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S.1Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
WARNING : Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL proceedings.
S e t - 1S o l u t i o n s
Code No: 09A50102/R09
III B.Tech. I Semester Examinations
May/June - 2012
DESIGN OF REINFORCED CONCRETE STRUCTURES
( Civil Engineering )
Time: 3 Hours Max. Marks: 75
Answer any FIVE Questions
All Questions carry equal marks
- - -
1. (a) What are the code recommendations for limit state design? (Unit-I, Topic No. 1.1)
(b) Draw the characteristic and design stress curves for concrete and steel (Fe415
). (Unit-I, Topic No. 1.1)
(c) Calculate stress block parameters of concrete. [15] (Unit-I, Topic No. 1.2)
2. Determine the moment of resistance of a T-beam section with the details given below.
bw= 250 mm, d = 550 mm, b
f= 1200 mm, D
f= 100 mm, A
st= 2946 mm2, f
ck= 25 MPa, f
y= 415 MPa. [15]
(Unit-II, Topic No. 2.2)
3. Design the reinforcement in a rectangular beam of section 300 mm 600 mm subjected to an ultimate twisting moment
of 50 kN-m combined with an ultimate bending moment of 120 kN-m. Assume M25
concrete and Fe415
steel. [15]
(Unit-III, Topic No. 3.1)
4. Design a slab for a two-room masonry building with internal dimension of each room as 4.00 m 5.00 m with one long
edge continuous. The masonry wall are of 300 mm thick. Assume a live load of 4.0 kN/m2and a finish load of 1.5 kN/
m2. Assume that the slab corners are not free to lift up. Consider M20
concrete and Fe415
steel. [15]
(Unit-IV, Topic No. 4.1)
5. Design an isolated footing for a column with an axial force of 2000 kN under working load. The size of the column is
300 mm 600 mm. Consider S.B.C of soil as 250 kN/m2. Consider M20
Concrete and Fe415
steel. Assume mild exposure
condition. [15] (Unit-V, Topic No. 5.1)
6. A corner column 300 mm 450 mm located in the multi storey of a system of braced frames, is subjected to factored
loads Pu
= 1600 kN, Mux
= 175 kNm and Muy
= 100 kN-m. The unsupported length of the column is 3.0 m. Design the
reinforcement in the column, assuming M30
concrete and Fe415
steel. [15] (Unit-VI, Topic No. 6.1)
7. (a) Distinguish between short and long term deflection and what are the limits prescribed by the code IS:456.
(Unit-VII, Topic No. 7.1)
(b) Explain effective and cracked moment of inertia. [7+8] (Unit-VII, Topic No. 7.1)
8. Design a suitable dog-legged stair in a public building, to be located in a staircase room 6 m long, 3.0 m wide and
the floor height is 3.5 m. Use M20
concrete and Fe415
steel. [15] (Unit-VIII, Topic No. 8.1)
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SOLUTIONS TO MAY/JUNE-2012, SET-1, QP
Q1. (a) What are the code recommendationsfor limit state design?
Answer : May/June-12, Set-1, Q1(a)
Indian standard code has prescribed the following
recommendations for limit state design,
(i) Characteristic strength and loads (consider-
ations)
(ii) Partial safety factors for loads and materials
(iii) Design stress-strain curve for concrete
(iv) Design stress-strain curve for reinforcing steel.
(i) Characteristic Strength and Load Considerations
A characteristics load can be defined as the load which
has 95% probability of not exceeding its defined value dur-
ing the entire life of the structure.
In case, if the statistical data for calculating charac-
teristic strength or load is not available, the design charac-
teristic load values for dead, live and wind loads are recom-
mended to follow the code from IS - 875 : 1987 whereas the
seismic load follows IS : 1893 : 2002 standard.
(ii) Partial Safety Factor For Loads and Materials
In material, the partial factor of safety is employed so
as to reduce the characteristic strength value and obtain the
design strength values.
In case of loads, the partial safety factor is used to
increase the characteristic load and obtain the design load.
The partial safety factor commonly used in limit state method
are,
c= Partial safety factor for concrete = 1.5
s= Partial safety factor for steel = 1.15
f= Partial safety factor for loads = 1.5
Where, c=
dc
ck
f
f,
s=
ds
y
f
f,
f=
df
f
Where,fde
,fds
are design characteristic strengths.
fd= Design load.
(iii) Design Stress-strain Curve for Concrete
The curves shown in figure (1) adopted by IS-456 :
2000 for concrete under flexural compression. The maximum
stresses in characteristic and design curves are limited to
0.67fck
and 0.45fck.
0 0.001 0.002 0.003
0.45 fck
0.003
0.67 fck
Idealized curve
Characteristic curve
Design curve for concrete
Stress
fck
Strain
0 0.001 0.002 0.003
0.45 fck
0.003
0.67 fck
Idealized curve
Characteristic curve
Design curve for concrete
Stress
fck
Strain
Maximum characteristic stress,fc max
=5.1
ckf = 0.67fck
Maximum design stress, fd max
=5.15.1
ckf= 0.44f
ck
Where, 1.5 is a partial safety factor.
Figure (1): Stress Strain Curve of Concrete
(iv) Design Stress-strain Curve for Reinforcing Steel
0.0000
500.25 f
y
1000.50 f
y
150
2000.75 f
y
250 fy
0.87fy
Characteristic curve
Design curve m
fy
m= 1.15
Designyiel
dstrain
Es
0.87fy
y =
0.002 0.004 0.006 0.008
Strain
Stress
0.0000
500.25 f
y
1000.50 f
y
150
2000.75 f
y
250 fy
0.87fy
Characteristic curve
Design curve m
fy
m= 1.15
Designyiel
dstrain
Es
0.87fy
y =
0.002 0.004 0.006 0.008
Strain
Stress
0.0000
500.25 f
y
1000.50 f
y
150
2000.75 f
y
250 fy
0.87fy
Characteristic curve
Design curve m
fy
m= 1.15
Designyiel
dstrain
Es
0.87fy
y =
0.002 0.004 0.006 0.008
Strain
Stress
Figure (2): Stress Strain Curve for Mild Steel
The curve was given by IS - 456 : 2000. The charac-teristic strength is considered equal to the yield stress, while
the design strength is obtained by using a partial safety
factor of 1.15. As mild steel comprises of a well defined yield
point, it is linearly elastic to design stress 0.87fy
and hence
the design stress is considered to be constant.
The high yield steel bars do not have a well defined
yield point and hence it is assumed that the linear elastic
behaviour changes to linear inelastic behaviour at a strength
of 0.8fyin the characteristic curve. The design curve is as-
sumed to transform from elastic to inelastic state at a stress
level of 0.8fyd
.
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S.3Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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Where,fyd
= Design yield strength.
0.001 0.002 0.003 0.00S0.004
0.00070.00030.0001
500
400
300
200
100
0
Fe500
Fe415
0.001 0.002 0.003 0.00S0.004
0.00070.00030.0001
500
400
300
200
100
0
Fe500
Fe415
Figure (3): Stress-Strain Curve for Steel (Fe415
and Fe500
)
(b) Draw the characteristic and designstress curves for Concrete and Steel(Fe
415).
Answer : May/June-12, Set-1, Q1(b)
Stress-Strain Curve for Concrete
For answer refer May/June-12, Set-1, Q1(a), Topic:
Figure (1).
Stress-strain Curve for Steel
For answer refer May/June-12, Set-1, Q1(a), Topic:
Figure (2).
(c) Calculate stress block parameters ofconcrete.
Answer : May/June-12, Set-1, Q1(c)
For answer refer Unit-I, Q8.
Q2. Determine the moment of resistance of aT-beam section with the details given below,
bw= 250 mm, d = 550 mm, b
f= 1200 mm,
Df= 100 mm, A
st= 2946 mm2, f
ck= 25 MPa,
fy= 415 MPa.
Answer : May/June-12, Set-1, Q2 M[15]
Given that,
Width of web, bw= 250 mm
Width of flange, bf= 1200 mm
Depth of flange,Df= 100 mm
Depth of web, d= 550 mm
Area of steel,Ast= 2946 mm2
M25
grade,fck
= 25 MPa = 25 N/mm2
Fe415
steel,fy= 415 MPa = 415 N/mm2
bf= 120 mm
d=
550
mm
450mm
250 mm
Df= 100 mm
Ast
= 2946 mm2
(1 MPa = 1 N/mm2)
bf= 120 mm
d=
550
mm
450mm
250 mm
Df= 100 mm
Ast
= 2946 mm2
(1 MPa = 1 N/mm2)
Figure
Let the neutral axis falls in the flange (i.e.,x4is less
than depth of flange Df)
ux
d=
dbf
Af
fck
sty
36.0
87.0
ux
d=
55012002536.0
294641587.0
ux
d= 0.179
xu
= 0.179 d
xu
= 0.179 550
xu = 98.450 mm xu(Hence safe)
The beam is under-reinforced
The moment of resistance,MR
MR
= 0.87fy
Ast
d
dbffA
fck
yst1
Moment of resistance,MR
= 0.87 415 2946 550
550120025
41529461
MR
= 541.662 106N-mm
MR
= 541.662 kN-mm
Therefore, the moment of resistance of T-beam,
MR
= 541.662 kN-m.
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Q3. Design the reinforcement in a rectangularbeam of section 300 mm 600 mm subjectedto an ultimate twisting moment of 50 kN-m
combined with an ultimate bending momentof 120 kN-m. Assume M
25concrete and Fe
415
Steel.
Answer : May/June-12, Set-1, Q3 M[15]
Given that,
Width of the beam, b= 300 mm
Overall depth,D= 600 mm
Ultimate twisting moment, Tu= 50 kN-m
Ultimate bending moment,Mu= 120 kN-m
For M25
concrete,fck
= 25 N/mm2
For Fe415
concrete,fy= 415 N/mm2
Provide an effective cover of 40 mm all around the
beam
Effective depth, d = 600 40 = 560 mm
Design of Longitudinal Reinforcement
Equivalent bending moment due to torsion,
Mt= 1.7
uT
+b
D1
=7.1
50
+
300
6001
Mt= 88.235 kN-m
Total bending moment for design,Meq
Meq
=MtM
u
= 88.235 120
Meq
= 208.235 kNm or 31.765 kN-m
Meq1
= 208.235 kNm () orMeq2
= 31.765 kN-m ()
The momentMeq1
= 208.235 kNm is consider for pro-
viding flexural steel at top, while the momentMeq2
= 31.765
kNm is considered for providing flexural steel at bottom.
At Top
R1
=1
2
eqM
bd=
2
6
560300
10235.208
R1
= 2.213 N/mm2
100
tP=
bd
Ast=
y
ck
f
f
.2
ckf
R1598.411
100
tP=
4152
25
25
213.26.411 = 0.006929510
100
tP ~ 0.007
Area of steel required,
(Ast1
)req
=100
tP bd
= 0.007 300 560
= 1176 mm2
For a 20 mm bars, we get
Number of bars, n =1st r eqA
A=
u
220
1176
= 3.743
n ~ 4
Provide4 bars of 20 mm at top.
At Bottom
R2
= 22
bd
Meq= 2
6
560300
10765.31
R2
= 0.338 N/mm2
100
tP=
bd
Ast=
y
ck
f
f
.2
ckf
R26.411
100
tP=
4152
25
25
338.06.411
100
tP= 0.0009516600 ~ 0.001
(Ast2
)req
=100
tP= bd
= 0.001 560 300 = 168 mm2
Minimum steel =stA
bd=
0.85
yf
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S.5Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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(on (Ast)
min=
415
85.0 300 560)
Ast,min
= 344.096 mm2> (Ast2
)req
Provide minimum area of steel.
(Ast2
)req
= 344.096 mm2
For a 12 mm bars, we get,
Number of bars =2( )st reqA
A=
4/)12(
096.3442
= 3.042 4
Provide 4 bars of 12 mm
At Side
Since overall depthD> 450 mm, torsional reinforce-
ment is provided at sides
(Ast3
)req
= 0.001 bD
= 0.001 300 600
(Ast3
)req
= 180 mm2
Provide minimum area of steel (Ast3
)req
=Amin
.
Provide 4 bars of 12 mm
40 mm
40 mm 4 bars of 12 mm
4 bars of 12 mm
4 bars of 20 mm
600mm
300 mm
40 mm
40 mm 4 bars of 12 mm
4 bars of 12 mm
4 bars of 20 mm
600mm
300 mm
Figure: Reinforcement in a Rectangular Beam
Q4. Design a slab for a two-room masonry build-ing with internal dimension of each room as4.00 m 5.00 m with one long edge continu-ous. The masonry wall are of 300 mm thick.Assume a live load of 4.0 kN/m2and a finishload of 1.5 kN/m2. Assume that the slab cor-ners are not free to lift up. Consider M
20con-
crete and Fe415
steel.
May/June-12, Set-1, Q4 M[15]
Answer :
Given that,
Internal dimension of slab = 4.00 m 5.00 mThickness of walls = 300 mm
Live load on slab = 4.0 kN/m2
Floor finish load = 1.5 kN/m2
For M20
concrete,fck
= 20 kN/mm2
For Fe415
steel,fy= 415 N/mm2
One long edge is continuous.
Design Constants
For M20
concrete,
, m a xuxd
=41587.01100
700+
= 0.479
Ru= 0.36f
ck.
, m a xux
d
d
x max,.416.01 4
= 0.36 20 0.479 (1 0.416 0.479)
Ru= 2.762
Loading Calculations
Consider the span/depth ratio as the average of sim-
ply supported slab and continuous slab. As high strength
bars of steel 415 are used multiply the value with a factor of0.8.
Span
Depth, x
l
d=
35 400.8
2
+ [Qfrom figure (1)]
d
4300= 30
d=30
4300
d= 143.333 mm
Assume a nominal cover of 20 mm and using 8 mm bars
Overall depth,D= 143.333 + 20 + 4 = 167.333 mm
Take overall depth,D= 170 mm
Effective depth, d = 170 20 4 = 146 mm
Dead load of slab = 0.17 1 1 25 = 4.25 kN/m2
Live load on slab = 4 kN/m2
Floor finish load = 1.5 N/m2
Total load = 9.75 kN/m2
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Factored load, wu
= 1.5 9.75 = 14.625 kN/m2.
Effective length, lex
= 4300 + 146 = 4446 mm
ley
= 5300 + 146 = 5446 mm
r =ey
ex
l
l=
4446
5446= 1.225 1.225 < 2
The slab is a two way slab.
Moment in shorter span,Mux
= x. w
u. l
x
2
Moment in longer span,Muy
= y. w
u. l
x
2
Where, x,
yare obtained (from table 26 of IS 456 :
2000)
For one long edge continuous, we get,
x +ve
at mid span = 0.054
x veat continuous edge = 0.072y + ve
at mid span = 0.043
y ve
at continuous edge = 0
Muxve
= xve
wu l
x
2
= 0.072 14.625 4.4462
Muxve
= 20.815 kN-m
Negative moment at shorter span, M
ux ve= 20.815 kN-m
Mux + ve
= x + ve
wu l
x
2
= 0.054 14.625 4.4462
Mux + ve
= 15.611 kN-m
Positive moment at shorter span,
Mux +ve
= 15.611 kN-m
Mux + ve
= y + ve
14.625 4.662
= 0.043 14.625 4.4462
Muy + ve
= 12.431 kN-m
Positive moment at longer span,
Muy +ve
= 12.431 kN-m
Width of Strips
(i) Short Span
Width of middle strip =43 l
ey=
43 5.446 = 4.085 m
Width of edge strip =2
1(5.446 4.085) = 0.681 m
(ii) Long Span
Width of middle strip =4
3. l
ex=
4
3 4.446 = 3.335 m
Width of edge strip =2
1(4.446 3.335) = 0.556 m
Effective Depth Calculations
Effective depth from moment consideration is given
by,
d=bR
M
u
veux
.
=
1000762.2
10815.206
d= 86.811 mm
Therefore, provide depth according to deflection
considerationD= 170 mm
Effective depth in short span,dx= 170 20 4 = 146 mm
Effective depth in long span, dy= 146 8 = 138 mm
Area of Steel
Short Span
For an under reinforced section, area of steel at con-
tinuous edge.
vexstA 1 =y
ck
f
f5.0 2
4.61 1
.
ux ve
ck x
M
f bd
bd
=415
205.0
6
2
4.6 20.815 101 1
20 1000 146
1000 146
vexst
A1
= 420.159 mm2.
Minimum steel,
1 minst
A =100
12.0 1000 146
= 175.2 mm2 < vexstA 1
Provide 8 mm bars at spacing, s =1st x ve
A
A
1000
=159.420
4
82
1000
= 119.634 ~ 115 mm
Provide 8 mm bars at spacing of 115 mm c/c.
vexstA +2 =y
ck
f
f5.0 2
4.61 1 ux ve
ck x
M
f bd
+
bdx
=415
205.0
6
2
4.6 15.611 101 1
20 1000 146
1000 146
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S.7Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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vexstA +2 = 309.951 mm2 >Amin
Provide 8 mm bars at spacing of,
S=951.309
4
82
1000
S= 162.172 mm 160 mm. Provide 8 mm bars at 160 mm c/cLong Span
Ast,y+ve
=y
ck
f
f5.0
+
2
6.411
yck
veuy
bdf
Mbd
y
veystA +, =415
205.0
26
138100020
10431.126.4
11
1000 138
veystA +, = 259.765 mm2>A
stmin.
Provide 8 mm bars at spacing,
S=765.259
4
82
1000 = 193.504
S~ 190 mm.
Provide 8 mm bars at 190 mm c/c.
Check for ShearShort Span
Shear force at edge of longer span
Vx
= wu.l
ex.
r
r
+2
= 14.625 4.446 225.12
225.1
+ V
x= 24.699 kN
Shear stress, vx
=xV
bdx=
324.699 10
1000 146
vx= 0.169 N/mm2Allowable shear stress in concrete,
bdx
A prvexst )(100 1 =
1461000
10001154
8100
2
=43709.115
146000= 0.299
From IS - 456 : 2000, table no: 19 ;c= 0.384 N/mm2>
vx
Hence safe.
Long Span
Shear force at edges of shorter span
Vy=y3wu. lex= 3
1
14.625 4.446
Vy= 21.674 kN
Shear stress, vy
=.
yV
b dy
=1381000
10674.213
vy
= 0.157 N/mm2
Allowable shear stress in concrete,
bdy
A prvesty )(100 + =1381000
1000190
4
8
100
2
= 0.192
From IS- 456 : 2000, table no : 19 ; c= 0.314 N/mm2>
vy
Hence safe.
Design of Torsional Reinforcement
Since the ends of the slabs are not allowed to lift up,
torsional reinforcement is required size of mesh,
lm
=5
exl=
5
446.4= 0.889 m from the centre of sup-
port.
or lm
= 0.889 +2
300.0= 1.039 1m from edge
Area of torsional steel,
Ast1,t
=4
3. vexstA 1
=4
3 420.159
Ast,t
= 315.119 mm2 Provide 8 mm bars at spacing of,
S=119.315
4
82
1000 = 159.513 mm
S 155 mm Provide 8 mm bars at 155 mm c/cNote
1. The area of steel vexstA 1 (with obtained spacing) at
continuous edge and at middle strip of length 4.085
m provide vexstA 2
2. Similarly, the area of steel vexstA +1 is provided only at
middle strip of length 3.335 m. The edge strip lengthof 0.556 m is provided with an area of steel equal to(1.2 D), whereD= Overall depth of slab.
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8/10/2019 Hyd Civil 3 1 Drcs Set 1
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S.9Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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M20
grade concrete,fck
= 20 N/mm2
Fe415
steel,fy= 415 N/mm2
d
xu max = 0.48
(For Fe415
d
xu max = 0.48from IS-456 : 2000 clause 38.1,
Pg.No : 70)
Ru
= 0.36fck
d
xu max,
d
xu max,416.01
Ru
= 0.36 20 0.48 (1 0.416 0.48)
Ru
= 2.766
(i) Size of Footing
Assume, weight of footing, w' = 10% of P
=100
10 2000
= 200 kN
Area of footing,A ='P w
SBC
+
=250
2002000 +
Area,A = 8.8 m2(L b)
Let,b
L=
600
300= 0.5 = 1/2
1/2LL= 8.8 L2 = 17.6
4.195mL = 4.2 m
B = 1/2 l
B = 0.5 4.2
2.1mB =
Provide footing of size,L B = 4.2 m 2.1 m
Net Upward Pressure Intensity (NUPI)
Pu
=BL
P
Pu=
1.22.4
2000
Pu= 226.757 kN/m2
(ii) Design of Section
On The Basis of Bending
Bending moment on the face AC of column,
M1
=8
UBP(la)2
=8
1.2757.226 (4.2 0.6)2
= 771.427 kNm
0.9 m
0.3 m
0.9 m
2.1 m
1.8 m 1.8 m0.6 m
A
4.2 m
A
D C
B
B
D C
0.9 m
0.3 m
0.9 m
2.1 m
1.8 m 1.8 m0.6 m
A
4.2 m
A
D C
B
B
D C
Figure (1)
M1
= 771.427 106 N/mm
Bending moment on the faceAB,
M2
=8U LP (B b)2
=8
2.4757.226 (2.1 0.3)2
M2
= 385.714 kNm
M2
= 385.714 106Nmm
Ultimate Moments
Mu1
= 1.5 771.427 106
Mu1
= 1157.141 106Nmm
Mu2
= 1.5 385.714 106
Mu2
= 578.571 106Nmm
Depth of Footing (d)
d =1u
u
M
R b(M
u1>M
u2)
d =300766.2
10141.11576
d = 1180.882
d 1181 mm
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Nominal cover = 40 mm
Assuming 12 mm main bars
Overall depth,D= 1181 + 40 + 12/2 D= 1227 mm
Provide over all depth of D 1230 mm
Effective depth, d= 1230 40 6
d= 1184 mm
This depth is provided near the face of the column.
As the bending moment and shear force changes abruptly
near the edge.
Provide edge thickness of 0.2 d
= 0.2 1184
= 236.8 mmProvide over all depth of 285 mm
Effective depth of edge, de= 285 40 6 = 239 mm >
236.8 mm
Hence safe.
Checks
(i) Check for One-Way Shear
Critical section for one-way shear acts at a distance
of (d = 1184 mm) from the face of the column.
Shear force at critical section,
V= PuB{ 1/2 (L a) d }
= 226.757 2.1 { 1/2 (4.2 0.6) 1.184 m}
V= 293.333 kN
Vu= 1.5 V= 1.5 293.333
Vu= 440 kN
Effective Depth of Footing at Critical Section, d'
d'= de+
a
dd e
(a' d)
= 239 + 1800
2391184 (1800 1184)
d'= 562.4 mm
Top width of Footing at Critical Section, b'
b'= a +a
aL
d
b'= 600 +
1800
6004200 1184
b'= 2968 mm
Assumeux
d= 0.4 (For under reinforced section)
xu
= 0.4 562.4
xu= 224.960 mm
Width at neutral axis,( )
( )n u
e
L bb b x
d d
= +
nb = 2968 +(4200 2968)
(562.4 239)
224.960
nb = 3824.990
nb ~ 3825 mm Nominal shear stress,
v
v=
u
n
Vb d
v=
4.5623825
104403
v= 0.205 N/mm2
Assuming,P = 0.7% (For an under reinforced section)
c= k permissible stress
k = 1 (QD> 300 mm)
Permissible stress for m20
concrete ~ 0.35 N/mm2
(From IS 456 : 2000 Table no:23 Pg . No: 84)
c = 1 0.35
c= 0.35 N/mm2>
v
Hence safe.
(ii) Check for Two-Way Shear
Perimeter of (.PQRS)critical section of two-way shear,
P = 2 [(a+ d) + (b+d)]
= 2 [a+ b+ 2d ]
= 2 [600 + 300 + 2 1184]
P = 6536 mm
A
D C
B
d/2d/2
d/2
d/2
P Q
SR
D C
A B
1.273 m 0.527 m 0.527 m 1.1273 m0.6 m
4.2 m
A
D C
B
d/2d/2
d/2
d/2
P Q
SR
D C
A B
1.273 m 0.527 m 0.527 m 1.1273 m0.6 m
4.2 m
Figure (2)
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S.11Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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Area of section PQRS,
A1
= (a+ d) (b+ d)
= (0.6 + 1.184) (0.3 + 1.184)
A1
= 2.647 mm2
Punching shear, Vp
Vp
=Pu [AA
1]
Vp
= 226.757 [(4.2 2.1) 2.647]
Vp
= 1399.771
Vp
~ 1400 kN
Shear stress, vp
= 1.5 .
pV
Pd
= 11846536
1014005.13
vp
= 0.271 N/mm2.
Allowable shear stress,
c= 0.25 ckf = 0.25 20
c= 1.118 N/mm2
Ks= (0.5 +
c) =
+
600
3005.0 = 1
Permissible stress = Ks
c
= 1 1.118
= 1.118 N/mm2> vp
Hence safe.
Reinforcement Design
As the section is an under reinforced section (dpr
> dreq
) the area of steel is given by,
Ast1
= 0.5 y
ck
f
f
2
16.411bdf
M
ck
ubd
= 0.5 415
20
2
6
118430020
10141.11576.411 300 1184
Ast1
= 3372.751 mm2
Provide 16 mm bars
Number of bars, n=A
Ast1=
4/)16(
751.33722
= 16.775 17 no.s
Therefore, provide 18 bars of 16 mm , which should be spaced uniformly along the length of 4.2 m(iii) Along Width
Ast2
= 0.5 y
ck
f
f
2
2
..
6.411
daf
M
ck
u a d
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S.13Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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Q6. A corner column 300 mm 450 mm locatedin the multi storey of a system of bracedframes, is subjected to factored loads P
u=
1600 kN, Mux = 175 kNm and Muy = 100 kN-m.The unsupported length of the column is 3.0m. Design the reinforcement in the column,assuming M
30concrete and Fe
415steel.
Answer : May/June-12, Set-1, Q6 M[15]
Given that,
Size of column, bD= 300 450 mm
Factored load, Pu= 1600 kN
Factored moment in x-direction,Mux
= 175 kNm
Factored moment in y-direction,Muy
= 100 kNm
Unsupported length, l = 3.0 m = 3000 mmM
30concrete f
ck= 30 N/mm2
Fe415
steel fy= 415 N/mm2
Slenderness ratios,
exl
b= k
x.
b
l= k
x.
300
3000= 10 . Kx
eyl
D= k
y.
D
l= k
x.
450
3000= 6.67 Kx
For braced columns, kx, k
y< 1
exl
b,
eyl
D< 12
Therefore, the column will be designed as a short
column.
Check for Minimum Eccentricities
Eccentricities of loads,
ex
=1600
101753
= 109.375 mm
ey = 1600
101003
= 62.50 mm
Minimum eccentricities,
minxe 1 = 500
l+
30
b=
500
3000+
30
300= 16 mm
minye 1 = 500
3000+
30
450= 21 mm
Since, the applied eccentricities are greater than the
minimum eccentricities, modification of moments is not re-
quired.
Equivalent moment,
Mu
= 1.15 2 2
ux uyM M+
Mu
= 1.15 22 100175 +
Mu
= 231.79 kNm ~ 232 kNm
Assume effective cover of 60 mm,
Dd =
450
60= 0.133 0.15
Pu=
Dbf
P
ck
u
..=
45030030
1016003
= 0.395 0.4
Mu = 2.. Dbf
Mck
u = 2
6
4503003010232
= 0.127 0.13
From sp:16 design charts; for Pu= 0.4,M
u= 0.13 and
Dd =0.15
ckf
p= 0.125
p = 0.125 30 = 3.75%
Asc,req
= 3.75 300 100
450= 5062.5 mm2
Provide 12 bars of 24 mm .
Asc1,pr
= 12 4
242
= 5428.672 mm2 5429 mm2
Check for Moment Capacity of Section
. .
u
ck
P
f b D= 0.4
ppr
=1st pr
A
b D 100
ppr
=450300
1005429
= 4.021%.
ck
pr
f
p=
30
021.4= 0.134
Actual d ' for a clear cover of 40 mm
d' = 40 + 8 +2
24= 60 mm
Dd ~ 0.15
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Referring to sp : 16 charts for Dd = 0.15,
ckfp
=
0.134 andpu
= 0.4
2
1
.. Dbf
M
ck
ux = 0.14
2
1
.. Dbf
M
ck
uy = 0.14
Mux1
= 0.14 30 300 4502= 255.15 106 Nmm >Mux
Muy1
= 0.14 30 300 4502= 255.15 106 Nmm >Muy
Hence safe.
Check for Safety Under bi-axial bending
n
ux
ux
M
M
1+
n
uy
uy
M
M
1< 1.0
Puz
= 0.45 .fck
Ag+ (0.75f
y 0.45f
ck) .A
sc
= 0.45 30 300 450 + (0.75 415 0.45 30)
5429
Puz
= 3438.985 kN ~ 3439 kN
u
uz
p
P=
3439
1600= 0.465 =x
Therefore, 0.2
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S.15Design of Reinforced Concrete Structures (May/June-2012, Set-1) JNTU-Hyderabad
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Where,
Mcr
= Cracking moment of inertia =c
crg
Y
fI .
M= Maximum moment developed due to service
load
Z = Lever arm = d x/3
d = Effective depth
x = Depth of neutral axis
bw
= Width of web
bf
= Width of flange
m = Modular ratio =c
s
E
E
d' = Effective cover to compression reinforcement
Asc= Area of steel in compression reinforcement
Ast
= Area of steel in Tension reinforcement
Ig
= Gross moment of inertia of section =12
3bD
D = Overall depth of rectangular section
fcr
= Modulus of rupture = 0.7 ckf
yc= Distance of extreme compression fibre from
N-A
The effective moment of inertia Ieff calculated byequation (1) should be greater than cracked moment of
inertia (Icr) and less than the gross moment of inertia (I
g),
i.e. geffcr III
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Weight of slab (on slope), w1
w' = Thickness Density
w' = 0.21 25
w' = 5.25 kN/m2
Dead weight horizontally (w1)
w1= w'
2 2R T
T
+
w1= 5.25
2 20.15 0.25
0.25
+
w1= 6.122 kN/m2
Dead weight of steps (w2)
w2
=b
R2
Density
=12
15.0
25
w2
= 1.875 kN/m Total death weight,
wd= 6.122 + 1.875
= 7.997 kN/m
wd 8 kN/m (Dead load per meter run, b= 1m)
Live load, wl= 3 kN/m2
Finishing load, wfl= 0.6 kN/m2
Total load, w = 8 + 3 + 0.6 w= 11.6 kN/m
Moment,M
M= 1.58
2effIW
M= 1.5 8
1.46.112
M= 36.562 KN/m
Design of Waist Slab
M20
concrete,fck
= 20 N/mm2
Fe415
Steel , fy= 415 N/mm2
For M20
concrete and Fe415
steel,
d
xu max,= 0.48
Ru= 0.36f
ck
d
xu max
,max1 0.416
ux
d
Ru
= 0.36 20 0.48 (1 0.416 0.48)
Ru
= 2.766
Depth or thickness of waist slab, (d)
d =bR
M
u
d =
636.562 10
2.766 1000
d = 114.971 ~ 115 mm
Provide overall depth of 150 mm with 10 mm barsand 30 mm nominal covers.
Effective depth,
d= 150 30 (10/2)
d= 115 mm
Reinforcement Design
As the provided depth is equal to the required depth,
the section is under reinforced.
Area of steel =y
ck
f
f
.2
2
6.411
bdf
M
ck
b d
=4152
20
2
6
115100020
10562.366.411
1000 115
Ast
= 1098.903 1099 mm2Provide 10 mm bars at spacing, S
S =st
A
A
1000
S =1099
4
102
1000
S = 71.465 mm
S 70 mm c/c Provide 10 mm bars at 70 mm c/c.Distribution Reinforcement
Provide a minimum reinforcement of 0.12% of cross
section.
(Ast)
d=
100
12.0 150 1000
(Ast)
d= 180 mm2
Provide 8 mm bars at spacing of
S=st
A
A
1000
S =180
4
82
1000
S = 279.253 mm
S 275 mm c/c Provide 8 mm bars at 275 mm c/c.