Mosque Design

8/9/2019 Mosque Design

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SSTTRRUUCCTTUURRAALLDDEESSIIGGNNOOFFMMOOSSQQUUEE

Designed By

Adnan Riaz (2001-civil-952)

Muhammad Yousaf (2001-civil-959)

Adnan Ahmed (2001-civil-962)

Imran Malghani (2001-civil-948)

PROJECT ADVISOR ---------------- Prof. Dr. Zahid Ahmed Siddiqi

EXTERNAL EXAMINER ----------------

DEPARTMENT OF CIVIL ENGINEERING

UNIVERSITY OF ENGINEERING AND TECHNOLOGY LAHORE


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IN THE NAME OF ALLAH,

THE MOST BENEFICENT,THE MOST MERCIFUL.

Read: In the name of your Lord

Who created, created man from

A clot.

Read: And your Lord is most bounteous.

Who taught by the pen.

Taught man which he did not know.

Al Quran


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DEDICATION

This project is dedicated to

Our beloved parents,

Respected teachers,

And sincere friends.

For their efforts and worthy encouragement.


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ACKNOWLEDGEMENT

All gratitude are due to Almighty Allah most gracious most

merciful, Who is the entire source of all knowledge and wisdom endowed to

mankind and who capacitate us to complete our project.

We are highly indebted to the honorable advisor, Prof. Dr. Zahid

Ahmed Siddiqifor his valuable guidance, and useful suggestion.

Our deepest gratitude to Multi Dimesional Consultants Lahore

who guided us a lot for the successful fulfillment of this task.

Authors


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DECLARATION

We hereby declare that we developed this project and this report entirely on

the basis of our personal efforts made under the sincere guidance of our project

supervisor.

It is further declared that no portion of the work presented in this report has

been submitted in support of any application for any other degree or qualification

of this or any other University or institute of learning.

We further declare that this project and all associated documents, reports

and records are submitted as partial requirement for the degree of B.S Civil

Engineering.

We understand and transfer copyrights for these materials to University of

Engineering and Technology Lahore.

Adnan Riaz (2001-civil-952)

Muhammad Yousaf (2001-civil-959)

Adnan Ahmed (2001-civil-962)

Imran Malghani (2001-civil-948)

Project Advisor

Signature ____________


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TABLE OF CONTENTS

Page No

ACKNOWLEDGEMENT

Chapter -1 AIMS AND OBJECTIVES

1.1 INTRODUCTON 1

1.2 SPECIAL MEMBERS IN THE STRUCTURE 2

1.3 OBJECTIVES 3

Chapter-2 ANALYSIS AND DESIGN TECHNIQUES

2.1 DOMES 5

2.2 BEAMS 12

2.3 SLABS 26

2.4 COLUMNS 31

2.5 RETAINING WALL 38

2.6 WIND ANALYSIS FOR MINARET DESIGN 44

CHAPTER-3 DESIGN OF SLAB PANELS AND STAIRS

3.1 SLABS 55

3.2 TWO-WAY JOIST SLAB 78

3.3 STAIR DESIGN 81

CHAPTER-4 ANALYSIS OF BEAMS AND COLUMNS

USING SAP 2000

4.1 JOINT REACTIONS 89

4.2 FRAME ELEMENT FORCES 90


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CHAPTER-5 DESIGN OF BEAMS 96

CHAPTER-6 DESIGNOF COLUMNS AND

RETAINING WALL6.1 COLUMN DESIGN 118

6.2 FOOTING DESIGN 121

6.3 RETAINING WALL DESIGN 126

CHAPTER- 7 DOMES AND MINARET

7.1 DESIGN OF DOMES 130

7.2 DESIGN OF MINARET 134

7.3 DESIGN OF FOOTING OF MINARET 135

CHAPTER- 8 DRAWINGS

CHAPTER- 9 CONCLUSIONS 139

REFRENCES


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CHAPTER No. 1

AIMS AND OBJECTIVES

1.1 INTRODUCTION

1.1.1 Architectural Consideration of Mosque

Architectural Design is the first step in any constructional work. For a mosque the

architectural consideration should be kept in mind.

Capacity of mosque should be consistent with the community requirements. It

should also depend on the cost of the land in the community that if land is costly the area

may be less and storey # can be increase.2nd

thing is the esthetic that should be look like a

mosque. It should be well lit and scene of wideness should be there. There should be

harmony in all the components of the mosque that arches and sofit of the arches should

be of same shapes. Dome should also be of same shape.

1.1.2 Structural Design

Next thing of structural after architectural design is the structural design. This the

actual work of the Civil Engineer. It is starts from top to the bottom. Structural design

mean to make the architectural design feasible to construct and applicable with durability,

reliability, and safety with economy.

For a project the basic objective is the safety and economy and to meet these

requirements side by side a complete systematic procedure is adopted consisting of

following steps.


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1.2.2 Minarets:

Greenhouses, lath houses, Radio towers and other towers of trussed construction

shall be designed and constructed to withstand wind pressures specified.

For the purpose of wind-resistant design, each structure shall be placed in one of

the occupancy categories.

1.3 Objectives

1.3.1. To gain knowledge about practical design:

Up to the bachelor level of Engineering, the knowledge about the design of

domes, minarets and two way joist slab was poor but after this project Structural

Design Of Mosque, we found ourselves in a much better and stronger position

for such design.

1.3.2. To Study The Design And Construction Of Domes:

The design of domes is different from design of slabs. Domes are designed

against meridional thrust and tangential stresses. At crown magnitude of each of

the stresses is Wr/2 and at base it is Wr. Meridional steel can be curtailed at

different location according to the magnitude of stresses acting. Tangential

stresses are zero at an angle of 51048 from the Zenith (crown). Tangential

reinforcement is maximum at base and decreases upward and becomes zero at an

angle of 51048 from the crown, then again starts increasing. Meridional stresses

are maximum at base and minimum at crown.

1.3.4. To Study The Design And Construction Of Minaret:

The design of minaret is different from simple column design. Two square

minarets are designed, each of size 9ft 9ft. Height of minaret is kept 100 ft each

minaret consist of four columns. Since the wind effect is also there and wind


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pressure at the top of minaret is quite high, so we had to provide reinforcement

throughout the column same.

To provide enough stiffness, ring beams are provided at constant spacing

throughout the minaret. Wall panels are provided between columns of minaret.

Footing of minaret is set at the depth of height / 7 from the ground level.

The base slab of footing was designed against overturning, shear and bearing

capacity of soil.

1.3.6. To Study The Design Of Two-Way Joist:

Design of two-way joist is different from simple flat slab, because in this

case depth of slab is reduced and hence economy is achieved, infact by removing

the chunk of concrete from bottom, as steel in joist is enough to bear the tension at

bottom of equivalent flat slab. The joists are designed just like T-beams in both the

directions to provide sufficient stiffness to the slab system. The slab system in

prayer hall of basement is two-way joist.

1.3.7. Preparation Of Structural Drawings:

In the construction phase of any Civil Engineering projects, structural

drawings play an important role. It is easier for the Site Engineer to continue his

work with elaborated drawings. So the drawings must be clear and easy to

understand. Manual drawings are not comparable to the computer aided drawings.

In our project all the structural drawings are prepared using computer software

Auto-CAD. The advantages of using this software are enormous.

The main advantage to us is that we are now well aware of the detailing and

curtailment of reinforcement.


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Chapter 2:

ANALYSIS AND DESIGN TECHNIQUES

2.1. Domes

A member made in the form of a thin shell whose shape is a surface of revolution,

the axis of revolution being vertical is called dome. This type of structure may resist

applied loads by a series of pure tension and compression under certain conditions given

below :

1. That the surface is supported at as horizontal section.

2. That all loading must be symmetrical about the axis of revolution.

Consider a spherical dome under vertical loads as represented by the fig 2.1. At

the crown dome is carrying a point load F along with its own weight, the surface of dome

is considered to have uniform thickness very less in magnitude compared with other

dimensions.

Let

F = Point load at crown

W = Self wt per unit area

t = Thickness of dome (uniform very small)

r = Radius of the spherical surface Fig. 2.1

N = Intensity of direct stress as shown

T = Intensity of hoop tension or compression (as that of force T)

The dome is supported along the circular perimeter EF of a horizontal cross-

section of dome. The direction of supporting force is, by first assumption, tangent to the

surface.


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Take an annular ring DCKG bounded by two horizontal cross-sections, separately

shown in fig.2.2, the length of ring DC equal to r dand is the angle measured from the

pole or crown to point C. The stresses on the ring are.

rd

+ d

1.

A compressive stress N along its upper surface

2. A compressive stress N + dN along its lower surface

3. Self wt

4. Hoop tensile stress T acting perpendicular to the plane of

paper

Fig. 2.2

The horizontal radius of ring at horizontal section through C is CK = r sin

Now at upper perimeter through point C

Total resisting force = N Area

= N t 2 r sin

Total vertical component = 2r N t sin. sin

= 2r N t sin2

This must be equal to the total wt. above the section.

Wt. of dome itself = W Area of surface

= W (BK) 2r

= W 2r r (1-cos)

Total downward force

= 2r2(1-cos)W +F

Total downward force = total resisting force


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7

2r2(1-cos) W + F = 2r N t Sin2

N = Wr (1-cos) + F 2.1t sin

2 2r t sin2

Force per unit length of upper surface is N t

The outward horizontal component is N t cos

As the horizontal radius of ring is r sin. This would cause, if acting alone , a

hoop tension equal to

N t cosr sin

The stresses on lower surface would cause a hoop compression

(N + dN) t cos (+ d) r sin (+ d)

The difference would be total hoop tension which would be

T (t. r. d)

Nt cosr sin =

sin

cos

(r

2

(1-cos)W + 2

F

Putting value of N from eq. 2.1 we get :

+

=

22

2

cos2sin

coscos1' ec

rt

F

t

WrT ....2.2

2.1.2. Segmental Dome

A dome having a height less than the radius of the dome is called a segmental

dome.

Consider :

T = circumferential force (in horizontal plane) in unit strip at S.


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N = Meridional thrust (acting tangentially).

W = Uniform load per unit area of surface of dome.

T =

+

Cos

CosCosWr1

1

2

2.3

N =

2

1

Sin

CosWr

=

+ Cos

Wr1

1 ..2.4

AT CR N:OW

5148

Fig. 2.3

INTO THT(PAPER)

T = N = Wr (Compression)

LOAD F CONCENTRATED AT CROWN

T =

2

2Cos

r

F 2.5

22 rCosFN = .....2.6

Not applicable alues of for small v

.1.3. Shallow Segmental Dome1

APPROX. METHOD:

r = ad 12

a 28

pports = R = 2Total load on su r Fig. 2.5


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9

At springing = N =2.

2Rr

d

ING BEAM AT SPRINGING:TENSILE FORCE IN R

Tr =d

arR )(

.

xample:

e a hemispherical dome of 50 ft diameter, having a wall thickness of 5 in.

An external load of 10,000 lbs is applied over the crown spread over a circle of 5 ftdiamete

e = 50ft

me

5in

E

Analyz

r.

Dia of dom

Fig. 2.5

Hemispherical Do

Thickness =

So wt = 63 lb /sq ft

We have

22 2

)1(

rtSin

F

tSin

CosWr+

N =

= 2sin2)cos1( rt

F

t

Wr+

+

Suppose

= 10,000 lbs & spread over a circle of 5 ft diameterF


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10

So total force

= sqftlb /512)5(

4100002 =

Wi = 63 + 512

ft= 575 lb/sq

r = 25

t = 25/12 ft

N = )1(

.rWi

Cost +

= 542 Cos = 0.595

er exceeds 542

N = 17300 lb/sqft

Tension T does not exist as nev

Taking the whole dome

22)1( rtSin

F

Cost

WrN = +

+

= /2 cos = 0 sin = 1

N =15252

110000122563 2

51

+

r

= 3780 + 153

= 3933 lb/sq ft


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11

Hoop tension

T =

22

cos2cos1

coscos1ec

rt

F

t

Wr+

+

= 3933 lb/sqft

12

53933Tension per ft. height =

1630 lbs

0

=

For steel having Fy = 40,000 Psi

fs = 20,00 Psi

As =000,20

1630

= 0.1 Sq in

aking 8 bars f C/C both in horizontal & vertical planes and are placed at middle of

the section i.e., centre of slab.

T 3/


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2.2. Beams

2.2.1. Strength Design Of Rectangular Beams For Moment

From the basic principles and equations established in the preceding sections we

now develop a procedure for designing a beam with a rectangular cross section. Since

most reinforced concrete beams used in construction are rectangular, this procedure will

be used repeatedly by the designer. All steps-are consistent with the requirements of .the

current ACI Code 318-99.

All beams are designed to ensure that the moment produced by factored loads

does not exceed the available flexural design strength of the cross section at any point

along the length of the beam. If the flexural design strength Mnjust equals the required

flexural strengthMu (which ensures the most economical design), the criterion for design

can be stated as

Mu = Mn (2.21)

where 0.9 and Mn is the nominal moment capacity of the cross section.

This criterion can be developed into a design equation if we express Mn in terms of the

material and the geometric properties of a rectangular cross section (Fig. 2.7d). If we sum

moments about the centroid of the tension steel,Mncan be expressed as

=2

adCMn

(2.22)

where C is the resultant of the compressive stresses and a is the depth of the

rectangular stress block. As indicated in Fig. 2.20 d, C = 0.85fcab. Substituting this value

of C into Eq. (2.12)


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Fig2.20 (a) state of stress in an

underreinforced beam at failure;

(b) strains, (c) stresses, (d)

internal cou le

=2

'85.0a

dabfMn c (2.23)

To express a in Eq. (2.13) in terms of the dimensions-of the cross section and the

properties of the material, andfy, we set T = C and solve for a, to give

c

ys

fb

fAa

'85.0(= (2.24)

Multiplying both top and bottom of Eq. (2.14) by d and settingAJbd = p leads to

)'85.0( c

ys

fb

dfAa= =

c

y

f

df

'85.0

(2.25)

Substituting Eq. (2.15) into (2.13) and simplifying gives

=

c

yf

fybdfMn

'7.112

(2.26)

Finally, Eq. (2.15) is substituted into Eq. (2.11) to give the basic beam design equation

=

c

yffybdfMn

'7.112 (2.27)

where must not be greater than b, or less than minassociated withAs,min. The first

requirement ensures that the beam will be underreinforced and will fail in a ductile

manner; the second requirement prevents a brittle failure, i.e., the rupture of the steel

when the beam cracks initially.


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Equation (2.17) can be used either to investigate the capacity of a cross section if

the dimen sions and material properties are known or to design a cross section (i.e., to

establish the width b, the depth d, and the area of steel As) if the value of the factored

moment A is specified. Although Eq. (2.17) can be used to establish the flexural design

strength of a cross-section since all terms on the right side of-the equation are known, the

designer may prefer to work directly with the internal forces on the rectangular cross

section to evaluate Mnbecause of the simplicity, of the calculations. In the latter

procedure, T =Asfyis first evaluated, then the depth of the stress block a is computed by

equating T = C, and finally the internal couple is evaluated by multiplying Tby the arm

d - a/2between T and C.

2.2.2. Design Of Beams With Compression Steel

If a beam designed in accordance with the ACI Code is reinforced with tension

steel only, the maximum flexural capacity the cross section can develop is achieved when

an area of steel equal to three-fourths of the balanced steel area is used. When restrictions

are placed on the dimensions moment capacity of a member (even when reinforced with

three-fourths of the balanced steel area) may not be adequate to supply the required

moment capacity. Undo such conditions, additional moment capacity can be created

without producing a brittle, over reinforced beam by adding additional reinforcement to

both the tension and compression sides the cross section. As shown by Eq.(As,max.(3/4

Cc+Cs)/fy), the maximum area of tension steel that can be used to reinforce a cross section

is a direct function of both the strength of the concrete compression zone and the area of

the compression steelA's.

Figure 2.21 illustrates two situations in which compression steel can be usedadvantageously. In Fig. 2.21a compression steel is used to increase the flexural capacity

of the compression zone of a prefabricated beam whose sides have been cut back to

provide a seat to support beams framing in from each side". Figure 2.21b shows a

common design situation it which compression steel is used to reduce the size of a

continuous T-beam of constant cross section by adding flexural capacity in the region

where the effective cross section is smallest and the moment greatest. Near midspan of a


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continuous beam (see section 1 of Fig. 2.21b), where the positive moment creates

compression in the flange, the beam behaves as if it were rectangular beam with a width

equal to that of the flange. Even if the beam is shallow, the large compression zone

supplied by the flange provides the potential for a large moment capacity. If the moment

produces tension in the flange and compression in the web (the situation at the supports

where negative bending occurs), the beam, which now behaves like a narrow rectangular

beam with a width equal to that of the web, has a much smaller flexural capacity than the

flanged section at midspan. If compression steel is added to the compression zone (see

section 2 of Fig. 2.21), the flexural strength can be substantially raised without increasing

the width of the web or the depth of the cross section. By using compression steel to raise

the capacity of the compression zone the dead weight can be reduced and the headroom

increased.

To be most effective, compression steel should be placed where the compressive

strains at greatest, i.e., as far as possible from the neutral axis. If compression steel is

positioned near the neutral axis, the compressive strains may be too small to stress the

steel to its full capacity. Under this condition the compression steel has little influence on

the flexural strength or behavior of the member.

FIGURE 2.21 Examples

of beams reinforced with

compression steel; (a)

precast inverted T-beam,

(b) continuous beam with a

portion of the positive steel

extended into the supports

to be used as compression

steel


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FIGURE 2.22Load-deflection curvesShowing the improvement in ductility

and toughness produced by the

addition of compression steel in an

underreinforced beam.

Besides increasing the flexural capacity of a cross section, compression steel

produces a marked improvement in behavior by raising the amount of compressive strain

the concrete can sustain before crushing and by reducing the tendency of the concrete to

break down at high levels of strain. Stabilizing the compression zone of a highly stressed

beam with compression steel reduces creep and increases ductility. Comparing the load-

deflection curves of two under-reinforced beams of identical proportions (except for the

presence of compression steel in one). Fig. 2.22 illustrates the improvement in ductility

afforded by the addition of compression steel. As indicated in Fig. 2.22, the flexural

capacity is not increased significantly by the addition of compression steel to an

underreinforced beam because the magnitude of, the internal couple is controlled by the

area of the tension steel.

Recognizing the beneficial effect of compression steel on bending behavior, many

building codes require that all flexural members of structures located in seismic zones be

reinforced with a minimum area of compression steel, even when the design calculations

indicate that compression steel is not required for strength. The addition of compression

steel produces tough ductile members that can withstand the large bending deformations

and repeated reversals of "stress produced in building members by cyclic, earthquake-

induced ground motions.

Recognizing that an improvement in the strength and ductility of concrete in

compression can be achieved by providing lateral confinement of the concrete, ACI Code

7.11.1 requires that compression steel be enclosed by closely spaced ties throughout the

region in which it is used. By providing a certain limited amount of lateral confinement to

the concrete in the compression zone, ties increase the ultimate strain required to produce


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a compression failure and also reduce the rate at which heavily compressed concrete

strained into the inelastic region- breaks down.

When no. 10 (no. 30 metric) or smaller bars are used as compression steel, ACI

Code 7.10.5.1 specifies that ties be at least | in (11.3 mm) in diameter If bundled bars or

no. 11 (no. 35 metric) or larger bars are used as compression steel, ties must be at least in

(16 mm) in diameter. In accordance with ACI Code 7.10.5.2. the maximum spacing of

ties is not to exceed the smallest of the following distances:

1. Sixteen bar diameters of the compression steel

2. Forty-eight tie diameters

3. The least dimension of the cross section

Although inserting compression steel into a cross section permits the use of large

areas of tension steel, the designer must verify (1) that the steel can be fitted into the

tension zone while maintaining the required spacing between bars and the minimum

concrete cover sped-' fled by the ACI Code and (2) that the limit on crack width as

measured by the ACI expression z = 0.6 fy 3 Adc can be satisfied. While the use of a

small number of large-diameter bars increases the spacing between bars, the second

requirement , the control of crack width, is most easily satisfied by specifying a large

number of small-diameter bars.

FIGURE 2.23Moment capacity of a beam with compression steel; (a) cross section with A s=

As1+ As2, where As1 = Asb; (b) Strain distribution at failure based on the cross section

reinforcement with As1 only; (c) concrete couple M1= T1(d- a/2); (d) steel couple M2= T2(d-d)


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18

When designing a beam with compression steel, it is convenient to break the total

internal moment into two couples. The first coupleM (Fig. 2.23c) represents the nominal

flexural strength of the cross section reinforced with Asb where Asb, applies to the

section without compression steel. The second couple M2represents the nominal flexural

strength produced by the forces in the compression steel and in the additional tension

steel Asbwhich is added to balance the force in die compression steel (Fig. 2.23d). The

total moment capacity Mnof the cross section can then be expressed as

Mn = (M1+M2)

where = 0.9 the concrete couple is M1= T1(d - a/2), and the steel couple is

M2= T2(d - d')

2.2.2.1. Design Procedure

Step l. Determine the moment M\ that the beam can carry using ASl ^Asf,, where

A,t, represents balanced steel for the cross section without compression steel (see Fig.

2.23c).

Step 2. The excess moment Mi, the difference between the required flexural strength

Mu and the flexural design strength of the concrete couple


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19

Where 12 MMu

M =

Step 4.Compute the areas of the additional tension steel A,2and the compression steel

ASAt bottom:

y

sf

TA 2

2=

s

Sf

CsA

'' =

Where Cs = T2

2.2.3. Design Of T-Beams

2.2.3.1. Introduction

Beams with T-shaped cross sections are used extensively as components of

concrete structures. They occur most frequently when concrete beams are poured

monolithically with slabs to form the floors of buildings and the roadways of bridges.

Rigidly joined together by reinforcement, a portion of the slab acts with the beam to

produce a T-shaped flexural member. The slab is termed the flange, and the portion of the

beam that projects below the slab is called the stem (see Fig. 2.24).

Although isolated T-beams of poured-in-place concrete are uncommon, large

quantities of T-beams and double T-beams are produced by the precast concrete industry

for use as components of prefabricated buildings (see Fig. 2.25). These members are

typically placed side by side with their flanges joined to form a floor. Since precast

beams of the same nominal depth differ slightly in height as a result of the manufacturing

process, several inches of concrete topping are often placed on top of the flanges to form

a level-surface. Light reinforcement, such as welded-wire mesh or small-diameter


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deformed reinforcing bars, is added to the topping to provide continuity and reduce

cracking.

2.2.3.2. State of Stress at Failure

A T-shaped cross section is most efficiently used when the flange is placed in

compression. The wide flange not only permits a large compression force to develop but

also maximizes the arm of the internal couple by positioning the resultant of the

compression stresses near the compression surface (see Fig. 2.26). The elimination of

concrete from the tension zone, where only the steel reinforcement is effective in carrying

tension, reduces the dead weight but does not influence the bending strength of the cross

section. For long-span beams, where a large percentage of the design moment is

produced by the dead weight of the member, use of the T-shaped section will result in a

considerable reduction in weight, which in turn will permit the design of smaller and

lighter members.

FIGURE 2.25 Precast beams; (a) T-beam, (b)

double T-beam,FIGURE 2.24Floor System with T-beams

Since the flange of the typical T-beam is wide, the depth of the stress block will

normally be small. As a result, when failure occurs, the position of the neutral axis will

usually be located in the flange near the compression surface. As shown in Fig. 2.26, the

strains in the steel failure will be many times greater than those in the concrete because of

the elevated position of the neutral axis; therefore a ductile mode of failure associated

with large deflections and extensive stretching of the steel is assured.


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FIGURE 2.27

FIGURE 2.26 State of stress in a T-beam at failure; (a)

cross section,(b) strain, (c) stresses, (d) internal couple

2.2.3.3. Effective Width of Flange

In beams with a compact cross section, stress in the compression zone is assumed

to be constant in magnitude across the width of the beam (Fig. 2.27). In T-beams with

long thin flanges, the stresses vary across the flange width because of the sheardeformations of the flange. The approximate variation of stress in the flange is shown in

Fig. 2.28a.

To simplify the design of T-beams, the variable stress distribution acting over the

full width of flange is replaced by an equivalent uniform stress, which is assumed to act

over a reduced width beff, selected so that the uniform stress acting over the reduced

width produces the same resultant compression force in the flange as the actual stress,

which varies over the full width b. To establish the effective width of slab that acts as the

compression zone for a beam that is a component of T-beam-and-slab construction, ACI

Code 8.10 gives the following criteria.

Figure 2.28variation of compressive stresses

in the flange of a T-beam;(a) actual, (b)

simplified


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Figure 2.29Effective flange width

CASE 1: FLANGES ON EACH SIDE OF WEB. The effective width (see Fig. 2.29a) is

given by the smallest value of

1. One-fourth of the beam's span length

2. The stem width plus a flange overhang of eight times the slab thickness on each side of

the stem

3. The stem width plus a flange overhang not greater than half the clear distance to the

next beam.

CASE 2: BEAM WITH AN L-SHAPED FLANGE. The width of flange (see Fig. 2.29b)

is to be taken as the-stem width plus a flange overhang equal to the smallest of

1. One-twelfth the beam's span length

2. Six times the thickness of the slab

3. One-half the clear distance to the next beam

2.2.3.4. Distribution of Flexural Reinforcement in the Flanges

2.2.3.4.1. Longitudinal Reinforcement

When the flanges of T-beams carry tensile stresses (for example, in regions of

negative moment), ACI Code 10.6.6 requires that part of the main reinforcement be

spread over a width equal to the smaller of the effective flange width (Fig. 2.29) or a

width equal to one-tenth of the span. Further, if the effective flange width exceeds one-

tenth of the span, some reinforcement should be placed in the outer sections of the flange.

This provision will ensure that many fine cracks rather than a few wide cracks

perpendicular to the span of the beam will develop on the ' top surface.


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2.2.3.4.2. Transverse Reinforcement

Load applied directly to the flange of a T-beam will cause the flanges to bend

downward (Fig. 2.30). To prevent a bending failure of the flange, transverse

reinforcement must be added to the top of the flange overhangs. This reinforcement can

be sized by treating the flange over hangs as cantilevers fixed at the face of the stem and

having a span equal to the length of the flange "overhang (fig. 2.30b). ACI Code

8.10.5.2 requires that the spacing of the transverse reinforcement not exceed five times

the slab thickness or 18 in (500 mm). Additional longitudinal steel will be required in the

flange to hold the transverse steel in position when concrete is poured.

Figure 2.30 (a) transverse bending of T-

beam flange, (b) shear and moment curves

for flange overhange.

2.2.3.5. T-Beam Design

Most T-beams occur as part of continuous floor systems (see Fig. 2.24). The

dimensions of these beams are normally determined by the strength required to carry the

shear and moment at the supports, where the compression zone is at the bottom of the

stem and the member acts as a rectangular beam whose width is equal to that of the web

(see Fig. 2.26). To ensure that deflections are not excessive, the designer should verify

that the depth of the T-beam is not less than the minimum values specified in Table 2.2.2.

In regions of positive moment, where the flange is in compression, the designer

has only to select the area of the flexural steel and verify that it can be placed in the webwith the required spacing between bars. The minimum area of flexural steel equals, but

cannot be less than 2006K. In T-beam design, as in rectangular-beam design, the

maximum area of steel to be used as flexural reinforcement is equal to, (the procedure for

computingAs is given in Sec. 2.13).


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CASE 1: STRESS BLOCK CONFINED TO FLANGE

The design procedure for determining the moment capacity of a particular T-

shaped cross section will depend on the position of the bottom of the stress block. If the

stress block lies completely in the flange, the most common case, the beam is designed

exactly like a rectangular beam (Fig. 2.31). On the other hand, if the bottom of the stress

block falls in the web, the stress block must be divided into known areas, the forces on

these areas computed, and the moment capacity of the cross section established by

summing the forces in the compression zone about the centroid of the tension steel (Fig.

2.32). As an alternative, the trial method discussed in Sec. 2.12 can be used to select-the

reinforcement.

CASE 2: STRESS BLOCK EXTENDED INTO STEM.

Break the total internal moment capacity into two couples (Fig. 2.32). One couple

MI represents the moment capacity of the flange overhangs, and the second MI represents

the moment capacity of the rectangular beam portion. The total moment capacity is

Mn = (M1+M2) (2.27)

=

2'85.01

ffc hdAfM (2.28)

=2

'85.02a

dAfM wc

(2.29)

Figure 2.31 (a) cross section, (b)

strains at failure, (c) stresses (a hf),

(d) internal couple.


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Figure 2.32 (a) cross section with As= Asf+ Asw, where Asfis the portion of As

used to balance compression force in flange overhangs and Aswis the portion of As

used to balance compression force in web; (b) stresses, (c) moment: flange

overhangs and Asf; (d) moment: web and Asw

Substituting Eqs.(2) and (3) into (1) gives

Mn = +

2

'85.0f

fc

hdAf

2'85.0

adAf wc (2.30)


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2.3. Slabs

slab is a broad flat plate usually horizontal with top and bottom surface parallel or

nearly so, it may be supported by reinforced concrete beams by masonry or reinforced

concrete walls, by structural steel members, directly by columns or continuously by the

ground. It used to provide a flat, useful surface. Slabs may be supported on opposite sides.

Slabs may be of different types such as, one way slabs, two way slabs, flat slab, flat plate

slab, grid slabs or waffle slab.

2.3.1 One Way Slab

These are the types of slabs which are only on two opposite sides as shown in

figure in which the structural action of the slab is essentially on e way the load is carried

by the in the direction perpendicular to the supporting beams. one way slab action is

produced if the ratio of length to width of one slab panel is lager then about two (2) most

of the load is carried in the short direction to the supporting beams and one way action is

obtained in effect, even though supports are provided on all sides.

S

L 1 m

Figure 2.31


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1.0 m

LS

Main reinforcement

Figure 2.32

2.3.2 Two Way Slab

if the length to width ratio is greater than two it is referred as two way slab

generally in this case load is being carried by the slab in the direction perpendicular to the

supporting beams. There may be beams on all four directions so that two way direction is

obtained

Figure 2.33


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2.3.3. Flat Slab

Concrete slabmay in some cases becarried directly by the columns without use of

beam and girder .such slabs are considered as and are commonly used whereare not

larger and loads are perpendiculare directly .it may be beamless but incorporates are

thickened slab region in the vicinity of column and often employs flared columns top

2.3.4. Waffled Slab

A slab supported directly on columns without beams and having recesses on the

soffit so that comprises a series of ribs in the two direction is known as Waffle Slab.

Generally the width of web in the waffle slab shall not be more then 3.5 inches and clear

spacing should not exceed 3 ft and dept should not be more then 3.5 times the minimum

width of web of the rib. It is provided with a solid head on the column.

Figure 2.34

Waffle slab is consists of series of small, closely spaced reinforced concrete Tee

beams framing into monolithically cast concrete girder, which are in turn carried by the

building columns. These beams are formed by the Void space in what other wise would

be a solid slab. Usually these voids are formed using special steel pans. For the most parts

the concrete remove is in tension and ineffective. So the lighter weight concrete floor

having same structural characteristics as the corresponding solid floor. Voids are usually

formed by using domes shaped steel pans that are removed for re use after the slabs have

hardened. Near the column the form work are removed so as to get the solid face on the


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2.4 Columns

Columns are the members that carry loads chiefly in compression. Usually

columns carry bending moments about one and both axis of the cross section as well as

the compressive forces. That bending may also creates tensile forces on the one face of

the column. Even in such cases columns are generally referred to as compression

member, because the compression forces domain their behavior. In addition to the most

common type if compression member i.e. vertical element in structure, compression

members includes columns, arches, or inclined members of the trusses etc. etc.

Compression members may of any type below.

1. member reinforced with longitudinal bars and lateral ties.

2. members reinforced with longitudinal bars and continuous spirals.

3.

composite compression members reinforced longitudinally bars with structural

steel shapes, pipes, or tubing with or without additional longitudinal bars, and

various types of lateral reinforcements

The main reinforcement in columns are longitudinal one, parallel to the direction of

the load and consists of the bars arrange in a square, rectangular, or circular pattern.

2.4.1. Design Of Column By Using Design Aids ( Charts )

The design of eccentrically loaded columns using the strain compatibility method of

analysis described requires that a trial column be selected. The trial column is then

investigated to determine if it is adequate to carry any combination ofPuandMuthat may

act on it should the structure be overloaded, i.e., to see if PuandMufrom the analysis of

the. Structure, when plotted on a strength interaction diagram, fall within the region

bounded by the curve labeled "ACI design strength." Furthermore," economical design

requires that the controlling combination ofPuand Ag be close to the limit curve. If these

conditions are not met, a new column must be selected for trial. While a simple computer

program can be written, based on the strain compatibility analysis, to calculate points on the

design strength curves, and even to plot the curve for ant trial column, in practice design

aids are used such as are available in hand books and special volumes published by the

American Concrete Institute and Concrete Reinforcing Steel Institute .They cover the most

frequent practical cases, such as symmetrically reinforced rectangular and square columns


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and circular spirally reinforced columns. There are also a number of commercially

available computer programs (e.g., PCACOL, Portland Cement Association, Skokie,

Illinois, and HBCOLUMN, Concrete Reinforcing Steel Institute* Schaumburg, Illinois).

The graphs are seen to consist of strength interaction curves labeled "ACI design

strength," i.e. the ACI safety provisions are incorporated. However, instead of plottingPuvs.

Mu, corresponding parameters have been used to make the charts more generally applicable,

i.e. load is plotted asPu/ Ag while moment is expressed asPu/ Ag (e/h). Families of curves are

drawn for various values of g= Ast / Ag. They are used in most cases in conjunction, with

the family of radial lines representing different eccentricity ratios e/h.

Charts such as these permit the direct "design of eccentrically loaded columns

throughout the common range of strength and geometric variables. They may be used in

one of two ways as follows, For a given factored loadPuand equivalent eccentricity.

1. (a) Select trial cross section dimensions b and h .

(b) Calculate the ratio based on required cover distances to the bar centroids, and

select the corresponding column design chart.

(c) CalculatePu/ A andMu / Ag h, whereAg= bh

(d) From the graph, for the values found in (c), read the required steel ratiopH.

(e)

Calculate the total steel areaAS = pgb h.

2. (a) Select the steel ratiopg.

(b)Choose a trial value of h and calculate e/h andy.

(c)From the corresponding graph, readPu/ Agand calculate the requiredAg:

(d) Calculate b =Ag/ h

(e) Revise the trial value of h if necessary to obtain a well-proportioned section.

(f) Calculate the total steel areaAst= g b h


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2.4.2. Column Footings

Footings are structural members used to support columns and walls and to

transmit and distribute their loads to the soil in such a way that the load bearing

capacity of the soil is not exceeded, excessive settlement, differential settlement, or

rotation are prevented and adequate safety against overturning or sliding is

maintained.

A building or bridge is generally considered to have two main portionsthe

superstructure and the substructure. The latter is often called as foundation.

It supports the superstructure, but it may contain various parts or units of its own.

There are many special types of foundations for which concrete is used. Scope

here is limited to RC footings.

The term foundation generally includes the entire supporting structure.

Sometimes, as in retaining walls, it is used to designate the material upon which

wall is supported.

It must not be confused with the word FOOTING, which is generally applied only

to that portion of the structure which delivers the load to the earth.

2.4.3.1 Types of Column Footing

2.4.3.1.1. Isolated Spread FootingIsolated or single footings are used to

support single columns. This is one of

the most economical types of footings

and is used when columns are spaced

at relatively long distances.

2.4.3.1.2. Combined Footings

Combined footingsusually support

two columns, or three columns not

in a row. Combined footings are

used when tow columns are so close

that single footings cannot be used

or when one column is located at or


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near a property line.

2.4.3.1.3. Cantilever Or Strap Footings

They consist of two single footings

connected with a beam or a strap and

support two single columns. This type

replaces a combined footing and is

more economical.

2.4.3.1.4. Raft Or Mat Foundation

They consists of one footing

usually placed under the entire

building area. They are used,

when soil bearing capacity is

low, column loads are heavy

single footings cannot be used,

piles are not used and differentialsettlement must be reduced.

2.4.4 Distribution of Soil Pressure

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 the soil is not uniform, but depends on may factors especially the

composition of the soil and degree of flexibility of the footing.

2.4.5 Design Considerations

Footings must be designed to carry the column loads and transmit them to the soil

safely while satisfying code limitations.

The area of the footing based on the allowable bearing soil capacity


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Two-way shear or punching shear.

One-way bearing

Bending moment and steel reinforcement required

Bearing capacity of columns at their base

Dowel requirements

Development length of bars

Differential settlement

2.4.5.1. Size Of Footing

The area of footing can be determined from the actual external loads such that the

allowable soil pressure is not exceeded.

( )pressuresoilallowable

weight-selfincludingloadTotalfootingofArea =

2.4.5.2. Two-Way Shear (Punching Shear)

For two-way shear in slabs (& footings) Vcis smallest of

dbfV 0cc

c 4

2

+=

where,

bc= long side/short side of column concentrated load or reaction area < 2

b0= length of critical perimeter around the column

When b >2 the allowable Vcis reduced.


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2.4.5.3. One-Way Shear Or Beam Shear

For footings with bending action in one

direction the critical section is located a distance d

from face of column

dbf 2 =V 0cc

The ultimate shearing force at section m-m

can be calculated if no shear reinforcement is to be

used, then d can be checked

= dcL

bqV22

uu

2.4.5.4. Flexural Strength and Footing

reinforcement

Another approach is to calculated Ru = Mu /

bd2

and determine the steel percentage required.

Determine As then check if assumed a is close to

calculated a


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Afy

bf0a

85. c

s=

The minimum steel percentage required in flexural members is 200/fy with

minimum area and maximum spacing of steel bars in the direction of bending shall be as

required for shrinkage temperature reinforcement.

The reinforcement in one-way footings and two-

way footings must be distributed across the entire

width of the footing.

2Reinforcem

1directionshortinentreinforcemTotal

widthbandinent

+=

footingofsideshort

footingofsidelong=

2.4.5.5 Bearing Capacity of Column at Base

The loads from the column act on the footing at the base of the column, on an

area equal to area of the column cross-section. Compressive forces are transferred to the

footing directly by bearing on the concrete. Tensile forces must be resisted by

reinforcement, neglecting any contribution by concrete.Force acting on the concrete atthe base of the column must not exceed the bearing strength of the concrete

( )1c1 85.0 AfN =

Where f = 0.7 and

A1 =bearing area of column

The value of the bearing strength may be multiplied by a factor for bearing

on footing when the supporting surface is wider on all sides than the loaded area.

0.2/ AA 12

The modified bearing strength

( )

( )1c2

121c2

85.02

/85.0

AfN

AAAfN


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2.4.5.6. Dowels In Footings

A minimum steel ratio r = 0.005 of the column section as compared to r = 0.01 as

minimum reinforcement for the column itself. The number of dowel bars needed is four

these may be placed at the four corners of the column. The dowel bars are usually

extended into the footing, bent at the ends, and tied to the main footing reinforcement.

The dowel diameter shall not =exceed the diameter of the longitudinal bars in the column

by more than 0.15 in.

2.4.5.7. Development Length Of The Reinforcing Bars

The development length for compression bars was given

cbyd /02.0 fdfl =

but not less thanin.8003.0 by df

Dowel bars must be checked for proper development length.

2.4.5.8. General Requirements For Footing Design

1. A site investigation is required to determine the chemical and physical properties

of the soil.

2. Determine the magnitude and distribution of loads form the superstructure.

3. Establish the criteria and the tolerance for the total and differential settlements of

e structure.th

4.

Determine the most suitable and economic type of foundation.

5. Determine the depth of the footings below the ground level and the method of

excavation.

6.

Establish the allowable bearing pressure to be used in design.7. Determine the pressure distribution beneath the footing based on its width

8. Perform a settlement analysis.


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2.5. Retaining Wall

2.5.1. General

Retaining walls are used to provide lateral support for a mass of earth or othermaterial the top of which is at a higher elevation than the earth or rock in front of the wall

as shown in Fig 1. Gravity retaining walls such as shown in Fig 2.50 (a) depend mostly

upon their own weight for stability. They are usually low in height and are expensive

because of their inefficient use of materials; sometimes they may be cheapened by using

cyclopean concrete.

In contrast to them, Fig. 2.50 (b) pictures an ordinary cantilever retaining wall.

Part of its stability is obtained from the weight of earth mass on heel, but the wall's

resistance to collapse depends upon the strength of its individual parts as cantilever beams.

(a) (b)

Figure 2.50(a) Gravity Retaining Wall,

(b) Cantilever retaining wall


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2.5.2. Definition Of Parts

The various portions of a typical reinforced concrete retaining wall are defined

as follows, using Fig.2.51 for reference:

Back fill

Key HeelToe

Front

Stem or Wall

Fig 2.51Definition of Parts of wall

2.5.3. Functions

Retaining walls are structures used to retain earth or other materials, which would

not be able to stand vertically unsupported. These walls are used to hold back masses of

earth or other loose material where conditions make it impossible to let those masses

assume their natural slopes. Such conditions occur when the width of an excavation, cut,

or embankment is restricted by conditions of ownership, use of structure, or economy.

For example, in railway or highway construction the width of the right of way is fixed

and the cut or embankment must be contained within that width. Similarly, the

basement walls of buildings must be located within the property and must retain the soil

surrounding the basement.


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2.5.4. Types Of Retaining Walls

There may be several types of retaining structures, the main types being:

(a)

Gravity wall, where stability is provided by the weight of concrete in the wall;

( ver andb)Cantilever wall, where the wall slab acts as a vertical cantile

stability is provided by the weight of earth on the base and/or the weight of

the wall;

(c)Counter fort and buttress walls, where the slab is supported on its sides by the

counter forts. Stability is provided by the weight of the structure in the case of

the buttress wall and by the weight of earth on the base as well in the counter

fort wall.

A gravity wall is usually of plain concrete and is used for walls updo about 10 ft

high. The cantilever is the most common type of retaining walls and is used in the range

of 10-25 ft in height. The main parts of a cantilever retaining wall are the stem, heel,

toe, wall slab and base slab.

2.5.5. Earth Pressures

It is every day knowledge that sound rock, evenly bedded, will stand with a sheer

face t consi s. This can be seen in artificial form in deep railway cuttings.o derable height

Soils on the other hand cannot stand more steeply than their natural angle of repose,

as evidenced again in railway work where engineers have cut back to slopes ranging

from about 1 in 1.5 to 1 in 3, depending on the nature of soil, and other physical

considerations. Where there are practical objections to sloping back in this manner (as

for example at basements to city buildings or where industrial materials like sand andstone have to be stored in limited areas), a wall has to be built to retain all material

required to lie above the natural angle of repose.

The main problem in designing the retaining walls lies in determining the

pressures on the back of the wall from the material to be retained and the capability of

the ground in front of the wall and under the base to resist the lateral and vertical


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forced arising from those pressures. Knowledge of the properties and behavior of soils is

therefore fundamental to the design of retaining walls and the problem is closely allied

to foundation engineering. References should be made to the literature related to

foundation design.

2.5.5.1. Variation Of Earth Pressures

In liquid retaining structures, the applied forces (pressures) are directly related to

the density of the liquid retained and the head at which it acts. This is because liquids are

both friction less and cohesion less. Soils behave differently. Sand, for example, when dry,

acts as a fictional material without cohesion and has a well-defined angle of repose. If

the same sand is now moistened, it develops a certain amount of cohesive strength and its

angle of repose increases, somewhat erratically. Further wetting will break down the

internal friction forces until the sand slumps and will hardly stand at any angle at all.

Clay on the other hand when first exposed in situ stands vertically to considerable

depths when reasonably dry, but after time will subside, depending on its moisture

content. And clay, in dry seasons, gives up its moisture to atmosphere with subsequent

shrinkage, so that at depths less than about 4 or 5 feet it may be unreliable as a stop to

react the forward movement of a retaining wall.

Thus the pressures from soils can vary very widely depending on the moisture

content. If a unit volume of soil is considered at a depth h below the free surface, the

lateral pressures can vary from about 30h to 90h in sands, and from Oh to about 90h

in clays. And within these ranges, the pressures behind retaining walls may vary due to

seasonal or other periodic changes. Indeed the construction of the retaining wall itself

may cause major changes in the ground conditions blocking a natural drainage path, or

exposing to shrinkage otherwise stable clay. Similarly the fictional resistance to sliding

under the base of a retaining wall is critical of moisture content. This is particularly true of

clay, which when dry can be rough and hard, but when wet can be smooth and slippery.

When the soil behind the wall is prevented from lateral movement (towards or

away from soil) of wall, the pressure is known as earth pressure at rest. This is the case

when wall has a considerable rigidity. Basement walls generally fall in this category. If a

retaining wall is allowed to move away from the soil accompanied by a lateral soil


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expansion, the earth pressure decreases with the increasing expansion. A shear failure

of the soil is resulted with any further expansion and a sliding wedge tends to move

forward and downward. The earth pressure associated with this state of failure is the

minimum pressure is known as active earth pressure. Contrary to that if the retaining wall

could be forced to move toward the soil causing the lateral contraction of soil, a state of

failure is reached with the formation of an upward and backward sliding wedge. The

earth pressure associated with this state of failure is the maximum and termed as

passive earth pressure.

2.5.5.1.2. Earth Pressure at Rest

pa = Ca w h 2 .51

Where w is unit weight of soil, h the depth at which pressure is estimated

and Cothe constant known as the coefficient of earth pressure at rest.

2.5.5.1.3. Active Earth Pressure

pa =Ca w h 2.52

2.5.5.1.4. Passive Earth Pressure

Pp = Cp w h 2.53

PRESSURE FORCE


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2.5.6. Stability And Safety Factor

The stability of a retaining wall is its ability to hold its position and to perform

its function safe gnitude of the forces that arely. The safety factor is a measure of the ma

required to caus that are really actinge failure of the structure compared with the forces

upon it. Thus, if failure. If, for anysafety factor is 1, the wall will be upon the point of

given design, it is 2, t rces may be doubledhen the overturning moment or horizontal fo

before the wall ependwill fail. The magnitude of safety factor to be used in design will d

upon the engineer's judgm ng code that is to beent, the specifications, or the buildi

followed. In general, it may vary from 1.5 to 2.

A retaining wall may fail in one of the four ways: by the collapse of its

component parts, by overturning about its toe, by excessive pressure upon its

foundation, or by sliding upon its foundation. In a well-balanced design, the wall should

be equally safe in all respects.

2.5.7. Critical Sections For Bending And Shear

The bending is critical at the junction of base and wall slabs. In normal

circumstance the shear should have been critical at a distance equal to effective depth of

base slab from face of wall. However, the shear is critical at the junction of heel and wallslabs, because both wall and heel slab are in tension at the junction.

2.5.8. The Critical Loading Conditions

The surcharge loading on the heel slab should not be considered effective

while stability -against sliding or overturning is being investigated. Care must be

exercised in considering the presence of earth in front of wall while stability against

sliding is checked. The pressure under the heel slab and burden over the toe slab can

be neglected while designing these slabs for strength.


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2.6. Wind Analysis For Minaret Design

2.6.1. General

Every building or structure and every portion thereof shall be determined and

constructed to resist the wing effects determined in accordance with the requirements of

this division. Wind shall be assumed to come from any horizontal direction. No reduction

in the wind pressure shall be taken for the shielding effect of the adjacent structures.

Structures sensitive to dynamic effects, such as buildings with the height to width

ratio greater than five, structures sensitive to wind-excited oscillations, such as vortex

edding and icing, and building over 400 feet in height, shall be, and any structure may

e designed in accordance with approved national standards.

Provision of this section do not apply to building and foundation systems in those

d wave action. Buildings and foundations

bject to such loads shall be designed in accordance with approved national standards.

2.6.2. D

having

XPOSURE D represents the most severe exposure in areas with basic wind

per hour (129 Km/h) or greater and has terrain that is flat and

unobstr km) or more in width with

sh

b

areas to scour and water pressures by wind an

su

efinitions

The following definitions apply only to this division:

BASIC WIND SPEED is the fastest mile wind speed associated with an annual

probability of 0.02 measured at a point 33 feet (10000 mm) above the ground for an area

exposure category C.

EXPOSURE B has terrain with buildings, forest or surface irregularities,

covering at least 20 percent of the ground level area extending one mile (1.61 Km) or

more from the side.

EXPOSURE C has terrain that is flat and generally open, extending half mile

(0.81 Km) or more from the side in any full quadrant.

E

speed of 80 miles

ucted facing large bodies of water over one mile (1.61


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relative

FASTEST-MILE WIND SPEED is the wind speed obtained from wind velocity

maps p

a fix point.

ll be considered as openings unless

such openings and their frames are specifically detailed and designed to resist the loads

on elem

dward projections.

E OR STOREY is a structure that has eighty five

percent or more openings on all sides.

to any quadrant of building side. Exposure D extends inland from the shoreline

mile (0.40 Km) or ten times the building height, which ever is greater.

repared by the National Oceanographic and Atmospheric Administration and is

the highest sustained average wind speed based on the time required for a mile-long

sample of air to pass

OPENINGS OR APERTURES OR HOLES in the exterior walls boundary of the

structures. All windows or doors or other openings sha

ents and components in accordance with the provisions of this section.

PARTIALLY ENCLOSED STRUCTURES OR STOREY is a structure or storey

that has more than fifteen percent of any windward projected area open and the area of

openings on all other projected areas is less than half of that on the win

SPECIAL WIND REGION is an area where local records and terrain features

indicate fifty year fastest-mile basic wind speed is higher.

UNENCLOSED STRUCTUR

2.6.3. Symbols And Notations.

The following symbols and notations apply to the provisions of this

division.

consideration as given in table 16-H

Qs =wind stagnation pressure at the standard height of 33 feet (10,000

mm) as set forth in Table 16-F.

Ce = combined height, exposure and gust factor coefficient as given in

table 16-GCq = pressure coefficient for the structure or portion of structure under

Iw = importance factor as set forth in table 16-K

P = design wind pressure


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2.6.4. Basic Wind Speed

The minimum basic wind speed at any site shall not be less than for those areas

designated as special wind regions and other areas where local records or terrain

indicate higher 50-year (mean recurrence interval) fastest- mile wind speeds, these

higher values shall be the minimum basic wind speeds .

2.6.5. E posure

xposure shall be assigned at each site for which a building or structure

is to b

2.6.6. Design

cordance with the following formula:

(2.60)

x

An e

e designed.

Wind Pressures

Design wind pressures for buildings and structures and elements therein shall

be determined for any height in ac

P = Ce.Cq.qs.Iw

2.6.7. Primary Frames And Systems

2.6.7.1. General

The primary frames or load-resisting system of every structure shallbe

designed for the pressures calculated using Formula (2.60) and the pressure

coefficients, Cq, of either Method 1 or Method 2. In addition, design of the overall

structure and its primary load-resisting system shall conform to Section 1605.

The base overturning moment for the entire structure, or for anyone of its

individual primary lateral-resisting elements, shall not exceed two-third of the dead

load-resisting moment. For an entire structure with a height-to-width ratio of 0.5 or less

in the wind direction and a maximum height of 60 feet, the combination of the effects

of uplift and overturning may be reduced by one-third. The weight of earth

superim alculate the dead-load-resisting moment.posed over footings may be used to c


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2.6.7.2. Method I (Normal Force Method)

Method 1 shall be used for thedesign of gabled rigid frames and may be used for

any structure. In the Normal Force Method, the wind pressures shall be assumed to act

simultaneously normal to all exterior surfaces. For structure pressures on roofs and

leeward walls, Ce, shall be evaluated at the mean roof height.

2.6.7.3. Method 2 (Projected area method)

Method 2 may be used for any structure less than 200 feet (60960 mm) in

height except those using gabled rigid frame. This method may be used in stability

determination for any structure less than 200 feet (60960 mm) high. In the projected

area method, horizontal pressures shall be assumed to act upon the full verticalprojected area of the structure, and the vertical pressure shall be assumed to act

simultaneously upon the full horizontal projected area.

2.6.7.4. Elements and Components of Structures

Design wind pressures for each element or component of a structure shall be

determined from Formula (2.60) and Cqvalues from Table 2.60-H, and shall be applied

perpendicular to the surface. For outward acting forces the. Value of Ce shall be obtained

from Table 2.60-G based on the mean roof height and applied for the entire height of

the structure. Each element or component shall be designed for the more severe of the

following loadings:

1.The pressures determined using Cq values for elements and components

acting over the entire tributary area of the element.

2.The pressures determined using Cq values for local areas at discontinuities

such as corners, ridges and eaves. These local pressures shall be applied over a

distance from a discontinuity of 10 feet (3048 mm) or 0.1 times the least width of the

structure, whichever is less.

The wind pressures from 2.6.7.3.and 2.6.7.4. need not be combined.


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2.6.8. Open-Frame Towers

Radio towers and other towers of trussed construction shall be designed and

constructed to withstand wind pressures specified in this section, multiplied by the

shape factors set forth in Table 2.60-H.

2.6.9. Miscellaneous Structures

Greenhouses, lath houses, agricultural buildings or fences 12 feet (3658 mm) or

less in height shall be designed in accordance with Chapter 16, Division 111. However,

three fourths of qs, but not less than 10 psf (0.48 kN/m2), may be substituted for qs in

Formula (2.60). Pressures on local areas at discontinuities need not be considered.

2.6.10. Occupancy Categories

For the purpose of wind-resistant design, each structure shall be placed in one of

the occ cy categories listed in Table 2.60-K. Table 2.60-K lists importance factors,upan

Iwfor each category.

TABLE 2.60-FWIND STAGNATION PRESSURE (qs) AT STANDARD HEIGHT OF33 FEET (10,058 mm)

Basic wind speed (mph)1 70 80

(x 1.61 for km/h)

90 100 110 120 130

Pressure qs(psf) 12

(x 0.0479 for kN/m2)

.6 16.4 20.8 25.6 31.0 36.9 43.3


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TABLE 2.60-GCOMBINED HEIGHT, EXPOSURE AND GUST FACTOR

COEFFICIENT (Ce)

HEIGHT ABOVE

D

(feet)

AVERAGE LEVEL OF

ADJOINING GROUN

x 304.8 for mm

EXPOSURE D EXPOSURE C EXPOSURE B

0-15

20

25

30

40

60

80

100

120

200

1.

1.45

1.50

1.54

1.62

1.73

1.81

1.88

1.93

2.10

2.23

2.34

6

1.13

1.19

1.23

1.31

1.43

1.53

1.61

1.67

1.87

2.05

2.19

0.67

.72

0.76

0.84

0.95

1.04

1.13

1.20

1.311.42

1.63

1.80

160 2.02 1.79

300

400

39 1.0 0.62

Values for intermediate heights above 15 feet (4572 mrn) may be interpolated.


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50

TABLE 2.6 PRESSURE COE ENTS {Cq)STRUCTURE DESCRIPTION Cq FACTOR

0-HOR PART THEREOF

FFICI

Method 1 (Normal forc od)

Walls:

indward wall

R

Wind perpendicular to

ard roof or flat

Windward roof

Less than 2:12(16.7

pe 2:12(16.7%)

t 12(75%)

Slope 9:12(75%) t 2

(

lope > 12:12 (10

Wind parallel to rigid an

r

0.8 inward

0.5 inward

0.7 outward

0.7 outward

0.9 outward inward

0.4 inward

0.7 inward

0.7 outward

e meth

W

Leeward wall

oofs1:

ridge

roofLeew

%)

Slo

han 9:

to less

o 12:1

100%)

S 0%)

d flat

oofs

or 0.3

1.Primary frame and systems

Method 2 (Projected area

method)

On vertical projected area:

Structure 40 feet(121 )

or less in height

Structure over 40 fee 92

mm) in height

On horizontal projected area1

1.3 horizon direction

1.4 horizon direction

0.7 upward

92 mm

t (121

tal any

tal any

Wall elements

All structures

Enclosed and unenclosed

structures

Partially enclosed structuresParapet walls

1.2 inward

1.2 outward

1.6 out ward1.3 inward or outward

2. Elem

ea of discontinuity

Roof elements3

Enclosed and unenclosed

structures

Slope< 7:12(58.3%)

Slope 7:12(58.3%) to

12:12(100%)

Partially enclosed structures

Slope < 2:12(16.7%)

%) to

7:12(58

1.3 outward

1.3outward or inward

1.7 outward

1.6 outward or 0.8 inward

.7 outward o

ents and components not in

ar 2

Slope 2:12(16.7

.3%)

Slope> 7:12(58.3%) to00%)12:12(1

1 r inward

3. Elements and components in area

of discontinuity2,4,5 dges

or 1.2 inward

.3 upward

Wall corners6

Roof eaves, rakes or r

verhangs6i

without o

Slope < 2:12(16.7%)

Slope 2:12(16.7%) to

7:12(58.3%)

Slope> 7:12(58.3%) to

12:12(100%)

1.5 outward

2

2.6 out ward

1.6 outward


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For slope


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TABLE 2.60-I SEISMIC ZONE FA

ZONE 1 3 4

CTOR Z

2A 2B

Z 0.075 0.15 0.2 0.3 0.4

TABLE 2.60-J SOIL PROFILE TYPES

Average Soil properties for top 100 feet (30480 mm)

for soil profile

SOIL

PROFILE

SOIL

PROFILETYPE NAME /

GENERIC

DESCRIPTIONShear wave

velocity, Vs

Feet/sec(m/sec)

Standard

Penetration test

N (Blows/Ft)

Undrained

Shear Strength ,

Su ,psf (KPa)

SA Hard rock > 5000 (1500)

SB Rock 2500 to 5000

(760 to 1500)

- -

S

and soft rock (360 to 760)

C Very dense soil 1200 to 2500 > 50 > 2000 (100)

S

(50 to 100)

D Stiff soil profile 600 to 1200

(180 to 360)

15 to 50 1000 to 2000

SE1 Soft soil profile 20 and wmc >= 40 %

Su< 500 psf (24 kPa). The Plasticity Index, PI, and the moisture content, wmc,

shall be determined in accordance with approved national standards.


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TABL

Occupancy

E 2.60-K O ANCY CA

Category

cy or Seismic Seismic

im

Wind

importance

r, IW

CCUP

Occupan

structure

TEGORY

function of

importance

factor, I fac

portance1

tor, IP facto

1 Essential

Facilities2

Group I,

cies h

ergens.

Fire and polic

Garages and shelters for

airc

an in

centers.Aviation contr

d in

required for emergency

response.

fire-suppression material or

equipment required for the

protection of Category 1, 2 or

3 structures.

25 1.15

Division 1

Occupan

and em

aving surgery

cy treatmentarea

e stations.

emergency

emergency

Structures

vehicles and

rafts.

d sheltersemergency-preparedness

ol towers.

Structures an equipmentgovernment communication

centers and other facilities

Standby power-generating

equipment for Category 1facilities.

Tanks or other structures

containing housing orsupporting water or other

1. 1.50

2 Hazardous

Facilities

Group H, Divisions 1, 2, 6and 7 Occupancies and

structures therein housing or

supporting toxic or explosivechemicals or substances.

or

ve substances that,

ained within a

building, would cause thatbuilding to be classified as a

1.25 1.50 1.15Nonbuilding structureshousing, supportingcontaining quantities of toxic

or explosi

if cont

Group H, Division 1, 2 or 7

Occupancy.


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3 Special

Occupancy

Structures3

A, Divisions 1,2 and

capacity

p Bcies used for college

h a

incapacitated

, Division 3

1.00 1.00 1.00

Group

2.1 Occupancies.

Buildings housing Group E,

Divisions 1 and 3Occupancies with a

greater than 300 students.Buildings housing GrouOccupan

or adult education wit

capacity greater than 500students.

Group I, Divisions 1 and 2

Occupancies with 50 or more

residentpatients, but not included in

Category 1.

Group IOccupancies.

All structures with an

occupancy greater than 5,000persons.

Structures and equipment inpower-generating stations,

and other public utility

facilities not included inCategory 1 or Category 2

above, and required for

continued operation.

4 Standard

Occupancy

Structures3

res

1.00 1.00 1.00

All structu housing

occupancies or having

functions not listed inCategory 1, 2 or 3 and Group

U Occupancy towers.

5 Miscellaneous

Structures1.00 1.00 1.00

Group U Occupancies except

for towers

1Th n of I ion 1633.2.4 shall be 1.0 for thee limitatio Pfor panel connections in Sect

enti .re connector

2Str servauctural ob tion requirements are given in Section 1702.

3For anchorage of required for life-safety systems, themachinery and equipment

value of IPshall be

taken as 1.5


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CHAPTER 3

DESIGN OF SLAB PANELS AND STAIRS

3.1 SLABS

Slab Panel S1 :

16x 22-6

m = 0.71

Slab thickness = perimeter/180

= 5.13 5.5

Slab wt = 5.5/12 150

= 69 psf

DEAD LOAD= 12012

55.1

+

= 65 psf

Live load = 40 + 20

= 60 psf

Wu = 1.2 Wu + 1.6 WLL

= 1.2 (69 + 65) + 1.6 60

= 257 psf

MOMENTS

Ma+

DL = 0.046 161 16212 = 22,751 lb-in

Ma+

LL = 0.057 9616212 = 16810 lb-in

Ma+ = 39561 lb-in

Ma

cout= 0.081 257 16212 = 63,950


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Ma

Dis

= 39561/3 = 13,187

Mb+

DL = 0.011 161 22.5212 = 10760 lb-in

Mb+

LL = 0.014 9622.52

12 = 8165 lb-in

Mb+ = 18925 lb-in

Mb

= 29665 lb-in

d = 4.75

SHORT SPAN

R = Mu/db2

Ra+ =

275.412

39561

= 146.12 lb-in/in2

= 0.0028

As = 0.0028 12 4.75

= 0.16 in2

#3 @ 8 C/C

As = 0.17

Ra

=2

75.412

63950

= 236.19 lb-in/in2

= 0.0047

As = 0.268 in2

# 3 @ 4 C/C


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As = 0.29 in2

LONG SPAN

d = 4.25 in

Rb+ =

225.412

18925

= 87.3 lb-in/in2

= 0.0018

As = 0.0918 in2

#3 @ 10 C/C

As = 0.13 in2

Rb

=225.412

29665

= 136.86 lb-in/in2

= 0.0028

As = 0.143 in2

# 3 @ 9 C/C


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SLAB PANEL S2 :

One way slab panel

16-6x 8

M+ =24

2

Wul

=24

1282572

M+ = 8,224 lb-in

M = 16448 l

R =

275.412

16448

= 60.75

= min = 0.0018

As = 0.11 in2

# 3 @ 10 C/C

+ve Steel = #3 @ 10 C/C

As = 0.13 in2

DISTRIBUTION STEEL

As = 0.0018 12 4.25

= 0.092 in2

# 3 @ 14 C/C

As = 0.094 in2


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Slab Panel S3 :

m = 0.85

16-6x14

WuDL = 161 psf

Wu LL = 96 psf

Wu = 257 psf

MOMENTS

Ma+

DL = 0.031 161 14212 = 11739 lb-in

Ma+LL = 0.041 9614212 = 9258 lb-in

Ma+ = 20997 lb-in

Ma

= 0.082 257 14212 = 49,566 lb-in

Mb+

DL

= 0.011 1 16.5212 = 5786 lb-in

Mb+

LL = 0.019 9616.5212 = 5,959 lb-in

Mb+ = 11745 lb-in

STEEL

SHORT SPAN

Ra+ =275.412

20997

= 77.55 lb-in/in2

= 0.0018

As = 0.10 in2

#3 @ 10 C/C


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Ra

=275.412

49566

= 183 lb-in/in

2

= 0.0036

As = 0.205 in2

SHORT SPAN

# 3 @ 6 C/C

LONG SPAN

# 3 @ 10 C/C

SLAB PANEL S4 :

12-6x 22-6

m = 0.55

Wu = 2.57 psf

WuDL = 161 psf

WuLL = 96 psf

MOMENTS

Ma+

DL = 0.031 161 12.5212 = 11471 lb-in

Ma+

LL = 0.063 9612.5212 = 11340 lb-in

Ma+ = 22812 lb-in

Ma

= 0.089 257 12.5212 = 42887 lb-in


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STEEL

SHORT SPAN

Ra

= 275.412

42887

= 158.4 lb-in/in2

= 0.003

As = 0.171 in2

#3 @ 7 C/C

LONG DIRECTION

+ve steel # 3 @ 10 C/C

12-6x 14

SLAB PANEL S5 :

m = 0.90

MOMENTS

Ma+

DL = 0.025 161 12.5212 = 7547 lb-in

Ma+

LL = 0.035 9612.5212 = 6300 lb-in

Ma+ = 13842 lb-in

Ma

= 0

Mb+

DL = 0.024 161 14212 = 9088 lb-in

Mb+

LL = 0.027 96 14212 = 6096 lb-in

Ma+ = 15185 lb-in

Ma

= 42313 lb-in


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STEEL

SHORT SPAN

Rb

= 275.412

42313

= 156.3 lb-in/in2

= 0.0030

As = 0.153 in2

#3 @ 8 C/C

S 6:

TWO WAY SLAB:

S612 x 12

LOADING:

DEAD LOAD:

Wt of slab = ?

h =180

Perimeter

=180

412 = 3.2

In this zone one slab have dimensions 16.5 16.5 taking that slab critical

h =180

45.16

= 4.4 = 5

Wt. of slab = 5/12 150

= 62.5 lb/sq ft


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4 earth filling = 4/12 120 = 15 lb/sq ft

Total dead load = 625 + 40 + 15 = 117.5 lb/sq ft

LIVE LOAD

100 lb sq ft

FACTORED LOAD

Wu = 1.4 WD+ 1.7 WL

= 1.4 117.5 + 1.7 100 = 335 lb/sq ft

Ratio of slab spans = m = La/Lb = 1

NEGATIVE MOMENT (LB) ONLY

ve

UbM = 0.071 2bWL

= 0.07 1 335 12212 = 41,101 lb-in

POSITIVE MOMENTS

LaSPAN

ve

UdlM

+ = 0.027 2aWL

= 0.027 164.5 12212 = 7675 lb-in

ve

alLM+ = 0.032 2

aWL

= 0.032 170 12212

ve

alLM

+ = 9400 lb-in

ve

aM+ = 7675 + 9400 = 17,075 lb-in


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LbSPAN

ve

bM+

dl = 0.033 WdlLa2

= 0.033 164.5 122

12 = 9381 lb in

ve

bM+

lL= 0.035 WlLLa2

= 0.035 170 12212 = 10282 lb in

ve

bM+ = 9381 + 10282 = 19663 lb-in

REINFORCEMENT

Lbspan

POSITIVE MOMENT

h = 5

d = 5-0.75-0.25 (= # 4 bar)

= 4

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

=

300085.0

4129.0

196632

114000

300085.0 2

= 0.0029

As = b d = 0.0029 12 4

As = 0.14 in2

# 4 bars @ 12 C/C


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S max = 2 h = 3 5

= 15

So S < Smax

OK

SHEAR CHECK

Vu =12

5335

2

1233515.1

= 2172 lbs

Vn = Vc= bdfc2 ,= 51230002

= 6573 lbs

Vc = 0.85 6573 = 5586 lbs

So Vu > Vc OK

LbSPAN ve MOMENT

Mu = 41,101 lb-in

d = 4

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.00625

As = b d = 0.00625 12 4

As = 0.200 in2

SPACING

# 3 bars @ 6 C/C


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La SPAN

For this span

d = 4 of bars

= 4 0.5 = 3.5

Mu = 17075 lb-in

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.0033

As = b d = 0.0033 12 3.5

As = 0.14 in2

# 3 bars @ 9 C/C

S7:

Two way slab :

12 x 12 La

L

LOADING

WdL = 117.5 lb/sq ft

Wu dL = 164.5 lb/sq ft

WlL = 100 lb/sq ft

Wu LL = 170 lb/sq ft

Wu = 335 lb/sq ft

h = 5


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MOMENTS

LaSPAN

ve

bM

+

dl= 0.018 W

dlL

a

2

= 0.018 164.5 12212 = 5117 lb in

ve

bM+

lL= 0.027 WlLLa2

= 0.027 170 12212 = 7932 lb in

ve

bM+ = 7932 + 5117 = 13049 lb-in

Ma = 0

LbSPAN

ve

bM+

dl = 0.027 WudlLb2

= 0.027 164.5 12212 = 7675 lb in

ve

bM+

lL= 0.032 WulLLb2

= 0.032 170 12212 = 9400 lb in

ve

bM+ = 7675 + 9400 = 17075 lb-in

Ma = 0.076 Wulb

2

= 0.076 335 122

12 = 43995 lb-in

REINFORCEMENTS

LbSPANve MOMENT

Mu = 41,101 lb-in

d = 4


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=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.00625

As = b d = 0.00625 12 4

As = 0.300 in2

SPACING

# 4 bars @ 7.5 C/C

La SPAN

For this span

d = 4 of bars

= 4 0.5 = 3.5

Mu = 17075 lb-in

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.0033

As = b d = 0.0033 12 3.5

As = 0.14 in2

SPACING

# 4 bars @ 12 C/C


76/136

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Ve MOMENTS

LbSPAN

LbSPANve MOMENT

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.00672

As = b d = 0.00627 12 4

As = 0.322 in2

SPACING

# 4 bars @ 7 C/C

S8:

Two way slab:

16.5 x 16.5 La

L

Size Of slab = 16.5 X 16.5

LOADING

WdL = 117.5 lb/ft2

WlL = 100 lb/ft2

Wu dL = 164.5 lb/sq ft

Wu LL = 170 lb/sq ft

WU = 335 lb/sq ft

h = 5


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ve MOMENTS:

LaSPAN

ve

aM

= 0.018 W

UL

a

2 = 0.05 335 16.5212

ve

aM = 54722 lb in

LbSPAN

ve

bM = 54722 lb-in

REINFORCEMENT

+ ve MOMENTS

LbSPAN

d = 4

= 0.00478

As = 0.229 in2

# 4 @ 9 C/C

LaSPAN

d = 3.5

Mu = 31812 lb-in

= 0.00632

As = 0.265 in2

# 4 @ 8 C/C


78/136

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ve MOMENTS

LbSPAN

d = 4

Mu = 54722 lb-in

= 0.00848

As = 0.407 in2

# 4 @ 5 C/C

As = 0.43 in2

LaSPAN

d = 3.5

Mu = 54722 lb in

= 0.0113

As = 0.476 in2

# 4 @ 5 C/C As = 0.47 in2

S8a:

12 x 16.5 Lb

La

La = 12

Lb = 16

m = 12/16

= 0.75

Slab thickness =180

2)1612( + = 5


79/136

72

MuLL = 102 psf (60 1.6)

Wu DL = 1.2 127.8 = 178.5 psf

Wu = 280.5 psf

Ma+

DL = 0.043 175 122 = 13300 lb in

Ma+LL = 0.052 102 122 = 9165 lb in lb in

Ma = 0.076 281 122 = 369.3 lb in

Mb+

DL = 0.013 179 162 = 7149 lb in

Mb+

LL = 0.016 102 162 = 5013.5 lb in

Mb

= 0.024 281 162 = 20718 lb in

REINFORCEMENT

SHORT SPAN

+ve STEEL

=

'85.0

/211

'85.0 2

fc

bdMu

fy

fc

= 0.00196

As = b d = 0.00196 12 5

As = 0.1 in2

# 3 @ 10 C/C

ve STEEL

M

a = 369093

= 0.00328

Mosque Design

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