Flexural design of Beam...PRC-I

Plain & Reinforced
Concrete-1
By Engr. Rafia Firdous
Flexural Analysis and
Design of Beams
(Ultimate Strength Design of Beams)
1

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

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
h
c
d
b 0.85fc
fs
0.85fc
a
Equivalent Stress
Diagram/
Whitney’s Stress
Diagram
a/2
fs
3

Actual Stress
Diagram
0.85fc 0.85fc
Equivalent Stress
Diagram/ Whitney’s
Stress Diagram
c
Cc 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
'0.72f
a
c
c
×=
cβa 1 ×= 4

Factor β1
β1 = 0.85 for fc’ ≤ 28 MPa
Value of β1decreases by 0.05 for every 7 MPa increase in
strength with a minimum of 0.65
0.65'0.00714f1.064β c1 ≥−=
85.0≤
5

Determination of N.A. Location at Ultimate Condition
CASE-I: Tension Steel is Yielding at Ultimate Condition
ys εε ≥ or ys ff ≥
CASE-II: Tension Steel is Not Yielding at Ultimate Condition
yf
yε sε
yε<sε or ys ff <
0.0015
200,000
300
E
f
ε
y
y === 0.0021
200,000
420
E
f
ε
y
y ===
For 300 grade steel For 420 grade steel
6

CASE-I: Tension Steel is Yielding at Ultimate
Condition
ysss fAfAT ×=×=
ab'0.85fC cc ××=
2
a
da −=l
abffA cys ××=× '85.0
For longitudinal Equilibrium
T = Cc
bf
fA
a
c
ys
×
×
=
'85.0 1β
a
c =and
Cc
T = Asfs
Internal Force Diagram
a/2
la
7

CASE-I: Tension Steel is Yielding at Ultimate
Condition (contd…)
Nominal Moment Capacity, Mndepending on steel = T x la






−××=
2
Mn
a
dfA ys
Design Moment Capacity






−××=
2
M bnb
a
dfA ysφφ
Nominal Moment Capacity, Mn depending on Concrete = Cc x la






−×××=
2
a
dab0.85fc'Mn






−×××=
2
a
dab0.85fc'M bnb φφ 8

Minimum Depth for Deflection Control
I
1
αΔ
3
(Depth)
1
αΔ
For UDL
4
ωLαΔ
( ) 3
LωLαΔ
Deflection Depends upon Span, end conditions, Loads and fy of
steel. For high strength steel deflection is more and more depth is
required. 9

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
10

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'0.85fCc ××=
ab = depth of equivalent rectangular stress block when balanced steel
ratio is used.
εcu=0.003
Strain Diagram
εy
cb
d- cb
11

Balanced Steel Ratio, ρb(contd…)
ybys fd)b(ρfAT ×××==
For the longitudinal equilibrium
cCT =
bcyb ab'0.85ffd)b(ρ ××=×××
d
a
f
'f
0.85ρ b
y
c
b = (1)
12

From the strain diagram εcu= 0.003
Strain Diagram
εy
cb
d- cb
B C
A
E D
ADE&ABCΔs
b
y
b cd
ε
c
0.003
−
=
y
b
ε0.003
0.003d
c
+
=
s
s
s
y
b
E
E
d
E
f
0.003
0.003
c ××
+
=
13

d
f600
600
c
y
b ×
+
=
d
f0.003E
0.003E
c
ys
s
b ×
+
=
b1b Cβa ×=As we know
d
f600
600
βa
y
1b ×
+
×= (2)
Put (2) in (1)
y
1
y
c
b
f600
600
β
f
'f
0.85ρ
+
××=
14

Types of Cross Sections w.r.t. Flexure at Ultimate
Load Level
1. Tension Controlled Section
A 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
−
≥
d
0.008
0.003
c ≤
d
8
3
c ≤ d
8
3
βa 1≤and
15

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- cd
8
3
βa 1> baa <
16

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
−
=
d
7
3
c = d
7
3
βa 1=
17

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.
18

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 >
19

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

Strength Reduction Factor (Resistance Factor), Φ
 Transition Section (contd…)
( )yt
y
εε
ε0.005
0.25
0.65Φ −
−
+=
For members with ties
For members with Spirals
( )yt
y
εε
ε0.005
0.15
0.75Φ −
−
+=
21

Maximum Steel Ratio, ρmax
For
T = C
ab'0.85ffA cys ××=×
ab'0.85ffbdρ cy ××=××
d
a
f
'f
0.85ρ
y
c
×=
For tension controlled section d
8
3
βa0.005ε 1s ==
So






×= 1
y
c
max β
8
3
f
'f
0.85ρ 22

Maximum Steel Ratio, ρmax(contd…)
For transition section
d
7
3
βa0.004ε 1s ==






×= 1
y
c
max β
7
3
f
'f
0.85ρ
23

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
c
min
f
1.4
4f
'f
ρ ≥=
24

Under-Reinforced Failure
Cc
T
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 Section
N.A. position is fixed, means “la”
remains constant. Only “T” and
“Cc” increase with the increase of
load
25

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
a/2
la
26

Derivation for ρ
Design Moment Capacity
abnb TM l×Φ=Φ






−×=
2
a
dfAΦMΦ ysbnb
For tension controlled section Φ = 0.9






−×=
2
a
df0.9AMΦ ysnb
And
b'0.85f
fA
a
c
ys
=
(1)
(2)
27

Put value of “a” from (1) to (2)






×
−=
b'0.85f2
fA
df0.9AMΦ
c
ys
ysnb






×
×
−××=
b'0.85f2
fρbd
dfρbd0.9MΦ
c
y
ynb
For economical design
unb MMΦ =






×
×
−××=
'0.85f2
fρ
1fρbd0.9M
c
y
y
2
u






×−×=
'0.85f
f
2
ρ
1f0.9ρ
bd
M
c
y
y2
u
28

Let
(MPa)R
bd
M
2
u
= ω
f
0.85fc'
y
=And
Hence






−×=
2ω
ρ
1f0.9ρR y






−=
2ω
ρ
1ρ
0.9f
R
y
2ω
ρ
-ρ
0.9f
R 2
y
=
0
0.9f
R2ω
ρ2ω-ρ
y
2
=
×
+×
0
fc'
fc'
0.85
0.85
0.9f
R2ω
ρ2ω-ρ
y
2
=××
×
+×
0
'0.3825f
Rω
ρ2ω-ρ
c
2
2
=
×
+×
2
0.3825fc'
ωR
44ω2ωρ
2
2 ×
×−±=
29

By simplification








−±=
0.3825fc'
R
11ωρ
We have to use –ve sign for under reinforced sections. So








−−=
fc'
2.614R
11ωρ
Reason
For under reinforced section ρ <ρb
If we use positive sign ρ will
become greater than ρb, leading to
brittle failure.
y
1
y
c
b
f600
600
β
f
'f
0.85ρ
+
×=
< 1.0
ω
ωρb < 30

Trial Method for the determination of “As”
b'0.85f
fA
a
c
ys
= (A)






−=
2
a
df0.9AM ysu (B)






−
=
2
a
d0.9f
M
A
y
u
s (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. 31

Is The Section is Under-Reinforced or NOT ?
1. Calculate ρ and if it is less than ρmax, section is
under reinforced
2. 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
2.614R
11ωρ
c
32

Is The Section is Under-Reinforced or NOT ?
(contd…)
1. For tension controlled section, εt = 0.005, d
8
3
βa 1=
Using formula of Mn from concrete side
acbnbu CΦMΦM l×==






−××=
2
a
dba'0.85f9.0M cu












−×





×=
2
d
8
3
0.85
dd
8
3
0.85b'0.85f0.9M cu
2
cu bd'0.205fM =
b'0.205f
M
d
c
u
min
×
=
If we keep d > dmin
the resulting section
will be under-
reinforced.
d > dmin means that
section is stronger in
compression.
33

Over-Reinforced Failure
Cc
T
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
34

Stage-III, Concrete reaches strain of 0.003 but steel not
yielding (contd…)






××=Φ
2
a
-dba'.85f00.65M cnb
aCcMnb l×=Φ
“a” is unknown as “fs” is not known
b'0.85f
fA
a
c
ss
=
(i)
(ii)
35

Stage-III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
εcu= 0.003
Strain Diagram
εs
c
B C
A
E D
d- c
Comparing ΔABC & ΔADE
c
cd
0.003
εs −
=
1
1s
a
d
0.003
ε
β
β
a−
=





 −
=
a
aβ
0.003ε 1
s
ss εEf ×=





 −
×=
a
aβ
0.003200,000f 1
s





 −
×=
a
aβ
600f 1
s
(iii)
(iv)
Eq # (iv) is applicable
when εs < εy
36

Stage-III, Concrete reaches strain of 0.003 but steel not yielding
(contd…)
Putting value of “fs” from (iv) to (ii)
b'0.85f
a
adβ
600A
a
c
1
s 




 −
×
= (v)
Eq. # (v) is quadratic equation in term of “a”.
Flexural Capacity






−Φ=





−Φ=Φ
2
'85.0
2
a
dbaf
a
dCM cbcbnb






−=





−=
2
a
dfAΦ
2
a
dTΦMΦ ssbbnb
Calculate “a” from (v)
and “fs” from (iv) to
calculate flexural
capacity from these
equations 37

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
≥ 0.004
Under-
Reinforced
(minimum strain
for beams)
0.65 to
0.9
≥ 0.005 Tension
Controlled
0.9
≥ 0.0075
Redistribution
is allowed
0.9








+
>
yf600
600








+
>
y
1
f600
600
β








+
>
yy
c
1
f600
600
f
'0.85f
β








+
≤
yf600
600








+
≤
y
1
f600
600
β 







+
≤
yy
c
1
f600
600
f
'0.85f
β
7
3
≤
7
3
β1≤
7
3
f
'0.85f
β
y
c
1 ×≤
8
3
≤
8
3
β1≤
8
3
f
'0.85f
β
y
c
1 ×≤
7
2
≤
7
2
β1≤
7
2
f
'0.85f
β
y
c
1 ×≤ 38

Capacity Analysis of Singly Reinforced
Rectangular Beam by Strength Design method
Data
1. Dimensions, b, h, d and L (span)
2. fc’, fy, Ec, Es
3. As
Required
1. ΦbMn
2. Load Carrying Capacity
39

Rectangular Beam by Strength Design method (contd…)
Solution
Step # 1Calculte the depth of N.A assuming the section as
under-reinforced
ys ff = ys εε ≥and
b'0.85f
fA
a
c
ys
=
1β
a
c =and
40

Solution
Step # 2 Calculate εs and check the assumption of step# 1
c
cd
0.003εε ts
−
== For extreme point
If 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'0.85f
a
adβ
600A
a
c
1
s 




 −
×
=
41

Solution
Step # 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






−Φ=Φ
2
a
dfAM ysbnb






−×Φ=Φ
2
a
dba'f85.0M cbnb
For under-reinforced Section
For over-reinforced Section 42

Alternate Method
Step # 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
43

Selection of Steel Bars for Beams
1. 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 market
6. Small diameter (as far as possible) bars are easy to cut
and bend and place.
44

Selection of Steel Bars for Beams (contd…)
7. ACI Code Requirements
There 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

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.
46

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.
47

Concrete Cover to Reinforcement (contd…)
ACI Code Minimum Clear Cover Requirements
1. Concrete permanently exposed to earth, 75 mm
2. Concrete occasionally exposed to earth,
 # 19 to # 57 bars 50 mm
 # 16 and smaller bars 40 mm
1. Sheltered Concrete
 Slabs and Walls 20 mm
 Beams and Columns 40 mm
48

Load Carried by the Beam
Beam Supporting One-way Slab
lx
lx
Exterior Beam
Interior Beam
Width of slab supported by interior beam = lx
Width of slab supported by exterior beam = lx/2 + Cantilever width
ly
(ly/lx > 2)

Load Carrie by the Beam
Beam Supporting Two-way Slab (ly/lx≤ 2)
lx
lx
ly ly
Exterior Long Beam
Interior Long Beam
Exterior Short Beam
Interior Short Beam
45o

Load Carrie by the Beam
Beam Supporting Two-way Slab (ly/lx≤ 2) contd…
45olx/2
lx/2
[ ] o
45cos
2/xl
[ ] 2
o
45cos
2/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

45o
45o
Equivalent Rectangular
Area
2
2
x
3
2
2
x
3
4
l
l
=×=
Factor of 4/3 convert this VDL into UDL.
Equivalent width supported
by interior short beam
lx
x
x
3
2
l
l 2
= x
3
2
l=
Equivalent width supported
by exterior short beam
+=
3
xl
Cantilever
x
3
2
l

Beam Supporting Two-way Slab (ly/lx≤ 2)contd…
lx
lx
ly
Exterior Long Beam
( )
2
2
x
xy
2
x






+−×=
l
ll
l
Supported Area
lx/2 lx/2ly - lx
4
x
2
x
2
yx 22
llll
+−=
4
x
2
yx 2
lll
−=






−=
2
x
yx
2
1 2
l
ll

Exterior Long Beam
2
R1
3
R1
F
2
−
−
=
y
x
R
l
l
=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

Exterior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
Length...Span
F)Supported..Area( ×
=
ly
y
1
2R1
3R1
2
x
yx
2
1 22
l
l
ll ×





−
−
×





−=






−
−
×





−=
2R1
3R1
2
x
1
2
x 2
lyll
( )



−= 3R1
2
x 2l
+ Cantilever (if present)

Interior Long Beam
Equivalent width
lx/2 lx/2ly - lx
Equivalent width
ly
( )3R1x 2
−l

Wall Load (if present) on Beam
tw (mm)
H
(m)
1000
81.9
1930
1000
tw
Hm1 ××





××=
UDL on beam
Htw019.0 ××= (kN/m)
Htw019.0 ××

Wall Load on the Lintel
Equivalent UDL on lintel if
height of slab above lintel
is greater than 0.866L 0.866L
60o
60o
L
Ltw11.0UDL ××= 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

Slab Load per Unit Area
Top Roof
Slab Thickness = 125 mm
Earth Filling = 100 mm
Brick Tiles = 38 mm
Dead Load
2
m/kg3002400
1000
125
=×=Self wt. of R.C. slab
Earth Filling
2
m/kg1801800
1000
100
=×=
Brick Tiles
2
m/kg741930
1000
38
=×=
554 kg/m2
Total Dead Load, Wd =

Slab Load per Unit Area (contd…)
Top Roof
Live Load
WL = 200 kg/m2
Ldu W6.1W2.1W +=
Total Factored Load, Wu
( )
1000
81.9
2006.15542.1Wu ××+×=
2
u m/kN66.9W =

Intermediate Floor
Slab Thickness = 150 mm
Screed (brick ballast + 25% sand) = 75 mm
P.C.C. = 40 mm
Terrazzo Floor = 20 mm
Dead Load
2
m/kg3602400
1000
150
=×=Self wt. of R.C. slab
Screed 2
m/kg1351800
1000
75
=×=
Terrazzo + P.C.C 2
m/kg1382300
1000
)4020(
=×
+
=
633 kg/m2
Total Dead Load, Wd =

Intermediate Floor
Live Load
Occupancy Live Load = 250 kg/m2
Moveable Partition Load = 150 kg/m2
WL= 250 + 150 = 400kg/m2
Ldu W6.1W2.1W +=
Total Factored Load, Wu
( )
1000
81.9
4006.16332.1Wu ××+×=
2
u m/kN73.13W =

Self Weight of Beam
Service Self Wight of Beam = b x h x 1m x 2400
2
L11.112400m1
18
L
12
L
=×××= Kg/m
Factored Self Wight of Beam
22
L131.0
1000
81.9
2.1L11.11 =××= 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.

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)

Design 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.

2. Using Trial Dimensions
I. 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
M
d
c
u
min =
hmin = dmin + 60 mm for single layer of steel
hmin = dmin + 75mm for double layer of steel
Round to
upper 75 mm

V. Decide the final depth.
minhh ≥ For strength
minhh ≥ For deflection
ahh ≈ Architectural depth
12
hh ≈
Preferably “h” should be multiple of 75mm.
Recalculate “d” for the new value of “h”

VI. Calculate “ρ” and “As”.








−−=
fc'
2.614R
11wρ
Four methods
y
c
f
'f
0.85w = 2
u
bd
M
R =
Design Table
Design curves
Using trial Method
a)
b)
c)
d)

VII. Check As ≥ Asmin.
Asmin=ρmin bd (ρmin= 1.4/fy to fc’ ≤ 30 MPa)
VIII. Carry out detailing
IX. Prepare detailed sketches/drawings.
X. Prepare bar bending schedule.

1. Using Steel Ratio
I. 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
ρf
1fρbd0.9MΦM
c
y
y
2
nbu

Flexural design of Beam...PRC-I

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