UNIT–2
PRINCIPLES OF LIMIT STATE DESIGN AND ULTIMATE
STRENGTH OF R.C. SECTION:
2.1 Introduction:
A beam experiences flexural stresses and shear stresses. It deforms and cracks are developed.
ARC beam should have perfect bond between concrete and steel for composites action. It is
primarily designed as flexural member and then checked for other parameters like shear, bond ,
deflection etc. In reinforced concrete beams, in addition to the effects of shrinkage, creep and
loading history, cracks developed in tension zone effects its behavior. Elastic design method
(WSM) do not give a clear indication of their potential strengths. Several investigators have
published behavior of RC members at ultimate load. Ultimate strength design for beams was
introduced into both the American and British code in 1950’s. The Indian code ES456
introduced the ultimate state method of design in 1964. Considering both probability concept
and ultimate load called as “Limit state method of design” was introduced in Indian code from
1978.

2.2 Behavior of Reinforced concrete beam
To understand the behavior of beam under transverse loading, a simply supported beam
subjected to two point loading as shown in Fig. 2.1 is considered. This beam is of rectangular
cross-section and reinforced at bottom.

11

When the load is gradually increased from zero to the ultimate load value, several stages of
behavior can be observed. At low loads where maximum tensile stress is less than modulus of
rupture of concrete, the entire concrete is effective in resisting both compressive stress and
tensile stress. At this stage, due to bonding tensile stress is also induced in steel bars.
With increase in load, the tensile strength of concrete exceeds the modulus of rupture of
concrete and concrete cracks. Cracks propagate quickly upward with increase in loading up ;to
neutral axis. Strain and stress distribution across the depth is shown in Fig 4.1c. Width of crack
is small. Tensile stresses developed are absorbed by steel bars. Stress and strain are
proportional till fc<





. Further increase in load, increases strain and stress in the section and

are no longer proportional. The distribution of stress – strain curve of concrete. Fig 41d shows
the stress distribution at ultimate load.
Failure of beam depends on the amount of steel present in tension side. When moderate
amount of steel is present, stress in steel reaches its yielding value and stretches a large
amount whth tension crack in concrete widens. Cracks in concrete propagate upward with
increases in deflection of beam. This induces crushing of concrete in compression zone and
called as “secondary compression failure”. This failure is gradual and is preceded by visible
signs of distress. Such sections are called “under reinforced” sections.
When the amount of steel bar is large or very high strength steel is used, compressive stress
in concrete reaches its ultimate value before steel yields. Concrete fails by crushing and failure
is sudden. This failure is almost explosive and occur without warning. Such reactions are called
“over reinforced section”
If the amount of steel bar is such that compressive stress in concrete and tensile stress in
steel reaches their ultimate value simultaneously, then such reactions are called “Balanced
Section”.
The beams are generally reinforced in the tension zone. Such beams are termed as “singly
reinforced” section. Some times rebars are also provided in compression zone in addition to
tension rebars to enhance the resistance capacity, then such sections are called “Doubly
reinforce section.
2.3 Assumptions
Following assumptions are made in analysis of members under flexure in limit state method
1. Plane sections normal to axis remain plane after bending. This implies that strain is
proportional to the distance from neutral axis.
2. Maximum strain in concrete of compression zone at failure is 0.0035 in bending.
12

3. Tensile strength of concrete is ignored.
4. The stress-strain curve for the concrete in compression may be assumed to be
rectangle, trapezium, parabola or any other shape which results in prediction of
strength in substantial agreement with test results. Design curve given in IS456-2000 is
shown in Fig. 2.2

5. Stress – strain curve for steel bar with definite yield print and for cold worked deformed
bars is shown in Fig 2.3 and Fig 2.4 respectively.

13

6. To ensure ductility, maximum strain in tension reinforcement shall not be less than



.

+ 0.002.

7. Perfect bond between concrete and steel exists.
2.4. Analysis of singly reinforced rectangular sections
Consider a rectangular section of dimension b x h reinforced with Ast amount of steel on
tension side with effective cover Ce from tension extreme fiber to C.G of steel. Then effective
depth d=h-ce, measured from extreme compression fiber to C.G of steel strain and stress
distribution across the section is shown in Fig.2.4. The stress distribution is called stress block.

From similar triangle properly applied to strain diagram

ε


=

ε

 − 

∈  =∈  ×

→ (1)

 − 
→ (2)


For the known value of x4 & 6cu the strain in steel is used to get the value of stress in steel
from stress-strain diagram. Equation 4.4-1 can be used to get the value of neutral axis depth as
 =

∈ 
∈ 
∈ 
× ( − ) =
×−
× 
∈
∈
∈

14

  1 +
 

Here

∈

∈∈

∈ 
∈ 
=
× 
∈
∈

∈ +∈ 
∈ 
=
×
∈
∈

∴ =

∈

∈∈

×  − (3)

is called neutral axis factor

For equilibrium Cu = Tu.
K1, k3 fcu b xu = fsAs

∴!

=

",$%
&'

∴ !

=

$$%


(

=

$$%
&
(

! = )1)3! ×

6

×





×

∈

×

=


 6

∈∈

+
,
*
×
-./ 0 = /..1 2345 =
+
* + * ,

52





$$%


− (4)

Value of fs can be graphically computed for a given value of P as shown in Fig2.6

15

After getting fs graphically, the ultimate moment or ultimate moment of resistance is calculated
as
Mu = Tu x Z = fsAs(d-k2 xu)
Mu= Cu x Z = k1k2fcubxu x (d-k2xu)
Consider
Mu = fsAs(d-k2 x
From (4)

∈

∈∈





∴ ; = !, <1 −

) = fsAsd(1-kz

=

fs 6

)

$$%
:

$
 6

$$%


$
 6

∈

∈∈

= − (5)

Here the term 1 – $$%
 is called lever arm factor
Using As=pbd in (5), the ultimate moment of resistance is computed as

@

&AB

= 0!  ?

; = !(0+) 1 −

!0
)2
)2!0
 -./ ? = (1 −
×
)
)1)3!
!
)1)3

Dividing both sides by fcu we get or

@

!
;
=0 ×
× ? − (6)

+
!

A graph plotted between 
&AB as shown in fig 2.7 and can be used for design

16

2.5 Stress Blocks
Stress blocks adopted by different codes are based on the stress blocks proposed by different
investigators. Among them that proposed by Hog nested and Whitney equivalent rectangular
block are used by most of the codes.
2.5.1 Stress block of IS456 – 2000

Stress block of IS456-2000 is shown in Fig 2.8. Code recommends ultimate strain εcu=0.0035
& strain at which the stress reaches design strength ε0=0.002. Using similar triangle properties
on strain diagram
0.0035
0.002
=

1

∴ = 0.57 → (7)

and x2 =xu-0.57xu=0.43xu

Area of stress block is A=A1+A2.
A = % × 0.45!) ∗ 0.57 + 0.45!) × 0.43


= 0.171 fckxu + 0.1935 fckxu.
A = 0.3645 fckxu 
 (8)
17

Depth of neutral axis of stress block is obtained by taking moment of areas about extreme
compression fiber.
∴ ̅ =

∑ HI'I
∑ HI

0.43
3
0.171!) < × 0.57 + 0.43= + 0.1935!) ×
8
2
̅ =
0.36!)

The stress block parameters thus are

̅ = 0.42 − (9)

K1 = 0.45
K2 = 0.42
K3 =

L.%MN
L.N

- (10)
= 0.81

4.5.5 Analysis of rectangular beam using IS456-2000 stress block
Case 1: Balanced section

18

Balanced section is considered when the ultimate strain in concrete and in steel are reached
simultaneously before collapse.
For equilibrium Cu=Tu
∴ 0.36fckxumax b = 0.87fy Astmax.
xumax =
':H'
A

L.OP
 ( QRST

L.%M
$

=

dividing both sides by

L.OP
 ( Q:H'

L.%M
$

∴0/:H' = <

0/:H' =

&

':H'

':H'
A

A

&A

= ×

× 0.414

From strain diagram

but

( Q :H'

L.%M
$
L.OP



$



−

&A

= ptmasc.

(11)

0.87!V
0.002 +
0.0035
*
=
U3
U3 − 
U3
0.0035
=
− (12)
0.87!V

+
0.0055
*

Values of

':H'
A

is obtained from equation (12). This value depends on grade of steel. Based on

grade of steel this value is given in note of clause 38.1 as (pp70)
Fy

Xumax/d

250

0.53

415

0.48

500

0.46(0.456)

ptmax given in equation (11) is called limiting percentage steel and denoted as pt lim.
To find moment of resistance, the internal moment of Cu & Tu is computed as
19

Mulim = Cu x Z = 0.36fck xulim b (d-0.42 xulim)
From equation (11)

':H'
A

= 2.42





$

ptmax

Mulim = Tu x Z = 0.87fyAst [d-0.42xulim]
Mulim = 0.87fyAst[d-0.42 x 2.42
= 0.87fyAst [1-





$





$

0/U3]

0/:H' ]

;14U
!V ,/
!V
= 0.87
(1 −
0/14U)

!)+
!) +
!)

!V
!V
;14U
=
0.87
0/14U
1
−
0/14U − (13)
!)
!)
!)+ 

From equation 4.5-5-2 ptlim can be expressed as

0/14U!V
U3
= 0.414
−→ (14)
!)


Values of

:]I:


$&AB

&

6Q]I:


$

For different grade of Steel is given in Table (page 10 of SP -16. This table is reproduced in table
2.1.
Table 2.1 Limiting Moment resistance & limiting steel
Fy

250

415


$&AB

:]I:

0.149

0.138

0.133


$

21.97

19.82

18.87

6Q]I: 


500

20

Where ptlim is in%

Now considering Mulim = Cu x Z.
Mulim = 0.36fckxulim b x (d-0.42xulim)

;14U
14U
14U
= 0.36 ×
^1 − 0.42
_ − 15
2
!)+



@]I:

Value of 
$&A is abailalbe in table C of SP16 & Value of
B

@]I:
&AB

for different grade of concrete

and steel is given in Tables. Value of pt lim for different grade of concrete and steel is given in
table E of SP –‘6’. Term
as Qlim

@]I:
&AB

is termed as limiting moment of resistance factor and denoted

∴ Mulim = Qlimbd2.

Case 2: Under reinforced section
In under reinforced section, the tensile strain in steel attains its limiting value first and at this
stage the strain in extreme compressive fiber of concern is less than limiting strain as shown in
Fig 2.10

21

The neutral axis depth is obtained from equilibrium condition Cu=Tu
∴ 0.36fckxub = 0.87fyAst
xu =
or

'
A

= 2.41

L.OP
( Q
L.%M
$ &

= 2.41


 ( Q


$ &A


 ( Q


$

&

− (16)

Moment of resistance is calculated considering ultimate tensile strength of steel ∴ MuR = Tu x Z
or Mur = 0.87fy Ast X (d-0.42xu)
'
A

= 0.87ftAst d(1-0.42

)


 ( Q

= 0.87fyAst d (1-0.42 x 2.41 
$
Considering pt = 100

( Q
&A

@`

Or





$

<

6Q

( Q
&A



= −

LL

)

expressed as % we get

MuR = 0.87fyAst d(1-1.0122
Or L.OP 
&AB =

&A




(

6Q

))


$ LL

(1-1.0122 
$ (LL) , taking 1.0122 ≈ 1

6Q

LL




+

6Q

;?
0/
!V 0/ 
=


−


0.87 !V+
100
!) 100

@`

L.OP 
&AB

= 0 − (17)

Equation (17) is quadratic equation in terms of (pt/100)
Solving for pt, the value of pt can be obtained as

bcb

Pt = 50 a

Pt = 50


$



d.efgh
ijklmB


/
$

^1 − c1 −

o

N.M@`

$&A B

_ − (18)
22

Let Ru =
Pt = 50

N.M@`

$&AB


$



/ℎ.q

r1 − √1 − ?uu

Case 3: Over reinforced section
In over reinforced section, strain in extreme concrete fiber reaches its ultimate value. Such
section fail suddenly hence code does not recommend to design over reinforced section.
Depth of neutral axis is computed using equation 4.5-6. Moment of resistance is calculated
using concrete strength.
∴ MuR = Cu x Z
= 0.36 fck xub (d-0.42xu) – 19
'
A

>

']I:
A

Position of neutral axis of 3 cases is compared in Fig. 2.11

23

Worked Examples
1. Determine MR of a rectangular section reinforced with a steel of area 600mm² on the tension
side. The width of the beam is 200mm, effective depth 600mm. The grade of concrete is
M20 & Fe250 grade steel is used.
Solve: fck = 20Mpa fy = 250Mpa, Ast = 600mm²
Step . 1 To find depth of NA

!V , Q
= 2.41

!$ +
xu = <2.41 ×

L
L

×

MLL

=600 = 90.375mm

LL ×MLL

Step 2 Classification
From clause 38.1, page 70 of IS456,
'
For Fe 250 gwxR
= 0.53, xulim = 0.53 X 600 = 318mm
A

xu < xulim. Hence the section is under reinforced.
Step . 3 MR for under reinforced section.
MR = 0.87fy Ast (d-0.42xu)
=

L.OP ×L ×MLL (MLLbL.N ×zL.%P)

=73.36kN-m.
2.

Le

yb::

LLL ×LLL

Determine the MR of a rectangular section of dimension 230mm X 300mm with a clear
cover of 25mm to tension reinforcement. The tension reinforcement consists of 3 bars of
20mm dia bars. Assume M20 grade concrete & Fe 415 steel.
If cover is not given, refer code – 456 → page 47
1 inch = 25mm → normal construction
Effective depth, d= 300-(25+10)
= 265mm.

Ast = 3 N × 20² = 942.48mm².
{

(1) To find the depth of N.A
xu =2.41×

= 2.41 ×





jk
N


×

×

(|}

×

&A
ML%.O
%L

= 104.916UU
24

ii) For Fe415,

'gwxR
A

= 0.48

xulim = 0.48 X 442 = 212.16mm
xu < xulim
The section is under reinforced.
iii) MR = 0.87fyAst (d-0.42xu)
= 0.87 X 415 X 603.18 (442-0.42X104.916)
= 86.66 kN-m.
3.

Find MR of the section with the following details.
Width of section: 230mm

i)

Overall depth of section: 500mm
Tensile steel: 3 bars of 16mm dia
Grade of concrete: M25
Type of steel : Fe 415
Environmental condition: severe
Solve: b= 230, h= 500mm,fck = 25, fy= 415
From table 16(page 47, IS 456-2000)
Min clear cover (CC) = 45mm
Assume CC 50mm.
Effective depth = 500-(50+8) = 442mm
{
Ast = 3 X N X 16² = 603.18mm²

ii)

For Fe 415,

To find the depth of N-A, xu =2.41
'gwxR
A

= 0.48

N
L

×

zN.NP

%L ×%L

= 204.915UU

xulim = 0.48 X 300 = 144mm.
xu > xulim ∴ over reinforced.

(1) MR = (0.36fckxub) (d-0.42xu).
=60.71kN-m.

4.

A R – C beam 250mm breadth & 500mm effective depth is provided with 3 nos. of 20mm
dia bars on the tension side, assuming M20 concrete & Fe 415 steel, calculate the following:

25

(i) N-A depth (ii) compressive force (iii) Tensile force (iv) ultimate moment (v) safe
concentrated load at mid span over an effective span of 6m.
Solve: d=500mm, b=250mm
{
Ast = 3 X N X20² = 942.48mm²
fck = 20Mpa fy = 415Mpa.

Step – 1

!V
,Q
= 2.41 ×
×

!$
+
= 2,41 ×

N
L

xu = 188.52mm

×

zN.NO

L ×L

Step – 2: For Fe 415

'gwxR
A

× 500

= 0.48 ; xulim = 0.48X 500
= 240mm

∴ xu < xulim, the section is under reinforced.
Cu = 0.36fckxub = 0.36X20X188.52 X 250/10³= 339.34kN.

Step – 3

MR for under reinforced section is

Mu = MR = 0.87fyAst (d-0.42xu)
=

L.OP ×N ×zN.NO (LLbL.N ×OO.)

= 143.1kN-m.

Le

Step – 4
Mu =

~g ×
N

=

~g ×M
N

Equating factored moment to MR
€g ×M
N

= 143.1

Wu = 95.5KN.
€

Safe load, W = .g load factor/factor of safety
= 63.67kN

26

Step .2 Tu =0.87! , Q =

L.OP ×N ×zN.NO
L

= 340.28kN.

Cu ≈ Tu.
5. In the previous problem, determine 2 point load value to be carried in addition to its self
weight, take the distance of point load as 1m.
Solve: Allowable moment,

@
.

=

N%.
.

= 95.4 ‚ƒ − U

Considering self weight & the external load,
M=MD+ML: MD = dead load moment, ML = live load moment, qd = self weight of beam
= volume X density: density = 25kN/m³ for R C C IS 875 –part – 1, plain concrete =
29kN/m³
Volume = b X h X 1m
Let CC= 25mm, Ce = 25 +

L


=3

H= 500+35 = 535mm
„A =

250 × 535
× 1 × 25 = 3.34‚ƒ/U.
1000

„A × 1 
3.34 × 6
; =
=
= 15.03‚ƒ − U.
8
8
M= 95.4= 15.03 + ML
ML = 80.37 kN-m
80.37 = WLX1
WL = 80.37kN
6. A singly reinforced beam 200mm X 600mm is reinforced with 4 bars of 16mm dia with
an effective cover of 50mm. effective span is 4m. Assuming M20 concrete & Fe 215 steel,
let the central can load p that can be carried by the beam in addition to its self weight max 5m
€.

=

N

,
{

Solve: Ast = 4 X N X 16² = 804.25mm²
B=200mm, d=550mm, h=600mm, fck = 20Mpa, fy = 250Mpa.
27

Step (1)

'g
A

xu = 2.41 ×

= 2.41
L
L

×




(|}


$ &A

OLN.
LL

=121.14mm

Step (2)
'gwxR
A

for Fe 250 us 0.53

∴ xulim = 0.53 X 550 = 291.5mm
xu < xulim .’. section is under reinforced.

Step – 3
Mu= MR = 0.87fy Ast[d-0.42xu]

0.87 × 250 × 804.25 †550 − 0.42(121.14)]
10M
Mu=87.308kNm.

M= Allowable moment = .g =
@

OP.%LO
.

= 58.20kN-m

M=MD+ML
qd=self weight of beam
qd = volume X density
density = 25kN/m³ for R C C
Volume = b X h X 2 = 200 X 600 X 2 = 2,40,000m³
qd= 0.2 X 0.6 X 1 X 25 = 3kN/m
qd =

NLLLL ×
(LLL)B

qd= 6KN/m
Md =

‡m ] B
O

28

3 × 4
=
8

Md = 6kN.m
M= 58.2 = 6 + ML
ML= 52.2kN-m
ML=

6.]
N

. ×N

P=

N

P=52.2kN.

7. Determine N-A depth & MR of a rectangular beam of section 300mm X 600mm. The
beam is reinforced with 4 bars of 25mm having an effective cover of 50mm, assume grade
of concrete & steel as M20 & Fe 415 respectively
Solve: b=300mm, h=600mm, d=550mm.
fck = 20Mpa fy=415Mpa, Ast= 1963mm²
Step- 1 Neutral axis depth


!V , Q
2.41 × 415 × 1963
= 2.41
=

!) +
20 × 300
xu = 327.21mm
Step – 2
'gwxR
A

Classification

= 0.48 → xulim = 0.48X550 = 264mm

xu > xulim. Hence the section is over – reinforced.

NOTE: Whenever the section is over reinforced, the strain in steel is less than the ultimate
strain (0.002 +

L.OP


). Hence actual N_A depth has to be computed, by trial and error

concept because in the above equation of xu, we have assumed the stress in steel as yield
stress & this is not true.
29

Step – 3

Actual N-A depth (which lies b/w xᵤ = 327 mm & xulim = 264mm)
Trial 1: Let xu =

ˆ
'g

‰ =

=

MN%P


= 295.5mm

ˆ|

LLb 'g

0.0035
(500 − 295.5)
295.5

= 0.00303
For HYSD bar, the stress strain curve is given in the code, however for definite strain values,
the stress is given in SP – 16

V
360.9 − 351.8
=
V  = 2.36
(0.00303 − 0.002 + 6)
0.0038 − 0.00276
fs=351.8+y1 = 351.8 + 2.36 = 3.54.16Mpa

Equating compressive force to tensile force,
C u = Tu
0.36fckxub= fsAst
 =

%N.M ×zM%

L.%M ×L ×%LL

= 321.86mm

Compared to the earlier computation, this value is less than 327. However to confirm we have to
repeat the above procedure till consecutive values are almost same.
Trial 2 Let xu =

z%.O


≈ 308mm

‹b
'g

= 317.7

Repeat the computation as in trial 1.

‰
‰
=

550 − 
‰ =

0.0035
(550 − 308)
308
30

For HYSD bars, the stress strain curve is given in the code, however for definite strain values,
the stress is given in SP-16.
V
351.8 − 342.8
=
0.00275 − 0.00241
0.00276 − 0.00241
V  = 8.74

fs= 342.8 + 8.74 = 351.54 Mpa

Equating compressive forces to tensile forces,
Cu=Tu
0.36fckb xu = fs Ast
0.36 X 20 X 300 X xu =351.54 X 1963.
xu=319.48mm
Compared to earlier computation this value is lesser than 321.8. However to confirm we have to
repeat the above procedure till consecutive values are almost same.
Trial 3 Let xu =

‰
‰
=

550 − 
‰ =

%LO%z.NO


≈ 313.74mm

0.0035
(550 − 313.74)
313.74

= 0.00263

For HYSD bars, the stress strain curve is given in the code, however for definite strain values,
the stress is given in SP-16.
! =

351.8 − 342.8
(0.00263 − 0.00241) + 342.8
0.00276 − 0.00241

! = 348.46

Equating compressive forces to tensile forces,

Cu=Tu
0.36fckxub = fs Ast
31

0.36 X 20 X 300 X xu =348.46 X 1963.
xu=316.7mm

Step – 4

Dia
(mm)
8
10
12
16
20
25

MR

Area
50
78.5
113
201
314
490

MUR=0.36fckxub(d-0.92xu)
=

0.36 × 20 × 300 × 375(550 − 0.42 × 315)
1 × 10M

=284.2kN-m

A rectangular beam 20cm wide & 40cm deep up to the center of reinforcement. Find the
reinforcement required if it has to resist a moment of 40kN-m. Assume M20 concrete & Fe 415
steel.
NOTE: When ever the loading value or moment value is not mentioned as factored load, assume
then to be working value. (unfactored).
Solve:

b=200mm, d=400mm, fck=20Mpa, fy = 415Mpa
M= 40kN-m, Mᵤ= 1.5X40= 60kN-m = 60X10⁶
Mu= MuR =0.87fy X Ast X (d-0.42xu) → ①

!V , Q

= 2.41
!$ +

 =

2.41 × 415 × , Q
= 0.25,
20 × 200

Q

Substituting in 1.

32

60 X 10⁶ = 0.87 X 415 X Ast (400-0.42 X 0.25 X Ast)
60 X 10⁶= 1,44,420 Ast-37.91 Ast²
Ast= 474.57mm² [ take lower value → under reinforced]

•

In beams, dia of reinforcement is taken above 12.
Provide 2-#16 & 1-#12
{

{

N

N

(Ast) provided = 2 X X 16² + 2 X X 12² = 515.2mm² > 474.57mm²

Check for type of beam
 = 2.41

!V , Q
!) +

= 2.41 ×

N
L

×

.
LL

= 128.82UU

From code, xumax = 0.48d = 192mm : xu < xumax
For M20 & Fe 415
∴ Section is under reinforced.
Hence its Ok . . . .

9. A rectangular beam 230mm wide & 600mm deep is subjected to a factored moment of 80kNm. Find the reinforcement required if M20 grade concrete & Fe 415 steel is used.

Solve: b=230mm, h=600mm, Mu=80kN-m,fck = 20Mpa, fy=415Mpa,Ce=50mm,
= 80X10⁶N-mm
Mu=MuR = 0.87fy Ast (d-0.42xu) →①


!V , Q
= 2.41

!) +
 =

2.41 × 415, Q
= 0.217,
20 × 230

Q

33

Substituting in ①
80 X 10⁶ = 0.87X415XAst ( 0.42 X 0.217Ast)

Procedure for design of beams
1. From basic equations.
Data required: a) Load or moment & type of support b) grade of concrete & steel.

Step . 1 If loading is given or working moment is given, calculate factored moment. (Mu)

Œ1 
; =
2

1.5(Œ + Œ1)1 
; =
8

Step – 2 Balanced section parameters
xulim, Qlim, ptlim, (table A to E SP 16)

@

Qlim = &AgB

Step – 3 Assume b and find
dlim = c

@g

wxR ×&

Round off dlim to next integer no.
h=d+Ce : Ce = effective cover : assume Ce.
∅

Ce= 25mm: Ce = Cc + 

34

Step – 4 To determine steel.
Find Pt = ? =
Ast =

( Q

6} &A
LL

0Q =

N.M@g


jk &AB

LL(|}
&A

: assume suitable diameter of bar & find out no. of bars required.

N=  : ast = area of 1 bar.
|}

Using design aid SP16
Step 1 & Step 2 are same as in the previous case i.e using basic equation
@

Step – 3 Find &AgB and obtain pt from table 1 to 4 → page 47 – 50
Which depends on grade of concrete.
From pt, calculate Ast as Ast =

6} .&.A
LL

Assuming suitable diameter of the bar, find the no. of re-bar as N=
where Qst = N ∅
{

(|}
|}

NOTE: 1 . To find the overall depth of the beam, use clear cover given in IS 456 – page 47
from durability & fire resistance criteria.

2. To take care of avoiding spelling of concrete & unforcy tensile stress, min. steel has to be
provided as given in IS 456 – page 47

(|

&A

=

L.O



If Ast calculated, either by method 1 or 2 should not be less than (As)min .
If

Ast < Asmin :

Ast = Asmin

1. Design a rectangular beam to resist a moment of 60kN-m, take concrete grade as M20 & Fe
415 steel.
Solve: M = 60 KN-m
Mu = 1.5 X 60 = 90kN-m

fck = 20Mpa , fy = 415 Mpa

35

Step 1: Limiting Design constants for M20 concrete & Fe 415 steel.
gRST
A

= 0.48 : From Table – c of SP-16, page 10

Qlim = 2.76
Pt lim = 0.96

column sizes 8 inches= 200mm
9 inches = 230mm

Step - 2 : &H] = c

@g

wxR ×&

Let b = 230mm
] = ‘

90 × 10M
= 376.5UU
2.76 × 230

Referring to table 16 & 16 A, for moderate exposure & 1½ hour fire resistance, let us assume
clear cover Cc=30mm & also assume 16mm dia bar ∴ effective cover Ce = 30 + 8 = 38mm
hbal =376.5 + 38 = 414.5

18 inches = 450mm

Provide overall depth, h= 450mm.
‘d’ provided is 450-38 = 412mm

Step .3 Longitudinal steel
Method . 1 → using fundamental equations
Let pt be the % of steel required
0Q =
=

L
$



L ×L
N

r1 − √1 − ?u; ? = 


N.M@

r1 − √1 − 0.53u

jk &A

B

=

N.M ×zL × Le

L ×%L × NB

= 0.53

= 0.758
→ using SP 16. Table – 2
Method .2
Specified in problem

page – 47

use this if it is not

36

;
90 × 10M
‚=
=
= 2.305
+ 
230 × 412
K=2.3 → pt = 0.757

K= 2.32 → pt = 0.765
For k = 2.305, pt= 0.757+

(L.PMbL.PP)
(.%b.%)

× (2.305 − 2.3)

= 0.759

Step 4 : detailing
Area of steel required, Ast =

6} &A
LL

=

L.PM ×%L ×N
LL

= 720mm²

M20 → combination 12mm & 20mm aggregate (As) size.
Provide 2 bars of 20mm & 1 bar of 12mm,
∴ Ast provided = 2 X ¾ + 113
= 741 > 720mm²
[ To allow the concrete flow in b/w the bars, spacer bar is provided]
•
•
•

1.Design a rectangular beam to support live load of 8kN/m & dead load in addition to its self
weight as 20kN/m. The beam is simply supported over a span of 5m. Adopt M25 concrete &
Fe 500 steel. Sketch the details of c/s of the beam.
Solve: qL =8kN/m b=230mm „A = 20kN/m fck = 25Mpa fy = 500Mpa. l= 5m=5000mm
Step .1: c/s

NOTE; The depth of the beam is generally assumed to start with based on deflection criteria of
]

serviceability. For this IS 456-2000 – page 37, clause 23.2-1 gives A = 20
with some correction factors. However for safe design generally l/d is taken as 12
]



= =

LLL


= 416mm. (no decimal)

Let Ce = 50mm, h=416+50 = 466mm

37

Step 2 Load calculation
i) Self weight =0.23 X 0.5 X 1 X 25 = 2.875kN/m =„A

ii) dead load given „A = 20kN

qd= „A + „A = 22.875kN/m

25kN/m [multiple of 5]

[Take dead load as x inclusive of dead load, don’t mention step . 2]

iii) Live load = 8kN/m

; =

‡m × ] B
O

Dead load

=

 × B
O

= 78kN-m(no decimal)

Live load moment =

‡w × ] B
O

=

O × B
O

= 25‚ƒ − U

Mu= 1.5MD+1.5ML =1.5 X 78 + 1.5 X 25 = 154.5kN-m

Step -3 Check for depth
dbal =c

@g

wxR ×&

From table.3, Qlim = 3.33 (page – 10) SP-16
(h bal < h assumed condition for Safe design)
154.5 × 10M
= ‘
= 449.13
3.33 × 230
Hbal = 449.13 + 50 = 499.13
Hence assumed overall depth of 500mm can be adopted.
Let us assume 20mm dia bar & Cc=30mm (moderate exposure & 1.5 hour fire resistance)
∴ Ce provided = 30+10 = 40, dprovided = 500-40 = 460mm
Step .4 Longitudinal steel
38

;
154.5 × 10M
=
= 3.17
+ 
230 × 460
Page 49 – SP-16

Mu/bd²

pt
3.15
3.20

@g

&AB

For

= 3.17, 0Q = 0.880 +

(L.OzObL.OOL)
(%.Lb%.)

0.880
0.898
× (3.17 − 3.15)

= 0.8872

,Q=

0Q +
0.8872 × 230 × 460
=
= 938.66UU
100
100

No. of # 20 =

z%O.MM
%N

= 2.98 ≈ 3 ƒ5.

(Ast)provided = 3 x 314 = 942mm2 > 938.66mm2.
Page – 47 – IS456
(Ast)min =

L.O&A



=

L.O ×%L ×NML
LL

= 179UU

∴( Ast)provided > (Ast)min . Hence o.k.
Step . 5 Detailing

3. Design a rectangular beam to support a live load of 50 kN at the free end of a cantilever beam
of span 2m. The beam carries a dead load of 10kN/m in addition to its self weight. Adopt M30
concrete & Fe 500 steel.
l=2m=2000mm, „A =10, fck= 30Mpa, fy=500Mpa, qL=50kN, b=230mm

Step – 1

c/s

39

NOTE: The depth of the beam is generally assumed to start with based on deflection criteria of
serviceability. For this IS 456-2000-Page – 37, clause 23.2.1 gives
]

A

]

= 5 with some correction factor.

= d → d=

LLL


= 400mm

Let Ce=50mm, h=400 + 50 = 450mm
However we shall assume h=500mm

Step – 2 Load calculation.
(i)
(ii)

(iii)

Self wt= 0.23 X 0.5 X 1 X 25 = 2.875kN/m = „A
Dead load given = „A =10kN/m

qd=„A + „A = 10+2.875 = 12.87kN/m ≈ 15kN/m [multiply of5]
Live load = 50kN

„ 1
15 × 2
; =
=
= 50‚ƒ − U
2
2
ML=W X 2 = 50 X 2 = 100kN-m
Mu = 1.5MD +1.5ML = 195kN-m = 88

Step – 3 Check for depth
dbal = c

@g

wxR ×&

From table – 3,SP-16 (page – 10) Qlim = 3.99

= c%.zz ×%L = 460.96mm
z × Le

hbal = 460.96+50= 510.96mm
hassumed = 550mm.
Let us assume 20mm dia bars & Cc=30mm constant
∴ Ce provided = 30+10 = 40, dprovided= 550-40 =510mm
Step – 4 Longitudinal steel
40

;
195 × 10M
=
= 3.26
+ 
230 × 510
Page – 49m SP-16m

Mu/bd²
3.25
3.30
For Mu/bd² = 3.26,
Pt = 0.916 +

(L.z%bL.zM)

= 0.9198.
Ast =

6} &A
LL

=

(Ast)Provided =
(Ast)min =

(%.%Lb%.)




× (3.26 − 3.25)

L.zzO ×%L ×L

L.O&A

LL

=

Step – 5 Detailing

pt
0.916
0.935

= 1078.9mm²
3-# 20, 1-#16 = 1143
2- #25 2=#20 = 1382

L.O ×%L ×L
LL

= 199.41UU

Design of slabs supported on two edges

Slab is a 2 dimensional member provided as floor or roof which directly supports the loads in
buildings or bridges.

In RCC, it is reinforced with small dia bars (6mm to 16mm) spaced equally.

41

Reinforcement provided in no. RCC beam → 1 dia width is very (12mm – 50mm) small
compared to length. Element with const. Width: fixed width.
It is subjected to vol.
RCC slabs → reinforcement are provided with equally spaced. No fixed width & length are
comparable, dia -6mm to 16mm. It is subjected to pressure.
Beams are fixed but slabs are not fixed. For design, slab is considered as a beam as a singly
reinforced beam of width 1m

Such slabs are designed as a beam of width 1m & the thickness ranges from 100mm to 300mm.
IS456-2000 stipulates that

]

A

for simply supported slabs be 35 & for continuous slab 40 (page –

39). For calculating area of steel in 1m width following procedure may be followed.
Ast = N ×

{
N

× ∅

; ,

Q

=

LLL × |}

ƒ=

LLL

For every 10cm there is a bar
H
S=(|} × 1000
|}

(The loading on the slab is in the form of pr. expressed as kN/m²)
As per clause 26.5.2-1 (page 48 min steel required is 0-15% for mild steel & 0-12% for high
strength steel. It also states that max. dia of bar to be used is 1/8th thickness of the slab. To
calculate % of steel we have to consider gross area is 1000Xh.
The slabs are subjected to low intensity secondary moment in the plane parallel to the span.
To resist this moment & stresses due to shrinkage & temp, steel reinforcement parallel to the
span is provided. This steel is called as distribution steel. Min. steel to be provided for
distribution steel.
Practically it is impossible to construct the
Slab as simply supported bozo of partial bond

42

b/w masonry & concrete, also due to the parapet wall constructed above the roof slab. This
induces small intensity of hogging BM. Which requires min. % of steel in both the direction
at the top of the slab as shown in fig. 2 different types of detailing is shown in fig.
Method – 1

Crank → for the change in reinforcement.

(1) Alternate cranking bars.
Dist. Steel is provided.
• To take care of secondary moment, shrinkage stresses & temp. stress steel is
provided parallel to the span.
Method – 2

1. Compute moment of resistance of a 1- way slab of thickness 150mm. The slab is reinforced
with 10mm dia bars at 200mm c/c. Adopt M20 concrete & Fe 415 steel. Assume Ce=20mm.
Solve: h=150mm,Ce=20mm, ф=10mm
d=150-20 =130mm, sx=200mm
Ast =

LLL

=

'

LLL
LL

×

×

{
N

{
N

× ∅

× 10
43

= 392.7mm²

44

Step - 1

N-A depth

xu = 2.41 ×

N


!V ,/
= 2.41
!) +

L

×

%z.P
LLL

=19.64mm.
xumax = 0.48d= 0.48X130 = 62.4mm.

Step – 2 Moment of resistance
xu < xumax. hence section is under reinforced.
MuR= 0.87fy Asta(d-0-42xu)

0.87 × 415 × 392.7 (130 − 0.42 × 19.64)
10M
=17.26 kN-m/m.

Doubly reinforced Beams.

Limiting state or Balanced section.
Cuc = 0.36fck bxulim
Tu1= 0.87fyAst
Mulim = 0.36fckbxulim(d-0.42xulim)
ptlim = 0.414

'gwxR
A

×


$



45

Ast =

“}wxR
LL

× +

Mu > Mulim

Mu = applied factored moment.

Mu₂= Mu-Mulim
For Mu₂ we require Ast in compression zone & Ast₂ in tension zone for equilibrium
Cus = Tu₂
Cus = fsc .Asc : Fsc is obtained from stress – strain curve of corresponding steel.
In case of mild steel, it is fsc =




.

= 0.87!V

In case of high strength deformable bars,
fsc corresponding to strain εsc should be obtained from table A-SP-16(Page – 6)
Z₂= d-d1

If εsc< 0.00109,
fsc = εsXεsc
else fst = fy/1.15

ˆ

'gwxR

=

ˆ 

'gwxR bA”

‰(]I: −  )
‰ =
]I]:

Tu2 = 0.87fyAst2 
 1

Couple Mu₂=Cus XZ₂ = Tu₂ Z₂
Mu₂ =(fsc Asc)(d-d1)
,



=

;
→2
!( −  )

Cus = Tu₂

fsc X Asc = 0.87fy Ast₂
,

Q

=

•,
→3
0.87!V
46

Procedure for design of doubly reinforced section.

Step. 1 Check for requirement of doubly reinforced section
1. Find xulim using IS456
2. Find Mulim for the given section as Mulim = QlimXbd² Refer table – D, page – 10,
of SP – 16 for Qlim
3. Find ptlim from table – E page 10 of SP – 16 & then compute.
0Q]I: × + × 
100
Step – 2 If Mu >Mulim then design the section as doubly reinforced section, else design as
singly reinforced section.

,Q=

Mu₂ = Mu - Mulim

Step – 3

Step – 4 Find area of steel in compression zone using the equation as
Asc = 


@gB

(AbA” )

fsc has to be obtained from stress – strain curve or from table – A, page – 6

of SP-16.

The strain εsc is calculated as

ˆjg ('gwxR bA” )
'gwxR

Step - 5 Additional tension steel required is computed as
,

Q

=


|j (|j

L.OP


∴ Total steel required Ast = Ast₁+Ast₂ in tension zone.

Use of SP-16 for design of doubly reinforced section table – 45-56, page 81-92 provides pt & pc:
pt =

( Q
&A

Pc =

× 100

( 
&A

X100 for different values of Mu/bd² corresponding to combination of fck & fy.

Following procedure may be followed.
Step – 1 Same as previous procedure.

47

Step – 2 If Mu > Mulim , find Mu/bd² using corresponding table for given fy & fck obtain pt &
pc . Table – 46.

NOTE: An alternative procedure can be followed for finding fsc in case of HYSD bars i.e use
table – F, this table provides fsc for different ratios of dᶦ/d corresponding to Fe 415 & Fe 500
steel
Procedure for analysis of doubly reinforced beam
Data required: b,d,d1,Ast(Ast₁+Ast₂), Asc,fck,fy
Step – 1 Neutral axis depth
Cuc+Cus = Tu
0.36fck < bxu+fsc Asc = 0.87fy. Ast.
xu =

L.OP
(|} b
|j (|j
L.%M
jk &

This is approximate value as we have assumed the tensile stress in tension steel is 0.87fy
which may not be true. Hence an exact analysis has to be done by trial & error. (This will be
demonstrated through example).
Step – 2: Using the exact analysis for N-A depth the MR can be found as.
Mulim = 0.36fck .b xu(d-0.4 2xu)+fst Asc X(d2-d1)
1. Design a doubly reinforced section for the following data.
b=250mm, d= 500mm, d1=50mm, Mu=500kN-m con-, M30, steel = Fe 500.
A”
A

= 0.1, fck = 30Mpa, fy= 500Mpa.

Mu = 500 X 10⁶ N-mm

Step – 1 Moment of singly reinforcement section.
'
Page – 70 –IS-456 gwxR
= 0.46
A
xulim = 0.46 X 500 = 230mm.

48

Mulim = Qlim X bd² =

%.zz ×L × LLB
Le

Page – 10 SP – 16
Ptlim = 1.13: Ast₁ =Astlim =

= 249kN-m.

.% ×L ×LL

Mu > Mulim

LL

= 1412.5mm²

Step – 2 Mu2= Mu-Mulim = 500-249 = 251kN-m.
Asc =

=

@gB


|j (AbA” )

 × Le

N(LLbL)

page – 13. SP-16. Table – F

= 1353.83mm²

From equilibrium condition,
! , 
412 × 1353.83
, Q =
=
= 1282.25UU
0.87!V
0.87 × 500
Ast = Ast₁ + Ast₂ = 1412.5 + 1282.25 = 2694.75mm²
Step – 3: Detailing.
Asc = 1353mm²
Ast = 2694mm²
Tension steel
Assume # 20 bars =

MzN
%N

= 8.5

However provide 2 - #25 + 6 - #20
(Ast)provided = 2 X 490 + 6 X 314 = 2864mm² > 2694mm²
Compression steel
Assume #25, No =

%%
NzL

= 2.7

Provide 3 - #25 (Ast)provided = 3 X 490 = 1470mm² > 1353mm²
Ce = 30 + 25 + 12.5 = 67.5mm

2. Design a rectangular beam of width 300mm & depth is restricted to 750mm(h) with a effective
cover of 75mm. The beam is simply supported over a span of 5m. The beam is subjected to
central con. Load of 80kN in addition to its self wt. Adopt M30 concrete & Fe 415 steel.
Wd = 0.3 X 0.75 X 1 X 25 = 5.625
MD =

€AB
O

= 17.6kN-m

49

ML=

~w ×]
N

= 100kN-m

Mu = 1.5(MD+ML) = 176.4

3. Determine areas of compression steel & moment of resistance for a doubly reinforced
rectangular beam with following data.
b= 250mm, d= 500mmm, d1=50mm, Ast = 1800mm², fck = 20Mpa, fy= 415Mpa. Do not neglect
the effect of
Compression reinforcement for calculating Compressive force.

Solve: Cc₁ → introduce a negative force
Note: Cu = Csc + Cc - Cc₁
= fsc Asc + 0.36fck bxu - 0.45fck X Asc
= 0.36fck bxu + Asc (fsc – 0.45fck)
For calculating compressive force then,
Whenever the effect of compression steel is to be considered.

Step – 1 Depth of N-A.
From IS – 456 for M20 concrete and Fe 415 steel is
'gRST
A

= 0.48 & xumax = 0.48 X 500 = 240mm

From table – 6 , SP – 16 page- 10, ptlim = 0.96
Ast₁ = Astlim =

L.zM ×L ×LL
LL

= 1200mm²

Ast₂ = Ast- Ast₁ = 1800 – 1200 = 600mm²

Step – 2 : Asc
For equilibrium, Cu = Tu.
In the imaginary section shown in fig.
50

Cu1 = Asc(fsc-0.45fck)
A”
A

=

LL
LL

= 0.1, Table – F, SP – 16, P – 13, fsc = 353Mpa

Ast₂
Asc(353 - 0.45 X 20) = 600 X 0.87 X 415
Asc = 629mm²
Asc(fsc - 0.45fck) (d – d1)

Step – 3 MR
Mur₂ = Cu1 x Z₂ =

Mz – (%%bL.N–L) – (LLbL)
L⁶

= 97.6kN-m
Mur₁ = Mulim = Qlimbd²
=

Qlim = 2.76
.PM–L–LL²
L⁶

= 172kN-m.

Mur = Mur1 + Mur2 = 97.6 + 172 = 269.8kN-m
Asc = 629mm², Mur = 269.8kN-m

4. A rectangular beam of width 300mm & effective depth 550mm is reinforced with steel of area
3054mm² on tension side and 982mm² on compression side, with an effective cover of 50mm.
Let MR at ultimate of this beam is M20 concrete and Fe 415 steel are used. Consider the effect
of compression reinforcement in calculating compressive force. Use 1st principles No. SP – 16.
Solve: Step: 1 N-A depth
˜™š›œ
A

= 0.48

xulim = 0.48 X 550 = 264mm

Assuming to start with fsc = 0.87fy = 0.87 X 415 = 361Mpa.
Equating the total compressive force to tensile force we get,
Cu = Tu = Cu = 0.36fck b xu + Ast(fsc – 0.45fck)
Tu = 0.87fy Ast.

51

xu =

L.OP – N – %LN – zO(%MbL.N – L)
L.%M – L – %LL

→1

=350.45 mm > xulim.
Hence the section is over reinforced.
The exact N-A depth is required to be found by trial & error using strain compatibility, for which
we use equation ① in which the value of fsc is unknown, hence we get,
xu =

žŸ – %LNbzO(žŸ¡bL.N–L)

=

L.%M – L – %LL

%LNžŸ – zOžŸ¡  OPPz
P

→②

Range of xu & 264 to 350.4
Cycle – 1 Try (xu)₁ =

MN%L.N


= 307mm
L

From strain diagram εsc = 0.0035(1- %LP) = 0.00293
εst =

L.LL% – L
%LP

(similar triangle)

– 1) = 0.00279

fsc = 339.3mm. The difference b/w (xu)₁ & (xu)₂ is large, hence continue cycle 2.
Cycle – 2 Try (xu)₃ =

%LP%%z.%


= 323mm
L

From strain diagram, εsc = 0.0035(1 - %%) = 0.00296.
L

εst =0.0035(%% – 1) = 0.00246
fsc =353.5, fst= 344.1, (xu)₄ = 328
The trial procedure is covering, we shall do 1 more cycle.
Cycle – 3 Try (xu)₅ =

%%%O


= 325.5mm.

From strain diagram, εsc= 0.0035(1 − %.) = 0.00296.
L

L

εst = 0.0035( %. – 1) = 0.00241
fsc = 353.5, fst = 342.8. (xu)₆ = 326.2
∴ xu = 326.2mm.
52

Step . 2
Mur =

(L.%M – L – %LL – %M) (L – L.N – %M)zO(%%. – L.N – L)(¢b¢ᶦ)(LbL)
L⁶

=463kN-m

5. Repeat the above problem for Fe 450.
Note: xu < xumax, hence cyclic procedure is not required.
xu = 211.5, Mur = 315kN-m.

6. A rectangular beam of width 300mm & effective depth of 650mm is doubly reinforced with
effective cover d1= 45mm. Area of tension steel 1964, area of compression steel = 982mm². Let
ultimate MR if M- 20 concrete and Fe 415 steel are used.
Ans: xu= 11734mm, Mur = 422kN-m
Flanged sections.
T- section

•
•
•

L- section

Concrete slab & concrete beam are
Cast together → Monolithic construction
Beam → tension zone, slab in comp. zone
Slab on either side → T beam
Slab on one side → 1 beam

•

bf → effective width bf>b

T- beam

53

bf =

š¤
M

+ bw + 6Df
•
•

lo=0.7le: continuous & frames beam.
A & B paints of contra flexure (point of zero moment)

L-Beam
bf =

š¤



+ bw + 3Df

Isolated T- beam

•
•

It is subjected to torsion & BM
If beam is resting on another beam it can be called as L – beam.
If beam is resting on column it cannot be called as L- beam. It becomes –ve
beam.

Analysis of T – beam :- All 3 cases NA is computed from Cu = Tu.

Case-1 – neutral axis lies in flange.

54

Case (ii) : NA in the web &




A

≤ 0.2

Whitney equivalent rectangular stress block.
Case (iii): NA lies in web &




A

> 0.2

1. All three cases NA is computed from Cu = Tu.
2. xulim same as in rectangular section.
• Depth of NA for balanced s/n depends on grade of steel.
3. Moment of resistance
Case . (1) Cu = 0.36fckbfxu
Tu = 0.87fy Ast
Mur = 0.36fck bf xu(d-0.42xu) or 0.87fyAst(d - 0.42xu)
Case – (ii) Cu = 0.36fckbwxu + 0.45fck(bf-bw)Df (d 1






)

2

Tu = 0.87fyAst
Mur = 0.36fck bw xu(d - 0.42xu) + 0.45fck(bf - bw)Df (d -

¥ž


)

Case (iii)
Cu = 0.36fck bwxu + 0.45fck(bf – bw)yf.
Tu = 0.87fyAst.
Page . 97,

Mur = 0.36fckbwxu(d-0.42xu) + 0.45fck(bf-bw) yf (d -

¦



)

Where, yf = 0.15 xu + 0.65Df
Obtained by equating areas of stress block.
55

•

When Df/xu ≤ 0.43 & Df/xu > 0.43 for the balanced section & over reinforced section use
xumax instead of xu.

Problem.
1.

Determine the MR of a T – beam having following data.
a) flange width = 1000mm = bf
b) Width of web = 300mm = bw
c) Effective depth = , d= 450mm
d) Effective cover = 50mm
e) Ast = 1963mm²
f) Adopt M20 concrete & Fe 415 steel.

Solve: Note:
In the analysis of T – beam, assume N-A
To lie in flange & obtain the value of
xu. If xu > Df then analyses as
case (2) or (3) depending on the
ratio of Df/d. If Df/d ≤ 0.2 case (2)
or Df/d > 0.2 case (3)

Step – 1
Assume NA in flange
Cu= 0.36fck bf xu Tu = 0.87fy Ast
C u = Tu .
0.36fck bf xu = 0.87fy Ast
xu =

L.OP – N – zM%
L.%M – L – LLL

= 98.4mm < Df = 1000mm.
Assumed NA position is correct i.e(case . 1)

Step – 2 : Mur = 0.36fck bf xu(d-0.42xu)
= 0.36 X 20 X 1000 X 98.4(450 - 0.42 X 98.4)/10⁶

56

= 290 kN-m
Or Mur = 0.87fy Ast (d - 0.42xu)
= 0.87 X 415 X 1963(450 - 0.42 X 98.4)
= 290kN-m
Use of SP 16 for analysis p: 93 - 95
For steel of grade Fe 250, Fe 415 & Fe 500, SP – 16 provides the ratio
of




A

&


&~

and

§™

ž¨© ª« ¢²

for combinations

using this table the moment of resistance can be calculated as Mu = KTfckbwd² where

KT is obtained from SP -16.
Solve:

¥ž
¢

=

ªž

ª¬

LL

NL

=

= 0.22 > 0.2

LLL
%LL

= 3.33

For Fe 415m P:94
ªž

ª¬

KT

3

4

0.309 0.395
ªž

KT for ª¬ = 3.3 ⇒ 0.309 +

(L.%zbL.%Lz)

= 0.337

Mulim =

L.%%P – L – %LL – NL²
L⁶

(Nb%)

X (3.3-3)

= 410kN-m.

This value corresponds to limiting value. The actual moment of resistance depends on quantity
of steel used.
2. Determine area of steel required & moment of resistance corresponding to balanced section of
a T – beam with the following data, bf = 1000, Df = 100mm, bw = 300mm, effective cover =
50mm, d= 450mm, Adopt M20 concrete & Fe 415 steel.
Use 1st principles.
Solve: Step –1
Step – 2

¥ž
¢

LL

= NL = 0.22 > 0.2 case (iii)

yf = 0.15 xumax + 0.65Df

Cu = 0.36fck bw xumax + 0.45fck(bf - bw)yf.
Tu = 0.87fy Astlim
For Fe 415, xumax = 0.48d = 216mm > Df.
57

Cu= 0.36 X 20 X 300 X 216 + 0.45 X 20(1000 - 300)97.4 = 1.0801 X 10⁶Nmm
Yf = 0.15 X 216 + 0.65 X 100 = 97.4mm
Cu=Tu
Cu= 0.87 X 415 Ast.
1. 0801 X 10⁶ = 0.87 X 415 Astlim
Astlim = 2991.7mm²

Step – 3 Mur = 0.36fck bw xumax(d - 0.42xumax) + 0.45fck(bf-bw)yf(d zP.N


¦ž


).

= 0.36 X 20 X 300 X 216 (450-0.42 X 216) + 0.45 X 20 (1000-300) 97.4 (450 -

)

Mur = 413.27kN-m

3. Determine M R for the c/s of previous beam having area of steel as 2591mm²
Step – 1

Assume N-A in flange

Cu= 0.36fckbfxu
Tu= 0.87 fy Ast
For equilibrium Cu = Tu
0.36 X 20 X 1000 xu = 0.87 X 415 X 2591
xu = 129.9mm > Df
∴ N-A lies in web
¥ž
¢

LL

= NL = 0.22 > 0.2

case (iii)

∴ yf = 0.15xu + 0.65Df = 0.15 X xu + 0.65 X 100 = 0.15 xu + 65
C u = Tu
Cu=0.36 fck bw xu + 0.45fck (bf- bw)yf =
Tu= 0.87fyAst = 0.87 X 450 X 2591 = 1014376.5
1014376.5 = 0.36 X 20 X 300 xu + 0.45 X 20(1000-300)84.485

58

xu = 169.398mm < xumax = 0.48 X 450 = 216mm
It is under reinforced section.

Step – 2 MR for under reinforced section depends on following.
(1)
(2)

¥ž

¢
¥ž

˜™

= 0.22 > 0.2 (case iii)
=

LL

Mz.%zO

= 0.59 > 0.43

Use yf instead of Df in computation of MR
∴ yf = 0.15xu + 0.65Df
= 0.15 X 169.398 + 0.65 X 100
= 90.409mm.
Mur = 0.36fck xu bw (d-0.42xu) +0.45fck (bf – bw) yf (d -

¦ž


)

= 0.36 X 20 X 169.398 X 300 (450-0.42 X 169.398) + 0.45 X 20 (1000-300) 90.409 (450-

zL.NLz


)

= Mur = 369.18kN-m

Design procedure
Data required:
1. Moment or loading with span & type of support
2. Width of beam
3. Grade of concrete & steel
4. Spacing of beams.

Step – 1 Preliminary design
From the details of spacing of beam & thickness of slabs the flange width can be calculated from
IS code recommendation.
bf =

š¤
M

+ bw + 6Df.

P-37 : IS 456

59

Approximate effective depth required is computed based on l/d ratio d ≈

š



to

]



Assuming suitable effective cover, the overall depth, h= d + Ce round off ‘h’ to nearest 50mm
integer no. the actual effective depth is recalculated, dprovided = h- Ce
Approximate area of steel i.e computed by taking the lever arm as Z= d -

¥ž


Mu = 0.87fyAst X Z
Ast =

§™

­®
B

L.OPž¦(¢b )

Using this Ast, no. of bars for assumed dia is computed.
Round off to nearest integer no. & find actual Ast.
NOTE: If the data given is in the form of a plan showing the position of the beam & loading on
the slab is given as ‘q’ kN/m² as shown in the fig.

W=q X S X 1 also, bf ≤ s

Step – 2

N –A depth

The N-A depth is found by trial procedure to start with assume the N-A to be in the flange. Find
N-A by equating Cu & Tu if xu < Df then NA lies in flange else it lies in web.

In case of NA in the web then find

¥ž
¢

. If

¥ž
¢

≤ 0.2, use the equations for

Cu & Tu as in case (II) otherwise use case (III).
Compute xulim & compare with xu. If xu > xulim increase the depth of the beam & repeat the
procedure for finding xu.

Step . 3

Moment of resistance
60

Based on the position of NA use the equations given in cases (I) or case (II) or case (III) of
analysis. For safe design Mur > Mu else redesign.

Step . 4

Detailing.

Draw the longitudinal elevation & c/s of the beam showing the details of reinforcement.
1.

Preliminary design.

Step . 1
bf =
=

š¤
M

+ bw + 6Df

LLL
š¯

Design a simply supported T – beam for the following data. (I) Factored BM =
900kN-m (II) width of web = 350mm (III) thickness of slab = 100mm (IV)spacing
of beams = 4m (V) effective span = 12m (VI) effective lover = 90mm, M20 concrete
& Fe 415 steel.

M

lo = le = 12000mm

+ 350 + 6 X 100 = 2950 < S = 4000
š¯

h ≈  to  (1000 to 800mm)
Assume h = 900mm
dprovided = 900-90 = 810mm
Approximate Ast =

@g

L.OP
(Ab

mi
B

)

=

zLL–L⁶

L.OP–N(OLb

”°°
)
B

= 3279mm²

Assume 25mm dia bar.
No. of bars =

%Pz
NzL

≈ 6.7

Provide 8 bars of 25mm dia
(Ast)provided = 8 X 491 = 3928mm²
xumax = 0.48 X 810 = 388.8
61

Step. 2 N-A depth, Assume NA to be in flange.
Cu = 0.36fck xu bf ; Tu = 0.87fy Ast.
0.36 X 20 X xu X 2950 = 0.87 X 415 X 3927.
xu = 66.75 < Df < xumax
hence assumed position of N-A is correct.
Step . 3 .

Mur = 0.36fckbfxu(d-0.42xu)
= 0.36 X 20 X 2950 X 66.75(810-0.42 X 66.75)
Mur = 1108.64kN-m > 9.01kN-m Mu
Hence ok

Step . 4

Detailing

2. Design a T-beam for the following data. Span of the beam = 6m (effective) & simply
supported spacing of beam -3m c/c, thickness of slab = 120mm, loading on slab -5kN/m²
exclusive of self weight of slab effective cover = 50mm, M20 concrete & Fe 415 steel. Assume
any other data required.

3. A hall of size 9mX14m has beams parallel to 9m dimension spaced such that 4 panels of slab
are constructed. Assume thickness of slab as 150mm & width of the beam as 300mm. Wall
thickness = 230mm, the loading on the slab (I) dead load excluding slab weight 2kN/m² (2) live
load 3kN/m². Adopt M20 concrete & Fe 415 steel. Design intermediate beam by 1st principle.
Assume any missing data.
* 1 inch = 25.4 or 25mm
* ‘h’ should be in terms of
multiples of inches.
62

Step . 1
S=

Preliminary design

N.%
N

= 3.55m

Effective span, le = 9 +
Flange width, bf =
h ≈

š¯



š¯



to

≈

š¤

z%L

M



 – L.%


= 9.23m

+ bw+6Df = 9.23/6+0.300+6X0.15 = 2.74m < S = 3.55
to

z%L


= 769.17 to 615.33

Let us assume h=700mm, Ce=50mm.
dprovided = 700-50 = 650mm
Loading
1. on slab
a) self weight of slab = 1m X 1m X 0.15m X 25 = 3.75kN/m²
b) Other dead loads (permanent)

= 2kN/m²

c) Live load (varying)
(It is known as imposed load)

= 3kN/m²
q = 8.75 9kN/m²

2. Load on beam
a> From slab = 9 X 3.55 = 31.95kN/m
b> Self weight of beam = 0.3 X 0.55 X 1 X 25 = 4.125kN/m
↓
↘depth of web
width of beam
w= 36.D75
M=

%M – z.%²
O

36kN/m

= 383.37kN-m

Mu= 1.5 X 383.4 = 575kN-m.
(Ast)app. =

§™

±
L.OPž¦(¢b ² )

=

B

P – L⁶

L.OP – N(MLb

”³°
)
B

= 2769mm² (Ast)actual = 491 X 6 = 2946
63

No. of # 25 bars =

PMz

3.28ft

NzL

= 5.65 ≈ 6

1ft of span → depth is 1 inch1m =

Provide 6 bars of 25mm dia in 2 rows.
Step .2

N-A depth

xumax = 0.48 X 650 = 312mm
Assuming N-A to lie in flange,
xu =

L.OPž¦ ´Ÿ

L.%Mž¡µ ªž

=

L.OP – N – zNM
L.%M – L – PNL

= 53.91mm < Df < xumax

Mur = 0.36 fckbf xu(d-0.42xu)
= 0.36 X 20 X 2740 X 53.91(650-0.42 X 53.91)/10⁶
= 667.23kN-m.
Design a T-beam for a simply supported span of 10m subjected to following loading as
uniformly distributed load of 45KN/m excluding self wt of the beam by a point load at mid-span
of intensity 50KN due to a transverse beam. Assume the width of the beam=300mm & spacing
of the beam=3m. Adopt M-20 concrete &Fe 415 steel.
Sol: M =

~]B
O

+

6 ×]
N

Self cut = 1 x 1 x 0.3 x 25 = 78.dKW/M.
W= 7.5+45 = 52.KN/M.

Shear, Bond & tension in RCC Beams
Shear

64

•
•
•

Types of cracks @ mid span → flexural crack beoz Bm is zero, SF is max
Type of crack away from mid span →shear f flexural crack.
Principal tensile stress at supports = shear stress

τ¶ =

·'(()
¸&

A= area above the point consideration

If (As) hanger<(Ast)min does not contribute to compression as in doubly reinforced beams.
(Ast)min =

L.O&A



RCC – Heterogonous material → Distribution of shear stress in complex

τ¹ =

·'(()
¸&


 Normal shear stress τ¹ =

¶'

&~A

Vu=Vcb+Vay+Vd+Vs
Vu = Vcu +Vs.

•

Shear reinforcement →Vertical stirrup & Bent - up bars

65

Truss analogy

Vcu = τcc x b x d

τc= Design shear stress

IS-456 P-73
τv ≤ τcmax
vertical stirrups
Vs

Inclined stirrups
Bent up bars.
L.N& ¶

(Asv)min ≥
L.OP

2) Vs=Vus =

52 (¹) max ≤

L.OP
( ¶ A
¶

L.OP
 ( ¶
L.N&

(sin ½ + 5½)

3) Vus = 0.87fyAsv sinα

No(6)
Whenever bent up bars are provided its strength should be taken as less than or equal to 0.5Vus
(shear strength of reinforcement).

66

Procedure for design of shear in RCC
Step; 1 From the given data calculate the shear force acting on the critical section where critical
section is considered as a section at a distance ‘d’ from the face of the support. However in
practice the critical section is taken at the support itself.
Step – 2 For the given longitudinal reinforcement calculate pt=
calculate from Pg. 73 τv=

·
&A

LL( Q
&A

, for this calculate Tc & Tv

; ¾ = 30014. ℎ.32 !52. calculated in step 1.

If τv > τcmax (page 73) then in crease ‘d’
Vcu = τcbd
If Vcu ≤ Vu, Provide min. vertical stirrup as in page 48, clause 26.5.1.6 ie(Sv)max ≤

0.87!V,¹
0.4+

Else calculate Vus=Vu-Vcu

Step 3 Assume diameter of stirrup & the no. of leg to be provided & accordingly calculates AsV
then calculate the spacing as given in P-73 clause 40.4(IS-456) This should satisfy codal
requirement for (Su)Max. If shear force is very large then bent-up bars are used such that its
strength is less than or equal to calculated Vus.
1.

Examine the following rectangular beam section for their shear strength & design shear
reinforcement according to IS456-2000. B=250mm, s=500mm, Pt=1.25, Vu=200kN,
M20 concrete & Fe 415 steel
Step 1: Check for shear stress
Nominal shear stress, τv =

LL × L
L ×LL

=

·
&A

= 1.6ƒ/UU

From table 19, p=73, τc = 0.67N/mm2.
From table 20, p=73, τcmax = 2.8N/mm2.
τc < τv < τcmax.
The depth is satisfactory & shear reinforcement is required.
Step 2. Shear reinforcement
Vcu = τcbd =

L.MP ×L ×LL
LLL

= 83.75‚ƒ.

Vus = Vu – Vcu = 200 – 83.75 = 116.25KN
67

{
N

Assume 2 leg -10mm dia stirrups, Asv =2 X X102 =157mm2.
Spacing of vertical stirrups, obtained from IS456-2000
Sv =

L.OP
( ¶A
·

=

L.OP ×N×P×LL
M.×L

Check for maximum spacing
i) Svmax =

L.OP( ¶

L.N&

=

= 243.8 = 240UU/

L.OP ×P×N
L.N ×L

= 566.8UU

ii) 0.75d = 0.75 x 500 = 375mm.
iii) 300mm
Svmax = 300mm(Least value)
2.

Repeat the previous problem for the following data
1) b=100mm, d=150mm, Pt=1%, Vu=9kN, M20 concrete & Fe 415 steel
2) b=150mm, d=400mm, Pt=0.75%, Vu = 150KN, M25 concrete & Fe 915 steel
3) b=200mm, d=300mm, Pt=0.8%, Vu=180kN, M20 concrete & Fe 415 steel.

3.

Design the shear reinforcement for a T-beam with following data: flange width =
2000mm. Thickness of flange = 150mm, overall depth = 750mm, effective cover =
50mm, longitudinal steel = 4 bars of 25mm dia, web width = 300mm simply supported
span=6m, loading =50kN/m, UDL throughout span. Adopt M20 concrete & Fe 415 steel
Step; [Flange does not contribute to shear it is only for BM]

Step -1 Shear stress
V=

L ×M


= 150KN

Vu= 1.5 X 150 = 225KN
τv = & gA =
¶

¿

 × L
%LL ×PLL

= 1.07

Ast = 4 × 491 = 1964mm2.
Pt =

LL ×zMN
%LL ×PLL

= 0.93

Pg – 73

0.75 
 0.56
1.00 
 0.62
68



τc = 0.56 +

(L.MbL.M)
(.LLbL.P)

(0.93 − 0.75)

= 0.6N/mm2.

From table 20, max = 2.8
∴τc < τv < τcmax = 2.8
Design of shear reinforcement is required
Step – 2 Design of shear reinforcement
Vcu = τcbwd =

L.M ×%LL ×PLL
LLL

= 126‚ƒ

Vus= Vu-Vcu=225-126=99KN
Assume 2-L, 8 dia stirrups Asu = 2 X
Spacing of vertical stirrups,
Sv =

L.O
(|g A
¶g|

=

L.OP ×N ×LL ×PLL
zz ×LLL

{
N

X 82 = 100mm2

= 255.28 = 250UU /

Check for Max spacing
i) Svmax =

L.OP
À (|g
L.N&¿

=

L.OP ×N ×LL
L.N ×%LL

= 300.87UU

ii) 0.75d = 0.75 × 700 = 525mm.
iii) 300mm
Sv < Svmax

∴ provide 2l -#8mm @ 250c/c

Step – 3 curti cement

From similar triangle

%

=

M
'

χ  = 1.68U = 1.6U.
69

∴ provide (i) 2L – 3 8@ 250 c/c for a distance of 1.4m
(ii) 2L-#8@300 c/c for middle 3.2m length
Step.4 Detailing
Use 2- #12mm bars as hanger bars to support stirrups as shown in fig

A reinforced concrete beam of rectangular action has a width of 250mm & effective depth of
500mm. The beam is reinforced with 4-#25 on tension side. Two of the tension bars are bent
up at 45ᵒ near the support section is addition the beam is provided with 2 legged stirrups of 8mm
dia at 150mm c/c near the supports. If fck =25Mpa & fy = 415Mp2. Estimate the ultimate shear
strength of the support s/n
(Ast)xx = 2 × N × 25 = 982mm2.
{

Pt = L ×LL == 0.78%
LL ×zO

Pt = 0.75 
 0.57
Pt = 1.00 
 0.64
For Pt = 0.78 ⇒ τc = 0.57 +

(L.MNbL.P)
(bL.P)

(0.78 − 0.75)

τc = 0.5784.

1) Shear strength of concrete
Vcu = τcbd =

L.PON ×L ×LL
LLL

= 72.3‚ƒ.

2) Shear strength of vertical stirrups
(Asv)stirrup = 2 × N × 8 = 100UU
{

70

(Vsu)st =
=

L.OP
(|g A


 = 150UU

L.OP ×N ×LL ×LL
L

= 120.35‚ƒ

3) Shear strength of bent up bars
(Asu)bent = 2 ×

{
N

× 25 = 982UU

(Vus)bent = 0.87fy(Asu)bent sin∝
=

L.OP ×N ×zO × IN
LLL

= 250.7KN.
Vu = Vcu +(Vus/st +(Vus)bent
= 72.3+120.35+250.7
Vu = 443.35KN

5. A reinforced concrete beam of rectangular s/n 350mm wide is reinforced with 4 bars of 20mm
dia at an effective depth of 550m, 2 bars are bent up near the support s/n. The beam has to carry
a factored shear force of 400kN. Design suitable shear reinforcement at the support s/n sing
M20 grade concrete & Fe 415 steel.
Vu = 400KN, b=350mm, d=550mm, fck=20Mpas
fy = 415Mpa, (Ast)xx = 2 x 314 = 628mm2.
Step – 1 Shear strength of concrete
Pt = %L ×L = 0.32%
LL ×MO

τc = &A =
·

•25U Ã3+1. − 19 Ä = 0.4;03.

NLL × L
%L ×L

= 2.07 Mpa
τc < τv < τcMax.
∴ Design of shear reinforcement is required.

71

Step -2 Shear strength of concrete
Vcu = τcbd =

= 77‚ƒ.

L.N ×%L ×L
LLL

Vus = Vu-Vcu = 323KN
Step -3 Shear strength of bent up bar.
(Asv)bent = 2 x 314 = 628mm2.
(Vus)bent = 0.87fy(Asv)bent sin∝.
=

L.OP ×N ×MO × IN
LLL

= 160.3KN Å<

·g|


= 161.5Ç

NOTE: If (Vus)bent >

·g|


then (Vus)bent =

·g|


Step – 4 Design of vertical stirrups
(Vus)st = Vus-(Vus)bent =323-160.3=162.7KN.
Assuming 2L-#8 stirrups

(Asv)st = 2 × N × 8 = 100UU
{

Sv =

L.OP
(|È A
(¶g| ) Q

=

L.OP ×N ×LL ×L
M.P ×LLL

= 122UU = 120UU



Provide 2L-#8@120 c/c
Svmax =

L.OP
( ¶
L.N&

= 257.89UU.

0.75d=412.5mm, 300mm.
Svmax = 258mm
Shear strength of solid slab
Generally slab do not require stirrups except in bridges. The design shear stress in slab given isn
table 19 should b taken as ‘kτc’ where ‘k’ is a constant given in clause 90.2.1.1
→ The shear stress τc < kτc hence stirrups are not provided.
→ Shear stress is not required Broz thickness of slab is very less.

72

Self study: Design of beams of varying depth Page: 72, clause 40.1.1

Use of SP-16 for shear design
SP – 16 provides the shear strength of concrete in table 61 Pg 178 table 62 (179) provides (Vus)st
¶
for different spacing of 2 legged stirrups of dia 6,8,10 & 12mm. Here it gives the value of g| in
A

kN/cm where ‘d’ is in cm. Table 63, Pg 179 provides shear strength of 1bent up bar of different
dia.
Procedure
Step- 1; Calculate

·

= τc = &Ag & obtain τc from table 61 & also obtain τcmax

from table 20, Pg

73 –IS956 If τc < τc < τcmax then design of shear reinforcement is necessary.
Step - 2 Vcu= τcbd
Vus = Vu-Vcu
Assuming suitable stirrup determine the distance for

¶g|
&

in kN/cm.

H.W Design all the problems using SP-16 solved earlier.

73

Bond & Anchorages
Fb = (πφ) (ld) (τbd)
Permitter
length

T=

{
N

stress

× ∅ × É

Fb = T

ld τbd = N × ∅ × É
{

Ê∅

ld = NË|

lm

τbd = Anchorage bond stress

τbf =flexural bond stress
For CTDs HYSD bars, flexural bond stress is ignored bozo of undulations on surface of steel.
∑ 5 = UU3/45q 5! 0.2U./.2 5! +32

·

τbf = ∑ ÌA

Z= lever arm
Codal requirement

Ê∅

Where ld = NË|

lm

;
+ 15 ≥ 1 → 0Î − 44 → 13. 26.2.3.3, Ï − 42
¾

; σs = tensile stress in steel
τbd = design bond stress.

The value of this stress for different grades of steel is given in clause 26.2.1.1 →Pg- 93 of code
for mild steel bar. These values are to be multiplied by 1.6 for deformed bars. In case of bar
under compression the above value should be increased by 25% σs = 0.87fy for limit state
design. If lo is insufficient to satisfy (1), then hooks or bents are provided. In MS bars Hooks
are essential for anchorage
74

Min = 4ø

K= 2 for MS bars
=4 for CTD bars

Hook for Ms bars
(K+1)ø

Standard 900 bond
Pg – 183 fully stressed =0.87fig
1.

Check the adequacy of develop. Length for the simply supported length with the following
data.
(IV) c/s = 25 x 50cm (ii) span=5m (iii) factored load excluding self wt =160KN/m. iv)
Concrete M20 grade, steel Fe 415 grade. (v) Steel provided on tension zone. 8 bars
of 20mm dia.

Solve: Ce = 50mm, h=500mm, d=450mm
qself = 0.25 x 0.5 x 1 x 25
= 3.125KN/m
quself = 1.5 x 3.125 = 4.6875KN/m.
Total load = 60 + 4.6875
= 64.6875KN/m
Vu=

MN.MOP ×


= 161.72KN

Xulim = 0.48 d=0.48 x 450=216mm
Mulim = 0.36fck xub(d-0.42xu)106
= 139.69kN-m
Let Ws = 300, lo = 150mm.

75

@”
·

+ lo =

ld =
@”
·

L ×L.OP ×N
N ×.z ×LLL

M.P

+ 0.15 = 1.01m
É = 0.87!V

∅ÊB

NÐlm

Ld =
τbd = 1.6 x 1.2 = 1.92

%z.Mz

Table R43.

= 0.94U.

+ 1 > 1 ; ℎ.q. 3!.

2. A cantilever beam having a width of 200mm & effective depth 300mm, supports a VDL hug
total intensity 80KN(factored) 4nos of 16mm dia bars are provided on tension side, check the
adequacy of development length (ld), M20 & Fe 415.
Design for torsion

¸

¸Ñ

=

Ґ
]

=




`

⇒ ! =

‹`
¸“

‹

τtmax = $&B

52

‹

$& B Ó

For the material like steel
Principal stresses all 4 faces reinforcement
Longitudinal bars

stirrups
Torsion

Secondary torsion
Nature of connection
Of constructions

Primary torsion
In analysis the effect of
torsion rigidity is taken
76

to account
Eg. Chejja or sunshade L-window

Cross – cantilever
Eg- for Secondary torsion

Eg. (1) Plan of Framed Structure
Primary torsion

(3) Ring beam in elevated tank

(2) Arc of a circle

BM(I)
RCC Beams

SF (Shear force)(II)
Torsion (III)

Combination of (I) – (II)& (I) –(II)
IS-456 →Pg – 79 Procedure
(1) Flexure & Torsion
Me1=Mu + MT
M T = Tu (

A/&
.P

) Ô=ℎ
77

D= Overall depth, b=breadth of beam
Provide reinforcement for Mt in tension side
If MT > Mu provide compression reinforcement for

(2) Shear & Torsion
Ve=Vu +
¶Õ

τv= &A

.M‹
&

; •52 3!. .4Îq Ķ > Ä .4Îq ℎ.32 2.4q!52.U.q/.

Shear reinforcement

g È
Asv = &A,L.OP

+

‹

=

(ËÈÕ b Ëj) & ¶

¶g È

A” (L.OP
)

L.OP


Pg.48 →clause 26.5.1.7
SVmax is least of
i) x1
ii)

'
N

iii) 300mm

Asw > LL × + × 
L.

1. Design a rectangular reinforced concrete beam to carry a factored BM of 200KN-m, factored
shear force of 120kN & factored torsion moment of 75 KN-m Assume M-20 concrete & Fe 415
steel
Sol: Mu = 20KN-m, Tu=75KN-m, Vu=120K.N.
Step -1= Design of BM & Torsion
Assume the ratio
MT =

Ö
l

‹g ( )
]× P

=

&

=2

P()
.P

= 132.35‚ƒ − U.
78

No compression reinforcement design is necessary
MT < Mu
Me1 = MT + Mu = 200 +132.35 = 332.35KN-m.
dbal = c

@Ք

ØgR &

= c

%%.% × Le
.PM ×%LL

= 633.5mm.

Assume b=300mm, θlim=2.76
Assuming overall depth as 700mm & width as 350mm & effective cover =50mm.
D Provided =700-50
=650mm
Area of steel required for under reinforced section,
Pt =

=

L
jk



L ×L
N

^1 − c1 
$&Aj”B _
N.M@

Ù1 − c1

N.M ×%%.% × Le
L ×%L × MLB

Ú

= 0.73 < 0.96
(pt lim)
∴ Ast =

L.P% ×%L ×ML
LL

= 1660.75UU

Assume 25mm dia bars =

MML.P
N.z

=3.38≈4

Provide 4 bars -#25 dia
(Ast) provided = 1963mm2

79

Step – 2 Design for shear force & torsion
Ve = Vu +

.M‹g

Ve = 120 +

τve =
Pt =

&

.M ×P
L.%

NM.OM × L
%L ×ML

N ×Nz

%L ×ML

= 462.86‚ƒ

= 2.03 < 2.8 (Tcmax) P-73

× 100 = 0.86

Ä¶Û > Ä

Table – 19 τc = 0.56 + (.LLbL.P) (0.86 − 0.56) = 0.58ƒ/UU
(L.MbL.M)

Assuming 2 – legged =12mm dia;
Asu = 2 x N ×122
{

= 226mm2

From IS-456;Pg-75
Sv =

L.OP
(|È

Üg
È
 g
l” m” B.³m”

b1=275mm, d1 = 600mm
y1 = 600+25+2 x 6 = 637mm
x1 = 275 + 25 + 2 x 6 = 312mm.
x1 y1 
 dist of centre of stirrups.
Provide 2 - #12@ top as hanger bars
Sv =

L.OP ×N ×M

ݳ × ”°e ”B° × ”°

Bݳ ×e°° B.³ ×e°°

= 152 ≈ 150/

Check

1. Asv >

(ËÈÕ b Ëj ) ¶&
L.OP


226> 210.84

=

(.L%bL.O)L ×%L

2. Svmax a) x1 = 312mm
d) 0.75d = 487.5

L.OP ×N

b)

'” ”
N

=

%M%P
N

= 237.25

c) 300mm

80

As D=h>450 , provide side face reinforcement
Asw =

L.

LL

x 350 x 650 = 227mm2

Provide 2-#16 bars (Ast = 400mm2) as side face

2. Repeat the same problem with Tu=150KN-m & other data remain same
Solve Mu = 200KN-m,Tu=150KN-m, Vu=120KN
Step – 1 Design of EM & torsion
Assume the ratio D/b = 2.
MT =

m
l

‹g ( )
.P

=

L()
.P

= 264.70KN-m

MT>Mu Compression reinforcement is required.
MG = MT+Mu = 264.7 + 200 = 4.64.7KN-m
= c .PM ×%LL = 749.15UU
&

@Ք

dbal = cØ

wxR

NMN.P × Le

Assume b= 300mm, θlim = 2.76.
Assuming over al depth as 800mm & width as 400mm & effective cover = 50mm.
D provided = 800-50 = 750mm
Area of steel required for under reinforced section,
Pt =

=

L
$



L ×L
N

^1 − c
$&AՔB _
N.M@

Ù1 − c L ×NLL × PLB Ú
N.M ×NMN.P × Le

Pt = 0.66 < 0.96(ptlim)

81

Assume 25mm dia bars =
Provide 6 bars - #25 dia

OL
Nz

= 5.86≈6.

Ast2= 255.99
Ast = 3135.99 ⇒ 7 - #25.
Me2 = 264.7 – 200 = 64.7KN-m.
= 260.95

@ÕB

|j (AbA)

Asc =
Ast1 =

6} &A
LL

, Ast2 =


|j (|j

L.OP


;,

Q

+ ,

! =

A”

Q

Q

A

= 0.066 !25U Î230ℎ

! = 354.2

= ,

Ast = kb of bars
Design for shear force torsion

Step – 2

.M‹g

Ve = vu +

= 120 +
τve =

&
.M ×L

PL × L
NLL ×PL

L.N

= 720‚ƒ.

= 2.4 < 2.8 (Ä:H' ) Ï − 73

Pt = 0.98
τc=0.61
Assuming 2 legged 12mm dia
Asv = 2× N × 12 = 226UU
{

From IS-456, Pg – 75
Sv =

L.OP
( ¶

Üg
Þg

l” m” B.³m”

Limit state of service ability

Deflection

Crack –width

Vibration

(Un factored load) Working load
1.

Deflection

Span to effective depth ratio
Calculation
82

ymax =



~]d

%ON

ymax ≤

]

¸

L

′ƒ& UU′

As control of deflection by codal provision for l/d ratio
Cause 23 .2.1 Pg – 37 of IS 456 – 2000
Type of beam
i)
ii)

Cantilever beam
Simply supported beam

iii)

Continuous beam

Effect on l/d ratio
1.

l/d ratio
Span, l ≤ 10m
Span>10m
7
Should be calculated
20 × 10
20
03q
26
× 10
26
03q

Tension reinforcement : > 1%
(( Q)àۇ

Pg – 38; fs = 0.58fy ((

Q)6à¶

SP-24 →Explanatory hand look
Mft = [ 0.225 + 0.003225fs + 0.625log10(pt)]-1 ≤ 2
2.

Compression reinforcement.
Mfc = Å

3.

Ç ≤ 1.5

.M6

6L.P

Flange action or effect
Mfl = 0.8 for

]

&


≤ 0.3

= 0.8 + P Å &
 − 0.3Ç !52


A

&~

&~

&~
&


> 0.3

= U
Q × U
 × U
] × <A= +34
]

Design
1. Flexure + torsion
2. Check for shear + Torsion, bond & Anchorage
3. Check for deflection
83

1.

Ast = 4 ×

Asc = 2 ×
Pt =

A simply supported R-C beam of effective span 6.5m has the C/S as 250mm wide by
400mm effective depth. The beam is reinforced with 4 bars of 20mm dia at the
tension side & 2-bars of 16mm dia on compression face. Check the adequacy of the
beam with respect to limit state deflection, if M20 grade concrete & mild steel bars
have been used.
B=250mm, d=400mm, fck=20Mpa, fy=250Mpa

{
N

{

× 20 = 1256 mU

N

× 16 = 402 mU

M ×LL
L ×NLL

= 1.256

Pl = L ×NLL = 0.402
NL ×LL

From Pg – 37, (l/d)basic = 20
fs = 0.58 x 250 x 1 = 145
mft = [0.225 + 0.00 3225 x 145 + 0.625log10(1.256)]-1 = 1.325
mfc = ÅL.NLL.PÇ = 1.12 (Î230ℎ)
.M ×L.NL

mfl = 1 (rectangular section)
( )0Renquired = 1.325  1.12  1  20 = 29.79
A
]

Check
(A) 025¹4. =
]

MLL
NLL

= 16.25 < 29.79; 3!.

2) Check the adequacy of a T-beam with following details (i) Web width (wb) = 300mm, (ii)
Effective depth (d) = 700mm (iii) flange width (bf) = 2200mm (iv) effective span of simply
supported beam(l) =8m (v) reinforcement a) tension reinforcement – 6bars of 25 dia b)
compression reinforcement – 3 bars of 20 dia (vi) Material M25 concrete & Fe 500 steel.
Deflection calculation

Short term deflection
Long term deflection (shrinkage, creep)

84

85

1.

Short term deflection

Ec= 5000á!) ; 0Î − 16, 1. 6.2.3.1.
Slope of tangent drawn @ origin →Tangent modulus
Slope of tangent drawn @ Specified point →secant Modulus 50% of Material
Igr=

&Ӂ


For elastic; Ief → cracked section

Pg. 88 Ieff =

¸à

fâ ã
T l¿
.
<b =
f m
m l

; äà ≤ äÛ

 ≤ äåà

Ir = Moment of inertia of cracked section
Mr = cracking Moment =
•
•
•
•

NA →stress is zero
CG→ It is point where the wt. of body is concentrated
Yt ≠x
M= Max. BM under service load: Z=lever arm
X= depth of NA : bw = width of web: b= width of compression face

(For flanged section b=bf)
For continuous beam, a modification factor xe given in the code should be used for Ir, Igr, & Mr.
The depth of NA ‘x’ & lever arm Z has to be calculated by elastic analysis is working stress
method explained briefly below.

Introduction to WSM

86

M=

|

j

=

'bA”

%Êjlj

; É& = Permissible stress P g.80

From property of similar triangles,

 − 
−
× ‰&‰ =
× ‰ → 1



‰ =

fc = Ec x ‰ c 
 2

!  = * × ∈ = U* ∈ → 3

fs = Es ∈s = mEc∈s 
 3a.

C = T.

Asc !  +  ! + = , Q ! + 1, 2 & 3


½ +  + †(U(, Q + ,  )] − U†(, Q  + ,   )] = 0 → 4
Eq (4) can also be written in the form of

+ 
+ (1.5U − 1),  ( −  ) = U, Q ( − ) → 5
2

In eq.5, the modular ratio for compression steel is taken as (1.5m)
Use of SP-16 for calculating Ief

1. Using Table – 91: Pg – 225-228, we can find NA depth for simply reinforced, Pc =0
2. Using Table – 87-90, find out cracked moment of inertia Ir
3. Ieff chart -89, Pg.216
Cracked moment of inertia can be found by the following equation.
For singly reinforced section
Ir =

&' 
%

+ U, Q ( − )

For doubly reinforced section,
Ir =

&' 
%

+ (U − 1),  ( −  )2

+ mAst(d-x)2

Shrinkage deflection
87

2. Long Term deflection
Creep effect deflection
a) shrinkage deflection
reduces stiffness (EI)
εsh=0.004 to 0.0007 for plain concrete
= 0.0002 to 0.0003 for RCC
Ysh = k2Ψcsl2
K3 = cantilever – 0.5
Simply supported member – 0.125.
Continuous at one end – 0.086 Pg-88
Full Continued – 0.063
Ψ∈ =

)N =

)N ∈
0.72(0Q − 0)
; )N =
≤ 1.0 !52 (0Q − 0)0.25 − 1.0
ℎ
á0/

0.65(0Q − 0)
á0/

≤ 1.0 !52 (0Q − 0) − 1.0

b) creep deflection → permanent


yscp =Initial deflection + creep deflection using Ecc in plane of Ec due to permanent Esc = 

j

Cc= Creep co-efficient
1.2 for 7 days loading
1.6 for 28 days loading
1.1 for 1 year loading
Ysp = Short per deflection using Ec
Ycp = Yscp –Ysp
A reinforced concrete cantilever beam 4m span has a rectangular section of size 300 X 600mm
overall. It is reinforced with 6 bars of 20mm dia on tension side & 2 bars of 22mm dia on comp.
side at an effective cover of 37.5mm. Compute the total deflection at the free end when it is
subjected to UDL at service load of 25KN/m, 60% of this load is permanent in nature. Adopt
M20 concrete & Fe 415 steel.
88

Sol:
{
N

Ast = 6 x

× 20 = 1885mU 1 = 4U, Œ =

Asc = 2 x N × 22 = 760mU
{

"y
:

.

fck = 200Mpa, fy = 415Mpa, Es = 2 x 105Mpa.
Ec = 5000√20 = 2.236 × 10N ;03
fcr = 0.07á!) = 3.13;03


m=
Ig =

=

 × L³

.%M × Ld

%LL × MLL


= 5.4 × 10z UUN =

= 300UU (

æç



Yt =

= 8.94
)

MLL


&L%


Step : 1 Short term deflection
yshort = O
Ieff =

j ¸Õii

Lb

Mcr =

M=

~]d


jâ ¸è
}

~]B


¸à

fjâ ã
T l¿
(b )
f m
m l

=

=

%.% ×.N × Lé
%LL

 × NB


=

M.%N × Le yb::
LM

= 56.34‚ƒ − ;

= 200‚ƒ − U.

;à
56.34
=
= 0.282
;
200

From equilibrium condition

+ 
+ (U − 1),  ( −  ) = U, Q ( − )
2
89

300 
+ (8.94 − 1)760( − 37.5) = 8.94 × 1885(600 − )
2
x2 #52.58x-64705.5=0

x = 189.28mm
Z ≈ d- = 562.5 −
'

%

Ir =

& × '
%

Oz.O
%

= 499.41UU.

+ (U − 1),  ( −  )2 +mAst(d-x)2

= 3.1645 x 109mm4
Ieff =

%.MN × Lé

LbL.O ×

déé.d”
”êé.Bê °°
<b
=
³eB.³
³eB.³ °°

= 3.01 x 109mm4
Yshort =

 × (NLLL)d

O ×.%M × Ld ×%.LM × Lé

= 11.31mm ; Icr ≤ Ieff ≤ Ig
∴ Ieff = Icr = 3.1645 x 109mm4.

Step – 2 long term deflection
a) Due to shrinkage
Ycs = k3Ψcsl2

0Q =

K3 = 0.5 for cantilever

OO ×LL

%LL ×M.

0 =

= 1.117

PML × LL

%LL ×M.

= 0.45

Pt – Pc = 1.117 – 0.45 = 0.667 < 1.0
K4 =

L.P (6} b6j )
√6Q

= 0.454

Ecs = shrinkage strain = 0.0003 (Assumed value)
Ψcs =

$d ∈j|
Ó

= 2.27 × 10bP =

L.NN ×L.LLL%
MLL

90

Ycs = 0.5 x 2.2 + 10-7 x 40002 = 1.82mm.

b) Due to creep
Ecc =
=



j

;  = 1.6 †!25U 5. !52 28 3V]

.%M ×Ld

Yscp =

.M
~Ñ × ]d

= 8600;03

O jj ¸Õii
L.M × × NLLLd

= O ×OMLL ×%.ML × Lé = 17.64

Ysp = 0.6 yshort = 6.8mm.
Ycp = 17.64 – 6.8 = 10.84
Y=yshort +ycs + ycp = 23.97
= 11.31 + 1.82 +10.84
Unsafe

Doubly reinforced section

Π=

¹51
0.3 × 0.75 × 1 × 25

Md =

~m ` B

Ml =

~A]

O

N

= 17.58‚ƒ − U.

=

OL ×
N

= 100‚ƒ − ë

Mu = 1.5(MD+Mc)=1.5(17.58+100)=176.37KN-M
Xulim = 0.48 x d = 324mm
Mulim = Qlimbd2
91

=

N.N ×%LL × MPB
Le

= 565.88KN-m.
Singly reinforced s/n Ast =
Ptlim = 7.43%

=

6QwxR &A
LL

.N% ×%LL ×MP
LL

6 -#25bar is taken.
(Ast)provided = 6 x 490 = 2940 > 2895

2b) b=150mm, d=400mm, pt=0.75%, vv=150KN, fck = 25, fy=415
τv =

L × L
L ×NLL

= ::B
.y

τc = 0.57N/mm2.
τcmax = 3.1N/mm2.
τc < τv < τcmac
vcu = τcbd =

L.P ×L ×NLL
L

vus = vu – vcu = 115.8KN.

= 34.2‚ƒ.

Assume 2L-10# Asu = 2 x N × 10
{

= 157.07mm2.

Sv =
=

L.OP
(|È A
¶g|

L.OP ×N ×P.LP ×NLL
.O × L

= 195.88mm

92

L.OP
À (|È
L.N&

Svmax =

= 945.16UU

0.75d = 300mm
300mm
∴ provide 2l-#10@ 195.99 c/c.
2c) b=200mm, d=300mm, pt=0.8%, vv=180KN, fck=20, fy=415.
τc =

¶

&A

=

OL × L
LL ×%LL

=

%y

::B

pt 0.75 


τcmax = 2.8

τc 0.56

1.0
Pf = 0.8; τ = 0.56 +

0.62

(L.MbL.M)
(bL.P)
(L.ObL.P)

τc = 0.572 N/mm2.
τv > τcmax ; unsafe, hence increased
Let ‘d’ be 350mm
τv = 2.57N/mm2.
∴ τc < τv < τcmax; safe sh.rei required.
Step 2 shear reinforcement.
Vcu = τcbd = 0.572 x 200 x 350 = 40KN.
Vus = vu-vcu = 180-40 = 140KN.
Assume 2L-#10; Asv=157mm2.
Spaces of vertical stirrups
Sv =

L.OP ×N ×P ×%L
NL ×LLL

= 141.71UU ≈ 140UU

Step 3: Check for max. spacing
1. Svmax =

L.OP ×P ×N
L.N ×LL

= 708.5UU.

2. 0.75d = 0.75 x 350 = 262.5mm
3. 300mm sv < sumax is 140 < 262.5mm
93

Hence provide 2L-#10@140 c/c.
4) le=10m
W=q1+q11 q1 = 45 KN/m.
bw = 300mm, s=3m, fck=20Mpa, fy=415Mpa
bf =

]
M

+ +Π+ 60!

LLLL
M

=

,U. Ô
 = 100UU

+ 300 + 6(100)

= 2566.67mm < 3000mm
h=

]



/5

† 8333.33 /5 666.67],U. Û = 50UU

]



h=750mm, d=700mm
b x h x l density
self = 0.3 x 0.65 x 1 x 25
q11 = 4.875 KN/m
W= 45 + 4.875 = 49.875 KN/m
M=

L.M × LB
O

+

L ×L
N

=

PNO.O"y
:

.

Mu = 1136.7 KN/m
(Ast)oppr =

@

L.OP
<Ab

L.OP
( Q

Step2: xu = L.%M


= 4843.56UU 10 − #25 = 9910UU

Öi
=
B

jk &i

= 95.92UU < Ô
 < ]I:

Mur = 0.36fck xubf(d-0.42xu) = 1169.4KN-m.
2. Step 2 : bf =

]

=
h =  /5
]

h=450mm

]



+ +Œ + 6Ô


M
MLLL
M

+ 300 + 6(120)

= 2020mm.

†500 /5 400] .1! Œ/ = 1  1  0.12  25 = 34/U2
q=8 5KN/m2

94

d=h-50=400mm
(Ast)app =

@

L.OP
<Ab

L. × Le

w-q x s x 1 = 8 x 3 x 1
Öi
=
B

”B°
L.OP ×N ÅNLLb
Ç
B

=

.1! Œ/ =

; =

~]B
O

N"y
:

, 0.3  0.33  1  25 = 24

= ë = 26.4, ; = 119.14 ; = 178.7

= 824.80mm2.
Provide 3-#20

∴ (Ast)provided 3 x 314 = 942 > 824.8

Step 2 : Assume NA to be on flange
Cu = 0.36fckbfxu
Tu= 0.87fyAst.
xu =

L.OP
(|}

L.%M
jk &i

= 23.38UU < Ô
 < 

xulim = 192mm.

Mur = 0.36fckbfxu (d-0.42xu)
= 0.36 x 20 x 2020 x 23.38 [400-0.42 x 23.38]
= 132.67KN/m < 101.25KN-m.

95