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Floor Slab Design

Introduction
PlastiSpan® insulation has provided designers and building owners with long-term thermal performance for over 45 years as a component in residential, commercial and industrial floor systems. Structural slab design is governed by the types and magnitude of loads on it which generally include wheel loads from forklifts or delivery vehicles, point loads from the legs of storage racks or distributed loads from product stored on the floor.

Review
PlastiSpan® insulation has provided designers and building owners with long-term thermal performance for over 45 years as a component in residential, commercial and industrial floor systems. The bulletin provides calculation examples to illustrate use of PlastiSpan insulation for structural floor slab applications.

Structural slab design is governed by the types and magnitude of loads on the concrete slab which may include wheel loads from forklifts or delivery vehicles, point loads from the legs of storage racks or distributed loads from product stored on the concrete slab. Often, selection of a sub-slab insulation product for structural slab-on-grade applications is based upon ability to sustain compressive loads transferred through the concrete slab, without full and accurate determination of the load distribution characteristics of the concrete slab.

In one design methodology, design calculations are based upon the assumption that loads distributed over a contact area on a concrete floor slab area can be "assumed" to be distributed by wheel contact area or base plate contact area through the concrete slab to a largely hypothetical bearing area on the top surface of the insulation. As illustrated in Table 1, the load exposure for the insulation (EPS compressive stress) calculated on this basis could dictate use of a high density, high compressive resistance insulation material increasing cost unnecessarily.

Table 1 - Examples of Typical Loads on Concrete Slabs

Example 1 - Forklift Wheel Load   Example 2 - Point Load (Storage Racks)
Wheel Load - F kN 35   Point Load - F kN 45  
lbf 7,875 lbf 10,125
Wheel Contact Area m 0.203 x 0.203 Base Plate Contact Area m 0.152 x 0.152
in 8.0 8.0 in 6.0 6.0
Stress Distribution Angle 45   Stress Distribution Angle 45  
Slab Thickness m 0.152 Slab Thickness m 0.152  
in 6.0 in 6.0
Loaded Area m2 0.26 Loaded Area m2 0.21  
in2 398 in2 322
EPS Compressive Load Exposure kPa 136 EPS Compressive Load Exposure kPa 240  
psi 20 psi 31


Another accepted design procedure to use for structural slab design with these types of loads is the theory of plates on elastic foundations. Using the theory of plates on elastic foundations design procedure, when a concrete slab is constructed over a compressible or elastic subgrade such as soil or rigid insulation, load distribution and transfer to the sub-slab insulation is controlled by the slab itself and its response to loads. Floor loads will cause concrete slab deflection as a function of both the concrete slab properties and the compressibility of the materials beneath it.

In order to use this method, designers use the insulation or subgrade response factor referred to as the modulus of subgrade reaction (k) or, in other cases, foundation modulus, k-modulus, k-value, etc. The use of k-values in the design of structural slabs as discussed in PCA Concrete Information reflects the response of the insulation and subgrade under temporary (elastic) conditions when small deflections occur.

Disclaimer Regarding Example Calculations
This bulletin provides examples of calculations to illustrate applying the theory of plates on elastic foundations procedure to concrete slab design based upon a hypothetical wheel load with a defined wheel contact area or point load supported on a defined base plate contact area. It must be stressed that in all cases, design calculations and details for specific applications must be prepared by a registered design professional to verify compliance with applicable codes for the jurisdiction in which the project is to be constructed.

Limitations of Use:

  1. It is not the intent of this bulletin to provide comprehensive design guidance. Concrete slab deflection for each
    application must be calculated by the design professional responsible for concrete slab design.
  2. Design approaches expressed in this bulletin are related specifically to the distribution of floor loads to subgrade
    insulation. Final slab design is generally controlled by flexural stresses to which the slab is exposed under long term
    or short term rolling load conditions.
  3. Relationships used to establish slab deflection assume load intensities on the subgrade insulation do not exceed
    the elastic limit of the insulation.
  4. Such factors as the bearing capacity and compressibility of subsoil and/or subgrade slabs below the insulation must
    be considered in the design of slab/subgrade insulation composites.
  5. These design considerations are applicable to concrete slabs-on-grade (insulation serving as grade) exposed to
    storage loads, storage rack post loads and vehicle axle or wheel loads causing limited slab deflection. High intensity
    column or wall loading on floor slabs requires further consideration.
  6. Soils, concrete, steel and subgrade insulation exhibit creep or cold flow under long-term load exposures. Such
    long-term load exposures must be considered in slab design in order to prevent objectionable slab settlement.

Calculation Examples Using Theory of Plates on Elastic Foundation Procedure
The following calculation examples use the hypothetical loads for the two load types provided in Table 1 to illustrate the theory of plates on elastic foundation procedure. Floor slab deflection establishes magnitude of unit load transferred to the subgrade material, in this case thermal insulation, based on slab-on-grade design using the theory of plates on elastic foundations . Slab deflection (W) is determined by load exposure, concrete slab strength and subgrade response (insulation and soil) to load transfer using the equation:



where:
            W         = slab deflection
            P          = applied load
            k          = modulus of subgrade reaction
           D          = Eh3/12(1-μ2)           where:           

E          = modulus of elasticity of concrete
h          = slab thickness, in.
μ          = Poisson’s ratio of concrete

Based upon the theory of plates on elastic foundation procedure, slab deflection and insulation compressive load are calculated using elastic foundation design analysis based upon the combined characteristics of the insulation and a subgrade material.

_______________________
1Portland Cement Association, Concrete Information, Packard, Robert G., Slab Thickness Design for Industrial Concrete Floors on Grade, 1996.

2Timoshenko, S. and Woinowsky-Kreiger, S., Theory of Plates and Shells, McGraw-Hill, 1959.

Assumptions for Theory of Plates on Elastic Foundation Calculations:

  1. Concrete strength () = 28 MPa (4000 psi)
  2. Concrete thickness (h) as noted in table 1
  3. Poisson's ratio for concrete = 0.15
  4. Insulation thickness = 76 mm (3")
  5. Subgrade k-value (ks) = 100 MN/m3 (368 pci)
  6. k-value Insulation and soil = 1/kT = 1/ki + 1/ks

E-modulus of Concrete (Ec):
In SI units:
= = 24,870 MPa (3.605 x 106 psi)


Compressive resistance at 1% strain, the industry accepted allowable stress for live and dead loads, is provided in Table 2 for a variety of PlastiSpan insulation types.

Table 2 - PlastiSpan Insulation Compressive Resistance @ 1% Strain

Units PlastiSpan HD
Insulation
PlastiSpan 25
Insulation
PlastiSpan 30
Insulation
PlastiSpan 40
Insulation
PlastiSpan 60
Insulation
kPa 45 60 75 103 124
psi 6.50 8.7 10.9 15.0 18.0


Modulus of subgrade reaction values (ki) expressed in units of MN/m3 or lbs/in3 (pci) for various PlastiSpan insulation types and thickness are provided in Table 3.

Table 3 - PlastiSpan Insulation Modulus of Subgrade Reaction (k)

PlastiSpan Insulation Types Units PlastiSpan Insulation Thickness - mm (in)
25(1") 50(2") 75(3") 100(4")
PlastiSpan HD Insulation MN/m3 176 147 111 92
pci 650 540 410 340
PlastiSpan 25 Insulation MN/m3 255 212 160 133
pci 940 780 590 490
PlastiSpan 30 Insulation MN/m3 299 247 187 157
pci 1100 910 690 580
PlastiSpan 40
Insulation
MN/m3 346 285 217 182
pci 1275 1050 800 670
PlastiSpan 60
Insulation
MN/m3 434 358 271 228
pci 1600 1320 1000 840


Step 1: Calculate the modulus of subgrade reaction for 76 mm (3") PlastiSpan insulation plus subgrade material. 1/kT= 1/ki + 1/ks

For floor slab designs incorporating multiple insulation layers and a subgrade material, k can be found by adding k values for each layer as follows: 1/kT = 1/k1 + 1/k2 + …1/kn

Table 4 - K-value - PlastiSpan Insulation Plus Subgrade Material (kT)

kT PlastiSpan HD
Insulation
PlastiSpan 25
Insulation
PlastiSpan 30
Insulation
PlastiSpan 40
Insulation
PlastiSpan 60
Insulation
MN/m3 53 62 65 68 73
pci 194 227 240 252 269

 

Step 2: Calculate slab deflection (W) due to load.


Table 5 - Slab Deflection (W)

Insulation Type PlastiSpan HD
Insulation
PlastiSpan 25
Insulation
PlastiSpan 30
Insulation
PlastiSpan 40
Insulation
PlastiSpan 60
Insulation
Example 1 - 152 mm (6") Concrete Slab Deflection (W) Under Wheel Load
mm 0.22033 0.20381 0.19806 0.19326 0.18707
in. 0.00867 0.00802 0.00780 0.00761 0.00737
Example 2 - 152 mm (6") Concrete Slab Deflection (W) Under Point Load
mm 0.28328 0.26204 0.25464 0.24848 0.24052
in. 0.01115 0.01032 0.01003 0.00978 0.00947


Step 3: Check compressive stress (F) in 76 mm (3") thick EPS insulation. The above slab deflection (W) will transfer load to the insulation material at intensity directly related to the insulation k-value from Table 3: F = KiW.

Table 6 - EPS Compressive Stress (F)

PlastiSpan Insulation Types Example 1 - Wheel Load Example 2 - Point Load
152 mm (6") Slab 152 mm (6") Slab
kPa psi kPa psi
PlastiSpan HD Insulation 25 3.56 32 4.57
PlastiSpan 25 Insulation 33 4.73 42 6.09
PlastiSpan 30 Insulation 37 5.38 48 6.92
PlastiSpan 40 Insulation 42 6.09 54 7.83
PlastiSpan 60 Insulation 41 5.89 52 7.58

 

Based upon the above calculation examples, the compressive load transferred to the PlastiSpan insulation is within the allowable stress range provided in Table 2 for all insulation types.

Step 4: Check concrete slab bending stress (fb).

fb - computation 

Where ƒb = Concrete bending stress, h = slab thickness and a = radius of load contact

   ; where Ac = load contact area


Table 7 - Concrete Bending Stress (ƒb)

Design Loads Wheel Load – Load Factor =1.5 Point load – Load Factor =1.25
kN lbf kN lbf
35 7,875 45 10,125
Radius of Contact (a) 115 mm (4.5") 86 mm (3.4")
PlastiSpan Insulspan Type 152 mm (6") Slab 152 mm (6") Slab
MPa psi MPa psi
PlastiSpan HD Insulation 2.88 417 3.70 536
PlastiSpan 25 Insulation 2.83 410 3.35 486
PlastiSpan 30 Insulation 2.81 407 3.33 484
PlastiSpan 40 Insulation 2.79 405 3.32 481
PlastiSpan 60 Insulation 2.77 402 3.30 478


Concrete Tensile Strength ():
MPa in SI Units:
Bending stress should not exceed the concrete tensile strength = 3.41 MPa (495 psi).

Selection of the PlastiSpan insulation type based upon concrete slab design using the theory of plates on elastic foundation would result in:

Calculation Example Summary:

  1. Under "assumed" or hypothetical load distribution patterns, the EPS compressive resistance requirement for the
    load exposure from the two load types typical for structural slabs would dictate use of a high strength
    thermal insulation.
  2. Using elastic foundation design theory, a much lower compressive load would be transferred to the sub-slab
    insulation based upon slab deflection under the two hypothetical load types allowing the use of a more
    cost-effective PlastiSpan insulation alternative.
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