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Name: Sadia Mahajabin ID : 10.01.03.098 4th year 2nd Semester Section : B Department of Civil Engineering Ahsanullah University of Science and Technology
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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