reinforced concrete ii_2013-2014.pdf

Collage of Engineering and Technology

Department of Civil and Architectural Engineering

Reinforced

Concrete

II

Dr. Nasr Younis Abboushi

2013-2014

Palestine Polytechnic University

Reinforced Concrete II Dr. Nasr Abboushi

II - ii

CONTENTS

CHAPTER 8 ONE-WAY SLABS 185

8.1 Types of slabs 185

8.2 Analysis of continuous beams and frames 187

8.3 Analysis and design of one-way solid slabs. ACI code limitations. 191

8.4 Minimum reinforcement ratio 195

8.5 Temperature and shrinkage reinforcement 196

8.6 Reinforcement details 197

8.7 One-way joist floors and one-way ribbed slabs 212

8.8 Design of one-way ribbed slab 216

CHAPTER 9 TWO-WAY SLABS 234

9.1 Introduction 234

9.2 Types of two-way slabs 234

9.3 Economical choice of concrete floor systems 237

9.4 Minimum thickness of two-way slabs 238

9.5 Slab reinforcement requirements 241

9.6 Shear strength of two-way slabs 243

9.6.1 Two-Way Slabs Supported on Beams 244

9.6.2 Two-Way Slabs Without Beams 244

9.6.3 Tributary Areas for Shear in Two-Way Slabs 247

9.6.4 Shear Reinforcement in Two-Way Slabs Without Beams 247

9.7 Analysis and design of two-way slabs 249

9.8 Slab analysis by the coefficient method 250

9.9 Slab analysis by the direct design method (DDM). 294

9.9.1 Limitations on the Use of the Direct-Design Method 294

9.9.2 Column and middle strips 295

9.9.3 Total Static Moment at Factored Loads 296

9.9.4 Assignment of positive and negative moments 296

9.9.5 Lateral Distribution of Moments (between Column Strips

and Middle Strips) 298


II - iii

9.10 Slab analysis by the equivalent frame method (EFM) 302

9.11 Shear design in flat plates 305

CHAPTER 10 STAIRS 317


10.2 Types of stairs 318

10.3 Slab type stairs. Structural system 327

CHAPTER 11 FOOTINGS AND FOUNDATIONS 341


11.2 Types of footings 342

11.3 Distribution of soil pressure. Gross and net soil pressures 343

11.4 Design considerations 346

11.4.1 Size of footings 346

11.4.2 One-way shear (Beam shear) 346

11.4.3 Two-way shear (Punching shear) 347

11.4.4 Flexural strength and footing reinforcement 348

11.4.5 Transfer of Load from Column to Footing 350

11.4.6 Bearing Strength 351

11.5 Spread (isolated) footings 353

11.6 Strip (wall) footings 360

11.7 Footings under eccentric column loads 363

11.8 Combined footings 372

11.9 Continuous footings 380

11.10 Mat foundations 386

CHAPTER 12 DEVELOPMENT, ANCHORAGE, AND SPLICING OF REINFORCEMENT 393


12.2 Flexural bond 393

12.3 Mechanism of bond transfer 394


II - iv

12.4 Development length 396

12.5 Hooked anchorages 405

12.6 Bar cutoffs and development of bars in flexural members 412

12.7 Development of positive moment reinforcement 417

12.8 Development of negative moment reinforcement 421

12.9 Reinforcement continuity and structural integrity requirements 425

12.10 Splices of reinforcement 446


185

CHAPTER 8 ONE-WAY SLABS

8.1 TYPES OF SLABS

Structural concrete slabs are constructed to provide flat surfaces, usually horizontal, in

building floors, roofs, bridges, and other types of structures. The slab may be supported by

walls, by reinforced concrete beams usually cast monolithically with the slab, by structural

steel beams, by columns, or by the ground. The depth of a slab is usually very small

compared to its span.

Structural concrete slabs in buildings may be classified as follows:

1. One-way stabs: If a slab is supported on two opposite sides only, it will bend or

deflect in a direction perpendicular to the supported edges. The structural action is one way,

and the loads are carried by the slab in the deflected short direction. This type of slab is

called a one-way slab.


186

If the slab is supported on four sides and the ratio

of the long side to the short side is equal to or

greater than ( ), most of

the load (about or more) is carried in the

short direction, and one-way action is considered

for all practical purposes. If the slab is made of

reinforced concrete with no voids, then it is called

a one-way solid slab.

2. One-way joist floor system: This type of

slab is also called a ribbed slab. It consists of a

floor slab, usually thick, supported

by reinforced concrete ribs (or joists). The ribs are usually tapered and are uniformly spaced

at distances that do not exceed . The ribs are supported on girders that rest on

columns. The spaces between the ribs may be formed using removable steel or fiberglass


187

form fillers (pans), which may be used many times. In some ribbed slabs, the spaces

between ribs may be filled with permanent fillers to provide a horizontal slab.

3. Two-way floor systems: When the slab is supported on four sides and the ratio of

the long side to the short side is less than ( ), the slab will deflect in

double curvature in both directions. The floor load is carried in two directions to the four

beams surrounding the slab. Other types of two-way floor systems are flat plate floors, flat

slabs, and waffle slabs. This chapter deals only with one-way floor systems.

8.2 ANALYSIS OF CONTINUOUS BEAMS AND FRAMES.

In reinforced concrete structures, as much of the concrete as is practical is placed in one

single operation. Reinforcing steel is not terminated at the ends of a member but is

extended through the joints into adjacent members. At construction joints, special care is

taken to bond the new concrete to the old by carefully cleaning the latter, by extending the

reinforcement through the joint, and by other means. As a result, reinforced concrete

structures usually represent monolithic, or continuous, units. A load applied at one location

causes deformation and stress at all other locations. Even in precast concrete construction,

which resembles steel construction in that individual members are brought to the job site

and joined in the field, connections are often designed to provide for the transfer of moment

as well as shear and axial load, producing at least partial continuity.


188

In statically determinate structures, such as simple-span beams, the deflected shape and

the moments and shears depend only on the type and magnitude of the loads and the

dimensions of the member. In contrast, inspection of the statically indeterminate structures

shows that the deflection curve of any member depends not only on the loads but also on

the joint rotations, whose magnitudes in turn depend on the distortion of adjacent, rigidly

connected members.

Continuous beams and frames can be analyzed using approximate methods or computer

programs, which are available commercially. Other methods, such as the displacement and

force methods of analysis based on the calculation of the stiffness and flexibility matrices,

may also be adopted. Slope deflection and moment-distribution methods may also be used.

These methods are explained in books dealing with the structural analysis of beams and

frames. However, the ACI Code, Section 8.3, gives approximate coefficients for calculating

the bending moments and shear forces in continuous beams and slabs. These coefficients

were given in Section 8.3 of this Chapter. The moments obtained using the ACI coefficients

will be somewhat larger than those arrived at by exact analysis. The limitations stated in the

use of these coefficients must be met.

In the structural analysis of continuous beams, the span length is taken from center to

center of the supports, which are treated as knife-edge supports. In practice, the supports

are always made wide enough to take the loads transmitted by the beam, usually the

moments acting at the face of supports. To calculate the design moment at the face of the

support, it is quite reasonable to deduct a moment equal to from the factored

moment at the centerline of the support, where is the factored shear and is the column

width.

According to ACI Code Section 8.9 – Span Length:

8.9.1 — Span length of members not built integrally with supports shall be considered as the

clear span plus the depth of the member, but need not exceed distance between

centers of supports.

8.9.2 — In analysis of frames or continuous construction for determination of moments,

span length shall be taken as the distance center-to-center of supports.

8.9.3 — For beams built integrally with supports, design on the basis of moments at faces of

support shall be permitted.

8.9.4 — It shall be permitted to analyze solid or ribbed slabs built integrally with supports,

with clear spans not more than 3 m, as continuous slabs on knife edge supports

with spans equal to the clear spans of the slab and width of beams otherwise

neglected.

The individual members of a structural frame must be designed for the worst combination of

loads that can reasonably be expected to occur during its useful life. Internal moments,

shears, and thrusts are brought about by the combined effect of dead and live loads, plus

other loads, such as wind and earthquake. While dead loads are constant, live loads such as

floor loads from human occupancy can be placed in various ways, some of which will result

in larger effects than others.


189

For structural analysis of continuous beam or rib to obtain the shear and moment diagrams,

it shall be permitted according to ACI code, 8.11.2, to assume that the arrangement of live

load is limited to combinations of:

a. Factored dead load on all spans with full factored live load on two adjacent

spans; and

b. Factored dead load on all spans with full factored live load on alternate spans.

Load Case 1: ACI-8.11.2-a

Load Case 2: ACI-8.11.2-a

Load Case 3: ACI-8.11.2-b

Load Case 4: ACI-8.11.2-b

Span 1 Span2 Span 3

LL

DL

DL

LL

DL

LL LL

DL

LL


190

From each case we get the Maximum moment:

• Maximum negative moment from load cases 1+2 (ACI-8.11.2-a)

• Maximum positive moment from load cases 3+4 (ACI-8.11.2-b)

• Envelope moment diagram from all possible load cases.

Moment Diagram from

Load Case 1

Moment Diagram from

Load Case 2

Moment Diagram from

Load Case 3

Moment Diagram from

Load Case 4


191

8.3 ANALYSIS AND DESIGN OF ONE-WAY SOLID SLABS. ACI CODE LIMITATIONS.

If the concrete slab is cast in one uniform thickness without any type of voids, it can be

referred to as a solid slab. In a one-way stab nearly all the loading is transferred in the short

direction, and the slab may be treated as a beam. A unit strip of slab, usually 1 m at right

angles to the supporting girders, is considered a rectangular beam. The beam has a unit

width with a depth equal to the thickness of the slab and a span length equal to the distance

between the supports. A one-way slab thus consists of a series of rectangular beams placed

side by side.

If the slab is one span only and rests freely on its supports, the maximum positive moment

for a uniformly distributed load of is , where is the span length

between the supports. If the same slab is built monolithically with the supporting beams or

is continuous over several supports, the positive and negative moments are calculated by

structural analysis or by moment coefficients as for continuous beams. The ACI Code,

Section 8.3, permits the use of moment and shear coefficients in the case of two or more

approximately equal spans.

The maximum positive and negative moments and shears are computed from the following

expressions:

Moment Diagrams of all Load cases

Envelope Moment Diagram from all Load cases


192

( )

(

)

where and are moment and shear coefficients given in table below and figure (page

193).

For all positive midspan moments, all shears and the negative moment at exterior

supports, , is for the span under consideration. For the negative moment at interior

supports, , shall be taken as ( )as defined in the figure above.


193


194

The conditions under which the moment coefficients for continuous beams and slabs should

be used can be summarized as follows:

1. Spans are approximately equal: Longer span (shorter span).

2. Loads are uniformly distributed.

3. The ratio (live load/dead load) is less than or equal to .

4. For slabs with spans less than or equal to , negative bending moment at

face of all supports is (

)

5. For an unrestrained discontinuous end, the coefficient is at end support and

(

) at midspan.

6. Shear force at C is and at the face of all other support is (

) .

7. The members are prismatic.

When these conditions are not satisfied, structural analysis is required. In structural analysis,

the negative bending moments at the centers of the supports are calculated. The value that

may be considered in the design is the negative moment at the face of the support, ACI

8.9.2, 8.9.3.

The following limitations are specified by the ACI code:

Atypical imaginary strip 1m wide is assumed.

The minimum thickness of one-way slabs using grade 420 steel can be defined

according to the ACI Code, 9.5.2.1, Table 9.5a, for solid slabs and for beams or ribbed

one-way slabs .

ACI 9.5.2.1 – Minimum thickness stipulated in Table 9.5(a) shall apply for one-way

construction not supporting or attached to partitions or other construction likely to be

damaged by large deflections, unless computation of deflection indicates a lesser thickness

can be used without adverse effects.


195

Deflection is to be checked when the slab supports are attached to construction

likely to be damaged by large deflections. Deflection limits are set by the ACI Code,

Table 9.5b.

ACI 9.5.2.2 – Where deflections are to be computed, deflections that occur immediately on

application of load shall be computed by usual methods or formulas for elastic deflections,

considering effects of cracking and reinforcement on member stiffness.

It is preferable to choose slab depth to the nearest .

Shear should be checked, although it does not usually control.

Concrete cover in slabs shall not be less than at surfaces not exposed to

weather or ground. In this case,

.

In structural slabs of uniform thickness, the minimum amount of reinforcement in

the direction of the span shall not be less than that required for shrinkage and

temperature reinforcement (ACI Code, Section 7.12).

The principal reinforcement shall be spaced not farther apart than three times the

slab thickness nor more than (ACI Code, Section 7.6.5).

Straight-bar systems may be used in both tops and bottoms of continuous slabs. An

alternative bar system of straight and bent (trussed) bars placed alternately may also

be used.

In addition to main reinforcement, steel bars at right angles to the main must be pro-

vided. This additional steel is called secondary, distribution, shrinkage, or

temperature reinforcement.

8.4 MINIMUM REINFORCEMENT RATIO.

For structural slabs and of uniform thickness, in the direction of the span shall be the

same as that required by 7.12.2.1 for temperature and shrinkage reinforcement (see section

8.5).


196

Maximum spacing of this reinforcement shall not exceed three times the thickness, nor

.

To limit the widths of flexural cracks in beams and slabs, ACI Code Section 10.6.4 defines

upper limit on the center-to-center spacing between bars in the layer of reinforcement

closest to the tension face of a member. The spacing limit is:

(

)

but

(

)

where is the least distance from

surface of reinforcement to the tension

face. It shall be permitted to take as

.

8.5 TEMPERATURE AND SHRINKAGE REINFORCEMENT.

Concrete shrinks as the cement paste hardens, and a certain amount of shrinkage is usually

anticipated. If a slab is left to move freely on its supports, it can contract to accommodate

the shrinkage. However, slabs and other members are joined rigidly to other parts of the

structure, causing a certain degree of restraint at the ends. This results in tension stresses

known as shrinkage stresses. A decrease in temperature and shrinkage stresses is likely to

cause hairline cracks. Reinforcement is placed in the slab to counteract contraction and

distribute the cracks uniformly. As the concrete shrinks, the steel bars are subjected to

compression.

Reinforcement for shrinkage and temperature stresses normal to the principal

reinforcement should be provided in a structural slab in which the principal reinforcement

extends in one direction only.

The ACI Code, Section 7.12.2, specifies that: area of shrinkage and temperature

reinforcement shall provide at least the following ratios of reinforcement area to gross

concrete area, but not less than :

For slabs in which grade 280 ( ) or 350 ( ) deformed bars

are used, .

For slabs in which grade 420 ( ) deformed bars or welded bars or

welded wire fabric are used, .

For Slabs where reinforcement with yield stress exceeding measured at a

yield strain of percent is used,

Shrinkage and temperature reinforcement shall be spaced not farther apart than five times

the slab thickness, nor farther apart than .

𝜌 ×

𝑓𝑦


197

For temperature and shrinkage reinforcement, the whole concrete depth exposed to

shrinkage shall be used to calculate the steel area.

8.6 REINFORCEMENT DETAILS.

In continuous one-way slabs, the steel area of the main reinforcement is calculated for all

critical sections, at midspans, and at supports. The choice of bar diameter and detailing

depends mainly on the steel areas, spacing requirements, and development length. Two bar

systems may be adopted.

In the straight-bar system: straight bars are used for top and bottom reinforcement

in all spans. The time and cost to produce straight bars is less than that required to

produce bent bars; thus, the straight-bar system is widely used in construction.

In the bent-bar, or trussed, system: straight and bent bars are placed alternately in

the floor slab. The location of bent points should be checked for flexural, shear, and

development length requirements. For normal loading in buildings, the bar details at

the end and interior spans of one-way solid slabs may be adopted as shown in

figures.


198


199


200

Example:

Design a simply supported one-way solid slab, span of , subjected to service live load

( ). Dead Load – own weight only. and .

Solution:

Minimum thickness (deflection requirements). For simply supported one-way solid

slab:

Assume bar diameter for main reinforcement.

Loads calculation:

( √

)

( √

)

Provide

Use then

𝑚 𝑠𝑡𝑟𝑖𝑝

𝑚𝑚

Co

ver

𝑚𝑚

𝑚𝑚

s

𝑚


201

Take .

Step ( ) is the smallest of:

1. ×

2.

(

) (

)

(

) (

)

–

( )

Take .


1. ×

2.

–

Example:

The cross-section of a continuous one-way solid slab in a building is shown below. The slabs

are supported by beams that span between simple supports. The dead load on the

slabs is that due to self-weight plus ; the live load is . Design the

continuous slab and draw a detailed section. Given: and .

𝑚𝑚

𝑚𝑚


202

Solution:

Minimum thickness (deflection requirements).

( )

( )

Take slab thickness


Loads calculation:


203

Check whether thickness is adequate for shear:

×

×

√

√

- for shear.

×

The thickness of the slab is adequate enough.

Even, if

for solid slabs, the thickness of the slab will be enough.

Factored moments at sections A, B, C, D, E:

For the negative moment at interior supports, , shall be taken as ( ).

Here ( )

Location

( )

A

B

C

( )

D

( )

E

Slab Design for the positive moments:

Midspan section B:

( √

)

( √

)


204

Use then

Take .


1. ×

2.

(

) (

)

(

) (

)

–

Midspan section E:

( √

)

( √

)

Use then

Take .

–

Slab Design for the negative moments:

Note that the second support has two values of moments by analysis, at section C and

section D. In construction, the provided reinforcement will be the same bar diameters on

opposite sides of the support, so the design may be done for the maximum moment of the

two moments at C and D (Only one design for Support section C).


205

Support section C:


( √

)

( √

)

Use then

Take .

–

Support section D (interior D supports):


( √

)

( √

)

Use then

Take .

–


206

Support section A:


( √

)

( √

)

Provide

Use then

Take .

–

Temperature and shrinkage reinforcement.

( )

Take .


1. ×

2.

–

Location

Required

Provided Reinforcement

A 3.3 ( )

B 3.3 ( )

C and first

interior D 3.3

( )

Interior D 3.3 ( )

E 3.3 ( )

Temperature and shrinkage reinforcement

( )


207

Example:

Design the one-way solid slab, which plan is shown below. The dead load on the slabs is that

due to self-weight plus weight of:

• Tiles, .

• Mortar, .

• Sand, .

• Plaster, .

• Partitions, .

The live load is .

Given:

and .

Solution:

Minimum thickness

(deflection requirements).

From the maximum span

length for one-end

continuous, we get

Take slab thickness


Loads calculation:

𝑚 𝑚 𝑚

𝑚 𝑚

𝑚


208

Quality Density

Material

× Tiles

× mortar

× Sand

× Reinforced Concrete solid slab

× Plaster

Partitions

Total Dead Load

Dead Load for 1 m strip of slab

Live Load for 1 m strip of slab


209

According to ACI 8.9.2 — In analysis of frames or continuous construction for determination

of moments, span length shall be taken as the distance center-to-center of supports.

According to ACI 8.9.3 — For beams built integrally with supports, design on the basis of

moments at faces of support shall be permitted.

Check whether thickness is adequate for shear:

√

√

- for shear.



210

Slab Design for the negative moment:


( √

)

( √

)

× ×

Use then

Take .


1. ×

2.

(

) (

)

(

) (

)

–

Check for strain (tension-controlled section - ):

(

) (

)


211

Slab Design for the positive moments:


( √

)

( √

)

Use then

Take

–


( √

)

( √

)

Use then

Take

–

For both positive moment designs


212


( )

Take .


1. ×

2.

–

8.7 ONE-WAY JOIST FLOORS AND ONE-WAY RIBBED SLABS.

A one-way joist floor system consists of hollow slabs with a total depth greater than that of

solid slabs. The system is most economical for buildings where superimposed loads are small

and spans are relatively large, such as schools, hospitals, and hotels. The concrete in the

tension zone is ineffective; therefore, this area is left open between ribs or filled with

lightweight material to reduce the self-weight of the slab.


213

The design procedure and requirements of ribbed slabs follow the same steps as those for

rectangular and T-sections explained in Chapter 4. The following points apply to design of

one-way ribbed slabs:

1. Ribs are usually tapered and uniformly spaced at about . Voids are

usually formed by using pans (molds) wide and deep,

depending on the design requirement. The standard increment in depth is 50 mm.

2. The ribs shall not be less than wide and must have a depth of not more than

times the width. Clear spacing between ribs shall not exceed (ACI Code,

Section 8.13).


214

3. Shear strength, , provided by concrete for the ribs may be taken greater than

that for beams. This is mainly due to the interaction between the slab and the closely

spaced ribs (ACI Code, Section 8.13.8).

4. The thickness of the slab on top of the ribs is usually and contains

minimum reinforcement (shrinkage reinforcement). This thickness over permanent

fillers shall not be less than (

) of the clear span between ribs or (ACI Code,

Section 8.13.5.2). When removable forms or fillers are used slab thickness shall be

not less than (

) of the clear distance between ribs, nor less than . (ACI

Code, Section 8.13.6.1)

5. The ACI coefficients for calculating moments in continuous slabs can be used for

continuous ribbed slab design.

If the live load on the ribbed slab is less than and the span of ribs exceeds ,

a secondary transverse rib (distribution rib) should be provided at midspan (its direction is

perpendicular to the direction of main ribs) and reinforced with the same amount of steel as

the main ribs. Its top reinforcement shall not be less than half of the main reinforcement in

the tension zone. These transverse ribs act as floor stiffeners. If the live load exceeds

and the span of ribs varies between and , one traverse rib must be provided,

as indicated before. If the span exceeds , at least two transverse ribs at one-third span

must be provided with reinforcement, as explained before.


215


216

Reinforcement for the joists usually consists of two bar in the positive bending region, with

one bar discontinued where no longer needed or bent up to provide a part of the negative

steel requirement over supporting girder. According to ACI Code section 7.13.2 at least one

bottom bar must be continuous over the support, or at non continuous supports, terminated

in a standard hook, as a measure to improve structural integrity in the event of major

structural damage.

The minimum thickness of beams or ribbed one way slabs depending on the support

conditions can be determined according to ACI Code 9.5.2. (see table 9.5(a), page 194).

8.8 DESIGN OF ONE-WAY RIBBED SLAB.

For the ribbed slab plan with section as shown below:

• Determine the total slab thickness.

• Design the topping slab.

• Design the rib for flexure and shear, the envelope moment and shear diagrams are

shown.

• Design the beams B1, B2 for flexure and shear, the envelope moment and shear

diagrams are shown.

• Take the material's density from the table below.


217

Compressive strength of concrete

Yield strength of steel,

Live Load, .

Quality Density

Material

Tiles

mortar

Sand

Reinforced Concrete

Hollow Block

Plaster

Partitions


218

Solution:

Rib 2

Rib 1

Bea

m 5

Bea

m 1

Bea

m 2

Bea

m 3

Bea

m 4

Bea

m 6

800 mm

800 mm 800 mm 800 mm 800 mm

800 mm


219

Minimum thickness (deflection requirements).

There are two groups of ribs and beams (Rib 1; Rib 2; Beam 1, 2, 3, 4; Beam 5, 6).

The thickness of the one-way ribbed slab without drop beams can be obtained according to

ACI code, table 9.5 (a).

The maximum span length for one-end continuous (for ribs): then

The maximum span length for both-ends continuous (for ribs): then

The maximum span length for one-end continuous (for Beams): then

The maximum span length for both-ends continuous (for Beams): then

The minimum ribbed slab thickness will be

Take slab thickness

( )

Topping Design.

Topping in One way ribbed slab can be

considered as a strip of 1 meter width and

span of hollow block length with both end

fixed in the ribs.

Dead Load calculations:

Dead Load from: × ×

Tiles × Mortar × Coarse Sand × Topping × Interior Partitions

Live Load calculations: ×

Total Factored Load:

𝑐𝑚

𝑤𝑙

𝑤𝑙

𝑤𝑙

𝑤𝑢 𝐷 𝐿


220

Strength condition, where for plain concrete.

√ (ACI 22.5.1, Equation 22-2)

where for rectangular section of the slab:

√ √ ×

NO Reinforcement is required by analysis. According to ACI 10.5.4., provide for slabs

as shrinkage and temperature reinforcement.

According to ACI 7.12.2.1, .

Try bars 8 with

Take in both directions.


1. ×

2.

(

) (

)

(

) (

)

Take in both directions. –

From practical concederation, the secondary reinforcement parallel to the ribs shall be

placed in the slab and spaced at distances not more than half of the spacings between ribs

(usually two bars upon each width block).

Load Calculations for Rib 1:

From the Geometry of T-section:

𝑈𝑛𝑖𝑡 𝑤𝑖𝑑𝑡 𝑚𝑚

𝑏𝑒

t =

80

mm

h=

250

mm

𝑏𝑤 𝑚𝑚


221


Live Load calculations:

×

Dead Load / rib:

Live Load /rib:

The Effective Flange width ( ) According to ACI 8.12.2 (see page 220):

is the smallest of:

( )

( )

( )

Take .

Structural Analysis of Rib 1. The envelope shear and moment diagrams (for all load

combinations).

Using the structural analysis and design programs, we obtain the Envelope Moment diagram

for Rib1.


Tiles × ×

Mortar × ×

Coarse Sand × ×

Topping × ×

RC Rib × ×

Hollow Block × ×

Plaster × ×

Interior Partitions ×


222


223

Design of Rib 1 for positive moments.

Assume bar diameter for main positive reinforcement.

The maximum positive moment in all spans of Rib 1

Check if

(

) (

) ×

The section will be designed as rectangular section with .

×

( √

)

( √

)

Check for

√

√

Use with

Check for strain:

(

) (

)

Usually, no reinforcement less than can be used. So, for all spans with positive

moments equal or less than , use for each rib span.


224

Design of Rib 1 for negative moments.


According to ACI 8.9.3 — for beams built integrally with supports, design on the basis of

moments at faces of support shall be permitted.

The maximum negative moment at the face of support

×

( √

)

( √

)

Check for

Use with

Check for strain:

(

) (

)

Usually, no reinforcement less than can be used. So, for all supports with negative

moments equal or less than , use for each rib support.

Design of Rib 1 for shear.

The maximum shear force at the distance from the face of support

Shear strength, , provided by concrete for the ribs may be taken greater than that for

beams. This is mainly due to the interaction between the slab and the closely spaced ribs

(ACI Code, Section 8.13.8).


225

( ) ×

√ ×

√

Minimum shear reinforcement is required except for concrete joist construction. So, No

shear reinforcement is provided.

Load calculations for Beam 4:

The distributed Dead and Live loads acting upon the Beam 4 can be defined from the

support reactions of the rib 1 and rib 2.


The maximum support reaction (factored) from Dead Loads for rib1 upon beam 4 is

. The distributed Dead Load from the Rib 1 on Beam 4:

Beam 4


226

Assume the width of the beam , then the own weight of the beam and the weight

of the floor layers within the beam width can be calculated:

The total factored Dead Load:


The maximum support reaction (factored) from Live Loads for rib1 upon beam 4 is .

The distributed Live Load from the Rib 1 on Beam 4:

The Live Load within the beam width ( ) can be calculated:

×

The total factored Live Load:

Important NOTE:

The dead and live loads acting within the beam width have been calculated twice actually.

That because the support reactions of ribs from the dead and live loads acting over the ribs

were calculated for full span of the ribs, measured center to center, which include the whole

beam width.

More accurately calculations to determine the loads on the beam could be done by taking

the loads that transfer from the rib to the beam which could be calculated as the sum of

shear values of the ribs at the face of support (beam) from each side and then adding the

dead and live loads acting directly on the beam within the beam width.

From the next envelope shear diagrams, the shear values at the face of beam 4 are:

From dead load: from the left, and from the right) which means

that the total dead load that transfers to the beam 4 is

The uniformly distributed dead load over the beam from the ribs only is


Tiles × ×

Mortar × × Coarse Sand × × RC Beam × × Plaster × × Interior Partitions ×


227

From live load: from the left, and from the right) which means

that the total live load that transfers to the beam 4 is

The uniformly distributed live load over the beam from the ribs only is

Shear diagram from Dead load only.

Envelope shear diagram from Live load only.



Beam 4

Beam 4


228

I ’ v h h f h h f l l . I ’ f l

to calculate the loads more accurate specially when the beam section is wide and the spans

are large. We can use the first method for simplicity, especially in this stage of calculations,

when the dimensions of the beam are not known yet.

Structural Analysis of Beam 4. The envelope shear and moment diagrams (for all load

combinations).

The Beam 4 is loaded from the ribs 1 (first two spans) and ribs 2 (last span). The load

transferred from ribs 1 to Beam 4 is calculated before. The load transferred from rib 2 to

Beam 4 will be obtained by analyzing the rib 2 as continuous beam as follows:

Beam 4


229


The maximum support reaction (factored) from Dead Loads for rib2 upon beam 4 is

. The distributed Dead Load from the Rib 1 on Beam 4:



The maximum support reaction (factored) from Live Loads for rib 2 upon beam 4 is .

The distributed Live Load from the Rib 2 on Beam 4:


Using the structural analysis and design programs, we obtain the Envelope Moment diagram

for Beam 4.


230

Design of Beam 4 for flexure.


The width of the Beam 4 can be defined from the maximum factored moment.

The maximum factored moment in Beam 4

Note that according to ACI 8.9.3 — for beams built integrally with supports, design on the

basis of moments at faces of support shall be permitted. Here the design will be done for the

moments at the center of supports.

Take f fl - ll

Assume .


231

𝑐𝑚

𝑐𝑚

Take ( )

(

)

(

)

(

) (

)

Usually in construction the maximum width of the beams is . Here, take

and no need to recalculate the loads acting on the beam.

Note that the factored moments of other supports and spans may be satisfied by the section

width of as a singly reinforced beam sections, but the support section with

may be designed as doubly reinforced section.

Check whether the section will be act as singly or

doubly reinforced section:

Maximum nominal moment strength from strain

condition

(

) (

) ×

Design the section as singly reinforced concrete section.

( √

)

( √

)


232

√

√

Take in one layer with

Check for strain:

(

) (

)

Check for bar placement:

× × ×

Design of Beam 4 for shear.

Critical section at distance from the face of support.

√

√

Check for section dimensions:

√

√

Find the maximum stirrups spacing:

√


233

√

√

Check for :

√

(

√

)

√

√

( )

( )

Or

Compute the stirrups spacing required to resist the shear forces:

Use stirrups 2U-shape (4 legs stirrups) with

Take 2U-shape (4 legs stirrups)

100 cm

25

cm

𝑚𝑚

𝑚𝑚

𝑚𝑚


234

Design the other beam sections for flexure (for positive and negative moments and caculate

the area of steel for each section).

Note that for shear design, it is obvious that, the stirrups cannot be less than two U-shape

stirrups and the step for all sections where stirrups are reqired.

So, for all sections the design for shear will be as the previous section ( 2 U-shape

).

An alternative design for the Beam 4 can be done on the basis of drop beam section, not as a

hidden beam as in the previous design.


234

CHAPTER 9 TWO-WAY SLABS

9.1 INTRODUCTION

When the slab is supported on all four sides and the length, , is less than twice the width,

, the slab will deflect in two directions, and the loads on the slab are transferred to all four

supports. This slab is referred to as a two-way slab. The bending moments and deflections in

such slabs are less than those in one-way slabs; thus, the same slab can carry more load

when supported on four sides. The load in this case is carried in two directions, and the

bending moment in each direction is much less than the bending moment in the slab if the

load were carried in one direction only.

9.2 TYPES OF TWO-WAY SLABS

Structural two-way concrete slabs may be classified as follows:

1. Two-Way Slabs on Beams: This case occurs when the two-way slab is supported by

beams on all four sides. The loads from the slab are transferred to all four supporting

beams, which, in turn, transfer the loads to the columns.

2. Flat Slabs: A flat slab is a two-way slab reinforced in two directions that usually does

not have beams or girders, and the loads are transferred directly to the supporting

columns. The column lends to punch through the slab, which can be treated by three

methods:


235

a. Using a drop panel and a column capital.

b. Using a drop panel without a column capital. The concrete panel around the column

capital should be thick enough to withstand the diagonal tensile stresses arising from

the punching shear.

c. Using a column capital without drop panel, which is not common.

3. Flat-Plate Floors: A flat-plate floor is a two-way slab system consisting of a uniform

slab that rests directly on columns and does not have beams or column capitals

(Fig. a). In this case the column tends to punch through the slab, producing diagonal

tensile stresses. Therefore, a general increase in the slab thickness is required or

special reinforcement is used.


236

4. Two-Way Ribbed Slabs and the Waffle Slab System: This type of slab consists of a

floor slab with a length-to-width ratio less than 2. The thickness of the slab is usually

5 to 10 cm and is supported by ribs (or joists) in two directions. The ribs are arranged

in each direction at spacings of about , producing square or

rectangular shapes. The ribs can also be arranged at or from the centerline

of slabs, producing architectural shapes at the soffit of the slab. In two-way ribbed

slabs, different systems can be adopted:

a. A two-way rib system with voids between the ribs, obtained by using special

removable and usable forms (pans) that are normally square in shape. The ribs are

supported on four sides by girders that rest on columns. This type is called a two-

way ribbed (joist) slab system.

b. A two-way rib system with permanent fillers between ribs that produce horizontal

slab soffits. The fillers may be of hollow, lightweight or normal-weight concrete or

any other lightweight material. The ribs are supported by girders on four sides,

which in turn are supported by columns. This type is also called a two-way ribbed

(joist) slab system or a hollow-block two-way ribbed system.

c. A two-way rib system with voids between the ribs with the ribs continuing in both

directions without supporting beams and resting directly on columns through solid

panels above the columns. This type is called a waffle slab system.


237

9.3 ECONOMICAL CHOICE OF CONCRETE FLOOR SYSTEMS

Various types of floor systems can be used for general buildings, such as residential, office,

and institutional buildings. The choice of an adequate and economic floor system depends

on the type of building, architectural layout, aesthetic features, and the span length

between columns. In general, the superimposed live load on buildings varies between 4 and

. A general guide for the economical use of floor systems can be summarized as

follows:

1. Flat Plates: Flat plates are most suitable for spans of and live loads

between and . The advantages of adopting flat plates include low-

cost formwork, exposed flat ceilings, and fast construction. Flat plates have low

shear capacity and relatively low stiffness, which may cause noticeable

deflection. Flat plates are widely used in buildings either as reinforced or

prestressed concrete slabs.


238

2. Flat Slabs: Flat slabs are most suitable for spans of and for live loads of

. They need more formwork than flat plates, especially for column

capitals. In most cases, only drop panels without column capitals are used.

3. Waffle Slabs: Waffle slabs are suitable for spans of and live loads of

. They carry heavier loads than flat plates and have attractive

exposed ceilings. Formwork, including the use of pans, is quite expensive.

4. Slabs on Beams: Slabs on beams are suitable for spans beiween and and

live loads of . The beams increase the stiffness of the slabs,

producing relatively low deflection. Additional formwork for the beams is

needed.

5. One-Way Slabs on Beams: One-way slabs on beams are most suitable for spans

of and a live load of . They can be used for larger spans

with relatively higher cost and higher slab deflection. Additional formwork for

the beams is needed.

6. One-Way Joist Floor System: A one-way joist floor system is most suitable for

spans of and live loads of . Because of the deep ribs, the

concrete and steel quantities are relatively low, but expensive formwork is

expected. The exposed ceiling of the slabs may look attractive.

9.4 MINIMUM THICKNESS OF TWO-WAY SLABS.

The ACI Code, Section 9.5.3, specifies a minimum slab thickness in two-way slabs to control

deflection. The magnitude of a slab's deflection depends on many variables, including the

flexural stiffness of the slab, which in turn is a function of the slab thickness, . By increasing

the slab thickness, the flexural stiffness of the slab is increased, and consequently the slab

deflection is reduced. Because the calculation of deflections in two-way slabs is complicated

and to avoid excessive deflections, the ACI Code limits the thickness of these slabs by

adopting the following three empirical limitations, which are based on experimental

research. If these limitations are not met, it will be necessary to compute deflections.

1. For

(

)

( )

but not less than .

2. For

(

)

but not less than .

3. For


239

where

clear span in the long direction measured face to face of columns (or face to face of

beams for slabs with beams).

the ratio of the long to the short clear spans.

the average value of for all beams on the sides of a panel.

the ratio of flexural stiffness of a beam section to the flexural stiffness of the slab

, bounded laterally by the centerlines of the panels on each side of the beam.

where , and are the moduli of elasticity of concrete in the beam and the slab,

respectively, and

the gross moment of inertia of the beam section about the centroidal axis (the beam

section includes a slab length on each side of the beam equal to the projection of the

beam above or below the slab, whichever is greater, but not more than four times the

slab thickness)

the moment of inertia of the gross section of the slab.

However, the thickness of any slab shall not be less than the following:

1. For slabs with then thickness

2. For slabs with then thickness

If no beams are used, as in the case of flat plates, then and . The ACI Code

equations for calculating slab thickness, , take into account the effect of the span length,

the panel shape, the steel

reinforcement yield stress,

, and the flexural stiffness

of beams. When very stiff

beams are used, Eq. ( )

may give a small slab

thickness, and Eq. ( ) may

control. For flat plates and

flat slabs, when no interior

beams are used, the

minimum slab thickness

may be determined directly

from Table 9.5(c) of the ACI

Code, which is shown here.


240

Other ACI Code limitations are summarized as follows:

1. For panels with discontinuous edges, end beams with a minimum equal to must

be used; otherwise, the minimum slab thickness calculated by Eqs. ( ) and ( ) must be

increased by at least (ACI Code, Section 9.5.3).

2. When drop panels are used without beams, the minimum slab thickness may be

reduced by . The drop panels should extend in each direction from the centerline of

support a distance not less than one-sixth of the span length in that direction between

center to center of supports and also project below the slab at least . This reduction

is included in Table 9.5(c).

3. Regardless of the values obtained by Eqs. ( ) and ( ), the thickness of two-way slabs

shall not be less than the following:

(1) for slabs without beams or drop panels, ;

(2) for slabs without beams but with drop panels, ;

(3) for slabs with beams on all four sides with , , and

for , (ACI Code, Section 9.5.3.).

The thickness of a slab also may be governed by shear. This is particularly likely if large

moments are transferred to edge columns and for interior columns between two spans that

are greatly different in length. The selection of slab thicknesses to satisfy shear requirements

will be discussed later. Briefly, it is suggested that the trial slab thickness be chosen such that

at edge columns and at interior columns.


241

9.5 SLAB REINFORCEMENT REQUIREMENTS.

Placement Sequence.

In a flat plate or flat slab, the moments are larger in the slab strips spanning the long

direction of the panels. As a result, the reinforcement for the long span generally is placed

closer to the top and bottom of the slab than is the short-span reinforcement. This gives the

larger effective depth for the larger moment. For slabs supported on beams having

greater than about , the opposite is true, and the reinforcing pattern should be reversed.

If a particular placing sequence has been assumed in the reinforcement design, it should be

shown or noted on the drawings. It also is important to maintain the same arrangements of

layers throughout the entire floor, to avoid confusion in the field. Thus, if the east–west

reinforcement is nearer the top and bottom surfaces in one area, this arrangement should

be maintained over the entire slab, if at all possible.

Concrete Cover.

ACI Code Section 7.7.1 specifies the minimum clear cover to the surface of the

reinforcement in slabs as for and smaller bars, provided that the slab is not

exposed to earth or to weather.

Spacing Requirements, Minimum Reinforcement, and Minimum Bar Size.

ACI Code Section 13.3.1 requires that the minimum area of reinforcement provided for

flexure should not be less than (see page 195):

For slabs in which grade 280 ( ) or 350 ( ) deformed bars

are used, .

For slabs in which grade 420 ( ) deformed bars or welded bars or

welded wire fabric are used, .

The maximum spacing of reinforcement at critical design sections for positive and negative

moments in both the middle and column strips shall not exceed two times the slab thickness

(ACI Code Section 13.3.2), and the bar spacing shall not exceed at any location (ACI

Code Section 7.12.2.2).

Although there is no code limit on bar size, the top steel bars abd steps in slab should be

enough to give adequate rigidity to prevent displacement of the bars under ordinary foot

traffic before the concrete is placed.

Bar Cutoffs and Anchorages

For slabs without beams, ACI Code Section 13.3.8.1 allows the bars to be cut off as shown in

the figure below (ACI Code Fig. 13.3.8). Where adjacent spans have unequal lengths, the

extension of the negative-moment bars past the face of the support is based on the length

of the longer span.

ACI Code Section 13.3.3 requires that the Positive moment reinforcement perpendicular to a

discontinuous edge shall extend to the edge of slab and have embedment, straight or

hooked, at least in spandrel beams, columns, or walls.


242

ACI Code Section 13.3.4 requires that all negative-moment steel perpendicular to an edge be

bent, hooked, or otherwise anchored in spandrel beams, columns, and walls along the edge

to develop in tension. If there is no edge beam, this steel still should be hooked to act as

torsional reinforcement and should extend to the minimum cover thickness from the edge of

the slab.

ACI Code Section 13.3.6 requires that at exterior corners of slabs supported by edge walls or

where one or more edge beams have a value of greater than , top and bottom slab

reinforcement shall be provided at exterior corners in accordance with 13.3.6.1 through

13.3.6.4.

13.3.6.1 — Corner reinforcement in both top and bottom of slab shall be sufficient to

resist a moment per unit of width equal to the maximum positive moment per unit width in

the slab panel.

13.3.6.2 — The moment shall be assumed to be about an axis perpendicular to the

diagonal from the corner in the top of the slab and about an axis parallel to the diagonal

from the corner in the bottom of the slab.

13.3.6.3 — Corner reinforcement shall be provided for a distance in each direction

from the corner equal to one-fifth the longer span.

ACI 318 - Fig. 13.3.8 — Minimum extensions for reinforcement in slabs without beams. (See 12.11.1 for reinforcement extension into supports).


243

13.3.6.4 — Corner reinforcement shall be placed parallel to the diagonal in the top of

the slab and perpendicular to the diagonal in the bottom of the slab. Alternatively,

reinforcement shall be placed in two layers parallel to the sides of the slab in both the top

and bottom of the slab.

9.6 SHEAR STRENGTH OF TWO-WAY SLABS.

In a two-way floor system, the slab must have adequate thickness to resist both bending

moments and shear forces at the critical sections. To investigate the shear capacity of two-

way slabs, the following cases should be considered.


244

9.6.1 Two-Way Slabs Supported on Beams

In two-way slabs supported on beams, the critical sections are at a distance from the face

of the supporting beams, and the shear capacity of each section is

√ . When the supporting beams are stiff and are capable of transmitting

floor loads to the columns, they are assumed to carry loads acting on floor areas bounded by

lines drawn from the corners, as shown in the figure below. The loads on the trapezoidal

areas will be carried by the long beams and , whereas the loads on the triangular

areas will be carried by the short beams and . The shear per unit width of slab is

highest between and in both directions, and , where is the uniform

factored load per unit area.

If no shear reinforcement is provided, the shearing force at a distance from the face of the

beam, , must be equal to

√

where

(

)

9.6.2 Two-Way Slabs Without Beams

In flat plates and flat slabs, beams are not provided, and the slabs are directly supported by

columns. In such slabs, two types of shear stresses must be investigated; the first is one-way

shear, or beam shear. The critical sections are taken at a distance from the face of the


245

column, and the slab is considered as a wide beam spanning between supports, as in the

case of one-way beams. The shear capacity of the concrete section is

√ .

The second type of shear to be studied is two-

way, or punching, shear, as in the design of

footings. Shear failure occurs along a truncated

cone or pyramid around the column. The critical

section is located at a distance from the

face of the column, column capital, or drop

panel. The ACI Code, Section 11.11.2 allows a

shear strength, , in slabs and footings without

shear reinforcement for two-way shear action,

the smallest of

(

) √

(

) √

√

where Ratio of long side to short side of the rectangular column.

For shapes other than rectangular, is taken to be the ratio of the longest dimension of the

effective loaded area in the long direction to the largest width in the short direction

(perpendicular to the long direction).

perimeter of the critical section taken at from the loaded area.

effective depth of slab.

for normal-weight concrete.

is assumed to be:

for interior columns,

for edge columns, and

for corner columns.


246

When openings are located at

less than times the slab

thickness from a column, ACI

Code Section 11.11.6 requires

that the critical perimeter be

reduced, as shown below.

ACI Commentary suggested

that the side faces of the

critical perimeter would extend

to the edge of the slab if the

distance from the face of the

column to the edge of the slab

does not exceed the larger of:

(i) four slab thicknesses, , or

(ii) twice the development

length, , of the flexural

reinforcement perpendicular

to the edge, shown by the

distances labeled A and B in

the Figures (b) and (c).

𝑑

𝑑

𝑑

𝑑 𝛼𝑆 𝑑

𝑑

𝑑 𝛼𝑆

Edge of the slab

𝑑

𝑑

𝛼𝑆

Edge of the slab

Edge

of

the

slab


247

9.6.3 Tributary Areas for Shear in Two-Way Slabs.

For uniformly loaded two-way slabs, the tributary areas used to calculate are bounded by

lines of zero shear. For interior panels, these lines can be assumed to pass through the

center of the panel. For edge panels, lines of zero shear are approximately at to

from the center of the exterior column, where is the span measured from center-to-

center of the columns. However, to be conservative in design, ACI Code Section 8.3.3

requires that the exterior supports must resist a shear force due to loads acting on half of

the span . Also, to account for the larger tributary area for the first interior support, ACI

Code Section 8.3.3 requires that the shear force from loads acting on half of the span must

be increased by . This essentially results in a tributary length of .

9.6.4 Shear Reinforcement in Two-Way Slabs Without Beams.

In flat-slab and flat-plate floor systems, the thickness of the slab selected may not be

adequate to resist the applied shear stresses. In this case, either the slab thickness must be

increased or shear reinforcement must be provided. The ACI Code allows the use of shear

reinforcement by shearheads and anchored bars or wires (see next figures).

The design for shear (punching) reinforcement in flat plates will be discussed later in details

(see section 9.11).

Critical sections and tributary areas for shear in a flat plate.


248


249

9.7 ANALYSIS AND DESIGN OF TWO-WAY SLABS.

An exact analysis of forces and displacements in a two-way slab is complex, due to its highly

indeterminate nature; this is true even when the effects of creep and nonlinear behavior of

the concrete are neglected. Numerical methods such as finite elements can be used, but


250

simplified methods such as those presented by the ACI Code are more suitable for practical

design. The ACI Code, Chapter 13, assumes that the slabs behave as wide, shallow beams

that form, with the columns above and below them, a rigid frame. The validity of this

assumption of dividing the structure into equivalent frames has been verified by analytical

and experimental research. It is also established that factored load capacity of two-way slabs

with restrained boundaries is about twice that calculated by theoretical analysis, because a

great deal of moment redistribution occurs in the slab before failure. At high loads, large

deformations and deflections are expected; thus, a minimum slab thickness is required to

maintain adequate deflection and cracking conditions under service loads.

The ACI Code specifies two methods for the design of two-way slabs:

1. The direct design method. DDM (ACI Code, Section 13.6), is an approximate

procedure for the analysis and design of two-way slabs. It is limited to slab systems

subjected to uniformly distributed loads and supported on equally or nearly equally

spaced columns. The method uses a set of coefficients to determine the design

moments at critical sections. Two-way slab systems that do not meet the limitations

of the ACI Code, Section 13.6.1, must be analyzed by more accurate procedures.

2. The equivalent frame method, EFM (ACI Code, Section 13.7), is one in which a three-

dimensional building is divided into a series of two-dimensional equivalent frames by

cutting the building along lines midway between columns. The resulting frames are

considered separately in the longitudinal and transverse directions of the building

and treated floor by floor, as will be shown later.

The systems that do not meet the requirements permitting analysis by the "direct design

method" of the present code, has led many engineers to continue to use the design method

of the 1963 ACI Code (The coefficient method) for the special case of two-way slabs sup-

ported on four sides of each slab panel by relatively deep, stiff, edge beams. It has been

used extensively here since 1963 for slabs supported at the edges by walls, steel beams, or

monolithic concrete beams having a total depth not less than about 3 times the slab

thickness. While it was not a part of the 1977 or later ACI Codes, its continued use is

permissible under the current code provision (ACI Code 13.5.1) that a slab system may be

designed by any procedure satisfying conditions of equilibrium and geometric compatibility,

if it is shown that the design strength at every section is at least equal to the required

strength, and that serviceability requirements are met.

9.8 SLAB ANALYSIS BY THE COEFFICIENT METHOD.

The coefficient method makes use of tables of moment coefficients for a variety of

conditions. These coefficients are based on elastic analysis but also account for inelastic

redistribution. In consequence, the design moment in either direction is smaller by an

appropriate amount than the elastic maximum moment in that direction. The moments in

the middle strips in the two directions are computed from


251

and

where tabulated moment coefficients.

uniform load,

length of clear span in short and long directions respectively.

The method provides that each panel be divided in both directions into a middle strip whose

width is one-half that of the panel and two edge or column strips of one-quarter of the panel

width (see figure below). The moments in both directions are larger in the center portion of

the slab than in regions close to the edges. Correspondingly, it is provided that the entire

middle strip be designed for the full, tabulated design moment. In the edge strips this

moment is assumed to decrease from its full value at the edge of the middle strip to one-

third of this value at the edge of the panel. This distribution is shown for the moments in

the short span direction in figure below. The lateral variation of the long span moments

is similiar.

The discussion so far has been restricted to a single panel simply supported at all four edges.

An actual situation is shown in next figure, in which a system of beams supports a two-way

slab. It is seen that some panels, such as , have two discontinuous exterior edges, while the

other edges are continuous with their neighbors. Panel has one edge discontinuous and

three continuous edges, the interior panel has all edges continuous, and so on. At a

continuous edge in a slab, moments are negative, just as at interior supports of continuous

beams. Also, the magnitude of the positive moments depends on the conditions of

continuity at kall four edges.


252

Correspondingly, Table 1 gives moment coefficients , for negative moments at continuous

edges. The details of the tables are self-explanatory. Maximum negative edge moments are

obtained when both panels adjacent to the particular edge carry full dead and live load.

Hence the moment is computed for this total load. Negative moments at discontinuous

edges are assumed equal to one-third of the positive moments for the same direction. One

must provide for such moments because some degree of restraint is generally provided at

discontinuous edges by the torsional rigidity of the edge beam or by the supporting wall.

For positive moments there will be little, if any, rotation at the continuous edges if dead load

alone is acting, because the loads on both adjacent panels tend to produce opposite

rotations which cancel, or nearly so. For this condition, the continuous edges can be

regarded as fixed, and the appropriate coefficients for the dead load positive moments are

given in Table 2. On the other hand, the maximum live load positive moments are obtained

when live load is placed only on the particular panel and not on any of the adjacent panels.

In this case, some rotation will occur at all continuous edges. As an approximation it is

assumed that there is restraint for calculating these live load moments. The

corresponding coefficients are given in Table 3. Finally, for computing shear in the slab and

loads on the supporting beams, Table 4 gives the fractions of the total load that are

transmitted in the two directions.


253


254


255


256


257

Example (Design of two-way edge-supported solid slab):

A monolithic reinforced concrete floor is to be composed of rectangular bays

measuring , as shown. Beams of width and depth are provided on

all column lines; thus the clear-span dimensions for the two-way slab panels are

. The floor is to be designed to carry a service live load and a dead load

on the slab due to self-weight plus weight of:

• Tiles, .

• Mortar, .

• Sand, .

• Plaster, .

• Partitions, .

Given:

and .

Find the required slab thickness and

reinforcement for the corner panel shown.

Solution:

1. Minimum thickness (deflection

requirements):

For slabs of this type the first trial

thickness is often taken equal to

Check for the minimum thickness of the slab:

Exterior beam:

(

)

𝑚

𝑚

𝑚

𝑚

𝑚

𝑚

Corner

Panel

𝑠𝑙𝑎𝑏

𝑚


258

Interior beam:

(

)

Slab section for Exterior beam:

Short direction

( )

Long direction

( )

Slab section for Interior beam:

Short direction

Long direction


259

∑

the minimum slab thickness will be:

(

)

( )

First trial thickness . Take slab thickness

2. Loads calculation:

Quality Density

Material

Tiles

mortar

Sand

Reinforced Concrete solid slab

Plaster

Partitions

Total Dead Load

Dead Load of slab ,

Live Load of slab ,

𝑚

𝑚

𝑚

𝑚

Corner

Panel

𝛼𝑓

𝛼𝑓

𝛼𝑓

𝛼𝑓


260

3. Moments calculations:

and

The moment calculations will be done for the slab middle strip.

Negative moments at continuous edges (Table 1):

(

) and (

)

(

) (

)

(

) and (

)

(

) (

)

Positive moments (Table 2 and Table 3):

(

) and (

)

(

) (

)

(

) and (

)

(

) (

)

Continuous Edge

Discontinuous Edge

Dis

con

tin

uo

us

Edge

Co

nti

nu

ou

s Ed

ge

Case 4

𝑚

𝑚

𝑚

𝑚

𝑚 𝑚

𝑚

𝑚

Column strip - short

Column strip - short

middle strip - short

Co

lum

n s

trip

- lo

ng

Co

lum

n s

trip

- lo

ng

Mid

dle

str

ip -

lon

g


261

(

) and (

)

(

) (

)

(

) and (

)

(

) (

)

Negative moments at Discontinuous edges (

):

𝑚

𝑚

𝐾𝑁 𝑚 𝑚 Moments

𝑎 short direction

𝑏

lon

g d

irec

tio

n


262

4. Slab reinforcement:

Short direction:


Midspan:

( √

)

( √

)

Provide

Use then

Take

Note that in the edge strips the positive moment, and the corresponding steel reinforcement

area, is assumed to decrease from its full value at the edge of the middle strip to one-third of

this value at the edge of the panel, which will not be provided.

Continuous edge:


( √

)

( √

)


263

Provide

Use then

Take

Discontinuous edge.

The negative moment at the discontinuous edge is one-third the positive moment in the

span.

Provide

Take

Long direction.

Design for positive and negative moment as in the short direction.

Note that the effective depth for the long direction will be

5. Check for shear:

(

) and (

)

(

) (

)

(

) and (

)

(

) (

)

The reactions of the slab are calculated from Table 4, which indicates that of the load is

transmitted in the short direction and in the long direction.

The total load on the panel being

The load per meter on the long beam is , and

The load per meter on the short beam is .

The shear to be transmitted by the slab to these beams is numerically equal to these beam

loads, reduced to a critical section a distance from the beam face. The shear strength of

the slab is


264

√

√

- for shear.


at the face of support, at distance from the face of support will be

smaller.

Even, if


Example:

A monolithic reinforced concrete solid slab is to be composed of rectangular bays as shown.

Beams of width and depth are provided on all column lines. The floor is to be

designed to carry a service live load and a dead load on the slab due to self-

weight plus weight of:

• Tiles, .

• Mortar, .

• Sand, .

• Plaster, .

• Partitions, .

Given: and .

Find the required slab thickness and reinforcement for the solid slab shown.

Solution:

1. Minimum thickness (deflection requirements):

For slabs of this type the first trial thickness is often taken equal to

Take As was required in the previous example.

Check for the minimum thickness of the slab (for

shaded panel):

Exterior beam:

(

)


265

6.6 m 6.6 m 6.6 m

8.6

m8.1

m8.1

m

6.0 m 6.0 m 6.0 m

8.0

m7.5

m7.5

m

B 1 (60 x 60 cm) B 2 (60 x 60 cm) B 3 (60 x 60 cm)

B 4 (60 x 60 cm) B 5 (60 x 60 cm) B 6 (60 x 60 cm)

B 7 (60 x 60 cm) B 8 (60 x 60 cm) B 9 (60 x 60 cm)

B 10 (60 x 60 cm) B 11 (60 x 60 cm) B 12 (60 x 60 cm)

B 1

3 (

60

x 6

0 c

m)

B 1

4 (

60

x 6

0 c

m)

B 1

5 (

60

x 6

0 c

m)

B 1

6 (

60

x 6

0 c

m)

B 1

7 (

60

x 6

0 c

m)

B 1

8 (

60

x 6

0 c

m)

B 1

9 (

60

x 6

0 c

m)

B 2

0 (

60

x 6

0 c

m)

B 2

1 (

60

x 6

0 c

m)

B 2

2 (

60

x 6

0 c

m)

B 2

3 (

60

x 6

0 c

m)

B 2

4 (

60

x 6

0 c

m)

𝐶𝑎 𝐷 𝐶𝑏 𝐷

𝐶𝑎 𝐿 𝐶𝑏 𝐿

𝐶𝑎 𝑛𝑒𝑔


𝑙𝑎 𝑙𝑏

𝐶𝑏 𝑛𝑒𝑔



𝐶𝑏 𝑛𝑒𝑔 𝐶𝑏 𝑛𝑒𝑔

𝑙𝑎 𝑙𝑏 𝑙𝑎 𝑙𝑏





𝑙𝑎 𝑙𝑏 𝑙𝑎 𝑙𝑏 𝑙𝑎 𝑙𝑏

𝐶𝑏 𝑛𝑒𝑔 𝐶𝑏 𝑛𝑒𝑔 𝐶𝑏 𝑛𝑒𝑔













𝑙𝑎 𝑙𝑏 𝑙𝑎 𝑙𝑏 𝑙𝑎 𝑙𝑏 𝐶𝑎 𝑛𝑒𝑔










𝛼

𝛼

𝛼

𝛼


266

Interior beam:

(

)


Short direction

( )

Long direction

( )


Short direction

Long direction

(

)


267

∑


(

)

( )

First trial thickness . Take slab thickness


Quality Density

Material

Tiles

mortar

Sand


Plaster

Partitions

Total Dead Load

Dead Load of slab ,

Live Load of slab ,

𝑚

𝑚

Corner

Panel

𝛼𝑓

𝛼𝑓

𝛼𝑓

𝛼𝑓


268


and

All negative and positive coefficients are shown on the slab plane (page 265).


):

6.6 m 6.6 m 6.6 m

8.6

m8.1

m8.1

m

6.0 m 6.0 m 6.0 m

8.0

m7.5

m7.5

m

B 1 (60 x 60 cm) B 2 (60 x 60 cm) B 3 (60 x 60 cm)

B 4 (60 x 60 cm) B 5 (60 x 60 cm) B 6 (60 x 60 cm)

B 7 (60 x 60 cm) B 8 (60 x 60 cm) B 9 (60 x 60 cm)

B 10 (60 x 60 cm) B 11 (60 x 60 cm) B 12 (60 x 60 cm)

B 1

3 (

60

x 6

0 c

m)

B 1

4 (

60

x 6

0 c

m)

B 1

5 (

60

x 6

0 c

m)

B 1

6 (

60

x 6

0 c

m)

B 1

7 (

60

x 6

0 c

m)

B 1

8 (

60

x 6

0 c

m)

B 1

9 (

60

x 6

0 c

m)

B 2

0 (

60

x 6

0 c

m)

B 2

1 (

60

x 6

0 c

m)

B 2

2 (

60

x 6

0 c

m)

B 2

3 (

60

x 6

0 c

m)

B 2

4 (

60

x 6

0 c

m)

+25.1

+16

.2

42.4

2

7.1

8.4

5

.4

Moments, 𝐾𝑁 𝑚 𝑚

+25.1

+16

.2

42.4

2

7.1

8.4

5

.4

+19.9

+11

.5

44.5

1

5.9

44.5

3

.8

+27.5

+14

.9

45.4

2

5.5

9.2

5

+27.5

+14

.9

45.4

2

5.5

9.2

5

+21.5

+9.6

46.6

1

4.9

46.6

3

.2

+21.5

+15

.3

32.9

3

8.3

7.2

3

8.3

+21.5 +1

5.3

32.9

38

.3

7.2

3

8.3

+18.6

+12

.2

38.8

2

5.2

38.8

2

5.2


269


Assume bar diameter .

Find corresponding to :

(

) (

)

Check for strain :

(

) (

)

Take


(

) (

)

Check for strain :

(

) (

)

Take


(

) (

)

Check for strain :


270

(

) (

)

Take

6.6 m 6.6 m 6.6 m

8.6

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

8.0

m7

.5 m

7.5

m

T 10@15 cm c/c T 12@15 cm c/c

L=500 cm

T 12@15 cm c/c

L=500 cm

T 10@15 cm c/c

L=300 cm

B 10@15 cm c/c

L=750 cm

B 10@15 cm c/c

L=750 cmB 10@15 cm c/c

L=750 cm

T 10@15 cm c/c

L=300 cm

T 12@15 cm c/c

L=500 cm

T 12@15 cm c/c

L=500 cm

T 10@15 cm c/c

L=300 cm

B 10@15 cm c/c

L=750 cm

B 10@15 cm c/c

L=750 cmB 10@15 cm c/c

L=750 cm

T 10@15 cm c/c

L=300 cm

T 14@15 cm c/c

L=500 cm

T 14@15 cm c/c

L=500 cm

T 10@15 cm c/c

L=300 cm

B 10@15 cm c/c

L=750 cm

B 10@15 cm c/c

L=750 cmB 10@15 cm c/c

L=750 cm

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

950 c

m

T

12@

15 c

m c

/c

L=

600 c

m

T

12@

15 c

m c

/c

L=

600 c

m

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

900 c

m

B

10@

15 c

m c

/c

L=

900 c

m

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

950 c

m

T

10@

15 c

m c

/c

L=

600 c

m

T

10@

15 c

m c

/c

L=

600 c

m

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

900 c

m

B

10@

15 c

m c

/c

L=

900 c

m

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

950 c

m

T

12@

15 c

m c

/c

L=

600 c

m

T

12@

15 c

m c

/c

L=

600 c

m

T

10@

15 c

m c

/c

L=

300 c

m

B

10@

15 c

m c

/c

L=

900 c

m

B

10@

15 c

m c

/c

L=

900 c

m

200

Typical section in solid slab

L=300 cm


271

5. Check for shear:

The maximum shear coefficient is (

) , from case 9.

The reactions of the slab are calculated from Table 4, which indicates that of the load is

transmitted in the short direction.


The load per meter on the long beam is ,

The shear to be transmitted by the slab to these beams is numerically equal to these beam

loads, reduced to a critical section a distance from the beam face. The shear strength of

the slab is

√

√

- for shear.


at the face of support, at distance from the face of support will be

smaller and can be calculated from

Another way (simply supported strip) to calculate the shear at distance from the face

of support

(

) (

)

Even, if


For the two methods of calculating the thickness of the slab is adequate enough.

6. Design of Beams 16-17-18 ( ):

The tributary load area for beams 16, 17, 18 from the solid slab is as shown (shaded area). In

addition to factored load acting on the beam from the solid slab, the own weight of the

beam must be added.

Load calculation for the beams 16-17-18:

No additional loads applied directly on the beams.

8.6 m 8.1 m 8.1 m

𝐷𝐿 𝐾𝑁 𝑚

𝐷𝐿 𝐾𝑁 𝑚 𝐿𝐿 𝐾𝑁 𝑚



B 16 B 17 B 18




272

Using software programs such as BEAMD to analyze the beam and to get shear and moment

envelopes, and then design it.

6.6 m 6.6 m 6.6 m

8.6

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

8.0

m7

.5 m

7.5

m

1.5

m

3.0

m3

.0 m

2.0

m

3.0

m3

.0 m

2.0

m

3.0

m3

.0 m

2.0

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

B 1

6 (

60

x6

0 c

m)

B 1

7 (

60

x6

0 c

m)

B 1

8 (

60

x6

0 c

m)

45°45° 45°45°

45°45°


273


274

Example (Design of two-way edge-supported ribbed slab):

A monolithic reinforced concrete ribbed slab is to be composed of rectangular bays as

shown. Beams of width and depth are provided on all column lines. The floor

is to be designed to carry a service live load and a dead load on the slab due to

self-weight plus weight of:

• Tiles, .

• Mortar, .

• Sand, .

• Plaster, .

• Partitions, .

Given: and .

Find the required slab thickness and reinforcement for the ribbed slab shown.

Solution:

1. Minimum thickness (deflection requirements): Assume the thickness for the shown

ribbed slab

Check for the minimum thickness of the slab (for the shaded corner slab):

All Exterior and Interior beams have rectangular section of 60 cm width and 52 cm

depth:


The moment of inertia for the ribbed slab is the sum of moment of inertia of T-section ribs

within a distance ( )

was defined as in one-way ribbed

slab design ( )


275

Typical section in ribbed slab

6.6 m 6.6 m 6.6 m

8.1

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

7.5

m7

.5 m

7.5

m

B 1

6 (

60x52 c

m)

B 1

7 (

60x52 c

m)

B 1

8 (

60x52 c

m)

B1 (60x52 cm) B2 (60x52 cm) B3 (60x52 cm)

B4 (60x52 cm) B5 (60x52 cm) B6 (60x52 cm)

B7 (60x52 cm) B8 (60x52 cm) B9 (60x52 cm)

B10 (60x52 cm) B11 (60x52 cm) B12 (60x52 cm)

B 1

3 (

60x52 c

m)

B 1

4 (

60x52 c

m)

B 1

5 (

60x52 c

m)

B 1

9 (

60x52 c

m)

B 2

0 (

60x52 c

m)

B 2

1 (

60x52 c

m)

B 2

2 (

60x52 c

m)

B 2

3 (

60x52 c

m)

B 2

4 (

60x52 c

m)


276

Short direction

Long direction


Short direction

(

)

(

)

Long direction

(

)

(

)

∑


(

)

( )

( )

7.5

m

6 m

Corner

Panel

𝛼𝑓

𝛼𝑓

𝛼𝑓

𝛼𝑓


277

First trial thickness .

Take slab thickness


Quality Density

Material

Tiles

mortar

Sand

Reinforced Concrete Topping

Reinforced Concrete Rib

Concrete Block

Plaster

Partitions

Total Dead Load,

Dead Load of slab:

Live Load of slab:


and

All negative and positive coefficients are shown on the slab plane (page 278).


):


278

Note that all moments and are for strip. For one rib ( )

6.6 m 6.6 m 6.6 m

8.1

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

7.5

m7

.5 m

7.5

m




















𝐶𝑎 𝐷

𝐶𝑏 𝐷


𝐶𝑎 𝐷

𝐶𝑏 𝐷


𝐶𝑎 𝐷

𝐶𝑏 𝐷















279

6.6 m 6.6 m 6.6 m

8.1

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

7.5

m7

.5 m

7.5

m

B 1 (60 x 60 cm) B 2 (60 x 60 cm) B 3 (60 x 60 cm)

B 4 (60 x 60 cm) B 5 (60 x 60 cm) B 6 (60 x 60 cm)

B 7 (60 x 60 cm) B 8 (60 x 60 cm) B 9 (60 x 60 cm)

B 10 (60 x 60 cm) B 11 (60 x 60 cm) B 12 (60 x 60 cm)

B 1

3 (

60 x

60 c

m)

B 1

4 (

60 x

60 c

m)

B 1

5 (

60 x

60 c

m)

B 1

6 (

60 x

60 c

m)

B 1

7 (

60 x

60 c

m)

B 1

8 (

60 x

60 c

m)

B 1

9 (

60 x

60 c

m)

B 2

0 (

60 x

60 c

m)

B 2

1 (

60 x

60 c

m)

B 2

2 (

60 x

60 c

m)

B 2

3 (

60 x

60 c

m)

B 2

4 (

60 x

60 c

m)

+13.71

+8.8

4

23.23

1

4.8

2

4.57

2

.95

Moments, 𝐾𝑁 𝑚 𝑟𝑖𝑏

+13.71

+8.8

4

23.23

1

4.8

2

4.57

2

.95

+10.85

+6.2

6

24.54

8

.69

24.54

2

.09

+13.71

+8.8

4

23.23

1

4.8

2

4.57

2

.95

+13.71

+8.8

4

23.23

1

4.8

2

4.57

2

.95

+10.85

+6.2

6

24.54

8

.69

24.54

2

.09

+11.72

+8.3

2

18

2

0.9

6

3.91

2

0.9

6

+11.72 +8

.32

18

20

.96

3.91

2

0.9

6

+18.09

+6.6

1

21.27

1

3.8

21.27

1

3.8


280


The design can be done directly for the negative moment or through section analysis with

assumed bar diameter and step.

Design for negative moment


( √

)

( √

)

Check for

√

√

Use with

Check for strain:

(

) (

)

Analysis the T-section rib for different two bars ( )

Assume bottom bars (+ve moment).


281

(

) (

)

Check for strain :

(

) (

)

Take

All panels will be reinforced with bottom bars in both directions except for the middle

panel in the short direction where will be used bottom bars as shown in the next

figure.

Assume top bars (–ve moment).

(

) (

)

Check for strain :

(

) (

)

Take


(

) (

)

Check for strain :

(

) (

)

Take


282

6.6 m 6.6 m 6.6 m

8.1

m8

.1 m

8.1

m6.0 m 6.0 m 6.0 m

7.5

m7

.5 m

7.5

m

L=500 cm L=500 cm L=300 cm

L=300 cm L=500 cm L=500 cm L=300 cm

L=300 cm

T 2 14

L=500 cm L=500 cm L=300 cm

B 2 10

L=750 cm

L

=6

00

cm

L

=6

00

cm

L

=3

00

cm

T 2

1

0

L

=3

00

cm B

2 1

0

L

=9

00

cm

L

=6

00

cm

L

=6

00

cm

L

=3

00

cm

L

=6

00

cm

T 2

1

2

L

=6

00

cm

L

=3

00

cm

Typical section in ribbed slab

32

cm

12 cm

24

cm

8 c

m

40 cm

8 @20 cm

2 10 or 12 or 14

2 10

T 2

1

0

L

=3

00

cm

T 2

1

0

L

=3

00

cm B

2 1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2

1

0

L

=9

00

cm

B 2 10

L=750 cm

B 2 10

L=750 cm

B 2 10

L=750 cmB 2 12

L=750 cm

B 2 10

L=750 cm

B 2 10

L=750 cmB 2 10

L=750 cm

B 2 10

L=750 cm

T 2 10

T 2 10

T 2 10 T 2 10

T 2 10

T 2 10

L=300 cm

T 2

1

0

T 2

1

0

T 2

1

0

T 2 14

T 2 14 T 2 14

T 2

1

2

T 2

1

2

T 2

1

2T

2 1

2

T 2

1

2T 2 12T 2 12


283


(

) (

)

Check for strain :

(

) (

)

Take

5. Design for shear:

Maximum shear coefficient will be in the short direction for the slab with boundary

conditions as in case 9.


The load per rib at face of the long beam is .

The shear critical section is at distance from the beam face:

The shear strength of one rib in the slab is

√

√

- for shear.

No need for shear reinforcement (exception for joist constructions)

The shear in the slab can be calculated using tributary

area for shear (as simply supported strip):

(

)

Design the rib for shear assuming that the critical shear

in the rib is .

3 m 3 m

3 m

3

m

1.5

m


284

√

√

( )

Provide minimum shear reinforcement.

Use for stirrups

(

)

(

)

Use for the distance of 1 m from the face of support, and

in the middle space. Note that the shear force at distance from the

face of support is , so no

shear reinforcement is required for the middle space.


The tributary load area for beams 16, 17, 18 from ribbed slab is as shown (shaded area). In

addition to factored load acting on the beam from the ribbed slab, the own weight of the

beam must be added.


Weight of the floor materials acting directly on the beam

Material

Tiles

mortar

Sand

Plaster

Partitions

Total Dead Load,

No additional loads applied directly on the beams (such as walls).


285

6.6 m 6.6 m 6.6 m

8.1

m8

.1 m

8.1

m

6.0 m 6.0 m 6.0 m

7.5

m7

.5 m

7.5

m

B 1

6 (

60

x5

2 c

m)

B 1

7 (

60

x5

2 c

m)

B 1

8 (

60

x5

2 c

m)

B1 (60x52 cm) B2 (60x52 cm) B3 (60x52 cm)

B4 (60x52 cm) B5 (60x52 cm) B6 (60x52 cm)

B7 (60x52 cm) B8 (60x52 cm) B9 (60x52 cm)

B10 (60x52 cm) B11 (60x52 cm) B12 (60x52 cm)

B 1

3 (

60

x5

2 c

m)

B 1

4 (

60

x5

2 c

m)

B 1

5 (

60

x5

2 c

m)

B 1

9 (

60

x5

2 c

m)

B 2

0 (

60

x5

2 c

m)

B 2

1 (

60

x5

2 c

m)

B 2

2 (

60

x5

2 c

m)

B 2

3 (

60

x5

2 c

m)

B 2

4 (

60

x5

2 c

m)

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

B 1

7 (

60

x5

2 c

m)

B 1

8 (

60

x5

2 c

m)

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

1.5

m

3.0

m3

.0 m

45° 45°


286

Total service on the beam and


Using software programs such as BEAMD to analyze and design the beams and to get

reinforcement details.


The tributary load area for beams 7, 8, 9 from ribbed slab is as shown (shaded area). In


beam must be added.


No additional loads applied directly on the beams (such as walls).





𝑚 𝑚 𝑚


𝐿𝐿 𝐾𝑁 𝑚








6.6 m 6.6 m 6.6 m







𝐿𝐿 𝐾𝑁 𝑚 𝐷𝐿 𝐾𝑁 𝑚



287


The tributary load area beams 22, 23, 24 from ribbed slab is as shown (shaded area). In


beam must be added and the load of the RC wall ( thick, height).


Note here that the width of the beam for the floor materials and live load on the beam will

be smaller because of the thickness of the exterior wall ( )

Weight of the floor materials acting directly on the beam

Material

Tiles

mortar

Sand

Plaster

Partitions

Total Dead Load,





𝑚 𝑚 𝑚







𝐿𝐿 𝐾𝑁 𝑚 𝐷𝐿 𝐾𝑁 𝑚



288


289


290


291


292


293


294

9.9 SLAB ANALYSIS BY THE DIRECT DESIGN METHOD (DDM).

The direct-design method also could have been called “the direct-analysis method,” because

this method essentially prescribes values for moments in various parts of the slab panel

without the need for a structural analysis. The reader should be aware that this design

method was introduced in an era when most engineering calculations were made with a

slide rule and computer software was not available to do the repetitive calculations required

to analyze a continuous-floor slab system. Thus, for continuous slab panels with relatively

uniform lengths and subjected to distributed loading, a series of moment coefficients were

developed that would lead to safe flexural designs of two-way floor systems.

9.9.1 Limitations on the Use of the Direct-Design Method.

The direct-design method is easier to use than the equivalent-frame method, but can be

applied only to fairly regular multipanel slabs. The limitations, given in ACI Code Section

13.6.1, include the following:

1. There must be a minimum of three continuous spans in each direction. Thus, a

ninepanel structure ( ) is the smallest that can be considered. If there are fewer

than three panels, the interior negative moments from the direct-design method

tend to be too small.

2. Rectangular panels must have a long-span/short-span ratio that is not greater than .

One-way action predominates as the span ratio reaches and exceeds .

3. Successive span lengths in each direction shall not differ by more than one third of

the longer span. This limit is imposed so that certain standard reinforcement cutoff

details can be used.

4. Columns may be offset from the basic rectangular grid of the building by up to 0.1

times the span parallel to the offset. In a building laid out in this way, the actual

column locations are used in determining the spans of the slab to be used in

calculating the design moments.

5. All loads must be due to gravity only and uniformly distributed over an entire panel.

The direct-design method cannot be used for unbraced, laterally loaded frames,

foundation mats, or prestressed slabs.

6. The service (unfactored) live load shall not exceed two times the service dead load.

Strip or checkerboard loadings with large ratios of live load to dead load may lead to

moments larger than those assumed in this method of analysis.

7. For a panel with beams between supports on all sides, the relative stiffness of the

beams in the two perpendicular directions given by ( ) (

) shall not be less

than or greater than . The term was defined in the prior section, and and

are the spans in the two directions.

Limitations and do not allow use of the direct-design method for slab panels that

transmit load as one-way slabs.


295

9.9.2 Column and middle strips.

In both direct design and equivalent frame methods, a typical panel is divided, for purposes

of design, into column strips and middle strips. A column strip is defined as a strip of slab

having a width on each side of the column centerline equal to one-fourth the smaller of the

panel dimensions and . Such a strip includes column-line beams, if present. A middle

strip is a design strip bounded by two column strips. In all cases, , is defined as the span in

the direction of the moment analysis and as the span in the lateral direction measured

center to center of the support. In the case of monolithic construction, beams are defined to

include that part of the slab on each side of the beam extending a distance equal to the

projection of the beam above or below the slab (whichever is greater) but not greater

than 4 times the slab thickness (see Section 9.4).


296

9.9.3 Total Static Moment at Factored Loads.

For purposes of calculating the total static moment in a panel, the clear span in the

direction of moments is used. The clear span is defined to extend from face to face of the

columns, capitals, brackets, or walls but is not to be less than . The total factored

moment in a span, for a strip bounded laterally by the centerline of the panel on each side of

the centerline of supports, is

The face of the support where the

negative moments should be

calculated is illustrated in the figure

below. The length is measured in a

direction perpendicular to and

equals the direction between center

to center of supports (width of slab).

The total moment calculated in

the long direction will be referred to

here as and that in the short

direction, as .

Once the total moment, , is

calculated in one direction, it is

divided into a positive moment,

, and a negative moment, ,

such that . Then

each moment, and , is

distributed across the width of the slab between the column and middle strips.

9.9.4 Assignment of positive and negative moments.

For interior spans, the total static moment is apportioned between the critical positive and

negative bending sections according to the following ratios:

Negative factored moment:

Positive factored moment:

The critical section for negative bending is taken at the face of rectangular supports, or at

the face of an equivalent square support having the same cross-sectional area as a round

support.

In the case of end spans, the apportionment of the total static moment among the three

critical moment sections (interior negative, positive, and exterior negative, as illustrated by

the figure below) depends upon the flexural restraint provided for the slab by the exterior

column or the exterior wall, as the case may be, and depends also upon the presence or

absence of beams on the column lines. ACI Code 13.6.3 specifies five alternative sets of

moment distribution coefficients for end spans, as shown in the next table and figure.


297


298

9.9.5 Lateral Distribution of Moments (between Column Strips and Middle Strips)

Having distributed the moment to the positive and negative-moment sections as just

described, the designer still must distribute these design moments across the width of the

critical sections. For design purposes, it is convenient to consider the moments constant

within the bounds of a middle strip or column strip unless there is a beam present on the

column line. In the latter case, because of its greater stiffness, the beam will tend to take a

larger share of the column-strip moment than the adjacent slab. The distribution of total

negative or positive moment between slab middle strips, slab column strips, and beams

depends upon the ratio , the relative stiffness of the beam and the slab, and the degree

of torsional restraint provided by the edge beam.

A convenient parameter defining the relative stiffness of the beam and slab spanning in

either direction is

in which and are the moduli of elasticity of the beam and slab concrete (usually the

same) and and are the moments of inertia of the effective beam and the slab.

Subscripted parameters and are used to identify computed for the directions of

, and , respectively.

The flexural stiffnesses of the beam and slab may be based on the gross concrete section,

neglecting reinforcement and possible cracking, and variations due to column capitals and

drop panels may be neglected. For the beam, if present, and for the slab, are

defined as in Section 9.4.

The relative restraint provided by the torsional resistance of the effective transverse edge

beam is reflected by the parameter , defined as

where , as before, is calculated for the slab spanning in direction , and having width

bounded by panel centerlines in the direction. The constant pertains to the torsional

rigidity of the effective transverse beam, which is defined according to ACI Code 13.7.5 as

the largest of the following:

1. A portion of the slab having a width equal to that of the column, bracket, or capital in

the direction in which moments are taken,

2. The portion of the slab specified in 1 plus that part of any transverse beam above and

below the slab,

3. The transverse beam defined as in Section 9.4.

The constant is calculated by dividing the section into its component rectangles, each

having smaller dimension and larger dimension , and summing the contributions of all

the parts by means of the equation

∑ (

)

The subdivision can be done in such a way as to maximize .


299

With these parameters defined, ACI Code 13.6.4 distributes the negative and positive

moments between column strips and middle strips, assigning to the column strips the

percentages of positive and negative moments shown in the next table. Linear interpolations

are to be made between the values shown.

Example (Calculation of Moments in an Interior Panel of a Flat Plate).

An interior panel of a flat-plate floor in an apartment building shown below. The slab

thickness is . The slab supports a design live load of and a superimposed

dead load of for partitions. The columns and slab have the same strength of

concrete. The story height is . Compute the column-strip and middle-strip moments

in the short direction of the panel.

Solution:

1. Compute the factored loads.

2. Compute the moments in the short span of the slab.

(a) Compute and and divide the slab into column and middle strips.

The column strip extends the smaller of or on each side of the column

centerline (ACI Code Section 13.2.1). Thus, the column strip extends on each

side of column centerline. The total width of the column strip is . Each half-middle strip

extends from the edge of the column strip to the centerline of the panel. The total width of

two half-middle strips is .


300

(b) Compute .

(c) Divide into negative and positive moments. From ACI Code Section 13.6.3.2,

Negative moment

Positive moment

This process is illustrated in the figure (a) below, and the resulting distribution of total

moments is shown in figure (b).


301

(d) Divide the moments between the column and middle strips.

Negative moments: From Table (page 299) for ( because there

are no beams between columns and in this panel),

Half of this

goes to each adjacent half-middle strip. Because the

adjacent bays have the same width, , a similar moment will be assigned to the other half of

each middle strip so that the total middle-strip negative moment is .

Positive moments: From Table (page 299), where

These calculations are illustrated in figure (a) below. The resulting distributions of moments

in the column strip and middle strip are summarized in figure (c).


302

In the next figure, the moments in each strip have been divided by the width of that strip.

3. Compute the moments in the long span of the slab. Although it was not asked for in

this example, in a slab design, it now would be necessary to repeat steps 2(a) to 2(d)

for the long span.

4. Design the middle and the column strips for these moments and find the

reinforcement area for each strip.

9.10 SLAB ANALYSIS BY THE EQUIVALENT FRAME METHOD (EFM).

The ACI Code presents two general methods for calculating the longitudinal distribution of

moments in two-way slab systems. These are the direct-design method (presented in the

previous section) and equivalent-frame methods, which are presented in this section.

Equivalent-frame methods are intended for use in analyzing moments in any practical slab–

column frame. Their scope is thus wider than the direct-design method, which is subject to

the limitations presented in Section 9.9.1 (ACI Code Section 13.6.1). In the direct-design

method, the statical moment, is calculated for each slab span. This moment is then divided

between positive- and negative-moment regions using arbitrary moment coefficients, which

are adjusted to reflect pattern loadings. For equivalent-frame methods, a stiffness analyses

of a slab–column frame is used to determine the longitudinal distribution of bending

moments, including possible pattern loadings. The transverse distribution of moments to

column and middle strips, as defined in the prior section, is the same for both methods.

The design requirements can be explained as follows.

1. Description of the equivalent frame: An equivalent frame is a two-dimensional

building frame obtained by cutting the three-dimensional building along lines

midway between columns (see the figure below). The resulting equivalent frames are

considered separately in the longitudinal and transverse directions of the building.


303

For vertical loads, each floor is analyzed separately, with the far ends of the upper

and lower columns assumed to be fixed. The slab-beam may be assumed to be fixed

at any support two panels away from the support considered, because the vertical

loads contribute very little to the moment at that support. For lateral loads, the

equivalent frame consists of all the floors and extends for the full height of the

building, because the forces at each floor are a function of the lateral forces on all

floors above the considered level. Analysis of frames can also be made using

computer programs.

2. Load assumptions: When the ratio of the service live load to the service dead load is

less than or equal to , the structural analysis of the frame can be made with the

factored dead and live loads acting on all spans instead of a pattern loading. When

the ratio of the service live load to the service dead load is greater than , pattern

loading must be used, considering the following conditions:


304

a. Only of the full-factored live load may be used for the pattern loading analysis.

b. The maximum negative bending moment in the slab at the support is obtained by loading only the two adjacent spans.

c. The maximum positive moment near a midspan is obtained by loading only alternate spans.

d. The design moments must not be less than those occurring with a full-factored live load on all panels (ACI Code, Section 13.7.6).

e. The critical negative moments are considered to be acting at the face of a rectangular column or at the face of the equivalent square column having the same area for nonrectangular sections.

3. Slab-beam moment of inertia: The ACI Code specifies that the variation in moment

of inertia along the longitudinal axes of the columns and slab beams must be taken

into account in the analysis of frames. The critical region is located between the

centerline of the column and the face of the column, bracket, or capital. This region

may be considered as a thickened section of the floor slab. To account for the large

depth of the column and its reduced effective width in contact with the slab beam,

the ACI Code, Section 13.7.3.3, specifies that the moment of inertia of the slab beam

between the center of the column and the face of the support is to be assumed equal

to that of the slab beam at the face of the column divided by the quantity

— , where is the column width in the transverse direction and is the

width of the slab beam. The area of the gross section can be used to calculate the

moment of inertia of the slab beam.

4. Column moment of inertia: The ACI Code, Section 13.7.4, states that the moment of

inertia of the column is to be assumed infinite from the top of the slab to the bottom

of the column capital or slab beams.


305

5. Column stiffness, , is defined by

∑

∑

where ∑ is the sum of the stiffness of the upper and lower columns at their ends,

∑

(

)

∑ (

)

6. Column moments: In frame analysis, moments determined for the equivalent

columns at the upper end of the column below the slab and at the lower end of the

column above the slab must be used in the design of a column.

7. Negative moments at the supports: The ACI Code, Section 13.7.7, states that for an

interior column, the factored negative moment is to be taken at the face of the

column or capital but at a distance not greater than from the center of the

column. For an exterior column, the factored negative moment is to be taken at a

section located at half the distance between the face of the column and the edge of

the support. Circular section columns must be treated as square columns with the

same area.

8. Sum of moments: A two-way slab floor system that satisfied the limitations of the

direct design method can also be analyzed by the equivalent frame method. To

ensure that both methods will produce similar results, the ACI Code, Section 13.7.7,

states that the computed moments determined by the equivalent frame method may

be reduced in such proportion that the numerical sum of the positive and average

negative moments used in the design must not exceed the total statical moment, .

9.11 SHEAR DESIGN IN FLAT PLATES.

Shear strength of two-way slabs was discussed in section 9.6.

Shear Reinforcement for Two-Way Slabs, two types of shear reinforcement will be discussed

only:

Stirrups:

ACI Code Section 11.11.3 allows the use of single-leg, multiple-leg and closed stirrups,

provided there are longitudinal bars in all corners of the stirrups.

Stirrups are allowed in slabs with effective depths, , that exceed the larger of or

16 times the stirrup diameter. The precision required to bend and place either the closed

stirrups or the multiple-leg stirrups makes these types of shear reinforcement labor-

intensive and expensive. As a result, shear reinforcement consisting of stirrups or bent

reinforcement is not used widely.


306

According to ACI Code Sections:

11.11.3.1 — shall be computed by Eq. (11-2) ( ), where shall not be taken

greater than

√ , and shall be calculated in accordance with 11.4.

In Eq. (11-15) (

), shall be taken as the cross-sectional area of all legs

of reinforcement on one peripheral line that is geometrically similar to the

perimeter of the column section.

11.11.3.2 — shall not be taken greater than

√ .

11.11.3.3 — The distance between the column face and the first line of stirrup legs that

surround the column shall not exceed . The spacing between adjacent stirrup legs in the

first line of shear reinforcement shall not exceed measured in a direction parallel to the

column face. The spacing between successive lines of shear reinforcement that surround the

column shall not exceed measured in a direction perpendicular to the column face

Headed Shear Studs:

The headed shear studs shown in the figure below are permitted by the ACI Code Section

11.11.5. They act in the same mechanical manner as a stirrup leg, but the head of the shear

stud is assumed to provide better anchorage than a bar hook.

Headed shear-stud reinforcement at a slab-column connection consists of rows of vertical

rods, each with a circular head or plate welded or forged on the top end, as shown. These

rows are placed to extend out from the corners of the column. To aid in the handling and


307

placement of the shear studs and to anchor the lower ends of the studs, they generally are

shop-welded to flat steel bars at the desired spacing. The vertical rods are referred to as

headed shear reinforcement or headed shear studs. The assembly of studs plus the bar is

called a stud rail.

ACI Section 11.11.5 — Headed shear stud reinforcement, placed perpendicular to the plane

of a slab or footing, shall be permitted in slabs and footings in accordance with 11.11.5.1

through 11.11.5.4. The overall height of the shear stud assembly shall not be less than the

thickness of the member less the sum of:

(1) the concrete cover on the top flexural reinforcement;

(2) the concrete cover on the base rail; and

(3) one-half the bar diameter of the tension flexural reinforcement.

Where flexural tension reinforcement is at the bottom of the section, as in a footing, the

overall height of the shear stud assembly shall not be less than the thickness of the member

less the sum of:

(1) the concrete cover on the bottom flexural reinforcement;

(2) the concrete cover on the head of the stud; and

(3) one-half the bar diameter of the bottom flexural reinforcement.

11.11.5.1 — For the critical section defined in 11.11.1.2, shall be computed using

Eq. (11-2) ( ), with and not exceeding

√ and

√ ,

respectively. shall be calculated using Eq. (11-15) (

) with equal to the cross-

sectional area of all the shear reinforcement on one peripheral line that is approximately

parallel to the perimeter of the column section, where is the spacing of the peripheral

lines of headed shear stud reinforcement. shall not be less than

√ .

11.11.5.2 — The spacing between the column face and the first peripheral line of

shear reinforcement shall not exceed . The spacing between peripheral lines of shear

reinforcement, measured in a direction perpendicular to any face of the column, shall be

constant. For prestressed slabs or footings satisfying 11.11.2.2, this spacing shall not exceed

; for all other slabs and footings, the spacing shall be based on the value of the shear

stress due to factored shear force and unbalanced moment at the critical section defined in

11.11.1.2, and shall not exceed:

(a) where maximum shear stresses due to factored loads are less than or

equal to √ ;

(b) where maximum shear stresses due to factored loads are greater than

√ .

11.11.5.3 — The spacing between adjacent shear reinforcement elements, measured

on the perimeter of the first peripheral line of shear reinforcement, shall not exceed .

11.11.5.4 — Shear stress due to factored shear force and moment shall not exceed

√

at the critical section located outside the outermost peripheral line of shear

reinforcement.


308

Example (Checking One-Way and Two-Way Shear at an Interior Column in a Flat Plate):

The figure below shows an interior column in a large uniform flat-plate slab. The slab is

thick. An average effective depth, , as shown, normally is used in shear strength

calculations for two-way slabs. Both one-way and two-way punching shear usually is checked

near columns where top reinforcement is used in both principal directions to resist negative-

bending moments. ACI Code Section 7.7.2(c) states that the minimum clear cover for slab

reinforcement is . Thus, assuming

bars are used as flexural reinforcement, the

average value for determining shear strength

of the slab is

The slab supports a uniform, superimposed dead load of and a uniform,

superimposed live load of . The normal-weight concrete has a compressive

strength of . Check whether the shear capacity is adequate.


309

Solution:

1. Determine the factored uniform load.

2. Check the one-way shear. One-way shear is critical at a distance from the face of

the column. Thus, the critical sections for one-way shear are A–A and B–B.

The loaded areas causing shear on these sections are cross hatched. Their outer boundaries

are lines of symmetry on which Because the tributary area for section A–A is larger,

this section will be more critical.

(c) Compute at section A–A.

(d) Compute for one-way shear. Because there is no shear reinforcement,

we have


310

where for one-way shear is

√

√

– the slab is OK in one-way shear.

3. Check the two-way punching shear. Punching shear is critical on a rectangular section

located at away from the face of the column. The load on the cross-hatched area

causes shear on the critical perimeter. Once again, the outer boundaries of this area

are lines of symmetry, where is assumed to be zero.

(a) Compute on the critical perimeter for two-way shear.

(b) Compute for the critical section. The length of the critical perimeter is

Now, is to be taken as the smallest of the following:

(

)√

(

)

(

)

(

)√

(

)

(

)

√


311

Therefore, the smallest value is

(

)√ √

So

Because exceeds the slab is OK in two-way shear.

Example (Design of an integral beam with vertical stirrups):

The flat plate slab with total thickness and effective depth is carried by

square columns on centers in each direction. A factored load of must be

transmitted from the slab to a typical interior column. Concrete and steel strengths used are,

respectively, and . Determine if shear reinforcement is required

for the slab; and if so, design integral beams with stirrups to carry the excess shear.

Solution:

The design shear strength of the concrete alone at the critical section from the face of

the column, is the smallest of the following:

(

)√

(

)

(

)

(

)√

(

)

(

)

√

where

Therefore, the smallest value is

√

√

This is less than , indicating that shear reinforcement is required. The effective

depth just satisfies the minimum allowed to use stirrup reinforcement, as

described in the beginning of this Section.

In this case, the maximum design strength allowed by the ACI Code is

√

√

satisfactorily above the actual . When shear is resisted by combined action of concrete and

bar reinforcement, the concrete contribution is reduced to

√

√


312

The vertical closed hoop stirrups will be used since must be times the stirrup

diameter (

) and arranged along four integral beams as shown below.

Thus, the provided is at the first critical section, a distance

from the column face, and the required spacing can be found from:

The spacing less than the maximum spacing of

, and stirrups

at a constant spacing of will be used. In other cases, stirrup spacing might be

increased with distance from the column, as excess shear is less, although this would

complicate placement of the reinforcement and generally save little steel.

The required perimeter of the second critical section, at which the concrete alone can carry

the shear, is found from the controlling equation (

√ ) as follows:

√

√

from which the minimum perimeter . Using this value of in the other

equations for :

(

)√

(

)√

(

)√

(

)√

The first equation governs. It is easily confirmed that requires a minimum

projection of the critical section past the face of

the column of .

( √ )

√ (

)

√ (

)

Five stirrups at a constant spacing will be

sufficient, the first placed at from

the column face, as indicated in the figure below.

This provides a perimeter at the second

critical section:

( √ )

which exceeds the requirement.

Note: the projection distance can be

calculated from the face of the column and assuming that is the column width.

( √ )


313

Four longitudinal bars will be provided inside the corners of each closed hoop stirrup,

as shown, to provide for proper anchorage of the shear reinforcement.

Note that the approach taken here is somewhat conservative because the portion of the slab

load applied inside the perimeter of the critical section does not act on that section and can

thus be subtracted from the factored load of .

Example (Design of an Interior Slab–Column Connection with Headed Shear Reinforcement

– Studs): A 20-cm-thick flat-plate slab with flexural reinforcement is supported by

square columns spaced at on centers N-S and on centers E-W. The

service loads on the slab are dead load (including self-weight) and live

load .The normalweight concrete has a compressive strength of

. Check the capacity of an interior slab-column connection. If necessary, design

shear reinforcement using headed-shear studs.

Solution:

1. Select the critical section for two-way shear around the column. At this stage in the

calculations, the designer does not know whether shear reinforcement will be

required.

We will assume it is not and will redo the calculations if we are wrong. Assuming

flexural reinforcement, the average effective depth of the slab is

The inner critical shear section for two-way shear in a flat plate extends around the column

at from the face of the column, as shown in the figure below.

The length of one side of the critical shear section around the column is

giving a shear perimeter

The area enclosed within the critical shear section is


314

2. Compute the shear acting on the critical shear section. The load combinations from

ACI Code Section 9.2 will be used to compute the total factored dead and live loads.

The basic combination for gravity load is

The factored shear force on the critical shear section is

Compute for the critical section:

Now, is to be taken as the smallest of the following:

(

)√

(

)

(

)

(

)√

(

)

(

)

√

Therefore, the last equation governs

√

√

So

Because less than , shear reinforcement is required at

the critical shear section. Using shear studs as shear reinforcement results in

√

√

This gives

The maximum value of allowed with headed-shear studs is

√

√

3. Lay out the punching shear reinforcement. Rows of shear studs welded to bars will

be placed parallel and perpendicular to the main slab reinforcement to cause the

least disruption in the placement of the main slab steel. Using a trial and error

process:

Try eight stud rails, each with seven -diameter studs ( ) with

-diameter heads, and (typical for shear studs).

The spacing to the first set of shear studs, , is to be taken less than or equal to . Thus,

select (see figure). Before proceeding to layout the subsequent rows of shear

studs, the shear strength should be checked at the inner critical section. In Equation

, the area provided by the inner row of shear studs is


315

Then, assuming only one line of shear studs are

crossed by the potential critical shear crack

nearest the column

,

This exceeds the required value of .

Try eight stud rails with the first stud located at from the column face.

Subsequent studs are at a spacing of with seven -diameter headed shear

studs per rail.

The outermost studs are at from the face of the column, and the

outer critical section is from the face of the column.


316

The outer critical-shear section is a series of straight line segments passing through points

located outside the outer shear studs, as shown in the figure. The perimeter of this

peripheral line is

( √ )

4. Check the shear stresses on the outer critical section. The factored shear force on the

concrete at the outer critical section is

and

The shear stress on the outer critical section is limited to

√

√

√

Because this is larger than , the design is OK.


317

CHAPTER 10 STAIRS

10.1 INTRODUCTION

Stairs must be provided in almost all buildings, either low-rise or high-rise, even if adequate

numbers of elevators are provided. Stairs consist of rises, runs (or treads), and landings. The

total steps and landings are called a staircase. The rise is defined as the vertical distance

between two steps, and the run is the depth of the step. The landing is the horizontal part of

the staircase without rises.


318

10.2 TYPES OF STAIRS

There are different types of stairs, which depend mainly on the type and function of the

building and on the architectural requirements. The most common types are as follows.

1. Single-flight stairs: The structural behavior of a flight of stairs is similar to that of a

one-way slab supported at both ends. The thickness of the slab is referred to as the

waist. When the flight of stairs contains landings, it may be more economical to

provide beams at and between landings (see next figure). If such supports are

not provided, which is quite common, the span of the staircase will increase by the

width of two landings and will extend between and . In residential buildings, the

landing width is in the range of , and the total distance between and

is about 6 m.


319

An alternative method of supporting a single flight of

stairs is to use stringers, or edge beams, at the two sides

of the stairs; the steps are then supported between the

beams.

2. Double-flight stairs: It is more

convenient in most buildings to build

the staircase in double flights between

floors. The types commonly used are

quarter-turn and closed-or open-well

stairs, as shown in figure. For the

structural analysis of the stairs, each

flight is treated as a single flight and is

considered supported on two or more

beams, as shown in the previous

figure (page 318). The landing extends

in the transverse direction between

two supports and is designed as a one-

way slab. In the case of open-well

stairs, the middle part of the landing

carries a full load, whereas the two

end parts carry half-loading only, as

shown in figure (d). The other half-

loading is carried in the longitudinal

direction by the stair flights, sections

A-A and B-B.

3. Three or more flights of stairs: In some

cases, where the overall dimensions of

the staircase are limited, three or four

flights may be adopted. Each flight will

be treated separately, as in the case of

double-flight staircases.


320

4. Cantilever stairs: Cantilever stairs are used mostly in fire-escape stairs, and they are

supported by concrete walls or beams. The stairsteps may be of the full-flight type,

projecting from one side of the wall, the half-flight type, projecting from both sides of

the supporting wall, or of the semispiral type. In this type of stairs, each step acts as a

cantilever, and the main reinforcement is placed in the tension side of the run and

the bars are anchored within the concrete wall. Shrinkage and temperature

reinforcement is provided in the transverse direction.


321

Another form of a cantilever stair is that using open-riser steps supported by a central beam,

as shown below. The beam has a slope similar to the flight of stairs and receives the steps on

its horizontally prepared portions. In most cases, precast concrete steps are used, with

special provisions for anchor bolts that fix the steps into the beam.

5. Precast flights of stairs: The speed of construction in some projects requires the use

of precast flights of stairs. The flights may be cast separately and then fixed to cast-

in-place landings. In other cases, the flights, including the landings, are cast and then

placed in position on their supporting walls or beams. They are designed as simply

supported one-way slabs with the main reinforcement at the bottom of the stair

waist. Adequate reinforcement must be provided at the joints.

Provisions must be made for lifting and handling the precast stair units by providing lifting

holes or inserting special lifting hooks into the concrete. Special reinforcement must be

provided at critical locations to account for tensile stresses that will occur in the stairs from

the lifting and handling process.


322

6. Free-standing staircase: In this type of stairs, the landing projects into the air without

any support at its end. The stairs behave in a springboard manner, causing torsional

stresses in the slab.


323

Three systems of loading must be considered in the design of this type of stairs, taking into

consideration that torsional moments will develop in the slab in all cases:

a. When the live load acts on the upper flight and half the landing only (Case 1), the

upper flight slab will be subjected to tensile forces in addition to bending moments,

whereas the lower flight will be subjected to compression forces, which may cause

buckling of the slab.

b. When the live load acts on the lower flight and half the landing only (Case 2), the

upper flight slab will be subjected to tensile forces, whereas the lower flight will be

subjected to bending moment and compression forces.

c. When the live load acts on both upper and lower flights, the loading of one flight will

cause the twisting of the other. The torsional stresses developed in the stairs require

adequate reinforcement in both faces of the stair slabs and the landing. Transverse

reinforcement in the slab and the landing must be provided in both faces of the

concrete in the shape of closed U-bars lapping at midwidth of the stairs. Typical

reinforcement details are shown in the figure below.


324

This type of stairs is favored by architects and sometimes called a pliers-shaped staircase or

jackknife staircase.

For practical design, the parameters may be chosen as follows: flight width between

and , horizontal span between and , total (light height between and

, and slab thickness between and .

The above information is a guide to help the designer to choose the right parameters for an

economical design.

7. Run-riser stairs: Run-riser stairs are stepped underside stairs that consist of a number

of runs and risers rigidly connected without the provision of the normal waist slab.

This type of stairs has an elegant appearance and is sometimes favored by architects.

The structural analysis of run-riser stairs can be simplified by assuming that the effect of

axial forces is negligible and that the load on each run is concentrated at the end of the run


325

(see next figure). For the analysis of a simply supported flight of stairs, consider a simple

flight of two runs, , subjected to a concentrated load at . Because joints and

are rigid, the moment at joint is equal to the moment at , or

where is the width of the run. The moment in rise, , is constant and is equal to .

When the rise is absent, the stairs, , act as a simply supported beam, and the maximum

bending moment occurs at midspan with value

For a flight of stairs that consists of a number of runs and risers, the same approach can be

used; the bending moment diagram is shown below. The moment in is constant and is

equal to the moment at joint , or . Similarly,

, and


326

8. Helical stairs (open-spiral stairs): A helical staircase is a three-dimensional structure,

which usually has a circular shape in plan. It is a distinctive type of stairs used mainly

in entrance halls, theater foyers, and special low-rise office buildings. The cost of a

helical stair is much higher than that of a normal staircase.

The stairs may be supported at some edges within adjacent walls or may be designed as a

free-standing helical staircase, which is most popular. The structural analysis of helical

staircases is complicated. Design charts for helical stairs are also prepared. Under load, the

flight slab will be subjected to torsional stresses throughout. The upper landing will be

subjected to tensile stresses, whereas compressive stresses occur at the bottom of the flight.

The forces acting at any section may consist of vertical moment, lateral moment, torsional

moment, axial force, shearing force across the waist of the stairs, and radial horizontal

shearing force. The main longitudinal reinforcement consists of helical bars placed in the

concrete waist of the stairs and runs from the top landing to the bottom support. The

transverse reinforcement must be in a closed stirrup form to resist torsional stresses or in a

U-shape lapped at about the midwidth of the stairs.

Based on many studies, the possible practical dimensions may be chosen as follows: Total

subtended arc between and , stair width between and , stairs slab

thickness between and and stair height between and .

The above information can be used as a guide to achieve a proper and economical design of

helical staircase.


327

An alternative method of providing a helical stair is to use a central helical girder located at

the midwidth of the stairs and have the steps project equally on both sides of the girder.

Each step is analyzed as a cantilever, and the reinforcement bars extend all along the top of

the run. Precast concrete steps may be used and can be fixed to specially prepared

horizontal faces at the top surfaces of the girder.

10.3 SLAB TYPE STAIRS. STRUCTURAL SYSTEM.

In the second and third types (slab type stairs), the main supporting element could be the

slab itself. The flight could be supported on the landing, which is in turn supported on the

supporting beams. From the structural point of view, it is better that the main supporting

element is spanning in the short direction. However, this depends on the surrounding

beams. If the beams exist around the perimeter of the stair well or at least along the long

sides, solution A in the figure below is more economical. If the supporting beams are only at

the short side, solution B is the only valid structural system.

Since the landing and the stairs are not straight, internal forces are generated in these

sloped elements. The two tensile forces and generated at the kink, producing a third

outward force as shown in the next figure. This force tends to cause splitting cracks if the

produced stresses exceed concrete tensile strength. Thus, tension reinforcement should be

extended from each side so that no outward force is generated.


328


329

Example

Design the staircase shown below, which carries a uniform live load of . Assume a

rise of and a run of . Use and .

Solution

1. Structural system: If no stringer beam is used, one of the four possible solutions

shown in figure (page 318) may be adopted. When no intermediate supports are

used, the flight of stairs will be supported at the ends of the upper and lower

landings. This structural system will be adopted in this example.

2. Minimum slab thickness for deflection is (for a simply supported one-way solid slab)

In the case presented here, where the slab ends are cast with the supporting beams and

additional negative reinforcement is provided, minimum thickness can be assumed to be

Take .

3. Loads: The applied live loads are based on the plan area (horizontal projection), while

the dead load is based on the sloped length. To transform the dead load into

horizontal projection the figure below explains how.

Flight Dead Load computation:

(

) (

)

𝐿

𝑞 𝐾𝑁 𝑚

𝜃

𝐿𝑜

𝑤 𝐾𝑁 𝑚

𝑤 ∙ 𝐿 𝑞 ∙ 𝐿𝑜

𝑤 𝑞 ∙𝐿𝑜𝐿

𝑞

cos𝜃


330

Quality Density

Material

(

) Tiles

(

) mortar

(

) Stair steps

cos


cos Plaster

Total Dead Load,

Landing Dead Load computation:

∙ ∙

Quality Density

Material

Tiles

mortar


Plaster

8.01 Total Dead Load

Live Load:

Total factored Load:

∙ ∙

∙ ∙

Because the load on the landing is carried into two directions, only half the load will be

considered in each direction .


331

4. Check for shear strenght:


Assume beam width

∙ ( )

√

∙ √ ∙ ∙ ∙

- for shear.

∙


5. Calculate the maximum bending moment and steel reinforcement:

(

) ∙ ∙ (

) ∙

∙

∙

∙


∙

∙

∙

( √

)

( √

∙ ∙

)

∙ ∙

∙ ∙

Use then

Take .


1.

2.


332

(

) (

) ∙

(

) (

)

s s –

6. Temperature and shrinkage reinforcement.

( ) ∙ ∙

Take .


1.

2.

s s –

If the slab will be cast monolithically with its supporting beams, additional reinforcement

must be provided at the top of the upper and lower landings. Details of stair reinforcement

are shown in the figure (page 333).

7. Design of landings: Considering a 1-m length of the landing, the load on the landing is

shown in the next figure. The middle will carry a full load, whereas the two

1.5-m lengths on each side will carry half the ultimate load.

(

) ∙ ∙ (

) ∙

∙

∙

∙


333


334

Assume bar diameter for main reinforcement. Because the bars in the landing will be

placed on top of the main stair reinforcement

∙

∙

∙

( √

)

( √

∙ ∙

)

∙ ∙

∙ ∙

then provide

Use then

Take .


1.

2.

(

) (

) ∙

(

) (

)

s s –

8. The transverse beams at the landing levels must be designed to carry loads from

stairs ( ) in addition to their own weight and the weight of the wall

above.


335

Example

Design the staircase shown below, which carries a uniform live load of . Assume a

rise of and a run of . Use and .

Solution

1. Minimum slab thickness for deflection is (for a simply supported one-way solid slab)

Take .

2. Loads:

Flight Dead Load computation:

(

) (

)

Quality Density

Material

(

) Tiles

(

) mortar

(

) Stair steps

cos


cos Plaster

Total Dead Load,


336

Landing Dead Load computation:

∙ ∙

Quality Density

Material

Tiles

mortar


Plaster

8.01 Total Dead Load

Live Load:

Total factored Load:

∙ ∙ ∙

∙ ∙ ∙

3. Design of slab S1:

Slab S1 is supported at the centerline of slabs S2 and S3.

The reaction at each end

∙

Check for shear strenght:


Take the maximum shear as the support reaction

√

∙ √ ∙ ∙ ∙

- for shear.

∙


337


Calculate the maximum bending moment and steel reinforcement:

( )

∙

∙


∙

∙

∙

( √

)

( √

∙ ∙

)

∙ ∙

∙ ∙

Use then

Take .


1.

2.

(

) (

) ∙

(

) (

)

s s –


( ) ∙ ∙


338

Take .


1.

2.

s s –


Slab S2 is supported on the beams located on axis 1,2 at the floor level. The reaction of the

slab S1 is applied at the centerline of the slab S2. Since the width of S2 is , the reaction

will be distributed along this width. Thus the load per meter equals

( )

or

( )

of the slab S1 is applied at the middle of the slab.


339

The reaction at each end

Check for shear strenght:


Take the maximum shear as the support reaction

√

∙ √ ∙ ∙ ∙

- for shear.

∙


Calculate the maximum bending moment at midspan and the steel

reinforcement:

(

) ∙

∙


∙

∙

∙

( √

)

( √

∙ ∙

)

∙ ∙

∙ ∙

Use then

Take .


1.


340

2.

(

) (

) ∙

(

) (

)

s s –


( ) ∙ ∙

Take .


1.

2.

s s –


Slab S3 is supported on the beams, the reaction of the slab S1 is applied at the middle of the

slab:

Design the slab S3 for flexure and shear as for slabs S1 and S2.


341

CHAPTER 11 FOOTINGS AND FOUNDATIONS

11.1 INTRODUCTION.

Reinforced concrete footings are structural members used to support columns and walls and

to transmit and distribute their loads to the soil. The design is based on the assumption that

the footing is rigid, so that the variation of the soil pressure under the footing is linear.

Uniform soil pressure is achieved when the column load coincides with the centroid of the

footing. Although this assumption is acceptable for rigid footings, such an assumption

becomes less accurate as the footing becomes relatively more flexible. The proper design of

footings requires that

1. The load capacity of the soil is not exceeded.

2. Excessive settlement, differential settlement, or rotations are avoided.

3. Adequate safety against sliding and/or overturning is maintained.

The most common types of footings used in buildings are the single footings and wall

footings. When a column load is transmitted to the soil by the footing, the soil becomes

compressed. The amount of settlement depends on many factors, such as the type of soil,

the load intensity, the depth below ground level, and the type of footing. If different footings

of the same structure have different settlements, new stresses develop in the structure.

Excessive differential settlement may lead to the damage of nonstructural members in the

buildings or even failure of the affected parts.

Vertical loads are usually applied at the centroid of the footing. If the resultant of the applied

loads does not coincide with the centroid of the bearing area, a bending moment develops.

In this case, the pressure on one side of the footing will be greater than the pressure on the

other side.


342

If the bearing soil capacity is different under different footings—for example, if the footings

of a building are partly on soil and partly on rock—a differential settlement will occur. It is

usual in such cases to provide a joint between the two parts to separate them, allowing for

independent settlement.

The depth of the footing below the ground level is an important factor in the design of

footings. This depth should be determined from soil tests, which should provide reliable

information on safe bearing capacity at different layers below ground level. Soil test reports

specify the allowable bearing capacity to be used in the design. In cold areas where freezing

occurs, frost action may cause heaving or subsidence. It is necessary to place footings below

freezing depth to avoid movements.

11.2 TYPES OF FOOTINGS.

Different types of footings may be used to support building columns or walls. The most

common types are as follows:

1. Wall footings are used to support structural walls that carry loads from other floors

or to support nonstructural walls. They have a limited width and a continuous length

under the wall. Wall footings may have one thickness, be stepped, or have a sloped

top.

2. Isolated, or single, footings are used to support single columns. They may be square,

rectangular, or circular. Again, the footing may be of uniform thickness, stepped, or

have a sloped top. This is one of the most economical types of footings, and it is used

when columns are spaced at relatively long distances. The most commonly used are

square or rectangular footings with uniform thickness.


343

3. Combined footings usually support two columns or three columns not in a row. The

shape of the footing in plan may be rectangular or trapezoidal, depending on column

loads. Combined footings are used when two columns are so close that single

footings cannot be used or when one column is located at or near a property line.

4. Cantilever, or strap, footings consist of two single footings connected with a beam or

a strap and support two single columns. They are used when one footing supports an

eccentric column and the nearest adjacent footing lies at quite a distance from it.

This type replaces a combined footing and is sometimes more economical.

5. Continuous footings support a row of three or more columns. They have limited

width and continue under all columns.

6. Raft, or mat, foundations consist of one footing, usually placed under the entire

building area, and support the columns of the building. They are used when:

a. The soil-bearing capacity is low.

b. Column loads are heavy.

c. Single footings cannot be used.

d. Piles are not used.

e. Differential settlement must be reduced through the entire footing system.

2. Pile caps are thick slabs used to tie a group of piles together and to support and

transmit column loads to the piles.

11.3 DISTRIBUTION OF SOIL PRESSURE. GROSS AND NET SOIL PRESSURES.

The figure below shows a footing supporting a single column. When the column load, P, is

applied on the centroid of the footing, a uniform pressure is assumed to develop on the soil

surface below the footing area. However, the actual distribution of soil pressure is not

uniform but depends on many factors, especially the composition of the soil and the degree

of flexibility of the footing. For example, the distribution of pressure on cohesionless soil

(sand) under a rigid footing is shown. The pressure is maximum under the center of the

footing and decreases toward the ends of the footing. The cohesionless soil tends to move

from the edges of the footing, causing a reduction in pressure, whereas the pressure


344

increases around the center to satisfy equilibrium conditions. If the footing is resting on a

cohesive soil such as clay, the pressure under the edges is greater than at the center of the

footing. The clay near the edges has a strong cohesion with the adjacent clay surrounding

the footing, causing the nonuniform pressure distribution.

The allowable bearing soil pressure, , is usually determined from soil tests. The allowable

values vary with the type of soil, from extremely high in rocky beds to low in silty soils.

Class of Material Maximum allowable

soil pressure [ ]

Rock of ultimate crushing strength

Compact coarse sand, compact fine sand, hard clay, or sand clay

Medium stiff clay or sandy clay

Compact inorganic sand and silt mixtures

Loose sand

Soft sand clay or clay

Loose inorganic sand-silt mixtures

Loose organic sand-silt mixtures, muck, or bay mud

The figure below shows a -thick spread footing with a column at its center and with its

top surface located below the ground surface. There is no column load at this stage.

The total downward load from the weights of the soil and the footing is . This is

balanced by an equal, but opposite, upward pressure. As a result, the net effect on the

concrete footing is zero. There are no moments or shears in the footing due to this loading.

When the column load is added, the pressure under the footing increases by

. The

total soil pressure is . This is referred to as the gross soil pressure and must

not exceed the allowable soil pressure, . When moments and shears in the concrete

footing are calculated, the upward and downward pressures of cancel out,

leaving only the net soil pressure, , to cause internal forces in the footing.

In design, the area of the footing is selected so that the gross soil pressure does not exceed

the allowable soil pressure. The flexural reinforcement and the shear strength of the footing

are then calculated by using the net soil pressure. Thus, the area of the footing is selected to

be


345

where and refer to the unfactored service dead and live loads.

Once the area of the footing is known, the rest of the design of the footing is based on soil

stresses due to the factored loads.


346

11.4 DESIGN CONSIDERATIONS.

Footings must be designed to carry the column loads and transmit them to the soil safely.

The design procedure must take the following strength requirements into consideration:

1. The area of the footing based on the allowable bearing soil capacity.

2. One-way shear.

3. Two-way shear, or punching shear.

4. Bending moment and steel reinforcement required.

5. Bearing capacity of columns at their base and dowel requirements

6. Development length of bars

7. Differential settlement

In this course we will consider the first 4 points only.

11.4.1 Size of footings.

The area of the footings can be determined from the actual external loads (unfactored

forces and moments) such that the allowable soil pressure is not exceeded. In general, for

vertical loads

where the total service load is the unfactored design load.

The area of footing can be determined from the net soil pressure by

Once the area is determined, a factored soil pressure is obtained by dividing the factored

load, , by the area of the footing. This is required to design the footing

by the strength design method.

The allowable soil pressure, , is obtained from soil test and is based on service load

conditions.

11.4.2 One-way shear (Beam shear).

For footings with bending action in one direction, the critical section is located at a distance

from the face of the column. The diagonal tension at section m-m in the figure of one way

shear can be checked as was done before in beams. The allowable shear in this case is equal

to

√ √

where width of section m-m. The factored shear force at section m-m can be

calculated as follows:


347

(

)

If no shear reinforcement is to be used, then can be determined, assuming :

√

11.4.3 Two-way shear (Punching shear).

Two-way shear is a measure of the diagonal tension caused by the effect of the column load

on the footing. Inclined cracks may occur in the footing at a distance from the face of

the column on all sides. The footing will fail as the column tries to punch out part of the

footing.

The ACI Code, Section 11.11.2 allows a shear strength, , in footings without shear

reinforcement for two-way shear action, the smallest of

(

) √

(

) √

√

where Ratio of long side to short side of the rectangular column.


348

For shapes other than rectangular, is taken to be the ratio of the longest dimension of the

effective loaded area in the long direction to the largest width in the short direction

(perpendicular to the long direction).

perimeter of the critical section taken at from the loaded area.

effective depth of footing.

for normal-weight concrete.

is assumed to be:

for interior columns,

for edge columns, and

for corner columns.

Based on the preceding three values of , the effective depth, , required for two-way

shear is the largest obtained from the following formulas assuming ):

( ) √

(

) √

√

11.4.4 Flexural strength and footing reinforcement.

The critical sections for moment occur at the face of the

column (section n-n). The bending moment in each

direction of the footing must be checked and the

appropriate reinforcement must be provided. In square

footings and square columns, the bending moments in

both directions are equal. To determine the

reinforcement required, the depth of the footing in each

direction may be used. Because the bars in one direction

rest on top of the bars in the other direction, the

effective depth, , varies with the diameter of the bars

used.

According to ACI, 7.7, minimum clear cover in cast-in-

place concrete against and permanently exposed to

earth (such as footings) should not be less than .

The ACI Code, Section 10.5, indicates that for structural

slabs of uniform thickness, the minimum area and


349

maximum spacing of steel bars in the direction of bending shall be as required for shrinkage

and temperature reinforcement (see section 8.4, 8.5 page 195).

The reinforcement in one-way footings

and two-way footings must be

distributed across the entire width of

the footing. In the case of two-way

rectangular footings, the ACI Code,

Section 15.4.4, specifies that in the long

direction, a portion of the total

reinforcement distributed

uniformly along the width of the

footing. In the short direction, a certain

ratio of the total reinforcement in this

direction must be placed uniformly

within a bandwidth equal to the length

of the short side of the footing

according to

The bandwidth must be centered on the centerline of the column. The remaining

reinforcement in the short direction must be uniformly distributed outside the bandwidth.

This remaining reinforcement percentage shall not be less than that required for shrinkage

and temperature.

When structural steel columns or masonry walls are used, then the critical sections for

moments in footings are taken at halfway between the middle and the edge of masonry

walls and halfway between the face of the column and the edge of the steel base place (ACI

Code, Section 15.4.2).


350

11.4.5 Transfer of Load from Column to Footing

The column applies a concentrated load on the footing. This load is transmitted by bearing

stresses in the concrete and by stresses in the dowels or column bars that cross the joint.

The design of such a joint is considered in ACI Code Section 15.8. The area of the dowels can

be less than that of the bars in the column above, provided that the area of the dowels is at

least times the column area (ACI Code Section 15.8.2.1) and is adequate to transmit

the necessary forces. Such a joint is shown below.

Generally, the column bars stop at the bottom of the column, and dowels are used to

transfer forces across the column–footing joint. Dowels are used because it is awkward to

embed the column steel in the footing, due to its unsupported height above the footing and

the difficulty in locating it accurately. The above figure a shows an column

with and bars. The column is supported on a footing made of

concrete. There are Grade-420 dowels in the connection. The dowels

extend into the footing a distance equal to the compression development length of a

bottom bar in concrete ( ) and into the column a distance equal to the

greater of

1. the compression lap-splice length for a bar in concrete ( ), and

2. the compression development length of a bar in concrete ( ).


351

This joint could fail by reaching various limit states, including

1. crushing of the concrete at the bottom of the column, where the column bars are no

longer effective,

2. crushing in the footing below the column,

3. bond failure of the dowels in the footing, and

4. failure in the column of the lap splice between the dowels and the column bars.

11.4.6 Bearing Strength.

The total capacity of the column for pure axial load is , of which is carried

by the steel and the rest by the concrete, as shown in the previous figure b. At the joint, the

area of the dowels is less than that of the column bars, and the force transmitted by the

dowels is , where is the area of the dowels and is that for tied columns. As a

result, the load carried by the concrete has increased. In the above figure, the dowels are

hooked so that they can be supported on and tied

to the mat of footing reinforcement when the

footing concrete is placed. The hooks cannot be

used to develop compressive force in the bars

(ACI Code Section 12.5.5).

The maximum bearing load on the concrete is

defined in ACI Code Section 10.14 as

. If the load combinations from ACI

Code Section 9.2 are used, ACI Code Section

9.3.2.4 gives for bearing and is the

area of the contact surface.

When the supporting surface is wider on all sides

than the loaded area, the maximum bearing load

may be taken as

√

Where √

not more than , where is the

area of the lower base of a right pyramid or cone

formed by extending lines out from the sides of

the bearing area at a slope of 2 horizontal to 1

vertical to the point where the first such line

intersects an edge. This is illustrated in next

figure. The first intersection with an edge occurs

at point B, resulting in the area shown

crosshatched.


352

If dowels are needed, the required area of dowels

Two distinct cases must be considered:

(1) joints that do not transmit computed moments to the footing and

(2) joints that do.

These will be discussed separately.

No Moment Transferred to Footing

If no moments are transmitted, or if the eccentricity falls within the kern of the column,

there will be compression over the full section. The total force transferred by bearing is

then calculated as times the smaller of the bearing stresses allowed on the

column or the footing, where is total area of the column and is the area of the bars

or dowels crossing the joint. Any additional load must be transferred by dowels.

Moments Are Transferred to Footing

If moments are transmitted to the footing, bearing stresses will exist over part, but not all,

of the column cross section. The number of dowels required can be obtained by considering

the area of the joint as an eccentrically loaded column with a maximum concrete stress

equal to the smaller of the bearing stresses allowed on the column or the footing. Sufficient

reinforcement must cross the interface to provide the necessary axial load and moment

capacity. Generally, this requires that all the column reinforcement or similar-sized dowel

bars must cross the interface. This steel must be spliced in accordance with the

requirements for column splices.

Practical Aspects

Three other aspects warrant discussion prior to the examples. The minimum cover to the

reinforcement in footings cast against the soil is (ACI Code Section 7.7.1). This

allows for small irregularities in the surface of the excavation and for potential

contamination of the bottom layer of concrete with soil. Sometimes, the bottom of the

excavation for the footing is covered with a lean concrete seal coat, to prevent the bottom

from becoming uneven after rainstorms and to give a level surface for placing

reinforcement.

The minimum depth of the footing above the bottom reinforcement is for footings

on soil and for footings on piles (ACI Code Section 15.7). ACI Code Section 10.6.4,

covering the distribution of flexural reinforcement in beams and one-way slabs, does not

apply to footings.


353

11.5 SPREAD (ISOLATED) FOOTINGS.

Spread footings are square or rectangular pads that spread a column load over an area of

soil large enough to support the column load. The soil pressure causes the footing to deflect

upward, causing tension in two directions at the bottom. As a result, reinforcement is placed

in two directions at the bottom.

Example (isolated concentrically loaded square column):

Determine the base area and overall thickness for a square spread footing with the following

design conditions:

Service dead load

Service live load

Service surcharge

Permissible (allowable) soil pressure

Soil density

, .

Solution:

Calculating the weight of footing,

soil, and the surcharge floor load:

weight of footing (assume

:

weight of soil:

Total surcharge load on foundation:

Net soil pressure, :


354

Required sizes of footing:

√ √

Take

Depth of footing and shear design:

One-way shear (Beam shear)

at distance from the face of support:

(

) (

)

Let

√

√

(

)

√

Assume cover 75 mm, and steel bars of

Generally, the thickness of aspread footing is governed by two-way shear. The shear will be

checked on the critical perimeter at from the face of the column and, if necessary, the

thickness will be increased or decreased. Because there is reinforcement in both directions,

the average will be used:


355

Two-way shear (Punching shear)

Let

( )

( )

(

)√

(

)

(

)

(

)√

(

)

(

)

√

Take √ √

Try

( )

( )

√ √

Design for flexure in long direction:

Take steel bars of

.


356

( √

)

( √

)

Take with

Using bars of instead of as assumed before makes the effective depth larger.

So no need to check for .

Step is the smallest of:

1.

2.


357

Design for flexure in short direction:

Take steel bars of

.

( √

)

( √

)

Take with




358

Check for the development length as will be discussed later (see chapter 12.)

Design the column–footing joint.

The column-footing joint is shown here. The factored load at the base of the column is

𝑚

0.5

5 m

𝑚

𝑚

𝑚


359

The maximum bearing load on the bottom of the column (ACI

Code Section 10.14.1) is where is the area

of the contact surface between the column and the footing

and is for the column. When the contact supporting

surface on the footing is wider on all sides than the loaded

area, the maximum bearing load on the top of the footing

may be taken as

√

where √

, is the area of the lower base of a right

pyramid or cone as defined earlier. ACI Code Section 9.3.2.4 defines equal to 0.65 for

bearing.

The allowable bearing on the base of the column is

Note: here for column.

The allowable bearing on the footing is

By inspection, √

for the footing exceeds 2; hence, the

maximum bearing load on the footing is

√

Note: here for footing.

Thus, the maximum load that can be transferred by bearing is , and dowels are

NOT needed.

The minimum area of dowels

Note: If dowels are needed, the required area of dowels

Where has been used. This is the value from ACI Code Sections 9.3.2.2(b) for

compression-controlled tied columns and 9.3.2.4 for bearing.

Use or use the same reinforcement as in the column if larger.

The dowels must extend into the footing a distance equal to the compression-development

length ( and concrete). The bars will be extended down to the level of the

main footing steel and hooked . The hooks will be tied to the main steel to hold the

dowels in place. The dowels must extend into the column a distance equal to the greater of

a compression splice for the dowels or the compression development length of the column

bars. Dowel each corner bar.

2.8 m

2.8

m


360

11.6 STRIP (WALL) FOOTINGS.

A wall footing cantilevers out on both sides of the wall as shown. The soil pressure causes

the cantilevers to bend upward, and as a result, reinforcement is required at the bottom of

the footing. The critical sections for design for flexure and anchorage are at the face of the

wall (section A–A). One-way shear is critical at a section a distance from the face of the

wall (section B–B).

The presence of the wall prevents two-way shear.

Example (strip or wall footings):

A thick concrete wall carries a service (unfactored) dead load of and a

service live load of . From the geotechnical report, the allowable soil pressure,

, is for shallow foundations. Design a wall footing to be based below

the final ground surface, using normal-weight concrete and .

The density of the soil is .

Solution:

Estimate the size of the footing and the factored net pressure. Consider a 1- strip of

footing and wall. Allowable soil pressure is . Because the thickness of the

footing is not known at this stage, it is necessary to guess a thickness for a first trial.

Try a - -thick footing.

Therefore,

and we have:

Take .



361


Shear usually governs the thickness of footings. Only one-way shear is significant in a wall

footing. at distance from the face of column:

(

) (

)

Let

√

√

(

)

√

Assume cover , and steel bars of

Design for flexure:

Take steel bars of

( √

)

( √

)

Or Take with




1.


362

2.

Select the minimum (temperature) reinforcement. By ACI Code Section 7.12.2.1 we

require the following reinforcement along the length of the footing.

The maximum spacing is or . Provide 9 ( ) for shrinkage

reinforcement, placed as shown.


363

11.7 FOOTINGS UNDER ECCENTRIC COLUMN LOADS.

When a column transmits axial loads only, the footing can be designed such that the load

acts at the centroid of the footing, producing uniform pressure under the footing. However,

in some cases, the column transmits an axial load and a bending moment, as in the case of

the footings of fixed-end frames. The pressure q that develops on the soil will not be uniform

and can be evaluated from the following equation:

where

vertical load, positive in compression,

area of the contact surface between the soil and the footing,

moment of inertia of this area,

moment about the centroidal axis

of the footing area

distance from the centroidal axis to

the point where the stresses are being

calculated.

The moment, , can be expressed as

, where is the eccentricity of the

load relative to the centroidal axis of the

area . The maximum eccentricity for

which the previous equation applies is

that which first causes at some

point. Larger eccentricities will cause a

portion of the footing to lift off the soil,

because the soil – footing interface

cannot resist tension. For a rectangular

footing, this occurs when the

eccentricity exceeds

This is referred to as the kern distance.

Loads applied within the kern, the

shaded area will cause compression

over the entire area of the footing, and

the previous equation can be used to

compute .

Various pressure distributions for

rectangular footings are shown in the

figure below. If the load is applied

concentrically, the soil pressure is


364

. If the load acts through the kern point, at one side and at the

other. If the load falls outside the kern point, the resultant upward load is equal and

opposite to the resultant downward load, as shown in next figure (d). Generally, such a

pressure distribution would not be acceptable, because it makes inefficient use of the

footing concrete, tends to overload the soil, and may cause the structure to tilt.


365

The different design conditions are summarized as follows:

1. When , the soil pressure is trapezoidal (figure b, page 364).

2. When , the soil pressure is triangular (figure c, page 364).

3. When , the soil pressure is triangular (figure d, page 364).

(

)

( )

Example:

Determine the dimensions of an isolated footing subjected to the loads:

Permissible net (allowable) soil pressure

Column dimensions

Solution (1) rectangular footing with uniform pressure:

Use

Solution (2) rectangular footing with nonuniform pressure:

Use

No tension zone.

𝑀 𝐾𝑁 𝑚

𝑃 𝐾𝑁

𝑚𝑚

𝑃 𝐾𝑁

𝑚

𝑚 𝑚 𝑚

𝑚𝑚

𝑚 𝑚


366

. Take

Solution (3) Square footing with nonuniform pressure:

Assume

Design the footing for shear and flexure as usual.

Example:

Determine the dimension of an isolated footing subjected to the loads:

Permissible net (allowable) soil pressure

Column dimensions

Solution:

There is a tension zone.

(

)

Take

( )

( )

𝑃 𝐾𝑁

𝑚𝑚

𝑚

𝑒 𝑚

𝑒 𝑚 𝑥 𝑚

𝐾𝑁 𝑚


367

Example (Design a Rectangular Footing for a Column Subjected to Axial Load and Bending): The rectangular footing has a width of and it will be supporting a column

that is carrying the following loads.

Service surcharge

Permissible (allowable) soil pressure

Soil density ( soil depth

The length and depth of the footing are to be determined.

Design the footing assuming and .

Solution:

1. Factored loads.

2. Estimate footing length and depth. Estimate the overall thickness, , of the footing to

be between 1.5 and 2.0 times the size of the column, so select .

The net permissible bearing pressure is:

Assume the soil pressure distribution at the base of the footing is trapezoidal (no tension

zone). Assuming a footing width, , and a footing length, , an expression for the maximum

bearing pressure under the footing is:

Setting we can use the following expression to solve for the required footing

length, .

or

√

Using the positive solution, select a footing length of .

3. Calculate factored soil pressures. (Note: In the prior step the acting loads were used

with the allowable soil bearing pressure to select the size of the footing. Now, the

factored loads will be used to determine the soil pressures that will subsequently be

used to determine the factored moment and shear used for the design of the

footing.)


368

4. Check for one-way shear. The critical section for checking one-way shear strength is

shown. To simplify this check, it is conservative to assume that the maximum

factored soil pressure of acts on the entire shaded region. Thus, the

factored shear force to be resisted at the critical section is (Take steel bars of ):



369

(

)

Let

√

√

Thus, the footing is OK for one-way shear.

5. Check footing thickness for two-way shear. The critical shear perimeter is located

away from each column face, as shown. Assume the average effective depth for

the footing is (Take steel bars of ):

Thus, each side of the critical shear perimeter has a length of . Using the average

factored shear stress inside the critical perimeter

(

) the net factored shear to be transferred across

the critical perimeter is:


370

(

)√

(

)

(

)

(

)√

(

)

(

)

√

Take

√

√

Thus, the footing depth satisfies the strength requirement for the net shear force.

6. Design for flexure in long direction. Take steel bars of :


371

( √

)

( √

)

Take with




1.

2.

The design procedure for flexural reinforcement in the transverse direction would be Similar.

For the arrangement of the bars in the transverse direction (ACI Code Section 15.4.4.2).


372

11.8 COMBINED FOOTINGS.

Combined footings are used when it is necessary to support two columns on one footing.

When an exterior column is so close to a property line that a spread footing cannot be used,

a combined footing is often used to support the edge column and an interior column.

The shape of the footing is chosen such that the centroid of the area in contact with soil

coincides with the resultant of the column loads supported by the footing.

Sometimes, a combined footing will be designed as two isolated pads joined by a strap or

stiff beam. Here, the exterior footing acts as a wall footing, cantilevering out on the two

sides of the strap. The interior footing can be designed as a two-way footing. The strap is

designed as a beam and may require shear reinforcement in it.


373

The soil pressure is assumed to act on longitudinal beam strips, A–B–C. These transmit the

load to hypothetical cross beams, A–D and B–E, which transmit the upward soil reactions to

the columns. For the column placement shown, the longitudinal beam strips would deflect

as shown, requiring the reinforcement shown.

For the transverse direction its

assumed that the column load is

spread over a width under the

column equal to the column

width plus on each side,

whenever that is available. In

other words, the column load

acts on a beam under the column

within the footing, which has a

maximum width of ( ) and a

length equal to the short side of

the footing.


374

Example (combined footing): A combined footing supports a column carrying a service dead load of

and a service live load of , plus a column

carrying service loads of dead load and live load. The

distance between the columns is , center to center. The allowable net soil bearing

pressure is .

Design the footing, assuming that

(normal-weight concrete) and .

Solution:

Footing dimensions.

∑

𝑃 𝐿 𝐾𝑁

𝑃 𝐷 𝐾𝑁

𝑐𝑚

𝑚

𝑐𝑚

𝑃 𝐿 𝐾𝑁

𝑃 𝐷 𝐾𝑁

𝑐𝑜𝑙 𝑐𝑜𝑙

𝑞

𝑃 𝐾𝑁

𝑐𝑚

𝑚

𝑐𝑚


𝑚

𝑚

𝑚 𝑚

𝑚 𝑚


375


𝑐𝑚

𝑚

𝑐𝑚


𝑚

𝑚

𝑚 𝑚

𝑚 𝑚

𝐾𝑁 𝑚

𝑥

𝑚

𝑀𝑚𝑎𝑥 𝐾𝑁

𝑃 𝑢 𝐾𝑁

𝑃 𝑢 𝐾𝑁


376



Assume and steel bars of

√

√

At column 1 :

At column 2 :

The thickness is adequate enough.


At column 1, :

(

)

(

)

(

)√

(

)

(

)

(

)√

(

)

(

)

√

√

√



377

At column 2, :

Check for two options of punching action

(

)

(

)

(

)√

(

)

(

)

(

)√

(

)

(

)

√

√

√


Now, we can design the footing for flexure. By calculating the negative and positive

moments or by analyzing the combined footing using BeamD software we plot the moment

diagram. The moment diagram is plotted for the full 3.3 m width of the footing.

Design the flexural reinforcement in the longitudinal direction.

Design for Negative moment (midspan) :

Take steel bars of

.


378

( √

)

( √

)

Provide

Take with




1.

2.

Design for Positive moment (under columns) at the face of

column 2 and at the face of column 1.

The is control here.

Take with

Design the flexural reinforcement in the transverse direction (transverse beams).

Transverse strips under each column will be assumed to transmit the load from the

longitudinal beam strips into the column. The width of the beam strips will be assumed to

extend on each side of the column. The actual width is unimportant, because the

moments to be transferred are independent of the width of the transverse beams.

The factored load on column 1 is .

This is balanced by an upward net force of

The maximum moment in this transverse beam at the face of column 1 is

(

)


379

The Band width under column 1 is (

)

Take the Band width

( √

)

( √

)

Take with

The factored load on column 2 is .

This is balanced by an upward net force of

The maximum moment in this transverse beam is

(

)

The Band width under column 1 is

Take the Band width

( √

)

( √

)

Take with


380

Select the minimum (temperature) reinforcement. By ACI Code Section 7.12.2.1 we

require the following reinforcement along the length of the footing.

The maximum spacing is or .

Provide 7 ( ) or for shrinkage reinforcement, placed as

shown.

11.9 CONTINUOUS FOOTINGS.

Example

A continuous footing supports three columns carrying a service dead load and a

service live load as shown. The length of the footing is . The allowable net soil bearing

pressure is .

Design the footing, assuming that and .

𝑐𝑚

𝑐𝑚


𝑚

𝑚

𝑚 𝑚

𝑚

𝑐𝑚

𝑚

𝑚

𝑐𝑜𝑙

𝑃𝐷 𝐾𝑁 𝑃𝐿 𝐾𝑁




381

Solution:

1. Footing width.

∑

The eccentricity of the resultant force is

Note

No tension or zero pressure.

2. Calculate factored soil pressures. .

𝑐𝑚

𝑐𝑚


𝑚

𝑚

𝑚 𝑚

𝑚

𝑐𝑚

𝑚

𝑚

𝑐𝑜𝑙




𝐾𝑁 𝑚

𝐾𝑁 𝑚


382

3. Depth of footing and shear design:

Assume and steel bars of

The beam load:


√

√

At column 1, :


(

)

At column 2, :

at distance from the Left face of support:

(

)

at distance from the Right face of support:

(

)

Or from the right:

(

)

At column 3, :

𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝑥


383


(

)



At column 1, :

(

)

(

) (

)

(

)√

(

)

(

)

(

)√

(

)

(

)

√

√

√

The thickness is NOT adequate enough.

Take ,

(

)


384

(

) (

)

√

√


At column 2, :

(

)

(

)√

(

)

(

)

(

)√

(

)

(

)

√

√

√


At column 3, :

(

)


385

(

) (

)

(

)√

(

)

(

)

(

)√

(

)

(

)

√

√

√


Now, we can design the footing for flexure. By calculating the negative and positive

moments or by analyzing the combined footing using BeamD software we plot the moment

diagram. The moment diagram is plotted for the full width of the footing.

A

A

1 1

A

A

2 2

A

A

3 3

A

A

4 5 4

0.3 0.3 0.3 1.7 0.3 0.3 1.7 0.3 0.3 0.3

0.45 2. 2. 0.45

load group no. 1 Total loads Units:kN,meter

-2127.7 -2070.5

0.45 2. 2. 0.45

-2070.5 -1816.3

0.45 2. 2. 0.45

-1816.3 -1562.1

0.45 2. 2. 0.45

-1562.1 -1504.9

0.45 2. 2. 0.45

213.5

-15.9 95.2

816.2

527.2 506.4

154.3 68.3

-25.4

-482.3 -387.7

0.14 0.46

0.51 0.13


386

11.10 MAT FOUNDATIONS.

When the bearing capacity of the soil is low, isolated footings are replaced by a raft

foundation. In such a case, a solid reinforced concrete rigid slab is constructed under the

entire building as shown below. Structurally, raft foundations resting directly on soil act as a

flat slab or a flat plate, upside down, i.e., loaded upward by the bearing pressure and

downward by the concentrated column reactions. The raft foundation develops the

maximum available bearing area under the building. If the bearing capacity of the soil is so

low that even this large bearing capacity is insufficient, deep foundations such as piles must

be used. Apart from developing large bearing areas, another advantage of raft foundations is

that their continuity and rigidity that helps in reducing differential settlement of individual

columns relative to each other, which might be caused by local variations in the quality of

subsoil, or other causes.

The design of raft foundations may be carried out by one of two methods:

• T conv ntional igid m t od and;

• T finit l m nt m t od utilizing comput p og ams.

The conventional method is easy to apply and the computations can be carried out using

hand calculations. However, the application of the conventional method is limited to rafts

with relatively regular arrangement of columns.

In contrast, the finite element method can be used for the analysis of raft regardless of the

column arrangements, loading conditions, and existence of cores and shear walls.

Commercially available computer programs can be used. The user should, however, have

sufficient background and experience.

Conventional rigid method:

The raft foundation shown below has dimensions . Columns' service loads are

indicated as , , , ... etc. The application of the conventional method can be

summarized as follows:


387

∑

Where

area of the raft

moment of inertia of the raft about axis

moment of inertia of the raft about axis

moment of the applied loads about the axis,

moment of the applied loads about the axis,

Where and , are the eccentricities of the resultant from the center of gravity of the raft.


388

The coordinates of the eccentricities are given by:

Where are the coordinates of

Where are the coordinates of

To draw the shear and moment diagrams we can divide the raft into several strips in the

direction (B1, B2, B3) and in the direction (B4, B5, B6, B7). The soil pressure at the

center-line of the strip is assumed constant along the width of the strip.

Design each strip for shear and flexure as in the continuous footing design.

Example

Design the mat foundation, assuming that and . The allowable

net soil bearing pressure is . Columns dimensions are .


389

Solution:

0.9 m

9 m

9 m

9 m

0.9 m

0.6 m 0.6 m

7.3 m 7.3 m 7.3 m

X

Y

A B C D

E F G H

I J K

K

L

M N O P

445 KN 267 KN (712) KN (961) KN

801 KN 534 KN (1335) KN (1816) KN

845 KN 534 KN (1379) KN (1868) KN

489 KN 311 KN (800) KN (1084) KN

801 KN 534 KN (1335) KN (1816) KN

1601 KN 890 KN (2491) KN (3345) KN

1779 KN 1112 KN (2891) KN (3914) KN

890 KN 534 KN (1424) KN (1922) KN

845 KN 578 KN (1423) KN (1939) KN

1779 KN 1068 KN (2847) KN (3844) KN

1957 KN 1334 KN (3291) KN (4483) KN

890 KN 534 KN (1424) KN (1922) KN

534 KN 311 KN (845) KN (1138) KN

801 KN 534 KN (1335) KN (1816) KN

801 KN 534 KN (1335) KN (1816) KN

534 KN 311 KN (845) KN (1138) KN


390

∑

P M

D A

𝑚

𝑚

𝑒𝑦 𝑚

𝑒𝑥 𝑚

B

N


391

The maximum soil pressure ill b at point “P” and is l ss t allo abl soil

pressure .

Design the strip - B F J N

∑

961 1816 1868 1084 5729

1816 3345 3914 1922 10997

1939 3844 4483 1922 12188

1138 1816 1816 1138 5908

34822

Check

∑

(

)


392

The mat foundation will be divided into several strips in both directions. The stress

distribution will be defined for each strip as was done for the previous strip.

Each strip will be treated as continuous footing with flexure along the strip only. Analysis and

design of the strip will be analoge to the analysis an design of continuous footing.

𝑚

𝑚 𝑚

𝐾𝑁

𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝐾𝑁

𝐾𝑁

𝐾𝑁

𝑚 𝑚

𝑚

B F J

N

𝑚


393

CHAPTER 12 DEVELOPMENT, ANCHORAGE, AND SPLICING OF REINFORCEMENT.

12.1 INTRODUCTION

In a reinforced concrete beam, the flexural compressive

forces are resisted by concrete, while the flexural tensile

forces are provided by reinforcement, as shown in the

figure (a) right. For this process to exist, there must be a

force transfer, or bond, between the two materials. The

forces acting on the bar are shown in the figure (b). For the

bar to be in equilibrium, bond stresses must exist. If these

disappear, the bar will pull out of the concrete and the

tensile force, , will drop to zero, causing the beam to fail.

Bond stresses must be present whenever the stress or

force in a reinforcing bar changes from point to point

along the length of the bar. This is illustrated by

the free-body diagram in the figures. If is

greater than bond stresses, , must act on

the surface of the bar to maintain equilibrium.

Summing forces parallel to the bar, one finds

that the average bond stress, , is

( )

( )

And taking ( ) gives

12.2 FLEXURAL BOND

In a beam, the force in the steel at a crack can

be expressed as

where is the internal lever arm and is the

moment acting at the section.


394

If we consider a length of beam between two cracks, as shown in the above figures, the

moments acting at the two cracks are and . If the beam is reinforced with one bar of

diameter , the forces on the bar are as shown. Summing horizontal forces gives

( )

( )

( )

( )

If there is more than one bar, the bar perimeter is replaced with the sum of the perimeters

∑ , giving

∑

The last two equations give the average bond stress between two cracks in a beam. The

actual bond stresses vary from point to point between the cracks.

12.3 MECHANISM OF BOND TRANSFER

A smooth bar embedded in concrete develops bond by adhesion between the concrete and

the bar and by a small amount of friction. Both of these effects are quickly lost when the bar

is loaded in tension, particularly because the diameter of the bar decreases slightly, due to

Poisson’s ratio. For this reason, smooth bars are generally not used as reinforcement. In

cases where smooth bars must be embedded in concrete (anchor bolts, stirrups made of

small diameter bars, etc.), mechanical anchorage in the form of hooks, nuts, and washers on

the embedded end (or similar devices) are used.

Although adhesion and friction are present when a deformed bar is loaded for the first time,

these bond-transfer mechanisms are quickly lost, leaving the bond to be transferred by

bearing on the deformations of the bar as shown in figure (a). Equal and opposite bearing

stresses act on the concrete, as shown in figure (b). The forces on the concrete have both a

longitudinal and a radial component (figures c and d). The latter causes circumferential

tensile stresses in the concrete around the bar. Eventually, the concrete will split parallel to

the bar, and the resulting crack will propagate out to the surface of the beam. The splitting


395

cracks follow the reinforcing bars along the bottom or side surfaces of the beam,

as shown in figure below. Once these cracks develop, the bond transfer drops rapidly unless

reinforcement is provided to restrain the opening of the splitting crack.


396

The load at which splitting failure develops is a function of:

1. the minimum distance from the bar to the surface of the concrete or to the next bar

— the smaller this distance, the smaller is the splitting load;

2. the tensile strength of the concrete; and

3. the average bond stress — as this increases, the wedging forces increase, leading to a

splitting failure.

Typical splitting failure surfaces are shown in the above figure. The splitting cracks tend to

develop along the shortest distance between a bar and the surface or between two bars.

The circles touch the edges of the beam where the distances are shortest.

If the cover and bar spacings are large compared to the bar diameter, a pull-out failure can

occur, where the bar and the annulus of concrete between successive deformations pull out

along a cylindrical failure surface joining the tips of the deformations.

12.4 DEVELOPMENT LENGTH

Because the actual bond stress varies along the length of a bar anchored in a zone of

tension, the ACI Code uses the concept of development length rather than bond stress. The

development length, , is the shortest length of bar in which the bar stress can increase

from zero to the yield strength, . If the distance from a point where the bar stress equals

to the end of the bar is less than the development length, the bar will pull out of the

concrete. The development lengths are different in tension and compression, because a bar

loaded in tension is subject to in-and-out bond stresses and hence requires a considerably

longer development length. Also, for a bar in compression, bearing stresses at the end of the

bar will transfer part of the compression force into the concrete.

The development length can be expressed in terms of the ultimate value of the average

bond stress by setting in equation for equal to :

Here, is the value of at bond failure in a beam test.

1. Tension-Development Lengths

Analysis of Bond Splitting Load

Although the code equations for bond strength were derived from statistical analyses of test

results, the following analysis illustrates the factors affecting the splitting load.

Consider a cylindrical concrete prism of diameter , containing a bar of diameter as

shown in figure (a) below. The radial components of the forces on the concrete, shown

below in figures b and c, cause a pressure on a portion of the cross section of the prism, as

shown in (b). This is equilibrated by tensile stresses in the concrete on either side of the bar.

In figure (c), the distribution of these stresses has arbitrarily been assumed to be triangular.

The circular prism in the figure below represents the zones of highest radial tensile stresses,


397

shown by the larger circles in the previous figure. Splitting is assumed to occur when the

maximum stress in the concrete is equal to the tensile strength of the concrete, . For

equilibrium in the vertical direction in a prism of length equal to

(

)

where is the ratio of the average tensile stress to the maximum tensile stress and equals

for the triangular stress distribution shown in figure “c” below. Rearranging gives

(

)

If the forces shown in the figures b and c on page (395) are assumed to act at , the

average bond stress, , at the onset of splitting is equal to . Taking √ gives

√ (

)

The length of bar required to raise the stress in the bar from zero to is called the

development length,

Substituting equation for gives

√ (

)

Arbitrarily taking (for the reasons to be given in the derivation of Equations for

Development Lengths later in the chapter) and rearranging yields

√

Basic Tension-Development Equation.

The ACI Code simplified the design expressions, in two stages as will be explained later. First,

a basic expression was developed for the development length, given in ACI Code Section

12.2.3 as

√

(

) ( )


398

where the confinement term (

) is limited to 2.5 or smaller, to prevent pull-out bond

failures, and the length is not taken less than . Also,

is the development length, .

is the bar diameter, .

is a bar-location factor given in ACI Code Section 12.2.4.

is an epoxy-coating factor given in ACI Code Section 12.2.4.

is a bar-size factor given in ACI Code Section 12.2.4.

is the lightweight concrete factor defined in ACI Code Section 12.2.4(d).

is the smaller of:

(a) the smallest distance measured from the surface of the concrete to the center of a

bar being developed, and

(b) one-half of the center-to-center spacing of the bars or wires being developed.

is a transverse reinforcement factor given in ACI Code Section 12.2.3.

Values for these factors will be presented later.

The second stage in the derivation of the basic expression for was to substitute common

values of and into the basic expression for (equation ( )) as described next.

Simplified Tension-Development-Length Equation.

The basic equation for was simplified by substituting lower limit values of and for

common design cases, to get widely applicable equations that did not explicitly include these

factors. For deformed bars or deformed wire, ACI Code Section 12.2.2 defines the

development length, as given in the following Table.

The Cases 1 and 2 described in the top row of the Table are illustrated in the next figures

(a and b). The“code minimum”stirrups and ties mentioned in case 1 correspond to the

minimum amounts and maximum spacings specified in ACI Code Sections 11.4.5, 11.4.6.3,

and 7.10.5 ( , and ties spaces and bars arrangement).

and Smaller Bars

and Deformed Wires and Larger Bars

Case 1: Clear spacing of bars being developed

or spliced not less than , and stirrups or ties

throughout not less than the code

minimum

√

( )

√

( ) or

Case 2: Clear spacing of bars being developed

or spliced not less than and clear cover

not less than

Other cases

√

( )

√

( )

- The length computed using these equations shall not be taken less than .


399

Bar-Spacing Factor,

The factor is the smaller of two quantities:

1. In the first definition, is the smallest distance from the surface of the concrete to the

center of the bar being developed. (See figure (a) on page 397).

ACI Code Section 7.7.1(c) gives the minimum cover to principal reinforcement as .

For a beam stem not exposed to weather, with bars enclosed in stirrups or ties,

will be

( ) (

)

2. In the second definition, is equal to one-half of the center-to-center spacing of the

bars.

ACI Code Section 7.6.1 gives the minimum clear spacing of parallel bars in a layer as , but

not less than . For bars, the diameter is , giving the center-to-center

spacing of the bars as

and

The smaller of the two values discussed here is . The minimum

stirrups or ties given in ACI Code Sections 7.10.5, 11.4.5, and 11.4.6.3 correspond to

between and , depending on a wide range of factors. Thus, for this case,

( ) . Substituting this and the appropriate bar size factor, , into equation

( ) gives equations ( ) and ( ) in the above table.

For Case 2 in the table, consider a slab with clear cover to the outer layer of bars of and a

clear spacing between the bars of . It is assumed that splitting in the cover will be

restrained by bars perpendicular to the bars being developed. As a result, is governed by

the bar spacing. If the clear spacing is , the center-to-center spacing is . Thus,

. Substituting this into Eq. ( ) and taking gives Eqs. ( ) or ( ) in

the previous Table.

For the situation where the minimum clear cover to the bar being developed is and

the minimum clear spacing is is the smaller of and . Substituting and

into Eq. ( ), assuming that gives Eqs. ( ) and ( ), which apply to

cases other than 1 and 2, as shown in the Table.

Typically, is about for to bottom bars and about for and

larger bottom bars.


400

Factors in Equations ( ) through ( )

The Greek-letter factors in Equations ( ) through ( ) are defined in ACI Section 12.2.4 as

follows:

bar location factor

Horizontal reinforcement so placed that more than of fresh

concrete is cast in the member below the development length or

splice............................................................................................................ 1.3

Other reinforcement…................................................................................. 1.0

coating factor

Epoxy-coated bars or wires with cover less than or clear spacing less

than ....................................................................................................... 1.5

All other epoxy-coated bars or wires........................................................... 1.2

Uncoated and galvanized reinforcement .................................................... 1.0

The product of ( ) need not be taken greater than 1.7.

bar-size factor

and smaller bars and deformed wires .............................................. 0.8

and larger bars.................................................................................... 1.0

Comparison of equation ( ) with a large collection of bond and splice tests showed that a

shorter development length was possible for smaller bars.

lightweight-aggregate-concrete factor

When any lightweight-aggregate concrete is used ................................... 0.75

However, when the splitting tensile strength is specified, shall be

permitted to be taken as √ but not more than.......................... 1.0

When normal-weight concrete is used ........................................................1.0

transverse reinforcement index

where

total cross-sectional area of all transverse

reinforcement within the spacing , which

crosses the potential plane of splitting along

the reinforcement being developed within

the development length, .(illustrated in

right figure).

maximum center-to-center spacing of transverse reinforcement within ,

number of bars or wires being developed or spliced along the plane of splitting.

ACI Code Section 12.2.3 allows to be taken equal to zero to simplify the calculations,

even if there is transverse reinforcement.


401

IMPORTANT. The basic rule governing the development and anchorage of bars is as follows:

The calculated tension or compression in reinforcement at each section of reinforced

concrete members shall be developed on each side of that section by embedment length,

hook, headed deformed bar, or mechanical anchorages, or a combination thereof. (ACI Code

Section 12.1)

Example:

A 400-mm-wide cantilever beam frames into the edge of a 400-mm-thick wall similar to

figure below. To reach , the three bars at the top of the beam are stressed to their

yield strength at point at the face of the wall. Compute the minimum embedment of the

bars into the wall and the development length in the beam. The concrete is sand/low-

density concrete with a strength of . The yield strength of the flexural reinforcement

is . Construction joints are located at the bottom and top of the beam, as shown.

The beam has closed stirrups with at a spacing of throughout

its length. (The stirrups are not shown.) The cover is 40 mm to the stirrups. The three

bars are inside vertical steel in each face of the wall.

Solution:

Anchorage of a traight bar (in a wall).

We shall do this problem twice, first using ACI Code Section 12.2.2 (Table – page 398) and

then using ACI Code Section 12.2.3, Eq. ( ).

1. Find the spacing and confinement case for bars anchored in wall. The clear side

cover to the bars in the wall is ( ). The clear spacing of

the bars is

Because the clear spacing between the bars is not less than and the clear cover to the

bars exceeds , this is Case 2, and for bars, Eq. ( ) applies

2. Compute the development length. From Eq. ( ),

1.5 m

Construction joint

Wall

A

B

Construction joint

400 mm

@ 𝑚𝑚 O.C.

45

0 m

m

38

7.5

mm

𝑙𝑑


402

√

where

because there will be more than 300 mm of fresh concrete under the bar when

the concrete in the beam covers the bars ( ).

because the bars are not epoxy coated.

because the concrete has low-density aggregates

√

The bars must extend 2184 mm into the wall to develop the full yield strength.

Extend the bars 2.2 m into the wall.

3. Compute the development length for bars in the beam using Eq. ( ).

√

(

)

Where and are the same as in step 2. For a bar, . is the smaller of:

(a) the distance from the center of the bar to the nearest concrete surface; measuring

from the side face to the center of the bar is

, and

(b) half the center-to-center spacing of the bars; (

)

Thus, .

The transverse reinforcement index is:

where

is the spacing of transverse reinforcement along the development length,

which is equal to in this example (vertical steel bars in each

face of the wall).

The number of bars being anchored.

is the area of transverse reinforcement crossing the

potential splitting plane within the spacing (one vertical bar @ 300

mm in each face of the wall). Thus,

Then,

Thus, this term is set equal to 2.5 in Eq. ( ) to get

√

( )


403

Thus, and the bars must extend 1310 mm into the wall. Extend the bars 1.35

m into the wall.

In this case, there is a large difference between the computed from equation ( ) and

the computed from equation ( ). This is because equation ( ) was derived by using

(

) equal to 1.5. In this case it is actually equal to 2.5.

Development of a bar in a cantilever.

The cantilever extends 1.5 m from the face of the wall. The bars shown are stressed to their

yield strength at the face of the wall. Is there adequate development length for bars in

the span? If not, what is the largest-size bar that can be used?

The point of maximum bar force occurs at the face of the wall (point A). The bar must be

developed on each side of this point. To accomplish this, the bar must extend a minimum of

into the support (previous solution) and a minimum into the span.

1. Find the spacing and confinement case. From the previous example, the clear

spacing between the bars in the beam is . for the 175-mm stirrup

spacing, the minimum area of stirrups, by section 11.4.6.3, is

√

√

√

The stirrups provided are double-leg stirrups, for which .

The spacing does not exeed .

Since the clear spacing between the bars is at least and the stirrups exceed the minimum

amount required, this is case 1, and for bars, equation ( ) applies.

2. Compute the development length. From the previous example, the development

length for a top bar is .

Since the bar extend into the beam from the face of the wall, there

is insufficient length to develop a bar. We must use smaller bars, with shorter , or

hook the bars at . Try with ( compared with for

). For the new bar arrangement, we must start over.

1. Find the spacing and confinement case for bars anchored in the beam. The clear side

cover to the bars in the beam is ( ). The clear spacing

of the bars is

Since the stirrups exceed the ACI Code minimum and the clear spacing between the bars is

not less than and the clear cover to the bars exceeds , this is Case 1, and for

bars, Eq. ( ) applies


404

2. Compute the development length. From Eq. ( ),

√

√

The bars must extend 1258 mm into the beam. Since , bars can

be developed whithout hooks at the free end. Use bars.

ACI Section 12.2.5 allows the development length to be reduced by multiplying by the

ration ( ) ( ) . We will not take advantage of

this, because the reduced anchorage length will not allow the bars to be used for their full

capacity in the event of a change in use of structure.

Compression-Development Lengths

Compression-development lengths are considerably shorter than tension-development

lengths, because some force is transferred to the concrete by the bearing at the end of the

bar and because there are no cracks in such an anchorage region (and hence no in-and-out

bond). The basic compression-development length is (ACI Code Section 12.3)

√

where the constant has units of . The development length in compression

may be reduced by multiplying by the applicable modification factors given in ACI Code:

Section 12.3.3 for excess reinforcement……………………………. ( )

Reinforcement enclosed within spiral reinforcement……………………………………....( )

The resulting development length shall not be less than .

Example:

Determine the required development length of column steel reinforcements which are

embedded in a footing a s shown below in the figure.

Take , .

Solution:

√

√

The available development length:

𝑚𝑚


405

12.5 HOOKED ANCHORAGES

Behavior of Hooked Anchorages.

Hooks are used to provide additional anchorage when there is insufficient straight length

available to develop a bar. Unless otherwise specified, the so-called standard hooks

described in ACI Code Section 7.1 are used. Details of and standard hooks and

standard stirrup and tie hooks are given in the next figure. It is important to note that a

standard hook on a large bar takes up a lot of room, and the actual size of the hook is

frequently quite critical in detailing a structure.

OR

(a) Standard hooks—ACI Code Sections 7.1 and 7.2.1


406

A hook loaded in tension develops forces in the manner shown in the figure below. The

stress in the bar is resisted by the bond on the surface of the bar and by the bearing on the

concrete inside the hook. The hook

moves inward, leaving a gap between

it and the concrete outside the bend.

Because the compressive force inside

the bend is not collinear with the

applied tensile force, the bar tends to

straighten out, producing compressive

stresses on the outside of the tail.

Failure of a hook almost always

involves crushing of the concrete inside

the hook. If the hook is close to a side

face, the crushing will extend to the

surface of the concrete, removing the

side cover. Occasionally, the concrete

outside the tail will crack, allowing the

tail to straighten.

Design of Hooked Anchorages.

The design process described in ACI Code Section 12.5.1 does not distinguish between

and hooks or between top and bottom bar hooks. The development length of a hook,

(illustrated in the figure (a), page 405), is computed using the following equation, which

may be reduced by appropriate multipliers given in ACI Code Section 12.5.3, except as

limited in ACI Code Section 12.5.4. The final development length shall not be less than

or , whichever is greater. Accordingly,

(b) Stirrup and tie hooks—ACI Code Section 7.1.3


407

√ (

)

where for epoxy-coated bars or wires and 1.0 for galvonized and uncoated

reinforcement, and is the lightweight-aggregate factor given in ACI Code Section 12.2.4(d).

Multipliers from ACI Section 12.5.3

The factors from ACI Code Section 12.5.3 account for the confinement of the hook by

concrete cover and stirrups. Confinement by stirrups reduces the chance that the concrete

between the hook and the concrete surface will spall off, leading to a premature hook

failure.

For clarity, ACI Code Section 12.5.3(a) has been divided here into two sentences. The factors

are as follows:

12.5.3(a1) for hooks on and smaller bars with side cover (normal to the plane of

the hook) not less than 65 mm ..........................................................................

12.5.3(a2) for hooks on and smaller bars with side cover (normal to the plane of

the hook) not less than 65 mm and cover on the bar extension (tail) beyond the

hook not less than 50 mm .................................................................................

The multipliers in ACI Code Section 12.5.3(b) and (c) reflect the confinement of the concrete

outside the bend.

12.5.3(b) for hooks on and smaller bars that are either

enclosed within ties or stirrups perpendicular to the bar being developed, spaced

not greater than along the development length, , of the hook, as shown in

Fig. (a) below, or

enclosed within ties or stirrups parallel to the bar being developed, spaced not

greater than along the length of the tail extension of the hook plus bend, as

shown in Fig. (b)........................................................................................ , except

as given in ACI Code Section 12.5.4.

12.5.3(c) for hooks on or smaller bars enclosed within ties or stirrups

perpendicular to the bar being developed, spaced not greater than along the

development length, , of the hook………………………………………………. except

as given in ACI Code Section 12.5.4.

12.5.3(d) where anchorage or development for is not specifically required, reinforcement

in excess of that required by analysis ………………….… ( ) ( ).

ACI Code Section 12.5.4 states that for bars being developed by a standard hook at

discontinuous ends of members with both side cover and top (or bottom) cover over a hook

of less than 65 mm the hooked bar shall be (must be) enclosed within ties or stirrups

perpendicular to the bar being developed, spaced not greater than along the

development length of the hook, . In this case, the factors of ACI Code Section 12.5.3(b)

and (c) shall not apply. ACI Code Section 12.5.4 applies at such points as the ends of simply

supported beams (particularly if these are deep beams), at the free ends of cantilevers, and

at the ends of members that terminate in a joint with less than 65 mm of both side cover


408

and top (or bottom) cover over the hooked bar. Hooked bars at discontinuous ends of slabs

are assumed to have confinement from the slab on each side of the hook; hence, ACI Code

Section 12.5.4 is not applied.

Concrete cover per 12.5.4.


409

Table for Hook Lengths, from equation ( √ ) Times Factors

from ACI Code Sections 12.5.3 and 12.5.4, but Not Less than or .

Location Type

Hooked

Bar

size

Side

Cover,

mm

Top or

Bottom

Cover,

mm

Tail

Cover Stirrups or ties Factor

1. Anywhere,

12.5.3(a) Any Any Not required

2. Anywhere,

12.5.3(a) Any Not required

3. Anywhere,

12.5.3(b) Any Any Any

Enclosed in stirrups or ties perpendicular to hooked bar, spaced

along

Except as in line 6

4. Anywhere,

12.5.3(b) Any Any Any

Enclosed in stirrups or ties parallel to hooked

bar spaced

along

Except as in line 6

5. Anywhere,

12.5.3(c) Any Any Any


along

Except as in line 6

6. At the ends

of members,

12.5.4 or

Any


is the diameter of the bar being developed by the hook

The first stirrup or tie should enclose the hook within of the outside of the bend.

If two or more factors apply, is multiplied by the product of the factors

Line 6 (ACI Code Section 12.5.4) applies at the discontinuous ends of members

The previous Figures (page 408) shows the meaning of the words “ties or stirrups parallel

to” or “perpendicular to the bar being developed” in ACI Code Sections 12.5.3(b) and (c), and

12.5.4. In ACI Code Sections 12.5.3 and 12.5.4, is the diameter of the hooked bar, and the

first tie or stirrup shall enclose the bent portion of the hook, within of the outside of the

bend.

If a hook satisfies more than one of the cases in ACI Code Section 12.5.3, from Equation

is multiplied by each of the applicable factors. Thus, if a hook satisfies both the covers

from ACI Code Section 12.5.3(a) and the stirrups from 12.5.3(b), is the length from the


410

first part of Equation for multiplied by from 12.5.3(a) and 0.8 from 12.5.3(b), making

the total reduction

Finally, the length of the hook, shall not be less than or ,

whichever is greater, after all the reduction factors have been applied.

Hooks may not be used to develop bars in compression, because bearing on the outside of

the hook is not efficient.

Example (Hooked Bar Anchorage into a Column)

The exterior end of a 400-mm-wide-by-600-mm-deep continuous beam frames into

a 600 mm-square column, as shown below. The column has four longitudinal bars. The

negative-moment reinforcement at the exterior end of the beam consists of four bars.

The concrete is normal-weight concrete. The longitudinal steel strength is

. Design the anchorage of the four bars into the column. It should be clear

(from the previous examples) that a straight bar cannot be developed in a 600-mm-

deep column. Thus, assume a hooked bar anchorage is required.


411

Solution:

Compute the development length for hooked beam bars. The basic development length for a

hooked bar is

√

Therefore,

√

Assume that the four bars will extend into the column inside the vertical column bars,

as shown in the above figure. ACI Code Section 11.10.2 requires minimum ties in the joint

area. The required spacing of closed ties by ACI Code Section 11.10.2 (Except for

connections not part of a primary seismic load-resisting system that are restrained on four

sides by beams or slabs of approximately equal depth, connections shall have lateral

reinforcement not less than that required by Eq. (11-13) within the column for a depth not

less than that of the deepest connection of framing elements to the columns. See also 7.9.) is

computed as:

√

The second expression governs, so

The side cover to hooked bars is determined as:

(

)

This exceeds 65 mm and is therefore O.K. The top cover to the lead-in length in the joint

exceeds 65 mm because the joint is in the column.

The cover on the bar extension beyond the hook (the tail of the hook) is

12.5.3.2(a): The side cover exceeds 65 mm, and the cover on the bar extension is equal to

50 mm; therefore, the 0.7 reduction factor could be used. Note, if we used ties in the

joint, the cover on the bar extension after the hook would be less than 50 mm, and thus, the

multiplier = 1.0.

ACI Code Section 12.5.4 does not apply because the side cover and top cover both exceed

65 mm. Therefore, only the minimum ties required by ACI Section 11.10.2 are required:

ties at 300 mm. These are spaced farther apart than therefore,

ACI Code Section 12.5.3(b) does not apply, and so the multiplier is 1.0. Thus,

Take

The hook-development length available is


412

Because 600 mm exceeds 350 mm, the hook development length is O.K.

Check the vertical height of a standard hook on a bar. From Fig. a (page 410), the

vertical height of a standard hook is . This will

fit into the joint.

Therefore, anchor the four bars into the joint, as shown in the previous figure.

12.6 BAR CUTOFFS AND DEVELOPMENT OF BARS IN FLEXURAL MEMBERS

Why Bars Are Cut Off

In reinforced concrete, reinforcement is provided near the tensile face of beams to provide

the tension component of the internal resisting couple. A continuous beam and its moment

diagram are shown below. At midspan, the moments are positive, and reinforcement is

required near the bottom face of the member. The opposite is true at the supports. For

economy, some of the bars can be terminated or cut off where they are no longer needed.

The location of the cut-off points is discussed in this section.

Four major factors affect the location of bar cutoffs:

1. Bars can be cut off where they are no longer needed to resist tensile forces or where

the remaining bars are adequate to do so. The location of points where bars are no

longer needed is a function of the flexural tensions resulting from the bending

moments and the effects of shear on these tensile forces.

2. There must be sufficient extension of each bar, on each side of every section, to

develop the force in that bar at that section. This is the basic rule governing the

development of reinforcement, presented in Section 8-6 (ACI Code Section 12.1).

3. Tension bars, cut off in a region of moderately high shear force cause a major stress

concentration, which can lead to major inclined cracks at the bar cutoff.

4. Certain constructional requirements are specified in the code as good practice.


413

Generally speaking, bar cut offs should be kept to a minimum to simplify design and

construction, particularly in zones where the bars are stressed in tension.

In the following sections, the location of theoretical cut-off points for flexure, referred to as

flexural cut-off points, is discussed. This is followed by a discussion of how these flexural

cut-off locations must be modified to account for shear, development, and constructional

requirements to get the actual cut-off points used in construction.

Location of Flexural Cut-Off Points

The calculation of the flexural cut-off points will be illustrated with the simply supported

beam shown next. At midspan, this beam has reinforcing bars, shown in section

(figure c). At points and two of these bars are cut off, leaving bars in the end

portions of the beam, as shown in figure b.

The beam is loaded with a uniform factored load of including its selfweight, which

gives the diagram of ultimate moments, , shown in figure d. This is referred to as the

required-moment diagram, because, at each section, the beam must have a reduced

nominal strength, at least equal to . The maximum required moment at midspan is

Assuming concrete, Grade-420 reinforcement and a tension-controlled section

so , the moment capacity, , of the section with bars is ,

which is adequate at midspan. At points away from midspan, the required is less than

, as shown by the moment diagram in figure d. Thus, less reinforcement

(less ) is required at points away from midspan. This is accomplished by “cutting off” some

of the bars where they are no longer needed. In the example illustrated in the figure below,

it has been arbitrarily decided that bars will be cut off where they are no longer

needed. The remaining bars give a reduced nominal strength

. Thus, the two bars theoretically can be cut off when

because the remaining three bars will be strong enough to resist . From an equation for

the required-moment diagram (figure d), we find at from each support.

Consequently, the two bars that are to be cut off are no longer needed for flexure in the

outer of each end of the beam and theoretically can be cut off at those points, as

shown in figure e.

Figure f is a plot of the reduced nominal moment strength, , at each point in the beam

and is referred to as a moment-strength diagram. At midspan (point E in figure e), the beam

has five bars and hence has a capacity of . To the left of point , the beam

contains three bars, giving it a capacity of . The distance represents the

development length, , for the two bars cut off at . At the ends of the bars at point ,

these two bars are undeveloped and thus cannot resist stresses. As a result, they do not add

to the moment capacity at . On the other hand, the bars are fully developed at , and in

the region from to ’ they could be stressed to if required. In this region, the moment

capacity is .


414


415

The three bars that extend into the supports are cut off at points and . At and ,

these bars are undeveloped, and as a result, the moment capacity is at and .

At points and the bars are fully developed, and the moment capacity

In figure g, the moment-capacity diagram from figure f and the required moment diagram

from figure d are superimposed. Because the moment capacity is greater than or equal to

the required moment at all points, the beam has adequate capacity for flexure, neglecting

the effects of shear.

The ACI Code does not explicitly treat the effect of shear on the tensile force. Instead, ACI

Code Section 12.10.3 arbitrarily requires that longitudinal tension bars be extended a

minimum distance equal to the greater of or past the

theoretical cut-off point for flexure.

Development of Bars at Points of Maximum Bar Force. For reinforcement and concrete to

act together, each bar must have adequate embedment on both sides of each section to

develop the force in the bar at that section. In beams, this is critical at

1. Points of maximum positive and negative moment, which are points of maximum bar

stress.

2. Points where reinforcing bars adjacent to the bar under consideration are cut off or

bent (ACI Code Section 12.10.2).

Thus, bars must extend at least a development length, , each way from such points or be

anchored with hooks or mechanical anchorages.

ACI Code 12.10 requires special precautions, specifying that no flexural bar shall be

terminated in a tension zone unless one of the following conditions is satisfied:

1. The shear is not over two-thirds of the design strength .

( )

2. Stirrups in excess of those normally required are provided over a distance along each

terminated bar from the point of cutoff equal to

. These "binder" stirrups shall

provide an area . In addition, the stirrup spacing shall not exceed

, where is the ratio of the area of bars cut off to the total area of bars

at the section.

3. The continuing bars, if or smaller, provide twice the area required for flexure at

that point, and the shear does not exceed three-quarters of the design strength

( )


416


417


418

12.7 DEVELOPMENT OF POSITIVE MOMENT REINFORCEMENT.

The concept of requiring the development of reinforcement on both sides of a section where

the bars are to be fully stressed may also be applied to the continuation of positive moment

tension reinforcement beyond either the centerline of a simple support or a point of

inflection.

Simple Supports

Referring to the next figure, consider the point A on the factored moment curve near a

simple support, where the factored moment equals the moment capacity of the

bars continuing into the support. The distance from point A to the end of the bars must be at

least equal to the required development length as computed from ACI-12.2. This

requirement given in ACI-12.11.3 is that the available embedment length must equal or

exceed , or

( )

Where nominal flexural strength of all reinforcement at the section assumed to be

stressed at (

)

factored shear at the support, and

the straight embedment length beyond the centerline of support to the end of

the bars [When the bars are hooked, the equation ( ) is considered automatically satisfied.]

In the next figure, the area of the shaded portion of the shear diagram equals the change in

moment between the center of support and point A; thus the ordinate on the factored

moment diagram at A is

Thus, the distance between the point A and the centerline of support is approximately

equal to

( )

Comparing Eq. ( ) and Eq. ( ), it is seen that Eq. ( ) is identical to

or

The factor accounts for the fact that bars extending into a simple support have less

tendency to cause splitting when confined by a compressive reaction. When the beam is

supported in such a way that there are no bearing stresses above the support, the factor

becomes (as eq. ( )) [The ACI Code allows a increase in the value of or

in

such cases [see Eq. ( )].


419

Inflection Points

Since an inflection point is a point of zero moment located away from a support (refer to the

next figures), bars in that region are not confined by a compressive reaction; therefore the

factor is interpreted as not to apply. In this case, the embedment length that must

exceed the required development length ; (ACI-12.11.3) may be stated as

[

]

( )

where and refer to the nominal flexural strength and the factored shear at the point

of inflection.


420

The limitation of the usable to or the has been

applied because, according to the ACI Commentary-R12.11.3, there is no experimental

evidence to show that long anchorage length will be fully effective in developing a bar in a

short length between the point of inflection and a point of maximum stress.

Additional development of reinforcement at the face of support is required by ACI-12.11.2

when the flexural member is part of the primary lateral load resisting system.

Equations ( ) and ( ) are written in terms of rather than because the

development length equations were derived on the basis of developing in the bars, not

. These Equations are not applied in negative-moment regions, because the shape of the

moment diagram is concave downward such that the only critical point for anchorage is the

point of maximum bar stress.

The consequence of these special requirements at the point of zero moment is that, in some

cases, smaller bar sizes must be used to obtain smaller , even though requirements for

development past the point of maximum stress are met.

ACI Code 12.11.1 requires that at least one-third the positive moment reinforcement in

simple members and one-fourth the positive moment reinforcement in continuous members

shall extend along the same face of member into the support. In beams, such reinforcement

shall extend into the support at least .


421

Example

Check the Code requirement of development length at support of the beam shown.

Take .

Solution:

𝐾𝑁 𝑚

𝑚𝑚

𝑚 𝑚𝑚

𝑚𝑚

𝑚𝑚


422

(

)

Using equation ( ) from the table on page 398 (Case 1 with minimum stirrups).

√

√

The available length satisfies the Code requirement.

12.8 DEVELOPMENT OF NEGATIVE MOMENT REINFORCEMENT.

According to ACI Code 12.12, at least one-third of the total reinforcement provided for

negative moment at the support must be extended beyond the extreme position of the

point of inflection a distance not less than one-sixteenth the clear span ( ), or , or

, whichever is greatest.

Requirements for bar cutoff or bend point locations are summarized in the next figure. If

negative bars are to be cut off, they must extend a full development length beyond the

face of the support. In addition, they must extend a distance or beyond the

theoretical point of cutoff defined by the moment diagram. The remaining negative bars

(at least one-third of the total negative area) must extend at least beyond the theoretical

point of cutoff of bars and in addition must extend , , or (whichever is

greatest) past the point of inflection of the negative-moment diagram.

If the positive bars are to be cut off, they must project past the point of theoretical

maximum moment, as well as or beyond the cutoff point from the positive-moment

diagram. The remaining positive bars must extend past the theoretical point of cutoff of

bars and must extend at least into the face of the support.


423

When bars are cut off in a tension zone, there is a tendency toward the formation of

premature flexural and diagonal tension cracks in the vicinity of the cut end. This may result

in a reduction of shear capacity and a loss in overall ductility of the beam.

As an alternative to cutting off the steel, tension bars may be anchored by bending them

across the web and making them continuous with the reinforcement on the opposite face.

Although this leads to some complication in detailing and placing the steel, thus adding to

construction cost, some engineers prefer the arrangement because added insurance is


424

provided against the spread of diagonal tension cracks. In some cases, particularly for

relatively deep beams in which a large percentage of the total bottom steel is to be bent, it

may be impossible to locate the bend-up point for bottom bars far enough from the support

for the same bars to meet the requirements for top steel. The theoretical points of bend

should be checked carefully for both bottom and top steel.

It may be evident from review of the previous sections that the determination of cutoff or

bend points in flexural members is complicated and can be extremely time-consuming in

design. It is important to keep the matter in perspective and to recognize that the overall

cost of construction will be increased very little if some bars are slightly longer than

absolutely necessary, according to calculation, or as dictated by ACI Code provisions. In

addition, simplicity in construction is a desired goal, and can, in itself, produce compensating

cost savings. Accordingly, many engineers in practice continue all positive reinforcement into

the face of the supports the required 150 mm and extend all negative reinforcement the

required distance past the points of inflection, rather than using staggered cutoff points.

Example

Locate the flexural cut-off for the beam shown below.

Take .

Solution:

(

)

(

) (

)

𝐾𝑁 𝑚

𝑚𝑚

𝑚 𝑚𝑚

𝑚𝑚

𝑚𝑚


425

√

For :

(

)

√

√

The length of from symmetrical center of beam

Take

The total length of is .

𝑤 𝐾𝑁 𝑚

𝐴𝑦 𝐾𝑁

𝑥

𝑀𝑢

𝑉

𝑚

𝑥 𝑚 𝑚

𝑑 𝑚𝑚

𝑚𝑚


426

For :

The total length of is ( ) .

12.9 REINFORCEMENT CONTINUITY AND STRUCTURAL INTEGRITY

REQUIREMENTS.

The primary purpose for both the continuity and structural-integrity reinforcement

requirements is to tie the structural elements together and prevent localized damage from

spreading progressively to other parts of the structure. However, because of the limited

amount of calculations required to select and detail this reinforcement, structures satisfying

these requirements cannot be said to have been designed to resist progressive collapse.

Continuity Reinforcement

Requirements for continuity reinforcement in continuous beams are given in ACI Code

Sections 12.11.1 and 12.11.2 for positive-moment (bottom) reinforcement and in ACI Code

Sections 12.12.1 through 12.12.3 for negative-moment (top) reinforcement.

These requirements are summarized in next figures for positive-moment reinforcement and

for negative-moment reinforcement respectively.

𝐵𝐵 𝐿

𝐵𝐵 𝐿


427

ACI Code Section 12.11.1 requires that at least one-third of the positive-moment

reinforcement used at midspan for simply supported members and at least one-fourth of the

positive-moment reinforcement used at midspan for continuous members shall be

continued at least 150 mm into the supporting member (Fig. a). Further, if the beam under

consideration is part of the primary lateral load-resisting system, ACI Code Section 12.11.2

requires that the bottom reinforcement must be continuous through interior supports and

fully anchored at exterior supports (Fig. b).

ACI Code Section 12.12.1 requires that negative-moment reinforcement must be continuous

through interior supports and fully anchored at exterior supports (next figure).

ACI Code Section 12.12.2 requires that all of the negative-moment reinforcement must

extend the development length, , into the span before being cut off. Finally, ACI Code

Section 12.12.3 requires that at least one-third of the negative-moment reinforcement

provided at the face of the support shall be extended beyond the point of inflection a

distance greater than or equal to the largest of , , and . Theoretically, no top

steel should be required beyond the point of inflection, where the beam moment changes

from negative to positive. The minimum extension given by the ACI Code accounts for


428

possible shifts in the theoretical point of inflection due to changes in the loading and for the

effect of shear on longitudinal steel requirements.

Structural-Integrity Reinforcement

Requirements for structural-integrity reinforcement in continuous floor members first

appeared in the 1989 edition of the ACI Code. These requirements, which are given in ACI

Code Section 7.13, were clarified and strengthened in the 2002 and 2008 editions of the ACI

Code. The structural-integrity requirements are supplemental to the continuity

requirements discussed previously and were added to better tie the structural members

together in a floor system and to provide some resistance to progressive collapse. Because

the ACI Code Committee was concerned that a significant number of structural engineers

using Chapter 12 were not aware of the structural-integrity requirements, ACI Code Section

12.1.3 was added in 2008 to specifically direct the designer’s attention to the need to satisfy

ACI Code Section 7.13 when detailing reinforcement in continuous beams. Structural-

integrity requirements for reinforced concrete (nonprestressed) continuous slabs are given

in ACI Code Chapter 13, for precast construction in Code Chapter 16, and for prestressed

two-way slab systems in Code Chapter 18. Those requirements will not be discussed here.

The structural-integrity requirements in ACI Code Section 7.13 can be divided into

requirements for joists, perimeter beams, and interior beams framing into columns. For joist

construction, as defined in ACI Code Sections 8.13.1 through 8.13.3, ACI Code Section

7.13.2.1 requires that at least one bottom bar shall be continuous over all spans and through


429

interior supports and shall be anchored to develop at the face of exterior supports.

Continuity of the bar shall be achieved with either a Class B tension lap splice or a

mechanical or welded splice satisfying ACI Code Section 12.14.3. Class B lap splices are

defined in ACI Code Section 12.15.1 as having a length of (but not less than ).

The value for the development length, , is to be determined in accordance with ACI Code

Section 12.2, which has been given previously ( see pages 397 and 398 ). When determining

, the minimum does not apply, and the reduction for excessive reinforcement

cannot be applied.

ACI Code Section 7.13.2.2 states that perimeter beams must have continuous top and

bottom reinforcement that either passes through or is anchored in the column core, which is

defined as the region of the concrete bounded by the column longitudinal reinforcement.

The continuous top reinforcement shall consist of at least one-sixth of the negative-moment

(top) reinforcement required at the face of the support, but shall not be less than two bars.

The continuous bottom

reinforcement shall

consist of at least one-

fourth of the positive-

moment (bottom)

reinforcement required at

midspan, but not less than

two bars. At

noncontinuous supports

(corners), all of these bars

must be anchored to

develop at the face of

the support. Also, all of

the continuous

longitudinal bars must be

enclosed by closed

transverse reinforcement


430

(ACI Code Section 7.13.2.3), as specified for torsional transverse reinforcement in ACI Code

Sections 11.5.4.1 and 11.5.4.2, and placed over the full clear span at a spacing not exceeding

. As before, reinforcement continuity can be achieved through either the use of Class B

tension lap splices or a mechanical or welded splice.

For interior beams framing between columns, ACI Code Section 7.13.2.5 defines two ways to

satisfy the structural-integrity requirements for continuous longitudinal reinforcement. If

closed transverse reinforcement is not present, then structural integrity must be achieved by

continuous bottom reinforcement similar to that required for perimeter beams (see next

Figure a). As before, this reinforcement must pass through or be fully anchored in the

column core, and reinforcement continuity can be achieved through either a Class B tension

lap splice or a mechanical or welded splice. For interior beams that are not part of the

primary system for resisting lateral loads, the bottom reinforcement does not need to be

continuous through interior supports or fully anchored at exterior supports, and structural

integrity can be achieved by a combination of bottom and top steel that is enclosed by

closed transverse reinforcement (Fig. b). The top steel must satisfy the requirements of ACI

Code Section 12.12 and must be continuous through the column core of interior supports or

fully anchored in the column core of exterior supports. The bottom steel must satisfy the

requirements given in ACI Code Section 12.11.1. The closed transverse reinforcement (not

shown in Fig. b) must satisfy ACI Code Sections 11.5.4.1 and 11.5.4.2 and must be provided

over the full clear span at a spacing not exceeding . How continuity and structural-

integrity requirements affect the selection of cut-off points and longitudinal reinforcement

detailing are given in the examples.


431

Example

The beam shown below is constructed of normal-weight concrete, , and Grade-

420 reinforcement, . It supports a factored dead load of and a

factored live load of . The cross sections at the points of maximum positive and

negative moment, as given in figures below, are shown next.


432

Solution:

1. Locate flexural cut offs for positive-moment reinforcement. The positive moment in

span AB is governed by the loading case in

Fig. a below .From a free-body analysis of a

part of span AB (Fig. d), the equation for

at a distance from A is

At midspan, the beam has two plus two

bars. The two bars will be cut off. The

capacity of the remaining bars is calculated as

follows:

(

)

(

) (

)

Therefore, flexural cut-off points occur where

.

Setting the equation for

can be rearranged to


433

√

Thus or from A. These flexural cut-off points are shown in figure c for

bending moment diagram and figure a for location of positive moment cut-off points. They

will be referred to as theoretical flexural cut-off points E and F. In a similar fashion, by

setting the flexural cut-off point G is found to be from A or from B.

2. Compute the development lengths for the bottom bars.

Because the bar spacing exceeds for both the and bars, and because the beam

has minimum stirrups, the bars satisfy Case 1 in the Table on page 398.

√

√

Thus, for the bars, and for the bars

.

3. Locate actual cut-off points for positive-moment reinforcement. The actual cut-off

points are determined from the theoretical flexural cut-off points using rules stated

earlier. Because the location of cut offs G and D are affected by the locations of cut

offs E and F, the latter are established first, starting with F. Because the beam is

simply supported it is not included in the ACI Code listing of members that are

susceptible to actions requiring structural integrity.

a) Cutoff F. Two bars are cut off. They must satisfy rules for anchorage, extension

of bars into the supports and effect of shear on moment diagrams.

Extension of bars into the supports. At least one-third of the positive-moment

reinforcement, but not less than two bars, must extend at least into the

supports. We shall extend two bars into each of the supports A and B.

Effect of shear. Extend the bars by the larger of or

. Therefore, the first trial position of the actual cutoff is

at from the center of the support at A, say, . (see point

F’ in next Fig. b).

Anchorage. Bars must extend at least past the points of maximum bar stress. For

the bars cut off at F’, the maximum bar stress occurs near midspan, at from

A. The distance from the point of maximum bar stress to the actual bar cutoff is

. for the bars is . The distance available is

more than —therefore OK. Cut off two bars at from A (shown as point

F’ in Fig. b).


434

b) Cutoff G. Two bars are cut off; we must consider the same three items as

covered in step (a), plus we must check the anchorage at a point of inflection using

Eq. ( ) on page 419.

Extension of bars into simple supports. In step (a), we stated the need to extend two

bars into support B. Thus, G’ is at .

Effects of shear. Because the cut off is at the support, we do not need to extend the

bars further.

Anchorage. Bars must extend at least past actual cut offs of adjacent bars. for

bottom bar . Distance from F’ to G’

. Bar does not extend , therefore extend the bars to

say, .

Anchorage at point of inflection. Must satisfy Eq. ( ) on page 419 at point of

inflection (point where the moment is zero). Therefore, at G, . The

point of inflection is from the support (see figure c). At this point is

(Fig. b) and the moment strength for the bars in the beam at the point

of inflection (two bars) is

larger of ( ) or but not more than the actual

extension of the bar past the point of inflection ( ). Therefore,

and

OK, because this length exceeds . Cut off two bars ( )

from B (shown as point G’ in Fig. b).

c) Cutoff E. Two bars are cut off; we must check for the effect of shear and

development length (anchorage).

Effects of shear, positive moment. Extend the bars past the flexural

cutoff point. Therefore, the actual cutoff E’ is at

from A.

Anchorage, positive moment. The distance from the point of maximum moment to

the actual cutoff exceeds therefore OK. Cut off two bars at

from A (point E’ in Fig. b; note that this is changed later).

d) Cutoff D. Two bars are cut off; we must consider extension into a support,

extension beyond the cut-off point and development of bars at a simple support

using Eq. ( ) on page 418.

Extension into simple support. This was done in step (a).

Bars must extend from the actual cutoff E’, where ( bars).

The maximum possible length available is . Because

this is less than , we must either extend the end of the beam, hook the ends of the


435

bars, use smaller bars, or eliminate the cutoff E’. We shall do the latter. Therefore,

extend all four bars 150 mm past support A.

Development of bars at simple support. We must satisfy Eq. ( ) on page 418 at the

support.

( )

( ) (

)

Because this exceeds , development at the simple support is satisfied. The actual cut-off

points are illustrated in figure below.

4. Locate flexural cutoffs for negative-moment reinforcement. The negative moment is

governed by the loading case in the next figure a. The equations for the negative

bending moments are as follows:

Between A and B, with measured from A


436

and between C and B, with measured from C,

Over the support at B, the reinforcement is two bars plus two bars. The two

bars are no longer required when the moment is less than

(strength of the beam with two bars).

So, between A and B,

from A.

Therefore, the theoretical flexural cut-off point for two top bars in span AB is at

from A. Finally, between B and C,

from C

Therefore, the theoretical flexural cut-off point for

two top bars in span BC is at from C.

The flexural cut-off points for the negative-

moment steel are shown in the figure a below and

are lettered H, J, K, and L.

5. Compute development lengths for the top

bars. Because there is more than

of concrete below the top bars,

. Thus, for the bars,

, and for the bars,

.

6. Locate the actual cut-off points for the

negative-moment reinforcement.

Again, the inner cutoffs will be considered first,

because their location affects the design of the

outer cutoffs. The choice of actual cut-off points is

illustrated in Fig. b below.

a) Cutoff J. Two bars are cut off;

Effects of shear. Extend bars by

past the theoretical flexural

cut off. Cut off at

from A and from B, say, .


437

Anchorage of negative-moment steel. The bars must extend from the point of

maximum bar stress. For the two top bars, the maximum bar stress is at B. The

actual bar extension is . This exceeds therefore OK.

Cut off two bars at from B (point J’ in Fig. b).

b) Cutoff H. Two bars cut off.

Anchorage. Bar must extend past J’, where . Length available

therefore OK. Extend two bars to

from the end of the beam (point H’ in Fig. b).

c) Cutoff K. Two bars are cut off. The theoretical flexural cut off is at from C

( from B).

Effect of shear. Extend bars . The end of the bars is at

from B, say, .

Anchorage. Extend past B. for a top bar . The extension of

is thus not enough. Try extending the top bars past B to point K’.

Therefore, cut off two bars at from B (point K’ in Fig. b). Note that this is

changed in the next step.


438

d) Cutoff L. Two bars are cut off.

Anchorage. The bars must extend past K’. For a top bar

. The available extension is which is less than

therefore, not OK. Two solutions are available: either extend all the bars to the end

of the beam, or change the bars to six bars in two layers. We shall do the

former.

The final actual cut-off points are shown in the next figure.

7. Check whether extra stirrups are required at cutoffs. ACI Code Section 12.10.5

prohibits bar cutoffs in a tension zone, unless

12.10.5.1: at actual cutoff

( ) at that point, or

12.10.5.2: Extra stirrups are provided at actual cut-off point, or

12.10.5.3: The continuing flexural reinforcement at the flexural cutoff has twice the required

and ( ).

Because we have determined the theoretical cut-off points on the basis of the continuing

reinforcement having 1.0 times the required it is unlikely that we could use ACI Code

Section 12.10.5.3, even though the actual bar cut-off points were extended beyond the

theoretical cut-off points. Further, because we only need to satisfy one of the three sections

noted, we will concentrate on satisfying ACI Code Section 12.10.5.1.

As indicated from the beam section in Fig. b on page 432, the initial shear design was to use

double-leg stirrups throughout the length of the beam. We can determine the value

for

( )

√

√

( )

From this,

. So, if at the actual cut-off point exceeds , we will

need to modify the design for the stirrups to increase , and thus, .

a) Cutoff F’. This cutoff is at from B, which is in a flexural tension zone for the

bottom reinforcement. From Fig. b (page 432), for load Case 1, the shear at F’ is:

where the sign simply indicates the direction of the shear force. This value for exceeds

so we must either decrease the spacing for the stirrups or change

to a larger ( ) stirrup. We will try stirrups at a spacing:

and then,


439

( )

This value exceeds so the modified stirrup design is OK. This tighter spacing should start at

the cut-off point and extend at least a distance toward the maximum positive-moment

region. For simplicity, use a stirrup spacing from the center of support B for 1.65

in. toward midspan of span A–B (see next Figure).

b) Cutoff J’. The cutoff is located at from B. The flexural tension that occurs in

these bars is due to load Case 2 (Figure on page 437). By inspection, at J’ is

considerably less than

. Therefore, no extra stirrups are required

at this cut off.

The final reinforcement details are shown in next figure. For nonstandard beams such as this

one, a detail of this sort should be shown in the contract drawings.

The calculations just carried out are tedious, and if the underlying concepts are not

understood, the detailing provisions are difficult to apply. Several things can be done to

simplify these calculations. One is to extend all of the bars past their respective points of

inflection so that no bars are cut off in zones of flexural tension. This reduces the number of

cutoffs required and eliminates the need for extra stirrups, on one hand, while requiring

more flexural reinforcement, on the other. A second method is to work out the flexural cut-

off points graphically.

Example

Design the simply supported beam shown below. The dead load is not

including the weight of the beam. The live load consist of a concentrated load of at

midspan.

Take .

Use @ for vertical stirrups. Assume for self-weight.


440

Solution:

( ) .

( ) .

Take

At midspan ( )

(

)

(

) (

)

( )

For ( )

(

)

(

) (

)

( )

𝑚

𝑐𝑚

𝑚𝑚

𝑚𝑚


441

√

For :

√

(

)

(a)

, and

(b) half the center-to-center spacing of the bars; one bar only (no spacing)

Thus, .

(

) (

) (

)

√

For :

(a)

, and

(b) half the center-to-center spacing of the bars;

( )

Thus, .

(

) (

) (

)

√

NOTE: according to ACI 12.2.3 It shall be permitted to use as a design simplification even if transverse reinforcement is present.


442

For :

The length of from symmetrical center of beam

Take

The total length of is .

For :

The total length of is ( ) .

𝐵𝐵 𝐿

𝐵𝐵 𝐿

𝑚

𝑥 𝑚 𝑚

𝑑 𝑚𝑚

𝑙𝑑 𝑚𝑚

𝑙𝑑 𝑚𝑚

𝑚

𝑚𝑚


443

Example

Design the beam shown below. The total factored load is

Take .

Stirrups @

Solution:

Point of inflection (when )

𝑐𝑚

𝑐𝑚

𝑚 𝑚

𝐴𝑦 𝐾𝑁 𝐵𝑦 𝐾𝑁

𝑤𝑢 𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝑚

𝑐𝑚 𝑐𝑚


444

For ( ) ( )

(

)

(

) (

)

For ( )

(

)

(

) (

)

√

Development length:

√

(

)

For :

(c)

, and

(d) half the center-to-center spacing of the bars; one bar

only (no spacing)

Thus, .

(

) (

) (

)

Splitting plane


445

√

For

For :

(c)

, and

(d) half the center-to-center spacing of the bars;

( )

Thus, .

(

) (

) (

)

√

For (outer or corner bars):

(a)

, and

(b) half the center-to-center spacing of the bars;

(

)

Thus, .

(

) (

) (

)

√

Splitting plane

Splitting plane


446

The final length of bars will be calculated and checked as discussed in the example

(page 431). These bars are illustrated in the above figure.

𝐴𝑦 𝐾𝑁 𝐵𝑦 𝐾𝑁

𝑤𝑢 𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝐾𝑁 𝑚

𝑚

𝑚

𝑚

𝐾𝑁 𝑚

𝑑 𝑚 𝑑 𝑚

𝑙𝑑 𝑚 𝑙𝑑 𝑚 𝑙𝑑 𝑚

𝑙𝑑 𝑚

𝑚

𝑚

𝑚 𝑚

𝐵 𝐵 𝐿

𝐵 𝐵 𝐿

𝑐𝑚

𝑇 𝐵 𝐿


447

12.10 SPLICES OF REINFORCEMENT.

Frequently, reinforcement in beams and columns must be spliced. There are four types of

splices: lapped splices, mechanical splices, welded splices, and end-bearing splices. All four

types of splices are permitted, as limited in ACI Code Sections 12.14, 12.15, and 12.16.

Tension Lap Splices

In a lapped splice, the force in one bar is transferred to the concrete, which transfers it to

the adjacent bar. The force-transfer mechanism shown in Fig. a is clearly visible from the

crack pattern sketched in Fig. b. The transfer of forces out of the bar into the concrete

causes radially outward pressures on the concrete, as shown in Fig. c; these pressures, in

turn, cause splitting cracks along the bars similar to those shown in Fig. a (page 395). Once

such cracks occur, the splice fails as shown in the next photo. The splitting cracks generally

initiate at the ends of the splice, where the splitting pressures tend to be larger than at the

middle.

As shown in Fig. b, large transverse cracks occur at the discontinuities at the ends of the

spliced bars. Transverse reinforcement in the splice region delays the opening of the splitting

cracks and hence improves the splice capacity.

ACI Code Section 12.15 distinguishes between two types of tension lap splices, depending on

the fraction of the bars spliced in a given length and on the reinforcement stress at the

splice. Table R12.15.2 of the ACI Commentary is reproduced below. The splice lengths for

each class of splice are as follows:

Class A splice:

Class B splice:


448

Because the stress level in the bar is accounted for in Table above, the reduction in the

development length for excess reinforcement allowed in ACI Code Section 12.2.5 is not

applied in computing for this purpose.

The center-to-center distance between two bars in a lap splice cannot be greater than one-

fifth of the splice length, with a maximum of . (ACI Code Section 12.14.2.3). Bars

larger than cannot be lap spliced, except for compression lap splices at footing-to-


449

column joints (ACI Code Section 15.8.2.3). Lap splices should always be enclosed within

stirrups, ties, or spirals, to delay or prevent the complete loss of capacity indicated in the

previous photo. As indicated in ACI Code Sections 12.2.2 and 12.2.3, the presence of

transverse steel may lead to shorter and hence shorter splices. ACI Code Section 21.5.2.3

requires that tension lap splices of flexural reinforcement in beams resisting seismic loads be

enclosed by hoops or spirals.

For lap splices of slab and wall reinforcement, effective clear spacing of bars being spliced at

the same location is taken as the clear spacing between the spliced bars (R12.15.1). This

clear spacing criterion is illustrated in (a). Spacing for noncontact lap splices (spacing

between lapped bars not greater than ( ) lap length nor ) should be considered

the same as for contact lap splices. For lap splices of column and beam bars, effective clear

spacing between bars being spliced will depend on the orientation of the lapped bars; see

Fig. (b) and (c), respectively.

The designer must specify the class of tension lap splice to be used. The class of splice

depends on the magnitude of tensile stress in the reinforcement and the percentage of total

reinforcement to be lap spliced within any given splice length as shown in Table above. If the

area of tensile reinforcement provided at the splice location is more than twice that required

for strength (low tensile stress) and or less of the total steel area is lap spliced within the

required splice length, a Class A splice may be used. Both splice conditions must be satisfied,

otherwise, a Class B splice must be used. In other words, if the area of reinforcement


450

provided at the splice location is less than twice that required for strength (high tensile

stress) and/or more than 1/2 of the total area is to be spliced within the lap length, a Class B

splice must be used.

Compression Lap Splices

Reinforcing bars in compression are spliced mainly in columns, where bars are most often

terminated just above each floor or every other floor. This is done partly for construction

convenience, to avoid handling and supporting very long column bars, but it is also done to

permit column steel area to be reduced in steps, as loads become lighter at higher floors.

Compression bars may be spliced by lapping, by direct end bearing, or by welding or

mechanical devices that provide positive connection. The minimum length of lap for

compression splices is set according to ACI Code 12.16:

For bars with

For bars with ( )

but not less than . For less than , the required lap is increased by one-

third. When bars of different size are lap-spliced in compression, the splice length is to be

the larger of the development length of the larger bar and the splice length of the smaller

bar. In exception to the usual restriction on lap splices for large-diameter bars, and

bars may be lap-spliced to and smaller bars.

End-Bearing Splices

Direct end bearing of the bars has been found by test and experience to be an effective

means for transmitting compression. In such a case, the bars must be held in proper

alignment by a suitable device.

Section 12.16.4 specifies the requirements for end-bearing compression splices. End-bearing

splices are only permitted in members containing closed ties, closed stirrups or spirals

(12.16.4.3). Section R12.16.4.1 cautions the engineer in the use of end-bearing splices for

bars inclined from the vertical. End-bearing splices for compression bars have been used

almost exclusively in columns and the intent is to limit use to essentially vertical bars

because of the field difficulty of getting adequate end bearing on horizontal bars or bars

significantly inclined from the vertical. Mechanical or welded splices are also permitted for

compression splices and must meet the requirements of 12.14.3.2 or 12.14.3.4, respectively.

Column Splices

Lap splices, butt-welded splices, mechanical connections, or end-bearing splices may be used

in columns, with certain restrictions. Reinforcing bars in columns may be subjected to

compression or tension, or, for different load combinations, both tension and compression.

Accordingly, column splices must conform in some cases to the requirements for

compression splices only or tension splices only or to requirements for both. ACI Code 12.17


451

requires that a minimum tension capacity be provided in each face of all columns, even

where analysis indicates compression only. Ordinary compressive lap splices provide

sufficient tensile resistance, but end-bearing splices may require additional bars for tension,

unless the splices are staggered.

Type of lap splice to be used will depend on the bar stress at the splice location, compression

or tension, and magnitude if tension, due to all factored load combinations considered in the

design of the column. Type of lap splice to be used will be governed by the load combination

producing the greatest amount of tension in the bars being spliced. The design requirements

for lap splices in column bars can be illustrated by a typical column load-moment strength

interaction as shown.

Bar stress at various locations along the strength interaction curve define segments of the

strength curve where the different types of lap splices may be used. For factored load

combinations along the strength curve, bar stress can be readily calculated to determine

type of lap splice required. However, a design dilemma exists for load combinations that do

not fall exactly on the strength curve (below the strength curve) as there is no simple exact

method to calculate bar stress for this condition.

A seemingly rational approach is to consider factored load combinations below the strength

curve as producing bar stress of the same type, compression or tension, and of the same

approximate magnitude as that produced along the segment of the strength curve

intersected by radial lines (lines of equal eccentricity) through the load combination point.


452

This assumption becomes more exact as the factored load combinations being investigated

fall nearer to the actual strength interaction curve of the column. Using this approach, zones

of “bar stress” can be established as shown in the previous figure.

For factored load combinations in Zone 1, all column bars are considered to be in

compression. For load combinations in Zone 2 of the figure, bar stress on the tension face of

the column is considered to vary from zero to in tension. For load combinations in

Zone 3, bar stress on the tension face is considered to be greater than in tension. Type

of lap splice to be used will then depend on which zone, or zones, all factored load

combinations considered in the design of the column are located. The designer need only

locate the factored load combinations on the load-moment strength diagram for the column

and bars selected in the design to determine type of lap splice required. Use of load-moment

design charts in this manner will greatly facilitate the design of column bar splices. For

example, if factored gravity load combination governed design of the column, say Point A in

the figure above, where all bars are in compression, but a load combination including wind,

say Point B, produces some tension in the bars, the lap splice must be designed for a Zone 2

condition (bar stress is tensile but does not exceed in tension).

The design requirements for lap splices in columns are summarized in the Table below. Note

that the compression lap splice permitted when all bars are in compression (see 12.17.2.1)

considers a compression lap length adequate as a minimum tensile strength requirement.

For lap splices, where the bar stress due to factored loads is compression, column lap splices

must conform to the requirements presented in the previous section for compression

splices. Where the stress is tension and does not exceed , lap splices must be Class B if

more than one-half the bars are spliced at any section, or Class A if one-half or fewer are

spliced and alternate lap splices are staggered by . If the stress is tension and exceeds

then lap splices must be Class B, according to ACI Code.

If lateral ties are used throughout the splice length having an area of at least in

both directions, where is the spacing of ties and is the overall thickness of the

member, the required splice length may be multiplied by but must not be less than


453

. If spiral reinforcement confines the splice, the length required may be multiplied

by but again must not be less than .

End-bearing splices, as described above, may be used for column bars stressed in

compression, if the splices are staggered or additional bars are provided at splice locations.

The continuing bars in each face must have a tensile strength of not less than times

the area of reinforcement in that face.

As mentioned in before, column splices are commonly made just above a floor. However, for

frames subjected to lateral loads, a better location is within the center half of the column

height, where the moments due to lateral loads are much lower than at floor level. Such

placement is mandatory for columns in "special moment frames" designed for seismic loads.

Splices of welded deformed wire reinforcement in tension

For tension lap splices of deformed wire reinforcement, the code requires a minimum lap

length of , but not less than

Lap length is measured

between the ends of each

reinforcement sheet. The

development length is the

value calculated by the provisions

in 12.7. The code also requires

that the overlap measured

between the outermost cross

wires be at least . The

next Figure shows the lap length requirements.

If there are no cross wires within the splice length, the provisions in 12.15 for deformed wire

must be used to determine the length of the lap.


454

𝑙 𝑠

@ 𝑐𝑚 𝑐 𝑐

Splices of welded plain wire

reinforcement in tension

The minimum length of lap for

tension lap splices of plain wire

reinforcement is dependent upon

the ratio of the area of

reinforcement provided to that

required by analysis. Lap length is

measured between the outermost

cross wires of each reinforcement

sheet. The required lap lengths are

shown in Figure below.

Example

Compute the minimum lap length of the reinforcement at the retaining wall joints shown

below.

Take .

Solution:

Clear cover

(a)

, and

(b) half the center-to-center spacing

of the bars;

Thus, .

(

) (

) (

)

√

Class B lap splice Take .


455

Example

An axially loaded reinforced concrete column of dimension with six bars of

is intended to lapped at a floor level. Determine the lap length of the reinforcement.

Take .

Ties @

Solution:

√

√

In the 300 mm direction:

In the 450 mm direction:

Take

𝑙 𝑠

RC Slab

𝑚𝑚

𝑚𝑚


456

Example

Design the tension lap splices for the grade beam shown below.

Take .

Dimensions of the Beam section .

bars top and bottom (continuous).

Stirrups @ (entire span)

Bending moments @ @

IMPORTANT: Preferably, splices should be located away from zones of high tension. For a typical

grade beam, top bars should be spliced under the columns, and bottom bars about midway between

columns. Even though in this example the splice at A is not a preferred location, the moment at A is

relatively small. Assume for illustration that the splices must be located as shown.

Solution:

Assuming all bars are spliced at the same location.

(c)

, and

(d) half the center-to-center spacing of the bars; ( )

Thus, .


457

√

(

)

Lap Splice of Bottom Reinforcement at Section B

(

) (

) (

)

√

@

(

) (

)

( √

)

( √

)

( @ )

( )

Class B lap splice required

Note: Even if lap splices were staggered ( ), a Class B splice must be used

with (

)

Class B Splice

It is better practice to stagger alternate lap splices. As a result, the clear spacing between

spliced bars will be increased with a potential reduction of development length.

Clear spacing

Center-to-center spacing of bars being developed

Distance from center of bar to concrete surface

Thus,


458

(

) (

) (

)

√

Class B Splice

Use 1.1 m lap splice @ B and stagger alternate lap splices.

Lap Splice of Top Reinforcement at Section A

As size of top and bottom reinforcement is the same, computed development and splice

lengths for top bars will be equal to that of the bottom bars increased by the 1.3 multiplier

for top bars.

In addition, because positive and negative factored moments are different, the ratio of

provided to required reinforcement may affect the type of splice as demonstrated below.

@

(

) (

)

( √

)

( √

)

( @ )

( )

If alternate lap splices are staggered at least a lap length (As spliced = 50%):

Class A splice may be used

If all bars are lap spliced at the same location (within required n lap length):

Class B splice must be used

Assuming splices are staggered, the top bar multiplier will be 1.3.

Class A splice

Use 1.1 m lap splice @ A also, and stagger alternate lap splices.


459

Example

Design the lap splice for the tied column detail shown.

Continuing bars from column above ( )

Offset bars from column below ( )

Take .

Dimensions of the column section

.

bars (above and below floor level)

Ties @

Lap splice to be designed for the following factored load

combinations:

1.

2.

3.

Solution:

1. Determine type of lap splice required.

Type of lap splice to be used depends on the bar stress at

the splice location due to all factored load combinations

considered in the design of the column. For design

purposes, type of lap splice will be based on which zone, or


460

zones, of bar stress all factored load combinations are located on the column load-moment

strength diagram. The load-moment strength diagram (column design chart) for the

column with bars is shown above, with the three factored load

combinations considered in the design of the column located on the interaction strength

diagram.

Note that load combination (2) governed the design of the column (selection of

bars).

For load combination (1), all bars are in compression (Zone 1), and a compression lap splice

could be used. For load combination (2), bar stress is not greater than (Zone 2), so a

Class B tension lap splice is required; or, a Class A splice may be used if alternate lap splices

are staggered. For load combination (3), bar stress is greater than (Zone 3), and a Class

B splice must be used.

Lap splice required for the bars must be based on the load combination producing

the greatest amount of tension in the bars; for this example, load combination (3) governs

the type of lap splice to be used.

Class B splice required


461

2. Determine lap splice length

.

a) distance from center of bar being developed

to the nearest concrete surface

, and

b) half the center-to-center spacing of the bars.

Clear spacing between bars being developed is large

and will not govern.

Thus, .

√

(

)

(

) (

) (

)

√

Class B splice

Use lap splice for the bars at the floor level indicated.

reinforced concrete ii_2013-2014.pdf

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