Presentation on theme: "Design of Slabs-on-Grade"— Presentation transcript:
1 Design of Slabs-on-Grade CE A433 – RC DesignT. Bart Quimby, P.E., Ph.D.Spring 2007
2 IntroductionSlabs on grade are PAVEMENTS not generally structural elementsPavements pass loads through compression to the supporting soilAs long as the soils deformations are low, there is negligible bending in the slabSlabs 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 soilA 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 discontinuityOften occur when a floor is over loadedShrinkageShrinkage 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 failureThe failure may result from one or more of the following conditionsThe subgrade is improperly designed or preparedThe slab thickness is too thin for applied loads and the stiffness of the subgradeThe concrete does not have sufficient strengthIt 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 CracksCracks 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 soilSlab stiffness is a function of slab thicknessSlab cracking strength is a function of concrete strength and slab thickness
9 PCI MethodA series of charts for various loading conditions (wheels, racks, posts, etc)Example of slab thickness determination for a wheeled vehicle:Data for lift truckAxle load = 25 kWheel spacing = 37 inNumber of wheels = 2Tire inflation pressure = 110 psi
10 PCI Example Continued Contact area = wheel load/inflation pressure Contact area = (25,000 lb / 2 wheels) / 110 psi = 114 in2Subgrade and Concrete DataSubgrade Modulus, k = 100 pciConcrete 28-day strength, f’c = 7,000 psiConcrete 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 permittedConcrete working stress = MR/FSWS = MR/FS = 640 psi / 2 = 320 psiSlab stress per 1,000 lb of axial loadWS / 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 loadUse 8” SlabSubgrade ModulusWheel 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 dryingNormal concrete can only stretch about inches per foot without rupturingNormal shrinkage is about (+25%) inches per footIf the slab is restrained against movement then cracking is inevitable
15 Minimizing Shrinkage Cracking Shrinkage cracking can be minimized byReducing the shrinkage characteristics of the concrete mixReducing restraint on the slabShrinkage cracking can be controlled byEncouraging cracks to appear at predetermined locationsThe use of reinforcing steel
16 Reducing Shrinkage Characteristics of the Concrete Mix Reduce the volume of water in the mixThe challenge is to limit the amount of water in the mix while maintaining workability and finishability without excessive use of water reducersUse coarser ground cementUse the largest sized aggregate permitted by designUse shrinkage compensating concrete
17 Reducing Shrinkage Characteristics of the Concrete Mix (Cont.) Use proper curing techniquesProper curing keeps water in the concrete until it has achieved sufficient tensile strength before shrinkage occursProper curing allows drying to occur more evenly through the slab thickness
18 CurlingDifferential 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 slabBearing on other features (walls, foundation, drain pipes, columns, etc)Attachment to other 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 JointsThe 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 JointsThese 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
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 jointsReinforced SlabsUse 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/ftwfs = Allowable steel stress (20,000 psi or 24,000 psi)W = Dead weight of slab, psfm = 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 concreteJoint should extend at least ¼ slab thickness through the slabVertical 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 SteelSmaller bar sizes are better choices than large diametersThis steel “should be positioned one-fourth the slab thickness below the top surface up to 2.0 in maximum.” ACI 302.1R, pg 5Minimum cover of the steel is controlled by ACITop 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 weather3” clear between bars and the ground.
34 Is Reinforcement Needed? Concrete Floors on GroundBy Portland Cement AssociationSecond 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 rebarVariablesfs = 20,000 psim = 2.0 (assume that gravel surface has some interlock with the slab)L = 25 ftW = 5” (150 pcf / 12”) = 62.5 psf
36 Calculation Continued From drag equation:Req’d As = in2/ftwSpacing Calcs:#3 bar: s < (.11 in2/bar)(12”/ft)/(.0781 in2/ft) = 16.9 in#4 bar: s < 30.7 in6x6 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.