Transcript Slide 1

Design and drawing of RC
Structures
CV61
Dr. G.S.Suresh
Civil Engineering Department
The National Institute of Engineering
Mysore-570 008
Mob: 9342188467
Email: [email protected]
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Portal frames
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Learning out Come
• Introduction
• Procedure for design of Portal frames
• Design example
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Books for Reference
N.Krishna Raju Advanced Reinforced concrete Design
Jaikrishna and O.P.Jain Plain and reinforced concrete Vol2
B.C.Punmia Reinforced Concrete Structures Vol2
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INTRODUCTION
• A portal frame consists of vertical member
called Columns and top member which may
be horizontal, curved or pitched.
• Rigidly connected
• They are used in the construction of large
sheds, bridges and viaducts.
• The base of portal frame may be hinged or
fixed.
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INTRODUCTION
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For Shed
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Inside View of Shed
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For Rectangular Buildings
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For Bridges
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For Viaduct
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INTRODUCTION
• The portal frames have high stability
against lateral forces
• A portal frame is a statically indeterminate
structure.
• In the case of buildings, the portal frames
are generally spaced at intervals of 3 to 4m
• Reinforced
concrete
slab
cast
monolithically between the frames
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INTRODUCTION
• Frames
used for ware house sheds and
workshop structures are provided with sloping of
purlins and asbestos sheet roofing between the
portal frames.
• The base of the columns of the portal frames are
either fixed or hinged.
• Analysis of frames can be done by any standard
methods
• Columns are designed for axial force and bending
moment, whereas beam is designed for bending
moment and shear force
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INTRODUCTION
• Step1: Design of slabs
• Step2: Preliminary design of beams and
columns
• Step3: Analysis
• Step4: Design of beams
• Step5: Design of Columns
• Step6: Design of footings
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Problem 1
•
The roof of a 8m wide hall is supported on a
portal frame spaced at 4m intervals. The height
of the portal frame is 4m. The continuous slab
is 120 mm thick. Live load on roof = 1.5 kN/m2,
SBC of soil = 150 kN/m2. The columns are
connected with a plinth beam and the base of
the column may be assumed as fixed. Design
the slab, column, beam members and suitable
footing for the columns of the portal frame.
Adopt M20 grade concrete and Fe 415 steel.
Also prepare the detailed structural drawing.
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Data given:
• Spacing of frames = 4m
• Span of portal frame = 10m
• Height of columns = 4m
• Live load on roof = 1.5 kN/m2
• Thickness of slab = 120mm
• Concrete: M20 grade
• Steel: Fe 415
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Step1:Design of slab
• Self weight of slab = 0.12 x 24 = 2.88 kN/m2
• Weight of roof finish = 0.50 kN/m2 (assumed)
• Ceiling finish
= 0.25 kN/m2 (assumed)
• Total dead load wd
= 3.63 kN/m2
• Live load wL
= 1.50 kN/m2 (Given in the
data)
2
2
w
L
w
L
• Maximum
service
load moment at interior support
d
L
  
8
.
5
kN
m/m
10
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•Mu=1.5 x 8.5 = 12.75 kN-m/m
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Step1:Design of slab (Contd)
• Mulim=Qlimbd2 (Qlim=2.76)
• = 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m
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M
.
75
x
10
u 12


1
.
275
2
2
bd
1000
x
100
• From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x
100)/100= 384 mm2
• Spacing of 10 mm dia bars = (78.54 x 1000)/384=
204.5 mm c/c
• Provide #10 @ 200 c/c
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Step1:Design of slab (Contd)
• Area of distribution steel
Adist=0.12 x 1000 x 120 / 100 = 144 mm2
• Spacing of 8 mm dia bars
= (50.26 x 1000)/144= 349 mm c/c
• Provide #8 @ 340 c/c.
• Main and dist. reinforcement in the slab is
shown in Fig.6.3
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Step1:Design of slab (Contd)
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Step2: Preliminary design of beams and columns
Beam:
• Effective span = 8m
• Effective depth based on deflection criteria
= 8000/12 = 666.67mm
• Assume over all depth as 700 mm with effective
depth = 650mm, breadth b = 400mm
Column:
• Let column section be equal to 400 mm x 600 mm.
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Step3: Analysis
Load on frame
i) Load from slab = (3.63+1.5) x 4 =20.52 kN/m
ii) Self weight of rib of beam
= 0.4x0.58x24
= 5.56 kN/m
Total
 27.00 kN/m
• The portal frame subjected to the udl considered
for analysis is shown in Fig. 6.4
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Step3: Analysis (Contd.)
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Step3:Analysis(Contd)
• The moments in the portal frame fixed at
the base and loaded as shown in Fig. 6.4
are analysed by moment distribution
• IAB = 400 x 6003/12 = 72 x 108 mm4,
• IBC= 400 x 7003/12 = 114.33 x 108 mm4
• Stiffness Factor:
• KBA= IAB / LAB = 18 x 105
• KBC= IBC / LBC = 14.3 x 105
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Step3:Analysis(Contd)
• Distribution Factors:
5
K
18

10
BA
D

 5

0
.
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BA
5
K
18

10

14
.
3

10

BA
5
K
14
.
3

10
BC
D

 5

0
.
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BC
5
K
18

10

14
.
3

10

BC
• Fixed End Moments:
• MFAB= MFBA= MFCD= MFDC 0
wL 27
x
8


• MFBC= - 12 12 =-144 kN-m
27
x8

• and MFCB= wL
=144 kN-m
12
12
2
2
2
2
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Step3:Analysis(Contd) Moment Distribution Table
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Step3:Analysis(Contd) Bending Moment diagram
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Step3:Analysis(Contd) Design moments:
• Service load end moments: MB=102 kN-m,
MA=51 kN-m
• Design end moments MuB=1.5 x 102 = 153 kN-m,
MuA=1.5 x 51=76.5 kN-m
• Service load mid span moment in beam
= 27x82/8 – 102 =114 kN-m
• Design mid span moment Mu+
=1.5 x 114 =171 kN-m
• Maximum Working shear force (at B or C) in beam
= 0.5 x 27 x 8 = 108kN
• Design shear force Vu = 1.5
x
108
=
162
kN
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Step4:Design of beams:
• The beam of an intermediate portal frame is
designed. The mid span section of this beam is
designed as a T-beam and the beam section at the
ends are designed as rectangular section.
Design of T-section for Mid Span :
• Design moment Mu=171 kN-m
L
• Flange width bf= 6 b 6D
• Here Lo=0.7 x L = 0.7 x 8 =5.6m
• bf= 5.6/6+0.4+6x0.12=2m
o
w
f
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Step4:Design of T-beam:
•bf/bw=5 and Df /d =0.2
Referring to table 58 of SP16, the moment
resistance factor is given by KT=0.459,
•Mulim=KT bwd2 fck = 0.459 x 400 x 6002 x 20/1x106 =
1321.92 kN-m > Mu Safe
•The reinforcement is computed using table 2 of
SP16
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Step4:Design of T- beam:
•Mu/bd2 = 171 x 106/(400x6002)1.2 for this
pt=0.359
•Ast=0.359 x 400x600/100 = 861.6 mm2
•No of 20 mm dia bar = 861.6/(x202/4) =2.74
•Hence 3 Nos. of #20 at bottom in the mid
span
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Step4:Design of Rectangular beam:
•Design moment MuB=153 kN-m
•MuB/bd2= 153x106/400x6002 1.1 From table 2 of
SP16 pt=0.327
•Ast=0.327 x 400 x 600 / 100 = 784.8
•No of 20 mm dia bar = 784.8/(x202/4) =2.5
•Hence 3 Nos. of #20 at the top near the ends for a
distance of o.25 L = 2m from face of the column as
shown in Fig 6.6
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Step4:Design of beams Long. Section:
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Step4:Design of beams Cross-Section:
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Step4:Check for Shear:
•Nominal shear stress =
pt=100x 942/(400x600)=0.390.4
•Permissible stress for pt=0.4 from table 19
c=0.432 < v
•Hence shear reinforcement is required to be
designed
•Strength of concrete Vuc
=0.432 x 400 x 600/1000 = 103 kN
•Shear to be carried by steel Vus=162-103 = 59 kN
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Step4:Check for Shear:
•Spacing 2 legged 8 mm dia stirrup sv=
0
.
87
f
A
d
0
.
87

415

2

50

600
y
sv


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3
V
59

10
us
•Two legged #8 stirrups are provided at 300
mm c/c (equal to maximum spacing)
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Dr. G.S.Suresh
Civil Engineering Department
The National Institute of Engineering
Mysore-570 008
Mob: 9342188467
Email: [email protected]
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