Lecture 07 Design of Reinforced Concrete Slabs By: Prof Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk 1 Topics Addressed Introduction Analysis and Design of slabs Strip Method of Analysis for One-way Slabs 2 1
Introduction In reinforced concrete construction, slabs are used to provide flat, useful surfaces. A reinforced concrete slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so. It may be supported by reinforced concrete beams (and is usually cast monolithically with such beams), by masonry or reinforced concrete walls, by structural steel members, directly by columns, or continuously by the ground. 3 Introduction Beam Supported Slabs Slabs may be supported on two opposite sides only, as shown in Figure, in which case the structural action of the slab is essentially one-way, the loads being carried by the slab in the direction perpendicular to the supporting beams. 4 2
Introduction Beam Supported Slabs Slabs may be supported by beams on all four sides, as shown in figure, so that two-way slab action is obtained. 5 Introduction Flat Plate Concrete slabs in some cases may be carried directly by columns. Punching shear is a typical problem in flat plates. 6 3
Introduction Flat Slab Flat slab construction is also beamless but incorporates a thickened slab region in the vicinity of the column and often employs column capital. Drop Panel: Thick part of slab in the vicinity of columns. Column Capital: Column head of increased size. Punching shear can be reduced by introducing drop panel and column capital. 7 Introduction One-way Joist Joist construction consists of a monolithic combination of regularly spaced ribs and a top slab arranged to span in one direction or two orthogonal directions. Rib Peshawar University Auditorium 8 4
Introduction Two-way Joist 9 Analysis and Design of Slabs Analysis Unlike beams and columns, slabs are two dimensional members. Therefore their analysis except one-way slab systems is relatively difficult. Design Once the analysis is done, the design is carried out in the usual manner. So no problem in design, problem is only in analysis of slabs. 10 5
Analysis and Design of Slabs Analysis Methods Analysis using computer software (FEA) SAFE, SAP 2000, ETABS etc. ACI Approximate Method of Analysis Strip Method for one-way slabs Moment Coefficient Method for two way slabs Direct Design Method for two way slabs 11 Analysis and Design of One way Slabs Only one way slabs will be discussed in the next slides 12 6
One way Slabs Definition of One way Slab Case 1 (Slab supported on two opposing sides): If a slab is supported on two opposing sides, bending in the slab will be produced only along the side perpendicular to the direction of supports. In this case the slab will be called as one way slab. 13 One way Slabs Definition of One way Slab Case 2 (Slab supported on all sides): If a slab is supported on all sides and the ratio of length to width is equal to or larger than 2, major bending in the slab will be produced along the short direction and the slab will be called as one way slab. If the ratio is less than 2, bending will occur in both directions and the slab will be called as two way slab. One-Way Slab Two-Way Slab 14 7
One way Slabs Reason for more Demand (Moment) in short direction Δ central Strip = (5/384)wl 4 /EI Consider two strips along the long and short direction as shown in the figure. As these imaginary strips are part of monolithic slab, the deflection at any point, of the two orthogonal slab strips must be same: Δ a = Δ b (5/384)w a l a4 /EI = (5/384)w b l b4 /EI w a /w b = l b4 /l a4 w a = w b (l b4 /l a4 ) Thus, larger share of load (Demand) is taken by the shorter direction. 15 Analysis of One-way Slabs Strip method of analysis: For the purpose of analysis and design, a unit strip of one way slab, cut out at right angles to the supporting beams, may be considered as a rectangular beam of unit width, with a depth h and a span l a as shown. The method is called as strip method of analysis. 16 8
Analysis of One-way Slabs Strip method of analysis: The strip method of analysis and design of slabs having bending in one direction is applicable only when: Slab is supported on only two sides on stiff beams or walls, Slab is supported on all sides on stiff beams or walls with ratio of larger to smaller side greater than 2. Note: Not applicable to flat plates etc., even if bending is primarily in one direction. 17 Basic Steps for Structural Design Selection of Size Calculation of Loads Analysis Design Drafting 18 9
Sizes: ACI table 7.3.1.1 gives the minimum one way slab thickness. l = Span length, defined on the next slide. For f y other than 60,000 psi, the expressions in Table 7.3.1.1 shall be multiplied by (0.4 + fy/100,000). 19 Sizes (Definition of Span Length, l) Slab Beam h Wall l n l n lc/c lc/c 1) l = l n ; for integral supports such as beams and columns with l n 10 2) l = Minimum of [(l n +h) or c/c distance] ; for non-integral supports such as walls with any distance & for integral supports (beams and columns) with l n > 10 l (span length) is used in calculating depth of members. l n (clear span) is used for determining moments using ACI coefficients. l c/c is (center to center distance) is used for analysis of simply supported beam. 20 10
Loads: One way slabs are usually designed for gravity loading. As slabs are two dimensional elements, loads are calculated per unit area. Ultimate Load is calculated as follows: w u = 1.2w D + 1.6w L w u = load per unit area (small letter) W u = load per unit length (capital letter) 21 Analysis: The analysis is carried out for ultimate load including self weight obtained from size of the slab and the applied dead and live loads. The maximum bending moment value is used for flexural design. 22 11
Design: Capacity Demand Capacity or Design Strength = Strength Reduction Factor (f) Nominal Strength Demand = Load Factor Service Load Effects Bar spacing (in inches) = A b /A s 12 (A b = area of bar in in 2, A s = Design steel in in 2 /ft) 23 Design: Flexural Reinforcement (ACI 7.6.1.1): Minimum reinforcement Requirement For Grade 40, A smin = 0.0020 A g For Grade 60, A smin = 0.0018 A g 24 12
Design: Maximum Spacing Requirement: Main Reinforcement Least of 3h or 18 (ACI 7.7.2.3) 25 Design: Shrinkage Reinforcement: Concrete shrinks as it dries out. It is advisable to minimize such shrinkage by using concretes with the smallest possible amounts of water and cement compatible with other requirements, such as strength and workability, and by thorough moist-curing of sufficient duration. However, no matter what precautions are taken, a certain amount of shrinkage is usually unavoidable. 26 13
Design: Shrinkage Reinforcement: Usually, however, slabs and other members are joined rigidly to other parts of the structure and cannot contract freely. This results in tension stresses known as shrinkage stresses Since concrete is weak in tension, these temperature and shrinkage stresses are likely to result in cracking 27 Design: Shrinkage Reinforcement: In one-way slabs, the reinforcement provided for resisting the bending moments has the desired effect of reducing shrinkage and distributing cracks. However, as contraction takes place equally in all directions, it is necessary to provide special reinforcement for shrinkage and temperature contraction in the direction perpendicular to the main reinforcement. This added steel is known as temperature or shrinkage reinforcement, or distribution steel. 28 14
Design: Minimum reinforcement Requirement for shrinkage and Temperature reinforcement Same as main reinforcement requirement (ACI R7.6.1.1) Reinforcement is placed perpendicular to main steel to control shrinkage and temperature cracking. 29 Design: Maximum Spacing Requirement: Shrinkage Reinforcement Least of 5h or 18 (ACI 7.7.2.4) 30 15
Design 12 feet simply supported slab to carry a uniform dead load (excluding self weight) of 120 psf and a uniform live load of 100 psf. Concrete compressive strength (f c ) = 3 ksi and steel yield strength (f y ) = 60 ksi. 12 inches h Slab 9 11.25 9 12 31 Slab Design Solution: Step No. 01: Sizes From ACI table 7.3.1.1 For 12 length, h min = l/20 l = span length, minimum of l n + h or l c/c Take ln = 11.25 and h = 6 l n + h = 11.25 + 6/12 = 11.75 or l c/c = 12 Therefore l = 11.75 h min = 11.75 x 12/20 = 7.05 rounded to 7.5 32 16
Slab Design Solution: Step No. 02: Loads Self weight of slab = (7.5 / 12) x 150 = 93.75 psf SDL = 120 psf LL = 100 psf w u = 1.2 (self weight + SDL) + 1.6 LL w u = 1.2 (93.75 + 120) + 1.6 x 100 w u = 416.5 psf For 1 foot strip width, W u = 416.5 psf x 1ft = 416.5 lb/ft 33 Slab Design Solution: Step No. 03: Analysis For unit strip width, (01 foot of slab): M u = W u l 2 / 8 = 416.5 x 12 2 / 8 = 7497 / 1000 = 7.497 ft-kip M u = 7.497 x 12 = 89.96 in-kip 34 17
Slab Design Solution: Step No. 04: Design Main Reinforcement: h = 7.5, d = 7.5 1 = 6.5 A s = M u / {Φf y (d a/2)} Clear cover for slab is usually taken 0.75. h = d y If #4 (dia 0.5 ) bar is to be used y = 0.75 + 0.5/2 y = 1 d Calculate A s by trial and success method 35 Slab Design Solution: Step No. 04: Design Main Reinforcement: First Trial: Assume a = 0.2h = 0.2 x 7.5 = 1.5 A s = 89.96 / [0.9 60 {6..5 (1.5/2)}] = 0.29 in 2 a = A s f y / (0.85f c b w ) = 0.29 60/ (0.85 3 12) = 0.57 inches 36 18
Slab Design Solution: Step No. 04: Design Main Reinforcement: Second Trial: A s = 89.96 / [0.9 60 {6..5 (0.57/2)}] = 0.27 in 2 a = A s f y / (0.85f c b w ) = 0.27 60/ (0.85 3 12) = 0.53 inches Close enough to the previous value of a so that A s = 0.27 in 2 O.K 37 Slab Design Solution: Step No. 04: Design Main Reinforcement: Minimum reinforcement check: A smin = 0.0018A g = 0.0018 bh A smin = 0.0018 x 12 x 7.5 = 0.162 in 2 As the design A s = 0.27 in 2 > 0.162 in 2 Therefore A s is ok. 38 19
Slab Design Solution: Step No. 04: Design Main Reinforcement: A b = Area of bar in in 2, A s = Design steel area in in 2 /ft Bar Placement: Bar spacing (in inches) = (A b / A s ) 12 Using #4 bars with A s = 0.20 in 2 Spacing = (0.20 / 0.27) x 12 = 8.98 say 8.5 39 Slab Design Solution: Step No. 04: Design Main Reinforcement: Maximum Spacing Requirement Least of 3h or 18, 3h = 3 x 7.5 = 22.5 Provided spacing is OK 40 20
Slab Design Solution: Step No. 04: Design Shrinkage/ Reinforcement: A smin = 0.0018A g = 0.0018 bh A smin = 0.0018 x 12 x 7.5 = 0.162 in 2 Using #4 bars, Spacing = (0.20 / 0.162) x 12 = 14.81 say 14.5 41 Slab Design Solution: Step No. 04: Design Shrinkage/ Reinforcement: Maximum Spacing Requirement Least of 5h or 18, 5h = 5 x 7.5 = 37.5 Provided spacing is OK 42 21
Step No. 05: Drafting 43 Step No. 05: Drafting 44 22
Slab Design Placement of reinforcement: Main reinforcing bars are placed in the direction of flexure stresses and placed at the bottom(at the required clear cover) to maximize the d, effective depth. 45 Practice Design 10 feet simply supported slab to carry a uniform dead load (excluding self weight) of 40 psf and a uniform live load of 120 psf. Concrete compressive strength (f c ) = 3 ksi and steel yield strength (f y ) = 40 ksi. 46 23
Assignment # 03 Submit # 02 of lecture 06-Design of RC Beam for Shear in the next class. Take the length of beam equal to 20 feet. 47 Quiz # 03 A short quiz will be taken in Lecture 07-Design of RC Slab in the next class 48 24
References Design of Concrete Structures 14 th / 15 th edition by Nilson, Darwin and Dolan. ACI 318-14 49 25