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30 RESK)ENTIAL SLABS ON GROUND procedures. All this information is intended to amplify and clarify the recommendations set forth in Section II. PART A: Selection and Design of Slabs 1. O FUNCTIONS OF SLABS- ON- GROIJND Slabs-on-ground constitute an element of residential construction performing in at least the first and frequently both of the following two capacities: as a separating element between the ground and habitable space; as a structural element receiving part or all of the loads of and on the superstructure and transmitting such loads to the foundation soil. While slabs-on-ground act in the former capacity at all times, the degree to which they function in the latter capacity depends upon engineering definition. Wherever a slab-on-ground acts simply as a separator between living space and ground, it carries no load-bearing or large load- producing elements of the superstructure. In this case, satisfactory performance may be defined to include no unsightly cracks, and no large differential settlements which may be functionally or aestheti- cally objectionable, such as a noticeable "out-of-plumb" condition affecting trafficability or equipment and building elements supported on the slab, an exposure of the foundation, and/or an effect upon the performance of any mechanical/electrical services which pass through the slab. Such a slab, however, can, with proper details, be allowed to settle to some degree without detriment either to struc- ture, services, or aesthetic considerations. Wherever a slab also acts as a structural element, transferring the substantive superstructure loads to the foundation soil, of neces- sity it must be able to do so satisfactorily. In this case, the differ- ential settlement of the slab, unless confined within prescribed limits, may have critical consequences for the superstructure-for example, if a wall is supported on a slab which settles unevenly, it may rack or crack. Therefore, a slab which receives superstructure loads should be made stiff enough to support such loads without excessive deflection or uneven settlements; this requires that it
SUPP LE ME NTARY INFORMATION 31 behave as nearly as necessary like a monolithic rigid body which, if it should settle, will do so uniformly. Stating these two situations in another way, slabs to be treated solely as separators must be founded on firm ground and soil not subject to substantial changes in volume as a result of changes in water or moisture content. The separate structural supports must similarly be well founded. In those areas where either or both are not possible, slabs may as well be made monolithic with the founda- tion and thus act both as separators and as receivers of all imposed loads for transmittal to the ground. Even though those instances where a structural or monolithic slab and foundation system will be needed are limited in number, they are the most demanding of attention and design effort. This circumstance is reflected in this report in the disproportionately large number of pages devoted to the analysis and design of slabs which act both as separators and as structural elements, even though the need for such slabs is limited to a small percentage of all residential construction in the nation. 2.0 FUNDAMENTAL FACTORS OF SLAB DESIGN AND CONSTRUCTION The design of slabs-on-ground consists of three basic operations, namely: a. Selection of slab type to be used b. Dimensioning the slab (layout) c. Reinforcing the slab (wherever necessary). To perform these operations successfully under a specific set of conditions, the designer must analyze many factors which directly or indirectly influence his decisions. Those assuming dominant importance in the great majority of cases are: a. Soil properties of the ground on which the slab is to be supported b. Climate at the building site
32 RESH)ENTIAL SLABS ON GROUND c. Type of superstructure (for slabs which transmit super- structure loads to the foundation) d. Quality control in materials use and in construction. These four principal factors are, for this report, the bases on which procedures are developed for selection, and specification or design, of slabs-on-ground. The first three (soil, climate, and superstructure) are presented and analyzed below, in relation to slab selection, and specification or design; the fourth and equally important factor of quality control will be presented independently in Part B of this section. 3.0 SELECTION OF SLAB TYPE The slab appropriate to any given set of conditions should be ade- quate in terms of performance and economy. Below is a descrip- tion of each of the four types, under one or another of which almost all slabs encountered in practice can be classified. Selection of the appropriate type to be applied in each case depends on only two of the four fundamental factors-soil and climate. The impact of these factors on slab-type selection is analyzed following the descriptions. 3.1 Types of Slab- On- Ground 3.1.1 Slab Type I This 4-inch-thick slab, intended for use on firm ground which will develop no change in volume with time, is cast directly on a properly prepared building site and slab base and carries no rein- forcement over its entire area. Its use is limited to that of separa- tor between ground and living space. Its maximum dimensions are limited by the need to avoid shrinkage cracking. Successful per- formance depends on compliance with a set of specifications. 3.1.2 Slab Type II Also limited to the function of separating ground from living space, this 4-inch-thick slab, which may be of larger dimension
SUPPLEMENTARY INFORMATION 33 than Type I, is applicable to ground which may undergo small movements (shrinkage and expansion) with weather changes and under loading. To withstand these small movements as well as to accommodate the stresses of drying shrinkage and thermal change without serious damage, it is provided with light reinforce- ment. Successful performance depends on compliance with a set of specifications. 3.1.3 Slab Type III Unlike Types I and II, this slab receives and transmits all super- structure loads to the foundation soil. It is used with soils which in all likelihood will undergo substantial volume change with time. Use of spread footings for the foundation is not advisable on such ground; therefore, loads are distributed by the slab over its entire ground-support area. This reduces the bearing stresses on the ground and also forces the foundation, the slab, and the superstruc- ture to act as a monolithic structure (somewhat like a rigid boulder in a soft mass of ground). To assure that the slab will actually be- have in this manner, the designer must impart to the slab the necessary rigidity and strength. Hence, slabs of this type need to be carefully analyzed and designed so that dimensions (for stiffness) and reinforcement (for strength) will be accurately determined and provided. 3.1.4 Slab Type rv This slab also receives and transmits all superstructure loads to the foundation soil. Unlike Type III, however, this slab does not it- self rest on the ground. Rather, it is supported on beams which are in turn carried by caissons, piles, footings, or similar special foun- dations carrying the loads to solid ground well below the level of the slab. It is used on very poor soils which are extremely sensitive to weather, have negligible bearing capacity, or are high in organic- materials content. This type is designed in the same manner as structural floor slabs of concrete, in accordance with the ACI code. Each of the four types discussed above is considered minimal for the condition described. Obviously, a slab type of greater capability can be selected-e.g., Type II instead of Type I; however, any de- cision in this respect should be predicated on the desire to improve quality of performance within predetermined economic limits.
34 RESIDENTIAL SLABS ON GROUND 3.2 Soil Investigation The importance of determining the nature and properties of the soil on the site where a residential slab is to be used cannot be over- emphasized; proper identification of the foundation soil is a critical factor in slab selection. For purposes of this report, the Unified Soil Classification Sys- tem has been adopted.1 Details and specifics relating to soils are provided in Part C of this section; here, only the basic specifica- tions on minimum requirements for soil investigation are given. Unless competent engineering advice indicates otherwise (see also Step 1, p. 10), it is advisable to perform at least one test boring on each slab site. When the boring reveals unusual conditions, such as organic soils, soft or loose soils, highly plastic soils, or rock, additional borings should be made. These test borings can be made with simple tools, the important thing being to determine soil types and extent of each to a depth of at least 15 feet, or to a solid layer of rock.2 A record of the class of soil, its depth, consistency, and moisture content should be kept. Where CL, OL, CH, or OH soils are encountered, it is also necessary, for the appropriate selection of slab type, to determine the unconfined compressive strength (qu). It may be helpful In the site investigation to examine existing resi- dences in the immediate area, but it must first be determined that the same conditions prevail with respect to soil type, topography, and construction type; also, that the existing structures examined are old enough to have experienced the design range of climatic variations likely to occur in the area. 3.3 Climatic Rating Along with soil classification, climate is the other important factor in the selection of slab type. Climate affects the behavior of a slab- on-ground primarily through changes in the moisture content of the soil underlying the slab. If there are wide variations in the amount of moisture in the supporting soil, and if this soil is water-content sensitive, expanding as it absorbs water and shrinking with its loss, then the slab is subjected to a sequence of uplift (as the soil swells) 1See Appendix D, p. 289. 2Bucket augers or helical-blade augers are usually satisfactory, since pieces of undisturbed soil can often be obtained.
SUPPLEMENTARY INFORMATION 35 and settling (as the soil shrinks). Whenever a time of high water content is followed by drought, the moisture at and beneath the perimeter of the slab will generally evaporate much more rapidly than that under its center, where it is trapped and sealed from direct exposure. Moisture will often remain under the slab center even after extended periods of drought (one year or several years), and/or accumulate there due to capillary action as well as migra- tion, even though the soil around the periphery has dried to a con- siderable depth. A similar but opposite phenomenon develops when the ground moves from low to high moisture content. Particularly if prolonged periods of alternating drought and wetting occur, con- siderable difference in moisture content can develop between one and another of the various points underlying a slab. If the soil happens to be such that substantial change in volume will occur with change in moisture content, one of the following two conditions may ensue: a. If the slab is relatively flexible, it will follow the uneven con- tour of the soil which will result from the uneven change of volume; the superstructure, if it rests on the slab, then will be exposed to distortions which may cause damage. b. If the slab is sufficiently rigid, it will refuse to follow the uneven contour of the ground. As a result, higher soil pressure will develop on the slab over the high plateaus, with greatly re- duced pressure over the valleys. The slab will be subjected to bending as it endeavors to accommodate to the uneven contour, and the soil may deform in areas of high bearing pressure, trans- ferring load to adjoining areas. If the slab carries the superstruc- ture, the latter thus will be provided protection against damage. Obviously, it is difficult to assign exact values to the amount of precipitation, its variation in occurrence, or its effect on soil under- lying slabs-on-ground. The important consideration is whether or not climatic conditions will be likely to change the moisture content of the soil during and after construction. Involved may be such matters as freezing, which, in some soils, will cause volume change through the formation of ice lenses; or the presence of trees and shrubbery in the immediate proximity of the slab perimeter, which will affect soil moisture content by providing a shield from natural precipitation and by extracting moisture during growth. Studies of weather data disclose at least five variables affecting consistency of climate. They are:
36 RESIDENTIAL SLABS ON GROUND a. Yearly annual precipitation b. Degree of uniformity through the year in distribution of precipitation c. Number of times precipitation occurs d. Duration of each occurrence e. Amount of precipitation at each occurrence. ~ a study of drought hazards to crops,1 a relationship was noted between soil grain size and moisture availability as affected by rain- fall. Even though the principal concern of this study was something other than soil moisture retention, its findings bear out the accepted premise that, the finer the soil grain size, the slower the loss or gain of total moisture. U.S. Weather Bureau studies have further disclosed a strong ~n- verse correlation between two factors: the amount of rainfall for any particular period and the number of occurrences. Without de- tailed explanation of how these values are obtained, it suffices, for purposes of this report, that the frequency function provides an excellent measure of the potential for soil activity; for it gives a sound indication of the likelihood of extended periods during which the normal soil-moisture balance may be upset through evaporation by reason of low rainfall, or through concentration in fewer-than- normal occurrences. ~ either instance, cohesive soils can be expected to shrink during dry periods. Upon restoration of the normal rainfall pattern, cohesive soils can be expected to swell. The rate at which moisture is lost or gained by soils is not at this time thoroughly understood. It is generally accepted, however. that air movement accelerates loss of soil moisture. Since air movement is independent of rainfall, it can be assumed to increase loss of soil moisture, especially during extended periods of little or no precipitation. While it is recognized that other factors such as temperature and relative humidity also influence loss or gain of soil moisture, the effects exerted are comparatively unimportant. On the basis of U.S. Weather Bureau data, a climatic rating (Cw) has been assigned to all points in the continental United States, 1Gerald L. Barger and H. C. S. Thom, "Evaluation of Drought Hazard," Agronomy Journal, Vol. 41, No. 11, November 1949, pp. 519-526.
SUPPLEMENTARY INFORMATION 37 as shown in Fig. 1, p. 38. The Cw for any particular locality not directly on an isoline can be determined simply by interpolation to the nearest whole number; for example, Jackson, Mississppi, would be assigned a Cw of about 37, while for Columbia, Missouri, Cw would be about 33. 3.4 Correlation of Climate and Soil for Selection of Slab Once the foundation soil of a slab is classified, and the severity of the climate at the site is identified with the help of Fig. 1, the proper slab type can be selected. When the soil is basically cohesionless, selection of slab type depends exclusively on the density and con- sistency of the foundation soil, without regard to climate. Thus, a Type I slab can be successfully used on all gravelly soils (GW, GP) under all climatic conditions. It can also be used on all sandy soils with or without silts and clays (GM, GC), as well as on silts (ML, MH), provided they are classified as medium or dense. Table VI, p. 142, provides a quantitative measure of the various densities of cohesionless soils In terms of the number of blows required to drive a standard 2-~nch OD sampler 1 foot into the ground by means of a standard 140-lb hammer falling 30 inches.1 Whenever cohesionless soils or soils of low plasticity (GM, GO, SW, SP, SM, SC, ML, MH) are present in loose condition, a Type II slab is the more suitable (regardless of climatic conditions), since such soils in loose condition are subject to a limited degree of uneven settlement after the erection of the superstructure. Light reinforcement, therefore, will be required to protect the slab from cracking. A Type II slab can also be used over clay (CL) or organic soils (OL) when the plasticity index rating is less than 15 and the ratio qu/w (where w is the average total slab dead and live load, and qu is the unconfined compressive strength of the soil) is more than 7.5, thus permitting superstructure loads to be supportable directly on spread footings. Where PI ~ 15 but the soil is relatively firm (qu/w > 7.5), a Type II slab is still adequate provided the climate is optimum, i.e., the climatic rating (Cw) is at least 45. Type HI slabs have a limited application, in the sense that they are needed only where clays or organic soils (CL, OL, CH, OH) 1ASTM Designation D 1586-64T (or most recent edition), Standard Pene- tration Test. Philadelphia: American Society for Testing and Materials.
(A W\,o S: \ . ~ ~ ~ Q O ~ ~ ~ ~ it ~ §'! ~ o 3 5] - ; u, o ~ 1 ~ f(_: if: / ,_ of_ ~ ~1 ~ _ ~° ~ 0 CO C. U] I: Ct % ~ C) O 3 U] ._ 8 c) _ ·_ ·_. ~4 ~ o~
SUPPLEMENTARY INFORMATION 39 occur in localities having a climate which is less than ideal (i.e., Cw < 45), or where the average load (w) is high relative to the un- confined strength (qu) of the supporting soil (qu/w ~ 7.53. When CL, OL, CH, and OH soils having a low compressive bear- ing capacity (qu/w < 2.5) are encountered, a Type IV slab resting on special foundations should be used. Table I, p. 11, correlates the various combinations of soil type and climate and classifies them with respect to the type of slab recommended for use. 4.0 CRITERIA FOR TYPE I SLABS 4.1 General Type I slabs (Fig. 2, p. 40) are not affected either by the type of superstructure or by the climate at the construction site. The superstructure is supported directly on footings, and the soils on which Type I slabs are founded are practically unaffected by cli- mate and water content changes. This type of slab, by its very nature, possesses only limited capabilities; specifically, it has only compressive strength and cannot tolerate appreciable amounts of tension or warping. It may crack during drying, but, when used under controlled conditions, such cracks as do occur should not become excessively wide nor prove a detriment to the serviceability of the slab. The controlling factors in the successful performance of these slabs are the quality of materials and construction, size, and cer- tain other basic details. Aspects of quality control are described fully in Part B. pp. 126-136, and the other factors are discussed below. 4.2 Site Any site upon which this slab is to be placed should be well drained and properly graded.1 The soil should be one of those appropriate for supporting a slab of this type, and should be uniformly and 1This report, Part B. pp. 126-136.
40 RESIDENTIAL SLABS ON GROUND . . . · ~ . ~ i'< -Insulation or Expansion Joint - ~ . . a, ., ~ . _ = Insulation or Expansion Joint_ ~! . ~ ' ~ 'I a\ c. :~ Insulation or b b. ~ . Expansion Joint Grooved to Typical Separate Weakened Aggregated ~PIane Jointed . ~Z2, . ~ , . . `, ~ . ^' :~ 1~ ~ 1/3 1 o. Add Non-cellulosic: Strip Separator Insulation or Expansion Joint - Note: This type of construction, entailing ledge support di rec tly under slab, with or without insulation or expansion loins, es nor recommended. d. FIG. 2 Typical Type I Slabs adequately compacted] to provide the support necessary to ensure that warping and tensile stresses which contribute to cracking are not induced in the slab. 1This report, Part B. Fig. 22, p. 128.
SUPPLEMENTARY INFORMATION 41 4.3 Dimensions This slab is one of uniform 4-inch thickness, rectangular or square in outline. The maximum dimension is 32 feet, in order to minimize shrinkage during drying and subsequent thermal volume change. 4.4 Irregular Shapes The slab cross section should not contain irregularities in its hori- zontal plane, lest the stresses incident to drying shrinkage or ther- mal change induce cracking at such irregularities. Whenever the top surface of the slab must be interrupted (e.g., lowered to permit proper installation of ceramic tile), the uniform 4-inch thickness should be maintained by lowering the slab underside, beginning at least 24 inches away. In addition, when the vertical displacement is greater than 1-1/2 inches, 6x 6 - 6/6 WWF should be placed mid- way in the slab and for 25 inches on either side of the point of dis- placement. When the slab has an irregular outline, such as a T- or L-shape, weakened planes (see Fig. 2d, p. 40) should be provided in a manner which will divide the slab into squares or rectangles; otherwise, objectionable diagonal cracks are likely to occur at the junction. 4.5 Weakened- Plane Joints A useful method of predetermining the location of shrinkage cracks due to drying and temperature change in such unreinforced slabs is to create weakened-plane joints (Fig. 2d). Since this slab type is expected to receive uniform nonyielding support, the only considera- tion is judicious placement of joints to induce cracks at the least objectionable location-e.g., under partitions. To provide controlled cracks, it is necessary to reduce the effective cross section of the slab, e.g., by Sawing a grove in the slab after initial set Grooving the moist concrete with a T-bar or jointing tool Placing continuous strips of a noncellulosic material on the slab site, prior to placing the slab. Whichever method is used, the effective thickness of the slab at the weakened plane should be reduced by a quarter to a third of the normal slab thickness. ~ this manner, the tensile stresses generated
42 RESIDENTIAL SLABS ON GROUND will find relief at the weakened plane, reducing the likelihood of cracks occurring where they might be objectionable. Control joints should be placed no more than 32 feet apart and preferably closer, In order to minimize width (4.3 above). ~ this connection, however, consideration should be given to the possibility of other deleterious effects, such as creating a pathway for termites or moisture. 4.6 Embedment In Slab Heating coils, pipes, or conduits, such as are shown In Fig. 4b, p. 44), should not be embedded In an unreinforced slab, since these will induce tangential stresses. Heating ducts may be embedded, pro- vided they are completely encased In not less than 2 inches of con- crete, and the slab over the duct is reinforced with 6x6 - 6/6 WWF (similar to Fig. 4c). This reinforcement should extend for 19 inches on either side of the centerline of the duct or to the slab edge, whichever is closer. 4.7 Loads The slab should not be subjected to partition loads greater than 500 plf, nor to equivalent concentrated loads such as chimneys. The slab cannot tolerate loadings of this magnitude without undue deflection. A method of supporting partition or concentrated loads in excess of 500 plf is shown in Fig. 3. When this method is used, care should be , \1 ID . j ' D 1. ·D-. ., . ~ ~d 7 1 Expansion Joint I., / At/ I 1< ~ ~ - -a 7 1^ .C? o. . °; ~ a ~ Note: Reinforcement can I be omitted whenever / h/a 2 2. Firm Sal I ~ FIG. 3 Method of Supporting Partition Loads in Excess of 500 PLF
SUPPLEMENTARY INFORMATION 43 exercised to ensure that the footing rests on undisturbed soil, that the slab does not lose support near the partition, and that the load- bear~ng element is isolated from the slab. 4.8 Openings Openings in the slab should be kept to a minimum, since they intro- duce nonuniform stresses. Wherever a 12-inch or wider opening must be included, the slab should be reinforced with 25-inch-width 6x6 - 10/10 WWF all around the opening. 5.0 CRITERIA FOR TYPE II SLABS 5.1 General Type II slabs (Fig. 4, p. 44) need not be adjusted for various types of superstructure, because the latter will be directly supported on spread footings; nor will such slabs be influenced by climate, since soils on which they are founded are not significantly affected by climate changes. Reinforcement of Type II slabs provides for control of crack size only. Although cracking is almost certain to occur, cracks will, with reinforcement, be held tightly closed and will not be objectionable. However, this slab cannot be expected to bridge a void, since the amount of reinforcement provided is based on the premise that the entire slab will remain uniformly supported by underlying soil. The conditions of loading, site drainage, and grad- ing are identical with those of Type I, except that less desirable soils are acceptable as support. The controlling factors for success are quality of materials and construction, reinforcement, size, and other basic details. Aspects of quality control are described fully in Part B. pp. 126-136, and other criteria follow. 5.2 Dimensions This slab is basically one of uniform 4-inch thickness and may be of dimensions up to 75 feet. These slabs are expected to deflect with slight movements of the soil, and should be free of foundation
44 RESIDENTIAL SLABS ON GROUND 1 c, . ,, _~_ Rei nforceme nt ~a. O p . O D: Insulation or Expansion Joint Asbestos cement Reinforcement j~71 Note: . ' -~ ~ . lo. ~1 · Or I D · / L Q; Insulation or Expansi on Joi nt ~ Pipe or Coil lo. ' / 2-in. Min. ~ a./ 0 I l Although not meeting all criteria for Type I slabs (i .e. ~ with re- spect to independent foundation support of exteri or wa I I s) / when reinforced in accordance with the requi remeets of Tabl e 11 r P · 45 such monolithic slab and grade beam construction has been found to perform satisfactori 1 y in a structural sense where a rectan- gular Type I slab is recommended and all other Type l-slab criteria are observed. d. FIG. 4 Typical Type II Slabs
SUPPLEMENTARY INFORMATION 45 walls, piers, or footings; otherwise, the slab is likely to experience flexural stresses which it is not capable of resisting without cracking. 5.3 Reinforcement In order to control the size of cracks, the slab should have a speci- fied minimum WWF reinforcement over its entire area, located midway between top and bottom of the slab and supported on chairs. Placement of the fabric in the center of the slab is to ensure that cracks will be kept tightly closed. The sizes recommended are based on the "drag" theory as follows: As= FLwd/2fs where As F L wd area of steel required per foot of width (in.2) coefficient of friction = 1.2 5 longest dimension of slab (it) weight of slab = 12.5 pounds per inch of thickness (psf) allowable steel design stress (45,000 psi for Type II slabs). Typical reinforcement needs are shown in Table II. TABLE II Minimum WWF Reinforcement between Design Joints for Type II Slabs Maximum Dimension Wire Spacing Wire Gauge (ft) (in.) (no.) Up to 45 6 x 6 10/10 45 to 60 6 x 6 8/8 60 to 75 6 x 6 6/6 5.4 Embedment in Slab Heating coil may be embedded, since a slab of this type is rein- forced over its entire area. The reinforcement provided can also accommodate the thermal stresses induced due to heating, but
46 RESIDENTIAL SLABS ON GROUND only if coils or ducts are completely encased in at least 2 inches of concrete (Fig. 4b and c). 5.5 Irregular Shapes The same restriction as for slabs of Type I apply to planes or cross sections of irregular shape, with the added requirement that reinforcement be continuous across the weakened plane to eliminate the possibility of vertical displacement between adjacent sections. 5.6 Loads Since it is reinforced, this slab can accommodate partition loads of up to 500 plf; however, when this loading is exceeded, an addi- tional layer of reinforcement should be provided to extend at least 25 inches on either side, in order to distribute the added stresses. Further, concentrated loads in excess of this amount (500 plf), should be supported on independent footings. 5.7 Openings Openings in the slab should be kept to a minimum and should be less than 12 inches in width. Openings greater than 12 inches in width should be provided with an additional layer of 6x6 - 6/6 WWF rein- forcement all around to prevent the concentration of stresses which could crack the slab at these points. 6.0 EXAMPLE OF PROCEDURE FOR DETERMINING WHETHER A TYPE I OR II SLAB IS APPROPRIATE, AND APPLICATION OF CRITERIA 6. 1 General The procedures which follow demonstrate the application on more stable soil types of the criteria recommended in pare. 1.0, 10-20.
SUPPLEMENTARY INFORMATION 47 Location: McAlester, Oklahoma Floor plan and outside dimensions: 4 2-O l _ , 0 1 . Type of construction: Solid masonry with plaster-on-lath interior and ceramic tile bathroom floor Total weight of superstructure: 143 Rips Method of heating: Warm air, with ducts in attic, i.e., none in slab Partitions: Non-loadbearing, weight 100 plf Openings through slab: None greater than 8 inches Concentrated loads: None. 6.2 Determination of Slab Type1 Step 1-Soil investigation results show Type of soil: SM win a PI = 2 Thickness: 15 It Relative density: 0-4 It = 25%, and 4-15 It = 50%. Step 2-Determine appropriate slab type From Table I, it is noted that for a loose SM soil the recom- mended slab is Type II (or Type I if the soil is compacted to a dense state). 1See pare. 1.1, pp. 10-12.
48 RESIDENTIAL SLABS ON GROUND 6.3 Procedure1 Since the slab is not to be heated, the decision is to density the soil and use a Type I slab as follows: Step 1-Check method of heating. No coils or ducts are to be embedded; therefore, special re- inforcement is not required. Step 2 Check maximum slab dimensions. The slab is L-shaped, and two dimensions exceed 32 feet; there- fore, two weakened planes will be needed. Step 3-Check partition loads. All partitions are non-loadbearing, and none exceeds the 500 plf maximum allowable; therefore, partitions may rest on the slab. Step 4-Check openings through slab. All openings in the slab are less than 12 inches wide; therefore, no special reinforcing is required. Step 5-Check slab shape. _ The slab is irregular in shape; however, Me two weakened planes (Step 2) can be conveniently located under the edge of parti- tions, dividing the plan into three segments (12 x 18 feet, 18 x 24 feet, and 24 x 24 feet) as follows: 1, 'S-0 L 24 1 1 See pare. 1.2, pp. 12-13.
SUPPLEMENTARY INFORMATION 49 Step 6-Check irregularities in horizontal plane of slab. Ceramic tile is to be used for the bathroom floor; therefore, the slab surface under the tile will be 1 inch lower than the remaining slab, and the underside of the slab must be sloped in order to main- tain the 4- inch slab thickness. Step 7-Check for concentrated loads. There are no concentrated loads involved, and the roof load is supported on the exterior walls; therefore, there is no need for independent footings through the slab. Step 8-Locate necessary weakened planes. The two weakened plane joints, located in Step 5, satisfy all requirements. Step 9-Provide for soil densification. The slab site will be compacted to 95% of that obtainable by the standard Proctor density test. 7.0 DESIGN OF TYPE m SLABS Unlike Types I and II, these slabs are affected in their design (dimensions and reinforcement) both by the type of superstructure and by soil properties as affected by climate. The superstructure, depending on its rigidity, imposes limits on the maximum slab de- flection that can be tolerated. The climate, on the other hand, affects the pattern of distribution of the soil bearing stresses; hence, the distribution and intensity of stresses, and resultant deforma- tion of the slab (as explained in some detail below). 7.1 Effect of Superstructure Since, with this slab type, the superstructure is supported on the slab itself, the walls and other load-carrying elements of the super- structure will tend to follow any slab deformations. The deforma- tions that can be tolerated, before undesirable effects develop, such as cracking, or window- and door-opening distortions, depend upon the materials and nature of superstructure construction used. It is estimated that wood-frame construction can sustain deformations
50 RESIDENTIAL SLABS ON GROUND about twice as great as concrete block construction before the effects of warping and cracking create mechanical and aesthetic problems. This tolerance is due to the fibrous nature of wood and the plastic nature of member connections, as against the compara- tive brittleness of cementitious materials-concrete, stucco, plaster, mortar-which can tolerate only small deformations before actual cracking develops. Thus, for various types of superstructure, particularly for me wall and wall-finishing materials, a limit must be imposed on the magnum differential vertical displacement of any two points on the slab, in order to protect against objectionable distortions and/or cracks. Table m presents, for certain commonly used superstructure types, recommended limits in terms of A/L, to be measured between any two points on the slab along one or another of its principal rectangular axes. 7.2 Effect of Soil Behavior Some fundamental assumptions relating to the mode of slab support must be made for analysis of Type III slabs. The assumptions adopted for purposes of this report are described in the following paragraphs and lead to the definition of the Support Index (C) used in the analysis. TABLE III Permissible Differential Settlements for Stiffened Slabs to Minimize Utility Damage to Superstructure Typical Types of Superstructure Wood Unplastered masonry or gypsum wallboard Stucco or plaster Maximum Permissible Deflection Ratio2 (~/L) 1 in 200 = 1/200 1 in 300= 1/300 1 in 360=1/360 The deflection ratios are determined by the weakest exposed finish mate- rial in the superstructure; therefore, if the superstructure contains one or more of the listed materials as an exposed finish, the most severe Corresponding AL will apply. The listed values are based in part on information contained in U.S. Department of Commerce, National Bureau of Standards, BMS 109, Strength of Houses (Washington: 1948), Table 19, p. 84, and Fig. 42, p. 71.
SUPPLEMENTARY INFORMATION 51 As pointed out previously, soil properties and climate determine the magnitude of potential soil shrinkage or swelling, which, in turn, directly affects the pressure distribution beneath the slab. Thus, if either a very flexible or a very rigid slab rests on level ground, and the soil beneath the slab shrinks or swells unevenly because of change in soil moisture brought about by climate variation, one of two extreme situations results. The very flexible slab will deform to follow the changing ground contour, resulting in differential settle- ments in the slab but no stress. On the other hand, the infinitely rigid slab, retaining its plane, will be supported on the high points of the soil surface without differential settlements but with stress induced in the slab. The Type III slab under consideration is both stiffened (through deep beams) and strengthened (through reinforcement); as such, its behavior falls between that of a very flexible and that of an infinitely rigid slab. Although differential settlements will occur, they will be substantially smaller than those of the supporting ground, and stresses will develop because of the accompanying uneven ground reactions. The degree of stiffening should be sufficient to reduce the differential settlements to a level which can be accommodated by the superstructure, and the reinforcing should be sufficient to impart to the slab the strength necessary to withstand the resulting stresses from the uneven reactions. Actually, there will generally be continuous contact between the soil and slab along the entire surface of the slab; but, because of slab stiffness, the soil does not develop the differential settlement that would occur if the slab were flexible and capable of following soil movements point by point. Instead, it develops a variable bear ing intensity, leading to the consolidation of soils in the higher- pressure areas. This maintains the continuity of soil-to-slab con- tact under most conditions. Obviously, slight tilting of the structure as a rigid body and the phenomena related to such tilting do not influence the safety of slabs- on-ground or the structures supported thereon; consequently, tilting has been ignored in the above discussion. For this reason, "differen- tial settlement of the slab" as used here must be recognized as re- ferring to the deflection of the slab rather than to all differential settlements, which would also include those due to tilting. - 7.3 Support Index Since the exact shape of the pressure distributions is not known, the assumptions that follow have been introduced to permit develop
52 RESIDENTIAL SLABS ON GROUND ment of a rational procedure for analysis and design of a Type III slab. A study of the various patterns of uneven bear~ng-stress dis- tribution has shown that the most severe deflections will result from one or the other of two principal forms of bearing support (Fig. 5~: a. When the maximum bearing stresses develop under the center of a rectangular slab of dimensions L by L' (center support); and b. when the maximum bearing stresses develop on two diagonally opposite corners of the rectangular slab (diagonal support). It is assumed that, whenever either of these two critical bearing conditions develops, bearing stresses of uniform intensity will occur only under the cross-hatched areas of the slab. This is, of course, not a statement of fact, but a supposition which corresponds to some true conditions giving the same deflection or the same maximum stresses in the slab. As such, therefore, it can be treated as an equivalent mode of support. Since climate and sensitivity of soil affect the extent and inten- sity of stresses under the slab, the coefficient C (Fig. 6, p. 53), which defines the boundaries of the supported fraction of slab, is defined as the support index of the slab; this varies with the cli- matic rating of the site and the plasticity index of the soil upon which the slab rests. For the most favorable combination of climate and soil, the value of C is taken equal to unity (i.e., it is assumed that the slab is uniformly supported over its entire area). For the combination of most severe climate and soil conditions, C obtains the value of 0.6 (i.e., the slab is supposedly supported over 36% of its total area, at the center or at diagonally opposite corners). I - L: N _ \ j_ ~ _~_ ~ 1 ~ . -_ J -2~ cL '2~,1'~ - ~ , a. Center Support FIG. 5 support Assumptions J CL (I-C)L . ~ . i L b. D;agona I Support
/ / - > to to · o o- c~ · · a) to - c ~- .... totototo xapul ~Joddns l //l 77 1 ~ I// ~oi - ~ hi: red / ~ 7 // I/ 1 1 11 / . o CO o o o on o to Cat o x C a) 3 to on 3 V o ._ · ~ to .~ ._ C) ._ ._ ._ ~2 U. - ._ o o ~ ~o ._ sat ._ o o o o - o Cat m c) - x o
54 RESIC)ENTLAL SLABS ON GROUND Figure 6 defines the support index (C) in terms of the climatic rating for the slab site and the criterion used to evaluate sensitivity of foundation soil to moisture charge. This criterion may be the PI of the soil, PVC-meter reading for the soil, or percentage swell of the soil. The potential volume change of a particular soil is best determined by a swell test on an undisturbed sample of the soil subjected to probable loading and moisture changes which will likely be experienced beneath the structure. However, in the absence of such a test or tests, PI and PVC-meter readings may be used as an indication of potential volume change in determining the support in- dex (C) from Fig. 6. The effective value of PI, PVC-meter reading or percentage swell to be used in connection with Fig. 6, is obtained as specified under pare. 7.8.1, pp. 65-68 (see also, Fig. 7, p. 553. In the ensuing text, all references to PI should be construed to refer generally to the soil sensitivity criterion-whether PI, PVC- meter reading, or percentage swell-which is used by the designer in each case of slab design. 7.4 Increase of Support Index (C) It is evident, from the way the support index (C) is defined, that it serves as a parameter which permits the setting up of an idealized model (defined by the support assumptions of Fig. 5) for analysis. It is recognized that, under actual conditions, the loads and the re- action pressures on the slab vary greatly from those stipulated by the support index (C) and the support assumptions of Fig. 5. How- ever, the diminishing values of C along with the climatic rating (Cw), and increasing PI values (Fig. 6, p. 53), indicate that, if slab dimensioning and reinforcement are based on analysis of the model resulting from the support assumption of Fig. 5, the slab will be- come progressively stronger and stiffer as C becomes smaller- i.e., as PI increases or Cw is lowered, a stiffer and stronger slab (smaller C) is needed for the satisfaction of similar criteria of strength and deflection. Therefore, adoption of this model as rep- resentative of the actual slab satisfies the qualitative requirements of the actual slab, i.e., that it be stronger and stiffer as the climate becomes more adverse and as the plasticity index of the supporting soil increases. As discussed in pare. 3.3, the climate rating (Cw) is indicative of the intensity of dry-moisture cycles in the supporting soil. The upper limit (Cw = 45) indicates more stable conditions in the soil- moisture balance. As the Cw-value becomes smaller, it represents
~ r ~ . ~ l l : I ~ 1 1 \ CO ~ 6u!pDa~ WOW DAd i \ l ~. ~ , ~Cal o o Cot lo o o ~ To Cd C) Pi 1 of ~ o ED o ~ ~ ._ cd C) Cat . C) - o Cal -
56 RESIDENTIAL SLABS ON GROUND more severe climatic conditions which tend to upset the soil-mois- ture balance. It is conceivable, however, that in an area with a low Cw (i.e., with a climate tending to upset the soil-moisture balance), stable soil-moisture conditions may exist or be achieved by artifi- cial means1-in which case, soils with high PI will be subject to less swelling and shrinkage than in the absence of such artificial means. To compensate for the favorable effects of soil-moisture balancing, the support index (C) can be taken as larger than the value specified in Fig. 6; however? this increase In the absence of justifying quanti- tative data should not exceed that provided by the empirical expression Cm = 0~5 (1 + C) where Cm is the modified value of the index C. 7.5 Determination of Support Index (C) for Compressible Soils (2.5< qu/w< 7.5) When the ratio of unconfined compressive strength (qu) of the soil to average total load (w) on the slab is in the range 2.5 < qu/w <7.5, the soil is herein characterized as compressible. For such soils, a reduced support index (Cr) is substituted for the support index (C). The value of Cr is defined as follows: For C > 0.65 - wc/w Cr = (2.5 - qu/w) [0.13 - 0.2 (wc/w + C)] + (0.65 - wc/w) (7.5a) and for C s 0.65 - wc/w or = C (7.5b) where w = W/LL' = total slab dead and alive load (W)-including its own weight-averaged over the area LL' of the slab; and WC = WC/LL = total concentrated dead and live loads (We), applied Such a case would mist if the water table is, or is maintained, at a suffi- ciently high level to produce a relatively constant soil moisture condition, e.g., as is accomplished by a system of drainage and controlled pumping in the New Orleans (Louisiana) area.
SUPPLEMENTARY INFORMATION 5 7 within the central 50% in any direction (L or L') of the slab and averaged over the entire area of the slab (LL'). With this definition, it is possible that the support index (C) will assume one value for design of the slab in the long direction and another in the short direction. For example, for the slab of Fig. 8, the reduced support indices do differ and are arrived at as follows: _Bearino Wall :-1 Ol o - 1 1 I r 1 1 1 1 1 1 1 1 1 1 1 1 1 1 LJ 20'-0" ~ t L ' 50'-0" | ° ~ran Tohl load tW) = 300 hips Tool concentrated bearing wall load = We ~ 30 hips qua, ~ 1 200 psf C (Fig. 6)-0.90 FIG. 8 Determining Cr for Slab on Compressible Soil a. In the long direction w = W/LL' = 300,000/50~30) = 200 psf qu/w = 1200/200 = 6, which is within the range 2.5<qU/w<7.5. Since We acts as a concentrated load with respect to the long (L) dimension of the slab and is located within the central 50% of the L-dimension We = 30,000/50~30) = 20 psf and, using equation 7.5a, p. 56, Cr = (2.5 - 1200/200) [0.13 - 0.2~20/200+0.9~+ (0.65- 20/200) = (-3.5) (-0.07) + 0.65 - 0.1 = 0.795.
58 RES1DENTLAL SLABS ON GROUND b. In the short direction We = 0 (due to the fact that in the short direction, the load We is not acting as a concentrated load) and Cr = (2.5 - 6) [0.13 - 0.2 (0.9)] + 0.65 = (-3.5)(-0.05)+0.65 = 0.825. Equation 7.5a is a linear expression of the ratio qu/w. Consequently, Cr changes linearly with qu/w between a low limit of 0.65 - wc/w and a high limit of C for a corresponding variation of qu/w between 2.5 and 7.5. This is shown graphically in Fig. 9. - This distance varies with the Blue of C from fig. 6, and the ratio wJw. 2.5 FIG. 9 Variation of Or with Ratio qu/w 7.6 Typical Stiffened-Slab Cross Section 7. 5 qu/w There are several cross sections which might be used in designing a stiffened slab; however, the procedures which follow apply only to the waffle or ribbed slab (Fig. 10, p. 59), and to flat slabs.
SUPPLEMENTARY INFORMATION 59 l_ j , it, 1 1 1 ~ 1 1 E_d _ '~ ~<156-08' `28~' ~ r ~f W~ 11: 1 1 1 1 _d . 'Zi . .. _ ~ . FIG. 10 Typical Type III Slab Section A special concern in connection with the waffle slab is uniformity of distribution of the rib-stiffening effect along the length of the stiffening beams. IN this connection, the following conditions should be met: a. All interior stiffening beams should be made continuous and "dead-ended" at the perimeter beams. b. Beams should be equally spaced along each side, unless special analysis clearly demonstrates that another spacing will provide a slab of uniform stiffness along its width. 7.7 Analysis of Type III Slabs As a basis for analysis, slabs are taken to be rectangular in shape with sides L and L '. Whenever a given slab is not rectangular, it is assumed to be composed of an aggregation of rectangular com- ponent slabs, each of which is independently analyzed in a manner which will be explained in the design examples which follow. For purposes of analysis, the basic support condition assumed (7.3 above) is retained. Further, rather than make a two-dimen- sional analysis, each slab is analyzed for deflection and stress dis- tribution as two one-dimensional cases, one each for the directions L and L '. Comparison of results obtained by the two one-dimensional analyses with those from a rigorous two-dimensional analysis by electronic computation indicates that: a. Smaller limiting moments are derived from the one-dimen- sional than from the two-dimensional analysis.
60 RES1I)ENTLAL SLABS ON GROUND b. Larger deflections are derived from the one-dimensional than from the two-dimensional analysis. For each one-dimensional analysis, it is assumed that the slab is deformed to a cylindrical surface. The two-dimensional effect is obtained by combining the two cylindrical deformations, one in each direction. This necessitates examining two limiting cases (Fig. lla and b) of support corresponding to the support assump- tions defined in Fig. 5, p. 52. For each of these limiting cases, two one-dimensional analyses are required for support conditions specified in Fig. 11, below. Such assumptions, while erroneous, simplify analysis and de- sign, and are fully justified in view of the fact that the error in- volved in one-dimensional analysis is smaller than the inherent error in the various assumptions made in establishing the problem parameters. In addition, whereas the ultimate capacity of concrete provides ample margin of safety against failure due to bending moment, there is little ability to accommodate deflections greater ///////D L _ ~ V _ ~ ~ A J_ -Gil vl~ a . Canti lever b. Simple Beam Jo (I-C)L 2 ~-1 (I-C)L . CL ~ 2 ~ 1 ~ ~L~ CL CL is 2 `, (1-C)L J. 2 ' FIG. 11 Assumed Ground Reaction Conditions to Permit One-Dimensional Slab Analysis 1 i
SUPPLEMENTARY INFORMATION 61 than those allowed. Therefore, analysis in one dimension provides a compensation in that it automatically increases safety against excess deflection, while reducing the existing margins of safety against bending moment failure. Loading on Type m slabs is assumed to be uniformly distributed over the entire slab area. With these basic assumptions made, the analysis of Type m slabs reduces to the analysis of two basic cases, center support (Fig. 12a-cantilever) and end support (Fig. 12b-simple beam). w 1 1 1 1 1 1 llillLl _ l __ _ ~tc ~,CL~ w/c (1 CAL CL 2 t CL ~2 w . . 1 1 1 1 1 :llLIl~ , ~ - CL t-- 2 ~ W/C ~ Lit CL _ ~ 2 a. Canti lever b. Simple Beam FIG. 12 Limiting Cases of Support for Type III Slab Analysis Analyzing slabs having the above two limiting cases of support for stress and deflection Cantilever Maximum moment (Oman) wL2L/(1 - C)/8 Maximum shear (Vmax) - wLL / (1 - C) /2 = Maximum deflection (~m ax) = wL4L' (3 _ 4C2 + C3) /384E HI Simple Span wL2L'(1 - C)/8 wLL'(1 - C)/2 wL4L'(5- 6C + C3) /384E 'I where w is the per-square-foot average total load on the slab; L and L' are the slab length and width, respectively; I is the moment of inertia of the slab cross section; and E ' is the effective modulus of elasticity, i.e., 1.5 (10~6 psi for concrete under sustained loading. Since slabs-on-ground are likely to remain in maximum deflected condition for long periods, the creep modulus of elasticity of con
62 RES1DENTLAL SLABS ON GROUND Crete under sustained loading is considered to be the "effective modulus." The creep modulus of elasticity (E) for concrete is of the order of one half the modulus of elasticity, i.e., 3~10~6 psi, for concrete with f'c = 3000 psi. Introducting the conventional concrete design modulus of elas- ticity (E) into the above formula Cantilever Max = wL4L'(3_4c2 + C3~/192 EI Simple Beam wL4L / (5-6C + C3~/192 EI For values of 0.6 ~ Cs 1.0, the linear function 4~1-C) provides a suitable approximation for both functions 3_4c2 ~ C3 and 5-6C + C3. Therefore, with this approximation introduced, MmaX = wL2L' (1 - C) /8 and V =wLL'(1 C)/2 In the range 0.6 s C s t.o Amax = wL4L' (1 - C) /48 EI (7.7a) (7.7b) and, by solving for the deflection ratio (~/L), this equation becomes &/LmaX = wL3L' (1 - C) /48 EI. (7.7c) For a reinforced rectangular concrete beam of width b and depth d, the moment of inertia (I) after cracking of the section, is considered to be 1/3 k3bd3 ~ np (1-k)2bd3 where k is the ratio of the depth of compression zone to the effec- tive depth of the slab, n is the ratio of the moduli of elasticity for steel and concrete, and p is the steel ratio, i.e., ratio of steel cross- sectional area (As) to concrete cross-sectional area Qad). Introducing the moment of inertia factor Z - 1/3 k3+np (1 _k)2 I =bd3Z
SUPPLEMENTARY INFORMATION 63 and, by substituting in equation 7.7c, ~/LmaX = wL3L'(1 - C)/48 Ebd3Z. (7.7d) In the elastic range, the maximum bending moment resisted by the internal stresses of the beam section ¢bd) based on the steel reinforcement is ASfSjd where Is = allowable unit tensile stress of steel (20,000 psi)1 j = ratio of internal moment arm to depth (an average 0.865) As = cross-sectional area of tensile reinforcement (pbd). Therefore, the maximum allowable bending moment (M), should be fSidAS = fSjpb2 (7.7e) Similarly, if VC is the maximum allowable shear stress for con- crete, then the maximum shear capacity (V) of the slab will be vcm. (7.7f) Considering the pairs of equations 7.7b and I, and 7.7a and e, plus equation 7.7d, and designating the magnum permissible differential deflection ratio to be A/L, the following condition equations provide the design criteria for Type III slabs: Vma6X 5 VmaXaJlow Mma,X ~ (~/L)maxallow (~/L)max S (~/L)maxal1Ow or or w(1-C)LL' /2 ~ VC Bd or w(1-C)L2L i/8 ~ fSjpBd2 w(1-C)L3L //48 EBd3Z 5 ~/L 1Values of Is for A-432 and higher strength steels, as recommended in the 1963 American Institute of Steel Construction (AISC) Specification, can also be used. The value used here (20,000 psi) is for A-15 steel.
64 RESIDENTIAL SLABS ON GROUND where B = sum of the web widths of beams running parallel to the L-dimension of the slab. These condition equations can be rewritten In more convenient form as follows: u1~1-C)(L'/B)(L/d) ~ kl, where k1 = 2vc = 150 psi=21,600 psf l-C)(L'/B)(L/d)2 5 peg, where k2 = 8jfS = 2~10~7 psf (7.7h) w(1-C)(L'/B)(L/d)3 ~ Zk3, where k3= 48E(~/L) =207~10~8~/L)psf. (7.7i) Obviously, if the analysis is conducted along the width (L') of the slab, the above relationships (7.7g, h, and i) will have to be modified by interchanging L with L' and substituting B' for B; i.e;, B' = sum of web widths of beams running parallel to dimension L . Defining as the slab dimension (L or L') along which the slab analysis is conducted ' as the slab dimension (L' or L) normal to the direction along which the analysis is conducted b as the aggregate width of stiffening beams along the dimen- sion ~ for which the design is conducted conditions 7.7g, h, and i obtain the following general form, which is adaptable to the analysis or design of the slab along either of its principal dimensions (L or L '): w(1 - C) (a '/b) (~/d) ~ k1 = 2vc = 150 psi = 21,600 psf (7.7g ') w(1 - C) (a '/b) (Q/d)2 s pk2, where k2 = 8jfS = 2(10~7 psf (7.7h ') w(1 - C) (1'/b) (l/d)3 5 Zk3, where k3 = 48E (~/L) = 207(10(/L psf. Conditions 7.7g', h', and i' are necessary and sufficient to safeguard the slab against excessive shear failure, bending moment, and differential settlement, respectively.
SUPPLEMENTARY INFORMATION 65 In expression 7.7f, the maximum shear force (Vmax) is taken equal to vcbd in accordance with the 1963 ACI Code. 1h 7.7g, VC was assumed equal to 75 psi. The 1963 ACI Code allows, as a measure of diagonal tension, a maximum shear stress (vc) equal to 60 psi at a distance d from the face of the support, i.e., 60/vc = 0.5 wL'(1-C)L-wL'd/0.5 wL'(1-C)L = 1-2d/(1-C)L or However, and Therefore, and and 7.8 Design of Type III Slabs 2d/(1-C)L = 1- 60/Vc 1/1-C 2 1/0.4 or 1/1-C 2 2.5 d/L ~ 1/24 or 2d/L > 1/12. 2d/(1-C)L 2 2.5/12 or 2d/(1-C)L 2 5/24 1 - 60/VC 2 5/24 or 60/VC 5 1 - 5/24 or 60/VC ~ 19/24 VC 2 24(60)/19 or VC 2 75 psi. 7.8.1 Determination of Effective PI, PVC-Meter Reading, or Percentage Swell of Soil An essential design parameter, in addition to the average dead and live load on the slab, is the support index (C). As observed in pare. 7.3, this index depends on the climatic rating and the PI of the soil.
66 RES~ENTLAL SLABS ON GROUND a. When the soil has a uniform PI for a depth of at least 15 fees, 1 this PI is considered to be the effective PI for design pur- poses. However, when different soil strata possess different PI's, the determination of a PI value to be used for design purposes is made as follows: If the PI at the layer immediately below the lowest elevation of the slab is the maximum PI of all soil layers within a depth of 15 feet below the lowest elevation of the slab, then the PI of the top layer is considered applicable to the entire foundation soil. In all other cases, the PI to be used for design purposes is taken as the weighted average of PI's within the top 15 feet of the soil immediately below the slab stiffening beams. Weight factors (3, 2, and 1, respectively) are used for soil strata within the top, middle, and bottom 5 feet of soil below the stiffening beams. For example, for the hypothetical soil profile of Fig. 13, below, PI=1/30ft(~) Elev. 0.00-Underside of Slab Stiffening Beams 1 Pl ~30 r Elev. 3.00 ~m , So - 1 Pl =70 ' r Elev.-9. 00 Pl =60 `~' - Elev. 15.00 Wt. factor = 3 _O 1 mu _ . _ . . . . _ . . _ L~' , Elev. 10.00 I Wt. factor =1 ~o ~ Elev.-S.Oo Wt . factor = 2 1 - Elev. 15.00 FIG. 13 Hypothetical Soil Profile with Variable PI 1All soil measurement depths are taken from below the bottom of the slab stiffening beams.
SUPPLEMENTARY INFORMATION 67 where E= sum of [(weight factor)(thickness of stratum)(PI of stra- tum)], and 30 It = 5 It times the sum of weight factors = 5~3+2+1~. Referring to Fig. 13, p. o6 ~ = 3~3 ft)30+3~2 ft)70+2~4 ft)70+2~1 ft)60+1~5 ft)60 = 1670 It then PI= 1670/30 = 55.7 or 56. b. When the determination of support index (C) is based on values of Cw and a PC-meter reading, the effective value to be used for the PC-meter reading should be derived- in a manner similar to that given above, i.e., if the PVC-meter reading for the soil stra- tum immediately below the bottom of the slab1 is higher than the reading obtained in underlying strata within a depth of 15 feet below the slab, 1 then this reading should be considered applicable to the entire foundation soil. Otherwise, a weighted average value of PVC- meter readings should be computed for the soil within 15 feet from the bottom of the slab,1 i.e., weight factors of 3, 2, and 1, respectively, should be used for the soil strata within the top, middle, and bottom 5 feet of soil below the stiffening beams. (See also Fig. 7, p. 55.) c. When the determination of support index (C) is made in terms of Cw and percentage swell of the foundation soil, the percentage swell value also should be derived in a manner similar to that given above; i.e., the percentage swell of the top layer should be considered applicable to the entire foundation soil if it is the highest within a depth of 15 feet below the bottom of the slab.1 Otherwise, the effective percentage swell should be obtained from the percent- age swell of the soil strata within 15 feet below the bottom of the slab1 as a weighted average using weight factors 3, 2, and 1, re- spectively, for the top, middle, and bottom 5 feet of soil below the bottom of the slab.1 The percentage swell for a specific soil stratum should be ob- tained through swell tests using conventional consolidometer test equipment on undisturbed soil samples under pressure correspond- ing to the in situ overburden pressure plus the average total dead and live load (w) on the slab (see also, pare. 7.3, p. 51~. The 1 See footnote' p. 66.
68 RESIDENTIAL SLABS ON GROUND undisturbed samples should be obtained under soil moisture con- ditions representative of conditions prevailing at the time of construction. 7.8.2 Effective Load (w) on Slab It has been assumed that the slab is uniformly supported on the ground over at least 60% (C 2 0.6) of its area in each of its princi- pal directions. It has also been assumed that the maximum un- supported length in each direction is proportional to the length (L) or width (L ') of the slab. While both these assumptions are reason- able when L = L', they become progressively less so as the ratio L/L' becomes greater than unity. As the ratio L/L' increases above unity, the design becomes more conservative for the long dimension (L) of the slab as compared to the design for the short dimension (L '). For this reason, the load acting along the long dimension is reduced by a coefficient up which is equal to unity when L = L', and which obtains a minimum value of 0.5 when L ~ 2.25L', varying linearly between these limits; thus up = 1.4 - 0.4 (L/L') 2 O. 5. With this adjustment, and in consideration of expressions 7.7g', h', and i', the effective load (w) acting on the slab can be obtained from the equation w = w(1 -cap where = 1.4 - 0.4 (L/L'), but not less than 0.5, for the analysis in the long direction (L), i.e., when Q = L' = 1 for the analysis in the short direction (L'), i.e., when Q = L. When the symbol for the effective load (w) is introduced in expressions 7.7g', h', and i', they become wit' /b) (Q/d) s k1, where k1 = 21,600 psf (7.8a) we ' /b) (Q/dJ2 ~ pk2, where k2 = 2~10~7 psf (7.8b) wit' /b) (Q/d)3 ~ Zk3, where k3 = 207~10~8 (~/L) psf. (7.8c)
SUPPLEMENTARY INFORMATION 69 7 .8 .3 Determination of Effective Load (w) Once the effective PI of the soil is determined and the climatic rating of the site is known, the value of the support index (C) can be obtained from the curves of Fig. 6, p. 53. The average load (w) on the slab consists of two parts Wd = dead load of the slab itself WS = dead and live load of the superstructure. Of these, wd is estimated by the empirical formula Wd =~2L+30)psf where L = the long dimension of the slab in feet. The slab dead load (Wd) empirically obtained as above, must obviously be checked for accuracy after the actual dimensions of the slab, and therefore loads, become known. The value of WS is derived from the total of actual superstructure dead load plus live load calculated on the basis of 30 psf on each floor and 10 psf on the roof, with no reduction for total area. With Wd and WS computed, the effective load (w) is finally estimated as follows: w = Wd + WS w = w(1-C)(o = w (1-C)(1.4-0.4 L/L') 2 0. 5 w(1-C), for ~ = L w = w(1-C), for ~ = L ' . 7.8.4 Limiting Values of Variables The allowable range of the steel ratio (p) is 0.003 s p ~ 0.020. For this range of p, the function Z = 1/3 k3 + np (1 - k)2 varies in the range of 0.0224 ~ Z ~ 0.0908. For extreme cases (the heaviest possible two-story residence built on the worst type of soil, in the most sensitive climatic region), the quantity w becomes of the order of 150 psf. Under usual design conditions, the ratio L/L' varies between 1 and 2, i.e., 1 ~ L/L' ~ 2, except for overlapping slab portions which belong to a slab of irregular shape.
70 RESH)ENTIAL SLABS ON GROUND The width of waffle slab stiffening beam webs usually varies between 8 and 12 inches and rarely reaches 14 inches. The spacing varies at most between 8 and 15 feet. An economical depth of the stiffening beam is usually between 1/14 and 1/24 of the long dimension of the slab, i.e., 14 ~ L/d ~ 24. For average effective loads, i.e., w = w (1-C)`p = 25 to 50 psf, a good design depth (d) is 1/20 of the long dimension (L). From condition equation 7.8a, i.e., w(t'/b)~/d) s k1, it be- comes evident that the maximum allowable value for w (! '/b) will be obtained for the minimum t/d ratio, i.e., for t/d = 7. For this value of t/d w(t'/b) ~ 21,600/7 = 3086 or 3100 psf. For the maximum condition (~/d = 24), 7.8a is reduced to wtQ'/b) s 21,600/24 = 900 psf. Therefore, if * (t'/b) is less than 900, the shear condition (7.8a) is invariably satisfied. By dividing respective sides of equation 7.8b by equation 7.8a, and equation 7.8c by equation 7.8b, two expressions are obtained for the ratio t/d t/d = k2/k1 (p) = 925 p and (7 .8d) t/d = k3/k2 (Z/p) = 1035 (~/L) Z/P. (7 . Be) By equating the right sides of these two equations Z/p = 0.894 (L/~)p. (7.8f) With Z a function of p, it is obvious that for each of the dis- crete values of L/6, i.e., 200, 300, or 360, there is a unique value of p which satisfies equation 7.8f. At this steel-ratio value, all three conditions, i.e., shear (7.8a), moment (7.8b), and deflection (7.8c), are simultaneously satisfied. Therefore, this value of p is required for balanced design of the slab. Corresponding values of p and Q/d for the various deflection ratios are
SUPPLEMENTARY INFORMATION 71 L/6 = 200; Pbalanced = 0.002 (less than the minimum required) Lie= 300; Pbalanced = 0.0177 (corresponding to t/d = 16.4) L/^ 360; Pbalanced = 0.0154 (corresponding to t/d = 14.25) However, the steel ratio for balanced design is not associated with the most economical design. Inasmuch as the shear condition (7.8a) is more easily satisfied than the conditions for bending mo- ment and deflection, it is not necessary to design for a state of stress for which all three conditions become concurrently critical. In order to define the boundaries for which the dominating con- ditions of bending moment and deflection become simultaneously critical, equation 7.8c is divided by equation 7.8b, or t/d = 1035 (~/L) Z/p. if (7.8g) For a given fixed value of ~/L, and for each value of t/d, there .s one specific value of the Z/p ratio for which both conditions 7.8b and c are simultaneously satisfied. Values of the steel ratio (p) above the value corresponding to the Z/p ratio for which equation 7.8g is satisfied, tend to become uneconomic because steel, rather than concrete, is used to impart stiffness to the slab in order to prevent excessive deflection. If the steel ratio for the long dimen- sion of a rectangular slab is selected so that critical conditions for bending moment and deflection develop simultaneously, this slab will be stiffer than necessary in the short direction. Thus, in actual design practice, economic design of rectangular slabs often requires that some steel be used to impart stiffness in the long dimension. The significance of this practice will become more evident in the examples of design, and the pertinent graphs to aid the design process, which follow. 7.8.5 Use of Charts for Design To facilitate design, Figs. 14, 15, and 16, have been prepared for the discrete values L/200, L/300, and L/360 (pp. 72-74~. The charts enable one to select the steel ratio (p) in terms of the load index, i.e., the amount w (Q'/b) for specific values of the depth ratio (~/d) at increments of one.
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SUPPLEMENTARY ~FOR~TION 7 5 For a given t/d, the minimum required steel ratio (p) increases linearly with wit '/b), in accordance with the bending moment con- dition (7.8b). This linear dependence is graphically expressed in the curves of Figs. 14, 15, and 16. However, in accordance with shear condition 7 .8a, the product t/d times w(t '/b) cannot exceed k1 = 21,600 psf unless the stiffening beams are reinforced with stirrups for shear. The limits, therefore, up to which the linear function 7.8b is valid for use of stiffening beams without stirrups, are defined in the charts by the curve marked VC = 75 psi. This means that if, for a selected combination of t/d and w (Q'/b), the corresponding value of minimum steel ratio (p) is above the shear line VC = 75 psi, the stiffening beam then designed should be reinforced with stirrups to accommodate shear stresses in excess of 75 psi. This practice is not recommended. Instead, if this happens, changes should be made in b or d, to effect a reduction below the shear line of the minimum required steel ratio (p). If, however, for any reason this is not done, the designer should compute the necessary steel for stirrups on the basis of a maximum shear given by equation 7.7b, and modified by the load coefficient, i.e., VmaX= 0~5W(LL') = 0.5 W(1-C)LL. The deflection condition 7.8c requires that, for a given deflec- tion ratio (~/L), the minimum required Z be 1/k3 (w)t'/b(Q/d)3 Table rv provides values of Z for discrete values of p, at incre- ments of 0.1% in the range 0.003 s p ~ 0.02. Fig. 17, p. 77, charts the function Z versus p. A four-degree polynomial curve, fitted by the method of least squares, provides p as a function of Z in accordance with p = -0.088092446 + 0.17827805(Z) - 0.5542073(10)-2 (Z)2 + 0.17393919(10)-2 (Z)3 - 0.55434473(10)-4 (Z)4. This shows that p increases with Z. and that the minimum re- quired p increases nonlinearly with the load index [w(t '/b] in accordance with 7.7i. The curve designated as M-D (for moment-deflection) in Figs. 14, 15, and 16, represents the values of p for which the deflection and bending moment conditions are simultaneously satisfied. Above
76 RES~ENTLAL SLABS ON GROUND TABLE IV Values of the Steel Ratio Function (Z) P zl k2 0.003 0.0224 0.216 0.004 0.0277 0.246 0.005 0.0332 0.270 0.006 0.0384 0.291 0.007 0.0433 0.309 0.008 0.0479 0.327 0.009 0.0523 0.344 0.010 0.0565 0.359 0.011 0.0605 0.373 0.012 0.0644 0.386 0.013 0.0681 0.398 0.014 0.0717 0.410 0.015 0.07 52 0.421 0.016 0.0785 0.432 0.017 0.0817 0.442 0.018 0.0848 0.452 0.019 0.0879 0.460 0.020 0.0908 0.464 1 z = 1/3k3 + np(1 - k)3 2 k =42np + n2p2 _ up n = 10 this curve, the required steel to satisfy the deflection condition 7.8c is greater than the steel ratio required to satisfy the bending moment condition 7.8b-this derivation is as described above. If the value of the minimum required steel ratio 0?) is, for a given set of w(t'/b) and t/d values, above the M-D curve, the deflection condition 7.7i controls, and the value of p is obtained by nonlinear extension of the curves as shown in Figs. 14, 15, and 16. For reasons explained above, the broken-line curve marked in the chart as "limiting steel" should be regarded as the curve de- fining the maximum allowable steel ratio (p) for any combination of w({'/b) and t/d. If, for a given combination of w(i'/b) and t/d, the minimum required p is above this boundary line, ~ change should be made in the values of d, b, or both, to effect a reduction of the minimum required steel ratio ~) below this line. 7.8.6 Design Sequence The designer usually bases design on the following given conditions:
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78 RESI:C)ENTIAL SLABS ON GROUND a. Slab dimension L and L' b. Support coefficient (C), which depends on slab site, and PI and qu of the soil c. Effective load coefficient (`p), which depends on the ratio L/L' d. Dead and live load (w) acting on the slab. From these quantities, the designer must determine values for the bottom and top steel ratios (along the two dimensions of the slab), in terms of widths b and b' and depth d of the stiffening beams. These variables must be so defined as to ensure economy of materials and construction, and to satisfy the shear, bending moment, and deflection criteria given by expressions 7.8a, b, and c, respectively. To achieve these goals, the following guidelines and procedure are recommended: Step 1-Estimate the total average dead and live load on the slab, i.e., a. First compute the total superstructure load (ws), allowing for a live load of 30 psf of floor area and 10 psf of roof area, and introducing all superstructure dead loads at true value b. Next, estimate the per-square-foot dead load (wd) of the slab itself from the empirical formula Wd = (2L + 30) psf where L is the long side of the rectangular slab in feet c. Finally, set the average total load, i.e., w = wd + ws. Step 2 - (A the slab site, determine a. The lowest unconfined compressive strength (qu) from un- disturbed samples within the top 15 feet of soil immediately below the lowest point of the slab (if qu obviously exceeds 7.5w, i.e., 7.5 times the average slab dead and live load, its determination
SUPPLEMENTARY INFORMATION 79 is not necessary); verify that qu/w 2 2.5-if not, a Type IV slab will be necessary (Table I, p. 11). b. The plasticity index (PI), as specified in pare. 7.8.1, pp. 65-68. Step 3-Determine (Fig. 6, p. 53) the support index (C) in terms of the climate rating (Cw) as obtained from Fig. 1, p. 38 and the PI as determined in Step 2 above. Equate C to the value of the modified support index (Cm) as given by the empirical equa- tion of pare. 7.4, pp. 54 and 56, whenever-because of special cir- cumstances-conditions are such that variations in soil moisture are not reasonably expected to the extent stipulated by the climatic rating (Cw), pare. 7.4. Wherever 7.5 < qu/w ~ 2.5, equate C to the value of the reduced support index (Cr) as given in pare. 7.5, pp. 56-58. In computing the ratio qu/w, use a tentative empirical value for wd obtained from the relationship Wd = 2L + 30 psf where L is the long dimension of the slab in feet. Step 4-From Table m, p. 50, determine the maximum al- lowable deflection ratio (~/L) for the contemplated type of superstructure. Step 5-Determine outside dimensions of slabs to be designed.] Divide slabs of irregular shape into overlapping rectangles in such fashion that the resulting exterior boundary provides complete congruence with the slab perimeter-e.g., En': ~ --t-- . ~ KI: WE W: Design each of the composing rectangles thus derived as in Step ~ below. Step 6-Id terms of the ratio L/L', average load (w), and support index (C), compute the effective average load (w) on the 1Seealso5.0,pp.2l_22.
80 RESIDENTIAL SLABS ON GROUND slab for design in the long and short dimensions from the follow- ing equations: w = w~l-C) up for the long direction w = w(1-C) for the short direction where up = 1.4 - 0.4 (L/L ') ~ 0. 5. Step 7-Divide both sides of the slab with stiffening beams spaced at equal distances. Select a beam spacing not greater than 15 feet, nor smaller than 8 feet-preferably 9 to 12 feet. Select a beam width between 8 and 14 inches, preferably 8 to 10 inches. In practice it is often difficult to open an 8-inch-w~de trench in the ground, especially a deep one, i.e., 20 inches or more. However, it must be recognized that selection of wider beams does not lead to savings in beam depth or in the amount of reinforcement unless a. Shear stresses are high and an increase of B or B' may reduce the load factor tw (L '/b) ~ b. The steel required for deflection control is substantially more than the steel required for bending moment-a condition which is more likely to occur in the long direction. Consequently, it is recommended that design be based on narrow beams which can be widened if it becomes evident that such widen- ing will reduce the steel or ease shear stresses. Once designed, the beams can of course be built wider in the field if doing so appears to offer advantages in construction. Within the range of values set here, a closer spacing of beams and a greater width are recommended for comparatively greater values of the effective load. (Usually beams 8 inches wide at about 12-foot spacing will be adequate for effective loads not exceeding 30 to 35 psf.) If L and L' are, respectively, the long and short dimensions of the slab, compute the load factors in the long direction, wit /b) = w(1 - C) ~ (L/B)
SUPP LE ME NTARY INFORMATION 8 1 and, in the short direction, w (t '/b) = w (1-C) (L/B). In both cases, the load index must be smaller than 3100. It is recommended that the load index in the long direction be kept below 1200, and, in the short direction, below 2000, by adjusting the spac- ing and/or width of the stiffening beams. Step 8-Select a beam depth (d). This depth is determined in- directly from the depth ratio (L/d), i.e., long dimension (L) over beam depth (d). The L/d ratio is selected in terms of the effective load (w) in the long direction and the deflection ratio (~/L). The following rule of thumb can be used for selecting a trial value for the depth ratio (L/d): For O sew s25psf, L/d 220 For 25 < w s 50 psf, try 20 2 L/d 2 17 For 50 < w ~ 150 psf, try 18 2 L/d 2 14. Usually a comparatively lower range of L/d values is recom- mended, i.e., greater depth (d) for L/~= 360. The higher range of values is more appropriate for L/~200. Square slabs require com- paratively lower values of L/d. Step 9-For the given value of ~/L (depending on type of super structure, Table III, p. 50), enter the appropriate chart, i.e., Fig. 13, 14, or 15. Then proceed as follows: a. When designing for the long (L) direction, find the value of load index w (t'/b) = w (1-C) up (L'/B) and determine p for the value corresponding to the ratio L/d. If p is too far below the M-D curve, it means that beams selected were too deep, i.e., d is too large. If p is above the "limiting steel" curve, it means that beams selected were too shallow and probably too narrow. In both cases, an adjust- ment should be effected in the value of either B or d, or both, and the design repeated from Step 7. If the minimum required steel ratio (p) lies above the 75-psi curve, it means that the selected width (B) is too small. An increase should be made and the slab redesigned from Step 7.
82 RESIDENTIAL SLABS ON GROUND If, in an exceptional case, it is desired to restrict beam dimen- sions, the beams should be provided with stirrups to accommodate the excessive shear stresses. In this case, the ACI Code should be used, and the maximum effective shear force should be estimated from VmaX = 0.5 (1-C) LL' This shear force will be assumed to be distributed among all the beams in the direction for which the design is executed. A good selection of values for B and d will usually (with some exception for very lightly or very heavily loaded slabs) result in values of p in the neighborhood of the M-D curve b. When designing for the short (L') direction, find the value of load index wit '/b) = w(1 - C) (L/B ' ), and determine p for the value corresponding to the depth ratio t/d = L'/d. If the p-value is less than 0.003, ascertain whether B ' can be reduced Tut not below a value corresponding to a beam width of less than 8 inches). If B' cannot be reduced, check the previously determined steel ratio (p) for the long direction. If p for the long direction is below the M-D curve, reduce the depth (d) of the beams and redesign from Step 7. If p in the long direction is above the M-D curve, the width of beams running along the short direction is 8 inches, and p in the short direction is less than 0.003, then use 0.003 for p in the short direction (the minimum required steel ratio). c. For reinforcement, if bs and b's are, respectively, the width of each stiffening beam along the long and short dimensions of the slab, the bottom steel (As) in each beam is determined by As = PbS for the beams in the long direction, and by As = pb'Sd for the beams in the short direction, where p is the steel ratio for the long and short dimensions, respectively. The top steel (A's) in each beam is given by A'S =AS-0.65in.2 This top steel is to be placed one inch clear of the slab top, and the bottom steel is to be placed two inches clear of the beam bottoms. d. Adjustments for unequal spacing should be made if, as often
SUPPLEMENTARY INFORMATION 83 happens in slabs of irregular shape, the beams in one direction of a particular rectangular portion of the slab are not all equally spaced. Since the above design assumes equal spacing of all beams in any one direction of the slab, an adjustment is needed for the steel. Whether this adjustment will influence only the steel, or both the steel and beam width, depends on how critical the shear cri- terion is for the beams concerned. If, in determining the steel ratio (p) for average spacing, the point in the appropriate chart (Fig. 14, 15, or 16) corresponding to this ratio is found to be very close (within 0.15%) to the VC = 75-psi curve, then the shear cri- terion must be considered close to critical. In such case, both the width (bs) and the steel (As) are increased in all beams at more than average spacing by a factor equal to the ratio of the actual over the average spacing of the beams. True Spacing of an inter- mediate beam at unequal distances from adjacent beams to either side is considered equal to the average value of these two unequal distances; spacing of an end beam is considered equal to its dis- tance from the first interior beam. If the shear criterion is not close to critical (ratio p not within 0.15% of the vc = 75-psi curve), the only adjustment required is for steel. In this case (which is the most commonly encountered), only the steel in beams spaced at more than the average spacing is in- creased, i.e., by the ratio of the actual to the average beam spacing. 7.8.7 Top Slab The stiffening beams of a waffle slab divide the slab into rec- tangular or square bays. If the beam spacing is 12 feet, then a top slab 4 inches thick is adequate, provided no unusual concentrated loads (such as chimneys or heavy equipment) are acting on the slab. This 4-inch slab is to be reinforced with No. 3 bars, 12 inches o.c. each way, placed at one third the thickness of the slab from the top. If the maximum clear dimension of some rectangular or square bays exceeds 12 feet, the designer then has the choice of using either a slab with No. 3 bars at 10 inches o.c. each way, or a 5-inch slab with No. 3 bars at 12 inches o.c. each way. ~ either case, the steel is placed at one third the thickness of the slab from the top. These provisions for the top slab are valid with the limitation that no unusual concentrated loads are acting on the slab.
84 RESIDENTIAL SLABS ON GROUND 7.8.8 Excluded Parameters Obviously, in the analytical procedure thus far presented, there are numerous parameters which have not been entered in the mathe- matical expressions or reflected in the conclusions of this report. Among these parameters are the actual moment of inertia of the concrete T-beam which deflects because of both negative and posi- tive moment, and the contribution of the stiffness of superstructure to the stiffness of the slab. The problem is sufficiently complex when analyzed on the basis of the parameters which have been given consideration. While a rigorous mathematical procedure has been provided here, this analysis is based in several critical instances upon assumptions which of necessity limit the precision of the solution that can be obtained in each instance. Since the additional parameters would influence design results to a degree less than the measure of accu- racy provided by the considered basic assumptions, their inclusion would complicate the problem of analysis without making a mean- ingful contribution. Thus, besides adding to the difficulties of analysis, it would mislead by implying an accuracy beyond that inherent in the assumptions which have been made. It is felt that, in its present form, the analytical method pro- vided herein gives due and adequate consideration to the principal parameters involved. These parameters, as well as the pertinent assumptions, are clearly identified; thus experience plus future case histories of designs based on this analysis can be expected ultimately to provide data needed for reevaluation of the basic assumptions and for any necessary corrections. As the degree of approximation in present assumptions can be more accurately gauged, other parameters can be worked into the design procedures. This should happen as soon as the order of accuracy associated with the omitted parameters is comparable with the overall accu- racy of the developed analysis. Only the properties of the soil itself and its tendency to heave under varying climatic conditions of moisture have been considered. Excluded is any design reference to frost action. Where frost is a problem, local codes usually require that the perimeter wall, even with a slab-on-ground, be carried below the frost line. Where such is the case and the slab is to be stiffened as well, interior cross beams need be made only as deep as needed to conform to the re- quirements established in this report, not to the greater depth which may be required for exterior beams to reach below the frost line.
SUPPLEMENTARY INFORMATION 85 7.8.9 Use of 24,000-psi Steel (ASTM A-432) or WWF Whenever 24,000-psi reinforcing bars are used, the cross- sectional area of required steel, as determined from the formulas in this report, can be modified as follows: a. Bottom reinforcement of beams can be reduced by one sixth, provided it is not less than 3% of the effective beam concrete area (bd). b. Slab reinforcement cannot be reduced unless use is made of WWF. In this case, 6x6 - 3/3 WWF may be substituted for No. 3 bars at 12 inches o.c. and 6x6 - 2/2 WWF may be substituted for No. 3 bars at 10 inches o.c. c. Top reinforcement of beams will not be reduced unless re- inforcing bars used in the top slab comply with ASTM A-432 or consist of WWF. In this event, top reinforcement will be equal to bottom reinforcement (reduced as specified in a above), less 0.55 in.2. 7.9 Design of Type m Slabs on Compressible Soils (7-5>qU/W2 2-5) Compressible soils usually have an unconfined compressive strength (qu) of less than 2000 psf. The average dead and live load on slabs-on-ground supporting residential construction of no more than two stories varies in the range of 200 psf s w ~ 400 psf. There- fore, by defining compressible soils as those for which the ratio qu/w ~ 7.5, it is assumed that a hazard resulting from settlement due to soil compressibility becomes likely when the unconfined strength of the soil (qu) is less than a limit which varies between 200 (7.5) = 1500 psf and 400 (7.5) = 3000 psf. Since it is the ultimate settlement, rather than the slope of the stress-settlement curve, which defines the damage potential of compressible soil, a given soil can be considered "acceptable" with respect to compressibility when subjected to relatively small loads (w), and unacceptable when subjected to larger loads (w). This explains why a lower limit LOU = 1500 psc defines compressible soils when subjected to light slab loads of approximately 200 psf, while the higher limit (qu = 3000 psf) is used to define compressibility for soils subjected to higher loads, i.e., approximately 400 psf.
86 RESH)ENTLAL SLABS ON GROUND 7.9.1 Behavior of Compressible Soils under Uniform Load (w) Ln describing the behavior of compressible clay soils under the application of uniformly distributed loads (w), it is useful to review two theoretically extreme cases. The first (Fig. 18a), graphically represents an absolutely flexible slab. With such a slab, loads (w) are balanced by equal soil reactions (w) from below, since the slab cannot develop or transmit shear forces or bending moments (a bedsheet provides a simple analogy). If a slab of this kind receives a uniform load (w), it will have to develop support stresses exactly equal to w, in lieu of shear or bending moment. The result will be uniform loading (w) of the soil over its entire surface. When the soil is loaded uniformly at its surface, it will develop nonuniform stresses at a depth below the surface in accordance with the w loads w 111, reactions w z reactions at depth z below slab a reactions at depth z below slab b FIG. 18 Stress Pattern in Clay Compressible Soils under Absolutely Flexible and Absolutely Rigid Slabs 't111111111111 loads w w, 2 1 11~ ~ ~3 Hi reactions w1 << w We >> w J we z
SUPPLEMENTARY INFORMATION 87 elasticity analysis of Boussinesque, or the Westergaard theory, for imperfectly elastic soils. According to these theories, higher stresses will develop under the center of the slab than under the perimeter as the depth below the slab increases. Thus, the soil will undergo a greater total settlement under the center than under the edges, and the perfectly flexible slab will settle unevenly, "dishing" as shown in Fig. 18a. The second extreme case, shown graphically in Fig. 18b, is an absolutely rigid slab, i.e., one having an infinite modulus of elas- ticity or rigidity and hence one which will not deform under any combination of loads. If such a slab is placed over a compressible clay soil and loaded uniformly with load w, it will have to settle uniformly because of its rigidity. Therefore, it will force a distri- bution of stresses on the soil, such that the soil surface will settle uniformly. By applying either the Boussinesque or the Westergaard theory for analysis of soil settlement, substantially higher stresses (w2) must be applied on the soil surface below the perimeter of the slab than the stresses (wl) under the center of the slab, in order to bring about uniform settlement of the soil surface under the entire area of the slab. Under such an unequal distribution of stresses at the slab undersurface, the stresses which will develop at some depth below the slab will tend to be equal over the full extent of the slab; eventually, at a sufficient depth below the slab, stresses will tend to be higher below the center than below the perimeter of the slab. Since settlement will be equal at all points on the slab, it follows that the integrals of all settlement increments over all increments of depth are identical at all points, even though the contact pressure immediately beneath the slab is higher at the perimeter than over the interior area under the slab. From study of these two extreme cases (Fig. 18), it becomes evident that in a regular, beam-stiffened slab-on-ground, the soil reactions must be uneven and substantially higher under the perim- eter of the slab if excessive differential settlements of the slab are to be prevented. Therefore, to have a slab-on-ground which will not settle unacceptably, sufficient strength and stiffness must be im- parted to the slab to resist the bending moments resulting from un- even support pressures within the limits of the allowable A/L ratio. 7.9.2 Assumed Reactions and Corresponding Support Index (C) for Slabs on Compressible Soils The assumed soil reactions on a slab supported on compres- sible soils are in accordance with the discussion in the preceding
we ~ 8 8 RE SIDE NTIA L SLABS ON GROUND paragraph, and provide for a pressure intensity along the perimeter of the slab four times as high as the pressures under the center. To determine qualitatively the support index (C) corresponding to a slab on compressible soil, consider the idealized general case illustrated In Fig. 19. This slab is loaded with a total weight, which includes (W) dead and live loads, the weight of the slab itself and all concentrated loads, and is analyzed along its long dimension (L). Any substantial concentrated load (Wc), such as bearing walls, act- ing normal to the L-direction of the slab, within the central 50% of its length (L), are assumed concentrated at the center of the slab and uniformly distributed along its least dimension (width L'). The part of the total load (W- We) which is not assumed con- centrated at the centerline of the slab, is assumed uniformly Area within which concentrated loads of total intensity We are acti ng 0.25L , O.5L i, O. 25L i, O.5L , we) , W111~1191111 IllL:LIll _ __ ~ ~ `~: tittll}~ITf' wl we = 4W ~0.2L i, 0.6L i, 0.2L ,~ W = Tota I average load on slab OF we = Wc = Tow I concentrated load acting within {F critica I area of slab al ong L direction, and averaged over tool slab area FIG. 19 Loading and Reaction Assumptions for Slab Supported on Compressible Soils
SUPPLEMENTARY INFORMATION 89 distributed over the entire slab. Both these assumed loads and reactions are shown. Summarizing, w1 = reaction intensity under the slab center W2 = 4w1 = reaction intensity along end strips of width 0.2L w = W/LL' = average total load intensity over the entire slab, including concentrated loads Wc = WC/LL' = total concentrated load (Wc) averaged over the entire slab area. The conditions and equations for these forces are derived from statics, i.e., w-wc = intensity of uniform load acting on the slab W - (w-wc) LL ' + Wc = downward forces (7.9a) W = 0.4LL'w2 + 0.6LL'w1 = 1.6LL'w1 + 0.6LL'wl=2.2LL W1. Then, w1 = W/2.2LL' = w/2.2 = upward forces. (7.9b) The maximum bending moment occurs at the slab center. Therefore, MmaX = 0.2LL (4w1) 0.4L + 0. 3LL' w1 (O. l SL) - 0. 5LL' (w-wc) 0.2 5L = 0.32w1 L2L' + 0.045w1 L2L' - 0.125(w-wc) L2L' = 0.365w1 L2L' - 0.125(w-wc) L2L . Incorporating equation 7.9b, MmaX = (0.365/2.2~(wL2L')- 0.125wL2L'+0.125wcL2L' = 0.041wL2L' + 0.125wcL2L' = (L2L'/8)(0.33W + WC). (7.9C)
90 RES0)ENTLAL SLABS ON GROUND The maximum bending moment (based on the criteria developed for expansive soils) for a slab supported by a soil for which qu/w 2 7.5 is given by equation 7.7a, p. 62, and is 1/8 (L2L ') (1-C) w. Equating this maximum bending moment to the righthand side of equation 7.9c permits an evaluation of the support index (Cr) corresponding to compressible soils, i.e., (L2L'/8~0.33w + wc) = (L2L'/8~1-Cr~w (7 .9d) where C = reduced support index for compressible soils (2.5 qu/w < 7.5~. Solving equation 7.9d for Cr. Cr = 1-0~33_ wc/w = 0.67- wc/w ~ 0.65- wc/w. Since compressibility is a relative rather than an absolute soil property, it is assumed that the support index (C) for soils with qu/w ~ 7.5 is reduced linearly to the value 0.65 - wc/w, as the ratio qu/w moves from 7.5 to the limiting value of 2.5. With this assump- tion, the reduced support index (Cr) becomes a linear function of the ratio qu/w, and is expressed mathematically by the relationship (equation 7.5a, p. 56) Cr = (2.5- qu/w)~0.13- 0.2 (wc/w) - 0.2C] +0.65- (wc/w) for C 2 0.65 - (wc/w). (7.9e) Obviously, for the case C ~ 0.65 - (wc/w) (i.e., when the value of C for noncomoressible soils is already smaller than the limiting value 0.65 - ~ , (wc/w) of the reduced support index), the value of Cr is equated to that of C, so that forC<0.65 - (wc/w). Cr = C (7.9f) It is easy to see that for slabs with no concentrated loads (Wc), equations 7. 9e may be simplified, i. e.,
SUPPLEMENTARY INFORMATION 91 Cr = [2~5 - (qu/w)] (0.13 - 0.2C)+ 0.65 (7 gg) for C ~ 0.65. 7.9.3 Design Sequence The design sequence for Type III slabs on compressible soils is identical with that for Type III slabs on soils for which qu/w 2 7.5, with the following modifications: a. The design is carried out both for the initial value of the sup- port index (C), and for the reduced value of the support index (Cr) as obtained from equations 7.9e and I. b. If the soil is not compressible (i.e ., if qu/w 2 7. 5), the appli- cable support index would be C and the amount of steel (As) result- ing from Step 9c, p. 82, would be placed at the bottom of each beam. A corresponding amount of steel (A's = As - 0.65 in.2) would be necessary at the top of each beam. C)n the contrary, if the slab is to be placed on compressible soil (i.e., 2.5 ~ qu/w ~ 7.5), the appli- cable support index has the value Cr. and the reinforcement per beam (As) computed on this basis is necessary only at the bottom of each beam; i.e., slabs on compressible soils deform with the tensile side on the bottom (pare. 7.9.1, pp. 86-87~. c. Since the transition from soils which are not classified com- pressible (qu/w ~ 7.5) to those which are compressible is gradual rather than sudden, it is not reasonable to shift at the boundary of qu/w - 7.5 from a slab with beams reinforced at top and bottom to beams reinforced only at the bottom. For this reason, if the bottom steel (As) computed on the basis Cr is greater than or equal to As + A's (sum of bottom and top steel computed on the basis of C for a slab with identical dimensions), then no top reinforcement is placed, and the bottom reinforcement is equal to the reinforce- ment (As) computed on the basis of Cr. If, on the other hand, bottom steel (As) computed on the basis Cr is less than As + A's (sum of bottom and top steel computed on the basis of C), then the slab is designed with bottom steel equal to As as computed on the basis of Cr. and with top steel equal to the difference of this steel (As) and the sum of As + A's as computed on the basis of C. This design procedure will be illustrated in pare. 7.13, pp. 115-122.
92 RESIDENTIAL SLABS ON GROUND 7 . 10 Related Design Considerations 7.10. 1 Concentrated Loads on the Slab Every effort should be made to ensure that concentrated loads such as chimneys are so located that these loads are transferred to the stiffening beams. It is particularly important that concen- trated loads in excess of 100 psf per bay area (including slab load) be so located as to rest on at least two stiffening beams. In the case of smaller concentrated loads, a structural analysis of the supporting slab is necessary to assure its ability to transfer such loads to adjacent stiffening beams. Such analysis should be conducted on the assumption that the particular portion of the sub involved will be called on to act as a framed slab supported directly on the stiffening beams, and hence that ACI provisions apply 7.10.2 Limiting Reinforcement It is expected that a Type ID slab will be subjected to a variety of support conditions; consequently, deflections will occur in the slab. To ensure that the slab will be able to accommodate these deflections and accompanying stresses, it is essential that a steel ratio no less than 0.003 nor greater than 0.02 be provided. ~ addition, the need for shear reinforcement will usually be dictated by whether the slab itself possesses sufficient inherent shear strength. Every effort should be made to eliminate the need for special shear reinforcement. Nonetheless, in all cases, No. 2 stirrups at 5 feet o.c. should be provided as a minimum, properly to position and maintain the spacing of longitudinal reinforcement. The distribution and location of reinforcement is also important, a. bottom reinforcement should be positioned 2 inches clear of the bottom of the beam b. top reinforcement should be positioned at least 1 inch clear of the top of the slab c. slab reinforcement for a 4-inch slab should be positioned one third the slab thickness clear of the top of the slab and should be uniformly distributed; for slabs of greater thickness, clearance
SUPPLEMENTARY INFORMATION 93 should be increased so that the bottom of the lower crossbars is at the slab center line d. reinforcing bars for the top slab should be spaced not closer than 4 inches o. c., nor farther than 12 inches o. c. If the selected beam dimensions are larger than those required by structural considerations incorporated in the design procedure of this report, the limiting reinforcement ratio should be considered applicable to the dimensions used for the structural design, instead of the dimensions specified for construction. However, in this case, the contribution to strength and/or stiffness of dimensions specified for construction will be ignored throughout the design. 7.10.3 Embedment of Conduits in the Slab Since movements in Type III slabs are to be expected, embed- ment In the slab of utility lines, heating pipes, ducts, and the like is highly inadvisable. Special care must be taken to prevent break- ing of such pipes by isolation from the slab whenever passage through the slab is necessary. Pipes should pass through the slab vertically and be provided with expansion joints; service lines entering the house should pass beneath the stiffening beams. 7.10.4 Other Stiffened-Slab Designs The design procedures in previous paragraphs are not to be construed as precluding use of other stiffened-slab designs, nor of precast and prestressed slab designs. Such designs should be con- sidered acceptable for Type III slabs, provided that the design con- cepts incorporated in this report are also incorporated in the alternate design. ~ the case of prestressed concrete, the design should be based on the provisions of the ACI Code. The moment of inertia of the slab should be based on the complete uncracked section rather than the cracked section as recommended earlier, and the computation of deflection on an effective modulus of elasticity (E') of creeping concrete which is equal to one third the value of the modulus of elasticity (E) for concrete provided in the ACI Code for the type of concrete to be used.
94 RESIDENTIAL SLABS ON GROUND 7.11 Example 1-Design of Type III Slabs Supported on Expansive Soils The procedures which follow demonstrate the application on expan- sive soils of the criteria recommended in 1.4, pp. 14-20. 7.11.1 Determination of Slab Type1 Slab type determinations are made in three steps corresponding to the criteria in pare. 1.1, pp. 10-12. Once it has been determined on the basis of this investigation that a Type III slab is required, the design proceeds as is illustrated in the ten steps which follow (pare. 7.11.2~. Location: San Antonio, Texas. Floor plan and outside dimensions: L 42'-0" l , ~ 18-0 ~ -o Type of construction: solid masonry with plaster-on-lath interior Total weight of superstructure = all dead and live loads = 185kips Method of heating: warm-air with ducts in attic Partitions: non-load bearing, weighing 100 plf Openings through slab: none greater than 8 inches; all having expansion joints Concentrated loads: none Concrete: 2 500 psi; VC = 7 5 ps See pare. 1.1, pp. 10-12.
SUPPLEMENTARY INFORMATION 9 5 Step 1-Summarize soil investigation results. a. Soil type: CH with PI = 41 to a depth of 10 ft. and GW with PI = 0 from 10-20 It in depth b. Consistency of CH soil: Stiff, a.u = 2800 psf c. Relative density of GW = 65%. Step 2-Determine climatic rating. Referring to Fig. 1, p. 38, Cw = 17 for San Antonio (by interpolation). Step 3-Determine appropriate slab type. Since the soil to a depth of 10 feet is CH with PI ~ 15, and qu/w is obviously > 7.5, it is determined from Table I that a Type m slab is required. 7 .11.2 Application of Type m Procedures Step 1-Determine total average load. a. Compute superstructure load per square foot of slab area [total slab area = 42~24) + 12~18) = 1224 ft2] Ws = 185,000/1224 = 151 psf. b. Compute estimated dead weight of slab Wd = 2L + 30 = 2~42) ~ 30 = 114 psf (est.) c. Compute total superstructure and slab dead load w = wd + WS = 114 ~ 151 = 265 psf. Step 2-Establish controlling soil properties. a. qu/w = 2800/265 ~ 7.5 b. In accordance with the provisions of pare. 7.8.1, p. 65, PI is equal to the PI of the top 5 feet of that soil within the top 15 feet of the soil;in this ease, PI=41. See pare. 7. 8.6, pp. 76-83.
96 RESIDENTIAL SLABS ON GROUND Step 3-Determine support index. Referring to Fig. 6, p. 53, for Cw = 17 and PI = 41, the support index (C) is found to be 0.72. Since no special circumstances pre- vent or diminish the expected variations in soil moisture, the support index (C) should not be increased to the value of Cm. Since qu/w > 7.5, the support index (C) should not be reduced to the value of Cr. Therefore, C = 0.72. Step 4-Ascertain the deflection ratio. From Table III, p. 50, the permissible ^/L= 1/360. Step 5-Determine outside slab dimensions. It is necessary to - plan for two slabs, i.e., slate one (L1 L'1) =42 by24ft slate two (L2 L'2) = 36by 18 ft. Step 6-Determine effective loads for slabs one and two. a. Coefficient `,o for the long direction of slab one, is 1.4-0.4(L1/L'l) = 1.4-0.4~42/24) = 0.7. b. Effective loads for slab one are w = (1-0.72) 265 = 74.2 psf in the short direction, and in the long direction. w= 0.7 (74.2) = 51.9 psf c. Coefficient ~¢ for the long direction of slab two is 1.4-0.4~36/18) = 0.6. d. Effective loads for slab two are w= (1-0.72) 265 = 74.2 psf in the short direction, and
SUPPLEMENTARY INFORMATION 97 w = 0.6(74.2) = 44.5 psf in the long direction. Step 7-Develop layout of stiffening beams. The long beams of slab two must coincide with the short beams of slab one. The arrangement of beams which follows is selected to satisfy this condition. L 9-0 l 9-0 ~12-0 ~12'-0" -O. -, ~ -° ' - -1-- 1 1 l 1 1 1 1 , ~ _ _ _ _ ~ _ _ _ _ 1 1 1 - 7.11.3 Design of Slab C)ne-Analytical and Chart Procedures This design will be carried out using two different procedures for purposes of illustration, i.e., through an analytical procedure, and through use of charts. The analytical procedure will be used first. Step 8-Select tentative design values for d, B. and B'. Since w is approximately equal to 50 psf, an L/d ratio of 18 is selected (Step 8, p. 17 and p. 81. On this basis, d= 42~12~/18 = 28 in. A good initial value for the width of each beam is 8 inches (see discussion, Step 7, pp. 16-17 and pp. 80-81~. On this basis, B = 3~8) = 24 in. B'=5~8~-40in.
98 RESlOENTIAL SLABS ON GROUND Step 9-Execute recommended design computations. a. Depth ratios are b. Load indices are In the long direction, and In the short direction. c. Shear criteria are L/d= 18 L'/d = 24(12)/28 = 10.3. w(L'/B) = 51.9 [24~12~/24] = 623 psf w(L/B') - 74.2 [42~12~/40] = 935 psf w(L'/B) L/d = 623~18) = 11,200 < 21,600 psf in the long direction, and w(~/B') L'/d = 935~10.3) = 9,630 < 21,600 psf in the short direction. d. Moment criteria are p = w(L'/B)(L/d)2/2 (10~7 = 623~18~2/2 (10~7 = 0.0101 in the long direction, and p = W(L/B,)(Li/d)2/2 (10~7 = 935~10.3~2/2 (10~7 = 0.00496 in the short direction. e. in the long direction. Deflection criterion is w(L'/B) ( L/d)3 207(1038(~/L) 623(18~3 (360? = 0.0632 207(10) 8
SUPPLEMENTARY INFORMATION 99 Note: From Fig. 17, p. 77, for Z = 0.0632, Pi= 0.01175. This value is compared with the value of p obtained for the moment criterion. If Pi > p, and Pi does not exceed p by more than 0.0015, PI is the controlling steel ratio. If PI > P and P1-P > 0.0015, a considerable percentage of steel is needed to impart stiffness rather than strength to the beam. This means that the dimensions of the beam must be increased in order to reduce P1-P. Returning to this example, P1-P = 0~01175 - 0.0101 = 0.00165' 0.0015. Therefore, a wider beam is needed in the long direction. Each beam will be made 10 inches wide in the long direction, and the design procedure repeated from Step 8 onward. Revised Step 8-Select basic dimensions. d = 28 in. B = 3(10) = 30 in. B'=5(8)=40in. Revised Step 9-Design computations. a. Depth ratios are b. Load indices are in the long direction, and in the short direction. L/d = 18 L'/d = 10.3 w(L'/B) = 51.9 [24(12~/30] = 498 psf w (L/B') = 74.2 [42 (12)/40] = 935 psf
100 RESIDENTIAL SLABS ON GROUND c. Shear criteria These need not be revised; in fact, the new dimensions have helped. d. Moment of criteria Since the beam depth was not increased, the total steel required to resist maximum moment is unchanged. Consequently, the steel ratio has changed in inverse proportion to the cross-sectional area of the beam. Thus, in the long direction p= 0.0101 (24/30) = 0.0081, and in the short direction, e. Deflection criteria p = 0.00496 (same as before). The value of Z will be different for the new load index (w). Thus, in the long direction, Z = (498/623) 0.0632 = 0.051. The corresponding steel ratio (Pi) from Fig. 17, p. 77, is PI = 0.0087 and P1-P= 0.0087 - 0.0081 = 0.0006 < 0.0015. Therefore, the steel ratio in the long direction is P = Pi = 0.0087. In the short direction, Z - w(L/B')(L'/d)3 207 (10~8 1/L 935 (10.3) 360 207 (10~8 = 0.0178. The corresponding steel ratio (Pi), from Fig. 17, is less than 0.003. Therefore, P > Pi, and the bending moment controls in the short direction, for which the required steel ratio (p) is 0.00492
SUPPLEMENTARY INFORMATION 101 f. Reinforcing required per beam in the long direction is as follows: Bottom steel As= 0.0087 (10) 28 = 2.44 in.2 Use four No. 7 bars per beam (area = 2.4 in.23. Top steel A's= 2.44 - 0.65= 1.79 in.2 Use three No. 7 bars per beam (area = 1.8 in.2~. Reinforcing per beam required in the short direction is as follows: Bottom steel As = 0.00492 (8) 28 = 1.1 in.2 Use two No. 7 bars per beam (area = 1.2 in.2~. Top steel As= 1.1 - 0.65 = 0.45 in.2 Use four extra No. 3 bars per beam (area = 0.44 in.2), placed as top reinforcement in excess of the No. 3 bars at 12 inches o.c. placed as reinforcement in the slab. Charts will now be used for the design. Step 8-This step remains the same as that for the analytical procedure above, i.e., d = 28 in., B = 30 in., B' = 40 in. Step 9-Select basic parameters (from revised Step 9, pp. 99- 101~. a. Depth ratios are in the long direction, and Q/d= 18
102 RESIDENTIAL SLABS ON GROUND ted = 10.3 in the short direction. b. Load indices are in the long direction, and in the short direction. w(t'/b) = 498 psf w(t'/b) - 935 psf c. Steel ratios (p), using Fig. 16, p. 74, are determined as follows: For 6/L = 1/360, ordinate w~Q'/b) = 498 psf. For this value and for the curve t/d = 18, the required steel ratio (p) is 0.00875. This steel ratio is above the M-D curve but is not above the broken-line curve nor the VC = 75-psi curve. Therefore, deflection controls the steel ratio and the shear condition is satisfied. Note that if the initial width selected for the long beams had been 8 inches, the load index [w(t'/b)] would be 623 psf (as computed previously). For this value and for t/d = 18, the required steel ratio obtained from the chart would be 1.18%, and would be located above the limiting-line curve. Therefore, this value of p would be un- acceptable and an increase in the dimensions of the beam would be necessary, exactly as in the analytical determination. From b above, the ordinate for the design along the short direc- tion is 935 psf. For this value and for the corresponding curve t/d = 10.3, the steel ratio (p) in the short direction (by interpola- tion between curves t/d = 10 and Q/d = 11) is 0.006. The steel ratio in the short direction is obviously controlled by bending moment, as the corresponding point in the chart is below the M-D curve. For obvious reasons, the shear criterion is satisfied. A comparison of values thus obtained with those obtained analyti- cally indicates a high degree of accuracy for the chart method. d. Reinforcing steel computations are identical with those result- ing from the purely analytical procedure.
SUPPLEMENTARY INFORMATION 103 7.11.4 Design of Slab Two-Chart Procedure Only Step 8-Select basic dimensions. Beam depths equal to the depth of those for slab one will be selected, because the long beams of slab two coincide with some of the short beams of slab one. In this case, the length does not vary substantially between slab one and slab two (42 vs. 36 feet). If the difference in length were substantial (e.g., 60 vs. 30 feet), the alternative of designing slab two beams shallower than those of slab one could be considered. In this case, the basic dimensions are chosen as follows: d = 28 in. B =3~8~=24in. B' = 4 (10) = 40 in. (Three of the four beams require a 10- ~nch width because they belong to slab one also. The fourth beam is made 10 inches wide for uniformity.) Step 9-Execute design computations. a. Depth ratios are b. Load indices are L/d = 12 (36~/28 = 15.5 L'/d = 12~18~/28 = 7.7. w(L'/B) = 44.4 [12 (18~/24] = 400 psf in the long direction (note that the effective load in the long direc- tion of slab two is only 44.4 psf), and w(L/B') = 74 [12~36~/40] = 800 psf in the short direction. c. Determine steel ratios (p). From Fig. 16, p. 74, for a 400-psf load index and a depth ratio t/d = 15.5, by interpolation, the ratio is
104 RES~ENTLAL SLABS ON GROUND p= 0.48%- 0.0048 in the long direction. Since the corresponding point is below the M-D line, the normal criterion controls, obviously, the shear condition is easily satisfied. In the same chart (Fig. 16) for an 800-psf load index and a depth ratio Q/d = 7.7, the required steel is less than 0.3%. Therefore, In the short direction, the minimum value of p is used, i.e., p= 0.3%= 0.003. d. Required reinforcing per beam in the long direction is as follows: Bottom steel As= 0.0048 (8) 28 = 1.08 in.2 This area of steel is greater than that required for the short beams of slab one but less than the steel specified (i.e., two No. 7 bars); therefore-, all of those beams of slab one in the short direction which coincide with those of slab two will have 1.20 in.2 of steel-use two No. 7 bars per beam. Top steel A's = 1.08 - 0.65 = 0.43 in.2 Use four extra No. 3 bars per beam (area = 0.44 in.2~. Required reinforcing per beam in the short direction is as follows: Bottom steel As = 0.003 (10) 28 = 0.84 in.2 This area of steel is less than that required in the long beams of slab one. Therefore, the only short beam to be reinforced with this amount of steel is the one end beam which does not participate in the long direction of slab one. Use two No. 7 bars in this beam (area- 1.2 in.2~.
SUPPLEMENTARY INFORMATION 105 Top steel AS - 0.84 - 0.65 = 0.19 in.2 Use two extra No. 3 bars in this beam, i.e., in addition to the No. 3 bars at 12 inches o.c. used in the slab. e. Adjust for unequal beam spacing. The beams along the short direction of slab two are equally spaced at 12 feet o.c. This, however, is not the case with the five beams along the short direction of slab one. The average spacing for these beams is 42/4=lO.Sft. 1 L 9-0" i 9-0 i 12'-0" i 12'-0" i 1 ll Beams~and~are spaced at 9 feet o.c., i.e., at less than the average, and thus need no adjustment. Beam(~)is spaced at an average (9 + 123/2 = 10.5 feet. This beam needs no adjustment because its spacing is equal to the average spacing of all beams. Beams(~)and(~)are spaced at 12 feet o.c. (which is greater than the 10.5-foot average); therefore, these require adjustment. How- ever, no adjustment is needed for beam width because the shear criterion is satisfied by a wide margin. The bottom steel of these beams was found to be
106 RESIDENTIAL SLABS ON GROUND As= 1.10 in.2 The adjusted steel will have to be As = 1.10 (12/10.5) = 1.26 in.2 Note that beams,, and~were reinforced with two No. 7 bars (area = 1.2 in.2) because of the controlling requirement of the long beams of slab two. The same reinforcement is used for beams and@)to satisfy the value of the adjusted steel. The small inadequcy in steel area is not considered significant. Top steel for this beam, however, will be 1.26- 0.65=-0.61 in.2 Therefore, one extra No. 7 bar will be used. f. Select top slab reinforcement. Since all squares of the top slab have clear spans smaller than 12 feet, the top slab is designed 4 inches thick, with No. 3 bars at 12 inches o.c. each way placed at one third the slab thickness clear from the top. Figure 20, pp. 107-108, shows a schematic layout of slab and reinforcement as designed above. Step 10-Check slab dead load. Lineal feet of 10-inch beam Lineal feet of 8-inch beam 3~42) + 18 = 144 it 2 (24) + 3~36) - [~5/6) (18) ~ = 141 it Adjusted beam depth is 28 + 3 - 4 = 27 in. (28-inch depth plus 3-inch steel coverage minus 4-inch slab thickness). Volume of concrete in the beams is
SUPPLEMENTARY INFORMATION 107 144 (5/6) 2.25 + 141 (2/3) 2.25 = 481.5 ft.3 Slab area is 42(24) + 18(12) = 1224 ft.2 Slab thickness equivalent to beam volume is 481.5(12)/1224= 4.72 in. Equivalent uniform slab thickness is 4 + 4.72 = 8.72 in. Average slab load is w = 8.72(150)/12 = 109 psf. This load is less than the 114 psf estimated in Step 3. Estimated total average load is w = 265 psf. Actual total average load is w=265 - 114+ 109=260 psf. Ratio of estimated load to actual load is 260/265 = 0.98 > 0.95. Therefore, no adjustment of the design is required. 7.11.5 Discussion of Design For a load (w) equal to 74 psf, it is recommended in this report that the depth ratio (Q/d) be selected between the values of 14 and 18. If this recommendation had been adopted for the design example, then the beam depth would have been greater than the 28 inches chosen. If, for example, a depth of 30 inches were chosen, it would not be necessary to widen the long beams of slab one to 10 inches;
108 RESIDENTIAL SLABS ON GROUND ,, 9'-0" ~, 9'-0" ~1 2'-0" ~ 12 '-0" ~l 1 r ~n r- -~r~ ~r-- r ! ! . 31 " ~ r ~ r - - -~- _ _ _ ~ I ' I I . . _8" l~r8" 3 L''J3 _wr81' 5~ a' 11 _ I J L ~J L _ _ ~ _ _ _J L__ _ i~L _ _ J ~r ~r~~ - ' ---~r~~~~~~~~~ 1 1 1 2 L' 'd 2 1 1 ~ _ - t _ _ ~ ~ _ _~ _ ~ ~ r~-r--~r- -~- -q I I 1 1/3 slab2 _J1 ~l 3 t_' 23 thickne i ~ ~ ~ __~_ ~ ~o a. Slab Layout l , 11 l l J'B.. 4~1_14 J 11 . ~1~ _ _ _ ~ _ _ J 1~_ _-] _ _ ' .J .. 1 Note: Slab 4 in. thick with 15 #3atl2in. o.c. both ways (in addition to top beam _8" reinforce- ment shown bel ow) . / ,#3at12in. o.c.,: b. Slab Section 4 #3 bars 4 #3 bars j:314"~= ::I-~" ~ ,~- 8", ~ 8" ~,~, 8 Section 1-1 Section 2-2 & 3'-3'Sectior' 3-3 & 5-5 c de Note: All stirrups shown are placed for the purpose of positioning reinforcing steel, and are a minimum No. 3 at 5'-O"o. c . _ 2#7bars FIG. 20 Slab Layout and Reinforcement-Para. 7.11 Design Example
SUPPLEMENTARY INFORMATION 109 _ 3 t7 bars 31 " 1~ 3#7bars 3 Debars , .~ , ~, r ~ ~ t~ l4" 5 ~ Art ... 1'. 1 ''I L] 10" 4 #7 bars - 2~" hi Section 7-7 Section 7'-7' & UQ-8 Section 9-9 f 9 ~ 2 #3 bars /1 #7 bar 318'i 8" ~ Section 6-6Section 4-4 ii Note: All stirrups shown are placed for the purpose of positioning reinforcing steel, and are a minimum No. 3 at 5'-O"o.c. Q'. U or ~ h 31 " 1 t ~ 10~ 1~ 31 " FIG. 20 (Cont.) Slab Layout and Reinforcement-Para. 7.11 Design Example
110 RESIDENTIAL SLABS ON GROUND also, the steel would be less. On the other hand, a depth of 30 inches would cause very low depth ratios for slab two, which is smaller in dimension. Thus, the design for slab one would be more economical; however, more concrete than necessary would be used in slab two. In any event, the overall difference in materials and cost between the two designs would not be large. It is up to the designer, therefore, to exercise judgment on the basis of the pre- vailing special conditions associated with each slab under design, when deciding on the choice of basic parameters. ~ a report such as this, only guidelines can be provided for achievement of efficiency in design; it is both impossible and undesirable to set forth rigid and absolute rules which will be valid under all possible conditions. 7.12 Example 2-Design of Type III Slabs Supported on Expansive Soils: Shallow Beams 7.12.1 Determination of Slab Type For purposes of this example, it will be assumed that the slab is the same as that designed in pare. 7.11, subjected to the same loads, and founded on similar soil, but in a different geographic location, i.e., one having a climatic rating (Cw) = 39. 7.12.2 Application of Type III Procedure The design through Step 2 of the Type III Procedure of the pre ceding example remains the same and provides the following values. w = 265 psf PI = 41. Step 3-Determine support index. Referring to Fig. 6, p. 35, for Cw = 39 and PI = 41, the support index (C) = 0.94 ~ 0.90. Step 4-Ascertain the deflection ratio. From Table III, p. 50, the permissible A/L = 1/360.
SUPPLEMENTARY INFORMATION 111 Step 5-Determine outside slab dimensions Slab one= (L1L'l) =42by24ft Slab two = (L2 L'2) = 36 by 18 ft. Step 6-Determine effective loads for slabs one and two. a. Coefficient up for the long direction of slab one is 1.4 - 0.4 (42/24) = 0.7. b. Effective loads for slab one are w = 0.1w = 0.1 (265) = 26.5 psf in the short direction, and in the long direction. w = 0.1w~p= 26.5 (0.7) = 18.6 psf c. Coefficient up for the long direction of slab two is 1.4 - 0.4 (36/18) = 0.6. d. Effective loads for slab two are w = 0.1w = 0.1 (265) = 26.5 psf in the short direction, and in the long direction. w = 0.1w~p= 26.5 (0.6) = 15.9 psf Step 7a-Design for d, bs, and As for both slabs using Steps 6- 10, pp. 16-19. Step 7-Develop layout of stiffening beams. Use same layout as shown for the example of pare. 7.11.
112 RESIDENTIAL SLABS ON GROUND 7.12.3 Design of Slab One Step 8-Select tentative design values for d, B. and B' Try d = 23.5 in. B = 3~8) = 24 in. B'=5~83=40~n. Step 9-Execute recommended design computations. a. Depth ratios are b. Load indices are in the long direction, and in the short direction. Lid = 42~12),/23.5 = 21.4 L '/d = 24 (12)/23.5 = 12.2 . w (L'/B) = 18.6 [24~12)/24] = 223 psf w (L/B') = 26.5 [42~12)/40] = 334 psf c. Steel ratios (p) for ~/L = 1/360, using Fig. 16, p. 74, are 0.006 in the long direction, and in the short direction. 0.003 (min.) d. Reinforcing required per beam is as follows: Bottom steel in the long direction As= 0.006 (23.5) 8 = 1.13 in.2
SUPPLEMENTARY INFORMATION 113 Bottom steel in the short direction As= 0.003 (23.5) 8 = 0.56 in.2 Step 8a-Determine d, bs, As, and A's per beam. bs = 8 in. for all beams d =2+ 10 (1-0.94)(d-2)=2+ 10(0.06)21.5= thin. As (long dimension) = 101-C) As = 10~1-0.94) 1.13 = 0.68 in.2 As (long dimension) = 10(1 -C) As - 0.65 = 0.68 -0.65 = 0.03 in.2, or effectively zero. As (short dimension) = 10~1-C) As = 10~1-0.94) 0.56 = 0.34 in.2 < 0.65 in.2 Step 9a-Reinforcement used in the long direction is as follows: Bottom steel, 2 No. 6 bars per beam (area = 0.88 in.2) Top steel, No. 3 bars in the slab at 12 in. o.c. Reinforcement used in the short direction is as follows: Bottom steel, 1 No. 6 bar per beam (area = 0.44 in.2) Top steel, No. 2 bars in the slab at 9 in. o.c. (area = 0.067 in.2/ft) for a requirement of 0.18 (0.34) = 0.061 in.2/ft. 7.12.4 Design of Slab Two Step 8-Select tentative design values for d, B. and B' Try d = 23.5 in. B = 3~8) = 24 in. B'=4~8~=32in.
1 14 RE SIOENTIA L SLABS ON GROUND Step 9-Execute recommended design computations. a. Depth ratios are b. Load indices are in the long direction, and in the short direction. L/d = 36(12)/23. 5 = 18.4 L'/d = 18(12)/23. 5 = 9.2 w(L'/B) = 15.9 [18(12)/24] = 143 psf w(L/B) = 26.5 [36(12)/32] = 358 psf c. Steel ratios (p) for ~/L= 1/360, using Fig. 16, are 0.003 (min) in the long direction, and in the short direction. 0.003 (min) d. Reinforcing required per beam is as follows: Bottom steel in both the long and short directions As = 0.003 (23.5) 8 = 0.56 in.2 Step 8a-Determine d, Us, As, and A's per beam. bs = 8 in. for all beams d = 2 + 10~1-0.94~(d-2) = 2 + 10 (0.06) 21.5 = 15 in. Bottom steel is the same in both directions; therefore As= 10 (1-0.94) 0.56= 0.34 in.2 < 0.65 in.2 Since As ~ 0.65 in.2, A's = 0
SUPP LE ME NTARY INFORMATION 1 15 Step 9a-Reinforcement used in both directions is as follows: Bottom steel, 1 No. 6 bar per beam (area = 0.44 in.2) Top steel, No. 2 bars in the slab at 9 in. o.c. (area = 0.067 in.2/ft) for a requirement of 0.18 (0. 34) = 0.061 in.2 /ft. Note: Slab reinforcement (No. 2 bars at 9 in. o.c.) exceeds the WAIF reinforcement specified for the corresponding Type II slab. Steps lea and lla-Not applicable. The full slab layout is shown in Fig. 21, p. 116. No steel adjust- ment is made for unequal beam spacing, because the steel provided in excess of the minimal steel required is ample compensation. Since beams are shallow, the use of stirrups (No. 3 at 5 ft-O in.) is optional-bottom steel can be easily placed and secured by other means. However, if stirrups are not used, chairs or other means should be provided to assure that bottom steel will be held clear a minimum of 2 inches from the soil as recommended herein. It should be noted too, that for smaller slabs or for slabs on less active soils or in less unfavorable climates, the depth of beams would be even less, approaching a flat slab or Type II slab. 7.13 Example 1-Design of Type III Slabs Supported on Compressible Soils The procedures which follow demonstrate the application on com- pressible soil of the criteria recommended in pare. 1.4, Step 9c, 1.14, and amplified in pare. 7.9-7.9.3, pp. 85-91. 7. 13.1 Given Conditions Location: Alexandria, Louisiana Floor plan and outside dimensions. Bear~ngWa11; total load at base = 15^ - 1 i . 18'- 0" ~24'- 0" l L 42'- 0" ,
116 RESIDENTIAL SLABS ON GROUND ~ 9~-08' ~ 96-0~' ~128-0~. ~12'-0" ~' N -2 1 , C~ , ' I .. J _ J-8" 1kg" 3' L~.'J3' I I I I I ~ L ~J L~ JL_ _~. r ~ r~ ,~ ~ 11 a' 2L!'~t2 ~J~8 l ~ _ ~_ _ ~ ~ _~ _ _ ~ ~ _ _ r-~r~~ =~ ' ~~] 1 L ,_~1 , 1 3 L . I ~o I I = I L_~__] L =~__u ~-t 1 1 I ~ Oe, 1: -i''-0 J ~ I 1 1 _ 5 t 11 -1 J5 l 8$' J l 1 _ J I ~#3at 12in. o.c. inE-Wdirection thickness2I :_~:y,~ #2 at 9 in. o.c. in N-S direction a. Slab Layout b. Slab Section i/r :] i6~ ~ 1 \7i " 8" 8" 8" ~ ~. ~Y .H ~ Section 1-1 & 6-6 Section 2-2, 3'-3' & 4-4 Section 3-3 & 5-5 c d 176 " a a .~/ D i 2 #6 bars ·- II-2t Q.' ~,, Section 7-7 f ~2116~ ~ '~ 8" ~ ~ 8" `, Section 7'-7' 8' 8-8 Section 9-9 9 h FIG. 21 Slab Layout and Reinforcement-Para. 7.12 Design Example 1 7ji"
SUPPLEMENTARY INFORMATION 117 Type of construction: wood frame; masonry veneer and plaster- board interior Total weight of superstructure = all dead and live loads, includ- ing concentrated loads = 140 kips Openings through slab: none greater than 8 inches; all having expansion joints Concentrated loads: one bearing wall, with a total dead and live load of 15 Rips, located as shown above. Step 1-Summarize soil investigation results. a. Soil type: CH with PI = 35 to a depth of 8 It and OH with PI = 44 from 8-20 It in depth b. Consistency of CH soil: qu = 1200 psf. Step 2-Determine climatic rating. Referring to Fig. 1, p. 38, Cw = 35 for Alexandria, Louisiana. Step 3-Determine appropriate slab type. Since the soil is CH and OH, PI > 15, and Cw = 45, a Type Ill slab is required unless qu/w < 2.5, In which case a Type IV slab would be needed (Table I, p. 11~. 7.13.2 Application of Type III Procedure Step 1-Determine total average load. a. Compute psf-superstructure load. ws= 140,000/24~42) = 139 psf b. Compute estimated dead weight of slab. wd= 2L+ 30 =2~42) + 30 = 114 psf c. Compute total superstructure and slab dead load. w = wd + WS = 114 + 139 = 253 psf Resee pare. 7.9.3, p. 91.
1 18 RE SIDENTIA L SLABS ON GROUND Step 2-Establish controlling soil properties. a. The minimum qu in the top 15 feet of the soil immediately below the bottom of the slab stiffening beams is the qu for the CH soil stratum, i.e ., qu = 1200 psf. Therefore qu/w= 1200/253 = 4.75 and 2.5 ~ qu/w ~ 7.5. b. ~ accordance with the provisions of 7.8.1a, p. 66, PI of the soil is determined as follows: The top 3 It are devoted to the depth of stiffening beams From 3 to 8 ft. PI = 35 (total depth = 5 It and weight factor = 3) From 8 to 13 ft. PI = 44 (total depth = 5 It and weight factor = 2) From 13 to 18 ft. PI = 44 (total depth = 5 It and weight factor = 1) From which PI = 1/30 [3~5) 35 + 2~5) 44 + 1~5) 44] = - 5/30 (105 + 88 + 44) = 1/6 (237) = 39.5 Step 3-Determine support index. From Fig. 6, p. 53, for PI = 39.5 and Cw = 35' C = 0.91. No special circumstances prevent or diminish the expected varia- tions in soil moisture; therefore Cm = C = 0.9. Since 2.5 ~ qu/w < 7. 5, the support index (C) must be reduced and equated to Cr. in accordance with 7.5, p. 56 and 7.9, p. 85, and, since C > 0.65, Cr is determined from the equation qu/w= 4.75.
SUPPLEMENTARY INFORMATION 119 Total superstructure load (W) is w (24) 42 = 0.253 (24) 42 = 255 Rips Wc = 15 Rips and w W c c - w W Therefore, in the long direction = 2~55= 0.059. Cr = (2.5 - 4.75~0.13 - 0.2 (0.059) - 0.2 (0.91~] + (0.65 - 0.059) = -2.25 (-0.064) + 0.591 = 0.735. Because the concentrated load is uniformly distributed along the short direction, Wc = 0 for the short direction, and Cr = (2.5 - 4.75)[0.13 - 0.2 (0.91)] + 0.65 = O.767. Step 4-E stablish deflection ratio . From Table III, p. 50, allowable l`/L = 1/300. Step 5-Determine outside slab dimensions. L =42ft L' = 24 It Step 6-Determine effective loads on the slab. Then ~ = 1.4 - 0.4 (L/L') = 1.4 - 0.4 (42/24) = 0.7. w= (1-Cr) w = (1.0 - 0.7673~255) = 59.4 or 59 psf
120 RES~ENTL9^L SLABS ON GROUND in the short direction, and w = (1-Cr~w~p= (1.0 - 0.735~255) 0.7 = 47.3 psf in the long direction. The initial value of the support index is C = 0.91, and the effective load in the short direction is w = 255 (1.0 - 0.91) = 23 psf, and the effective load in the long direction is w = 23ro psf = 23 (0.7) = 16.1 or 16 psf. Step 7-Layout of the slab Three stiffening beams will be placed along the 42-foot dimen- sion at 12 feet o.c., and five stiffening beams along the short dimension at approximately 10 feet o.c. 1 1 1~1' ~L l l 10' - 5" 10' - 5" 10' - 5'il0' - 5"l Step 8-Select basic beam dimensions. d = 28 in. B = 3 (8) = 24 in. B'= 5 (8) = 40 in. Step 9-Select basic parameters. a. Depth ratios are - ~ ' ' cow .
SUPPLEMENTARY INFORMATION 121 L/d = 42 (12)/28 = 18 L'/d = 24(12)/28 = 10.3. b. Load indices are in the long direction, and in the short direction. w(L'/B) = 47.3 t24(12~/24] = 568 psf w(L/B') = 59 [42(12~/40] = 743 psf For the initial value of the support index (C = 0.91), the load indices are in the long direction, and in the short direction. w(L'/B) = 16 t24(12~/24~= 192 psf w(L/B') = 23 [42(12~/40] = 290 psf c. Determine steel ratios (p). Referring to Fig. 15 for A/L - 1/300, ordinate w(t'/b) is 568 (p= 0.95%) for Q/d= 18 743 (p= 0.39%) for t/d= 10.3 192 ~=0.31%)forQ/d= 18 290 (Pmin= 0.3%) for Q/d= 10~3. d. Reinforcing steel required per beam in the long direction is As= 0.009 (28) 8 = 2.13 in.2 (bottom) and in the short direction is As= 0.0039 (28) 8 = 0.87 in.2 (bottom).
122 RESIl)ENTIAL SLABS ON GROUND For the initial value of the support index (C = 0.91), the required steel in the long direction is As= 0.0031 (28) 8 = 0.69 in.2 (bottom) As = 0.69 - 0.65 = 0.0 in.2 (top) and in the short direction is As = 0.003 (28) 8= 0.67 in.2 (bottom) A's = 0.67 in.2 _ 0.65 in.2 = 0.02 in.2 (top). Compare require 2ents in the long direction. Since the 2.13 in. bottom reinforcement exceeds the sum of bottom plus additional reinforcement obtained for the initial value, C = 0.91 (i.e., since 2.13 ~ (0.69 + 0.04) in.2, no additional top reinforcement is required). Compare requirements in the short direction. 0.87 ~ (0.67 + 0.02) in.2 Therefore, no additional top reinforcement is needed in the short direction either. 7.14 Example 2-Design of Type III Slabs Supported on Compressible Soils Assuming that the slab of the preceding example (pare. 7.12) was to be applied in Dallas, Texas, instead of Alexandria, Louisiana, the design would have been affected as follows: Cw for Dallas (Fig. 1, p. 38) would have been 20 From Fig. 6, p. 53, for PI= 39.5 and Cw = 20, the value of C would have been 0.775. Continuing with step 3 of the previous example and referring to equation Age, p. 90, the value of Cr in the long direction is
SUPP LE ME NTARY INFORMATION 12 3 \ = (2.5 - qu/w)[O. 13 - 0.2 (wc/w) - 0.2C ] + (0.65 - wc/w) (2.5 - 4.75)[0.13 - 0.2(0.059) - 0.2(0.775)] + 0.65 - 0.059 = 0.674. In the short direction, WC = 0 (because the concentrated load W is uniformly distributed along the short direction), and Cr = (2.5 - 4.75)[0.13 - 0.2 (0.775)] + 0.65 = 0.706. Steps 4 and 5 remain unchanged from the preceding example. Step 6-In determining effective loads on the slab, lo = 0.7 as . before; however, the effective loads for the reduced value of C are w = (1 .0 - 0.706) (2 53) = 74.4 psf in the short direction, and in the long direction. w = (1.0 - 0.674)(0.7)(253) = 57.8 psf Effective loads for the initial value C = 0.775 are w = 2 53(1.0 - 0.77 5) = 57 psf in the short direction, and in the long direction. w = 57 lo = 57 (0.7) = 40 psf Steps 7 and 8 remain unchanged from the preceding example. Step 9-Select basic parameters. a. Depth ratios are L/d = 18 L'/d = 10.3.
124 RESIDENTIAL SLABS ON GROUND b. Load indices are w(L'/B) = 57.8 [24(12)/24~ = 694 psf for the reduced value of C in the long direction, and w(L/B') = 74.4 [42~12~/40] = 937 psf in the short direction. For the initial value C = 0.775, load indices are w(L'/B) = 40 t24~12~/24] = 480 psf in the long direction, and in the short direction. w(L/B ') = 57 [42 (12~/40 ~ = 718 psf c. Steel ratios (p), Fig. 15, p. 73, are for the reduced value Cr; therefore wtQ,/b) = 694 (p= 1.12%) for t/d= 18 937 (p= 0.49%) for I/d= 10.3. = For the initial value C = 0.755 w(L'/b) = 480 (p = 0.78%) for L/d = 18 = 71 (p = 0.39%) for L/d = 10.3. d. Reinforcing steel required per beam for the reduced value of Cr is in the long direction, and in the short direction. A s = 0.011 (28) 8 = 2.47 In .2 (botto m) As = 0.0049 (28) 8 = 1.10 in.2 (bottom)
SUPPLEMENTARY INFORMATION 125 Reinforcing steel required per beam for the initial value C = 0.775 is As = 0.0078 (28) 8 = 1.75 in.2 (bottom) A' = 1.75- 0.65= 1.0 in.2 (top) in the long direction, and AC! = 0.0039 (28) 9 = 0.985 in.2 (bottom) ~7 A's = 0.985 - 0.65= 0.335 in.2 (top) in the short direction. Compare requirements in the long direction. 2.47 < (1.75 + 1.0) in.2 Therefore, additional top reinforcement is needed, i.e., A's = (1.75 + 1.0) - 2.47 = 0.28 in.2 (top). Compare requirements in the short direction. 1.10 ~ (0.985 + 0.335) in.2 Therefore, additional top reinforcement is needed, i.e., A's = (0.985 + 0.335) - 1.10 = 0.22 in.2 (top). Summarizing, in the long direction, and in the short direction. As = 2.47 in.2 (bottom) As = 0.28 in.2 (top) As = 1.10 in.2 (bottom) A's = 0.22 in.2 (top)
