Dr. B. Rajeevan gave a lecture on design for flexure in reinforced concrete structures. The key points covered included: - Type II design problems involve determining cross-sectional dimensions and reinforcement area where no unique solution exists. - Requirements for flexural reinforcement include nominal cover, clear cover, and effective cover to protect concrete. - Spacing of reinforcing bars is limited to a minimum for constructability and a maximum to control cracking and ensure bond. - Design of singly and doubly reinforced beams was demonstrated through examples, determining reinforcement areas based on flexural strength requirements. - Design of flanged beams like T-beams and L-beams was discussed, highlighting differences in reinforcement

1. Lecture Notes 15 July 2010
Dr. B. Rajeevan 1
Dr. B. Rajeevan
Senior Lecturer
Department of Civil Engineering
Govt. College of Engineering Kannur
E-mail: rajeevan@gcek.ac.in
Mob: 9495 333 088
ContactTime: 4 pm – 5 pm
Design for Flexure
• Type II Design Problems
– Determine the c/s dimensions and area of
reinforcement
– No unique solution exists
15 July 2010 Dr. B. Rajeevan 2
• Given,
• Span, loads and material properties
• Limit states to consider are ultimate limit states of flexure, shear,
torsion and bond
• Serviceability limit states of cracking, deflection and durability
Requirements of flexural
reinforcement
• Cl. 26.4.1
– Nominal cover
– Clear cover
– Effective cover
Cover protects concrete from corrosion, fire and
gives bond to the surrounding concrete.
15 July 2010 Dr. B. Rajeevan 3
Table 16/ IS 456
15 July 2010 Dr. B. Rajeevan 4
Exposure Nominal cover
(mm)
Min Grade
Mild 20 M 20
Moderate 30 M 25
Severe 45 M 30
Very severe 50 M 35
Extreme 75 M 40
Also refer Table 3 & 5 of IS 456
Spacing of reinforcing bars
• Minimum limit
– To ensure easy placing of concrete
• Maximum limit
– Controlling crack width and bond
15 July 2010 Dr. B. Rajeevan 5
Beam width – multiple of 25 mm or 50 mm
Slab thickness – multiple of 5 mm
Spacing of reinforcing bars
15 July 2010 Dr. B. Rajeevan 6

2. Lecture Notes 15 July 2010
Dr. B. Rajeevan 2
Spacing of reinforcing bars
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Spacing of reinforcing bars
15 July 2010 Dr. B. Rajeevan 8
16
Reinforcement - Beam
tension reinforcement
– To ensure crack control
– To take care of unforeseen loads
– To control shrinkage and temperature variations
15 July 2010 Dr. B. Rajeevan 9
,min 0.85
y
Ast
bd f
.26.5.1.1( )Cl a
Reinforcement - Beam
tension/compression
reinforcement
– To avoid congestion
15 July 2010 Dr. B. Rajeevan 10
,max 0.04stA bD
.26.5.1.1( )
.26.5.1.2
Cl b
Cl
Requirements- Slab
• Maximum diameter of bars (Cl. 26.5.2.2)
– Not greater than one eighth of the thickness of
slab.
– For crack control
• Minimum reinforcement (Cl. 26.5.2.1)
– 0.15bD
– 0.12bD for HYSD bars
15 July 2010 Dr. B. Rajeevan 11
Deflection Control
15 July 2010 Dr. B. Rajeevan 12
4
2
max
2
max 2 2 2
3
2
4
24
3
5
384
8
8 8 8
6
12
8
5
65
384
384
1
const
2
5
; conan stan
42
t t
w
EI
w
M
bD
w M Z
bD
I
bD
E b
D E
w
D
E
I

3. Lecture Notes 15 July 2010
Dr. B. Rajeevan 3
Codal provisions for deflection control
15 July 2010 Dr. B. Rajeevan 13
max
.23.2.1
ba
t c
sic
k k
d
Cl
d
7 cantilever
20 simplysupported
26 continuous
basicd
Refer Figs. 4 and 5 of code to get andt ck k
.22.2
/ distance between supports
clear span +
EffectiveSpan Cl
c c
smaller of
d
Selection of Member sizes
• Beams
– Placing of concrete
– Increase depth than width
• Provides better control of deflection and crack
15 July 2010 Dr. B. Rajeevan 14
,lim0.5 0.8
/ 1.5 2
200 ,250 300
230
/ 10 16
t tAssume p to p
D b to
Beamwidth mm mmand mm
Masonry wall mm
Span D to
Selection of Member sizes
• Slabs
– Based on deflection criteria
15 July 2010 Dr. B. Rajeevan 15
,lim0.4 0.5 1.25
/ 25
/ 32
415
t t tAssume p p k
d span ss slab
span continuous slab
Fe steel normallyused
clearcover + ( 5 10 )
2
D d round tonearest mmor mm
Design of singly reinforced rectangular
beams
15 July 2010 Dr. B. Rajeevan 16
,max
,lim2
0.87 1 , 1.1( )
100
Refer
1
6
00
1
yu t t
y t t
ck
u uR u uM M with
fM p p
R
x
f p p AnnexG b
bd f
x
SP
/ 8,
/ 25,
/ 32,
u
l cantilever
ss
continuous
M
d
Rb
Trial valuesof d beam
3
clearcover +
2
. 19.2.1 25 /
/ 1.5 2
clearco
,
ver
2
tie
tie
Exceedin
D d
Cl w kN m
D b g changebto
d D
Design Example
15 July 2010 Dr. B. Rajeevan 17
Solution
15 July 2010 Dr. B. Rajeevan 18
Inside beam – coastal are – Table 3, 5 & 16
Moderate exposure – clear cover = 30 mm and grade of concrete M25
Determine uM
250 ; 600 ( /10)
50 550
6
6 0.23 0.55 6.32
6
25 0.25 0.6 25 3.75 /
5 3.75 8.75 / ; 10 /
, 1.5 1.5 8.75 10 28.
Trialsection
.22
1
.2
DL LL
u DL LL
b mm D mm span
d D mm
m
m
m
s
Cl
elfweight bD kN m
w kN m w kN m
Factored l
Loads
oad w w w /kN m

4. Lecture Notes 15 July 2010
Dr. B. Rajeevan 4
15 July 2010 Dr. B. Rajeevan 19
2 2
,lim
26
min
lim ,lim
lim 2
/ 8 28.1 6 / 8 126
250
0.138
126 10
( )
3.45 250
0.138 3.45
382
382 3
Momen
6
t
0
,
1
,
u
u u
u
u ck
u
ck
M w kNm
Let b mm
M
Get from
M f bd
sameasinitially a
d
R b M
R f
M
Fixi
bd
mm
Then D
ng
ssumed
SP Tab
b
leC
d and D
Assuming 25 &
450 ( )
25
8 ( )
2
432.5 ( )
25
450 30 8 399
8
minimu
2
m
.5
m
mmdiabars mms
mm say
m
d m
tirrups
m
15 July 2010 Dr. B. Rajeevan 20
6
2 2
,
2
,
,
126 10
3.158
250 399.5
1.082 1.061
1.061 3.158 3.15 1.064
3.2 3.15
1062.67
100
1062.67
. 25 , 2.16
490.87
Provide1-25 +2-
3/ 16
95 / 16,
u
t
t
st reqd
st reqd
b
st reqd
M
bd
p
p
A
Table S
bd mm
A
No of mmbars n
A
P
Ta SP
A
ble
, ,
20
491 628 1119st prov st reqdA A
15 July 2010 Dr. B. Rajeevan 21
,
max
max
.4, 4
100 100 1119
1.12
250 399.5
228.4
1.014; 1
/ 1.014 1 20 20.28
/ 6000 / 399.5 15 /
5
,
6
st
t prov
s
t c
provided
DesignCheck for deflect
A
p
bd
f
k k
l d
l d
i
l d He
on
nce OK
Fig IS
15 July 2010 Dr. B. Rajeevan 22
Homework
15 July 2010 Dr. B. Rajeevan 23
Example 2
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5. Lecture Notes 15 July 2010
Dr. B. Rajeevan 5
15 July 2010 Dr. B. Rajeevan 25
2 2
Moment,
4160
/ 25 4000 / 25 160
160 40 200
4000 230 4230
4000 160 4160
25 0.2 1 1 1 6 /
4 1 4 /
, 1.5 15 /
/ 8 15 4.16 / 8 32.45 /
D
u
L
LL
u DL LL
u u
Assumed span mm
D mm
mm
mm
w kN m
w kN m
Factored Load w w w kN m
M w kNm m
M
m
etrewid
m
thof slab
15 July 2010 Dr. B. Rajeevan 26
2
Moderateexposure 25
Main bars,
1.268; 3/ 16, 0.375
ck
u
t
st
f MPa
M
R Table SP p
bd
A
2
600 /
100
Using10 , , 100 131
. 3 3 160 480 300 300
Provide10
.26.3.3. (1)
@130 /
t
st
st
p bd
A mm metrewidthof slab
A
bars spacing s mm
A
Max Spacing Cd mmor mm m
m
m b
mc c
15 July 2010 Dr. B. Rajeevan 27
2
0.12 240 /
100
Using8 , , 100 209
. 5 5 160 800 450 45
Provide8 @200 /
.26.3.3. (2)0
st dist
st
bD
A mm metrewidthof slab
A
bars spacing s mm
A
M
Distributors
mmc cas distri
ax Spacing d mmor mm m
buto
Cm
r
b
s
2
,lim
200 30 10 / 2 165
1000 1000 78.54
604.15
130
10 ,0 0.37 1.1 69, 1
actual
st prov
prov
st prov
t t
Stre
d mm
A
A mm
ngthCheck
Detai
s
A
p p
b
Tab
ling
le
d
E SP
15 July 2010 Dr. B. Rajeevan 28
Design of Doubly Reinforced Beams
• Depth of beam is restricted, but moment is
higher
• Bending moment changes in sign
• To increase the ductility
• To reduce long term deflections due to
shrinkage and creep
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6. Lecture Notes 15 July 2010
Dr. B. Rajeevan 6
15 July 2010 Dr. B. Rajeevan 31
/
0.0035 1cs
u
d
x
/
0.36 0.416
uc us
u uc c us s
u ck u u sc sc
C C C
M C z C z
M f bx d x f A d d
15 July 2010 Dr. B. Rajeevan 32
15 July 2010 Dr. B. Rajeevan 33
,lim 2
,lim ,max ,max 1
2 ,lim
/ 2
2 /
2
1
2
1 2
2
0.36 0.416 0.87
0.87
0.87
0.87
,
,
0.87
u u u
u ck u u y st
u u u
st
st
st
st s
u
u y st
y
y
t st
sc
sc
s
sc sc y st
M M M
M f bx d x f A
M M M
M
M f A d d
f
TensileSteel Area A
A
A
A A A
Compression Steel Area A
d d
f A
f A f A A
2t
scf
Example
• Design a doubly reinforced beam with the following
data to carry a factored moment of 1000 kNm
15 July 2010 Dr. B. Rajeevan 34
2 2 /
400 ; 550 ; 30 / ; 415 / ; 60ck yb mm d mm f N mm f N mm d mm
Solution
15 July 2010 Dr. B. Rajeevan 35
2
,lim 0.138
5
( / 16)
01
u ck TM f bd ableC
m
SP
kN
,lim
2
1
1.43
1.43
400 550 3146
100
( / 16)t
st
Table E SPp
A mm
2 ,lim
22
2 /
2
1 2
1000 501 499
2820.6
0.87
5967
u u u
u
st
y
st st st
M M M kNm
M
A mm
f d d
A A A mm
Solution
15 July 2010 Dr. B. Rajeevan 36
/
2
6
22
/
(
60 / 550 0.1
353 /
499 10
2885
353 550 60
6)
,
/ 1sc
u
sc
sc
sc
Tab
d
d
f N mm
M
A mm
f d d
Areaof Compression Steel A
le F SP

7. Lecture Notes 15 July 2010
Dr. B. Rajeevan 7
Solution by SP16
15 July 2010 Dr. B. Rajeevan 37
Homework
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Design of Flanged Beams
15 July 2010 Dr. B. Rajeevan 39
T-beam under positive moment
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T-beam under negative moment
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L-beam under positive moment
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8. Lecture Notes 15 July 2010
Dr. B. Rajeevan 8
L-beam under negative moment
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Example – Design of T beam
15 July 2010 Dr. B. Rajeevan 44
Take Finish load = 1.72 kN/m2
Solution
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9. Lecture Notes 15 July 2010
Dr. B. Rajeevan 9
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RC SOILD SLAB
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RC Solid Slab
1-way 2-way Flat Slab Flat Plate
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DESIGN OF ONE WAY SLAB
15 July 2010 Dr. B. Rajeevan 53
Loads
• Self weight at 25 kN/m3 for reinforced concrete
• Finish and partition – 1.5 kN/m2
• Live load
– Based on usage
– Roof
• 1.5 kN/m2 for roof with access
• 0.75 kN/m2 for roof without access
– Floors
• 2 kN/m2 for residential floors
• 3 kN/m2 for office floors
15 July 2010 Dr. B. Rajeevan 54

10. Lecture Notes 15 July 2010
Dr. B. Rajeevan 10
One way slabs
15 July 2010 Dr. B. Rajeevan 55
 Simply supported
Continuous
Simplified design using moment and shear coefficients
Tables 12 and 13
Moment and Shear Coefficients
15 July 2010 Dr. B. Rajeevan 56
Design Considerations
• Span to Effective depth ratio,
– Simply supported – 25
– Continuous – 30
• Concrete Cover
– Tables 3, 5 & 16
15 July 2010 Dr. B. Rajeevan 57
/ d
Design Considerations
• Calculation of steel area based on UR section
• Diameter of bar not to exceed 1/8th of thickens of
slab
• Steel area
• Spacing of main steel not greater than 3d or 300
mm
• Spacing of secondary steel not greater than 5d or
450 mm
15 July 2010 Dr. B. Rajeevan 58
0.87u y stM f A lever arm
0.12 ( 0.15 )stA bD or bD
Design Steps
– Design as beams of one metre width, with span as short dimension
– Assume depth based on deflection criteria
• Min depth 100 mm
– Determine factored loads. Compute factored moment and shear. Use
coefficients given in Table 7 for continuous one way slab
– Find effective depth required based on moment
– Add cover and find total depth of slab(thickness)
– Check for shear (k c) using Table 13 and Clause 39.2
– Revise depth, if necessary
– Determine
15 July 2010 Dr. B. Rajeevan 59
stA
Design Check
– Check for minimum percentage of steel
– Check for spacing
– Recheck for shear based on the actual steel
– Check for deflection
– Provide secondary reinforcement
– Detail
15 July 2010 Dr. B. Rajeevan 60

11. Lecture Notes 15 July 2010
Dr. B. Rajeevan 11
STRUCTURAL DETAIL
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Design Example 1
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12. Lecture Notes 15 July 2010
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Design Example 2

13. Lecture Notes 15 July 2010
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14. Lecture Notes 15 July 2010
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