Effective depth of footing = 517mm Selfweight of the footing = 60 kN Area of Footing, A= 4m2 Sectional modulus of footing about X axis = 1.33 m3 Sectional modulus of footing about Y axis = 1.33m3 SAFE Maximum bearing pressure = [533.88 + 65.325 +16.785] 4 1.33 1.33 = 195.206kN/m2 < 200 kN/m2 Minimum bearing pressure = [533.88 − 65.325 − 16.785 ] 1.33 1.33 4 = 71.733kN/m2 > 0 SAFE Calculation for bottom steel Net soil pressure at ultimate loads = 533.88 2∗2 Mu about YY axis at column face = 133.47∗2∗0.975∗0.975 = 126.87 kN/m 2 R = Mu/ bd2 = 126.8*106/ (2000*5172) = 0.237 MPa Pt/100 = 0.5 fck (1- √(1 − 4.6������/������������������)) fy = 0.055 % Ast min = 1440 mm2 Pt min = 0.139 % Since Pt min > Pt Provide minimum reinforcement = 1440 mm2 Spacing = 135 mm c/c Hence provide 16mm dia bar @135 mm c/c parallel to length of footing at bottom Mu about XX at column face = 133.47∗2∗075∗0.75 = 75.07 kNm 2
R = 0.1404 MPa Pt = 0.032 % Pt min = 0.134 % Provide min reinforcement = 1440 mm2 Spacing = 135 mm c/c Hence provide 16mm dia bar @135 mm c/c parallel to breadth of footing. Fig.8.8 Footing plan details (F16) Fig.8.9 Footing section details (F16)
Check for one way shear One way shear at critical section 1-1: The critical section at a distance of effective depth ‘d’ from column face Shear force , Vu = 55.52 kN Shear stress, τv = Vu/bd = 0.053 N/mm2 τc = 0.29 N/mm2 τv <τc Hence ok. One way shear at critical section 2-2: Shear force , Vu = 62.19 kN Shear stress, τv = Vu/bd = 0.06 N/mm2 τc = 0.29 N/mm2 τv <τc Hence ok. Fig.8.10 Footing in one way shear
Check for two way shear Allowable shear stress, τv allowable = ks τc ks = (0.5 + βc) should not exceed 1 ks = (0.5 + 0.9) = 1.4 >1 Therefore ks = 1 τc = 0.25 √������������������ = 1.36 N/mm2 ks τc = 1.36 N/ mm2 Shear force = 133.47(22-1.067*1.017) = 389.04 kN Length of critical section = 4168 mm Area of critical section = 2154856 mm2 Shear stress, τv = 0.18 N/mm2 τv< τv allowable Hence ok Fig.8.11 Footing in two way shear
8.5 Analysis and Design of a stair case The structure consists of an open well staircase consisting of three flights and is designed as waist slab type stair case. The entire stairs have a tread of 300 mm and rise of 150 mm. From the deflection criteria, the depth of waist slab and landing slab is assumed to be 200 mm. Fig.8.12 Plan of staircase Consider a one meter wide strip. Loads acting are as follows: Live load on the stair = 3 kN/m2 Finishing load = 1 kN/m2 Weight of waist slab on slope = .2*1*1*25*335.410/300 = 5.59 kN/m2 Weight of stairs = .5*.15*1*25 = 1.875 kN/m2 Total factored load on going= (3+1+5.59+1.875)*1.5 = 17.19 kN/m Weight of landing slab = .2*1*25= 5 kN/m2 Total factored load on landing slab = (5+5+1)*1.5 = 13.5 kN/m
Flight 1 Consider the flight as supported on beam on first riser and at the end of landing, Span = 0.2 + 1.8 + 0.9 = 2.9 m Fig.8.13 Loads acting on flight 1 Flight 2 Consider the flight as supported on end of landing beam on one side and beam at the end of landing on other side. Span = 0.115+1.5+1.8+1.65+0.15 = 5.215 m Fig.8.14 Loads acting on flight 2 After analysis, Mmax = 60.18 kN/m Flight 3 Consider the flight as supported on end of landing beam on one side and beam at the end of stairs on other side. Span = 0.9+ 1.8+0.2 = 2.9 m Fig.8.15 Loads acting on flight 3
Consider flight 2 Factored Bending Moment (Mu) = 60.18 kNm Factored Shear Force (Fu) = 39.92 kN Clear Cover =20 mm Diameter of Main Reinforcement = 12 mm Effective Depth, d = 200 – 20 – 6 = 174 mm Main reinforcement R = Mu/bd2 = (60.18 * 106)/ (1000*1742) = 1.98 MPa Ast = 0.5fck b������ (1- √(1 − 4.6������/������������������)) = 863.87 mm2 fy Required spacing = 130mm c/c Provide 12mm dia bars @ 130mm c/c Distribution reinforcement = 240 mm2 Provide 8mm dia bars @ 200mm c/c as distribution reinforcement Check for shear Vu = 39.92kN Shear stress, τv = 0.229Mpa τcmax = 3.5MPa τc = 0.5 MPa τv < τc Therefore slab is safe in shear. Fig.8.16 Detailing of flight 2
CHAPTER 9 CONCLUSION The inplant training was to model, analyse, design and detail a proposed residential building from the provided architectural plans. Through this 45 days inplant training I got exposed to the exact work environment. The training has helped me to become knowledgeable for the essential structural engineering softwares. The analysis and design processes were done in ETABS which was really time saving. The manual calculations were also performed to get a more practical design values. Throughout the internship period I was able to correlate theoretical knowledge which I have gained earlier, with the practical scenario. Overall this internship has given me a great experience with abundant knowledge.
REFERENCES 1. S. Unnikrishna Pillai and Devdas Menon “Reinforced Concrete Design” McGraw- Hill Publishing Company Limited, New Delhi, 2003. 2. IS: 875 (Part I)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Buildings and Structures”. 3. IS: 875 (Part II)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Buildings and Structures”. 4. IS: 875 (Part III)-2015, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Buildings and Structures”. 5. IS: 875 (Part V)-1987, “Indian Standard Code of Practice for Design Loads (Other than earthquake) for Buildings and Structures”. 6. IS: 1893 (Part I) 2016-, “Criteria for Earthquake Resistant Design of Structures”. 7. IS: 13920 – 2016, “Ductile Design and Detailing of Reinforced Concrete Structures Subjected to Seismic Forces – Code of Practice”. 8. IS: 456 – 2000, “Indian Standard Plain and Reinforced Concrete – Code of Practice”. 9. SP: 34 – 1987, “Handbook on Concrete Reinforcement and Detailing”. 10. SP: 24 (S&T) – 1983, “Explanatory Handbook on Indian Standard Code of Practice for Plain and Reinforced Concrete”. 12. SP: 16 – 1980, “Design Aids for Reinforced Concrete to IS: 456-1978”.
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