Design and drawing of RC Structures CV61

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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

• Introduction

• Procedure for design of Portal frames

• Design example

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Books for ReferenceN.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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8.00m

4.00m

4.00m

4.00m

4.00m

XY

Z

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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/m-kN 5.8

910

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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

275.1100x1000

10x75.12

bd

M2

6

2u

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• 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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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: AnalysisLoad on frame

i) Load from slab = (3.63+1.5) x 4 =20.52 kN/m

ii) Self weight of rib of beam

= 0.4x0.58x24 = 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:

• Fixed End Moments:

• MFAB= MFBA= MFCD= MFDC 0

• MFBC= - =-144 kN-m

• and MFCB= =144 kN-m

55.0103.141018

1018

K

KD

55

5

BA

BABA

45.0103.141018

103.14

K

KD

55

5

BC

BCBC

12

8x27

12

wL 22

12

8x27

12

wL 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 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

• 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

6

L

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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)=0.390.4

•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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Step4: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)

3671059

60050241587.0

V

dAf87.03

us

svy

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Dr. G.S.Suresh

Civil Engineering Department

The National Institute of Engineering

Mysore-570 008

Mob: 9342188467 Email: [email protected]

Documents

Design and drawing of RC Structures CV61