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Design of Slabs-on-Grade

CE A433 RC Design

T. Bart Quimby, P.E., Ph.D.Spring 2007

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Introduction

Slabs on grade are PAVEMENTS notgenerally structural elements

Pavements pass loads through compression tothe 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

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Pavement

Apply load to top of slab

Since the slab is stiffer than the soilthe load is distributed over a larger

area of soil

A thicker slab is stiffer anddistributes the load over a

larger area of soil

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Types of Cracks

Structural 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, widthor number after the drying process iscompleted.

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Causes of Structural Cracking

Virtually all structural cracks are the result ofsubgrade failure

The failure may result from one or more of thefollowing 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 expectedloads so that the proper slab thickness can bedetermined

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Structural Cracks

Cracks form when the Moment exceeds the Cracking Moment

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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 theload, the stronger and stiffer the slabmust be to spread the load over more ofthe supporting soil Slab stiffness is a function of slab thickness Slab cracking strength is a function of

concrete strength and slab thickness

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Thickness Design Procedures

PortlandCement

Association Wire

ReinforcingInstitute

Corp. ofEngineers

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PCI Method

A series of charts for various loadingconditions (wheels, racks, posts, etc)

Example of slab thickness determinationfor a wheeled vehicle: Data for lift truck

Axle load = 25 k

Wheel spacing = 37 in

Number of wheels = 2

Tire inflation pressure = 110 psi

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PCI Example Continued

Contact area = wheel load/inflationpressure

Contact area = (25,000 lb / 2 wheels) / 110psi = 114 in2

Subgrade and Concrete Data Subgrade Modulus, k = 100 pci

Concrete 28-day strength, fc = 7,000 psi Concrete flexural strength, MR ~ 7.5sqrt(fc) ~ 640

psi

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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.

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PCI Example Continued

Slab Stress per1,000 lb of axle

load

EffectiveContactArea

Wheel SpacingSubgradeModulus

Use 8Slab

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PCI Chart for Racks

Need to matchcriteria for the

chart Read the

instructions for

each chart!

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Causes of Shrinkage Cracking

Shrinkage cracking occurs due to thenormal volumetric changes associated

with drying Normal concrete can only stretch about

0.002 inches per foot without rupturing

Normal shrinkage is about 0.006 (+25%)inches per foot

If the slab is restrained against movementthen cracking is inevitable

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Minimizing Shrinkage Cracking

Shrinkage cracking can be minimizedby

Reducing the shrinkage characteristics of the

concrete mix Reducing restraint on the slab

Shrinkage cracking can be controlledby

Encouraging cracks to appear atpredetermined locations

The use of reinforcing steel

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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 andfinishability without excessive use of waterreducers

Use coarser ground cement

Use the largest sized aggregate permittedby design

Use shrinkage compensating concrete

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Reducing Shrinkage Characteristics

of the Concrete Mix (Cont.) Use proper curing

techniques

Proper curing keeps water

in the concrete until it hasachieved sufficient tensilestrength before shrinkageoccurs

Proper curing allows drying

to occur more evenlythrough the slab thickness

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Curling

Differential shrinkage due to drying can result in curling of the slabedges, resulting in an induced moment in the slab.

When the moment equals the cracking moment a crack forms,

redistributing the stress

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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

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Friction Restraint

Axial Stress Diagram

Tensile Capacity

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Shrinkage Cracks

Axial Stress Diagram

Tensile Capacity

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Restraint byFeatures

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Locating Cracks

Control and construction joints are placesof intentional weakness. They are placed

close enough together to keep tensilestresses in the slab below the tensilerupture strength of the concrete

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Control Joints

The purpose of these joints is to predeterminethe 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, theresulting induced crack may offer no structural

advantage over a randomly occuring shrinkagecrack.ACI 302.1R, pg 6

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Construction Joints

These joints are placed in a slab wherethe concreting operations are concluded

for the day, generally in conformity with apredetermined joint layout. If at any timeconcreting is interrupted long enough for

the placed concrete to harden, aconstruction joint should be used. ACI302.1R pg 6

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ControlJointDetails

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Construction Joints

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Joint Spacing

Unreinforced Slabs

2 to 3 ft for each inch of slab thickness.

Smaller aggregate size, higher watercontents, and local experience may dictateuse of closer joints

Reinforced Slabs Use a subgrade drag equation to computejoint spacing (See ACI 360R 6.3)

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Drag Equation

Where:

L = distance between joints, ft As = Area of steel per foot width of slab, in

2/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)

W

fA

L

StrengthfAL

WFriction

ss

allowabless

m

m

2

2

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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 toconcrete

Joint should extend at least slabthickness through the slab

Vertical load transfer across constructionjoints can be accomplished with plain barsor properly designed keyed joints.

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No Vertical Load Transfer

Joints have verticaltransfer but allow in plane

shrinkage movement

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Controlling Shrinkage Cracking with

Reinforcing Steel Reinforcement serves to restrain the shrinkage,

effectively subdividing the slab and hence

distributing the crack area more evenly. Thisproduces smaller and more numerous cracksthan would occur in an unreinforced slab of thesame dimensions. The actual crack area

remains essentially the same. Fricks, T.J. Cracking in Floor Slabs, reprinted in ACI

SCM-25 (92), pg 122.

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Reinforcing Steel

Smaller bar sizes are better choices than largediameters

This steel should be positioned one-fourth the

slab thickness below the top surface up to 2.0 inmaximum. ACI 302.1R, pg 5

Minimum cover of the steel is controlled by ACI318 7.7.

Top cover inch clear cover for slabs protectedfrom the weather, 1.1/2 for #5 or smaller bars and2 for larger bars exposed to weather

3 clear between bars and the ground.

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Is

ReinforcementNeeded?

Concrete Floors on Ground

By Portland CementAssociation

Second Edition

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Sample Slab Reinforcing Calculation

Determine the reinforcing steel requirement foran outdoor, 5 thick concrete slab with control

joints spaced 25 ft apart. The slab is cast on a

compacted gravelly soil surface. Use 40 ksirebar Variables

fs = 20,000 psi

m = 2.0 (assume that gravel surface has someinterlock with the slab) L = 25 ft W = 5 (150 pcf / 12) = 62.5 psf

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Calculation Continued

From drag equation: Reqd As = 0.0781 in

2/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 = 0.080 in2/ftw.

ACI 318 7.6 limits spacing to min(3h, 18)

Decision: Use #3 bars 15 O.C. each way. Placewith a clear cover of 1 below top of slab.