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Transcript of 1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National...
1
Design and drawing of RC Structures
CV61
Dr. G.S.Suresh
Civil Engineering Department
The National Institute of Engineering
Mysore-570 008
Mob: 9342188467Email: [email protected]
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Portal frames
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Learning out Come
• Review of Design of Portal Frames
• Design example Continued
4
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 2
6
PROBLEM 2
A portal frame hinged at base has following data:
Spacing of portal frames = 4m
Height of columns = 4m
Distance between column centers = 10m
Live load on roof = 1.5 kN/m2
RCC slab continuous over portal frames. Safe bearing capacity of soil=200 kN/m2
Adopt M-20 grade concrete and Fe-415 steel. Design the slab, portal frame and foundations and sketch the details of reinforcements.
7
Data given:
• Spacing of frames = 4m
• Span of portal frame = 10m
• Height of columns = 4m
• Live load on roof = 1.5 kN/m2
• Concrete: M20 grade
• Steel: Fe 415
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9
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Step1:Design of slab• Assume over all depth of slab as 120mm and effective
depth as 100mm• 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)• Maximum service load moment at interior support =
= 8.5 kN-m9
Lw
10
Lw 2L
2d
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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
• 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
275.1100x1000
10x75.12
bd
M2
6
2u
12
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
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Step1:Design of slab (Contd)
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Step2: Preliminary design of beams and columns
• Beam:
• Effective span = 10m
• Effective depth based on deflection criteria = 10000/13 = 769.23mm
• Assume over all depth as 750 mm with effective depth = 700mm, breadth b = 450mm and column section equal to 450 mm x 600 mm.
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Step3: AnalysisLoad on frame• i) Load from slab = (3.63+1.5) x 4
=20.52 kN/m• ii) Self weight of rib of beam = 0.45x0.63x24 =
6.80 kN/m• Total 28.00 kN/m• Height of beam above hinge = 4+0.1-075/2 =3.72 m• The portal frame subjected to the udl considered
for analysis is shown in Fig. 6.10
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Step3: Analysis (Contd.)
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Step3:Analysis(Contd)
• The moments in the portal frame hinged at the base and loaded as shown in Fig. is analised by moment distribution
• IAB = 450 x 6003/12 = 81 x 108 mm4, IBC= 450 x 7503/12 = 158.2 x 108 mm4
• Stiffness Factor:
• KBA= IAB / LAB = 21.77 x 105 KBC= IBC / LBC = 15.8 x 105
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Step3:Analysis(Contd)
• Distribution Factors:
Fixed End Moments:
• MFAB= MFBA= MFCD= MFDC 0
• MFBC= -=-233 kN-m and MFCB= =233 kN-m
5.0108.151077.21
1077.21
K
KDD
55
5
BA
BABCBA
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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=156 kN-m,
• Design end moments MuB=1.5 x 156 = 234 kN-m,
• Service load mid span moment in beam= 28x102/8 – 102 =194 kN-m
• Design mid span moment Mu+=1.5 x 194 = 291 kN-m
• Maximum Working shear force (at B or C) in beam = 0.5 x 28 x 10 = 140kN
• Design shear force Vu = 1.5 x 140 = 210 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=291 kN-m• Flange width bf= • Here Lo=0.7 x L = 0.7 x 10 =7m• bf= 7/6+0.45+6x0.12=2.33m
fwo D6b
6
L
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Step4:Design of T-beam:
•bf/bw=5.2 and Df /d =0.17 Referring to table 58 of SP16, the moment resistance factor is given by KT=0.43,
•Mulim=KT bwd2 fck = 0.43 x 450 x 7002 x 20/1x106 = 1896.3 kN-m > Mu Safe
•The reinforcement is computed using table 2 of SP16
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Step4:Design of T- beam:
•Mu/bd2 = 291 x 106/(450x7002)1.3 for this pt=0.392
•Ast=0.392 x 450x700/100 = 1234.8 mm2
•No of 20 mm dia bar = 1234.8/(x202/4) =3.93
•Hence 4 Nos. of #20 at bottom in the mid span
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Step4:Design of Rectangular beam:
•Design moment MuB=234 kN-m
•MuB/bd2= 234x106/450x7002 1.1 From table 2 of SP16 pt=0.327
•Ast=0.327 x 450 x 700 / 100 = 1030
•No of 20 mm dia bar = 1030/(x202/4) =3.2
•Hence 4 Nos. of #20 at the top near the ends for a distance of o.25 L = 2.5m from face of the column as shown in Fig
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Step4:Design of beams Long. Section:
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Step4:Design of beams Cross-Section:
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Check for Shear:
•Nominal shear stress =
•pt=100x 1256/(450x700)=0.390.4
•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 450 x 700/1000 = 136 kN
•Shear to be carried by steel Vus=210-136 = 74 kN
67.0700450
10x210
bd
V 3u
v
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Check for Shear:•Nominal shear stress =
pt=100x 942/(400x600)=0.390.4
•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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Check for Shear:
•Spacing 2 legged 8 mm dia stirrup
•sv=
•Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)
53.3411074
70050241587.0
V
dAf87.03
us
svy
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Step5:Design of Columns:
• Cross-section of column = 450 mm x 600 mm
• Ultimate axial load Pu=1.5 x 140 = 210 kN (Axial load = shear force in beam)
• Ultimate moment Mu= 1.5 x 156 = 234 kN-m ( Maximum)
• Assuming effective cover d’ = 50 mm; d’/D 0.1
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07.060045020
10234
bDf
M2
6
2ck
u
04.060045020
10210
bDf
P 3
ck
u
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Step5:Design of Columns:
• Referring to chart 32 of SP16, p/fck=0.04; p=20 x 0.04 = 0.8 %
• Equal to Minimum percentage stipulated by IS456-2000 (0.8 % )
• Ast=0.8x450x600/100 = 2160 mm2
• No. of bars required = 2160/314 = 6.8
• Provide 8 bars of #20
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Step5:Design of Columns:
8mm diameter tie shall have pitch least of the following
• Least lateral dimension = 450 mm
• 16 times diameter of main bar = 320 mm
• 48 times diameter of tie bar = 384
• 300mm
Provide 8 mm tie @ 300 mm c/c
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600
Tie #8 @300 c/c8-#20
450
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Step6:Design of Hinges:• At the hinge portion, concrete is under
triaxial stress and can withstand higher permissible stress.
• Permissible compressive stress in concrete at hinge= 2x0.4fck =16 MPa
• Factored thrust =Pu=210kN
• Cross sectional area of hinge required = 210x103/16=13125 mm2
• Provide concrete area of 200 x100 (Area =20000mm2) for the hinge
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Step6:Design of Hinges:• Shear force at hinge = Total moment in
column/height = 156/3.72=42• Ultimate shear force = 1.5x42=63 kN• Inclination of bar with vertical =
= tan-1(30/50) =31o
• Ultimate shear force = 0.87 fy Ast sin
• Provide 4-#16 (Area=804 mm2)
2o
3
st mm 33931sin41587.0
1063A
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Step7:Design of Footings:• Load:
Axial Working load on column = 140 kN
Self weight of column=0.45 x 0.6 x3.72x 24 = 24
Self weight of footing @10% = 16 kN
Total load = 180 kN
• Working moment at base = 42 x 1 =42 kN-m
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Step6:Design of Footings:
• Approximate area footing required = Load on column/SBC
= 180/200 =0.9 m2
• However the area provided shall be more than required to take care of effect of moment. The footing size shall be assumed to be 1mx2m (Area=2 m2)
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2mX
1m
0.7m
0.6m
X
0.45m
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Step6:Design of Footings:
• Maximum pressure qmax=P/A+M/Z = 180/2+6x42/1x22 = 153 kN/m2
• Minimum pressure qmin=P/A-M/Z = 180/2-6x42/1x22 = 27 kN/m2
• Average pressure q = (153+27)/2 = 90 kN/m2
• Bending moment at X-X = 90 x 1 x 0.72/2 = 22 kN-m
• Factored moment Mu33 kN-m
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Step6:Design of Footings:
• Over all depth shall be assumed as 300 mm and effective depth as 250 mm,
•
• Corresponding percentage of steel from Table 2 of SP16 is pt= 0.15% > Minimum pt=0.12%
528.02501000
1033
bd
M2
6
2u
43
Step6:Design of Footings:
• Area of steel per meter width of footing is Ast=0.12x1000x250/100=300 mm2
• Spacing of 12 mm diameter bar = 113x1000/300 = 376 mm c/c
• Provide #12 @ 300 c/c both ways
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Step6:Design of Footings:• Length of punching influence plane
= ao= 600+250 = 850 mm
• Width of punching influence plane = bo= 450+250 = 700 mm
• Punching shear Force = Vpunch
=180-90x(0.85x0.7)=126.5 kN
• Punching shear stress punch
=Vpunch/(2x(ao+bo)d)=126.5x103/(2x(850+700)250) = 0.16 MPa
• Permissible shear stress = 0.25fck=1.18 MPa > punch Safe
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Step6:Design of Footings:Check for One Way Shear• Shear force at a distance ‘d’ from face of
column• V= 90x1x0.45 = 40.5 kN
• Shear stress v=40.5x103/(1000x250)=0.162 MPa
• For pt=0.15 , the permissible stress c = 0.28 (From table 19 of IS456-2000)
• Details of reinforcement provided in footing is shown in Fig.
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Dr. G.S.Suresh
Civil Engineering Department
The National Institute of Engineering
Mysore-570 008
Mob: 9342188467 Email: [email protected]