The document discusses limit state design of reinforced concrete structures. It defines two main limit states - limit state of collapse and limit state of serviceability. Limit state of collapse deals with strength and stability under maximum loads, while serviceability deals with deflection, cracking and durability under service loads. Characteristic loads have a 95% probability of not being exceeded and are factored up using partial safety factors for design loads. Material strengths are reduced using partial safety factors to calculate design strengths. The document also derives design coefficients for moment of resistance, depth of neutral axis and percentage steel reinforcement for rectangular sections under flexure.

1. Unit - III Limit State Design
Objectives of Design:
Every material of the total structure takes part effectively for form, function,
aesthetics, strength as well as safety anddurability.
1. The structures should have an acceptable probability of performing
satisfactorily during their intended life:- There are uncertaintiesin the design
process both in the estimation of the loads likely to be applied on the structure
and in the strength of the material.
2. The designed structure should sustain all loads and deform within limits for
construction and use: - Adequate strengths and limited deformations are the
two requirements of the designed structure. The structure should have
sufficient strength and the deformations must be within prescribed limits due
to all loads during construction and use. However, has to ensure that the
failure of the structures in case of any over loading should give sufficient time
for the occupants to vacate. The structures, thus, should give sufficient
warning to the occupants and must not fail suddenly.
3 The designed structures should be durable.
4. The designed structures should adequately resist to the effects of misuse
and fire.
Characteristic load: Characteristic load is that load which has a 95%
probability of not being exceeded during the life of the structure. The loads
are predicted based on probabilistic / statistical approach, where it is assumed
that the variation of the loads acting on structures follows the normal
distribution. Characteristic load should be more than the average/mean load.
Characteristic load = Average/mean load + K x(standard deviation for load)
The value of K is assumed such that the actual load does not exceed the
characteristic load during the life of the structure in 95 per cent of the cases.
'-••• iiu..-
Ar.».»-0 45
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2
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Limit state of
collapse
It deals with strength and
stability of the structure
under maximum design load
it state of
serviceability
It deals with deflection and cracking
under service loads, durability under
working environment, fire resistance etc.
There are two main limit states: (i) limit state of collapse and (ii) limit state of
serviceability.
(i) Limit state of collapse deals with the strength and stability of structures
subjected to the maximum design loads out of the possible combinations of
several types of loads. Therefore, this limit state ensures that neither any part
nor the whole structure should collapse or become unstable under any
combination of expected overloads.
(ii) Limit state of serviceability deals with deflection and cracking of
structures under service loads, durability under working environment during
their anticipated exposure conditions during service, stabilityof structures as a
whole, fire resistance etc.
All relevant limit states have to be considered in the design to ensure adequate
degree of safety and serviceability. The structure shall be designed on the
basis of the most critical limit state and shall be checked for other limit states
Partial safety factors:
(i) Structures are subjected to overloading. Hence, structures should be
designed with loads obtained by multiplying the characteristic loads with
suitable factors of safety. These factors of safety for loads are termed as
partial safety factors (yf) for loads.
(Design load, Fd) = (Characteristic load F)x(Partial safety factor for load yf)
yf for DL and LL should be taken as 1.5 for limit state of collapse
and 1.0 for limit state of serviceability.
(ii) The characteristicstrengths of materials may differ from sample to
sample. Accordingly, the design strength is calculated dividing the
characteristic strength further by the partial safety factor for the material (ym).
(Design strength of material, fd) = (Characteristic strength of material, f ) /
(Partial safety factor of the material, ym)
ym for concrete and steel should be taken as 1.5 and 1.15, respectively
l | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

2. Assumptions:
a) Plane sections normal to the axis remain plane after bending.
b) The maximum strain in concrete at the outermost compression fibre is
taken as 0.0035 inbending.
c) The stress- strain relationshipof concrete may be assumed any shape in
substantial agreement with the results. An acceptable stress - strain curve is
given in Fig. The compressive strength of concrete in the structure shall be
assumed to be 0.67 fck. The partial safety factor ym = 1.5 .
'•k
d) Tensile strength of the concrete is ignored.
e) The stresses in the reinforcement are derived from representative stress-
strain curve for the type of steel used. Typical curves for mild and HYSD
steels are given in Fig. For design purposes the partial safety factor ym, equal
to 1.15 shall be applied.
mm
f) The maximum strain in the tension reinforcement in the section at failure
shall not be less than:-> ( fs/1.15 Es) + 0.002
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The stress-strain relations ship in elasto-plastic zone of HYSD bars -> Tabl
O.Sfyd
0.85fyd
0.9fyd
0.95fyd
0.975fyd
fyd
0
0.0001
0.0003
0
.
0
0
0
7
0.001
0.002
fy415
a
288.70
306.74
324.78
342.83
351.85
360.87
c
0
.
0
0
1
4
4
0.00163
0.00192
0.00241
0.00276
0.00380
fySOO
a
347.83
369.57
391.30
413.04
423.91
434.78
e
0.00174
0.00195
0.00226
0.00277
0.00312
0.00417
fee = 1000x0.446 fck [s - 250e2] if e < 0.002
fsc in case of doubly reinforced sections: (Table 2)
Fsc-
fy
415
500
d'/d
0.005
355
424
0.1
353
412
0.15
342
395
0.2
329
370
Working Stress Method
• The Stresses in an element is obtained from the working loads and
compared with permissible stresses.
• The method follows linear stress-strain behaviour of both the materials.
• Modular ratio can be used to determine allowable stresses.
• Material capabilitiesare under estimated to large extent. Factor of safety
are used in working stress method.
• Ultimateload carrying capacity cannot be predicted accurately.
• The main drawback of this method is that it results in an uneconomical
section.
Limit State Method
• The stresses are obtained from design loads and compared with design
strength.
• In this method, it follows linear strain relationshipbut not linear stress
relationship (one of the major difference between the two methods of
design).
• The ultimate stresses of materials itself are used as allowable stresses.
• The material capabilitiesare not under estimated as much as they are in
working stress method. Partial safety factors are used in limit state
method.
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3. 1
2
3
4
5
6
7
8
9
10
1
1
12
12
Linearity of
strain
Design
Philosophy
Design approach
Design
constraints
Stress- strain
relationship of
materials
Factor of safety
Major concerns
Stress
Strain
Compatibility
Loads
Economical
Area of section
%
Reinforcement
Tensile strength
of concrete
WSM LSM
Plane sections remains plane after bending
It is based on elastic
theory
It is based on
deterministic approach
Stresses in materials
are permitted to a
certain extent
Stress-strain diagram is
linear and It follows
Hooke's Law.
Factor of safety is used
to arrive at permissible
stresses
It's major concern is
only safety
The stresses are
compared with
permissible stresses
Modular ration can be
used to find stress, m=
280/3acbc
Strength is estimated
underworking
(Service) loads and
compared with
allowable stresses
Conservative
More
less
It is based on Elastic and
plastic theories
It is based on probabilistic
approach
Strain in material is
limited to a certain extent.
Stress-strain diagram is
non- linear. Reserve
strength is also considerd
Partial safety factor for
loads and materials are
used
Major concernis safetyas
well asserviceability
The stresses are
compared with design
stresses
Modular ratio is used only
for strain compatibility,
m= Es/Ec
Strength is estimated
under factored (ultimate)
loads and compared with
characteristic stresses
Economical
less
more
Tensile strength of concrete is ignored; Section is
analysed on cracked section basis
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a) Depth of Limiting NA, xu,lim
From strain diagram
xu.lim 0.0035
0.0035+- -+0.002
i.iSEs
For fy 250,xulim/d =0.53
For fy 415, xulim/d =0.48
For fy 500,xu lim/d =0.46
b) Depth of NA, xu
Area of stress block =
= 0.446 fckx (3xu/7)
+(2/3)0.446fckx(4xu/7)
= 0.19114+0.1699
= 0.362 fck xu
O_ ,
•c
'
d
'
c/
O
5 0.446 fck
20 -*4 C
fy/1.15E+0.002
>T
Comp force, C = area of stress block x b
From equilibrium : C=T -> :. (0.362 fckxu)b = Ast x 0.87 ry
Ast X 0.87 fy
[fy/U5 = 0.87fy] [Ga]
n xu
Hence, — =
d 0.36 x/cfc xb xd
c) Lever arm: Taking moments from the top of stress block (diagram)
0.19114(3/14) + 0.1699[3/7 + (3/8)4/7]
V = XU =
y 0.19114 + 0.1699
Lever armjd = d- 0.416xu
d) Moment of resistance: asC = T
MR; Mu = Cxjd = [0.362fckxu b ] x [d- 0.42 xu]
OR
MR, Mu = T xjd = [Ast x 0.87fy] x [d - 0.42 xu ]
Mu = Ast X0.87 fy Xd (l - 0.42
* Astx0.87/y xd (l - ***x fy
' J d
Mu i Ast x
0.87/yd = ^St
The under reinforced solution,
----- [Gb]
fy x b x d>
4.6 Mu
bd
fckbcl2
e) Percentage of steel for section with limiting (balanced section) case
as C = T; [0.362 fck xu b ] = [Ast x 0.87fy]
Ast 0.36 fck xu
3 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

4. 0,0035 .
3xu/7
xu ( .
I 4xu/7
0X532 + 0.87 f<
/ "
007
d
l r
^ do
StrainDiagram
Design constants for section with limiting NA
Q.4467fck
rf^
, 6 XL
-^?r-J-A
Cy
Id
.!_, »
Stress Diaeram
a) Depth of limitingNA, xu lim /d = 0.0035 / (0.0055+0.87 fy/ E)
b) Area of stress block = (17/21 bh) = 0.362 fck xu = 0.36 fck xu
c) Distance of resultant comp. from extreme comp. fibre =
9(3) + 8(6 + 3) /xux
y 1 1 fMlfi vii
17 VlJ °'41
d) Lever arm, jd = d-0.416 xu lim =d [ 1- 0.42 xu lim/d]
e) MR, Mu = (0.36 fck xu) b x d (1-0.42 xu lim/d) =
= 0.36 (xu lim/d)( 1-0.42 xu lim/d) fck x bd2 = Qbd2 —[Gc]
f) Percentage of steel, p = 100 x (0.36 fck/0.87fy) x (xu/d)
Mild/pla
in
Fe250
xu lim/d = xu max / d 0.5314
IfQ=qxfck; q 0.1496
MR Coeff, Q
% Steel, p
2.9933
1.77%
MR Coeff, Q
% Steel, p
3.7416
2.21%
Mild Tensi
steel
M20
M25
M30
MR Coeff, Q
% Steel, p
4.4899
2.65%
e HYSD/CTI
TOR Fe 41
0.4792
0.138
2.7745
0.96%
3.4682
1.2%
4.1618
1.44%
) HYSD500
5 Fe 500
0.4561
7 0.1336
2.6726
0.76%
3.3407
0.95%
4.0089 •
1.14%
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Question 1: Derive the Design coefficients(for NA, lever arm, MR and
percent steel) for a rectangular section with M25 and Fe 415.
Solutjpni
xu.lim 0.0035
a) Coefficient for depth of Limiting NA, k;
0.0035+ fy
1.1 sEs
= 0.48
+0.002
b) The resultant compressive force passes through the CG of Stress block @
0.42 xu from the extreme compression fibre
The lever armcoefficient,j = 1- 0.42 (xumax/d) = 1 - 0.42 x 0.48 = 0.798
c) Coefficient for ultimate Moment of resistance, Q = k xj x 0.36 fck
= 0.48(U042x0.48)x0.36 fck = 0.138 fck = 3.45
d) Coefficient for percentage of steel = percentage of steel, p
41.4(fck/fy)(xu max/d) = 41.4(25/415)(0.48) = 1.197%
Question 2: A RCC beam 230mmx 500mm is reinforced with 3 x 16mm dia
bars. Find moment of resist ance if the effective cover is 40mm.Assuming
M20 concrete & Fe415
Solution:
a) Depth of Limiting NA, xu max = 0.48 d = 0.48x460 = 220.8mm
b) Depth of NA, xu; from C = T -» 0.36 fck xu x b = (Ast) (0.87fy)
xu 0.87x/yx^st 0.87X415X603 _ ~o r o
— = = = U.toiJO
d 0.36x/ckxbXd 0.36X20x230X460
xu = 131.47 mm [it is not Over Reinforced]
c) MR, Mu = C x jd = (0.36 fck xu) b( d - 0.42 xu)
= 0.36 x20x!31.47x230[ 460 - 0.42 x!31.47 ] =88.13xl06Nmm=88.13 kNm
OR -»MR, Mu= T xjd = Ast x 0.87fy ( d - 0.42xu)
= 603 x 0.87 x 415 [460-0.42 xl31.47] = 88.13xl06Nmm =88.13 kNm
Question 3: Determinethe service load (udl) which can be carried by a
simply supported beam of 5m span having cross section 230 x 450
mm(effective), the beam is reinforced with 3 x 20 mm ty placed at an effective
cover of 50 mm. M20 and Fe 500.
Solution:
a) Depth of Limiting NA, xu max = 0.48 d = 0.48x450 = 216 mm
b) Depth of NA, xu =
0.87x/yx,4st 0.87x415x942
= 205.38 mm < xumax
0.36xfckxft 0.36x20x230
c) MR, Mu = T x jd = Ast x 0.87fy ( d - 0.42xu)
= 942x0.87x415 [450-0.42x205.38]= 123.71kNm
d) We know that, BM = MR; ^^ = 123.71
8
.-. Ultimate (factored)load, wu = 39.587 kN/tn
4 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

5. Service (working) load, w = wu/1.5 = 39.587/1.5 =26.39 kN /m
e) Superimposed Service load (Service load it can carry):
Self weight of beam/m = (1 x 0.23x 0.5)25 = 2.89kN/m
Hence it can carry, 26.39 -2.89= 23.5 kN/m
Question 4:Find the area of reinforcement required in a rectangular beam of
230 x 500, if it has to resist an ultimate moment of 125kNm. Use M 20 and
Fe4I5
Solution:
Effective cover = Nominal cover +<J)t + <j>m/2 *
Assume ~»30 + 8 + 20/2 = 48 [moderate exposure]
Effective depth = 500 - 48= 452 mm
[Gb] -> Mu = Ast x 0.87fy x d (
125x 106 = Ast x 0.87x 415 x 4521
fck xb xd,
Ast X 415
20 X230 X452,
1.99596 x 10~*Ast2 - Ast +765.9567 = 0 -> Ast=943.72 mm2
OR
Ast =
fck
- 1- 4.6 Mu
fck b d2
10
4l5
1- 1-
4.6 x 125 x 106
20 X 230 X 4522
bd
= 944mm2
Question 5: A singly reinforced beam of 6 m effective span has to carry a udl
of 24 kN/m (inclusive of self weight) under service conditions. The width of
beam is 250 mm and reinforce on tension side only. Design the smallest
section, calculate the depth of section and reinforcement. Use M20 & Fe 415.
Solution:
a) Factored load, wu = 1.5(DL+LL) = 1.5(24) = 36 kN/m
b) Maximum BM @ mid-span = 36 x 62/8 = 162 kNm
c) Depth of Beam :
'MU
We know that, BM s MR -> Mu = Qbd2; .-. d = [Q=kxjxO,36fck]
Effective depth required, d = f—-—-— = 484.5mm
V 2.76 x 250
Effective cover = 20 + 6 +20/2 = 36mm
Overall depth required = 520.5 = Say 525 mm
Hence provide D = 525 mm & /. Effective depth, d = 525-36 ~ 489 mm
d) Area of steel(reinforcement)
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Ast =
fck
2/y
i- i-
4.6 Mu
fck b d2
4.6 x 162 xlO6
20 x 250 x 4892
bd
= 1137.7 mm2 [0.0096x250x484.5 =1163]
Try 20 mm<J> -» No of bars, n = Ast/ast = 1137.7/314 = 3.62Nos
Provide 4 Nos of 20 mm<j) [Astprovided = 1256]
OR
Try 3 Nos of 20 mm <fr -> remainingsteel = 1137.7 - 942 = 195.7 mm2
No of 12 mm<f> bars required = 195.7/113 = 1.73 say 2 Nos
.'.Area of steel = 3 x 20<J. + 2 x 12<j> [Ast provided^1168]
Question 6: Determine Moment of Resistances of a rectangular section
reinforced with a steel of area 2600mm2 on the tension side. The width of the
beam is 300mm, effective depth 600mm. The grade of concrete is M20 &
Fe250 grade steel is used
Solution:
a) Depth of Limiting NA, xu max = 0.53 d = 0.53x600 = 318mm
b) Depth of NA, xu =
Q.87xfyXA$t 0.87X250X2600
= 261.8 mm <xumax
Q.36Xfckxb 0.36x20x300
c) MR,Mu= T xjd = Ast * 0.87fy ( d- 0.42xu)
= 2600 x 0.87x 250 [ 600 - 0.42 x261.8] = 277.12kNm
[ OR0.36x 20 x 261.8 x 300 ( 600 - 0.42x261.8 ) = 277.11 kNm ]
Question 7: A singly reinforced beam 300mmx600mm is reinforced with 4
bars of 20mm dia with an effective cover of 50mm. effective span is 4m.
Assuming M20 concrete & Fe415 steel, find the value of central load P that
can be carried by the beam
Solution:
a) Depth of Limiting NA, xu max = 0.48 d = 0.48x550 = 264 mm
u f x t A 0.87xfyxAst 0.87x415x1256
b) Depth ofNA, xu = Q36x/cfex, = 0.36X2QX300 = 209.9mm <xu max
c) MR, Mu = T xjd = Ast x 0.87fy ( d - 0.42xu)
= 1256 x 0.87x 415 [ 550 - 0.42 x209.9 ] = 209.44 kNm
[ OR 0.36x 20 x 209.9 x 300( 550- 0.42 x209.9 ) ]
d) We know that, BM = MR;
Self weight = (1 x 0.3x0.6) 25= 4.5 kN/m
/Wl w!2W 4.5x425 _ + — = 209.44 -> 1.5 —- + n
V 4 8 / V 4 8 J
.'. The valueof central load P that can be carried by the beam, W = 130.6 kN
5 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

6. Doubly reinforced sections:
Doubly reinforced sections are preferred, when the section dimensions are/is
limited (restricted), Or incase of restrained beams where both Sagging and
hogging moments occur in the beam
a) Depth of Limiting NA, xu,lim
From strain diagram
xu.lim 0.0035
0.0035+- fy -+0.002
l.lSEs
For fy 250, xulim/d = 0.53
For fy 415, xulim/d = 0.48
For fy 500, xu lim/d = 0.46
b) Depth of NA, xu
C ^ T
Cc + Cs =T
Cc = 0.36 fck xu b
Cs = Asc x (fsc-fcc)
T =Astx0.87fy
o fy/1.15E+0.002
0.446 fck
dc
fcc =0.446 fck, if ESC > 0.002
else fcc = 1000x0.446 fck [e - 250e2]
Note: fsc is always less than 0.87 fy in case of HYSD bars
It can be either found from strain diagram or the table 2
[ Initially fcc may be approximated to 0.446 fck
& fsc may be approximated to 0.87 fy ]
From equilibrium: C = T -» Cc + Cs = T
.-. (0.36 fck xu)b+Asc (fsc-fcc) = Ast x 0.87 fy d
.. xu Ast x 0.87 fy -Asc (fsc-fcc}
Hence, — = — —-—-
d 0.36 x/cfe xbxd
c) Lever arms: forCc ;jd 1= d - 0.42 xu
&forCs;jd2 = d-d?
If MR, Mu is to determined on the basis of Tension, jd = d- y
.... - Cc(0.42xu)+Cs(d')
Where, y = — —
J Cc+Cs
d) Moment of resistance: asC = T
MR; Mu = I (Cx jd ) = Cc xjdl + Cs x jd2
= [0.362 fck xu h ] [d - 0.42 xu] + [Asc(fsc-fcc)][d-d']
OR _
MR, Mu = T xjd - [Ast x 0.87 fy] x [d - y]
_Cs_
d -0.42xu
d - y
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Question 8: A rectangular beam 250 x 550 mm in M20 grade concrete is
provided with Fe 415 type reinforcement of 3 - 20 mm§ + 2-16 mm<}) on
tension side with an effective cover of 46 mm and 2 - 1 2 mm o on
compression side with an effective cover of 42 mm. Calculate the ultimate
moment capacity of the section.
Solution:
1. Depth of limiting NA, xu max = 0.48 d = 0.48x(550-46)=241.92 mm
2. Depth of NA, xu
Ast = 3x 314 +2x201 = 1344 mm2; Asc = 2x113 = 226 mm2.
As C=T -> Cc+Cs =T -> (b xu) 0.36 fck + Asc( fsc-fcc) = (Ast) ( 0.87fy);
Ast X O.o/ x fy —Asc(jsc — fcc)
u~
X~
0.36
Assuming fsc = 0.87 fy and fcc = 0.446 fck
1344 x 0.87 x 415 - 226(0.87 x 415 - 0.446x20)
MR
< 241.92
u 0.36x250x20
=225.372 mm
, Mu = I Cxjd = Cc x jdl + Cs x jd2
= 0.36 fck xu b [d-0.42 xu] + Asc( fsc-fcc) [d-d']
= 0.36 x 20 x 225.37 x 250 x (504-0.42x225.37)
+ 226 (0.87 x 415 - 0.446 x 20) x (504 - 42)
= 405666(504-0.42x225.37+79581.38 )x(504-42) = 202.824 kNm
OR
Ast x 0.87 x /y[0.42 xu] +Asc(fsc - fcc}[d]
y = -
x,, =
Ast x 0.87 x fy +Asc(fsc - /cc)
405666(0.42 x 225.37) + 79581.38 x 42
= 86.0198
405666 + 2958138
Mu = T xjd = [Ast x 0.87 fy] x [d - y] = 1344 x 0.87x 415[504-86.0198]
= 202.825 kNm
Question 9: Calculate the moment of resistance of a doubly reinforced RC
beam of rectangular section of size 300 x 500 mm reinforced with 4 - 25$
bars on tension side and 3 - 16$ bars on compression side. Use M20 and Fe
250. Assume an effective cover of 45 mm on both sides.
Solution:
1. Depth of limiting NA, max = 0.53 d = 0.53x(500-45}=241.2 mm
2. Depth of NA, xu
6 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

7. Cc+Cs =T -> (b xu) 0.36 fck + Asc( fsc-fcc) = (Ast) ( 0.87fy);
_ Ast X 0.87 x fy - Asc(fsc - fcc}
*u~ 0.36 xbxfck
Assuming fsc = 0.87 fy and fee = 0.446 fck
x,, =
4x491 x 0.87x 250 - 3x201 (0.87 x 250 - 0.446x20)
0.36 x 300 x 20
= 139.54 mm < 241,2
MR Mu = £ Cxjd = Cc x jdl + Cs x jd2
= 0.36 fck xu b [d-0.42 xu] + Asc( fsc-fcc) [d-d']
= 0.36 x 20 x 139.54 x 300 x (455-0.42x139.54)
+ 603 (0.87 x 250 - 0.446 x 20) x (455 - 45) = 171.04 kNm
Design of Doubly reinforced beam
A doubly reinforced beam is designed as Singly reinforced balanced section
PLUS Compression steel and additional tensile reinforcement to resist the
remaining moment.
Question 10 : A doubly reinforced beam of size 250 x 600 mm depth is
required to resist an ultimate moment of 310 kNm.Using M20 and mild steel
reinforcement, calculate the amount of steel required. The effective covers are
55 mm and 40 mm on tension and compression side respectively
Solution:
a) Ultimate moment of resistance of Singly reinforced balanced section =
Qbd2 = 2.76 x 250 x 5452 = 204.95 kNM
b) Area of steel required for SR Balanced section, Astl = 0.0096 x250 x545
= 1308mm2.
[OR Ast = l- i- 415 20
2/> fckb
C) Compressionreinforcement:
Balance Moment , Mubal = 310- 204.95 = 105.05kNm
Assuming fsc = 0.87 fy and fee = 0.446 fck
Mu bal = Asc( fsc-fcc) (d-d')
105.05 x 106 = Asc (0.87 x 415 - 0.446 x 20)(545 - 40) ; Asc = 590.75mm2
Try 20 mmfc No of bars = 590/201 = 2.9 say 3Nos
d) Additional tensile reinforcement
Mu bal= Ast2 x 0.87 fy (d-d') ,
105.05 x 106 = Ast2 (0.87 x 4 15)(545-40) ; Ast2 =576 mm2
e) Total tensile reinforcement
Ast = Astl + Ast2 = 1308 + 576 = 1884 mm2
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Try 25 mnn|) ; n= Ast/ast =1884/491 =3.83 say4Nos.
Question 11 :A doubly reinforced beam 230 x 500 mm size requiredto resist
an ultimate moment of 170 kNm and ultimate shear force of 80 kN. Using
M20 and Fe 415 steel, calculate the quantity of steel required and shear
reinforcement. The effective cover to tension steel is 50mm and compression
steel is 45 mm.
Solution:
a) Ultimate momentof resistance of Singly reinforced balanced section =
Qbd2 = 2.76 x 230 x 4502 = 128.55 kNM
b) Area of steel required for SR Balanced section, Astl = 0.0096 x230 x450
= 993.6 mm2.
[OR Ast = l- i- 415 20
c) Compression reinforcement:
Balance Moment , Mubal = 170 - 128.55 = 41.45 kNm
Assuming fsc = 0.87 fy and fee = 0.446 fck
Mu bal = Asc( fsc-fcc) (d-d')
41.45 x 106 = Asc (0.87 x 415-0.446 x 20)(450 -45) ; Asc = 290.65mm2
Try 12 mm<t>; No of bars = 290/113 = 2.6 say 3Nos
d) Additional tensile reinforcement
Mu bal = Ast2 x 0.87 fy (d-d')
41.45 x 106 = Ast2 (0.87 x 415)(450-45) ; Ast2 =283.47 mm2
.'. Total tensile steel, Ast = Astl + Ast2 = 993.6 + 283.47 = 1277 mm2
Try 25 mm$ ; n= Ast/ast =1277 7491 = 2.6 say 3 Nos.
e) Check for shear
Shear stress, tv = V/(bd) =80 x 103/(230 x 450) =0.773 MPa
Percentage of tensile steelt, p = 100 Ast/bd = 100xl277/(230x450) = 1.23%
Permissible shear stress, TC (Table 1 9- IS 456)
1% - - - 0.62
1.25--- 0.67
1.23 -» 0.62 + (0.05/0.25)(0.23) = 0.666 Mpa
iv> t c - Unsafe; Hence provide shear reinforcement
f) Design of shear reinforcement:
Design shear force, Vus = V - TC bd = 80x103 -0.666x230x450 =llxl03
Try 2 legged Vertical slirrups of 6 mm <jt , Asv = 2 x 28.3 = 58.6 mm2
4sv(0.87/»d 58.6(0.87 x 415)450
sv = -- -- = --—r-=- = 865 mmclc
Vus 11 x 103
Max permissiblespacing (smallerof the three)
7 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

8. i) 0.75 d = 0.75(450) = 337.5
ii) 300
....... . . ,- Asv(O.B7fy) 58.6(0.87X415) „„_
in) Minimum shear reinforcement, sv = —•—-*« = = = 229
0.46 0.4X230
/. Maximum permissible spacing is 229 mm c/c
Hence provide 6 mm<j> 2 legged vertical stirrups @ 200 c/c
J <J <J
:LQ c.
Singly reinforced sections
Question 12:Design a RC beam 350x700mm effective section, subjected to a
bending moment of 300kNm. Adopt M20concrete and Fe415 steel. Sketch
reinforcement details
Solution:
a) Design (Factored) Moment, Mu = 1.5 x 300 - 450 kNm
b) MR of a balanced section = Qbd2 = 2.76 x 350 x 7002 = 473kNm
Hence it is singly reinforced and under reinforced section
c) Area of tensile reinforcement
Ast =
fck
2fy
1- 1-
4.6 Mu
bd =
10
415
1-
4.6 x 450 x 106
20 x 350 x 7002
fck b d2
i
= 2186 mm2 [0.0096x350x700 =2352]
Try 25 mm<t>, n = 2186 7491 =4.45; provide 5 - 25$
bd
nooon
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Question 13: A reinforced concrete beam is supported on two walls 250mm
thick, spaced at a clear distance of 6m. The beam carries a super-imposed load
of 10 KN/m. Design the beam using M20 concrete and HYSD bars of Fe 415
grade.
Solution :
1. Assumption of dimensions: d *l/20k = 60007(0.75x20) = 400
Effective cover = 30 + 8 + 2072 = 48 mm -> say 50 mm
D = 450 mm, b » 0.5D =225 -» say 230mm
2. Effective span [22.2]
c7c of supports = 6+0.25 = 6.25m
Clear span + d = Lc + d = 6 + 0.4 = 6.4 m (Smaller)
Effective span = 6.25m
3. Loading: P""-^.
Selfwt = (lx0.23x0.45)25=2.5875kN7m ^^J
Superimposed load = 10 kN7m
Total, w= 12.5875 kN7m
Design load, wu = 1.5 x 12.5875 = 18.88 kN7m
4. Calculation of BM & SF
Max BM , Mu = wu x 1278 = 18.88x 6.25278= 92.19 kNm
Max SF, Vu = wu x 17 2 = 18.88 x 6.257 2 = 59 kN
5. Computation of effective depth, d =V(M7Qb) =V(92.19x10672.78x230)
= 379 mm < 400 mm Safe
[MR = C x jd = 0.36 fck b xc x (d-0.42 xc) = 0.138 fck bd2]
c • c Q.Sfck [, l^~ 4.6 Mu
6. Area of reinforcement, Ast = —7— 1- I I — - r
fy L fckbd2
10
415
1- 1-
4.6^92.19xl06
20 X 230 x 4002
230x400 = 774mm2
Try 16 mm $, No of bars = 7747201 = 4 bars
7. ShearReinforcement
Shear stress, TV= Vu7bd = 59,0007(230x400) = 0.64 N7mm2
Percentage of steel, p = 100 AsLOxi = 100x774 7(230x400)= 0.84%
0.75%-0.56
1.00%-0.62
TC = 0.84% - 0.56 +(0.0670.25)x0.09 =0.58 N7mm2.
TV > TC - unsafe - provide shear reinforcement
8 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

9. Design shear force, Vus = V - TC bd = 59,000 - 0.58(230x400) =
5640 N
Spacing of 6mm <f>, 2 legged vertical stirrups, sv = 0.87fy Asv d/Vus
= 0.87 x415 x (2 x28.27) x400/ 5640 = 1447 mm c/c
Maximum spacing [26.5.1.5]
(i) 0.75 d = 0.75(400) = 300
(ii) 300
(iii) Min. shear reinforcement 0.87fy Asv /0.4b
= 0.87x415x(2x28.27) /0.4x230 = 221
Provide 6 mm(|> 2 legged Vertical stirrups @ 200 mm c/c
Provide 2 nos of 6 mm <|> anchor bars to hold the stirrups
CHECK for 8
8. Check for bond
Development length, Id = <t>(0.87fy)/4Tbd = <t>(0.87x415)/(4xl.6xl.2)
= 47<[> = 47x16 = 752 mm
Ld/3 = 752/3 = 250 mm
Extend the 250 mm in side the support [26.2.3.3 a]
Ld<Ml/V +LO
[ The value of Ml/V in the above expression may be increased by 30% when
the ends of the reinforcementare confined by a compressive reaction]
Bent up 2 bars @ 0.15 1from support
Only 2 bars are available at support, Ml * M/2 =92.19/2 = 46.2 kNm
LO = 12$ or d (greater) 12<f> = 192 or 400 ; LO= 400
Ml/V + LO = 46.2x106/59xl03 + 400=1183 »752 Safe
9. Reinforcement Detailing
poor
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9 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

10. Winter 14
Q(5a)[ 14M]:
Q(5b)[ 14M]:
Q(6a)[ 14M]:
Q(6b)[ 14M] :
OR
Summer 15
Q (5a) [ 5M ]: Compare WSM and LSM
Q (5b) [ 9M ] : Derive Equation for limiting depth of neutral axis. Also sketch
the stress-strain diagram for beam in flexure. [ OR ]
Q (6a) [ 7M ]: Determine the service load which can be carried by a simply
supported beam of 5m span having cross section 230 x 450 mm effective, the
beam is reinforced with 3x16 mm § placed at an effective cover of 40 mm.
M20 and Fe 500.
Q (6b) [ 7M ]: Design a singly R/F rectangular beam for a clear span of 5 m
subjected to superimposed load of 12kN/m over the entire span. Use M20
and Fe415.
Winter 15
Q (5a) [ 6M ] : A RCC beam 230mmx 500mm is reinforced with 3 x 16mm
dia bars. Find moment of resistance if the effective cover is 40mm. Assuming
M20 concrete & Fe415
Q (5b) [ 7M ]: A singly reinforced beam of 4.5m span carries a udl of 30
kN/m inclusive of self weight. The width of beam is 230 mm and reinforce on
tension side only. Design the smallest section, calculate the depth of section
and reinforcement.Use M20 and Fe 415.
OR
Q (6a) [ 6M ]: Calculate the moment of resistance of a doubly reinforced RC
beam of rectangular section of size 300 x 450 mm reinforced with 6 - 20(|>
bars on tension side and 4 - 20<J> bars on compression side. Use M20 and Fe
250. Assume an effective cover of 35 mm on both sides.
Q (6b) [ 7M ]: A doubly reinforced beam of size 250 x 600 mm depth is
required to resist an ultimate moment of 310 kNm.Using M20 and mild steel
reinforcement, calculate the amount of steel required. The effective covers are
55 mm and 40 mm on tension and compression siderespectively
Summer 16
0 (5a) [ 6M ] : Explain : ^
i) Stress strain relationship for concrete,
ii) Stress strain relationship for steel in LSM.
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Q (5b) [ 7M ] : A rectangular beam is 20cm wide and 40cm deep up to the
centre of reinforcement. Find the area of reinforcement require if it has to
resist a moment of 25 kN/m. Use M 20 concrete mix and Fe 415 steel. Also
give check for sectors
Q (6a) [ 6M ]: Derive Equation for limiting Moment of resistances for
balanced, underreinforced and over reinforced section by LSM of singly
reinforced beam
Q (6b) [ 7M ]: A rectangular beam has a width of 250 mm and effective
depth of 500 mm. The beam is provided with tension steel of 5 -25 mm<t> and
compression steel of 2 - 25 mm <f>. The effective cover to the compression
steel being 50 mm. Calculate the ultimate moment capacity of the section of
fck = 20 MPa and fy = 250 MPa.
Winter 16
Q (5a) [ 7M ] : Design a RC beam 350- 700mm effective section, subjected
to a bending moment of 300kNm. Adopt M20concrete and Fe415 steel.
Sketch reinforcementdetails
Q (5b) [ 6M ] : Determine Moment of Resistances of a rectangular section
reinforced with a steel of area 2000mm2 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
OR
Q (6a) [ 7M ] : A singly reinforced beam 230mm • 600mm is reinforced with
4 bars of 16mm dia with an effective cover of 50mm. effective span is 4m.
Assuming M20 concrete & Fe415 steel, find the value of central load P that
can be carried by the beam
Q (6b) [ 6M ]: Derive Equation for limiting depth of neutral axis and moment
of resistances for balanced, under reinforced and over reinforced section by
using LSM
Summer 17
Q (5) [ 13M ]: Design a singly reinforced rectangularbeam for an effective
span of 5m subjected to a live load of 12 kN/m over the entire span. Calculate
main reinforcement and shear reinforcement. Give all necessary checks as per
IS 456. Draw the sketch of reinforcement details.
OR
Q (6) f 13M ] : A doubly reinforced beam 250 x 600 mm size required to
resist an ultimate moment of 310 kNm and ultimate shear force of 70 kN.
Using M20 and I'c 415 steel, calculate the quantity of steel required and shear
1 0 | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

11. reinforcement. The effective cover to tension steel is 55mm andcompression
steel is 40 mm.
JLV.k.1 1J -TV/ 1H111.
Recap from Structural Analysis
Support Reactions, SFD. BMD. 6, 8 & FEM
I ) Cantilever beam with concentrated load at free end
A
•
^- v
2) Cantilever beam with udl overthe entire span
. '' * . ' • • • ' ^ " '
* •*
3) SS Beam with concentrated load at mid-span
W
A | B
t
*- _ |_ — — — — -*
N/2 W/2|
W/2 +
W/2
SFD
D
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^^^-""<^r~
^^^^"^ " ^*^"wi^-^
BMD
:
t^^~^ -" t
Elastic Curve
GA = - GB = WL2/8EI
8c =WL3/48EI .
" FEM =WL/8
4) SS Beam with concentrated load at ccL from A
5) SS Beam with udl over the entire span
• ~n
4
?•<
*
*
•
l l | P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

12. aaaaaaaa
reinforced R.C. beam of Rectangular section of size 300 x 450 mm.
reinforced with 6-20 mm dia bars on tension side and 4-20 mmdia
bars on compression side.
Use M20 grade concrete and Fe 250 grade steel.
Assume eff. cover of 35 mm on both sides. (Use L. S.M.)
Q (6b) [ 7M]: A doubly reinforced beam of size 250mmx600mm
deep is required to resist on ultimate moment of 310 kNm. Using
concrete M 20 and mild steel reinforcement, Calculate the amount of
steel required. The effective cover to tension steel is 55mm whilethat
for compression steel is 40mm. (Use L.S. M.)
TABU A SAUCNT POINTS OH THE DC6KJN 6THC56 STRAIN OJHVC fOU
CnitVWORKED BARS
( fftiw I i S ;: i.
/, - MO N,'mm«
II)
«»/H
CHJ/H
VW/ni
at!/*
0 «T1 /,«
10/H
' Stnun
Ci
0-00144
OO31H
<KOI«
0OIK 41
now 7*
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Strut
01
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W4-I
M7I
!l 1
]60«
'smin
(4>
O001 74
00014)
(KXUM
OflCCTT
0001 11
MM I
I
Sttw
(i)
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)9C)
4110
4M-*
4U|
SJTT .. 1 irtrlr iM«f«UliM MJty M 04M l« iftlffflVOitlt HWtL
Winter 15
Q (5a) [ 6M ]: A RCC Beam 230 X500 mm is reinforced with
3-16mm dia bars. Find the moment of Resistance, if effective cover is
40 mm and effective span 3 m. Use M 20 concrete and Fe 415 steel.
(Use L.S.M.)
0 (5b) [ 7M ]: A simply supported beam of 4.5 m span carries
a udL of 30 kN/m inclusiveof self wt. The width of beam is 230 mm
and is reinforced on tension side only. Design the smallest section,
calculate depth of section and reinforcement. Use M20 concrete and (
Fe250 steel. (Use L. S. M.)
Q (6a) [ 6M ] : Calculate the moment of resistance of a doubly
TABLE E MAXIMUM PERCENTAGE c*
TENSILE REINFORCEMENT ft^ FOR
SINGLY RHWOROU1 RKTAWOUl-AR
15
j*
M
2)0
1-13
1'TC
2")0
1-44
411
»n
ii»
r4j
wo
0-J7
0-M
OM
I I )
C LIMITING MOMLNI OF
RF5ISTASCE AND RUM ORCIMI.ST
K)R ilNGLV
tilmaf
(ClMf JJ)
014*
:I*T
ulM 01)1
12 [ P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n g , N a g p u r

13. TABLE D LIMITING MOMtNT OF
R1SISTASO- rA4.TUR W,^_/W. S mm* TOR
SINGLY REINFORCFO RICIANCULAR
SECTIONS
A*.
II
jj
»
' 1)0
I 14
1-M
1TJ
4-47
£. N/rnm'
41}
107
216
1-4J
414
MO
JflO
I'M
J.j)
39»
TABLE F STRESS IN COMfUFSSKJN
RtlSfORCSMtNT/U, H'nxo* IN DOUBLV
-- - - MAM WITH rrwn.
WORKeO BARS
(f/«.i* ;j «
N/«>
41)
OOJ
JJ5
<HO
]»
411
en
J4J
m
Summer 15
( la)[7]Discuss the merits and demerits of Working Stress method
(Ib) [6] Derive design constants for neutral axis, lever arm, constant for
moment of Resistance and constant for percentage of steel in WSM
OR
(2a)[6]Explain under reinforce, over reinforced and balanced section in WSM
(2b)[7] A singly reinforced concrete beam is of width 400 mm and effective
depth 615 mm. It is reinforced with 8-20<J> mild steel bars. Assume M25 grade
concrete, determine its moment of resistance according to WSM.
(3a)[6] Explain the advantages of pre-stressed concrete over RCC
(3b)[7])Explain with the help of neat sketches any two of the followingpre-
stressing systems
(i) Freyssinet system
(ii) Magnel-Blaton system
(iii)Gifford Udall system
OR
(4X13]Explain
(i) Different types of losses in pre-stressed concrete beam
(ii) Pre tensioning and post tensioning
(iii) Application of Pre-stressed concrete.
(5a)[5] Compare WSM and LSM
[5b)[9]Derive equations for limiting depth of neutral axis. Also sketch the
stress and strain diagram for beam in flexure
Or
(6a)[7]Determine the service load which can be carried by Simply supported
beam of 5.0m span having 230 x 450 (eff), the beam is reinforced with 3-16<J>.
Eff cover is 40 mm. M20 and Fe 500 are used.
(6b)[7] Design a Singly reinforced rectangular beam for a clear span of 5.0m
subjected to a superimposed load of 12 kN/m over the entire span. Use M20
and Fe415
(7a)[6]Determine the MR of a T-beam from the following details
bf = 1000mm, Df = 120 mm, Ast = 6 -25<J> , d = 600 ,„ bw = 300 mm, M20
and Fe415
..
(7b)[7] Design a RCC Column of Rectangular section having unsupported
length of 3m subjected to an axial compressive laod of 1200kN using M20
and Fe415. One dimension is restricted to 400 mm.
OR
(8)[13] Design a rectangular pad footing for a column of size 300 x 500 mm
with compressive load of 1000 kN. Use M20 concrete with Fe 415 steel, the
density of soil is 21kN/m3 and SBC 150 kN/m2. Give all the necessary
checks as per IS 456 with neat sketch.
(9a)[7] What is limit state of serviceability? How it is ensured for beams?
(9b)[6]ExpIain in brief the various measures for deflection control as per IS
456-2000
OR
(10)[13]Design a rectangular beam section of size 250 x 450 mm subjecred to
a bending moment of 30 kNm, Shear force of 40 kN and torsional moment of
20 kNm at working condition. Use M20 And Fe 41 5.
(1 1)[14] Design a cantilever slab projecting 1.5 m from rhe face of column,
the slab carries live load of 1.5 kN/m2. Use M20 andf Fe 415. Sketch
reinforcement details.
OR
(12)[14]Design a slab panel for a hall of size 4 * 5.5msupported on 230 mm
thick brick wall all around, the slab carries superimposed load of 2.5 kN/m2
with floor finish of 0.75 kN/m2. Use M20 Grade of Concrete and Fe 415 type
13 ( P a g e R V R K P r a s a d , K D K C o l l e g e of E n g i n e e r i n a g p u r