1. 1 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: gss_nie@yahoo.com

2. 2 Portal frames

3. 3 Learning out Come Introduction Procedure for design of Portal frames Design example

4. 4 Books for Reference N.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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6. 6 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.

7. 7 INTRODUCTION

8. 8 For Shed

9. 9 Inside View of Shed

10. 10 For Rectangular Buildings

11. 11 For Bridges

12. 12 For Viaduct

13. 13 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

14. 14 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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16. 16 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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18. 18 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/m 2, SBC of soil = 150 kN/m 2. 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.

19. 19 Data given: Spacing of frames = 4m Span of portal frame = 10m Height of columns = 4m Live load on roof = 1.5 kN/m 2 Thickness of slab = 120mm Concrete: M20 grade Steel: Fe 415

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22. 22 Step1:Design of slab Self weight of slab = 0.12 x 24 = 2.88 kN/m 2 Weight of roof finish = 0.50 kN/m 2 (assumed) Ceiling finish= 0.25 kN/m 2 (assumed) Total dead load w d = 3.63 kN/m 2 Live load w L = 1.50 kN/m 2 (Given in the data) Maximum service load moment at interior support Mu=1.5 x 8.5 = 12.75 kN-m/m

23. 23 Step1:Design of slab (Contd) Mu lim =Q lim bd2 (Q lim =2.76) = 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m From table 2 of SP16 p t =0.384; A st =(0.384 x 1000 x 100)/100= 384 mm 2 Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c Provide #10 @ 200 c/c

24. 24 Step1:Design of slab (Contd) Area of distribution steel A dist =0.12 x 1000 x 120 / 100 = 144 mm 2 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

25. 25 Step1:Design of slab (Contd)

26. 26 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.

27. 27 Step3: Analysis Load 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

28. 28 Step3: Analysis (Contd.)

29. 29 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 I AB = 400 x 600 3 /12 = 72 x 10 8 mm4, I BC = 400 x 700 3 /12 = 114.33 x 10 8 mm4 Stiffness Factor: K BA = I AB / L AB = 18 x 10 5 K BC = I BC / L BC = 14.3 x 10 5

30. 30 Step3:Analysis (Contd) Distribution Factors: Fixed End Moments: MFAB= MFBA= MFCD= MFDC 0 MFBC= - =-144 kN-m and MFCB= =144 kN-m

31. 31 Step3:Analysis (Contd) Moment Distribution Table

32. 32 Step3:Analysis (Contd) Bending Moment diagram

33. 33 Step3:Analysis (Contd) Design moments: Service load end moments: M B =102 kN-m, M A =51 kN-m Design end moments M uB =1.5 x 102 = 153 kN-m, M uA =1.5 x 51=76.5 kN-m Service load mid span moment in beam = 27x8 2 /8 – 102 =114 kN-m Design mid span moment M u + =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 V u = 1.5 x 108 = 162 kN

34. 34 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 M u =171 kN-m Flange width b f = Here L o =0.7 x L = 0.7 x 8 =5.6m b f = 5.6/6+0.4+6x0.12=2m

35. 35 Step4:Design of T-beam: b f /b w =5 and D f /d =0.2 Referring to table 58 of SP16, the moment resistance factor is given by K T =0.459, M ulim =K T b w d2 f ck = 0.459 x 400 x 6002 x 20/1x106 = 1321.92 kN-m > M u Safe The reinforcement is computed using table 2 of SP16

36. 36 Step4:Design of T- beam: Mu/bd 2 = 171 x 106/(400x6002)  1.2 for this p t =0.359 A st =0.359 x 400x600/100 = 861.6 mm 2 No of 20 mm dia bar = 861.6/(  x202/4) =2.74 Hence 3 Nos. of #20 at bottom in the mid span

37. 37 Step4:Design of Rectangular beam: Design moment M uB =153 kN-m M uB /bd2= 153x106/400x6002  1.1 From table 2 of SP16 p t =0.327 Ast=0.327 x 400 x 600 / 100 = 784.8 No of 20 mm dia bar = 784.8/(  x20 2 /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

38. 38 Step4:Design of beams Long. Section:

39. 39 Step4:Design of beams Cross-Section:

40. 40 Step4:Check for Shear: Nominal shear stress = p t =100x 942/(400x600)=0.39  0.4 Permissible stress for pt=0.4 from table 19  c =0.432 <  v Hence shear reinforcement is required to be designed Strength of concrete V uc =0.432 x 400 x 600/1000 = 103 kN Shear to be carried by steel V us =162-103 = 59 kN

41. 41 Step4:Check for Shear: Spacing 2 legged 8 mm dia stirrup s v = Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

42. 42 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo.com