Design of One-Way Slab

DESIGN OF REINFORCED CONCRETE STRUCTURE
ONE WAY SLAB
Presented by -
MD. Mohotasimur Rahman (Anik)
24th batch, AUST

INTROCDUCTION
Slab in which the deflected surface is predominantly cylindrical termed as one-way slabs spanning in the direction of curvature. Curvatures, and consequently bending moments, of this slab shall be assumed same for all strips spanning in the shorted direction or in the direction of predominant curvature, the slab being designed to resist flexural stress in that direction only.1
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If a slab is supported on two opposite sides only, it will bent or deflect in a direction perpendicular to the supported edge. The structure is one way, and the loads are carried by the slab in the deflected short direction (fig-2a).2 If the slab is supported on four sides and the ration of the long side to the short side is equal to or greater than 2, most of the load (about 95% or more) is carried in the short direction, and one- way action is considered for all practical purposes (fig-2b).2
Figure 1. Deflection of One-Way Slab

DESIGN OF ONE-WAY SOLID SLABS
One-way slab may be treated as a beam. A unit strip of slab, usually 1 ft. (or 1 m) at right angles to the supporting girders, is considered a rectangular beam.3
One-way slabs shall be designed to have adequate stiffness to limit deflections or any deformations that affect strength or serviceability of a structure adversely.4
Minimum thickness stipulated in the table 1, shall apply for one-way slabs not supporting or attached to partitions or other construction likely to be damaged by large deflection, unless computation of deflection indicates that a lesser thickness can be used without adverse effects.4
The total slab thickness (푕) is usually rounded to the next higher 14 inch (5mm). For slab up to thickness 6 inch. Thickness and next higher 0.5 inch. (or 10 mm) for thicker slabs.5
Concrete cover in slabs shall not be less than 34 inch.6 (20 mm) at surface surfaces not exposed to weather or ground. In this case, 푑=푕 −34 −(푕푎푙푓 푏푎푟 푑푖푎푚푒푡푒푟).Where, 푑 is defined as distance from extreme compression fiber to centriod of tension reinforcement.7
Factored moment and shears in one way slab can be found either by elastic analysis or through the use of the same coefficients stated ACI 8.3.3. printed in table 2.8
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Table 1: Minimum Thickness of Non-Prestressed One-Way Slab (Normal weight concrete and Grade 60 (Grade 420) Reinforcement)ퟗ
Member
Minimum thickness, h
Simply supported
One end continuous
Both end continuous
cantilever
Solid one-way slab
푙 20
푙 24
푙 28
푙 10
Ribbed one-way slabs
푙 16
푙 18.5
푙 21
푙 8
Note:
(1)For 푓푦 other than 60,000 푝푠푖 () multiplied tabulated value by 0.4+(푓푦100,000) [for 푓푦 other than 420 푁푚푚2 multiplied tabulated value by 0.4+(푓푦700) ]
(2)For structural light weight concrete, multiply tabulated values by (1.65−0.005푤푐) but not less than 1.09. where 푤푐 is range in 90 푡표 115 푙푏푓푡3 .[For structural light weight concrete, multiply tabulated values by (1.65−0.003푤푐) but not less than 1.09. where 푤푐 is range in 1440 푡표 1840 푘푔푚3 .]
5

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Table:2 Approximate Moments and Shears in Continuous Beams.ퟏퟎ
Positive moment
End span
Discontinuous end unrestrained
푤푢푙푛 211
Discontinuous end integral with support
푤푢푙푛 214
Interior span
푤푢푙푛 216
Negative moments at exterior face of first interior support
Two spans
푤푢푙푛 29
More than two spans
푤푢푙푛 210
Negative moment at other faces of interior supports
푤푢푙푛 211
Negative moment at face of all supports for (1) Slabs with spans not exceeding 10 ft. and (2) beams where ratio of sum of column stiffness to beam stiffness exceeds eight at each end of the span.
푤푢푙푛 212
Continued

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Negative moment at interior face of exterior support for members built integrally with supports
Where support is spandrel beam
푤푢푙푛 224
Where support is a column
푤푢푙푛 216
Shear in end members at face of first interior support
1.15 푤푢푙푛2
Shear at face of all other supports
푤푢푙푛2
Where , 푤푢=푢푛푖푓표푟푚푙푦 푑푖푠푡푟푖푏푢푡푒푑 푙표푎푑 ; 푙푛=푠푝푎푛 푙푒푛푔푡푕
•If the slab rests freely on its supports the span length may be taken equal to the clear span plus depth of the slab but did not exceed the distance between centers of supports.11
•In analysis of frames or continuous construction for determination of moments, span length shall be taken as the distance center-to-center of supports.11
•It shall be permitted to analyze solid or ribbed slabs built integrally with supports, with clear spans not more than 10ft, as continuous slabs on knife edge supports with spans equal to the clear spans of the slab and width of beams otherwise neglected.11

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Figure 4: Summary of ACI Moment Coefficient: (a) beams with more than two span (b) beams with two spans only (c) slabs with spans not exceeding 10 ft. (d) beams in which the sum of column stiffness exceeds 8 times the sum of beam stiffness at each end of span.ퟏퟐ

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The conditions under which the moment coefficients for continuous beam and slabs given in table 2 should be used can be summarized as follows:10
•Spans are approximately equal, with the larger of two adjacent spans not greater than the shorter by more than 20 percent.
•There are two or more spans.
•Loads are uniformly distributed.
•Unit live load does not exceed three times unit dead load.
•Members are prismatic.
The maximum reinforcement ratio, 휌푚푎푥=0.85훽1 푓′ 푐 푓푦 휖푢 휖푢:휖푡 ; where, ∈푡= 0.004 a minimum set tensile strain at the nominal member subjected to axial loads less than 0.10푓′ 푐퐴푔 where 퐴푔 is the gross area of the cross section, provides the maximum reinforcement ratio.13 and 휖푢=Maximum usable strain at extreme concrete compression fiber shall be assumed equal to 0.003.14
The minimum required effective depth 푑푟푒푞= 푀푢 ∅휌푚푎푥푓푦푏(1;0.59휌푚푎푥 푓푦 푓′ 푐 ) ;15
Check 푑>푑푟푒푞; (okay) .15
Figure 5. distance from extreme compression fiber to centriod of tension reinforcement

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Reinforcement, 퐴푠= 푀푢 ∅푓푦푑;푎2 ; where, 푎= 퐴푠푓푦 0.85푓′ 푐 푏 ; at first assumed, 푎=1 to calculate 퐴푠; that value can be substituted in equation of 푎 to get a better estimate of 푎 and hence a new 푑−푎2 can be determined.16
For structural slabs of uniform thickness the minimum area of tensile reinforcement, 퐴푠푚푖푛 in the direction of the span shall be the same as temperature and shrinkage reinforcement area.17 In no case is the reinforcement ration to be less than 0.0014 .18
Table 3: Minimum Ratios of Temperature and Shrinkage Reinforcement in Slab based on Gross Area.ퟏퟖ
Slabs where Grade 40 (275) or 50 (350) deformed bars are used
0.0020
Slabs where Grade 60 (420) deformed bars or welded wire fabric (plain or deformed) are used
0.0018
Slabs where reinforcement with yield stress exceeding 60,000 psi (420 MPa) measured at a yield strain of 0.35 percent is used
0.0018 ×60,000 푓푦
In slabs, primary flexural reinforcement shall be spaced not farther apart than three times slab thickness, nor 18 inch (450 mm).19 and Shrinkage and temperature reinforcement shall be spaced not farther apart than five times the slab thickness, nor 18 in.18

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Check shear requirements. Determine 푉푢 at the distance from 푑 and calculate ∅푉푐= 2∅푓′ 푐푏푑. If 12∅푉푐>푉푢 the shear is adequate.20 Note that the provision of minimum area of shear reinforcement where 푉푢 exceeds 12∅푉푐 does not apply to slabs .21 If 푉푢> 12∅푉푐 , it is a common practice to increase the depth of slab .22 so 푑 can be determine, assuming 푉푢=∅푉푐=2∅푓′ 푐푏푑.23
Straight-bar systems may be used in both tops and bottom of continuous slabs. An alternative bar system of straight and bent (trussed) bars placed alternately may also be used.7
The choice of bar diameter and detailing depends mainly on the steel areas, spacing requirements, and development length.24
Figure 6. Reinforcement Details in Continuous One-Way Slab: (A) Straight Bars And (B) Bent Bars.ퟐퟒ

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QUESTION: The cross-section of a continuous one-way slab in a building is shown in figure 7. The slab are supported by beams that span 12 ft. between simple supports. The dead load on the slab due to self-weight plus 77psf; the live load psf. Design the continuous slab and draw a detailed section. Given 푓′ 푐=3 푘푠푖 and 푓푦=40 푘푠푖
Figure 7. Continuous One-Way Slab
SOLUTION:
Minimum depth,
푕푚푖푛=max 퐿 30= 15×1230=6", 퐿 10= 5×1210=6"=6"
Dead load = 612×0.15+0.077 푘푠푓=0.152 푘푠푓
Live load = 0.13 푘푠푓
Load, 푤=1.4 퐷퐿+1.7 퐿퐿=0.434 푘푠푓
Moment −푀푢=max− 푤푙212,− 푤푙212,(− 푤푙22)=max−8.13 ,−8.13,−5.42=−8.13 푘′/푓푡 Moment +푀푢=max+ 푤푙214=max +6.975=+6.975 푘′/푓푡
EXAMPLE OF ONE-WAY SOLID SLABS

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SFD
BMD

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To maintain ∅=0.9, reinforcement ratio corresponding to 휀푡=0.004 will be selected. 휌푚푎푥=0.85훽1 푓′ 푐 푓푦 휖푢 휖푢+휖푡 =0.85 ×0.85×360×0.0030.003+0.004=0.0155 푑푟푒푞= 푀푢 ∅휌푚푎푥푓푦푏(1;0.59휌푚푎푥 푓푦 푓′ 푐 ) = 8.13 ×120.9×0.0155×60×12(1;0.59×0.0155×603) =3.45"< 푑푝푟표푣푖푑푒푑=6−1"=5" (okay) −퐴푠= 푀푢 ∅푓푦푑;푎2 = 8.13 ×120.9×60×5;0.752 =0.39푖푛2푓푡 (푎푠푠푢푚푖푛푔 푎=1) ; 푎= 퐴푠푓푦 0.85푓′ 푐 푏 = 0.39×600.85×3×12=0.76푖푛 +퐴푠= 푀푢 ∅푓푦푑;푎2 = 6.975 ×120.9×60×5;0.642 =0.33푖푛2푓푡 ; 푎= 퐴푠푓푦 0.85푓′ 푐 푏 = 0.38×600.85×3×12=0.65 푖푛 Minimum Reinforcement 퐴푠푚푖푛=0.0018 ×12×5=0.11푖푛2푓푡 ; 퐴푠= −0.39 푖푛2푓푡 (controls) +0.33 푖푛2푓푡 (controls) ) Check 휌= 퐴푠 푏푑 = 0.3912×5=0.0065>휌푚푎푥; so ∅=0.9 (okay); At 푑 distance 푉푢= 푤푙 2−푤푑= 0.434×152−0.434×5=1.085 푘푖푝 12∅푉푐= 12×2∅푓′ 푐푏푑=0.85×3000×12×51000=2.79 푘푖푝>푉푢 ( so the provided depth is okay). Use ∅12 푚푚 @ 5"푐/푐 at two supports, ∅12 푚푚 @ 6"푐/푐 at mid span and ∅10 푚푚 @ 12"푐/푐 for temperature and shrinkage reinforcement.

1.BNBC-1993; pg. 6-146.
2.STRUCTURAL CONCRETE: THEORY & DESIGN 5th ed.; M. NADIM HASSOUN; pg. 300
4.BNBC-1993; pg. 6-155.
5.DESIGN OF CONCRETE STRENGTHS 13TH EDITION; ARTHUR H. NILSON; pg. 416
6.ACI 318-95; SECTION 7.7.1;pg. 318R318-68
9.NOTE ON ACI 318-08 BUILDING CODE REQUIREMENTS FOR STRUCTURAL CONCRETE 10th ed.; pg. 10-2
10.BNBC 1993; pg. 6-147
11.ACI 318-95; SECTION 8.7.4; pg.318/318R-82~83
14.ACI 318-11; SECTION 10.3.5.; pg. 138
16.BNBC-1993; pg. 6-148.
17.BNBC-2011; pg. 6-26.
REFERENCE
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