgrade and Concrete Data Subgrade Modulus, k = 100 pci Concrete 28-day strength, f’c = 7,000 psi Concrete flexural strength, MR ~ 7.5sqrt(f’c) ~ 640 psi

11 PCI Example Continued Use a factor of safety of 2.0
Choice depends of number of stress repetitions permitted Concrete working stress = MR/FS WS = MR/FS = 640 psi / 2 = 320 psi Slab stress per 1,000 lb of axial load WS / axle load, kips = 320/25 = 12.8 psi per 1,000 lbs.

12 PCI Example Continued Effective Contact Area
Slab Stress per 1,000 lb of axle load Use 8” Slab Subgrade Modulus Wheel Spacing

13 PCI Chart for Racks Need to match criteria for the chart
Read the instructions for each chart!

14 Causes of Shrinkage Cracking
Shrinkage cracking occurs due to the normal volumetric changes associated with drying Normal concrete can only stretch about inches per foot without rupturing Normal shrinkage is about (+25%) inches per foot If the slab is restrained against movement then cracking is inevitable

15 Minimizing Shrinkage Cracking
Shrinkage cracking can be minimized by Reducing the shrinkage characteristics of the concrete mix Reducing restraint on the slab Shrinkage cracking can be controlled by Encouraging cracks to appear at predetermined locations The use of reinforcing steel

16 Reducing Shrinkage Characteristics of the Concrete Mix
Reduce the volume of water in the mix The challenge is to limit the amount of water in the mix while maintaining workability and finishability without excessive use of water reducers Use coarser ground cement Use the largest sized aggregate permitted by design Use shrinkage compensating concrete

17 Reducing Shrinkage Characteristics of the Concrete Mix (Cont.)
Use proper curing techniques Proper curing keeps water in the concrete until it has achieved sufficient tensile strength before shrinkage occurs Proper curing allows drying to occur more evenly through the slab thickness

18 Curling Differential shrinkage due to drying can result in “curling” of the slab edges, resulting in an induced moment in the slab. When the moment equals the cracking moment a crack forms, redistributing the stress

19 Sources of Restraint Friction between the slab and the ground
As the slab shrinks, the friction resists the motion, causing tension in the slab Bearing on other features (walls, foundation, drain pipes, columns, etc) Attachment to other features

20 Friction Restraint Tensile Capacity Axial Stress Diagram

21 Shrinkage Cracks Tensile Capacity Axial Stress Diagram

22 Restraint by Features

23 Locating “Cracks” Control and construction joints are places of intentional weakness. They are placed close enough together to keep tensile stresses in the slab below the tensile rupture strength of the concrete

24 Control Joints The purpose of these joints “is to predetermine the location of cracks for esthetic and performance purposes.” ACI 302.1R, pg 6 “Unless the design provides for the specific supplemental reinforcing across the joint, the resulting induced crack may offer no structural advantage over a randomly occuring shrinkage crack.” ACI 302.1R, pg 6

25 Construction Joints These joints “are placed in a slab where the concreting operations are concluded for the day, generally in conformity with a predetermined joint layout. If at any time concreting is interrupted long enough for the placed concrete to harden, a construction joint should be used.” ACI 302.1R pg 6

26 Control Joint Details

27 Construction Joints

28 Joint Spacing Unreinforced Slabs Reinforced Slabs
2 to 3 ft for each inch of slab thickness. Smaller aggregate size, higher water contents, and local experience may dictate use of closer joints Reinforced Slabs Use a subgrade drag equation to compute joint spacing (See ACI 360R 6.3)

29 Drag Equation Where: L = distance between joints, ft
As = Area of steel per foot width of slab, in2/ftw fs = Allowable steel stress (20,000 psi or 24,000 psi) W = Dead weight of slab, psf m = Friction factor (1 to 2.5)

30 Important Concepts for Joint Details
Only reinforcement across the joint is to be used for vertical load transfer only. Use plain bars and coat to prevent bond to concrete Joint should extend at least ¼ slab thickness through the slab Vertical load transfer across construction joints can be accomplished with plain bars or properly designed keyed joints.

31 Joints have vertical transfer but allow in plane shrinkage movement
No Vertical Load Transfer

32 Controlling Shrinkage Cracking with Reinforcing Steel
“Reinforcement serves to restrain the shrinkage, effectively subdividing the slab and hence distributing the crack area more evenly. This produces smaller and more numerous cracks than would occur in an unreinforced slab of the same dimensions. The actual crack area remains essentially the same.” Fricks, T.J. “Cracking in Floor Slabs”, reprinted in ACI SCM-25 (92), pg 122.

33 Reinforcing Steel Smaller bar sizes are better choices than large diameters This steel “should be positioned one-fourth the slab thickness below the top surface up to 2.0 in maximum.” ACI 302.1R, pg 5 Minimum cover of the steel is controlled by ACI Top cover ¾” inch clear cover for slabs protected from the weather, 1.1/2” for #5 or smaller bars and 2” for larger bars exposed to weather 3” clear between bars and the ground.

34 Is Reinforcement Needed?
Concrete Floors on Ground By Portland Cement Association Second Edition

35 Sample Slab Reinforcing Calculation
Determine the reinforcing steel requirement for an outdoor, 5” thick concrete slab with control joints spaced 25 ft apart. The slab is cast on a compacted gravelly soil surface. Use 40 ksi rebar Variables fs = 20,000 psi m = 2.0 (assume that gravel surface has some interlock with the slab) L = 25 ft W = 5” (150 pcf / 12”) = 62.5 psf

36 Calculation Continued
From drag equation: Req’d As = in2/ftw Spacing Calcs: #3 bar: s < (.11 in2/bar)(12”/ft)/(.0781 in2/ft) = 16.9 in #4 bar: s < 30.7 in 6x6 W4.0xW4.0 wire mesh gives As = in2/ftw. ACI limits spacing to min(3h, 18”) Decision: Use #3 bars 15” O.C. each way. Place with a clear cover of 1” below top of slab.

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