portal framed

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]

1

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 = 4mHeight of columns = 4mDistance between column centers = 10mLive load on roof = 1.5 kN/m2RCC slab continuous over portal frames. Safe bearing

capacity of soil=200 kN/m2Adopt 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-m9Lw

1 0Lw 2L

2d +

10

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

2.11 0 0x1 0 0 01 0x7 5.1 2

b dM

2

6

2u ==

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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

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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

5.01 08.1 51 07 7.2 1

1 07 7.2 1K

KDD 555

B A

B AB CB A =+

==

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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 kN21

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

6L

++

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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 SafeThe 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=0.392Ast=0.392 x 450x700/100 = 1234.8 mm2

No of 20 mm dia bar = 1234.8/(x202/4) =3.93Hence 4 Nos. of #20 at bottom in the mid span

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Step4:Design of Rectangular beam:Design moment MuB=234 kN-mMuB/bd2= 234x106/450x7002 1.1 From table 2 of SP16 pt=0.327Ast=0.327 x 450 x 700 / 100 = 1030 No of 20 mm dia bar = 1030/(x202/4) =3.2Hence 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)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcement is required to be designedStrength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kNShear to be carried by steel Vus=210-136 = 74 kN

6.07 0 04 5 01 0x2 1 0

b dV 3u

v ===

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Check for Shear:Nominal shear stress =

pt=100x 942/(400x600)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcement is required to be designedStrength of concrete Vuc=0.432 x 400 x 600/1000 = 103 kNShear 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 300mm c/c (equal to maximum spacing)

5.31 07 4

7 05 024 1 58 7.0V

dAf8 7.03

u s

s vy =

=

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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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0.06 0 04 5 02 0

1 02 3 4b Df

M2

6

2c k

u =

=

0.06 0 04 5 02 0

1 02 1 0b Df

P 3

c k

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 300mmProvide 8 mm tie @ 300 mm c/c

34

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

s t m 3 33 1s i n4 1 58 7.01 06 3A =

=

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Step7:Design of Footings: Load:

Axial Working load on column = 140 kNSelf 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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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-m41

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%

5.02 5 01 0 0 0

1 03 3b dM

2

6

2u =

=

42

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.45

Dr. G.S.Suresh

Civil Engineering Department

The National Institute of Engineering

Mysore-570 008

Mob: 9342188467 Email: [email protected]

48

Design and drawing of RC StructuresCV61Portal frames Learning out ComeSlide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48

portal framed

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