l3 analysis and design of rect-section 32s-v2
DESCRIPTION
rc designTRANSCRIPT
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Analysis and Design of Rectangular Beam Sections
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Key differences with EC2
2013 Design code EC2 (National Annex) C50
Concrete Design based on:
Cube strength Cylinder Strength
Denoted by: fcu fckDesign strength 0.87fy, 0.45fcu 0.567fckStress block depth 0.9x 0.8x
redistribution dd
kk
xbal 45.0)(
2
1
Limiting x/d 0.5 0.45Limiting z/d 0.775 0.82
Limiting K or Kbal 0.156 0.167Limiting d/d(0% redistribution)
0.19 0.171
Moment x
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Analysis and Design of Rectangular Beam Sections
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Composite ActionThe strains at concrete and steel at the same level are the same !!
Effect of shrinkage and thermal movement on concrete causes cracking in concrete but strengthens the bond between concrete and steel bar
Effect of creep on concrete a change in the compressive strains transferred to reinforcing steel
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Analysis and Design of Rectangular Beam Sections
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Concrete stress-strain relationship
back
Design stress = 0.67fcu/1.5 = 0.447fcu 0.45fcu
(for fcu
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Analysis and Design of Rectangular Beam Sections
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Steel stress-strain relationship
back
For fy = 500 N/mm2
yy
sm
fE
m=1.15
Mild steel fy = 250 N/mm2High Yield steel fy = 500 N/mm2
Short-term design stress-strain curve for reinforcementFigure 4.2:
y k
m
f
S
t
r
e
s
s
N
/
m
m
2
Strain
Tension andcompression
200kN/mm2
y 002170y .
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Analysis and Design of Rectangular Beam Sections
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Distribution of strain and stress under bending
axis
triangular rectangularparabolic
equivalentrectangular
(a) (b) (c)
Section Strains Stress Blocks
st
cc
sc
Section with strain diagrams and stress blocks
As
A' s
Figure 4.3:
neutralx
d'
d
s = 0.8xs=0.9x
effective depth
C
T
(d)
force
l
e
v
e
r
a
r
m
back
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Analysis and Design of Rectangular Beam Sections
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Assumptions in the Reinforced Concrete Theory Concrete does not take any tension. (1~2 N/mm2) The stresses in concrete in compression follows the simplified
parabolic-rectangular stress-strain curve in (Concrete2013).
The ultimate limit of collapse in concrete is reached when the strain at the extreme compression fibre reaches 0.0035
The steel stresses follows the simplified short-term design stress-strain curve in (Concrete 2013).
A perfect bond exists between the steel bar and concrete. Where a section is designed for flexure only, the lever arm should
NOT be assumed to be greater than 0.95 times the effective depth.
Plane sections before bending remain plane after bending, i.e. the strain distribution of concrete in compression and steel in tension or compression are directly proportional to their distance from theneutral axis at which the strain is zero.
To
To
To
To
To
)( 2cu 60N/mmf
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Analysis and Design of Rectangular Beam Sections
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21
and
cu
stcust
dxxxd
0.5x d
st
cu2
scAs
A' s
x
d'
d
(
d
-
x
)
2
2
for 0.0035 (concrete crushes)and 0.00217 (steel yields)
( 500 / )
cu
st y
yf N mm
To ensure yielding of the tension steel 0.6170.0021710.0035
dx d
The tension steel yields when
RC Beam Section under bending moment
Concrete2013 however places limit at ----
--- Balanced Failure ---
back
Safe ???
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Analysis and Design of Rectangular Beam Sections
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st
cu2
scAs
A' s
x
d'
d
(
d
-
x
)
RC Beam Section under bending moment
Take high tensile bar for example
If x=0.617d, st= 0.002. Steel yields and concrete crushes at the same time
If x 0.002. Steel SURELY yields BEFORE concrete crushes
If x>0.617d, st
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Analysis and Design of Rectangular Beam Sections
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Equivalent Bending stress block for Design
axis
Section Strains Stress Block
st
Singly reinforced section with rectangular stress block
0.0035 0.85f / = 0.567f ckck c
Fst
Fcca
s=0.8x
Figure 4.4:
As
b
d
s/2
z = l d
xneutral
0.67fcu/m=0.45fcu
s=0.9x
< 0.95ddepth of stress block
They form a couple !!
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Analysis and Design of Rectangular Beam Sections
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and stress area 0.45 cuF f bscc (4.5) For equilibrium: zFzFM stcc
substitute 2/sdz (4.6) to give: 0 45 0.9 ( )cu cuM . f bs z f b d z z (4.7)
to give 2( / ) ( / ) / 0.9 0z d z d K Re-arrange the above and define
2/ cuK M bd f
We have ( / ) ( /1.15) 0.87st y m s y s y sF f A f A f A Solve for z/d to have [0.5 (0.25 /0.9)]z d K (4.8)
Substitute in (4.5) 0.87s yMAf z
(4.9)
K factor
Area of steel required
Input: M, b,d, fcu,fy
Output: As
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Analysis and Design of Rectangular Beam Sections
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K=0.156 corresponds to x=0.5d and z=d-s/2=0.775d, with Mu=0.156fcubd2=Kfcubd2
K
Mu is the max. moment provided by the concrete section limited byx=0.5d or it is the moment capacity of the concrete section
compression reinforcement required to provide BM larger than Mu
Obtain this curve by plotting Eq. (4.8)
l
e
v
e
r
a
r
m
f
a
c
t
o
r
upper limit or range
lower limit or range
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Analysis and Design of Rectangular Beam Sections
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The balanced sectionThe code specifies xbal = 0.5 d
Depth of the stress block, s=0.9 xbal = 0.45 d
Force in the concrete stress block,
Fccbal = 0.45 fcu bs = 0.2025 fcubdFor equilibrium of forces, Fstbal = 0.87 fy Asbal = Fccbal = 0.2025 fcubd
Asbal = 0.2328 fcubd/fywhich is the steel area in a balanced section
The ultimate moment of resistance of the balanced section,
Mbal = Fccbal zbal = 0.2025 fcubd (d - s/2) = 0.156 fcubd2
When the design moment Md = Mbal,bal2
cu
d K1560bdf
M .)( 2cu N/mm45f
)( 2cu N/mm45f
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Analysis and Design of Rectangular Beam Sections
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Singly reinforced beam design: Ex 4.1 - Determine As for given M
62185 10 1200
0.87 0.87 500 369s y
MA mmf z
As
b= 260
d
=
4
4
0
kNmM
fyfcu185
2
Lever Arm:
{0.5 0.25 }0.9
0.122440{0.5 0.25 } 3690.9
Kz d
mm
2
6
2
185 10 0.122 0.156260 440 30
Compression steel not required
cu
MKbd f
Checking the sufficiency of section !
Step 1:
Step 2:
Step 3:
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Analysis and Design of Rectangular Beam Sections
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Equilibrium : 0.45 0.87
0.45 25 300 0.87 500 1470 189.5 and / 0.9 189.5 / 0.9 210.5 0.5Hence steel has yielded
cc st
cu y s
F Ff bs f A
ss mm x s mm d
0.567f
ck
Fst
Fcc
s
b = 300
d
=
5
2
0
axisneutral
sA = 1470 mm
x
z
2
Singly reinforced beam design: Ex 4.2 Find the resistant moment of section given b, d and As
0.45fcu
fyfcu
To
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Analysis and Design of Rectangular Beam Sections
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6
Moment of Resistance of the section:0.87 ( / 2) 0.87 500 1470(520 189.5 / 2) 10271.9
st y sM F z f A d skNm
0.567f
ck
Fst
Fcc
s
b = 300d
=
5
2
0
axisneutral
sA = 1470 mm
x
z
2
0.45fcu
The moment of resistance can also be calculated as M = Fcc z
)( 2cu N/mm45f
Note that
x = (d z)/0.45
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Analysis and Design of Rectangular Beam Sections
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Rectangular section with compression reinforcement
Cross-section with compression reinforcementback
Mu
= Fst1+Fst2
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Analysis and Design of Rectangular Beam Sections
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Yielding of Asc will be checked separately !
For a singly reinforced section : 20.156u cuM f bd = K fcubd2
Even though the design (applied) moment exceeds this value, we still require x0.5d to ensure steel yielding and a ductile failure Therefore: For equilibrium: scccst FFF So with the compression reinforcement at yield:
'0.87 0.45 0.87y s cu y sf A f bs f A
and with 0.9 0.5 0.45s d d
'0.87 0.2025 0.87y s cu y sf A f bd f A
/ 2 0.9 / 20.9 0.5 / 2 0.775
z d s d xd d d
The lower limit of range for z
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Analysis and Design of Rectangular Beam Sections
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and taking moments about the centroid of the tension steel:
'
2 '
( ')0.2025 0.775 0.87 ( ')
0.156 0.87 ( ')
cc sc
cu y s
cu y s
M F z F d df bd d f A d d
f bd f A d d
Re-arrange to give the compression steel as:
2'
'
0.156 0.87 ( ')
cus
y s
M f bdAf A d d
2
' 0.1560.87 ( ')
cus
y
M f bdAf d d
(4.14)
(4.15)Mu
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Analysis and Design of Rectangular Beam Sections
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Multiply both sides of Eq. (4.13) by z=0.775d:
and with K =0.156 and K=M/bd2fcu:
20.156 0.87
'cus s
y
f bdA Af z
2' ( ')
0.87 ( ')cu
sy
K K f bdAf d d
2'
0.87'cu
s sy
K f bdA Af z
(4.16)
(4.17)
(4.18)
To
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Analysis and Design of Rectangular Beam Sections
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For 2500 /yf N mm and 0.002175sc y the
compression steel will have yielded only if:
2cuf 60N/mm' 0.0021751 0.38, when
0.0035 0.5dx x d
The Design code Concrete2013 specifies
d/x 0.38 to ensure the compression steel to yield before section fails. (The criteria can also be stated as d/d 0.19 )If d/x > 0.38, refer to Figure 3.9 in the Code for the compressive stress.
(4.20)
Check on yielding of compression steel -
,
d1
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Analysis and Design of Rectangular Beam Sections
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Doubly reinforced beam design: Ex 4.4 Find moment of resistance of section
fcu=25 fy=500
M=? kNm
= Fst1+Fst2
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Analysis and Design of Rectangular Beam Sections
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'
'
Equilibrium : 0.87 0.45 0.87
0.87 ( )
0.450.87 500(2410 628) 246.1
0.45 25 280with 246.1 , / 0.9 246.1/ 0.9 273.4 0.536 0.617
Hence tension steel has yi
st cc sc
y s cu y s
y s s
cu
F F Ff A f bs f A
f A As
f b
mm
s mm x s mm d d
elded
Also '/ 50 / 273.4 0.183 0.38Hence compression steel has yielded as assumed.
d x in real situation
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Analysis and Design of Rectangular Beam Sections
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'
6
Taking moments of compressive forces about the tension steel( / 2) ( ')
0.45 ( / 2) 0.87 ( ')0.45 25 280 246.1(510 246.1/ 2) 0.87 500 628(510 50)425.63 10 425.63
cc sc
cu y s
M F d s F d df bs d s f A d d
Nm kNm
Section Stress Block
0.567fck
Fst
Fscs=0.8x
A = 2410s
b = 280
d
=
5
1
0
FccA' = 628
s
d' = 50
0.45fcu
s=0.9x
= Fst1+Fst2
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Analysis and Design of Rectangular Beam Sections
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Doubly reinforced beam design: Ex 4.3 - Find As, As
As
b = 260
A's
d
=
4
4
0
d' = 50
25500
285
cu
y
ff
M kNm
2 2
2' 0.156 25 260 440 519.3 1842.70.87 0.87 500(0.775 440)
'cus s
y
K f bdA A mmf z
2'
22
( ')0.87 ( ')
(0.226 0.156)25 260 440 519.3
0.87 500 (440 50)
cus
y
K K f bdA
f d d
mm
K6
2 2
285 10 0.226 0.156260 440 25
Compression steel is required50 440 0 11 0 19
Compression steel has yielded
cu
MKbd f
d'/d / . .
Check the sufficiency of section !Step 1:
Step 2:
Step 3:
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Analysis and Design of Rectangular Beam Sections
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Design of tension and compression reinforcement - Ex. 7.3
The beam section shown below has characteristic material strengths of fcu = 25N/mm2 and fy = 500N/mm2 The ultimate moment is 165kN m. causing hogging of the beam. Check on sufficiency of concrete section
6
2 2
165 10 0.26 ' 0.156230 330 25
cuMK K
bd f
This means compression steel is required.
Step 1:
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Analysis and Design of Rectangular Beam Sections
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Design of tension and compression reinforcement
x=0.5d=165mm; d/x=50/165=0.303 < 0.38 (o.k.)
Therefore:
Compression steel:
Tension steel:
Provide two T20 bars for As=628mm2 and four T25 bars for As = 1960mm2.
0.87sc yf f
2 6 2'
2
( 0.156 ) (165 10 0.156 25 230 330 )0.87 ( ') 0.87 500 (330 50)
552.6
cus
y
M f bdAf d d
mm
2 2'
2
0.156 0.156 25 230 330 552.60.87 0.87 500 0.775 330
878.0 552.6 1430.6
cus s
y
f bdA Af z
mm
To
Step 2:
Step 3:
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Analysis and Design of Rectangular Beam Sections
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Evolution of Concrete Compressive Block with Moment of Resistancein Design Rectangular Beam Section
increasing applied bending moment on section
x0.5d x
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Analysis and Design of Rectangular Beam Sections
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Design Chart depends on the moment redistribution factor
over-reinforcedunder-reinforced
Margin not used in Code
Balanced point
back
For
For illustration ONLY !!
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Analysis and Design of Rectangular Beam Sections
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Example 1B=300, d=450, M=262.5 kNm
Solution
2
2
6
2
51849
371100chartfrom
324450300
105262
mm.A
.bdA,
..bdM
s
s
)1960(254Provide 2mmAT s To
For illustration ONLY !!
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Analysis and Design of Rectangular Beam Sections
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Example 2B=300, d=450, M=320 kNm
Solution
)2450(255
2093
551100chartfrom
,50100ofcurvenexttheUse
27545030010320
2
2
2
6
2
mmTmmA
.bdA,
.'/bdA
.bdM
s
s
s
)943(20367550100 22 mmT,mmA,.bdA '
'
s
s To
No Good !
x/d > 0.5
For illustration ONLY !!
~~~ End ~~~
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Analysis and Design of Rectangular Beam Sections
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To
Back
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Analysis and Design of Rectangular Beam Sections
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Distribution of strain and stress
axis
triangular rectangularparabolic
equivalentrectangular
(a) (b) (c)
Section Strains Stress Blocks
st
cc
sc
Section with strain diagrams and stress blocks
As
A' s
Figure 4.3:
neutralx
d'
d
s = 0.8x
Back
s=0.9x
< 0.95d