Design of rectangular beam by USD

PRE-STRESSED CONCRETE LABCE-416

Name: Sadia MahajabinID: 10.01.03.098

Section: BCourse Teacher:

Mr. Galib Muktadir & Sabreena N. Mouri Department of Civil Engineering

AHSANULLAH UNIVERSITY of SCIENCE AND TECHNOLOGY

(SINGLY & DOUBLY)

ULTIMATE STRENGHT DESIGN OF RECTANGULAR

BEAM

Based on the ultimate strength of the structure assuming a failure condition either due to concrete crushing or by yielding of steel.Additional strength of steel due to strain hardening is not encountered in the analysis or design.

Actual / working loads are multiplied by load factor to obtain the design loads.

ACI codes emphasizes this method.

Introduction:Ultimate Strength Design (USD)

Being used since 1957.

Historical Background:

1. Plane sections before bending remain plane after bending.

2. Strain in concrete is the same as in reinforcing bars at the same level, provided that the   bond between the steel and concrete is sufficient to keep them acting together under the different load stages i.e., no slip can occur between the two materials.

3. The stress-strain curves for the steel and concrete are known.

4.The tensile strength of concrete may be neglected.

5.At ultimate strength, the maximum strain at the extreme compression fiber is assumed equal to 0.003

Assumptions:There are five assumption that are made

n wUltimate Strength Design

designing beam .. .

DESIGN AND ANALYSIS

The main task of a structural engineer is the analysis and design of structures. The two approaches of design and analysis will be used

Design of a section:

This implies that the external ultimate moment is known, and it is required to compute the dimensions of an adequate concrete section and the amount of steel reinforcement. Concrete strength and yield of steel used are given.

 Analysis of a section:

This implies that the dimensions and steel used in the section (in addition to concrete and steel yield strengths) are given, and it is required to calculate the internal ultimate moment capacity of the section so that it can be compared with the applied external ultimate moment.

▫ Singly reinforced section

▫ Doubly reinforced section

Beam Types

▫ Singly Reinforced Beam

Ultimate Stress Design

A singly reinforced beam has only tension reinforcement.

Flexure Equations actual ACI equivalent

stress block stress block

University of Michigan, TCAUP Structures II

Slide 10/26

bd

As

Image Sources: University of Michigan, Department of Architecture

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

b = Asb / bd = 0.85fc’ ab / (fy. d)= β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)]

University of Michigan, TCAUP Structures II

Slide 12/26

Failure Modes

No ReinforcingBrittle failure

Reinforcing < balanceSteel yields before concrete fails ductile failure

Reinforcing = balanceConcrete fails just as steel yields

Reinforcing > balanceConcrete fails before steel yieldsSudden failure

bd

As

yf

200min

yy

cbal ff

f

87000

8700085.0 '1

bal 75.0max

!h!SuddenDeatmax

Source: Polyparadigm (wikipedia)

Rectangular Beam Analysis

Data: Section dimensions – b, h, d, (span) Steel area - As Material properties – f’c, fy

Required: Strength (of beam) Moment - Mn Required (by load) Moment – Mu Load capacitySteps:

1. Find = As/bd

(check min< < max)

2. Find a

3. Find Mn

4. Calculate Mu<= f Mn

5. Determine max. loading (or span)

nu MM

University of Michigan, TCAUP Structures II

Slide 13/26

'' 85.085.0 c

y

c

ys

f

dfor

bf

fAa

2

adfAM ysn

DLu

LL

LLDLu

wl

Mw

lwwM

4.18

7.1

8

)7.14.1(

2

2

Image Sources: University of Michigan, Department of Architecture

Rectangular Beam Analysis

Data: dimensions – b, h, d, (span) Steel area - As Material properties – f’c, fyRequired: Required Moment – Mu

1. Find = As/bd(check min< < max)

University of Michigan, TCAUP Structures II

Slide 14/26

Rectangular Beam Analysis cont.

2. Find a

3. Find Mn

4. Find Mu

University of Michigan, TCAUP Structures II

Slide 15/26

Rectangular Beam Design

Data: Load and Span Material properties – f’c, fy All section dimensions – b and hRequired: Steel area - AsSteps:1. Calculate the dead load and find Mu2. d = h – cover – stirrup – db/2 (one layer)

3. Estimate moment arm jd (or z) 0.9 d and find As

4. Use As to find a5. Use a to find As (repeat…)

6. Choose bars for As and check max & min

7. Check Mu< Mn (final condition)

University of Michigan, TCAUP Structures II

Slide 16/26

8

)7.14.1( 2lwwM LLDL

u

bf

fAa

c

ys

'85.0

2a

df

MA

y

us

2

adfAM ysn

Rectangular Beam Design

Data: Load and Span Material properties – f’c, fyRequired: Steel area - As Beam dimensions – b or dSteps:1. Choose (e.g. 0.5 max or 0.18f’c/fy)2. Estimate the dead load and find Mu3. Calculate bd2

4. Choose b and solve for db is based on form size – try several to find best

5. Estimate h and correct weight and Mu6. Find As= bd7. Choose bars for As and determine spacing and

cover. Recheck h and weight.

University of Michigan, TCAUP Structures II

Slide 17/26

8

)7.14.1( 2lwwM LLDL

u

'2

/59.01 cy

u

ffyf

Mbd

bdAs

Rectangular Beam Design

Data: Load and Span Material properties – f’c, fyRequired: Steel area - As Beam dimensions – b and d

1. Estimate the dead load and find Mu2. Choose (e.g. 0.5 max or 0.18f’c/fy)

University of Michigan, TCAUP Structures II

Slide 18/26

Rectangular Beam Design cont

3. Calculate bd2

4. Choose b and solve for db is based on form size.

try several to find best

University of Michigan, TCAUP Structures II

Slide 19/26

5. Estimate h and correct weight and Mu

6. Find As= bd7. Choose bars for As and

determine spacing and cover. Recheck h and weight.

University of Michigan, TCAUP Structures II

Slide 20/26

Rectangular Beam Design

Source: Jack C McCormac, 1978 Design of Reinforced Concrete, Harper and Row, 1978

Doubly Reinforced Rectangular Sections

21

Beams having steel reinforcement on both the tension and compression sides are called doubly reinforced sections. Doubly reinforced sections are useful in case of limited cross sectional dimensions being unable to provide the required bending strength even when the maximum reinforcement ratio is used

1- Reduced sustained load deflections. • transfer load to compression steel• reduced stress in concrete

2- Ease of fabrication• use corner bars to hold & anchor stirrups

3- Increased Ductility

reduced stress block depth → increase in steel strain larger

curvature are obtained.

4- Change failure mode from compression to tension

Reasons for Providing Compression Reinforcement

23

Doubly Reinforced Beams

Under reinforced Failure( Case 1 ) Compression and tension steel yields( Case 2 ) Only tension steel yields

Over reinforced Failure( Case 3 ) Only compression steel yields( Case 4 ) No yielding Concrete crushes

Four Possible Modes of Failure

Analysis of Doubly Reinforced Rectangular Sections

25

Analysis of Doubly Reinforced Rectangular Sections

26

Analysis of Doubly Reinforced Rectangular Sections

2s sT C

27

c sT C C

003.0s

c

dc

0.85s y c s sA f f ab A f

s 0.003s s s y

c df E E f

c

10.85 0.003s y c s s

c dA f f cb A E

c

200,000 MPasE

29,000 ksisE

Analysis of Doubly Reinforced Rectangular Sections

28

Analysis of Doubly Reinforced Rectangular Sections

Procedure:

10.85 0.003s y c s s

c dA f f cb A E

c

find c

s 0.003s s s y

c df E E f

c

s 0.003 0.005?c d

c

29

Ultimate Stress Design || Advantages

Advantages

▫ Better predicts strength▫ Requires lesser material▫ Easier to compute▫ More rational approach▫ Accounts for uncertainties in

load.

THANK YOU