Design and drawing of RC Structures CV 61

Design and drawing of RC Structures CV 61 Dr. G. S. Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo. com 1

Portal frames 2

Learning out Come • Review of Design of Portal Frames • Design example Continued 3

INTRODUCTION 4

For Shed 5

For Rectangular Buildings 6

For Bridges 7

For Viaduct 8

INTRODUCTION • Step 1: Design of slabs • Step 2: Preliminary design of beams and columns • Step 3: Analysis • Step 4: Design of beams • Step 5: Design of Columns • Step 6: Design of footings 9

PROBLEM 1 10

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Step 3: Analysis(Contd) Moment Distribution Table 13

Step 3: Analysis(Contd) Bending Moment diagram 14

Step 3: Analysis(Contd) Design moments: • Service load end moments: MB=102 k. N-m, MA=51 k. N-m • Design end moments Mu. B=1. 5 x 102 = 153 k. N-m, Mu. A=1. 5 x 51=76. 5 k. N-m • Service load mid span moment in beam = 27 x 82/8 – 102 =114 k. N-m • Design mid span moment Mu+ =1. 5 x 114 =171 k. N-m • Maximum Working shear force (at B or C) in beam = 0. 5 x 27 x 8 = 108 k. N • Design shear force Vu = 1. 5 x 108 = 162 k. N 15

Step 4: 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 k. N-m • Flange width bf= • Here Lo=0. 7 x L = 0. 7 x 8 =5. 6 m • bf= 5. 6/6+0. 4+6 x 0. 12=2 m 16

Step 4: Design of T-beam: • bf/bw=5 and Df /d =0. 2 Referring to table 58 of SP 16, the moment resistance factor is given by KT=0. 459, • Mulim=KT bwd 2 fck = 0. 459 x 400 x 6002 x 20/1 x 106 = 1321. 92 k. N-m > Mu Safe • The reinforcement is computed using table 2 of SP 16 17

Step 4: Design of T- beam: • Mu/bd 2 = 171 x 106/(400 x 6002) 1. 2 for this pt=0. 359 • Ast=0. 359 x 400 x 600/100 = 861. 6 mm 2 • No of 20 mm dia bar = 861. 6/( x 202/4) =2. 74 • Hence 3 Nos. of #20 at bottom in the mid span 18

Step 4: Design of Rectangular beam: • Design moment Mu. B=153 k. N-m • Mu. B/bd 2= 153 x 106/400 x 6002 1. 1 From table 2 of SP 16 pt=0. 327 • Ast=0. 327 x 400 x 600 / 100 = 784. 8 • No of 20 mm dia bar = 784. 8/( x 202/4) =2. 5 • Hence 3 Nos. of #20 at the top near the ends for a distance of o. 25 L = 2 m from face of the column as shown in Fig 6. 6 19

Step 4: Design of beams Long. Section: 20

Step 4: Design of beams Cross-Section: 21

Check for Shear: • Nominal shear stress = pt=100 x 942/(400 x 600)=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 Vuc=0. 432 x 400 x 600/1000 = 103 k. N • Shear to be carried by steel Vus=162 -103 = 59 k. N 22

Check for Shear: • Nominal shear stress = pt=100 x 942/(400 x 600)=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 Vuc=0. 432 x 400 x 600/1000 = 103 k. N • Shear to be carried by steel Vus=162 -103 = 59 k. N 23

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) 24

Step 5: Design of Columns: • Cross-section of column = 400 mm x 600 mm • Ultimate axial load Pu=1. 5 x 108 = 162 k. N (Axial load = shear force in beam) • Ultimate moment Mu= 1. 5 x 102 = 153 k. N-m ( Maximum) • Assuming effective cover d’ = 50 mm d’/D 0. 1 25

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Step 5: Design of Columns: • Referring to chart 32 of SP 16, p/fck=0. 03; p=20 x 0. 03 = 0. 6 • Minimum steel in column should be 0. 8 %, Hence min steel percentage shall be adopted • Ast=0. 8 x 400 x 600/100 = 1920 mm 2 • No. of bars required = 1920/314 = 6. 1 • Provide 8 bars of #20 27

Step 5: Design of Columns: 8 mm diameter tie shall have pitch least of the following • Least lateral dimension = 400 mm • 16 times diameter of main bar = 320 mm • 48 times diameter of tie bar = 384 • 300 mm Provide 8 mm tie @ 300 mm c/c 28

600 400 8 -#20 Tie #8 @300 c/c 29

Step 6: Design of Footings: • Load: • Axial Working load on column = 108 k. N • Self weight of footing @10% = 11 k. N • Total load= 119 120 k. N • Working load moment at base = 51 k. Nm 30

Step 6: Design of Footings: • Approximate area footing required = Load on column/SBC= 108/150 =0. 72 m 2 • However the area provided shall be more than required to take care of effect of moment. • The footing size shall be assumed to be 2 mx 3 m (Area=6 m 2) 31

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Step 6: Design of Footings: • Maximum pressure qmax=P/A+M/Z = 108/6+6 x 51/2 x 32 = 35 k. N/m 2 • Minimum pressure qmin=P/A-M/Z = 108/6 -6 x 51/2 x 32 = 1 k. N/m 2 • Average pressure q = (35+1) = 18 k. N/m 2 • Bending moment at X-X = 18 x 2 x 1. 22/2 = 25. 92 k. N-m • Factored moment Mu 39 k. N-m 33

Step 6: 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 SP 16 is pt= 0. 1%, Minimum pt=0. 12% 34

Step 6: Design of Footings: • Area of steel per meter width of footing is Ast=0. 12 x 1000 x 250/100=300 mm 2 • Spacing of 12 mm diameter bar = 113 x 1000/300 = 376 mm c/c • Provide #12 @ 300 c/c both ways 35

Step 6: Design of Footings: • • • Check for Punching Shear Length of punching influence plane = ao= 600+250 = 850 mm Width of punching influence plane = bo= 400+250 = 650 mm Punching shear Force = Vpunch =108 -18 x(0. 85 x 0. 65)=98 k. N Punching shear stress punch= Vpunch/ (2 x(ao+bo)d =98 x 103/(2 x(850+650)250) = 0. 13 MPa Permissible shear stress = 0. 25 fck=1. 18 MPa > punch Safe 36

Step 6: Design of Footings: • • • Check for One Way Shear force at a distance ‘d’ from face of column V= 18 x 2 x 0. 95 = 34. 2 k. N Shear stress v=34. 2 x 103/(2000 x 250)=0. 064 MPa Referring to table 19 of IS 456 this stress is very small and hence safe Details of reinforcement provided in footing is shown in Fig. 37

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Dr. G. S. Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: gss_nie@yahoo. com 40