Design of Reinforced Concrete Slabs 07...Analysis and Design of slabs Strip Method of Analysis for One-way Slabs Basic Design Steps Example 2. 7/9/2020 2 Department of Civil Engineering,

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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]



Topics Addressed

Introduction

Analysis and Design of slabs

Strip Method of Analysis for One-way Slabs

Basic Design Steps

Example

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Objectives

At the end of this lecture, students will be able to

Classify slab systems

Analyze one-way slabs using Strip Method of

Analysis for flexure

Design one-way slab system for flexure

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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 opposing beams.

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Introduction


Slabs may be supported by beams on all four sides, as shown in

figure, in which case the structural action essentially becomes

two-way.

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Introduction


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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 problem encounter in such kinds of slabs, can be reduced by

introducing drop panel and column capital.

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Introduction


Prof. Dr. Qaisar Ali CE-320: Reinforced Concrete Design-I 10

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

Jamia Haqqania Mosque at Akora Khattak



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

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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 long to short side is equal to or greater 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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One way & Two way Slabs

Case 1: One way Slab Case 2: Two way Slab



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 (lb

4/la4)

Thus, larger share of load (Demand) is

taken by the short direction.

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One way Slab

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Reason for more Demand (Moment) in short direction

Ma = wa la2/k ……… (1)

Mb = wb lb2/k ……… (2)

Substitute wa = wb (lb4/la

4) in Equ. 1

Ma = (wb lb4/la

4) la2/k

Finally, Ma = Mb x (lb/la)2

Thus, Bending in short direction is

more than Bending in long direction.

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One way Slab



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 opposing 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 Slab

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Strip method of analysis:

The strip method of analysis for slabs having bending in one direction is

applicable only when: a rectangular

Slab is supported on two opposing sides on stiff beams or walls,

Slab is supported on all sides on stiff beams or walls with ratio of long

to short side is equal or greater than 2.

Note: Strip method of analysis is not applicable to flat plates etc.,

even if bending is in one direction.

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Analysis of One-way Slab



Basic Design Steps

Basic Steps for Design

Selection of Size

Calculation of Loads

Analysis

Design

Drafting

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Basic Design Steps

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). Fy should be in psi.



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 +hf) 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/clc/c

ln ln

hf

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



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

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 or required steel in in2)



Basic Design Steps

Design:

Bar Placement

In the case of beam the No. of bars = n = As/Ab

While in slab we have

n = As/Ab ..................................(a)

For a unit width b = 12″ with center to center

spacing “s” then the above equation will become

b/s = As/Ab b = 12″

12/s = As/Ab

s(in) = (Ab/As) × 12

s(in) = (Ab/As) × 12

b

los

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Basic Design Steps

Design:

Flexural Reinforcement (ACI 7.6.1.1):

Minimum Main reinforcement Requirement:

Asmin = 0.0018 Ag (Ag = Grass Area of concrete = bhf)

Maximum Spacing Requirement (ACI 7.7.2.3):

Main Reinforcement

Least of 3hf or 18″



Basic Design Steps

Design:

Shrinkage Reinforcement:

Concrete shrinks as it dries out.

It is advisable to minimize such shrinkage by using concrete 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

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.



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

Design:

Minimum reinforcement Requirement for shrinkage and

Temperature reinforcement:

Same as main reinforcement minimum requirement (ACI 7.6.1.1)

Reinforcement is placed perpendicular to main steel to control

shrinkage and temperature cracking.

Maximum Spacing Requirement (ACI 7.7.2.4):

Shrinkage Reinforcement

Least of 5hf or 18”



Example

Design the given 12 feet simply supported slab carrying a uniform

service dead load (excluding self weight) of 120 psf and a uniform

service live load of 100 psf. Concrete compressive strength (fc′) = 3 ksi

and steel yield strength (fy) = 60 ksi.

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Slab

9″

9″

12”h

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Slab Design

Solution:

Step No. 01: Sizes

From ACI table 7.3.1.1

For 12′ length, hf,min = l/20

l = span length, minimum of (ln + hf) or lc/c

Take ln = 11.25′ and hf = 6″

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

Therefore l = 11.75′

hf,min= 11.75 x 12/20 = 7.05″ rounded to 7.5″

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Example



Example

Slab Design

Solution:

Step No. 02: Loads

Self-weight of slab = (7.5 / 12) x 0.150 = 0.09375 ksf

SDL = 0.120 ksf (SDL = Superimposed dead load)

LL = 0.100 ksf (LL = Live Load)

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

wu = 1.2 (0.09375 + 0.120) + (1.6 x 0.100)

wu = 0.4165 ksf

For 1 foot strip width, Wu = 0.4165 ksf x 1ft = 0.4165 kips/ft

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Example

Slab Design

Solution:

Step No. 03: Analysis

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

Mu = Wu lc/c2 / 8 = 0.4165 x 122 / 8

= 7.497 ft-kips/ft

Mu = 7.497 x 12 = 89.96 in-kip/ft



d

Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

hf = 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

ACI 20.6.1.3.1 Clear cover for slab is 0.75″.

d= hf – y

If #4 (dia 0.5″) bar is to be used

y = 0.75 + 0.5/2

y = 1″

hf

12”

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

First Trial:

Assume a = 0.2hf = 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/ft

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Example



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/ft

After Trails, As = 0.27 in2/ft

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Example

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Minimum reinforcement check:

Asmin = 0.0018Ag = 0.0018 bhf

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



Ab = Area of bar in in2,

As = Design steel area in in2/ft

Example

Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Bar Placement:

No of bars = n = As/Ab

Bar spacing, s (in inches) = b/n ; where b = 12 inches

Using #4 bar, with area Ab= 0.2 in2

s = 12n

= 12AsAb

= (Ab /As) x 12 = (0.2/ 0.27) x 12 = 8.88″ say 8.5″

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Slab Design

Solution:

Step No. 04: Design

Main Reinforcement:

Maximum Spacing Requirement

Least of 3hf or 18″,

3hf = 3 x 7.5 = 22.5″

Provided spacing is OK

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Example



Slab Design

Solution:

Step No. 04: Design

Shrinkage/ Reinforcement:

Asmin = 0.0018Ag = 0.0018 bhf

Asmin = 0.0018 x 12 x 7.5 = 0.162 in2

Using #4 bar, with area Ab= 0.2 in2

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

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Example

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Slab Design

Solution:

Step No. 04: Design

Shrinkage/ Reinforcement:

Maximum Spacing Requirement

Least of 5hf or 18″, 5hf = 5 x 7.5 = 37.5″

Provided spacing is OK

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Example



Slab Design

Step No. 05: Drafting

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Example

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Slab Design

Step No. 05: Drafting

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Example



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

Main reinforcement

Shrinkage reinforcementSpacer

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A

A

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Drafting in 3D

Section A-A


#4 bar @ 14.5″ c/c

Main Reinforcement

#4 bar @ 8.5″ c/c



Drafting in 3D

Section B-B

Main Reinforcement

#4 bar @ 8.5″ c/c


#4 bar @ 14.5″ c/c

B

B

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Practice Example

Class Activity: Design 10 feet simply supported slab to carry a uniform

service dead load (excluding self weight) of 40 psf and a uniform

service live load of 120 psf. Concrete compressive strength (fc′) = 3 ksi

and steel yield strength (fy) = 40 ksi.

Slab

10′

9.25′

9″

9″

12″h



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

and Dolan.

Building Code Requirements for Structural Concrete (ACI 318-19)

52

References

Documents

Design of Reinforced Concrete Slabs 07...Analysis and Design of slabs Strip Method of Analysis for One-way Slabs Basic Design Steps Example 2. 7/9/2020 2 Department of Civil Engineering,