Flexural design of Beam...PRC-I
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Plain & Reinforced Concrete-1
By Engr. Rafia Firdous
Flexural Analysis and Design of Beams
(Ultimate Strength Design of Beams)
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Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams(Strength Design of Beams)
Strength design method is based on the philosophy of dividing F.O.S. in such a way that Bigger part is applied on loads and smaller part is applied on material strength.
fc’
0.85fc’
Stress
Strain
Crushing Strength
0.003
favg
favg = Area under curve/0.003
If fc’ ≤ 30 MPa
favg = 0.72 fc’
β1 = Average Strength/Crushing Strength
β1 = 0.72fc’ / 0.85 fc’ = 0.85 2
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Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Cc
T = Asfs
la = d – a/2
N.A.
εcu=
0.003
Strain Diagram
Actual Stress
Diagram
Internal Force Diagram
In ultimate strength design method the section is always taken as cracked.
c = Depth of N.A from the extreme compression face at ultimate stage
a = Depth of equivalent rectangular stress diagram.
εs
hc
d
b 0.85fc
fs
0.85fca
Equivalent Stress
Diagram/ Whitney’s
Stress Diagram
a/2
fs
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Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Actual Stress
Diagram
0.85fc 0.85fc
Equivalent Stress Diagram/
Whitney’s Stress Diagram
cCc Cca
• The resultant of concrete compressive force Cc, acts at the centriod of parabolic stress diagram.
• Equivalent stress diagram is made in such a way that it has the same area as that of actual stress diagram. Thus the Cc, will remain unchanged.
a/2
ab'0.85fcbf cav
a'0.85fc'0.72f cc
c'0.85f
'0.72fa
c
c
cβa 1 4
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Plain & Reinforced Concrete-1
Ultimate Strength Design of Beams (contd…)
Factor β1
β1 = 0.85 for fc’ ≤ 28 MPa
Value of β1 decreases by 0.05 for every 7 MPa increase in strength with a minimum of 0.65
0.65 '0.00714f1.064β c1 85.0
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Plain & Reinforced Concrete-1
Determination of N.A. Location at Ultimate ConditionCASE-I: Tension Steel is Yielding at Ultimate Condition
ys εε orys ff
CASE-II: Tension Steel is Not Yielding at Ultimate Condition
yf
yεsε
ysε orys ff
0.0015200,000
300
E
fε y
y 0.0021200,000
420
E
fε y
y
For 300 grade steel
For 420 grade steel
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Plain & Reinforced Concrete-1
CASE-I: Tension Steel is Yielding at Ultimate Condition
ysss fAfAT ab'0.85fC cc
2
ada l
abffA cys '85.0
For longitudinal Equilibrium
T = Cc
bf
fAa
c
ys
'85.0
1β
ac and
Cc
T = Asfs
Internal Force Diagram
a/2
la
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Plain & Reinforced Concrete-1
CASE-I: Tension Steel is Yielding at Ultimate Condition (contd…)
Nominal Moment Capacity, Mn depending on steel = T x la
2Mn
adfA ys
Design Moment Capacity
2M bnb
adfA ys
Nominal Moment Capacity, Mn depending on Concrete = Cc x la
2
adab0.85fc'Mn
2
adab0.85fc'M bnb
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Plain & Reinforced Concrete-1
Minimum Depth for Deflection Control
I
1 α Δ
3(Depth)
1 α Δ
For UDL4ωL α Δ
3LωL α ΔDeflection Depends upon Span, end conditions, Loads and fy of steel. For high strength steel deflection is more and more depth is required. 9
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Plain & Reinforced Concrete-1
Minimum Depth for Deflection Control (Contd…)
ACI 318, Table 9.5(a)
Steel Grade
Simply Supported
One End Continuous
Both End Continuous
Cantilever
300 L/20 L/23 L/26 L/10
420 L/16 L/18.5 L/21 L/8
520 L/14 L/16 L/18 L/7
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Plain & Reinforced Concrete-1
Balanced Steel Ratio, ρb It is corresponding to that amount of steel which will cause yielding of steel at the same time when concrete crushes.At ultimate stage:
ys εε ys ff and
0.003εcu From the internal Force diagram
bc ab'0.85fC c
ab = depth of equivalent rectangular stress block when balanced steel ratio is used.
εcu=
0.003
Strain Diagram
εy
cb
d- cb
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Plain & Reinforced Concrete-1
Balanced Steel Ratio, ρb (contd…)
ybys fd)b(ρfAT
For the longitudinal equilibrium
cCT
bcyb ab'0.85ffd)b(ρ
d
a
f
'f0.85ρ b
y
cb (1)
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Plain & Reinforced Concrete-1
Balanced Steel Ratio, ρb (contd…) From the strain diagram εcu=
0.003
Strain Diagram
εy
cb
d- cb
B C
A
E D
ADE & ABC Δs
b
y
b cd
ε
c
0.003
yb ε0.003
0.003dc
s
s
s
yb E
Ed
Ef
0.003
0.003c
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Plain & Reinforced Concrete-1
Balanced Steel Ratio, ρb (contd…)
df600
600c
yb
df0.003E
0.003Ec
ys
sb
b1b Cβa As we know
df600
600βa
y1b
(2
)Put (2) in (1)
y1
y
cb f600
600β
f
'f0.85ρ
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Plain & Reinforced Concrete-1
Types of Cross Sections w.r.t. Flexure at Ultimate Load Level1. Tension Controlled SectionA section in which the net tensile strain in the extreme tension steel is greater than or equal to 0.005 when the corresponding concrete strain at the compression face is 0.003.
εcu=
0.003
Strain Diagram
εs=>0.005
c
d- c
cd
0.005
c
0.003
d0.008
0.003c
d8
3c d
8
3βa 1and
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Plain & Reinforced Concrete-1
Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…)
2. Transition Section The section in which net tensile strain in the extreme tension steel is greater than εy but less than 0.005 when
corresponding concrete strain is 0.003. εcu=
0.003
Strain Diagram
εy<εs<0.005
c
d- cd8
3βa 1 baa
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Plain & Reinforced Concrete-1
Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…)
2. Transition Section (contd…)
εcu=
0.003
Strain Diagram
0.004
c
d- c
To ensure under-reinforced behavior, ACI code establishes a minimum net tensile strain of 0.004 at the ultimate stage.
cd
0.004
c
0.003
d7
3c d
7
3βa 1
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Plain & Reinforced Concrete-1
Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…)
Both the “Tension Controlled Section” and “Transition Section” are “Under-Reinforced Section”
In Under-Reinforced Sections steel starts yielding before the crushing of concrete and:
ρ < ρ b
It is always desirable that the section is under-reinforced otherwise the failure will initiate by the crushing of concrete. As concrete is a brittle material so this type of failure will be sudden which is NOT DESIREABLE.
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Plain & Reinforced Concrete-1
Types of Cross Sections w.r.t. Flexure at Ultimate Load Level (contd…)
3. Compression Controlled Section (over-reinforced section)
The section in which net steel strain in the extreme tension steel is lesser than εy when corresponding concrete strain is 0.003.
Capacity of steel remain unutilized. It gives brittle failure without warning.
baa bCC
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Plain & Reinforced Concrete-1
Strength Reduction Factor (Resistance Factor), Φ It is reciprocal of minor part of overall factor of safety that is applied on strength of member to obtain its design strength.
Tension Controlled Section, Φ = 0.9 Compression Controlled Section
Member with lateral ties, Φ = 0.65
Members with spiral reinforcement, Φ = 0.75
Transition Section
For transition section Φ is permitted to be linearly interpolated between 0.65 or 0.75 to 0.9. 20
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Plain & Reinforced Concrete-1
Strength Reduction Factor (Resistance Factor), Φ Transition Section (contd…)
yty
εεε0.005
0.250.65Φ
For members with ties
For members with Spirals
yty
εεε0.005
0.150.75Φ
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Plain & Reinforced Concrete-1
Maximum Steel Ratio, ρmax
For T = C
ab'0.85ffA cys
ab'0.85ffbdρ cy
d
a
f
'f0.85ρ
y
c
For tension controlled section d8
3βa 0.005 ε 1s
So
1
y
cmax β
8
3
f
'f0.85ρ
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Plain & Reinforced Concrete-1
Maximum Steel Ratio, ρmax (contd…)
For transition section
d7
3βa 0.004 ε 1s
1
y
cmax β
7
3
f
'f0.85ρ
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Plain & Reinforced Concrete-1
Minimum Reinforcement of Flexural Members(ACI – 10.5.1)
The minimum steel is always provided in structural members because when concrete is cracked then all load comes on steel, so there should be a minimum amount of steel to resist that load to avoid sudden failure.
For slabs this formula gives a margin of 1.1 to 1.5. This formula is not used for slabs.
yy
cmin f
1.4
4f
'fρ
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Plain & Reinforced Concrete-1
Under-Reinforced Failure
Cc
T Internal Force Diagram
a/2
la
Stage-II, Cracked Section
When section cracks, N.A. moves towards compression face means “la” increases. “T” and “Cc” also increase.
Stage-I, Un-cracked SectionN.A. position is fixed, means “la” remains constant. Only “T” and “Cc” increase with the increase of load
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Plain & Reinforced Concrete-1
Under-Reinforced Failure (contd…)
Stage-III, Yielding in Steel Occur
T = Asfy remains constant and Cc also remains constant. “la” increases as the N.A. moves towards compression face because cracking continues.
Failure initiates by the yielding of steel but final failure is still by crushing of concrete
Cc
T Internal Force Diagram
a/2
la
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Plain & Reinforced Concrete-1
Under-Reinforced Failure (contd…)
Derivation for ρ Design Moment Capacity
abnb TM l
2
adfAΦMΦ ysbnb
For tension controlled section Φ = 0.9
2
adf0.9AMΦ ysnb
And
b'0.85f
fAa
c
ys
(1)
(2)
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Plain & Reinforced Concrete-1
Under-Reinforced Failure (contd…)
Put value of “a” from (1) to (2)
b'0.85f2
fAdf0.9AMΦ
c
ysysnb
b'0.85f2
fρbddfρbd0.9MΦ
c
yynb
For economical design unb MMΦ
'0.85f2
fρ1fρbd0.9M
c
yy
2u
'0.85f
f
2
ρ1f0.9ρ
bd
M
c
yy2
u
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Plain & Reinforced Concrete-1
Under-Reinforced Failure (contd…)
Let
(MPa) Rbd
M2u ω
f
0.85fc'
y
And
Hence
2ω
ρ1f0.9ρR y
2ω
ρ1ρ
0.9f
R
y
2ω
ρ-ρ
0.9f
R 2
y
00.9f
R2ωρ2ω-ρ
y
2
0fc'
fc'
0.85
0.85
0.9f
R2ωρ2ω-ρ
y
2
0'0.3825f
Rωρ2ω-ρ
c
22
20.3825fc'
ωR44ω2ωρ
22
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Plain & Reinforced Concrete-1
Under-Reinforced Failure (contd…)
By simplification
0.3825fc'
R11ωρ
We have to use –ve sign for under reinforced sections. So
fc'
2.614R11ωρ
Reason
For under reinforced section ρ <ρb
If we use positive sign ρ will become greater than ρb, leading to brittle failure.
y1
y
cb f600
600β
f
'f0.85ρ
< 1.0
ω
ωρb 30
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Plain & Reinforced Concrete-1
Trial Method for the determination of “As”
b'0.85f
fAa
c
ys (A)
2
adf0.9AM ysu (B)
2
ad0.9f
MA
y
us (C)
Trial # 1, Assume some value of “a” e.g. d/3 or d/4 or any other reasonable value, and put in (C) to get “As”
Trial # 2, Put the calculated value of “As” in (A) to get “a”. Put this “a” value in (C) to get “As”Keep on doing the trials unless “As” from a specific trial becomes equal to the “As” calculated from previous trial.
THIS VALUE OF AS WILL BE THE FINAL ANSWER.
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Plain & Reinforced Concrete-1
Is The Section is Under-Reinforced or NOT ?1. Calculate ρ and if it is less than ρmax, section is
under reinforced2. Using “a” and “d” calculate εt if it is ≥ 0.005,
section is under-reinforced (tension controlled)
3. If section is over-reinforced than in the following equation –ve term will appear in the under-root.
'f
2.614R11ωρ
c
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Plain & Reinforced Concrete-1
Is The Section is Under-Reinforced or NOT ?
(contd…)
1. For tension controlled section, εt = 0.005, d
8
3βa 1
Using formula of Mn from concrete side
acbnbu CΦMΦM l
2
adba'0.85f9.0M cu
2
d83
0.85dd
8
30.85b'0.85f0.9M cu
2cu bd'0.205fM
b'0.205f
Md
c
umin
If we keep d > dmin the resulting section will be under-reinforced.
d > dmin means that section is stronger in compression. 33
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Plain & Reinforced Concrete-1
Over-Reinforced Failure
Cc
T Internal Force Diagram
a/2
la
Stage-II, Cracked Section
These two stages are same as in under-reinforced section.
Stage-I, Un-cracked Section
Stage-III, Concrete reaches strain of 0.003 but steel not yielding
We never prefer to design a beam as over-reinforced (compression controlled) as it will show sudden failure.
Φ = 0.65 εs < εy fs<fy
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Plain & Reinforced Concrete-1
Over-Reinforced Failure
Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…)
2
a-dba'.85f00.65 M cnb
aCc Mnb l
“a” is unknown as “fs” is not known
b'0.85f
fAa
c
ss
(i)
(ii)
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Plain & Reinforced Concrete-1
Over-Reinforced Failure
Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…)
εcu=
0.003
Strain Diagram
εs
cB C
A
E D
d- c
Comparing ΔABC & ΔADE
c
cd
0.003
εs
1
1s
a
d
0.003
ε
a
a
aβ0.003ε 1
s
ss εEf
a
aβ0.003200,000f 1
s
a
aβ600f 1
s
(iii) (iv)Eq # (iv) is applicable when εs < εy
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Plain & Reinforced Concrete-1
Over-Reinforced Failure
Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…)
Putting value of “fs” from (iv) to (ii)
b'0.85fa
adβ600A
ac
1s
(v)
Eq. # (v) is quadratic equation in term of “a”.
Flexural Capacity
2'85.0
2
adbaf
adCM cbcbnb
2
adfAΦ
2
ad T ΦMΦ ssbbnb
Calculate “a” from (v) and “fs” from (iv) to calculate flexural capacity from these equations
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Plain & Reinforced Concrete-1
Extreme Tensile Steel Strain
εt
Type of X-
section c/d a/d ρmax Φ
< εy
Compression Controlled
0.65
≥ εy
Transition Section
(Under-Reinforced)
0.65 to 0.9
≥ 0.004Under-
Reinforced(minimum strain for beams)
0.65 to 0.9
≥ 0.005 Tension Controlle
d
0.9
≥ 0.0075Redistribution is allowed
0.9
yf600
600
y1 f600
600β
yy
c1 f600
600
f
'0.85fβ
yf600
600
y1 f600
600β
yy
c1 f600
600
f
'0.85fβ
7
3
7
3β1
7
3
f
'0.85fβ
y
c1
8
3
8
3β1
8
3
f
'0.85fβ
y
c1
7
2
7
2β1
7
2
f
'0.85fβ
y
c1 38
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Plain & Reinforced Concrete-1
Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design methodData
1. Dimensions, b, h, d and L (span)
2. fc’, fy, Ec, Es
3. As
Required1. ΦbMn
2. Load Carrying Capacity39
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Plain & Reinforced Concrete-1
Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…)
SolutionStep # 1Calculte the depth of N.A assuming the section as under-reinforced
ys ff ys εε and
b' 0.85f
fAa
c
ys1β
ac and
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Plain & Reinforced Concrete-1
Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…)
SolutionStep # 2 Calculate εs and check the assumption of step# 1
c
cd0.003εε ts
For extreme
pointIf εs ≥ εy, the assumption is correct
If εs ≤ εy, the section is not under-reinforced. So “a” is to be calculated again by the formula of over reinforced section
b'0.85fa
adβ600A
ac
1s
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Plain & Reinforced Concrete-1
Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…)
SolutionStep # 3 Decide Φ factor
For εs ≥ 0.005, Φ = 0.9 (Tension controlled section)
For εs ≤ εy, Φ = 0.65 (Compression controlled section) For εy ≤ εs ≤ 0.005, Interpolate value of Φ (Transition Section)
Step # 4 Calculate ΦbMn
2
adfAM ysbnb
2
adba'f85.0M cbnb
For under-reinforced Section
For over-reinforced Section
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Plain & Reinforced Concrete-1
Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…)
Alternate MethodStep # 1 to step # 3 are for deciding whether the section is over reinforced or under-reinforced. Alternatively it can be done in the following manner.
1. Calculate ρ and ρmax if ρ < ρmax section is under-reinforced.
2. Calculate dmin, if d ≥ dmin, section is tension controlled
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Plain & Reinforced Concrete-1
Selection of Steel Bars for Beams1. When different diameters are selected the
maximum difference can be a gap of one size.2. Minimum number of bars must be at least two,
one in each corner.3. Always Place the steel symmetrically.4. Preferably steel may be placed in a single layer
but it is allowed to use 2 to 3 layers.5. Selected sizes should be easily available in
market6. Small diameter (as far as possible) bars are
easy to cut and bend and place.44
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Plain & Reinforced Concrete-1
Selection of Steel Bars for Beams (contd…)
7. ACI Code RequirementsThere must be a minimum clearance between bars (only
exception is bundled bars). Concrete must be able to flow through the reinforcement. Bond strength between concrete and steel must be fully developed.
Minimum spacing must be lesser of the following
Nominal diameter of bars 25mm in beams & 40mm in columns 1.33 times the maximum size of aggregate used.
We can also give an additional margin of 5 mm. 45
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Plain & Reinforced Concrete-1
Selection of Steel Bars for Beams (contd…)
8. A minimum clear gap of 25 mm is to be provided between different layers of steel
9. The spacing between bars must not exceed a maximum value for crack control, usually applicable for slabs
What is Detailing? Deciding diameter of bars Deciding no. of bars Deciding location of bent-up and curtailment of bars making sketches of reinforcements.
25mm
Not O.K.
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Plain & Reinforced Concrete-1
Concrete Cover to Reinforcement Measured as clear thickness outside the outer most steel bar.
Purpose
To prevent corrosion of steel To improve the bond strength To improve the fire rating of a building It reduces the wear of steel and attack of
chemicals specially in factories.
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Plain & Reinforced Concrete-1
Concrete Cover to Reinforcement (contd…)
ACI Code Minimum Clear Cover Requirements
1. Concrete permanently exposed to earth, 75 mm2. Concrete occasionally exposed to earth,
# 19 to # 57 bars 50 mm # 16 and smaller bars 40 mm
3. Sheltered Concrete Slabs and Walls 20 mm Beams and Columns 40 mm
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting One-way Slab
lx
lx
Exterior BeamInterior Beam
Width of slab supported by interior beam = lx
Width of slab supported by exterior beam = lx/2 + Cantilever width
ly
(ly/lx > 2)
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Plain & Reinforced Concrete-1
Load Carrie by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)
lx
lx
ly ly
Exterior Long Beam
Interior Long Beam
Exterior Short Beam
Interior Short Beam
45o
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Plain & Reinforced Concrete-1
Load Carrie by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…
45olx/2
lx/2
o45cos
2/xl
2
o45cos2/x
lArea of
Square
Shorter Beams
For simplification this triangular load on both the sides is to be replaced by equivalent UDL, which gives same Mmax as for the actual triangular load.
2
2
xl
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…
45o45o
Equivalent Rectangular Area
22
x3
2
2
x
3
4l
l
Factor of 4/3 convert this VDL into UDL.
Equivalent width supported by interior short beam
lx
x
x32
l
l 2
x3
2l
Equivalent width supported by exterior short beam
3
xlCantilever
x3
2l
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)contd…
lx
lx
ly
Exterior Long Beam
2
2
xxy
2
x
l
lll
Supported Area
lx/2 lx/2ly - lx
4
x
2
x
2
yx 22 llll
4
x
2
yx 2lll
2
xyx
2
1 2lll
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)contd…
Exterior Long Beam
2R1
3R1
F
2
y
xR
ll
where
Factor F converts trapezoidal load into equivalent UDL for maximum B.M. at center of simply supported beam. For Square panel
R = 1 and F = 4/3
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2)contd…
Exterior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
Length...Span
F)Supported..Area(
lyy
1
2R1
3R1
2
xyx
2
1 22
ll
ll
2R1
3R1
2
x1
2
x 2lyll
3R1
2
x 2l+ Cantilever (if present)
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Plain & Reinforced Concrete-1
Load Carried by the BeamBeam Supporting Two-way Slab (ly/lx≤ 2) contd…
Interior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
ly
3R1x 2l
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Plain & Reinforced Concrete-1
Wall Load (if present) on Beam
tw (mm)
H (m)1000
81.91930
1000
twHm1
UDL on beam
Htw019.0 (kN/m)
Htw019.0
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Plain & Reinforced Concrete-1
Wall Load on the Lintel
Equivalent UDL on lintel if height of slab above lintel is greater than 0.866L
0.866L
60o 60o
LLtw11.0UDL kN/m
tw = wall thickness in “mm”
L = Opening size in “m”If the height of slab above lintel is less than 0.866L
Total Wall Load + Load from slab in case of load bearing wall
UDL = (Equivalent width of slab supported) x (Slab load per unit area)
= m x kN/m2 = kN/m
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Plain & Reinforced Concrete-1
Slab Load per Unit AreaTop Roof
Slab Thickness = 125 mmEarth Filling = 100 mmBrick Tiles = 38 mm
Dead Load 2m/kg3002400
1000
125Self wt. of R.C. slab
Earth Filling 2m/kg1801800
1000
100
Brick Tiles 2m/kg741930
1000
38
554 kg/m2Total Dead Load, Wd =
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Plain & Reinforced Concrete-1
Slab Load per Unit Area (contd…)
Top Roof
Live Load WL = 200 kg/m2
Ldu W6.1W2.1W
Total Factored Load, Wu
1000
81.92006.15542.1Wu
2u m/kN66.9W
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Plain & Reinforced Concrete-1
Slab Load per Unit Area (contd…)
Intermediate Floor Slab Thickness = 150 mmScreed (brick ballast + 25% sand) = 75 mmP.C.C. = 40 mmTerrazzo Floor = 20 mm
Dead Load2m/kg3602400
1000
150Self wt. of R.C. slab
Screed 2m/kg13518001000
75
Terrazzo + P.C.C 2m/kg13823001000
)4020(
633 kg/m2Total Dead Load, Wd =
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Plain & Reinforced Concrete-1
Slab Load per Unit Area (contd…)
Intermediate Floor Live Load
Occupancy Live Load = 250 kg/m2
Moveable Partition Load = 150 kg/m2
WL = 250 + 150 = 400 kg/m2
Ldu W6.1W2.1W
Total Factored Load, Wu
1000
81.94006.16332.1Wu
2u m/kN73.13W
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Plain & Reinforced Concrete-1
Slab Load per Unit Area (contd…)
Self Weight of Beam
Service Self Wight of Beam = b x h x 1m x 2400
2L11.112400m118
L
12
L Kg/m
Factored Self Wight of Beam
22 L131.01000
81.92.1L11.11 kN/m
Self weight of beam is required to be calculated in at the stage of analysis, when the beam sizes are not yet decided, so approximate self weight is computed using above formula.
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Plain & Reinforced Concrete-1
Design of Singly Reinforced Beam by Strength Method (for flexure only)
Data: Load, Span (SFD, BMD) fc’, fy, Es
Architectural depth, if any
Required: Dimensions, b & h Area of steel Detailing (bar bending schedule)
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Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method
(contd…)
Procedure:
1. Select reasonable steel ratio between ρmin and ρmax. Then find b, h and As.
2. Select reasonable values of b, h and then calculate ρ and As.
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Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method
(contd…)
2. Using Trial DimensionsI. Calculate loads acting on the beam.II. Calculate total factored loads and plot SFD
and BMD. Determine Vumax and Mumax.
III. Select suitable value of beam width ‘b’. Usually between L/20 to L/15. preferably a multiple of 75mm or 114 mm.
IV. Calculate dmin.b'f205.0
Md
c
umin
hmin = dmin + 60 mm for single layer of steel
hmin = dmin + 75mm for double layer of steelRound to upper 75 mm
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Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method
(contd…)
V. Decide the final depth.
minhh For strength
minhh For deflection
ahh Architectural depth
12hh
Preferably “h” should be multiple of 75mm.
Recalculate “d” for the new value of “h”
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Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method
(contd…)
VI. Calculate “ρ” and “As”.
fc'
2.614R11wρ
Four methods
y
c
f
'f0.85w 2
u
bd
MR
Design Table
Design curves
Using trial Method
a)
b)
c)
d)
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Plain & Reinforced Concrete-1Design of Singly Reinforced Beam by Strength Method
(contd…)
VII. Check As ≥ As min.
As min = ρmin bd (ρmin = 1.4/fy to fc’ ≤ 30 MPa)
VIII. Carry out detailing
IX. Prepare detailed sketches/drawings.
X. Prepare bar bending schedule.
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Plain & Reinforced Concrete-1
Design of Singly Reinforced Beam by Strength Method (contd…)
1. Using Steel RatioI. Step I and II are same as in previous method. III. Calculate ρmax and ρmin & select some suitable
“ρ”.IV. Calculate bd2 from the formula of moment
V. Select such values of “b” and “d” that “bd2” value is satisfied.
VI. Calculate As.VII. Remaining steps are same as of previous
method.
'1.7f
ρf1fρbd0.9MΦM
c
yy
2nbu