3. Chapter 3 - Analysis of Section
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Transcript of 3. Chapter 3 - Analysis of Section
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Consider a simply supported beam subjected to gradually increasing load. The load causes the beam to bend and exert a bending moment as shown in figure below.
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Behavior of Beam in Bendingwww.uthm.edu.my
The top surface of the beam is seen to shorten under compression, and the bottom surface lengthens under tension.
As the concrete cannot resist tension, steel reinforcement is introduces at the bottom surface to resist tension.
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For continuous beam, the loads also cause the to bend downward between the support and upward bending over the support.
This will produce tensile zone as shown in figure below. As the concrete cannot resist flexural tension, steel reinforcement would be introduced as detail in the figure.
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Behavior of Beam in Bendingwww.uthm.edu.my
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In the design of reinforced concrete beam the following assumptions are made (See EN 1991: Cl. 6.1 (2) P.) Plane section through the beam before bending remain plane
after bending. The strain in bonded reinforcement, whether in tension or
compression is the same as that in the surrounding concrete. The tensile of the concrete is ignored. The stresses in the concrete and reinforcement can be derived
from the strain by using stress-strain curve for concrete and steel.
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Basic Assumption in RC Designwww.uthm.edu.my
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Basic Assumption in RC Designwww.uthm.edu.my
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Figure below shows the cross section of a RC beam subjected to bending and the resultant strain and stress distribution in the concrete.
Top surface of cross section are subjected to compressive stresses while the bottom surface subjected to tensile stresses.
The line that introduced in between the tensile and compression zones is known as the neutral axis of the member.
Due to the tensile strength of concrete is very low, all the tensile stresses at the bottom fibre are taken by reinforcement.
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Distribution of Stresses and Strainwww.uthm.edu.my
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For fck < 50 N/mm2: η = 1 (defining the effective strength), εc = 0.0035,
αcc = 0.85, λ = 0.8, γc = 1.5,
fcd = 1.0 x 0.85 x fck / 1.5 = 0.567 fck
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Distribution of Stresses and Strainwww.uthm.edu.my
dh
b
x
εcc
εst
fcc
fst fyd
ηfcd
Fst
ηαccfck/γc
Fcc
z
s = λx
(1) (2) (3)
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Stress distribution in the concreteThe triangular stress distribution applies when the stress are very nearly proportional to the strain, which generally occurs at the loading levels encountered under working load conditions and is, therefore, used at the serviceability limit state.The rectangular-parabolic stress block represents the distribution at failure when the compressive strain are within the plastic range, and it is associated with the design for ultimate limit state.The equivalent rectangular stress block is a simplified alternative to the rectangular-parabolic distribution.
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Distribution of Stresses and Strainwww.uthm.edu.my
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The distribution of strains across the beam cross section is linear. That is, the normal strain at any points in a beam section is proportional to its distance from the neutral axis.
The steel strain in tension εst can be determined from the strain diagram as follows:
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Distribution of Stresses and Strainwww.uthm.edu.my
x
xd
xxd ccstccst
)(
cc
st
dx
1Therefore ;
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Since, εcc = 0.0035 for class ≤ C50/60 and
For steel with fyk = 500 N/mm2 and the yield strain is εst = 0.00217.
By substituting εcc and εst ,
Hence, to ensure yielding of the tension steel at limit state the depth of neutral axis, x should be less than or equal to 0.617d.
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Introductionwww.uthm.edu.my
dx 617.0
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As applied moment on the beam section increased beyond the linear elastic stage, the concrete strains and stresses enter the nonlinear stage.
The behavior of the beam in the nonlinear stage depends on the amount of reinforcement provided.
The reinforcing steel can sustain very high tensile strain however, the concrete can accommodate compressive strain much lower compare to it.
So, the final collapse of a normal beam at ultimate limit state is cause by the crushing of concrete in compression, regardless of whether the tension steel has yield or not.
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Type of RC Beam Failurewww.uthm.edu.my
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Depending on the amount of reinforcing steel provided, flexural failure may occur in three ways: Balanced : Concrete crushed and steel yields simultaneously at the ultimate limit
state. The compressive strain of concrete reaches the ultimate strains εcu and the tensile strain of steel reaches the yield strain εy simultaneously. The depth of neutral axis, x = 0.617d.
Under-reinforced : Steel reinforcement yields before concrete crushes. The area of tension steel provided is less than balance section. The depth of neutral axis, x < 0.617d. The failure is gradual, giving ample prior warning of the impending collapse. This mode if failure is preferred in design practice.
Over-reinforced : Concrete fails in compression before steel yields. The area of steel provided is more than area provided in balance section. The depth of neutral axis, x > 0.617d. The failure is sudden (without any sign of warning) and brittle. Over-reinforced are not permitted.
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Type of RC Beam Failurewww.uthm.edu.my
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For a singly reinforced beam EC2 limits the depth to the neutral axis, x to 0.45d (x ≤ 0.45d) for concrete class ≤ C50/60 to ensure that the design is for the under-reinforced case where failure is gradual, as noted above. For further understanding, see the graph shown below .
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Introductionwww.uthm.edu.my
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Section 6.1 EN 1992-1-1, deal with the analysis and design of section for the ultimate limit state design consideration of structural elements subjected to bending.
The two common types of reinforced concrete beam section are: Rectangular section : Singly and doubly reinforced Flanged section : Singly and doubly reinforced
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Analysis of Sectionwww.uthm.edu.my
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Beam cross section, strains and stresses distribution at ULS of singly reinforced rectangular beam
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Singly Reinforced Rectangular Beamwww.uthm.edu.my
dh
b
x
0.0035
εstFst
0.567fck
Fcc
z = d – 0.5s
s = 0.8x
Neutral axis
Notation:h = Overall depth d = Effective depthb = Width of section s = Depth of stress blockAs = Area of tension reinforcement x = Neutral axis depthfck = Characteristic strength of concrete z = Lever armfyk = Characteristic strength of reinforcement
As
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Tension force of steel, Fst
Compression force of concrete, Fcc
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Singly Reinforced Rectangular Beamwww.uthm.edu.my
sky Af 87.0Fst = Stress x Area
Fcc = Stress x Area bxfxbf ckck 454.0)8.0(567.0
For equilibrium, total force in the section should be zero.
stcc FF
sykck Afbxf 87.0454.0
bf
Afx
ck
syk
454.0
87.0
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Moment resistance with respect to the steel
Lets; Therefore;
Moment resistance with respect to the concrete
Area of tension reinforcement,
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Singly Reinforced Rectangular Beamwww.uthm.edu.my
zFMcc
xdbxfM ck 4.0454.0
zFMst
xdAfM sky 4.087.0
Kd
x
d
x
4.0
1454.0
2..4.0454.0
dbfd
xd
d
xck
2... dbfKM ck
)4.0.(.87.0 xdf
MA
yks
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To ensure that the section designed is under-reinforced it is necessary to place a limit on the maximum depth of the neutral axis (x). EC2 suggests:
Then ultimate moment resistance of singly reinforced section or Mbal can be obtained by;
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Singly Reinforced Rectangular Beamwww.uthm.edu.my
22 [email protected] dbfKdbfM ckbalckbal
xdbxfM ckbal 4.0454.0
)]45.0(4.0)].[45.0(..454.0[ dddbfM ckbal
)82.0).(...2043.0( ddbfM ckbal
x ≤ 0.45d
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Therefore;– M = K.fck.b.d2
– Mbal = Kbal.fck.b.d2
where; Kbal = 0.167
If;– M ≤ Mbal or K ≤ Kbal : Singly reinforced rectangular beam
(Tension reinforcement only)– M > Mbal or K > Kbal : Doubly reinforced rectangular beam
(Section requires compression reinforcement)
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Singly Reinforced Rectangular Beamwww.uthm.edu.my
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The cross section of rectangular beam is shown in figure below. Using stress block diagram and the data given, determine the area and the number of reinforcement required.
Data:Design moment, MED = 200 kN.m
fck = 25 N/mm2
fyk = 500 N/mm2
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Example 3.1www.uthm.edu.my
b = 250 mm
d = 450 mm
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Calculate the ultimate moment resistance of section, Mbal
Singly reinforced sectionNeutral axis depth, x
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Solution of Example 3.1www.uthm.edu.my
2...167.0 dbfM ckbal
)450)(250)(25(167.0 2
kNmMkNm 20036.211
xdbxfM ck 4.0454.0
)4.0450)()(250)(25(454.010200 6 xx
045.17621111252 xx
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x = 188 mm @ 937 mm
Use x = 188 mm
Checking;
Lever arm, z = (d – 0.4x)
= (450 – 0.4(188)) = 374.8 mm
Area of reinforcement, As
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Solution of Example 3.1www.uthm.edu.my
45.042.0450
188
d
x
zf
MA
yks 87.0 2
6
1227)8.374)(500(87.0
10200mm
)1257(204Pr 2mmAHovide sprov
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Figure below shows the cross section of a singly reinforced beam. Determine the resistance moment for that cross section with the assistance of a stress block diagram. Given fck= 25 N/mm2 and fyk = 500 N/mm2.
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Example 3.2www.uthm.edu.my
450 mm
250 mm
2H25
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A stress block diagram is drawn with the important values and notations.
For equilibrium;
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Solution of Example 3.2www.uthm.edu.my
d = 450
b = 250
x
Fst=0.87fykAs
0.567fck
Fcc=0.454fckbx
z = d – 0.4x
s = 0.8xNeutral axis
As = 982 mm2
stcc FF
sykck Afbxf 87.0454.0 bf
Afx
ck
syk
454.0
87.0
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Checking;
Moment resistance of section;
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Solution of Example 3.2www.uthm.edu.my
mmx 151)250)(25(454.0
)982)(500(87.0
45.034.0450
151
d
x
zFMcc
xdbxfM ck 4.0454.0 ))151(4.0450)(15125025454.0( M
kNmM 167
zFMst@
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When the load applied increases gradually and it will reach a state that the compressive strength of concrete is not adequate to take additional compressive stress.
Compression reinforcement is required to take the additional compressive stress.
This section is named as doubly reinforced section.
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
dh
b
d’
As
As’
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Strain and stress block diagrams of doubly reinforced beam.
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
dh
b
x
0.0035
εst
Fst
0.567fck
Fcc
z = d – 0.4x
s = 0.8x
Neutral axis
d’εsc
Fsc
As
As’
z1 = d – d’
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Internal force;
Lever arms;
For equilibrium of internal force;
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
bxfF ckcc 454.0
sykst AfF 87.0 '87.0 syksc AfF
xdz 4.0 '1 ddz
scccst FFF '87.0454.087.0 sykcksyk AfbxfAf
and
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Taking moment about the centroid of the tension steel,
For design purpose, x = 0.45d
The area of compression reinforcement, As’
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
1.. zFzFM sccc )').('87.0()4.0).(454.0( ddAfxdbxfM sykck
)').('87.0()]45.0(4.0).[454.0( ddAfddbxfM sykck
)').('87.0(167.0 2 ddAfbdf sykck )').('87.0( ddAfM sykbal
)'(87.0
)('
ddf
MMA
yk
bals
)'(87.0
)('
2
ddf
bdfKKA
yk
ckbals
or
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The area of Tension reinforcement, As
Multiplied equilibrium internal force equation by z ,Limiting x = 0.45d and z = d – 0.4(0.45d) = 0.82d
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
zAfbxzfzAf sykcksyk '87.0454.087.0
zAfddbfzAf sykcksyk '87.0)82.0)(45.0(454.087.0 zAfbdfzAf sykcksyk '87.0167.087.0 2
'87.0
167.0 2
syk
cks A
zf
bdfA '
87.0
2
syk
ckbals A
zf
bdfKA or
![Page 31: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/31.jpg)
Stress in compression reinforcement. The derivation of design formula for doubly reinforced section
assumed that the compression reinforcement reaches the design strength of 0.87fyk at ultimate limit state.
From the strain diagram as shown in figure below.
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
dh
b
x
0.0035
εst
d’εsc
As
As’
xdxsc 0035.0
)'(
0035.0
)'( sc
x
dx
0035.01
' sc
x
d
![Page 32: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/32.jpg)
For the design strength 0.87fyk to be reached, εsc = 0.87fyk / Es
Therefore, if d’/x < 0.38 the compression reinforcement can be assumed reach the design strength of 0.87fyk. If d’/x > 0.38, a reduced stress should be used.
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Doubly Reinforced Rectangular Beamwww.uthm.edu.my
002175.010200
)500(87.087.03
s
yksc E
f
38.00035.0
002175.01
'
x
d
scssc Ef .
)/'1)(0035.0(10200 3 xdfsc
)/'1(700 xd
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The cross section of rectangular beam is shown in figure below. Using the data given, determine the area and the number of reinforcement required.
Data:Design moment, MED = 450 kN.m
fck = 25 N/mm2
fyk = 500 N/mm2
d’ = 50 mm
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Example 3.3www.uthm.edu.my
b = 250 mm
d = 500 mm
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Ultimate moment resistant of section, Mbal
Compression reinforcement is required
Area of compression reinforcement, As’
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Solution of Example 3.3www.uthm.edu.my
2...167.0 dbfM ckbal
kNmMkNm 45094.260 )10)(500)(250)(25(167.0 62
)50500)(500(87.0/10)94.260450( 6
)'(87.0/)(' ddfMMA ykbals
2966mm
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Checking d’/x ratio
Compression steel achieved it design strength at 0.87fyk
Area of tension steel, As
Provide 2H25 (As’ Prov. = 982 mm2) – Compression reinforcement
5H25 (As Prov. = 2454 mm2) – Tension reinforcement
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Solution of Example 3.3www.uthm.edu.my
mmdx 225)500(45.045.0 38.022.0225/50/' xd
966)50082.0(50087.0
1094.260 6
22429mm
'87.0 s
yk
bals A
zf
MA
![Page 36: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/36.jpg)
Calculate moment resistance of the doubly reinforced section shown in figure below. Given fck= 30 N/mm2 and fyk = 500 N/m2.
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Example 3.4www.uthm.edu.my
d = 500 mm
b = 250 mm d’ = 50 mm
3H20
5H25
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A stress block diagram is drawn with the important values and notations
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Solution of Example 3.4www.uthm.edu.my
0.8xx Fcc = 0.454fckbx
Fst = 0.87fykAs
Z
Neutral Axis
b = 250 mm
Fsc = 0.87fykAs’
Z1
d’ = 50 mm
3H20
5H25
![Page 38: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/38.jpg)
Reinforcement used 3H20, As’ = 943 mm2 & 5H25, As = 2455 mm2
Neutral axis depth, x
Checking the stress of steel
Steel achieved it design strength 0.87fy as assumed
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Solution of Example 3.4www.uthm.edu.my
)250)(30(454.0
)9432455)(500(87.0
454.0
)'(87.0
bf
AAfx
ck
ssyk
mmx 193
38.026.0193/50/' xd
45.039.0500/193/ dx
![Page 39: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/39.jpg)
Moment resistance of section, M
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Solution of Example 3.4www.uthm.edu.my
zFzFM ccsc .. 1
)4.0(454.0)'('87.0 xdbxfddAf cksyk
)50500)(943)(500(87.0 610))193(4.0500)(193)(250)(30(454.0
kNm462
![Page 40: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/40.jpg)
Flanged beams occur when beams are cast integrally with and support a continuous floor slab.
Part of the slab adjacent to the beam is counted as acting in compression to form T- and L-beams as shown in figure below.
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Flange Beamwww.uthm.edu.my
Where; beff = effective flange widthbw = breadth of the web of the beam.hf = thickness of the flange.
beff beff
bw bw
hf
h
T-Beam L-Beam
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The effective width of flange, beff is given in Sec. 5.3.2.1 of EC2.
beff should be based on the distance lo between points of zero moment as shown in figure below.
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Flange Beamwww.uthm.edu.my
![Page 42: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/42.jpg)
The effective flange width, beff for T-beam or L-beam may be derived as:
Where and
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Flange Beamwww.uthm.edu.my
bbbb wieffeff ,
ooiieff llbb 2.01.02.0, iieff bb ,
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Based on figure below, determine the effective flange width, beff of beam B/1-3.
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Example 3.5www.uthm.edu.my
3000 4500
2500
4000
1 2 3
A
B
C
200 x 500
200 x 500 200 x 500
200 x 500 200 x 500
200
x 50
020
0 x
500
200
x 50
020
0 x
500
FS1 (150 thk.)
FS2 (150 thk.) FS3 (150 thk.)
![Page 44: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/44.jpg)
D
lo (distance between points of zero moment)
Effective flange width, beff
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Solution of Example 3.5www.uthm.edu.my
3000 mm 4500 mm
lo = 0.85 x 3000 = 2550 mm
lo = 0.85 x 4500 = 3825 mm
lo = 0.15 x (3000 + 4500) = 1125 mm
1 2 3
bbbb wieffeff ,
![Page 45: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/45.jpg)
Span 1-2
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Solution of Example 3.5www.uthm.edu.my
2500 4000
b1 = 2500/2 = 1250 b2 = 4000/2 = 2000
beff,1 beff,2
bw = 200 mm
beff
b = 1250 + 2000 = 3250 mm
beff1 = 0.2(1250) + 0.1(2550) = 505 mm < 0.2lo = 510 mm < b1 = 1250 mm
beff2 = 0.2(2000) + 0.1(2550) = 655 mm > 0.2lo = 510 mm < b2 = 2000 mm
beff = (505 + 510) + 200 = 1215 mm < 3250 mm
A B C
![Page 46: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/46.jpg)
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Solution of Example 3.5www.uthm.edu.my
Span 2-3
2500 4000
b1 = 2500/2 = 1250 b2 = 4000/2 = 2000
beff,1 beff,2
bw = 200 mm
beff
b = 1250 + 2000 = 3250 mm
beff1 = 0.2(1250) + 0.1(3825) = 632.5 mm < 0.2lo = 765 mm < b1 = 1250 mm
beff2 = 0.2(2000) + 0.1(3825) = 782.5 mm > 0.2lo = 765 mm < b2 = 2000 mm
beff = (632.5 + 765) + 200 = 1597.5 mm < 3250 mm
A B C
![Page 47: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/47.jpg)
Span 1-2 Span 2-3
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Solution of Example 3.5www.uthm.edu.my
beff = 1215mm
bw = 200 mm
hf = 150 mm
beff = 1598 mm
bw = 200 mm
hf = 150 mm
![Page 48: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/48.jpg)
The design procedure of flange beam depends on where the neutral axis lies. The neutral axis may lie in the flange or in the web.
In other word, there are three cases that should be considered. Neutral axis lies in flange (M < Mf)
Neutral axis lies in web (M > Mf but < Mbal)
Neutral axis lies in web (M > Mbal)
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Flange Beamwww.uthm.edu.my
beff beff
![Page 49: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/49.jpg)
Neutral axis lies in flange (M < Mf)
This condition occur when the depth of stress block (0.8 x) less then the thickness of flange, hf as shown in figure below.
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Flange Beamwww.uthm.edu.my
beff
bw
hf
d
0.8x Fcc
Fst
Z = d – 0.4x
x
As
0.567fck
![Page 50: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/50.jpg)
Moment resistance of section, M
For this case, maximum depth of stress block, 0.8x are equal to hf
Where, Mf = Ultimate moment resistance of flange.
Therefore, if M ≤ Mf the neutral axis lies in flange and the design can be treated as rectangular singly reinforced beam.
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Flange Beamwww.uthm.edu.my
zFMcc
xdxbfM effcu 4.08.0567.0
2/567.0 ffckf hdbhfMM
)4.0(87.0 xdf
MA
yks
zf
MA
yks 87.0 or
![Page 51: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/51.jpg)
The T-beam with dimension as shown in figure below is subjected to design moment, M = 250 kNm. If fck = 30 N/mm2 and fyk = 500 N/mm2 have been used, determine the area and number of reinforcement required.
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Example 3.6www.uthm.edu.my
beff = 1450 mm
bw = 250 mm
hf = 100 mm
d = 320 mm
As
![Page 52: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/52.jpg)
Moment resistance of flange, Mf
Since M < Mf, Neutral axis lies in flange
Compression reinforcement is not required
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Solution of Example 3.6www.uthm.edu.my
2/567.0 ffeffckf hdhbfM
kNmMkNm 2509.665 6102/100320100145030567.0
mmmmx [email protected]
xdbxfM ck 4.0454.0
)4.0320)()(1450)(30(454.010250 6 xx 02.316478002 xx
Use x = 41.74mm
![Page 53: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/53.jpg)
Checking
Lever arm,z
Area of tension reinforcement, As
Provide 4H25 (Asprov = 1964 mm2)
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Flange Beamwww.uthm.edu.my
45.013.0320
74.41
d
x
mmxdz 3.303)74.41(45.03204.0
21895mm
)3.303)(500(87.0
10250
87.0
6
zf
MA
yks
![Page 54: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/54.jpg)
Neutral axis lies in web (Mf < M < Mbal)
If the applied moment M is greater than Mf the neutral axis lies in the web as shown in figure below.
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Flange Beamwww.uthm.edu.my
beff
bw
hf
d0.8x Fcc1
Fst
z1
x
As
0.567fck
122 Fcc2
z2
![Page 55: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/55.jpg)
From the stress block, internal forces;
Lever arms, z
Moment resistance, M
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Flange Beamwww.uthm.edu.my
xbfxbfF wckwckcc 454.0)8.0.)(567.0(1
fweffckcc hbbfF ))(567.0(2
sykst AfF 87.0
xdz 4.01 fhdz 5.0
2
2211.. zFzFM
cccc
)5.0())(567.0(4.0454.0 ffweffckwck hdhbbfxdxbf
![Page 56: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/56.jpg)
Ultimate moment resistance of section, Mbal (When x = 0.45d)
Divide both side by fckbeffd2, then;
Therefore;
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Flange Beamwww.uthm.edu.my
)45.0(4.045.0454.0 dddbfM wckbal
)5.0())(567.0( ffweffck hdhbbf
d
h
b
b
d
h
b
b
dbf
M f
eff
wf
eff
w
effck
bal
211567.0167.0
2
feffck
bal
dbf
M 2
2dbfM effckfbal
![Page 57: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/57.jpg)
If applied moment M < Mbal, then compression reinforcement are not required. Area of tension reinforcement can be calculate as follows by taking moment at Fcc2.
Using; x = 0.45d
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Flange Beamwww.uthm.edu.my
)5.0(87.0
]5.04.0[454.0
fyk
fwkcs hdf
hxxbfMA
)]4.0()5.0[(454.0)5.0(87.0 xdhdxbfhdAf fwckfsyk
).(. 1212 zzFzFM ccst
)5.0(87.0
]36.0[1.0
fyk
fwkcs hdf
hddbfMA
![Page 58: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/58.jpg)
The T-beam with dimension as shown in figure below is subjected to design moment, M = 670 kNm. If fck = 30 N/mm2 and fyk = 500 N/mm2 have been used, determine the area and number of reinforcement required.
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Example 3.7www.uthm.edu.my
beff = 1450 mm
bw = 250 mm
hf = 100 mm
d = 320 mm
![Page 59: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/59.jpg)
Moment resistance of flange, Mf
Since M > Mf, Neutral axis lies in web
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Solution of Example 3.7www.uthm.edu.my
2/567.0 ffeffckf hdhbfM
kNmMkNm 6709.665 6102/100320100145030567.0
153.0)320(2
1001
1450
2501
320
100567.0
1450
250167.0
f
2dbfM effckfbal
kNmM bal 68210)320)(1450)(30(153.0 62
kNmMkNmM bal 670682
![Page 60: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/60.jpg)
Compression reinforcement is not required
Area of tension reinforcement, As
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Solution of Example 3.7www.uthm.edu.my
)50320)(500(87.0
]100)320(36.0)[320)(250)(30(1.010670 6
)5.0(87.0
]36.0[1.0
fyk
fwkcs hdf
hddbfMA
25736mm
Provide 8H32 (Asprov = 6433 mm2)
![Page 61: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/61.jpg)
Neutral axis lies in web (M > Mbal)
If the applied moment M is greater than Mbal the neutral axis lies in the web and the compression reinforcement should be provided. The stress block are shown in figure below.
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Flange Beamwww.uthm.edu.my
d
beff
bw
hf 0.8x Fcc1
Fst
z1
x
As
0.567fck
1
22 Fcc2
z2
As’ Fsc
z3
![Page 62: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/62.jpg)
From the stress block, internal forces;
Lever arms, z
Moment resistance, M
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Flange Beamwww.uthm.edu.my
xbfxbfF wckwckcc 454.0)8.0.)(567.0(1
fweffckcc hbbfF ))(567.0(2
sykst AfF 87.0
'87.0 syksc AfF
xdz 4.01 f
hdz 5.02
'3
ddz
32211... zFzFzFM
sccccc
![Page 63: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/63.jpg)
When x = 0.45d, then
Area of compression reinforcement, As’
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Flange Beamwww.uthm.edu.my
)5.0())(567.0(4.0454.0 ffweffcuwcu hdhbbfxdxbfM )'('87.0 ddAf syk
)'('87.0 ddAfMM sykbal
)'(87.0'
ddf
MMA
yk
bals
![Page 64: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/64.jpg)
For equilibrium of forces
Area of tension reinforcement, As
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Flange Beamwww.uthm.edu.my
scccccstFFFF
21
'87.0)(567.0)45.0(454.087.0 sykweffckwcksyk AfbbfdbfAf
'87.0
)(567.02.0s
yk
wefffckwcks A
f
bbhfdbfA
![Page 65: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/65.jpg)
Design Procedures for Rectangular SectionSupposed the design bending moment is M, beam section is b x d,
concrete strength is fck and steel strength is fyk, to determine the area of reinforcement, proceed as follows.
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Design Formulawww.uthm.edu.my
![Page 66: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/66.jpg)
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Design Formulawww.uthm.edu.my
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With Wisdom We Explore
Design Formulawww.uthm.edu.my
![Page 68: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/68.jpg)
Design Procedures for Flange SectionSupposed the design bending moment is M, beam section is b x d,
concrete strength is fck and steel strength is fyk, to determine the area of reinforcement, proceed as follows.
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Design Formulawww.uthm.edu.my
![Page 69: 3. Chapter 3 - Analysis of Section](https://reader033.fdocuments.us/reader033/viewer/2022061515/55cf9ace550346d033a37bb0/html5/thumbnails/69.jpg)
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Design Formulawww.uthm.edu.my
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Design Formulawww.uthm.edu.my
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www.uthm.edu.my