specific steps

8/19/2019 Specific Steps

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DEPARTMENT OF CIVIL ENGINEERING

CE-2105

DESIGN OF CONCRETE STRUCTURE-I

DESIGN OF BEAM

(AS PER ACI CODE)

Course Teacher- Presented By-

Dr. Md. Rezaul Karim Md. Jahidur Rahman

Associate Professor S.ID- 121041

Dhaka University of Engineering &

Technology, Gazipur-17001


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Content

2

USD(Ultimate Strength Design)

Classification of beam with respect to design system Assumptions

Evolution of design parameters Moment Factors Kn,

Balanced Reinforcement Ratio b Calculating Strength Reduction Factor

Calculating

Design procedure for Singly Reinforced Beam

Design procedure for Doubly Reinforced Beam

Design procedure for T-BeamAppendix

Jahidur Rahman


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Ultimate Strength Design(USD)

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Assuming tensile failure condition Additional strength of steel after yielding

ACI code emphasizes this method

Jahidur Rahman


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ASSUMPTIONS

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Plane sections before bending remain plane andperpendicular to the N.A. after bending

Strain distribution is linear both in concrete & steel and isdirectly proportional to the distance from N.A.

Strain in the steel & surrounding concrete is the same prior tocracking of concrete or yielding of steel

Concrete in the tension zone is neglected in the flexuralanalysis & design computation

εc=0.003

εs = fy / Es

h dc

0.85fc’

a a/2

d-a/2

b

C

T

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6

Concrete stress of 0.85fc’ is uniformly distributed over an

equivalent compressive zone.

fc’ = Specified compressive strength of concrete in psi.

Maximum allowable strain of 0.003 is adopted as safe limitingvalue in concrete.

The tensile strain for the balanced section is fy/Es

Moment redistribution is limited to tensile strain of at least0.0075

εsεy

f y

f s

Idealized

Actual

Es

1

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EVALUATION OF DESIGN PARAMETERS

Total compressive force -C = 0.85fc’

ba (Refer stress diagram)

Total Tensile force- T = As fy

C = T

0.85fc’

ba = As fy

a = As fy

/ 0.85fc’ b)

=

d fy / (0.85 fc

’)

[

= As / bd]

Moment of Resistance/Nominal Moment-

Mn = 0.85fc’ ba (d – a/2) or,

Mn = As fy (d – a/2)

= bd fy [ d – (dfyb / 1.7fc’) ]

=

fc’ [ 1 – 0.59

] bd2 = fy / fc’

Mn = Kn bd2 Kn = fc’ [ 1 – 0.59 ]

Ultimate Moment- Mu = Mn

= Kn bd2

( = Strength Reduction Factor)7 Jahidur Rahman


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Balaced Reinforcement Ratio ( b)

From strain diagram, similar triangles

cb / d = 0.003 / (0.003 + fy / Es); Es = 29x106 psi

cb / d = 87,000 / (87,000+fy)

b = A sb / bd

= 0.85fc’ ab / (fy. d)

= β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)]

Relationship b / n the depth `a’ of the equivalent rectangular stress block

& depth `c’ of the N.A. is

a = β1c

β1= 0.85 ; fc’ 4000 psi

β1= 0.85 - 0.05(fc’ – 4000) / 1000 ; 4000 < fc’ 8000

β1= 0.65 ; fc’> 8000 psi

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For beams the ACI code limits the max. amount of steel to 75% ofthat required for balanced section.

0.75 b

Min. reinforcement is greater of the following:

Asmin = 3fc’ x bwd / fy or 200 bwd / fy

min

= 3

fc’ /

fy or 200 / fy

For statically determinate member, when the flange is in tension,the bw is replaced with 2bw or bf whichever is smaller

The above min steel requirement need not be applied, if at every

section, Ast provided is at least 1/3 greater than the analysis

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Yes

No

Yes

No

Calculating


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Calculating


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Jahidur Rahman12

SINGLY REINFORCED BEAM

Beam is reinforced near the tensile faceReinforcement resists the tension.Concrete resists the compression.


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DESIGN PROCEDURE FORSINGLY REINFORCED BEAM

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1. Determine theservice loads

3. Calculate d= h – Effective cover

2. Assume `h` as perthe support conditions

[ As per ACI code intable 9.5(a)]

4. Assume the value of `b` by therule of thumb

5. Estimate self weight

6. Primary elastic analysis

and derive B.M (M), Shearforce (V) values

7. Compute min and b 8.Choose between min and b


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Note Below

Design the steel reinforcement arrangement with appropriate cover and

spacing stipulated in code. Bar size and corresponding no. of bars based on

the bar size n.

Check crack width as per codal provisions.

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9. Calculate , Kn

10. From Kn & Mcalculate `d’ required

11. Check the required `d’ with

assumed `d’

12. With the final values of , b, ddetermine the Total As required

OK


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DESIGN PROCEDURE FORSINGLY REINFORCED BEAM BY FLOWCHART

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Actual steel ratio Maximum steel ratio

Over

reinforced

beam

Yes Under

reinforced

beam

No

Ultimate moment


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Jahidur Rahman16

DOUBLY REINFORCED BEAM

Beam is fixed for Architectural purposes.Reinforcement are provided both in tension and

compression zone.Concrete has limitation to resist the total compressionso extra reinforcement is required.

DESIGN


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Actual steel ratio Maximum steel ratio

rectangular

beam

Yes Doubly

reinforced

beam

No

DESIGNPROCEDUREFORDOUBLYREINFORCED

BEAM

Concrete

section, Area

of steel are

known

YesNo Compression and

tension bar reach in

yielding

Compression bar

does not reach in

yielding

DESIGN


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DESIGNPROCEDUREFORDOUBLYREINFORCED

BEAM

Load or

ultimate

moment is

given

Maximum steel ratio

Reinforcement

required in comp.

zone too

rectangular

beam

YesDoubly

reinforced

beam

No


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Jahidur Rahman19

T- REINFORCED BEAM

A part of slab acts as the upper part of beam.Resulting cross section is T – shaped.The slab portion of the beam is flange.The beam projecting bellow is web or stem.


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DESIGN


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DESIGNPROCEDUREFORT-REINFORCED

BEAM

Moment or loading is

given

Total area of steel

Nominal moment

calculation

Rectangular

beam analysis

YesT - beam

analysis

No

Area of steel in flange

Area of steel in web

Additional moment

T section:

L section:

Isolated beam:

Check for


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Jahidur Rahman22

APPENDIX


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AS PER TABLE 9.5 (a)

Simply

Supported

One End

Continuous

Both End

ContinuousCantilever

L / 16 L / 18.5 L / 21 L/8

Values given shall be used directly for members with normal

weight concrete (Wc = 145 lb/ft3) and Grade 60 reinforcement

For structural light weight concrete having unit wt. In range

90-120 lb/ft3 the values shall be multiplied by

(1.65

–

0.005Wc) but not less than 1.09

For fy other than 60,000 psi the values shall be multiplied by

(0.4 + fy/100,000)

`h` should be rounded to the nearest whole number

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RULE OF THUMB

d/b = 1.5 to 2.0 for beam spans of 15 to 25 ft.

d/b = 3.0 to 4.0 for beam spans > 25 ft.

`b` is taken as an even number

Larger the d/b, the more efficient is the section due to lessdeflection

CLEAR COVER

Not less than 1.5 in. when there is no exposure to weather or

contact with the ground For exposure to aggressive weather 2 in.

Clear distance between parallel bars in a layer must not beless than the bar diameter or 1 in.

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BAR SIZE

n = n/8 in. diameter for n8.

Ex. 1 = 1/8 in.

….

8 = 8/8 i.e., I in.

Weight, Area and Perimeter of individual bars

inch mm3 0.376 0.375 9 0.11 1.178

4 0.668 0.500 13 0.20 1.571

5 1.043 0.625 16 0.31 1.963

6 1.502 0.750 19 0.44 2.356

7 2.044 0.875 22 0.60 2.749

8 2.670 1.000 25 0.79 3.142

9 3.400 1.128 28 1.00 3.544

10 4.303 1.270 31 1.27 3.990

11 5.313 1.410 33 1.56 4.430

14 7.650 1.693 43 2.25 5.319

18 13.600 2.257 56 4.00 7.091

Perimeter

(in.)

Stamdard Nominal DimensionsBar

No

Wt.per

Foot (lb)Diameter db C/S Area,

Ab (in2)

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