design of reinforced concrete slabs -...

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Department of Civil Engineering, University of Engineering and Technology Peshawar, Pakistan

Prof. Dr. Qaisar Ali CE-320 Reinforced Concrete Design-I

Lecture 07

Design of Reinforced Concrete Slabs

By: Prof Dr. Qaisar Ali

Civil Engineering Department

UET Peshawar

[email protected]

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Prof. Dr. Qaisar Ali CE-320 Reinforced Concrete Design-I 2

Topics Addressed

Introduction

Analysis and Design of slabs

Strip Method of Analysis for One-way Slabs

Basic Design Steps

Example

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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.

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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.

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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.

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Flat Plate

Concrete slabs in some cases may be carried directly by

columns. Punching shear is a typical problem in flat plates.

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Introduction

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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.

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Introduction



Rib

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.

Introduction

Peshawar University Auditorium

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Two-way Joist

Introduction



Analysis and Design of Slabs

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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.

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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

Analysis and Design of Slabs



Analysis and Design of One way Slabs

Only one way slabs will be

discussed in the next slides

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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.

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

One way Slabs

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Reason for more Demand (Moment) in short

direction

Δcentral Strip = (5/384)wl4/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:

Δa = Δb

(5/384)wala4/EI = (5/384)wblb

4/EI

wa/wb = lb4/la

4 wa = wb (lb4/la

4)

Thus, larger share of load (Demand) is taken by the shorter direction.

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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 la as shown.

The method is called as strip method of analysis.

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Analysis of One-way Slabs

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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.

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Analysis of One-way Slabs



Basic Design Steps

Basic Steps for Structural Design

Selection of Size

Calculation of Loads

Analysis

Design

Drafting

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Sizes: ACI table 7.3.1.1 gives the minimum one way slab thickness.

l = Span length, defined on the next slide.

For fy other than 60,000 psi, the expressions in Table 7.3.1.1 shall be multiplied by (0.4 +

fy/100,000).

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Basic Design Steps



Sizes (Definition of Span Length, l)

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1) l = ln ; for integral supports such as beams and columns with ln ≤ 10′

2) l = Minimum of [(ln +h) or c/c distance] ; for non-integral supports such as walls

with any distance & for integral supports (beams and columns) with ln > 10′

• l (span length) is used in calculating depth of members.• ln (clear span) is used for determining moments using ACI coefficients.

• lc/c is (center to center distance) is used for analysis of simply supported beam.

Beam

Slab

Wall

lc/c lc/c

ln ln

h

Basic Design Steps

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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:

wu = 1.2wD + 1.6wL

wu = load per unit area (small letter)

Wu = load per unit length (capital letter)

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Basic Design Steps



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.

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Basic Design Steps

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Design:

Capacity Demand

Capacity or Design Strength = Strength Reduction Factor (f)

Nominal Strength

Demand = Load Factor Service Load Effects

Bar spacing (in inches) = Ab/As × 12

(Ab = area of bar in in2, As = Design steel in in2/ft)

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Basic Design Steps



Design:

Flexural Reinforcement (ACI 7.6.1.1):

Minimum reinforcement Requirement

For Grade 40, Asmin = 0.0020 Ag

For Grade 60, Asmin = 0.0018 Ag

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Basic Design Steps

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Design:

Maximum Spacing Requirement:

Main Reinforcement

Least of 3h or 18” (ACI 7.7.2.3)

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Basic Design Steps



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.

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Basic Design Steps

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Design:


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

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Basic Design Steps



Design:


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.

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Basic Design Steps

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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.

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Basic Design Steps



Design:

Maximum Spacing Requirement:

Shrinkage Reinforcement

Least of 5h or 18” (ACI 7.7.2.4)

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Basic Design Steps

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Example

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 (fc′) = 3 ksi and steel yield strength

(fy) = 60 ksi.

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Slab

12′

h

12 inches

9″ 9″11.25′



Slab Design

Solution:

Step No. 01: Sizes

From ACI table 7.3.1.1

For 12′ length, hmin = l/20

l = span length, minimum of ln + h or lc/c

Take ln = 11.25′ and h = 6″

ln + h = 11.25 + 6/12 = 11.75′ or lc/c = 12′

Therefore l = 11.75′

hmin= 11.75 x 12/20 = 7.05″ rounded to 7.5″

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Example

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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

wu = 1.2 (self weight + SDL) + 1.6 LL

wu = 1.2 (93.75 + 120) + 1.6 x 100

wu = 416.5 psf

For 1 foot strip width, Wu = 416.5 psf x 1ft = 416.5 lb/ft

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Example



Slab Design

Solution:

Step No. 03: Analysis

For unit strip width, (01 foot of slab):

Mu = Wu l2 / 8 = 416.5 x 122 / 8

= 7497 / 1000

= 7.497 ft-kip

Mu = 7.497 x 12 = 89.96 in-kip

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Example

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

h = 7.5″, d = 7.5 – 1 = 6.5″

As = Mu/ {Φfy (d – a/2)}

Calculate “As” by trial and success method

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Example

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



Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

First Trial:

Assume a = 0.2h = 0.2 x 7.5 = 1.5″

As = 89.96 / [0.9 × 60 × {6..5 – (1.5/2)}] = 0.29 in2

a = Asfy/ (0.85fc′bw)

= 0.29 × 60/ (0.85 × 3 × 12) = 0.57 inches

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Example

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Second Trial:

As = 89.96 / [0.9 × 60 × {6..5 – (0.57/2)}] = 0.27 in2

a = Asfy/ (0.85fc′bw)

= 0.27 × 60/ (0.85 × 3 × 12) = 0.53 inches

“Close enough to the previous value of “a” so that

As = 0.27 in2 O.K

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Example



Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Minimum reinforcement check:

Asmin = 0.0018Ag = 0.0018 bh

Asmin = 0.0018 x 12 x 7.5

= 0.162 in2

As the design As = 0.27 in2 > 0.162 in2

Therefore As is ok.

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Example

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Bar Placement:

Bar spacing (in inches) = (Ab / As) × 12

Using #4 bars with As = 0.20 in2

Spacing = (0.20 / 0.27) x 12 = 8.98″ say 8.5″

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Ab = Area of bar in in2,

As = Design steel area in in2/ft

Example



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

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Example

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Slab Design

Solution:

Step No. 04: Design

Shrinkage/ Reinforcement:

Asmin = 0.0018Ag = 0.0018 bh

Asmin = 0.0018 x 12 x 7.5 = 0.162 in2

Using #4 bars,

Spacing = (0.20 / 0.162) x 12 = 14.81″ say 14.5″

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Example



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

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Example

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Step No. 05: Drafting

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Example



Step No. 05: Drafting

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Example

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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.

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Example



Practice Example

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 (fc′) = 3 ksi and steel yield strength

(fy) = 40 ksi.

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Assignment # 03

Submit Example # 02 of lecture 06-Design of RC Beam for Shear in the

next class.

Take the length of beam equal to 20 feet.



Quiz # 03

A short quiz will be taken in Lecture 07-Design of RC Slab in the next

class

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Design of Concrete Structures 14th / 15th edition by Nilson, Darwin

and Dolan.

ACI 318-14

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References

design of reinforced concrete slabs -...

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