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

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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 2

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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.

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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

• Concrete: M20 grade

• Steel: Fe 415

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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


• 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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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

33


• 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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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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• 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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• 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


• 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]

1 design and drawing of rc structures cv61 dr. g.s.suresh civil engineering department the national...

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