Portal Frame 2

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 ReferenceN.Krishna Raju Advanced Reinforced concrete DesignJaikrishna and O.P.Jain Plain and reinforced concrete Vol2B.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 ofpurlins and asbestos sheet roofing between theportal 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 bendingmoment, whereas beam is designed for bendingmoment 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) Maximum service load moment at interior support m-k N 5.8

91 0

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=+=LwLw Ld

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

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.6.3

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Step1:Design of slab (Contd)

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Step2: Preliminary design of beams and columnsBeam: 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: AnalysisLoad on framei) Load from slab = (3.63+1.5) x 4 =20.52 kN/mii) Self weight of rib of beam

= 0.4x0.58x24 = 5.56 kN/mTotal 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:

Fixed End Moments: MFAB= MFBA= MFCD= MFDC 0 MFBC= - =-144 kN-m and MFCB= =144 kN-m

5.01 03.1 41 01 8

1 01 8K

KD 555

B A

B AB A =+

==

4.0

1 03.1 41 01 81 03.1 4

KK

D 555

B C

B CB C =+

==

1 28x2 7

1 2w L 22

=

1 28x2 7

1 2w L 22

=

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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 kN33

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 Flange width bf= Here Lo=0.7 x L = 0.7 x 8 =5.6m bf= 5.6/6+0.4+6x0.12=2m

fwo D6b

6L

++

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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 SafeThe 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=0.359Ast=0.359 x 400x600/100 = 861.6 mm2

No of 20 mm dia bar = 861.6/(x202/4) =2.74Hence 3 Nos. of #20 at bottom in the mid span

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Step4:Design of Rectangular beam:Design moment MuB=153 kN-mMuB/bd2= 153x106/400x6002 1.1 From table 2 of SP16 pt=0.327Ast=0.327 x 400 x 600 / 100 = 784.8 No of 20 mm dia bar = 784.8/(x202/4) =2.5Hence 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)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < vHence 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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Step4: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)

31 05 9

6 05 024 1 58 7.0V

dAf8 7.03

u s

s vy =

=

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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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Design and drawing of RC StructuresCV61Portal frames Learning out ComeBooks for ReferenceSlide 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 42