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1.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1152 Planning, Analysis and Designing of Cantilever Residential Building M. Balachandran1, A. Anthonisamy2, V. Manikandan3 1,2,3Student, Department of Civil Engineering, Panimalar Engineering College, Tamil Nadu, India. -------------------------------------------------------------------***------------------------------------------------------------------------ Abstract - The project is about planning, analysis and designing of cantilever residential building. It is a framed structure, the residential building is located near to ooty. The total area of building is 100.3 sq.m. The plan and reinforcement details drawn by using Auto-CADD. Analysis done by Manual Calculation. Design are worked by referring code books. Key Words: Ground Floor size :100.35 sq.m, First Floor area :100.35 sq.m, No of Floors :G+1 1. INTRODUCTION The basics needs of human existences are food, clothing and shelter. From times immemorial man has been making efforts in improving their standard of living. The point of his efforts has been to provide an economic and efficient shelter. Every human has an inherent like on a peaceful environment for his pleasant living this objective is achieved by having a place for living, situated at a safe and convenient location. Such a place for comfortable and pleasant living is required to all. Construction is the important aspects of civil engineering. Mankind has been evolving different kinds of environment with the change in civilization and time. 2. METHODOLOGY Plan of the Building Design of Cantilever Slab Design of Cantilever Beam Design of Column Design of Footing Design of Staircase Conclusion 2.1 Plan of the building Ground Floor Plan First Floor Plan 2.2 Design of Cantilever Slab Span = 3 m Live Load = 3 KN/m² Floor Finish = 1 KN/m² M20 & Fe 415 grade Calculation of Depth of Slab: d = Span/7 = 3000/7
2.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1153 d = 428.57 mm = 460 mm Over all Depth = Depth of Slab + Cover Assume Cover = 20 mm D = 480 mm Load Calculation: Self Weight of Slab = 0.2x25 = 5 KN/m Floor Finish = 1 KN/m Live Load = 3 KN/m Total Load = 9 KN/m Ultimate Load = 13.5 KN/m Maximum Bending Moment, Mu = Wu l²/2 = 70.42 KN.m Maximum Shear Force, Vu = Wu l = 43.60 KN Calculation of Limiting Moment of Resistance: Mu limit = 0.138 fck bd² = 584.01 KNm Mu < Mu limit Hence it is Under Reinforced Section Calculation of Area of Steel: Mu = 0.87 fy Ast d (1 – fy Ast / fck bd) 70.42X10^6 = 0.87X415XAstX460 (1– 415Ast/20X1000X460 Ast = 432.43 mm² To Find Main Ast Provided: No of bars = Ast / ast Assume dia = 10 mm So ast = 78.53 mm² So No of bars = 6 Nos Ast Provided = 6 X π/4 X 10² = 471.23 mm² Ast Req < Ast Pro Hence Safe Calculate Spacing of Steel: Spacing = 1000 Ast Of 1 bar / Area of Steel Area of 1 bar = π/4 x d² Assume dia = 10 mm = 78.53 mm² Spacing = 190 mm Check for Spacing: 1) 300 mm 2) 3d = 3x460 = 1380 mm 3) 190 mm Hence Provide 8 mm dia bars @ 190 mm c/c Check Distribution Area of Steel: = 0.12 % of gross area = 552 mm² To Find Distribution Ast Provided: No of bars = Ast / ast Assume dia = 8 mm So ast = 50.26 mm² So No of bars = 12 Nos Ast Provided = 12 X π/4 X 8² = 603.16 mm² Ast Req < Ast Pro Hence Safe Calculate Spacing of Steel: Spacing = Area of one bar / Area of steel = 90 mm Check for Deflection: (L/d)max > (L/d)pro (L/d)max = B.V X Kt X Kc = 9.59 (L/d)pro = Span / Eff depth = 7.02 9.59 > 7.02 Hence Safe 2.3 Design of Cantilever Beam To Find Over all Depth: Length of cantilever = 3 m M.F = 1 eff depth = span/7 x MF = 3000/7 x 1 = 428 ~ 430 mm effective cover = 25 mm Overall depth of beam= 430 + 25 = 455 mm To Find Effective Depth: Provide an overall depth of 460 mm Effective depth d = 460 – 25 = 435 mm
3.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1154 To Find Dead Load Moment: Let the section of cantilever be, 230 mm x 435 mm at fixed end 2/3 x d = 2/3 x 435 = 290 mm 230 mm x 290 mm at free end D.L of the cantilever = (0.435 + 0.29/2) x 0.23 x 3 x 25000 = 6253.13 Load acts at a distance = {(0.435 + 2 x 0.29)/ (0.435 + 0.29)} x 3/3 = 1.4 m from fixed end D.L moment = 6253.13 x 1.4 = 8754.4 Nm To Find Live Load Moment: From Wall: Live load = 3 x 0.23 x 2 = 13.8 KN/m Total live load = 13.8 + 15.93 = 29.73 KN/m Live load moment= wl2/2 = 29.73 x 32/2 = 133.78 KNm To Find End Load Moment: End point load = 2.5 + 1.75 = 4.25 x 0.35 x 25 = 37.18 KN/m End load moment= W.L = 37.18 x 3 = 111.56 KN/m Total moment= E.M + L.L.M + D.L.M = 111.56 + 133.78 + 8.75 = 254.09 KNm To Find Limiting Moment: Total load = (29.73 + 8.75) x 3 + 37.18 = 152.62 KN Mu= 254.09 x 1.5 = 381.135 KNm Mu limit = 2.76 bd2 = 2.76 x 230 x 4602 =134.32 KNm To Find Area of Tension & Compression Reinforcement: Percentage of steel required, Pt= 25/460 = 0.05 Mu/bd2 = 381.135/0.23 x 0.462 = 6.35 Tension zone: Ast = 2.002/100 x 230 x 4602 = 974.3 mm2 Asc = 1.091/100 x 230 x 4602 = 530.91 mm2 To Find No of Bars: No of bars use 20 dia bars, Use 4 Nos of 20 dia @ tension zone No of bars at compression zone Use 16 mm dia No of bars = 530.91/ (𝜋/4 x 162) = 2.67 ~ 3 Nos To Find Check for Shear: τv = Vu/bd Vu= (152.62 x 103)/230 x 460 = 1.53 N/mm2 τc= 100 Ast/bd = (100 x 974.3) / (230 x 460) = 0.597 N/mm2 Kτc = 1 x 0.59 = 0.59 N/mm2 τcmax =2.8 N/mm2 τv < Kτc < τcmax Hence Safe 2.4 Design of Column To Find Axial Load: B = 300 mm D = 400 mm L = 3.5 m Mu1 = Wlx²/12 = 27.68x3.5²/12 = 37.44 KNm Mu2 = Wlx²/12 = 18.68x3²/12 = 14.01 KNm Mu = Mu1 + Mu2 = 51.45 KNm 51.45 = Wl²/8 = Wu x 6.5²/8 Wu = 9.74 KN/m Pu = 9.74x6.5 = 63.32 KN To Find Non-Dimensional Parameters: d' = 40 mm D = 400 mm = 0.1 From chart – 32, Mu/fck b D² = 51.45x10^6 / (20x300x400²) = 0.05
4.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1155 Pu/fck b D = 63.32x10^3 / (20x300x400) = 0.026 P/fck = 0.2 , P = 4 To Find Reinforcement: As = pbD/100 = 4x300x400/100 = 4800mm² Adopt 25 mm dia n = 4800 / (π/4x25²) = 8 Nos To Provide Lateral Ties: Condition: Diameter: 5) ¼ x 25 = 6.25 = 6 mm 6) Less than 16 mm Spacing: 7) LLD = 300 mm 8) 16x25 = 400 mm 9) 300 mm Take Spacing minimum value S = 300 mm 2.5 Design of Footing To Find Loads: Total load =383.91kN+[0.3x0.4x3x25] =392.91kN For 2 floor=2x392.91 P=785.82kN Pu=785.82x1.5 =1177.73~1200Kn To Find Area of Footing: SBC=300kN/m² Area of footing =1200/300 =4m² To Find Size of Footing: 4x x 3x=4 x²=0.33 x=0.577m 4x=4x0.577 =2.3m 3x=3x0.577=1.73~1.75m Size of footing = 2.3x1.75m Area =2.3 x1.75 =4.025 m² To Find Net Upward Soil Pressure: Net upward soil pressure(q) q=load on footing/A =1200/4.025 q=298.13kN/ m²<300kN/m² To Find Moment: Factored moment (Mud) Mmax occur at the face to the column for design purpose Consider only one overhang portion, Mud=1.5 [Upward soil pressure x hatched x distance between C.G for end and Face to column] =1.5[300 x (0.95x1.75) x 0.475] =355kN.m To Find Depth Upward for Moment: Equate Mud to Mulimit For M20 & Fe 415 From sp16 pg : 10, Table D Mulimit=2.76 bd² 355x10^6=2.76 x 1750 x d² d² =73573 d=271.24mm D=271.24+50 D=321.24mm~400mm d=400-50 =350mm Mulimit=2.76x1750x3502 =591.67kN.m Mulimit >Mud It is under reinforced section To Find Area of Main Reinforcement: Mud=0.87 x fy x Ast x[d-fy x Ast / fck x b] 355 x 106=126367.5 Ast-4.28 Ast2 Ast=3144.09mm2 To Find Area of Distribution Reinforcement: Astmin = 0.12% CSA =0.12/100 x 350 x 1750 Astmin=735mm2
5.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1156 To Find Spacing: Use 20 mm dia of bar S=ast/Ast x 1000 = (𝜋/4 x 202)/3144 x 1000 S=170 mm Provide 20 mm dia bar @ 170 mm C/C in both direction Check for Shear: Check for One Way Shear: Vu=1.5[q x hatched area] =1.5[300 x 1.75 x 0.55] Vu=433.12kN Resisting area for shear =b x d =1750 x 550 =962500 mm2 τv= Vu/bd =433.12 x 103/1750 x 550 =0.38 N/mm2 Astp =ast/s x 1750 = (𝜋/4 x 202)/170 x 1750 =3233.99~3234mm2 Pt = 100 Astp / bd =100 x 3234 / (1750 x 550 Pt =0.336 τc=0.40 k=1 kτc=0.40 τv < k τc 0.38 < 0.40 Hence it is safe. Check for Two Way Shear: (Or) Punching Shear: Side length of critical plane =size of column + d/2 + d/2 Short side = 300 + 300/2 +300/2 =600mm Long side = 400 + 400/2 + 400/2 = 800mm Punching critical plane, = 2(600) + (800) =2800mm Vu= q x hatched area x 1.5 = 300 x [(2.3 x 1.75) – (0.6 x 0.8)] x 1.5 Vu= 1595.25kN τv= Vu / bd =1595 x 103 / 2300 x 400 = 1.04 N /mm2 Permissible Shear Stress: βc= short side / long side = 300/400 βc= 0.75 2.6 Design of Staircase Design size = 2x3m Height of each flight =3/2= 1.5m Assume =150mm(Rise) Tread = 300mm Width of landing =400mm M20 @ Fe250 No.of rise =(1500/150)=10 No. Of treads of each flight = no.of rise -1 =9 Assume width of landing = 400mm Live load = 3KN/m To Find Effective Span: lₑ=(9x300)+(400/2)+(400/2)+(230/2) lₑ=3.125m To Find Depth of Waist Slab: Depth of wist slab = (effctive span / 20) = 3125/20 = 156.25mm~160mm D = 160 mm To Find Load Calculation: DL of slab (on slope) =0.16x1x25=4KN/m DL of slab on horizondal span = Ws√(R²+T²)/T = 4√(0.15²+0.3²)/0.3=4.47KN/m DL. of one step = (1/2)xrise xtread=(1/2)x0.15x0.3 =0.563 KN/m Load of steps per m length = 0.563x1/(1000/300) = 1.875 KN/m Load due to floor finish = 0.6 KN /m Total DL = 4.47+1.875+0.6+3 = 9.945 KN/m Total ultimate load = 9.945x1.5=14.91KN/m To Find Bending Moment: Maximum bending moment @ center in ultimate BM Mu = Wu x leff²/8 = 14.91x3.125²/8 = 18.20 KN/m
6.
International Research Journal
of Engineering and Technology (IRJET) e-ISSN: 2395-0056 Volume: 06 Issue: 03 | Mar 2019 www.irjet.net p-ISSN: 2395-0072 © 2019, IRJET | Impact Factor value: 7.211 | ISO 9001:2008 Certified Journal | Page 1157 Check for Depth: Mu = 0.149 xfck xbxd² = 0.149x15x1000x/d² d req = 90mm d req <d pro To Find Ast for Main Reinforcement: Mu = 0.87xfyxAstxd((1-fyxAst)/(fckxbxd)) 18.20x10⁶= 29145Ast-3.62 Ast² Ast = 684.76 mm² To Find Spacing: Use 12mm ϕ Spacing = (1000xast /Ast )= 165.15mm ~170 mm Provide 12mmϕ @spacing 170 mm c/c To Distribution Reinforcement: Ast min = 0.15%xbxD = (0.15/100)x1000x160 Ast min = 240 mm² Spacing = (1000 ast / Ast ) = (1000x78.53/240) Spacing = 300mm 3. CONCLUSION At the end of the site visit and experiencing real workingspace and the environment of construction,I realized that working at construction site is much more different and also more difficult, because of the weather conditions and the risks exit.This site visit gave me a chance to experience and learn what cannot be gained during lectures or tutorials. One of the first things that I have learnt is understanding the importance of safety which is a basic of construction, because of all dangers exist on site.Finally, this site visit helped for better understanding what I am going to face in future as a Quantity surveyor, and also giving me better understanding about construction to help me in my studying, especially for taking off in measurement. In overall, it was a great experience for me. REFERENCES [1] IS 456-2000 Plain and reinforced concrete code [2] SP 16 Design Aids for reinforced concrete [3] IS 875 part-II Code of practice for design loads for buildings and structures [4] Dr.P.Purushotama raj on “Design of reinforced concrete elements” by Lakshmi publications. BIOGRAPHIES Author 1: M.Balachandran Studying B.E Civil Panimalar Engineering College Chennai. Author 2: A.Anthonisamy Studying B.E Civil Panimalar Engineering College Chennai. Author 3: V.Manikandan Studying B.E Civil Panimalar Engineering College Chennai.
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