Design Of Reinforced Concrete - Icivil-hu.com

Design Of

Reinforced Concrete ACI 318-11 Code Edition

Anas G. Dawas

Hashemite University

ANAS

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1

ANAS DAWAS

This textbook presents an introduction to reinforced concrete design. I hope the material is written in such a manner as to interest students in the subject and to encourage them to continue its study in the years to come.

This textbook covers the following topics :

Design One Way Ribbed Slab

Design Two Way Slabs

Serviceability Design For Torsion

Design Footings

Design Columns

Sample Exams : Examples of sample exams are included for most topics in

the text. Problems in the back of each chapter are also suitable for exam questions

About the Author

I am currently a third year student in the CIVIL ENGINEERING at Hashemite University

Preface



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ANAS DAWAS

كلمةبسى هللا انشد انشدى خش كالو أبذأ ف سسانخ , انذذ هللا انزي

نعه سم انكثش انضع انخاضع فق قذس نا اكال زا

عهى انضيالء األفاضم أسأل هللا أ ك عها افعا نج خانصا .

يا أد انخأكذ عه أ زا انعم ي صع بشش خطئ

صب , أ انشاجع راحا انخ اعخذث عها ف جع انضع

فا اخطاء دسابت يخخهفت ز عهى أذي كباس انعهاء

, نزنك أحى ي صيالئ أ خذشا انعهيت أا كاج انغشب

فانخطأ اسد يا كاج دسجت انخشكض أ سايذ ا أخطأث ف

.أيش يا أ كا عا ن عهى حذقق زا انطشح قذس االيكا

أخشا , أحى ي هللا أ غفش نضيالئ انز افخى انت خالل

داح انجايعت انز كاا با طذ نا طخ ان ذ

ا ش ف أفسى يسخقبال يششقا نك هلل عاقبت األيس , أحى

ي كم شخص اخفع بزا انضع أ ذعا نى نجع انسه

انشدت , فذ صائش نا صاسا بانغفشة

فقكى هللا أدبخ



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ANAS DAWAS

ONE WAY JOIST

SLAB



4

ANAS DAWAS

One-Way Joist Floor System :

Long-span floors for relatively light live loads can be constructed as a series of closely

spaced, cast-in-place T-beams (or joists) with a cross section as shown in Fig. The

joists span one way between beams. Most often, removable metal forms referred to as

fillers or pans are used to form the joists. Occasionally, joist floors are built by using clay-tile

fillers, which serve as forms for the concrete in the ribs that are left in place to serve as the

ceiling (ACI Code Section 8.13.5).

When the dimensions of the joists conform to ACI Code Sections 8.13.1 to 8.13.3,

they are eligible for less cover to the reinforcement than for beams (ACI Code Section

7.7.2(c)) and for a 10 percent increase in the shear, carried by the concrete (ACI

Figure 1 Types of slabs



5

ANAS DAWAS

Code Section 8.13.8). The principal requirements are that the floor be a monolithic combination

of regularly spaced ribs and a top slab with

1. ribs not less than 100 mm. in width,

2. depth of ribs not more than 3.5 times the minimum web width, and

3. clear spacing between ribs not greater than 750 mm.

S

mm

fillerwith

mmfillerwithout

h f

12

1

50

40

Ribbed slabs not meeting these requirements are designed as slabs and beams.

Although not required by the ACI Code, load-distributing ribs perpendicular to the

joists are provided at the midspan or at the third points of long spans. These have at least

one continuous No. 4 (13 mm) bar at the top and the bottom. The CRSI Handbook [10-4] suggests

no load-distributing ribs in spans of up to 6 m, one at midspan for spans of 6 to 9 m , and

two at the third points for spans over 9 m .

For joist floors meeting the requirements of ACI Code Section 8.3.3, the ACI moment

and shear coefficients can be used in design, taking (Ln) as the clear span of the joists

themselves. For uneven spans, it is necessary to analyze the floor.

The negative moments in the ends of the joists will be underestimated if this is not done.

S



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ANAS DAWAS

Reinforcing Details

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.



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ANAS DAWAS

baiJ tgioJingiseDg

Design Example :

1. Calculating minimum depth according to ACI code (Table 9.5a)

mmthicknessslabtypicalusemm

continuousendonel

h n

250216216518

4004400

518

..

)(.

min

Design the joist solid slab for the floor system shown below .

Live load = 2KN/m2 , fc=25 Mpa and fy=420 Mpa. Cross section is shown in fig.2 above .

Figure 2 Design Example m520.



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ANAS DAWAS

2. Calculating ultimate load : a) Dead load for 1 m2 :

Slab weight = 0.07×1×25 1.75 KN/m2 Rib weight =0.14×0.18×25/0.52 1.2 KN/m2

Block weight =5×0.18/0.52 1.73 KN/m2 Plastering 0.025×22 0.55 KN/m2

Mortar 0.025×22 0.55 KN/m2 Sand fill 0.1×13 1.3 KN/m2

Tiles 0.025×22 0.55 KN/m2 Total Dead Load 7.63 KN/m2

b) Ultimate load = wu=1.2(7.63)+1.6(2)=12.35 KN/m2



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ANAS DAWAS

Notes :

Ultimate load on slab = 12 .35 KN/m2 but the Ultimate load on Joist = 12.35×0.52=6.422 KN/m .

Usually the number of blocks per meter in joist slab equal to 5 blocks with thickness equal to 20 cm and weight =0.18 KN per one .

Typical thickness depend on the depth of the blocks , for example , for 24 cm blocks ,the typical thickness equal to 240+70=310 mm

Ln used to calculate minimum thickness for joist slab is in the direction of the joist .

Primary Beam



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ANAS DAWAS

3. Calculating shear and moment subjected to the Joist :

mKNM

mKNM

mKNM

Ve

Ve

Ve

...

...

...

41119

4426

33714

4426

28424

4426

2

2

2

Effective depth =250-20-10-10/2=215 mm

For negative moment the rib will be designed as rectangular beam and

for positive moment will be designed as T-beam



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ANAS DAWAS

A. Design for positive moment :

10276250

86420

2151204141

91

2613215420900

1000337

62352025850

420510095215950420900

1000337215

520120

2

2

2

2

use

mmf

dbf

mmf

db

A

mmA

mmammA

mmd

mmbmmb

y

wc

y

w

s

s

s

fw

.

))((..

).

(.

.

..

.

..

.

,

(min)

B. Design for negative moment :



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ANAS DAWAS

102276250

2864202151204141

22149

2825215420900

10004411

4241202585042015722148215950420900

10004411

use

mmyf

dwbcf

mmyf

dwb

sA

mmsA

mmammsA

.

))((..

(min)

.)

.(.

.

..

...

.

4. Shear Design:

OKVV

KNdbf

V

KNlW

V

uc

wc

c

nuu

73172151206

2575011

611

76142

151

....

.)(.

5. Solid Slab Design :

blockormeshUse

mmA

mm

A

mKN

lwM

beamsimpleasitdesigncanwealsobut

beamfixefixeaspartthisdesigncanwe

s

s

nuu

/

))((.

..

.

..

..

.

(min)

10

12670100000180

15

4595042090

100024680

24680

8

403412

8

2

2

22

40cm



13

ANAS DAWAS



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ANAS DAWAS

Problems:

1-1 The one way joist slab shown below consists of 011 mm wide joists spaced at 510 mm , the web width of the spandrel beams is 400 mm and 500 for interior beams . Column dimension (400×400 mm).

A B C D

1

2

3

4



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ANAS DAWAS

The minimum required depth of the slab upon the ACI code requirement is most nearly

………… Calculate the Ultimate load /rib assuming that the Live load = 3 KN/m2,

25 mm Covering material with unit weight is 20 KN/m3 . blocks are 40×25×17 cm in dimension , each 17 Kg in weight . Use

fc =30 Mpa , fy=420 Mpa and 25 mm plaster with unit weight = 21 KN/m3.

The maximum ultimate negative moment in the joist ……… The maximum ultimate shear in the joist ……… The ultimate distributed load /m on the beam between column A and B is most nearly

………. The ultimate load carried by column A is most nearly…….. For the overhang slab , The minimum required depth of the slab upon the ACI code

requirement is………… The minimum required depth of the exterior beam upon the ACI code requirement is

……. The maximum ultimate negative moment in the overhang joist is most nearly …………. The maximum ultimate positive moment in the joist is most nearly ………. The largest ultimate load carried by the intermediate column is most ……… The minimum area of steel required to resist the maximum ultimate positive moment in

beam between column C2 and D2 upon ACI code is most nearly (assume the depth of the beam is 600 mm ) …………

Show by suitable sketch the reinforcements required to resist the maximum ultimate positive moment for any interior joist.

The minimum area of steel must be provided to solid slab part upon ACI code requirements is most nearly ……..



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ANAS DAWAS

1-2 For Simple supported square slab shown below is apart of a floor in a typical residential building and the live load = 2 KN/m2,fc=28 Mpa and fy=420 Mpa .use 12mm bars

The minimum required depth of the slab upon the ACI code requirement is most nearly………..

Calculate the Ultimate load /rib based on the typical section shown above……. Draw shear and moment diagram for the simple supported Joist …… Calculate the shear strength capacity for the joist upon ACI code……… Design the reinforcement in the joist to carry the positive moment, show suitable

sketch ……….. The minimum required depth of the primary beam upon the ACI code

requirement is most nearly………..



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ANAS DAWAS

Draw shear and moment diagram for the primary beam based on the minimum depth calculated above ………

1-3 The one way joist slab shown below consists of 150 mm wide joists spaced at 550 mm , the web width of the spandrel beams is 400 mm and 500 for interior beams . Column dimension (400×400 mm).

Assume that the dead load and live load on the slab is 7KN/m2 and 3 KN/m2:

m7

m6

m6

m7 m4 m7



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ANAS DAWAS

The minimum required depth of the slab upon the ACI code requirement is most nearly …………

The maximum ultimate negative moment in the joist ……… The maximum ultimate shear in the joist ……… The ultimate distributed load /m on the beam between column A3 and B3 is most nearly

………. ( neglect self weight of beam )

The minimum required depth of the exterior beam upon the ACI code requirement is …….

The maximum ultimate positive moment in the joist is most nearly ………. Show by suitable sketch the reinforcements required to resist the maximum ultimate

positive moment for any interior joist. The minimum area of steel must be provided to solid slab part upon ACI code

requirements is most nearly …….. Determine the design ultimate bending moment , ultimate shear force at the most

critical section for each for the simple supported joist ………..



19

ANAS DAWAS

TWO WAY SLAB



20

ANAS DAWAS

Two Way Slabs

Behavior , Analysis and Design

Two-way slabs are a form of construction unique to reinforced concrete among the major structural materials. It is an efficient, economical, and widely used structural system. In practice, two-way slabs take various forms. For relatively light loads, as experienced in apartments or similar buildings, flat plates are used. such a plate is simply a slab of uniform thickness supported on columns. In an apartment building, the top of the slab would be carpeted, and the bottom of the slab would be finished as the ceiling for the story below. Flat plates are most economical for spans from 4.5 to 6 m .



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ANAS DAWAS

In general, slabs are classified as being one way or two way. Slabs that primarily deflect in one direction are referred to as one way slab When slabs are supported by columns arranged generally in rows so that the slabs can deflect in two directions, they are usually referred to as Two way slabs .



22

ANAS DAWAS

Two-way slabs can be strengthened by the addition of beams between the columns, by thickening the slabs around the columns (drop

panels), and by flaring the columns under the slabs (column capitals). These situations are shown in Figure 16.1 and discussed in the next several paragraphs. Flat plates present a possible problem in transferring the shear at the perimeter of the columns. In other words, there is a danger that the columns may punch through the slabs. As a result, it is frequently necessary to increase column sizes or slab thicknesses or to use Shearheads .

Analysis of Two-Way Slabs:

A theoretical elastic analysis for such slabs is a very complex problem because of their highly indeterminate nature. Numerical techniques such as finite difference and finite elements are required, but such methods require sophisticated software to be practical in design. The methods described in this chapter can be done by hand or with simple spreadsheets, and are sufficiently accurate for most design problems. the design of two-way slabs is generally based on empirical moment coefficients, which, although they

might not accurately predict stress variations, result in slabs with satisfactory overall safety

factors. In other words, if too much reinforcing is placed in one part of a slab and too little

somewhere else, the resulting slab behavior will probably still be satisfactory. The total amount

of reinforcement in a slab seems more important than its exact placement.



23

ANAS DAWAS

DESIGN OF SLABS: Two procedures for the flexural analysis and design of two-way floor systems are presented in detail in the ACI Code:

Direct-design method—considered in the following section:

The calculation of moments in the direct-design method is based on the total statical moment

Equivalent-frame design method:

Here, the slab is divided into a series of two-dimensional frames (in each direction), and the positive and negative

moments are computed via an elastic-frame analysis.

THE DIRECT-DESIGN METHOD

Limitations on the Use of the Direct-Design Method:

The direct design method was developed from theoretical procedures for

determination of moments in slabs, requirements for simple design,

construction procedures, and performance of existing slabs. Therefore, the

slab system, to be designed using the direct design method, should conform

to the following limitations as given by ACI Code 13.6.1:

1. There must be three or more spans in each direction.

2. Slab panels must be rectangular with a ratio of longer to shorter span,

center-to-center of supports, not greater than 2.0.

3. Successive span lengths, center-to-center of supports, in each direction

must not differ by more than one-third of the longer span.

4. Columns must not be offset more than 10 % of the span in the direction

of offset from either axis between centerlines of successive columns.

5. Loads must be due to gravity only and uniformly distributed over the

entire panel. The live load must not exceed 2 times the dead load.

6. For a panel with beams between supports on all sides, the relative

stiffness of beams in two perpendicular directions is not

less than 0.20 and not greater than 5.0.

52002

12

221

l

l

f

f

.



24

ANAS DAWAS

2

2

4222

3111

ffff

ffff

s

b

scs

bcbf

f

l

l

I

I

IE

IE

of direction in

of direction in

slab of section gross of axis centroidalabout inertia ofmoment

beam of section gross of axis centroidalabout inertia ofmoment

beam the of side each on any, if , panelsadjacent of scenterlineby laterally bounded slab of width

a of stiffness flexuralto section beam of stiffness flexuralof ratio



25

ANAS DAWAS

Design Procedure:

1) Determination of the total factored static moment: Total factored static moment for a span is determined in a strip bounded laterally by centerline of panel

on each side of centerline of supports, as shown in Figure.

columnsbetweenspanclearl

striptheofwidthtransversel

areaunitperloadfactoredw

wherellw

M

n

u

nuo

2

22

8

:than less be tonot is direction each in moments factorednegative average and positive of sum Absolute

2) Distribution of the total factored static moment to negative and positive moments:



26

ANAS DAWAS

3) Distribution of the positive and negative factored moments to the column and middle strips:

- Column strips : Column strip is a design strip with a width on each side of a column

centerline equal to 0.25 l1 or 2 0.25 l2 whichever is less

- Middle strip : Middle strip is a design strip bounded by two column strips .



27

ANAS DAWAS

Factored moments in column strips: - According to ACI Code 13.6.4, column strip moments, as percentages of

total factored positive and negative moments

Aspect Ratio L2/L1

121 llf / 0.5 1.0 2.0 Negative moment at interior support 0 75 75 75

≥1.0 90 75 45 positive moment near mid span 0 60 60 60

≥1.0 90 75 45

Aspect Ratio L2/L1 121 llf / t 0.50 1.0 2.0

Negative moment at exterior support 0 0 100 100 100 ≥2.5 75 75 75

≥ 1.0 0 100 100 100 ≥2.5 90 75 45

Positive moment near mid span 0 60 60 60 ≥ 1.0 90 75 45

Negative moment at interior 0 75 75 75 ≥1.0 90 75 45

onelongest the isy , dimensionsmallest the is

supports of center

-centerto beam, of length span to equal slab of widtha of stiffness flexuralto

section beam edge of stiffness torsional of ratio as defined is

xwhereyx

y

xC

IE

CE

scs

cbt

t

:.

max

36301

2

3



28

ANAS DAWAS

Factored moments in beams caused by slab loads

Depth Limitations and Stiffness Requirements

It is obviously very important to keep the various panels of a two-way slab relatively level (i.e.,

with reasonably small deflections). Thin reinforced two-way slabs have quite a bit of moment

resistance, but deflections are often large. As a consequence, their depths are very carefully

controlled by the ACI Code so as to limit these deflections. This is accomplished by requiring

the designer to either (a) compute deflections and make sure they are within certain limitations

or (b) use certain minimum thicknesses as specified in Section 9.5.3 of the code. Deflection

computations for two-way slabs are rather complicated, so the average designer usually uses

the minimum ACI thickness values, presented in the next few paragraphs of this chapter.



29

ANAS DAWAS

1) Slabs without Interior Beams

For a slab without interior beams spanning between its supports and with a ratio of its long

span to short span not greater than 2.0, the minimum thickness can be taken from [Table 9.5(c) in the code]. The

values selected from the table, however, must

not be less than the following values (ACI 9.5.3.2):

Slabs without drop panels 125 mm .

Thickness of those slabs with drop panels outside the panels 100 mm

.

Very often slabs are built without interior beams between the columns but with edge beams running around the perimeter

of the building. These beams are very helpful in stiffening the slabs and reducing the deflections in the exterior slab

panels. The stiffness of slabs with edge beams is expressed as a function of αf , which follows.



30

ANAS DAWAS

2) Slabs with Interior Beams

To determine the minimum thickness of slabs with beams spanning between their supports on

all sides, Section 9.5.3.3 of the code must be followed. Involved in the expressions presented

there are span lengths, panel shapes, flexural stiffness of beams if they are used, steel yield

stresses, and so on. In these equations, the following terms are used:

panel a of sides all on stiffness slab-to-beam of ratios the of value average the

span clear short the to long the of ratio the

beams withslab for beams (b) and beams withoutslabs for columns (a) of

face, to face measured direction, long the in span clear the

fm

nl

mm

fl

h

For

mm

fl

h

For

For

yn

fm

fm

yn

fm

fm

909361400

80

02

125200536

140080

2002

20

).(

,.

).(

).(

,.

,.

) 16.1 tablesupports.( their between spanning beams interior

withoutslabs for they were as obtained are sthicknesse minimum the



31

ANAS DAWAS

SHEAR STRENGTH OF TWO-WAY SLABS A shear failure in a beam results from an inclined crack caused by flexural and shearing

stresses. This crack starts at the tensile face of the beam and extends diagonally to the compression zone .

One-way shear or beam-action shear (Fig. 4) involves an inclined crack extending across the entire width of the

structure. Two-way shear or punching shear involves a truncated cone or pyramid-shaped surface

around the column, as shown schematically in Fig. 3.

Once a punching-shear failure has occurred at a slab–column joint, the shear capacity of that particular joint is almost

completely lost. In the case of a two-way slab, as the slab slides down, the column load is transferred to adjacent column-

slab connections, thereby possibly overloading them and causing them to fail. Thus, although a two-way slab possesses

great ductility if it fails in flexure, it has very little ductility if it fails in shear

One Way Shear :

The critical section for one way shear is located at “d” from the face of the support or at “d” from the face of the drop panel or other change in thickness . The shear strength on the critical section is computed as for beams .

""datVbeshoulddbfV uwcc 6

1

Two Way Shear:

Location of the Critical Perimeter:

Two-way shear is assumed to be critical on a vertical section through the slab or footing

extending around the column. According to ACI Code Section 11.11.1.2, this section is

chosen so that it is never less than d/2 from the face of the column and so that its length ,bo ,

is a minimum.

Figure 4 Beam shear / one way Figure 3 Two way shear / punching shear



32

ANAS DAWAS



33

ANAS DAWAS



34

ANAS DAWAS

Tributary Areas for Shear in Two-Way Slabs

column corner for20

column edge for30

column interior for40

section critical the of perimeter the is

sideshort the to column of side long of ratio

s

o

c

b

dbfVc

dbf

b

dVb

dbfVa

wherecandbaofsmallesttheastakenisV

VslabmostIn

occ

oc

o

sc

oc

cc

c

s

3

1

122

12

42

0

)(

)(

)(

:)()(),(



35

ANAS DAWAS

Shear Reinforcement for Two-Way Slabs



36

ANAS DAWAS

elemasa1

Design the two-way flat plate with no edge (spandrel) beams, shown in Figure , given the following: interior columns are 40 cm × 40 cm, exterior columns are 30 cm × 30 cm, covering materials weigh 2.29 KN/m2

and the live load is 3 KN/m2 . Use fc= 28 Mpa and Fy= 420 Mpa .



37

ANAS DAWAS

Solution:

1) Evaluate slab thickness For corner panel Ln=7-0.3=6.7 m For Edge panel Ln=7-0.15-0.2=6.65 m For interior panel Ln=7-0.40=6.6 m For flat plates with no edge beams, minimum slab thickness = Ln/30 = 6700/30 = 223.33 mm

Take it as 30 cm .

2) Check limitations for slab analysis by the direct design method:

The first five conditions are satisfied, while the sixth condition does not apply due to the nonexistence of

beams.

3) Calculate the factored load on the slab:

Wu= 1.2 ( 0.3×25+2.29)+1.6(3)=16.55 4) Check slab thickness for shear:

a) Interior column :

davg = 300 – 20 -20 (use ø20 bar ) = 26 cm

x

y



38

ANAS DAWAS

- Punching shear

OKVVKNV

KN

KN

KN

ofsmallerV

KNV

mmb

ucc

c

u

o

3894750351192

7121026026402831

35119212

2602640282640

26040

181612

260264028142

01896876600765516

264040026042

...

.

.

.

.).)((.

)(

- Beam shear For section 1-1 :

OKKNV

KNV

mmx

c

u

810312606000286

1750

871301604035516

30402602

400

2

7000

..

.).(.

For section 2-2:

OK

KNV

KNV

mmy

c

u

12032607000286

1750

329454275516

25402602

400

2

6000

.

.).(.



39

ANAS DAWAS

b) Edge column :

You have to check the edge column as we did for interior column , by yourself.

c) Corner column Select tributary area as shown below then check beam and punching shear .

Figure 6 Edge column the short direction

Figure 5 Edge column in the long direction



40

ANAS DAWAS

When you select thickness for slab you have to check the shear strength , and you can increase thickness when “Not Ok”

I have assumed that the shear strength for edge and corner column is ok so h=30 cm 5) Calculate the factored static moment:

d



41

ANAS DAWAS

Strip in shorter direction

Width of intermediate strip = 7.0 m Width of the column strip is ( 6/4 +6/4)=3.0 m >>> Width of middle=7-3=4m ( 2m per half) Clear span for exterior panels = 600 – 15 – 20 = 565 cm Clear span for interior panels = 600 – 20 – 20 = 560 cm The larger of the two values will be used in moment calculations, or you can calculate static moment for each panel separately .

mKNMo ...

4638

65575516 2

6) Distribute the total factored static moment into positive and

negative moments:

KN.m 324.14630.70 support interior the onmoment negative

KN.m 120.384630.26support exterior the onmoment negative

KN.m 240.764630.52 moment positve : panel exterior For

KN.m 300.95463 0.65 moment negative

KN.m M 0.35 moment positive : panel interior the For o

05162463350 ..



42

ANAS DAWAS

05162.

95300. 95300.38120.

76240.

1324. 1324. 38120.

76240.

38120.

96

144

243

38120.243

14497

864.

81

Figure 7 Total Static moment

Figure 8 Column strip moment

Figure 9 Middle Strip moment



43

ANAS DAWAS

7) Distribute the positive and negative moments to the column and middle strips:

8) Design the reinforcement: Column strip reinforcement and Half middle strip reinforcement:

Design sections at maximum positive and negative moments as rectangular section :

Column Strip Half Middle Strip

Negative Positive Negative Positive

Moment(KN.m) 243 144 81/2= 40.5 96/2= 48

b (mm) 3000 3000 2000 2000

d (mm) 260 260 260 260

h (mm) 300 300 300 300

Fy ( Mpa) 420 420 420 420

Fc (Mpa) 28 28 28 28

As (mm2)

As-min (mm2) 0.0018bh 0.0018bh 0.0018bh 0.0018bh

Bar Dim Ø12 Ø12 Ø12 Ø12

No of bar

As Provided

øMn (KN.m)

Sample Of Calculation To Design Column Strip :

For Positive Moment

Slab Moment End span Interior span

Exterior

negative

Positive Interior

negative

Positive Negative

Total moment 120.38

240.76 324.1 162.05 300.95

Column

moment

120.38 0.6×240.76=

144.45

0.75×324.1=

243.075

0.6×162.05=

97.23

0.75×300.95=

225.7

Middle strip

moment

0 240.76-144.45=

96.31

324.1-243.075=

81.025

162.05-

97.23=64.82

300.95-225.7=

75.25



44

ANAS DAWAS

)(.).

(.

..

)(

)(

)(.

)(

.

)()()(.

.

).

(.

..

.

...

max

(min)

OKM

mmaa

dfAM

ScheckmmA

AbS

mmbarsuseBarNo

thisusemmA

mmA

mma

mmA

n

ysn

s

b

s

s

s

14451632

9792604201696900

979300028850

4201696

2

2001696

124

3000

16961215314

124

1620

1620300300000180

31491

2

1926042090

1000144

19300028850

42031542

3154226095042090

1000144

2

2

2

2

2

2

Strip in the long direction Width of intermediate strip = 600 cm and width of column strip is the smaller of (L1 / 2) and (L2 / 2), taken as (600/2) = 300 cm. Total factored static moment: Clear span for exterior panels = 700 – 15 – 20 = 665 cm Clear span for interior panels = 700 – 20 – 20 = 660 cm The larger of the two values will be used in moment calculations.

mKNM o ...

5498

65665516 2



45

ANAS DAWAS

You can design this strip by yourself now

elemasagE

For the two-way solid slab with beams on all column lines, shown in Figure , evaluate the moments acting on any of the internal beams, using the direct design method. All columns are 30 cm × 30 cm in cross section, all beams are 30 cm × 60 cm in cross section, slab thickness is equal to 14 cm, covering materials weigh 1.83 KN/m2 and the live load is 4 kN/m2. Use Use fc= 28 Mpa and Fy= 420 Mpa .



46

ANAS DAWAS

1) Evaluate the slab thickness For internal beams :



47

ANAS DAWAS

49923

23

1058910236353120437046030012

460300

2

14012041401220

12

1401220

12044603001401220

1402

460460300

2

1401401220

mm

I

mmy

b

...).)(())((

).)(())((

.)()(

))(())((

49923

23

1081084233423937046030012

460300

70423914076012

140760

4239300460140760

1402

46030046070140760

mm

I

mmy

b

)..().)(())((

).)(())((

.))(())((

))()(())()((

Beam Edge For



48

ANAS DAWAS

For Interior Slab :

293

10372112

1406000mmI s .

))((

For Exterior Slab:

49

3

10720

12

1401503000

mm

I s

.

))((

Torsional Constant

m03.

m03.



49

ANAS DAWAS

)(./))((

.

.

.

.

max

cminunits

cmC

cmC

cmC

t

B

A

472112146002

3403907

3403907

3403907

3305347

3

4

4

4

Calculating α :

Generally , you have to calculate αm for each span then calculate thickness for each one to

determine which values will control ……

Figure 10 Alpha Values

Okmmh

mm

fl

h

valuesall

cm

cm

yn

mf

mf

f

f

).(.)(

).(.

..

..

).(

.

.....

...

(.

(.

33139140331391936

1400

4208075

013006

3006

90936

140080

02

964

96969696

9584

96961111

1111720

8000

9861372

9850

2

1

2

4

4

1

beams) exterior For

beams) interior For



50

ANAS DAWAS

2) Factored load on the slab Wu=1.2 ( 0.14×25+1.83)+1.6(4)=12.8 KN/m2

3) Check slab thickness for shear Note : No need to check two way shear for two way slabs with beam , and you should check one way shear for interior and exterior beams but in this example they are the same so we will check for the interior one. x

OkVKNV

KNV

mm x

mmd

uc

u

64281086000286

750

521067422812

27421082

300

2

6000

1081220140

..

.).(.

section critical of Width



51

ANAS DAWAS

4) Total factored static moment

)(.

.)(.

.).(.

beforcalculated

l

l

mKNM

t

f

o

4721

9066

696

3128

756812

1

21

2



52

ANAS DAWAS

Figure 11 Positive / Negative Moment

Figure 12 Column Strip Moment

Figure 14 Beam’s Moment

Figure 13 Middle Strip Moment



53

ANAS DAWAS

stripcolumntheofresistBeamsl

l

ratiotheoninerpolatiby

For

xFor

For

f

t

t

t

%)(

%

%.

.

.

851

85

7552

4721

10000

1

21

Finally , you should design middle strip , column strip and beam between supports based on the maximum moment for each one .

Section across the strip : 1220 mm

6/4=1.5 m 6/4=1.5m

Middle Strip Column Strip Beam



54

ANAS DAWAS

Column Strip Middle Strip Beam

Negative Positive Negative Positive Negative Positive

Moment(KN.m) 139.8 133.35 54.9 44.49 139 113.34

b (mm) 1780 1780 3000 3000 300 300

d (mm) 108 108 108 108 108 544

h (mm) 140 140 140 140 600 600

Fy ( Mpa) 420 420 420 420 420 420

Fc (Mpa) 28 28 28 28 28 28

As (mm2)

As-min (mm2) 0.0018bh 0.0018bh 0.0018bh 0.0018bh

Bar Dim Ø12 Ø12 Ø12 Ø12 Ø12 Ø12

No of bar

As Provided

øMn (KN.m)

Notes :

Design the beam as a rectangle for negative moment and as T beam for a positive moment .

As (min) for a beam is not as slab ( column and middle strip ) You can design middle strip as a one part or you can design half middle strip based on

half value of the moment .



55

ANAS DAWAS

beoag etasg

Drop panels are thicker portions of the slab adjacent to the columns, as shown in Fig.

1. The minimum thickness of slab required to limit deflections may be reduced by 10 percent if the slab has drop panels conforming to ACI Code Section 13.2.5. The drop panel stiffens the slab in the region of highest moments and hence reduces the deflection.



56

ANAS DAWAS

2. A drop panel with dimensions conforming to ACI Code Section 13.2.5 can be used to reduce the amount of negative-moment reinforcement required over a column in a flat slab. By increasing the overall depth of the slab, the lever arm, jd, used in computing the area of steel is increased, resulting in less required reinforcement in this region.

3. A drop panel gives additional slab depth at the column, thereby increasing the area of the critical shear perimeter.

Example : Design a drop panel according to the ACI code requirements for the floor system shown below in the fig. Check two way shear . Interior columns are ( 500×300 mm) and exterior columns are (300×300 mm ), Live load = 2 KN/m2 and super imposed Dead load = 1.2 KN/m2.



57

ANAS DAWAS

mm2004

hh thickness panel drop The

m55. m6

mmhuse

mml

h

ml

mml

h

ml

n

n

n

n

160

15336

55

36

55506

15533

15

33

152

50

2

3055

2

1

.min

..

.min

...

.

0.0)( beams No

deflection control to thickness slab imum Min

m54.



58

ANAS DAWAS

The half width is at least ( L/6) in length 6000/6 =1000 mm 4500/6 =750 mm

This is minimum dimension for the panel for the interior column ,so you can increase this to 2500×1800 area with 200 thickness .

Calculating the ultimate load based on the slab’s thickness : Wu= 1.2( 0.16×24+1.2)+1.6(2)=9.25 KN/m2

Calculating the ultimate load based on the drop panel’s thickness : Wu= 1.2( 0.2×24+1.2)+1.6(2)=10.4 KN/m2

26394103

1259

3

2mKNWLoadUltimateAvg u /.).().(.

Now you have to check two way shear : For interior column :

mmb

KNV

mmd

oc

u

22726684682671300

500

22466680468054755639

1681220200

)(,.

.)....(.



59

ANAS DAWAS

OKVKNV

KNVc

KNVb

KNVa

uc

c

c

c

505248673750

2486731000

168227228

3

1

48341000

168

12

2272282

2272

16840

87391000

168

12

227228

671

42

..

.)(

.)()(

.).

()(



60

ANAS DAWAS

The punching shear at drop panel

uc

c

c

c

o

c

u

VKNV

KNVc

Vb

Va

mmb

KNV

mmd

15432057750

20571000

128911228

3

1

9112192826282

671300

500

42009281628254755639

1281220160

.

)(

)(

)(

)(

.

.)....(.

By the same way , check the two way shear for edge and corner columns.



61

ANAS DAWAS

MELBeReg

2-1 flat plate floor system with panel is 6.0 × 8.0 m (on centers of column ) is supported on 0.5 m interior columns and 0.40 m square exterior columns . Edge beams are not used along the exterior floor edges . Use fc=28 Mpa , fy=420 Mpa . (a) Using ACI code , determine the minimum slab thickness required for panels 1 and 3 based on deflection criterion . (b) Assume the floor is to support a service dead load ( including self-weight ) Of 10 KN/m2 and service live load of 4.0 KN/m2 . If slab thickness is 160 mm and ( d= 134 mm ) , then check punching shear for corner column . (c ) Use direct design method to calculate column and middle in the edge frame spanning 6.0 m in panel 2 . Use ultimate load of 15 KN/m2.



62

ANAS DAWAS

2-2 30 × 30 m a flat plate with no drop panel is shown below . (fy=420 Mpa) . Based on ACI code :

a) Determine the minimum permissible total thickness required for slabs

in panel 3 . b) Determine the minimum permissible total thickness required for slabs

in panel 2

Edge beam with (311×211) mm



63

ANAS DAWAS

c) The ACI code states that for slabs with beams between column along exterior

edges , the value αf for the edge beam shall not less than 0.8 . Determine αf if the slab is 200 mm in thickness and show if the provision is satisfied .

2-3 Consider the floor system shown below , columns 60×60 cm2 and beams as shown. Answer the following problems . Assume beams and slabs are monolithic And fy = 420 Mpa .



64

ANAS DAWAS

a) Compute α for interior beam in the short direction b) Compute α for exterior beam in the short direction c) Compute α for exterior beam in the long direction d) Compute βt for the exterior beam . e) Use the direct design method to calculate column and middle strip moments at the shaded

edge frame spanning . Use the ultimate load of 15 KN/m2.

f) Check one way shear for the beam between columns 6 and 7 .

2-4 The concrete for the slab and the beam shown below was placed in one pour . ( all dimensions in mm ):

1) Compute αf for the beam . 2) Compute the torsional constant for the edge beam shown .



65

ANAS DAWAS

2-5 For the flat plate shown in the figure below ( edge beam are not used ). Answer the following questions . The columns dimensions are (40×40 cm ) , fc=28 Mpa and fy=420 Mpa .

5 m

5 m

6 m

6 m 6 m 5.5 m

41×41 cm

A2

W3



66

ANAS DAWAS

a) The minimum required depth of the slab based upon ACI code requirement Is most nearly: If the ultimate load on the slab is Wu= 15 KN/m2 , slab thickness h= 15 cm and effective depth d = 12 cm :

b) The maximum one way shear Vu in the area around column A2 ……… c) The Ultimate punching shear Vu for the corner column W3. ……… d) The Ultimate punching shear Vu for the interior column A2. ………

For the shaded area ( column and middle strip) shown in the figure

e) The Total static moment in the W-E direction (hatched frame ) is ……… If the total static moment in W-E direction is 300 KN.m :

f) The positive moment in the frame ( hatched frame ) is : ……… g) The column strip positive moment in the frame (hatched frame ) is : ……… h) Design a drop panel for the interior column A2 ….. i) The ultimate punching shear Vu for the 2000 by 2000 mm interior drop panel ( use

ø12 bars and 20 mm cover ) .



67

ANAS DAWAS

2-6 A flat plate floor system with panels 7.0 by 8.0 m (on centers of columns ) is supported on 0.6 square interior columns and 0.40 m square exterior columns , all interior beams are 0.6 × o.6 m and exterior beams are 0.40 × 0.60 m , Use fc=21 Mpa Fy=420 Mpa and answer the following question :

a) For the exterior beam in the long direction , the span of the slab considered in α calculation is most nearly : :……….

7.0m

7.0m

8.0m 8.0m



68

ANAS DAWAS

b) For interior panel C if the average α is equal to 2.3 , then the minimum depth of the slab is most nearly : :……….

c) For the corner panel (A) , if the floor system is assumed to be without beams , then the minimum depth of slab is most nearly : :……….

d) If the ultimate slab load is 20 KN/m2, and effective depth d =190 mm , the one way shear in panel C is most nearly : :……….

e) For the exterior frame spanning in the long direction (hatched frame ) , the clear span Ln of the of the frame in panel A is most nearly : :……….

f) For the exterior frame spanning in the long direction (hatched frame ) , the transvers length L2 of the of the frame in panel A is most nearly: :……….

g) If the total static moment in panel (A) for the exterior frame spanning in the long direction (hatched frame ) is 300 KN.m , the exterior negative moment in the frame is most nearly : :……….

h) If the total static moment in panel (A) for the exterior frame spanning in the long direction (hatched frame ) is 300 KN.m , the exterior positive moment in the frame is most nearly : :……….

i) Assume the floor has no beams and the ultimate load is 20 KN/m2 , if d=190 mm then the ultimate punching shear Vu for an interior column is most nearly :……….

j) Assume the floor has no beams and the ultimate load is 20 KN/m2 , if d=190 mm then the ultimate punching shear Vu for an the corner column is most nearly :………. Notes :



69

ANAS DAWAS

Serviceability



70

ANAS DAWAS

yaelJbaeDJsJnS

Today the structural design profession is concerned with a limit states philosophy. The term limit state is used to describe a condition at which a structure or some part of a structure ceases to perform its intended function. There are two categories of limit states: strength and serviceability. Strength limit states are based on the safety or load-carrying capacity of structures and include buckling, fracture, fatigue, overturning, and so on. with the bending limit state of various members. Serviceability limit states refer to the performance of structures under normal service loads and are concerned with the uses and/or occupancy of structures. Serviceability is measured by considering the magnitudes of deflections, cracks, and vibrations of structures, as well as by considering the amounts of surface deterioration of the concrete and corrosion of the reinforcing. You will note that these items may disrupt the use of structures but do not usually involve collapse.

CRACKING This section presents a few introductory comments concerning some of the several types of cracks that occur in reinforced concrete beams. The remainder of this chapter is concerned with the estimated widths of flexural cracks and recommended maximum spacing of flexural bars to control cracks. Flexural cracks are vertical cracks that extend from the tension sides of beams up to the region of their neutral axes. Cracking starts when the tensile stress in the concrete reaches the tensile strength of the concrete at some point in the bar. When this occurs, the prism cracks.



71

ANAS DAWAS

Figure 15 types of cracks



72

ANAS DAWAS

Control of Flexural Cracks

Although cracks cannot be eliminated, they can be limited to acceptable sizes by spreading out or distributing the reinforcement. In other words, smaller cracks will result if several small bars are used with moderate spacing rather than a few large ones with large spacing.

.

ACI Code Provisions Concerning Cracks In the ACI Code, Sections 10.6.3 and 10.6.4 require that flexural tensile reinforcement be well distributed within the zones of maximum tension so that the center-to-center spacing of the reinforcing closest to a tension surface is not greater than the value computed with the following expression:

facetension the toent reinforcem the of surface fromdistanceLeast

MPa in stress bar load service

) center to (center mm in spacing bar

c

ys

s

c

s

C

ff

S

fC

fS

3

2

28030052

280380 .



73

ANAS DAWAS

elpmaxE

Is the spacing of the bars shown in the figures within the requirements of the ACI code from the standpoint of the cracking ? If fy= 420 Mpa.

e. acceptablspacing is mm, this less than

mm is f spacing oactual barSince the

mmmm

S

mmd

C bc

228

75

300228

4203

2

280300656052

4203

2

280380

65602

72875

275

)().(.)(

..



74

ANAS DAWAS

Skin Reinforcement

points above the reinforcement than they are at the level of the steel, as shown in Fig. 9-3b. To control the width of these cracks, ACI Code Section 10.6.7 requires the use of skin reinforcement that is distributed uniformly along both faces of the beam web for a distance of d/2 measured from the centroid of the longitudinal tension reinforcement toward the neutral axis. If h ≥ 900 mm , Skin reinforcement shall be used Askin=0.015bw S2 S2= smaller of ( d/6 or 300 mm ) Total Askin ≤ As/2

s

c

s fC

fS

28030052

280380 .



75

ANAS DAWAS

Deflections

Control of Deflections

One of the best ways to reduce deflections is by increasing member depths—but designers are always under pressure to keep members as shallow as possible. (As you can see, shallower members mean thinner floors, and thinner floors mean buildings with less height, with consequent reductions in many costs, such as plumbing, wiring, elevators, outside materials on buildings, and so on.) Reinforced concrete specifications usually limit deflections by specifying certain minimum depths or maximum permissible computed deflections.

Minimum Thicknesses

Table 9.5(a) of the ACI Code, provides a set of minimum thicknesses for beams and one-way slabs to be used, unless actual deflection calculations indicate that lesser thicknesses are permissible. These minimum thickness values, which were developed primarily on the basis of experience over many years, should be used only for beams and slabs that are not supporting or attached to partitions or other members likely to be damaged by deflections.

Maximum Deflections

If the designer chooses not to meet the minimum thicknesses given in Table 9.5(a) , he or she must compute deflections. If this is done, the values determined may not exceed the values specified in Table 6.1, which is Table 9.5(b) of the ACI Code.

Camber

The deflection of reinforced concrete members may also be controlled by cambering.



76

ANAS DAWAS

Common cases



77

ANAS DAWAS

Effective Moments of Inertia

Regardless of the method used for calculating deflections, there is a problem in determining the moment of inertia to be used. The trouble lies in the amount of cracking that has occurred.



78

ANAS DAWAS

Figure 16 : As moment subjected to the beam increases , the moment of inertia decreases because of cracks .



79

ANAS DAWAS

cr

cc

t

gr

cr

ff

fE

y

IfM

700

4700

.

Continuous-Beam Deflections

The following discussion considers a continuous T beam subjected to both positive and negative moments. As shown in Figure below , the effective moment of inertia used for calculating deflections varies a great deal throughout the member.

cr

a

crg

a

cre

ecra

gcra

IM

MI

M

MI

IIUseMM

IIUseMMfor

33 1 )()(

ng stageious loadit any prevputed or a being cominertia is

tof the momen for whichding stageat the loa

member nt in the imum momeM

fibertensionextremethetocentroidfromy

ruptureoff

inertiaofmomentgrossI

momentcracking M

a

t

r

g

cr

max

distance

Modulus



80

ANAS DAWAS

Figure 17 This formula to calculate deflection along continuous beam



81

ANAS DAWAS

Critical section shall be permitted to obtain deflection in ACI code :

Mid span for simple supported beam At support for cantilever Critical positive and negative moment sections for continuous beam

ACI code :( Ig )is the moment of inertia of the gross concrete section neglecting area of tension steel .

Transformed Section



82

ANAS DAWAS

Elastic Theory For Flexure

This theory helps us to calculate deflection with several assumptions must be taken in considered :

1) Plan sections remain plane after bending . 2) Linear stress – strain curves for steel and concrete . 3) Perfect bond between steel and concrete . 4) Concrete tension capacity is neglected .

This theory cannot be used when concrete stress are higher than cf60. .

elpmaxE

For the beam shown below , check if we can use Theory of elasticity or not at the given moments :

1- 28 KN.m 2- 113 KN.m

Transform section (un-cracked ) 2270003000110110 mmAnn s )()(,

Determine the centroid

mmyt 132627000300600

500270002

600300600

.)()(

)()()()(

As=3000 mm2, n=10



83

ANAS DAWAS

Determine the moment of inertia

Determine Cracking moment

mKNM cr ..

..65

10

10

1326600

346826

9

For M = 28 KN.m

For M = 113 KN.m you have to determine the location of the N.A based on the transformed cracked section .

49

223

10346

132650027000300132660030012

600300

mm

I g

.

).()().()(

okarechecksall

fMPaf

fMPaf

fMPaf

MM

rt

ys

cc

cr

2111010346

132660028

506471010346

13265002810

4504411010346

132628

6528

6

9

6

9

6

9

..

).(

...

).(

...

.



84

ANAS DAWAS

TheoryElasticusecanyou

okarechecksall

ff

fMPaf

mmI

mmyyy

y

ANaboutyA

ys

cc

cr

ii

501789101043

723150011310

5077101043

7231113

10437231500300003

7231300

72315003000102

300

00

6

9

6

9

4923

...

).(

...

.

.).(.

.)()(

..

Calculate The Deflection

Instantaneous Deflection : When a concrete beam is loaded, it undergoes a deflection referred to as an instantaneous deflection.



85

ANAS DAWAS

Sustained Load Deflection Under sustained loads, concrete undergoes creep strains and the curvature of a cross section increases.

If compression steel is present, the increased compressive strains will cause an increase in stress in the compression reinforcement, thereby shifting some of the compressive force from the concrete to the compression steel. As a result, the compressive stress in the concrete decreases, resulting in reduced creep strains.

.

,

creepinreductiongreaterthe

db

AsteelncompressioofratiogreaterThe s

The Total Long Time Deflection

duration load limited for factorTime

load sustained of duration infinite for factorTime

deflection load live Sustained

deflection load dead Immediate

deflection load live Immediate

t

SL

D

L

SLtDLLT

501



86

ANAS DAWAS

elpmaxE : Simple Supported Beam

A simple supported beam with the cross section shown in the figure , has a span of 6 m

And supports un-factored dead load of 30 KN/m2 including its self-weight plus an un-factored live load of 20 KN/m2 .

22000

20

30

420

28

mmA

mKNLL

mKNDL

MPaf

MPaf

s

y

c

/

/



87

ANAS DAWAS

1- Calculate the properties of the section a) Un-cracked section :

2000 mm2

6.0 m

KNM

Mpaf

Mpa

Gpan

MpaE

mmy

mmI

cr

r

c

t

g

88810

10

300

2773

732870

0825000

200

25000284700

3002600

102712

600400

6

9

293

...

..

.

/

.



88

ANAS DAWAS

b) Cracked section

2- Calculate ultimate moment at the critical section (mid span )

ecrLLDL

ecrDL

IIMmKNM

IIMmKNM

.)(

.

2258

62030

1358

630

2

2

3- Calculate effective moment of inertia

4- Calculate instantaneous deflection :

mmIE

wl

DLec

DL 12651095325000384

6000305

384

59

44

..

Section properties

Y before crack

Y after crack (Transformed)

gI

CrI

crM

At mid span 300 mm 168.8 mm 491027 mm. 4910672 mm. 88.8 KN.M

4923

2

106728168525200083

8168400

8168525200082

400

mmI

mmyyy

cr

.).()(.

.)()(

4933

4933

10952672225

888127

225

888

10953672135

888127

135

888

mmMonBasedI

mmMonBasedI

LLDLe

DLe

...

..

)(

...

..

)(



89

ANAS DAWAS

mm

mmIE

wl

DLLLDLLL

DLec

LLDL

314612654411

44111095225000384

6000505

384

59

44

...

..

5- Calculate long term deflection

mmLT 561600126523146

20501

2

....

)(

6- Compare with ACI code limits :

soandsospan

OKmmmmspan

exampleanjustthis

LT

LLi

480

316716360

6000

360

..

7- Summary

Load (KN)

Mm (KN.m) mid

I 4910 mm

( mm)

DL 30 135 3.95 5.126 DL+LL 50 225 2.95 11.44 DL+SL 30+0 No sus.L ---------- -------



90

ANAS DAWAS

elpmaxE : Continuous Beam

Determine the long term deflection at the mid span of the continues T beam shown above , The member supports a dead load including its self-weight of 16 KN/m and live load of 14 KN/m , Fc=28 Mpa , Fy = 420 Mpa . The beam cross section is shown below . Assume that 30% of the live load is sustained .

Duration LL

DL

SLt

LT

5 YEARS 2 2 11.44-5.126= 6.314 mm

(2)(5.126)= 10.252

________ 16.566 mm

3ø25

3ø25

6ø25

9.0 m



91

ANAS DAWAS

Notes : Firstly , you can check minimum thickness according to Table 9.5 (a) in ACI code Then , you can fill the following tables to get perfect solution : 1) You should calculate the properties for the sections at support and at mid span .

2) Calculate the moment caused by DL , (DL+LL) at the two edges of the member and at the mid span .

Section properties

Y before crack


gI

CrI

crM

At mid span At supports



92

ANAS DAWAS

3) Compare the moments from previous step with the cracking moment to determine which value of moment of inertia should be used .

4) Calculate Average value for moment of inertia using :

5) Calculate deflection using the following formula :

In this formula , you should substitute moments value with its sign

M1

Mm

M2



93

ANAS DAWAS

You can use these tables :

Note : In case of negative moment , consider the section as a rectangular beam and all calculation based on this consideration as shown below :

Load (KN)

M1 (KN.m)

Mm (KN.m)

M2 (KN.m)

I1 Imid

I2 Iavg

( mm)

DL DL+LL DL+SL

Duration LL DL

SLt LT



94

ANAS DAWAS

PROBLEMS

6-1 A simple supported beam with cross section shown below has a span of 6.0 m . The beam supports un-factored dead load of 24 KN/m , including its own self-weight plus an un-factored live load of 16 KN/m . The concrete compressive strength is 21 Mpa .

Compute the following : 1) Gross moment of inertia , cracking moment of inertia and cracking moment . 2) Immediate dead load deflection . 3) Immediate live load deflection . 4) Long term deflection at an age of 3 years , if 30% of live load is sustained .



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ANAS DAWAS

6-2 A Cantilever beam with cross section shown below has a span of 5.0 m . The beam

supports an un-factored concentrated live load of 60 KN and un-factored dead load of 15 KN/m . The concrete compressive strength is 28 Mpa and effective depth = 800 mm




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6-3 A simple supported beam with cross section shown below has a span of 6.0 m . The

beam supports un-factored dead load of 20 KN/m , including its own self-weight plus an un-factored concentrated live load of 40 KN. The concrete compressive strength is 28 Mpa.


6-4 For the cross sections shown below , determine whether the reinforcement satisfies the ACI code requirements for crack width control .

For section C design skein reinforcement Upon ACI code requirements .



97

ANAS DAWAS

A B

C D



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6-5 For 3.6 m span cantilever beam shown in the figure below , (Ignore weight in your calculation ) : 1) Determine instantaneous live load deflection at the free end of the beam due to the

load condition shown below . 2) Determine the Total long term deflection . Assume that only dead load is

Sustained. What code deflection criteria it meets and what limitations, if any , have to be placed on its use ?



99

ANAS DAWAS

6-6 The beam cross section shown is on a 8000 mm simple span and carries uniformly distributed service dead load of wD=15 kN/m and concentrated service live load of PL=30 kN. Use fc’ = 28 MPa, fy = 420 MPa and assume n = 9 (Ignore beam self-weight). Compute : 1. Cracked moment of inertia, 2. Cracking Moment, 3. Immediate deflection due to live load only, 4. Ultimate long-term deflection due to dead load

6-7 Continues beam with cross section shown below has a span of 9.0 m. The beam

supports un-factored dead load of 22 KN/m , including its own self weight , plus un-factored live load of 36 KN/m . The concrete compressive strength is 21 Mpa . Compute the following : 1) Gross moment of inertia for positive moment section . 2) Cracking moment for positive moment moment section . 3) Cracked moment of inertia for positive moment section



100

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4) Using the following assumption , Answers the followings :

Section properties

Y before crack


gI

CrI

crM

At mid span - - 2×1010 mm4 8×109 mm4 106 KN.m At supports - - 3×1010 mm4 7.4×109 mm4 76 KN.m



101

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A. Immediate dead load deflection B. Immediate live load deflection . C. Ultimate long term deflection of 30 % live load sustained .



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Torsion



103

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noeiJotg

The average designer probably does not worry about torsion very much. He or she thinks almost exclusively of axial forces, shears, and bending moments, and yet most reinforced concrete structures are subject to some degree of torsion. Until recent years, the safety factors required by codes for the design of reinforced concrete members for shear, moment, and so forth were so large that the effects of torsion could be safely neglected in all but the most extreme cases. Today, however, overall safety factors are less than they used to be and members are smaller, with the result that torsion is a more common problem. Appreciable torsion does occur in many structures, such as:

1) In the main girders of bridges, which are twisted by transverse beams or slabs. 2) In buildings where the edge of a floor slab and its beams are supported by a spandrel

beam running between the exterior columns. 3) Earthquakes can cause dangerous torsional forces in all buildings. 4) In curved bridge girders, spiral stairways, and balcony girders



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ANAS DAWAS

It should be realized that if the supporting member is able to rotate, the resulting

torsional stresses will be fairly small. If, however, the member is restrained, the torsional stresses can be quite large.

Torsion Cracks

Should a plain concrete member be subjected to pure torsion, it will crack and fail along 45o

spiral lines because of the diagonal tension corresponding to the torsional stresses. For a very effective demonstration of this type of failure, you can take a piece of chalk in your hands and twist it until it breaks. Although the diagonal tension stresses produced by twisting are very similar to those caused by shear, they will occur on all faces of a member. As a result, they add to the stresses caused by shear on one side and subtract from them on the other.



105

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Strength of Material Review

In a bar with a rectangular cross section, however, the torsional stresses vary from a maximum at the middle of the long sides of the rectangle to zero at the corners.

Torsional Reinforcing

Tests have shown that both longitudinal bars and closed stirrups (or spirals) are necessary to intercept the numerous diagonal tension cracks that occur on all surfaces of members subject to appreciable torsional forces.



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The normal -shaped stirrups are not satisfactory. They must be closed either by welding their ends together to form a continuous loop, as illustrated in Figure (a), or by bending their ends around a longitudinal bar, as shown in part (b) of the same figure.



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ACI Code Section 11.5.4.3 requires that longitudinal reinforcement for torsion be developed at both ends of a beam. Because the maximum torsions generally act at the ends of a beam, it is generally necessary to anchor the longitudinal torsional reinforcement for its yield strength at the face of the support. This may require hooks or horizontal U-shaped bars lap spliced with the longitudinal torsion reinforcement.



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The strength of closed stirrups cannot be developed unless additional longitudinal reinforcing is supplied. Longitudinal bars should be spaced uniformly around the insides of the stirrups, not more than 300 mm apart. There must be at least one bar in each corner of the stirrups to provide anchorage for the stirrup legs (Code 11.6.6.2); otherwise, if the concrete inside the corners were to be crushed, the stirrups would slip and the result would be even larger torsional cracks. These longitudinal bars must have diameters at least equal to 0.042 times the stirrup spacing. Their size may not be less than 10 mm .

DESIGN METHODS FOR TORSION

Skew bending theory developed: This theory assumes that some shear and torsion is resisted by the concrete, the rest by shear and torsion reinforcement.

Thin-walled tube/plastic space truss model Torsion is assumed to be resisted by shear flow, q , around the perimeter of the member as shown in Fig.. The beam is idealized as a thin-walled tube. After cracking, the tube is idealized as a hollow truss consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals approximately centered on the stirrups, as shown in Fig.



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Definitions To understand the subject , we should define some terms used a lot in our calculations :

1) cpcp PandA

2) hoh PandA : represents the gross area enclosed by the shear flow path around the

perimeter of the tube.



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Location of Critical Section for Torsion

The critical section for shear was found to be located at a distance d away from the face of the support. For an analogous reason, ACI Code Section 11.5.2.4 allows sections located at less than d from the support to be designed for the same torque, that exists at a distance d from the support. This would not apply if a large torque were applied within a distance d from the support.

Types Of Torsion

Equilibrium torsion For a statically determinate structure, there is only one path along which a torsional moment can be transmitted to the supports. This type of torsional moment, which is referred to as equilibrium torsion or statically determinate torsion, cannot be reduced by a redistribution of internal forces or by a rotation of the member. The edge beam must be designed to resist the full calculated torsional moment.



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Compatibility torsion The torsional moment in a particular part of a statically indeterminate structure may be substantially reduced if that part of the structure cracks under the torsion and “gives,” or rotates. The result will be a redistribution of forces in the structure. This type of torsion, which referred to as statically indeterminate torsion or compatibility torsion.



112

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Design Procedure For Members Subjected to Bending Moment , Shear and Torsion

1) . Draw the shear force, bending moment, and torque diagrams. 2) Select cross-sectional dimensions “b” and “h” based on factored bending moment,

and determine the required area of reinforcement.

)(.

)/(

).(

.,

Economic

ff

fK

K

Mdb

cy

cn

n

u

010

5901

9002

3) Check if torsion may be neglected. Torsion may be neglected if the torsion less than :

forceaxialWithfA

N

P

Af

forceaxialWithoutP

Af

cg

U

cp

cp

c

cp

cp

c

).

)((.

)(.

33010830

0830

2

2

If this is the case, proceed on with shear design and choose flexural and shear reinforcement.

Note : The critical section for torsion is located at distance d from the face of the

support if no torques are applied within this distance. If torques are applied within distance d from face of support, critical torsion is located at face of the support.

dbf

fA

c

dc

c

cd

y

cs

s

s

3190

0030

0030

.

)(.

)(.

(max)



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ANAS DAWAS

4) .Determine whether the case involves equilibrium or compatibility torsion.

forceaxialWith

cf

gA

UN

cpP

cpA

cf

uT

forceaxialWithout

cpP

cpA

cf

uT

).

)((.

)(.

.

3301

2

330

2

330

: followingthe to T reduce , torsionity compatibilFor B.

T fulluse , torsion mequilibriu For A.

u

u

5) Check the adequacy of the size of the cross section in terms of preventing brittle mode of failure resulting from diagonal compressive stresses due to shear and torsion combined.

SectionHollowcfdwb

cV

ohA

hPuT

dwb

uV

SectionSolidcfdwb

cV

ohA

hPuT

dwb

uV

).().

()(max

).().

()(max

6602

71

2

6602

271

2



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ANAS DAWAS

Note : If Eq. ( in step 5 ) is not satisfied, cross sectional dimensions need to be increased.

6) Determine the area of stirrups required for shear. To facilitate the addition of stirrups for shear and torsion, the area of shear reinforcement is expressed in terms:

enlarge.betoneedsSectioncrosstheVVif

dbfVandVV

Vbut

arealegonespstirruAdf

V

s

A

Cs

wccCu

s

v

yv

sv

4

6

1

2

Also, determine maximum stirrup spacing based on shear and if Vu less than 0.5Vc No need for stirrups .

7) Determine the required area of stirrups for torsion in terms of :

mm

p

ofsmaller

AAAf

T

s

A

h

oh

oyv

nt

300

8

8502

0

maxS

:torsion on based spacing stirrup maximum compute Besides,

.



116

ANAS DAWAS

yv

w

y

yv

ht

y

cpc

l

h

y

yvtl

f

b

f

fP

s

A

f

AfA

Pf

f

s

AA

1750420

2

.)(.

cot

min

s

A Where

t

8) Determine combined area of stirrups required for shear and torsion:

9) Select stirrup size, and compute stirrup spacing based on the amount determined in step 7.

calculated s

A(

size spstirru selected for

tv )

vA

S

NOTE Maximum stirrup spacing must not exceed the smaller of the two values

evaluated in steps 5 and 6.

10) Calculate the longitudinal reinforcement required for torsion.

yvf

wbcf

yvf

wb

ofs

tvABut

s

tA

s

vA

s

tvA

0620

350

2

.

.

maxmin)(



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ANAS DAWAS

11)

oho

cp

cpc

AA

P

AfTorsionCracking

850

3

2

.

elpmaxE

The cantilever beam shown below supports its own weight plus concentrated load . The beam is 1500 mm and concentrated load at a point 150 mm from the end of the beam and 150 mm away from the centroidal axis of the member . The un-factored concentrated load consists of a 85 KN dead load and a 85 KN live load . Use normal weight of concrete with compressive strength of 20 MPa and yielding strength for bars and stirrups equal to 420 Mpa .



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ANAS DAWAS

1) Assume a dimension for the section to calculate own weight ( d = 535 mm )

KNP

mKNwt

partmmforwt

beammajorforwt

ult

factored

23885618521

912676521

765246040150

765246040

2

)(.)(.

/...

...

...

2) Draw Shear , Moment and Torsion Diagram

400 mm

600

mm

mKN

mKNu

M

KNuV

uT ....

)..

....

..)..(

.....

( 8351502382

150150916

53303512382

251916351150916

424951916238150916



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ANAS DAWAS

3) Design for flexure .

ø Load 0.90 Moment 0.75 Shear 0.75 Torsion

Moment

Shear

Tortion

OKAAcheck

mmAmmbf

fAa

mmjdf

MA

sst

s

c

ys

y

us

(min)(max) ,,

..

.

.

...

.

2

2

18318110640020850

420861815

850

8618151000535900420900

10005330



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ANAS DAWAS

4) Check if torsion may be neglected:

5) The Torsion is needed for equilibrium , so design for full of 35 .8 KN.m

6) Check the adequacy of the size of the cross section (assume 40 mm cover and ø10 stirrups )

consideredbemustT

mKNP

AfT

mmhbP

mmhbA

u

cp

cp

cth

cp

cp

18835

18102000

2400002008307500830

200060040022

240000600400

622

2

..

..)(..)(.

)()(

Ok

Mpacfc

f

cfdwb

cV

MPa

ohA

hPuT

dwb

uV

mmyxP

mmyxA

mmy

mmx

ooh

oooh

o

o

77281

7726606

660

81

2

215810071

16406108352

535400

1000724522

71

2

164051031022

158100510310

51010402600

31010402400

2

..

.).().(

.max

.

..)

.()(max

)()(

)(

)(



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ANAS DAWAS

Note : Vu used above to check section size is at “d” distance from the support KNdV u 724553509164249 .).(..@

7) Determine the area of stirrups required for shear

mmmmdf

V

s

A

KNVV

V

KNdbfV

yv

sv

cu

s

wcc

/..

...

.

.

2

3

7480535420

10001168

11685159750

7245

515910535400206

1

6

1

8) Determine the required area of stirrups

mmmmAf

T

s

A

oyv

nt /.).(

.

.

23

3

4220101581008504202

750

10835

2

9) Determine combined area of stirrups required for shear and torsion:

22640

0620

23330

350

2591422027480

2

mm

yvf

wbcf

mm

yvf

wb

ofs

tvA

mmmms

tA

s

vA

s

tvA

..

..

maxmin)(

/.).(.



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ANAS DAWAS

10) Calculate the longitudinal reinforcement required for torsion

11) Spacing between stirrups :

mm

mmp

ofsmallerS

mmA

S

h

v

300

2058

1640

8

99591

104

2 2

max

.

)(

)

calculated s

A(

size spstirru selected for

tv

You can increase the diameter of the stirrups to increase the spacing (s).

12) Provide 3ø10 at the top , 3ø10 at the middle and the rest to the flexure steel:

.bottomatusemm

mmmm

25520522211831

221471692471106

2

22

2

22

381116404220420

24000020420

167017504220420

6921164014220

mmA

f

b

f

fP

s

A

f

AfA

mmPf

f

s

AA

l

yv

w

y

yvh

t

y

cpcl

h

y

yvtl

..

...)(.

.cot

min

mins

A Where

t



123

ANAS DAWAS



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ANAS DAWAS

elpmaxE

The one way joist system shown in figure , supports a total factored dead load of 7.5 KN/m2 and factored live load of of 8 KN/m2. Totaling 15.5 KN/m2 . Design the end span AB , of the exterior spandrel beam on grid line 1 . The factored dead load of the beam (i.e , self-weight) and the factored loads applied directly to it total 16 KN/m . The spans and loadings are such that the moments and shears can be calculated by using the moment coefficients from ACI section 8.3.3 . Use fy =420 MPa and fc = 30 MPa .



125

ANAS DAWAS

1) Initial choice of beam size . The beam overhangs the inside face of the columns by 50 mm. Effective length 405 mm and slab thickness is 110 mm .

2) Assume that the joists behave as a one unit with uniform load for simplicity , calculate the loads now :

Total load on the beam =load from the slab + load on it self

KNlw n 18816

2

3951516

2.

..

Compute the moment along the edge beam .

650 mm 47

0 m

m

600 mm

Spandrel beam Column

110 mm



126

ANAS DAWAS

KNlw

MnegativeendInterior

KNlw

MpositivespanMid

KNlw

MnegativeendExterior

nu

nu

nu

838310

66188

10

127414

66188

14

923916

66188

16

22

22

22

...

:

...

:

...

:

3) Check adequacy of the section size based on the maximum moment subjected to the edge beam. The section has adequate size for flexure .

4) Calculate area of steel for flexure :



127

ANAS DAWAS

Area of steel required (mm2)

Exterior end negative 1791 Mid-span positive 2046

First interior negative 2865

5) Draw final Shear and Torsion diagram for the edge beam .

KNVedgeatB

KNVedgeexterioratAForceShear

u

u

33347290151

72902

66188

....int)

...

):



128

ANAS DAWAS

mKNlt

T

mmKNt

KNlw

mKNlw

M

Torsion

nu

u

n

n

....

/..).(..

...

....

:

72612

66379

2

3793250172955

1722

39515

2

95524

39515

24

22

u

torsion a as beam to dTransferre isjoist theat Moment

KNdVu 7298405007887334 .).(..@

72.1 KN

55.9 KN.M

325

mm



129

ANAS DAWAS

6) Should torsion be considered ?

7) Equilibrium or Compatibility ?

The torque resulting from the moments at the ends of the joists exists only because the joint is monolithic and the edge beam has a torsional stiffness.

The torque resulting from the 25 mm. offset of the axes of the beam and column necessary for the equilibrium of the structure and hence is equilibrium torque. Note : You should calculate reduced value and equilibrium torsion at offset the design must be based on the maximum value .

Check equilibrium torsion

mKNT ...

.. 372

660250188

consideredbemustT

mKNP

AfT

mmP

mmA

u

cp

cp

cth

cp

cp

7137261

713102960

3451003008307500830

296036023601104706502

345100110360470650

622

2

..

..)(..)(.

)()(

)()(

mKN

cpP

cpA

cf

uT .)(..)(. 55

2960

234510030330750

2

330

support fromdat ValueReduced



130

ANAS DAWAS

Design should be based on T=55.1 KN

8) Check the adequacy of the section >>>>> OK

hP ohA nT oA 1868 mm 209989 mm2 73.5 KN.m 45 0.85 Aoh

88.1 KN

88.1 KN

25 mm



131

ANAS DAWAS

Table 1 These values are from book ( copy and paste ), Check it by yourself

max 1.781 MPa

sV sAv / 1.2076 mm2/mm

cV 240 KN sAt / 0.4902 mm2/mm )/( sA tv 2.188 mm2/mm

lA 916 mm2

min,lA 670 mm2 Stirrups , bars steel diameter



132

ANAS DAWAS

Redistribution for the moment along joist

Moment along the joist before and after redistribution :

mKNwl

edgeatM

mKNwl

midspanatM

mKNwl

edgeextatM

nu

nu

nu

....

.int

....

.

....

.

113410

39515

10

79514

39515

14

855524

39515

24

22

22

22

mKN

dl

Tt

n

u ./).(.

.

/)(19

24050266

155

22

55.85 KN.m 134.1 KN.m

95.7 KN.m

36.85 18.4



133

ANAS DAWAS

MELBeRyg

4-1 A simple supported precast T beam as shown in the sketch below . Use yielding strength of 420 MPa for all steel , compressive strength of concrete of 28 MPa . Assume 40 mm cover from all sides , ø12 stirrups then determine the following :

1) Compute the following section properties : Acp = …………………………. mm2 Aoh = …………………………. mm2 Ao = …………………………. mm2 Pcp = …………………………. mm Ph …………………………. mm

2) If the ultimate torque Tu @ d = 55 KN.m and Ultimate shear force at “d” distance from

face of the support is 285 KN , estimate the maximum shear stress in the section .

1200 mm

300 mm

800 mm

200 mm



134

ANAS DAWAS

3) Is the section adequate for shear design ? ( Yes/No) 4) Compute the cracking torqe for given section . 5) Compute the total area required for shear and torsion based on the values assumed in

(2).

)/( mmmms

A sv 2

6) Compute required area of longitudinal reinforcement for torsion if 26540 mm

s

tA. .

4-2 The 8 m span beam shown below is a part of a continuous frame that carries a cantilever

slab of 1.5 m width . The beam supports a service live load of 20 KN/m along the beam centerline plus 2 KN/m2 over the slab surface . Assume the effective depth of the beam is 530 mm and the clear cover to the stirrups is 40 mm . Including self-weight of the beam and slab in your calculation .

Draw the shear force and torsion moment diagrams for this beam . Determine the shear force and torsion moment at the most critical section . Design the reinforcement in the beam to carry the forces in previous part . Show a layout of the beam’s cross section with the reinforcement obtained .

From previous part .

MPaf

Mpaf

c

y

28

420



135

ANAS DAWAS

4-3 The figure below shows a cantilever beam subjected to a point load . Indicate the type of

torsion in this problem . Explain your answer .

4-4 A) Draw the Torsion moment diagram for part CD given the bending moment diagram for part AB . Determine this type of torsion with explanation .



136

ANAS DAWAS

B) Draw torsion moment diagrams for following cantilever .

4-5 The following figure shows a partial plan of precast roof system . Roof members are double tee units simple supported on precast beams .Continuity of spandrel beams is not provided . The roof is supporting a service live load of 1.5 KN/m2 and service dead load of 3.0 KN/m2 . Assume effective depth d =730 mm and ignore the self-weight in your calculation . a) Determine the factored load that will carried by the spandrel beam . b) Draw bending moment , shear force and torsion diagrams for the spandrel beam . c) Determine the design ultimate bending moment , ultimate shear and ultimate torsion

at most critical sections . d) Design shear reinforcement for the beam based on values from ( c ) .



137

ANAS DAWAS

4-6 Design the T beam shown in the figure below ,considering the contribution of the flanges for torsion if its subjected to a ultimate torsional moment of 18 KN.m . compressive strength of concrete is 30 MPa and yf =360 MPa for all steel .Assume

concrete cover of 40 mm to the centerline of the stirrups al around the cross section .



138

ANAS DAWAS

4-7 Architectural and clearance requirements call for the use of transfer girder , shown in the figure below , spanning 6 m between supporting columns. The girder must carry from above an ultimate concentrated column load of 90 KN at midspan , applied with eccentricity 600 mm from girder centerline . (Load factors are already included , as is an allowance for girder self-weight ). Flexural rigidity at the ends of the span can be assumed to develop 40 percent of the maximum moment that would be obtained if the girder were simple supported . Design both transvers and longitudinal steel for the beam . MPafandMPaf yc 42028

The member is to have dimensions , b=250 mm , h=500 mm , xo = 163 , yo=413 mm , d =450 mm .



139

ANAS DAWAS

4-8 Joint A is monolithically built with cross beam C-D. Beam A-B spans 8.0 meters between supports A and B while beam C-D spans 6.0 meters between supports C and D. Beam A-B carries a factored load of 120 kN/m and beam C-D carries a factored load of 85 kN/m (both are including self-weights). Use fy = 420 MPa for all steel and fc’ = 26 MPa and calculate the following:



140

ANAS DAWAS

1) Compatibility torsion applied to beam C-D: 2) Compatibility torsion that beam C-D should be designed for: 3) Design moments for beam A-B are :

4-9 The one way joist system shown in figure , supports a total factored dead load of 9

KN/m2 and factored live load of of 10 KN/m2. Totaling 19 KN/m2 . Design the end span AB , of the exterior spandrel beam on grid line 1 . The factored dead load of the beam (i.e , self-weight) and the factored loads applied directly to it total 18 KN/m . The spans and loadings are such that the moments and shears can be calculated by using the moment coefficients from ACI section 8.3.3 . Use fy =420 MPa and fc = 30 MPa .



141

ANAS DAWAS

it was found that joists with overall depth of 470 mm would be required .The slab thickness is 110 mm . The spandrel beam was made the same depth . The column supporting the beam are 600 mm square , the beam overhangs the inside face of the columns by 50 mm .Beam size is h= 470 mm , b=650 mm and d= 405 mm.

Answer the following : 1) Draw bending moment and shear forces diagrams for the beam. 2) Draw bending moment and shear forces diagrams for joist before

redistribution. 3) Calculate the torque resulting from the moments at the ends of the joist if the

joint is monolithic .Indicate this type of torsion and explain your answer. 4) Calculate the torque resulting from the 25 mm offset of the axis of the beam

and column . Indicate this type of torsion and explain your answer . 5) Calculate the design ultimate torsion at most critical section 6) Calculate all section properties :

ohcpohcp APPAA ,,,,

For the hollow section shown below , calculate the following : Assume the cover = 40 mm

1) Gross moment if inertia 2) Cracking moment of inertia 3) cpA

4) cpP

5) ohA 6) hP 7) oA



142

ANAS DAWAS

Footings



143

ANAS DAWAS

FOOTINGS

Footings are structural members used to support columns and walls and transmit their loads to the underlying soils. Reinforced concrete is a material admirably suited for footings and is used as such for both reinforced concrete and structural steel buildings, bridges, towers, and other structures.

Types of Footings

Among the several types of reinforced concrete footings in common use are the wall, isolated, combined, raft, and pile-cap types. These are briefly introduced in this section; the remainder of the chapter is used to provide more detailed information about the simpler types of this group.



144

ANAS DAWAS

1) A wall footing Wall footings are normally used around the perimeter of a building and perhaps for

some of the interior walls.

2) An isolated or single-column footing is used to support the load of a single column. These are the most commonly used

footings, particularly where the loads are relatively light and the columns are not closely spaced.

3) Combined footings Are used to support two or more column loads. A combined footing might be

economical where two or more heavily loaded columns are so spaced that normally designed single-column footings would run into each other.



145

ANAS DAWAS

4) A mat or raft or floating foundation Is a continuous reinforced concrete slab over a large area used to support many

columns and walls. This kind of foundation is used where soil strength is low or where column loads are large but where piles or caissons are not used. For such cases, isolated footings would be so large that it is more economical to use a continuous raft or mat under the entire area. The cost of the formwork for a mat footing is far less than is the cost of the forms for a large number of isolated footings. If individual footings are designed for each column and if their combined area is greater than half of the area contained within the perimeter of the building, it is usually more economical to use one large footing or mat. The raft or mat foundation is particularly useful in reducing differential settlements between columns—the reduction being 50% or more. For these types of footings, the excavations are often rather deep. The goal is to remove an amount of earth approximately equal to the building weight. If this is done, the net soil pressure after the building is constructed will theoretically equal what it was before the excavation was made. Thus, the building will float on the raft foundation.



146

ANAS DAWAS

SOIL PRESSURE UNDER FOOTINGS

The soil pressure at the surface of contact between a footing and the soil is assumed to be uniformly distributed as long as the load above is applied at the center of gravity of the footing. The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.



147

ANAS DAWAS

Allowable Soil Pressures

The allowable soil pressures to be used for designing the footings for a particular structure are preferably obtained by using the services of a geotechnical engineer.

Limit States for the Design of Foundations

1) Limit States Governed by the Soil: a bearing failure of the soil under the footing a serviceability failure in which excessive differential settlement between adjacent footings

excessive total settlement.

2) Limit States Governed by the Structure

flexural failure of the portions of the footing that project from the column or wall,

shear failure of the footing,

bearing failure at member interfaces, and

inadequate anchorage of the flexural reinforcement in the footing.



148

ANAS DAWAS

Elastic Distribution of Soil Pressure under a Footing

Generally , the stress under footing given by I

yM

A

Pq

calculated being are stress the epoint wher the to axis centroidal the fromdistance The

area footingthe of axis centroidal theabout Moment

area this of inertia ofMoment

footingthe and soil the between surfacecontact of Area

load Vertical

y

M

I

A

P

The loads within Kern distance : Loads applied within the “ kern “ , shaded

area in the figure alongside, will cause compression over the entire area of the footing .



149

ANAS DAWAS



150

ANAS DAWAS

Loads out of the “kern” distance This occurs when eccentricity in rectangular footing exceeds the shaded area of

the kern distance . A triangular stress distribution will develop over part of the base as shown below , applying the equilibrium equation gives :

eL

a

ba

Pf

Pabf

u

u

2

3

2

32

1

max

max )(



151

ANAS DAWAS

Net Soil Pressures(qn)

wtSurcharge wtconcretewtsoilqq an

Factored Net Soil Pressure

A

ploadnetfactoredq ultimatecolumn

nu

DESIGN PROCEDURE 1) Assume thickness of the footing then calculate area required .

it is necessary to guess a thickness for a first trial. Generally, the thickness will be 1 to 1.5 the wall thickness or to 2 times column thickness .

Thicknesses of wall footings are chosen in 25 mm. increments, widths in 50- or 75-mm. increments.

Surcharge

wtconcretewtsoilaqq

soilforq

LLDLloadServiceA

n

n

)(

2) Calculate factored net pressure .

A

ultimatecolumnploadnetfactorednuq

3) Check shear capacity

A. Punching shear



152

ANAS DAWAS

B. One way shear Calculate avarege effective length in the two directions

SHEARCHECKA

changenqchangeshouldyoudepthchangeyouif

depthincreasetoneednouVcVif

dobcf

dobcf

ob

ds

dobcf

c

ofsmallercV

,

)(

)(

3

1

122

12

42



153

ANAS DAWAS

thicknesschangeotherwiseOKuVcVif

dwbcfcV

bdhddeptheffective

)(.6

1750

-mm) (75 cover

4) Design flexural reinforcement The deflection for the spread footing

shown is in two directions , so the reinforcement should be provided in both the short and long directions .



154

ANAS DAWAS

mm

hofsmallerS

bhsA

bcf

yfSAa

jdyf

uMsA

fbnuquM

500

2

00180

850

2

2

max

.min

.,

500

500180

90000501

0030

hsmallestSbutbhsA

steeleTemperatur

SbeshouldS

SspacingtheandusedbeshouldbarsofNOfind

uMjdyfsAnMcheck

tifacc

cdt

max.

max

"".

?

..,)(.



155

ANAS DAWAS

5) Design column footing joint



156

ANAS DAWAS

6) Area Of dowels The purpose of the dowels is the same as the keyway. But dowels provide stronger

connection . You should place dowels in the

footing while the concrete is still wet .

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 .

650.



157

ANAS DAWAS

According to ACI code we can determine required area of dowels by

area section Column

1

1

650

0050

A

PPWhenf

PpA

PPWhenAA

nbuy

nbureqs

unbs

.;

.min

7) Check development length for dowels



158

ANAS DAWAS



159

ANAS DAWAS

8) Check development length for flexural reinforcement



160

ANAS DAWAS

elpmaxE (Design wall Footing )

Design a plain concrete footing to support a 400 mm thick concrete wall . The on the wall consist of 230 KN/m dead load (including self-weight) and a 146 KN/m live load KN/m .

The base of the footing is 1200 mm below final grade . ,, MPayfMPacf 42021 the

gross allowable soil pressure = 240 KN/m2 , and the soil density is 18 KN/m3 .

1) Estimate size of the footing and the factored net pressure. Assume depth of the footing =(1-1.5)×wall thickness =1.25×400=500 mm

2

92145021182550240 mKNnq /.)..()(.

m 1.75 footingthe of width strip m 1 assume

2

7519214

146230mA .

.

Use width = 1.80 m

2) Compute factored net soil pressure

2

1283801

1466123021mKNnuq /.

.

)(.)(.

3) Check one way shear :

OKcVuV

KNcV

KNmduV

mmdEstimate

24154121000216

1750

4811412502

4008011283

541251275500

..

....

.@

..



161

ANAS DAWAS

No need to check two way shear in the wall footings .

4) Design for flexure :

As,min=0.0018 bh not 0.0018bd (there is an error above , but its difficult to edit it )

mmSmUse

mmsA

mmsA

sA

mKNuM

2005

1000145

257421000541200180

2451

541295042090

1000469

4692

2

4081

2

1283

/

...min

...

.

..)..

(.



162

ANAS DAWAS

5) Check development length

6)

You should check this length

mm

cf

bd

yf

dhl

mml

mm

cf

bd

yf

dl

30821

14420240

240

625752

4001800

11552110

144209

10

9

.

.

hook 90 Use



163

ANAS DAWAS

elpmaxE (Square footing )

Design a square footing to support a 450 mm square tied interior column reinforced with 258 bars . The column carries an un-factored axial dead load of 1000 KN and an axial live

load of 900 KN. The base of the footing is 1200 mm below final grade and allowable soil pressure is 240 KN/m2 . Use MPacfMPayf 28420 ,

1) Estimate the footing size and factored net soil pressure .

23293

33

90061100021

0303

2878

5214

9001000

222141860211560240

600

2

mKNnuq

mUse

mA

mKNnq

mmhdepthfootingAssume

/.)(.)(.

..

..

/.))(..()(.

2) Check two way shear

OKcVuV

KNuV

KNcV

cV

mmob

mmavgd

323822

9375023932640

22394

375054874504

5487255175600

.).(.

.

).(

.)(.

c and b , a formulasthree the of value smaller the



164

ANAS DAWAS

3) Check one way shear

OKcVuV

KNcV

KNuV

798654873000286

1750

96923487502

450

2

33293

...

.)..

(.



165

ANAS DAWAS

4) Design for flexure

mmSmmUse

mmsA

mmsA

mKNuM

5237113

7523000263702513

23240600300000180

26325

112232

45033293

.)(

.min

.).

(.



166

ANAS DAWAS

5) Design column footing connection

OK

NuP

KNN

KNN

KNN

mmA

mmA

2640

6281

198652030

12382030286508502

6281203028650711

21238

221450212

220304504501

.

....

...

.)...(

...

6) Design area of dowels

21256204

210134504500050

mmuse

mmdowelsA

)(.

7) Check development length of dowels:

mmmm

yfbd

cf

yfbd

dhl

37038128

42020240

0440240

.

..

8) Check development length of reinforcement of footing

mm

cf

bdyf

dl 17862810

254209

10

9



167

ANAS DAWAS

mmmmdhlhookuse

mmlAvailable

120047628

4202524090

17861200752

4503000

.

13ø25

4ø20

3000 mm

600-75 mm

450 mm

75 mm



168

ANAS DAWAS

gomDJta geoonJt g

elpmaxE

224042020

1000

1300

2600600

2

650

890

2400600

1

mKNaqMPayfMPacf

loadsserviceareloadsAll

KNLL

KNDL

mm

COL

KNLL

KNDL

mm

COL

/,,



169

ANAS DAWAS

1) Determine required area

2418 m

nq

.

208.610001300650890requuired Area

2KN/m 208.65-1.2(22)-240

: thention,simplifica for just concreteand soil fordensity average the as 22 Take

2) Locate the point of application of column loads on the footing.

125 mm

600 mm 400 mm

6.0 m

Basement floor with thickness = 125 mm supporting an ultimate live load of 5 KN/m

1.2 m

x

Q1 Q2

mx

KNQKNQ

594.315402300

623002300100013002

154065089012Q1Q

L2Qx 3



170

ANAS DAWAS

3) Determine the dimension of the footing

mLAB

mmLxL

5.26.74.18

6.7588.7)2.0594.3(2)(2 2

4) Factored net pressure and factored loads

KNQKNQ

mKNq

uu

nu

3160)1000(6.1)1300(2.12108)650(6.1)890(2.1

/3.2775.26.7

)1000650(6.1)1300890(2.1

21

2

L/2

L3 L1 L2

Q1+Q2

Q1+Q2



171

ANAS DAWAS

5) Draw moment and shear diagrams Diagrams should be based on 277.3×2.5=693.25 KN/m

6) Check two way shear

165.14b perimeter shear critical of length

2536)5.15.1(3.2773160column interior For

9001000

o

c

um

KNV

mmdmmh

Try

2108 KN 3160 KN

13.8 KN.m

2782.6 KN.m

679.4 KN.m

138.65 KN

1969.35 KN

2190 KN

969.85 KN

693.25 KN/m



172

ANAS DAWAS

3220)8402)

6.3756)

5.1400600

2.35.1)85.0(24.1754)85.05.1(3.2772108

column interior For

4.603731)

1207412

)2()

905612

)42()

OKVVKNcKNb

KNaofsmallesV

mbKNV

OKVV

KNdbfc

KNdbf

bdb

KNdbf

a

ofsmallestV

uc

c

c

ou

uc

oc

oco

s

occ

c

7) Check one way shear

OKKNV

KNVmmdmmh

Try

thicknesstheincreaseKNdbfV

KN

c

u

wcc

5.139710002500206175.0

8.1288)0.13.0(25.693219010001100

7.12579002500206175.0

61

1.1358)9.03.0(25.6932190V ) column the of

center from mm 300d ( column interior the of face the fromd at critical is shearway One

u



173

ANAS DAWAS

8) Design the flexure reinforcement in the longitudinal direction

2

2min

2s

236

3

83402517

4950110025000018.00018.0

7651A iterationsmany after

81791010009.0104209.0

106.2782)2/(

mmUsemmbhA

mm

mmadfMA

s

yus



174

ANAS DAWAS

2min

2s

54012511

9.199510009.04209.0

1000679A

moment positive tosubjected column interior For

mmUse

Amm s

9) Design for transverse beam

2

2min

2

22

4418259

4158110021000018.0

mm 1588.4As iterationsmany after10009.04209.0

10004.570

/4.5702

95.012642

mmUsemmA

A

mKNwlM

s

s

u

B+1.5 d =2.1 m

B

0.75 d

0.95 m

3160 KN



175

ANAS DAWAS

2

2min

2

4.2454255

2277110011500018.0

/4.3802

95.02.843

:

mmUsemmA

mKNM

beamedgeFor

s

u

Check development length for all reinforcements required

2108 KN

2018/2.5=843.2 KN/m



176

ANAS DAWAS

NOTES

You can use these equations to calculate the ultimate bending moment , ultimate shear force and ultimate load subjected to footings with eccentricity at any point .

x

qmax

qmin

BLqqLqP

XBqqXBqV

qqqLxLqbut

XBqqXBqM

u

u

u

)2

)(()(

)2

)()(())((

)(

)3

)()(()2

()(

minmaxmin

max

minminmax

2max

2



177

ANAS DAWAS

Moment transfer from columns to footings depends on how the column–footing connection is constructed. Many designers treat the connection between columns and footings as a pinned connection. Others treat it as fixed, and still others treat it as somewhere in between. If it is truly pinned, no moment is transferred to the footing, and this section of the text is not applicable. If, however, it is treated as fixed or partially fixed, this section is applicable. If a column–footing joint is to behave as a pin or hinge, it would have to be constructed accordingly. The reinforcing in the column might be terminated at the column base instead of continuing into the footing. Dowels would be provided, but these would not be adequate to provide a moment connection. To provide continuity at the column–footing interface, the reinforcing steel would have to be continued into the footing. This is normally accomplished by embedding hooked bars into the footing and having them extend into the air where the columns will be located. The length they extend into the air must be at least the lap splice length; sometimes this can be a significant length. These bars are then lap spliced or mechanically spliced with the column bars, providing continuity of tension force in the reinforcing steel.



178

ANAS DAWAS

SMELBeRP

5-1 Design a square footing to support a 500 x 500 mm-square tied interior column reinforced with 825 bars. The column carries unfactored axial dead load of 1100 kN and unfactored axial live load of 750 kN. The base of the footing is 1200 mm below final grade and allowable NET soil pressure is 195 kN/m2. Use fc’ = 28 MPa and fy = 420 MPa. Assume d = h – 100mm.

5-2 A 4.0 x 4.0 m-square footing is

subjected to an ultimate (factored) axial load Pu = 2400 kN due to eccentric column as shown below (ignore footing self-weight). Assume effective depth d = 0.535 m.

Calculate maximum and minimum ultimate soil pressure. Calculate maximum ultimate flexural moment in the footing Calculate maximum one-way shear in the footing (Vu) Calculate the punching shear exerted by the 400 mm-square column on the footing (Vu).



179

ANAS DAWAS

5-3 A spread footing is subjected to 500×500 mm square tied interior column The base of the footing is 1500 mm below final grade and allowable soil pressure is 200 kN/m2, effective depth equal to (h-100 mm ). Answer the followings :

If the column carries unfactored axial dead load of 1300 kN and unfactored axial live load of 800 kN,. Assume thickness of footing = 700 mm , answer the following :

a) Determine the minimum dimension of the square footing required to support the loads . b) Calculate the ultimate design bending moment and ultimate design shear force at the

most critical sections for the 3.5×3.5 m2 footing . c) Calculate the maximum one way shear capacity( cV ) for 3.5×3.5 m2 square footing. d) Calculate the maximum two way shear capacity ( cV ) for the 3.5×3.5 m2 square

footing. e) Check development length for the 3.5×3.5 m2 square footing .

For a 4×6 m rectangular footing . Assume that the ultimate design bending moment in

the long direction is 900 KN.m , the ultimate design bending moment in the short direction is 600 KN.m and the ultimate one way shear force is 700 KN , answer the following :



180

ANAS DAWAS

f) Calculate the area of steel required in the long direction . g) Calculate the area of steel required in the long direction . h) Calculate the minimum area of steel in the long direction.

i) If we have to use 30ø12in the short direction , make a simple sketch showing the distribution of the bars in this direction .

5-4 The rectangular footing shown below is subjected to a concentrated factored column

load Pu=3000 KN and having an area 3.00×4.00 m2. Given the column size is 350×450 mm2 and effective depth of the footing is 610 mm :

a) Determine the value of the moment at the critical section(s) . b) Check the adequacy of the footing depth for one way shear . c) Check the adequacy of the footing depth for punching shear . d) Assume that the main reinforcement in the footing consists of 18-mm diameter bars

at 150 mm O.C and the dowels in the footing are 14- mm bars . Sketch the distribution of the reinforcement in the short direction .



181

ANAS DAWAS

5-5 For the wall footing shown below , draw the ultimate bearing stress distribution for the following cases and determine the values of ultimate shear and ultimate moment at their critical sections . Given effective depth of the footing is 310 mm . a) Pu= 800 KN b) Pu= 600 KN , Mu =100 KN.m c) Pu= 300 KN , Mu =100 KN.m d) Pu= 300 KN , Mu =200 KN.m

5-6 For the rectangular spread footing shown in the figure below , service dead load of 800 KN and service live load of 600 KN and the eccentricities along x-axis and y-axis equal to 0.20 m .Unit weight of the soil is 18 KN/m3. a) Calculate the soil pressure at

the four corners of the footing .

b) Calculate the maximum bearing stress capacity at the column-footing joint .

c) If the allowable soil pressure is 260 KN/m2 , show whether footing size is adequate or not ?

1.2 m

6.0 m

5.0 m

0.6 m



182

ANAS DAWAS

5-7 For the combined footing shown in the figure below . Allowable soil pressure is 250 KN/m2 at 1.5 m below surface , use 22avg KN/m3 for all materials (soil and concrete ) ,fc =21 MPa and fy =420 MPa .

Column size Service dead load Service live load 1 450×450 mm 700 KN 500 KN 2 600×600 mm 1000 KN 700 KN

COL2 COL1

6.0m 0.8m

= 0.60 m



183

ANAS DAWAS

a) Calculate the area required , and find the dimensions B and L . Assume that the dimensions of the footing are 2.5×9 m : b) Draw ultimate bending and shear diagrams for the footing . c) Calculate the required area of steel in the long direction for positive and negative

moments . d) Calculate the minimum area of steel in the long direction . e) Calculate the maximum one way shear (Vu) at the most critical section (s) . f) Calculate the required area of steel in the transverse direction .

g) Prepare neat design drawings showing footing dimensions and provided reinforcement.

5-8 A wall footing with 700 mm thickness and 4.0 m width , carries a service dead load of

250 KN/m and service live load of 180 KN/m . The base of the footing is 1300 mm under the soil with unit weight equal to 20 KN/m3 . Gross soil pressure is 240 KN/m2 , concrete compressive strength is 23 MPa and fy for all steel is 420 MPa .



184

ANAS DAWAS

a) Calculate the maximum bending moment at the most critical section(s) . b) Calculate the required area of steel in the short direction. c) Calculate the required area of steel in the long direction. d) How many bars per meter needed in the long and short directions . e) Calculate the ultimate shear force (Vu ) at the most critical section . f) Calculate the maximum shear strength capacity . g) Calculate the development length of the reinforcement used in the short direction .

Use ø25 M

5-9 Why it’s better to put the reinforcement in the long direction in a rectangular footing under the reinforcement in the short direction .



185

ANAS DAWAS

Biaxial Columns



186

ANAS DAWAS

Columns :Combined Axial Load and Bending

The majority of reinforced concrete columns are subjected to primary stresses caused by flexure, axial force, and shear. Secondary stresses associated with deformations are usually very small in most columns used in practice. These columns are referred to as "short columns." Short columns are designed using the interaction diagrams presented in this chapter. The capacity of a short column is the same as the capacity of its section under primary stresses, irrespective of its length. Many columns are subjected to biaxial bending, that is,

bending about both axes, such as : 1) Corner columns in buildings where beams and girders

frame into the columns from both directions are the most common cases.

2) where columns are cast monolithically as part of frames in both directions .

3) where columns are supporting heavy spandrel beams. Bridge piers are almost always subject to biaxial bending.

Most columns are subjected to significant bending in one direction, while subjected to relatively small bending moments in the orthogonal direction. These columns are designed by using the interaction diagrams discussed in RC1 course for uniaxial bending and if required checked for the adequacy of capacity in the orthogonal direction. However, some columns, as in the case of corner columns, are subjected to equally significant bending moments in two orthogonal directions. These columns may have to be designed for biaxial bending.



187

ANAS DAWAS

A circular column subjected to moments about two axes may be designed as a uniaxial

column acted upon by the resultant moment; (Why ?? ) Circular columns have polar symmetry and, thus, the same ultimate capacity in all

directions. The design process is the same, therefore, regardless of the directions of the moments. If there is bending about both the x- and y-axes, the biaxial moment can be computed by combining the two moments or their eccentricities as follows:

22

22

)()(

)()(

yx

uyuxu

eee

MMM

For shapes other than circular ones, it is necessary to consider the three-dimensional

interaction effects.

Whenever possible, it is desirable to make columns subject to biaxial bending circular in shape.

Several methods used to determine the nominal strength of the columns subjected to moments in the two directions :

1) static equations : - Such a procedure will lead to the correct answer, but the mathematics involved is so

complicated because of the shape of the compression side of the column that the method is not a practical one.



188

ANAS DAWAS

2) Computer programs 3) The strain-compatibility method 4) The equivalent eccentricity method. 5) Method based on 45 slice through interaction surface. 6) Bresler reciprocal load method : this will be discussed in this part .

Bresler reciprocal load method

The Bresler equation works rather well as long as Pn is at least as large as 0.10Po . Should Pn be less than 0.10Po it is satisfactory to neglect the axial force completely and design the section as a member subject to biaxial bending only. This procedure is a little on the conservative side. For this lower part of the interaction curve, it will be remembered that a little axial load increases the moment capacity of the section. The Bresler equation does not apply to axial tension loads. Professor Bresler found that the ultimate loads predicted by his equation for the conditions described do not vary from test results by more than 10%.

ACI Commentary Sections 10.3.6 and 10.3.7 give the following equation, originally presented by Bresler [11-18], for calculating the capacity under biaxial bending:

nnynxn pppp

1111



189

ANAS DAWAS

Procedure Used To Design Biaxial Columns

1) Select a trial section:

2) Select appropriate number of bars to be subjected to the column : A bar is adequately supported against lateral movement if its located at a corner of a tie and if the dimension “x” shown in the figure below is less than 15 cm .

. spiral for mm 300and mm 250 columntied a for dimension Min.)(5.0

)(4.0)(

%)55.2(%)21(

Columnspiralforgyfcf

up

ColumnTiedforgyfcf

up

trialgA

ColumnspiralforColumnTiedfor

gthatassume



190

ANAS DAWAS

3) Calculate Gamma

mmbddb bs 40cov2cov2

Figure 18 Ties shown dashed may be omitted if x < 150 mm

h

b

x x h



191

ANAS DAWAS

nxgy

y

nygxx

ny

nxuux

yu

uyx

oyx

Ple

Ple

ondependingPand

PfindtodiagramtheusethenPMeP

Me

eandeeCompute

,

,:

,

:,

spiralforP

TiedforPloadaxialMax

and

fAAAfP

PCompute

n

n

Ts

yssgcn

n

850

80

650750

850

.

..

..

))(.(

:

un

nnynxnpbeshouldp

pppp

1111

4) Compute ex , ey and eo

5) Compute oP

6) Solve for nP



192

ANAS DAWAS

Interaction Diagrams

The ACI column interaction diagrams are used in Examples to design or analyze columns for different situations. In order to correctly use these diagrams, it is necessary to compute the value of γ (gamma), which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by h, the depth of the column (both values being taken in the direction of bending). Usually the value of γ obtained falls in between a pair of curves, and interpolation of the curve readings will have to be made.

Caution 1) Be sure that the column picture at the upper right of the interaction curve being used

agrees with the column being considered. In other words, are there bars on two faces of the column or on all four faces? If the wrong curves are selected, the answers may be quite incorrect.



193

ANAS DAWAS

elpmaxE

Select column section size and reinforcement for a rectangular tied column with bars distributed along four faces, subject to biaxial bending. Given :

Pu Mux Muy fc fy 1900 KN 100 KN.m 190 KN.m 28 MPa 420 MPa

1) Estimate column size

sizesmallerselectcanYoummmmTry

ffPA

ygcu

g

222

g

200000400500mm 96.138483)420015.028(4.0

10001900A0.015 ratio steel gross assume )(4.0

2) Select appropriate number of ø25 bars



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3) Compute nxP (at ex=0.0)

units of takecare tonecessary its so unit Ksi with diagrams ninteractioused have I : note31201040050060.15

60.159.626.226.2 ioninterpolatby

3.275.0

???6875.0

2.260.0

6875.0400

25)10(2)40(2400

040.0025.0400500

254

10

132.0400

6.526.52

1900100

3

2

KNP

MPaKsibhP

KsibhP

for

forKsibh

Pfor

OKAA

he

mmmm

lemmP

Me

nx

n

n

n

gS

g

y

y

uux

y

2.2 Ksi



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4) Compute nyP ( at ey =0.0)

5) Compute noP You can calculate it by equation or by interaction diagrams ( independent of Gamma

value )

. diagrams using ofaccuracy less to due or units of because bemay ,P of values in difference largea is there , that note

4357420254

10254

104005002885.065.0

Equation Using34501040050025.17

25.175.20.0

025.0Diagrams Using

n

22

3

KNP

KNP

MPaKsibhP

he

on

on

ng

units of takecare tonecessary its so unit Ksi with diagrams ninteractioused have I : note26221040050011.13

11.139.69.1

9.175.0

75.0500

25)10(2)40(2500

040.0025.0400500

254

10

2.0500100

1001900190

3

2

KNP

MPaMPabhP

KsibhP

for

OKAA

he

mmmm

lemmP

Me

ny

n

n

gS

g

xx

u

uyx



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6) Solve for Bresler equation

OKKNPKNPP

un

n19002117

1072385.44357

12622

13120

11 4

In this chapter, Pn values were obtained only for rectangular tied columns. The same theory could be used for round columns, but the mathematics would be somewhat complicated because of the circular layout of the bars, and the calculations of distances would be rather tedious. Several approximate methods have been developed that greatly simplify the mathematics. Perhaps the best known of these is the one proposed by Charles Whitney, in which equivalent rectangular columns are used to replace the circular ones. This method gives results that correspond quite closely with test results.



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ANAS DAWAS

SMELBeRP

For all problems :

Cover Bars diameter Ties diameter cf yf 40 mm 25 mm 10 mm 28 MPa 420 MPa

6-1 The column shown below is to carry a factored load , Pu ,of 800 KN with eccentricities of ey = 90 mm and ex = 300 mm . Check the adequacy of the trial design by using Bresler method .

6-2 Check the adequacy of the circular column for the following axial load and biaxial bending moments. Use and procedure you feel is most suitable .

Pu Mux Muy 2200 KN 100 KN.m 175 KN.m



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6-3 For square column shown in the figure below , determine the maximum ultimate biaxial bending moment ( Mux and Muy ) that can be applied simultaneously with a load Pu = 3000 KN to column section . Assume ex = ey and =0.80

6-4 The section of a short tied column is 400×400 mm and is reinforced with 8ø25 bars as shown. Determine the allowable ultimate load on the section nP if its acts at ex = 108 mm and ey = 270 mm. Use Reciprocal Load Method: Bresler’s Formula and the interaction diagram given on Design Aids . Fill the table below .

noP

nxP

nyP

nP



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ANAS DAWAS



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Slender Columns



Slender

Columns

Reference

Chapter12: Reinforced Concrete MECHANICS AND

DESIGN 6th edition

Maher Fakhoury



Definition: A slender column is defined as a column that has a significant reduction in its axial-load capacity due to moments resulting from lateral deflections of the column. In the derivation of the ACI Code, “a significant reduction” was arbitrarily taken as anything greater than about 5 percent.

Most building columns fall in the short-column category. Exceptions occur in industrial buildings and in buildings that have a high first-floor story for architectural or functional reasons. An extreme example is shown in Fig. 12-7. The left corner column has a height of 50 times its least thickness. Some bridge piers and the decks of cable stayed bridges fall into the slender-column category.

Less than 10 % of columns in “braced” or “nonsway” frames and less than half of columns in “unbraced” or “sway” frames would be classified as “slender” following ACI Code Procedure.

Fig. 12-7

Bank of Brazil building,

Porto Alegre, Brazil. Each

Floor extends out over the

Floor below it. (Photograph

Courtesy of J. G. MacGregor.)



Concept of Slenderness Effect

An eccentrically loaded, pin-ended column is shown in Fig the

moments at the ends of the column are when the loads P are

applied, the column deflects laterally by an amount as shown.

Me = Pe

For equilibrium, the internal moment at mid height

Mc = P(e + δ)

The deflection increases the moments for which the column

must be designed. In the symmetrical column shown here, the

maximum moment occurs at mid height.

The slenderness of columns is based on their geometry and on

their lateral bracing. As their slenderness increases, their

bending stresses increase, and thus buckling may occur. If

they are “slender,” the moment for which the column must be designed is increased or magnified. Once

the moment is magnified, it’s very simple the column is then designed as a short column using the

increased moment.

The line O–A is referred to as a load–moment curve

for the end moment, while the line O–B is the load–

moment curve for the maximum column moment.

Failure occurs when the load–moment curve O–B for

the point of maximum moment intersects the

interaction diagram for the cross section. Thus the

load and moment at failure are denoted by point B in

Fig. Because of the increase in maximum moment

due to deflections, the axial-load capacity is reduced

from A to B. This reduction in axial-load capacity

results from what are referred to as slenderness

effects.



Buckling of Axially Loaded Elastic Columns Recall Leonhard Euler Equation from strength of materials laboratory,

Where EI = flexural rigidity of column cross section. L = length of the column. n = number of half-sine waves in the deformed shape of the column. *The effective length factor is K= 1/n



Computation of Effective Length Factor K:

In buildings, columns are restrained by beams or footings which always allow some rotation of the ends of the column. The effective lengths will be greater than the values for completely fixed ends. The actual value of k for an elastic column is a function

of the relative stiffnesses , of the beams and columns at each end of the column, where is

Where b and c refer to beams and columns, respectively, and the lengths

and are measured center-to-center of the joints.

Where Ic=0.70 Ig , Ib=0.35 Ig , Ec=4700

After Computing , the Effective Length Factor K is computed using Nomographs Fig12-6 (NEXT PAGE) A and B are top and bottom factors of columns. For a hinged end is

infinite or 10 and for a fixed end is zero or 1. -Assumptions for nomographs: 1. Symmetrical rectangular frames



2. Equal load applied at top of columns 3. Unloaded beams. 4. All columns buckle at the same moment

Calculation of, Column Footing Joints:

Where is the If moment of inertia of the contact area

between the bottom of the footing and the soil and Ks is

the coefficient of subgrade reaction which can be taken

from Fig. 12-27

Calculation of k from Tables (another way but NOT according to ACI CODE)

Table 12-2 can be used to select values of k for the design of nonsway frames. The shaded areas

correspond to one or both ends truly fixed. Because such a case rarely, if ever, occurs in practice, this

part of the table should not be used. The column and row labeled “Hinged”, “elastic” through to “fixed”

represent conservative practical degrees of end fixity.



Limiting Slenderness Ratios for Slender Columns Most columns in structures are sufficiently short to be unaffected by slenderness effects. To avoid checking slenderness effects for all columns, ACI Code Section 10.10.1 allows slenderness effects to be ignored in the case of columns in sway frames if,

*IF Then a second order analysis must be done.

Unsupported Lengths (Lu): The length used for calculating the slenderness ratio of a column, Lu, is its unsupported length. This length is considered to be equal to the clear distance between slabs, beams, or other members



that provide lateral support to the column. Summary of ACI Moment Magnifier Design Procedure for Columns in Nonsway Frames 1. Length of column. The unsupported length, Lu, is defined as the clear distance between members capable of giving lateral support to the column. For a pin-ended column it is the distance between the hinges. 2. Effective length. States that the effective length factor, k, can be computed from nangoraphs. 3. Radius of gyration. For a rectangular section and for a circular section, 4. Consideration of slenderness effects. For columns in nonsway frames, allows slenderness to be neglected if satisfies Eq. (12-20b). The sign convention for M1/M2 is given in Fig. 12-13. 5. Minimum moment. Requires that the maximum end moment on the column, not be taken less than M2,min = Pu(15 + 0.03h2) where h is in mm 6. Moment-magnifier equation. ACI Code Section 10.10.6 states that the columns shall be designed for the factored axial load, Where M2 is the larger end moment, and the magnified moment Mc, defined by:

*Extra Information



Example 1: Design of a Slender Pin-Ended Column (Nonsway)

Design a 6m-tall column to support an unfactored dead load of

400 KN and an unfactored live load of 334 KN. The loads act at

an eccentricity of 76 mm. at the top and 50 mm. at the bottom,

as shown in Fig. Use f’c= 28 MPa and Fy=420 MPa

1. Compute the factored loads Pu and moments M1/M2 and

Pu= 1.2 DL+1.6LL = 1.2 X 400+1.6 X 334 = 1014.4 KN

The Moment at the top = 1014.4 X 0.076 = 77.1 KN-m

The Moment at the bottom = 1014.4*0.05= 50.72 KN-m

By definition, M2 is the larger end moment in the column. Therefore,

M2= 77 KN-m and M1= 51.7 KN-m the ratio M1/M2 is taken to be

positive, because the column is bent in single curvature Thus

M1/M2=0.658

2. Estimate the column size. Use ρ=1.5%

= = 73935.86mm2

use bXh = 300 X 300 mm2

3. Is the column slender? a column in a nonsway frame is short if

K=1 because the column is pin ended, where r=0.3*300= 90 mm

=

For M1/M2=0.658

34 - 12 = 34-12 X 0.658= 26



Because 66.7 exceeds 26, the column is quite slender. This suggests that the Section 300mm by 300mm

probably is inadequate.

We shall select a 400mm-by-400mm.-section for the first trial.

4. Check whether the moments are less than the minimum. a braced column be designed for a

minimum eccentricity of 15+0.03 X 350 = 25.5 Because the maximum end eccentricity exceeds this,

design for the moments from step 1.

5. Compute EI.

Ec=4700 f’c= 4700 28=24870 MPa

Ig= bh3 /12 = 2.13X 109 mm4

The term Bdnsis the ratio of the factored sustained (dead) load to the total factored axial load:

Bdns = 1.2 X 400 / 1014.4 = 0.473

EI = = 1.44 X 1013 N-mm2

6. Compute the magnified moment.

= 0.6 + 0.4 x 0.658 =0.8632

= KN

1.3



Magnified moment (Mc) =1.3 X 77= 100.1 KN-m

7. Select the column reinforcement. We will use the tied-column interaction diagrams assuming an

equal distribution of longitudinal bars in two opposite faces of the column. The parameters required for

entering the interaction diagrams are:

Cover 50 mm and assume Ø25

From both and the required value for is less than ρ= 0.01. Therefore, to satisfy the

minimum column longitudinal-reinforcement ratio from ACI Code Section 10.9.1, use ρ= 0.01Thus,

As’ rqd = Ag * 0.015 = 400*400 * 0.01 = 1600 mm2

Use 8Ø16 with As= 1609mm2 for the 400x400 mm2 section. This section design would be very

conservative if we were designing a short column, but the slenderness of the column has required the use

of this larger section.



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