Project 2
1) Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows: Width 30m Length 50m Eave Height 9m Bay spacing 6m Soil type Medium Safe Bearing Capacity 200 kN/m2 Roof slope 1 in 12 Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875 1) Prepare a Design Basis Report for the project :- PDF formate attach below 2) Create structural model using STAAD Pro Connect Edition :- 3) Prepare DL, LL, WL and EQL load calculations as per IS 875 standards :- 1) Design Foundations using STAAD Foundation :- 4) Design using MS – Excel 1. a) Steel column :- Height of column h = 9m Yeild strength fy = 250N/mm^2 Elastic modulus E = 200000N/mm^2 Material factor γm = 1.1 Step :- 1 Section classification Properties of selected section ISHB 400 A 10466mm^2 h 400mm B 250mm t 10.6mm T 12.7mm rx 166.1mm ry 51.6mm Zp cm^3 Flange criterion b/T = 9.84 Web criterion d/t = 35.34 ε = 1 b/T < 10.5 d/t < 42 The section is classified as COMPACT Step :- 2 Effective length As both ends pin-jointed KLx = 9m KLy = 9m KLz = 9m Slenderness ratio KLx/rx = 54.18 KLy/ry = 174.42 Non-dimensional effective slenderness ratio λ = 1.964 Step :- 3 To calculate φ h/bf = 1.60 > 1.2 T = 12.7mm So, buckling class = b α = 0.34 θ = 0.5(1+α(λ-0.2)+λ^2) θ = 2.73 Step :- 4 Claculation of χ χ = 0.216N/mm^2 Step :- 5 Calculation of fcd fcd = χfy/γm = 49.17 Step :- 6 Factored Axial Load Pd = Afcd = 514.614kN 1. b) Rafter :- Span of I beam = 6m Dead load = 18kN/m Imposed load = 40kN/m Length of bearing = 100mm Yield strength = 250N/mm^2 Step :- 1 Design load calculation Factored load = 87kN/m Factored bending moment = 391.5kNm Step :- 2 Section modulus required Zreqd = 1722600mm^3 = 1722.60cm^3 Step :- 3 Section classification Properties of selected section ISMB 250 A 475.5mm^2 D 250mm B 125mm t 6.9mm T 12.5mm Izz 5131.6cm^4 Iyy 344.5cm^4 Zp 459.76cm^3 Flange criterion B/2T = 5 Web criterion (D-2T)t = 32.61 ε = 1 B/2T < 9.4 (D-2T)/t < 83.9 The section is classified as PLASTIC Step :- 3 Moment of resistance of the cross section Moment Md = (Zpχfy)/γm Md = 104.491kNm < 391.50 Hence section ISMB 250 is adequate for flexure Step :- 4 Shear resistance of the cross section Shear capacity Vc = 0.6fyAv/γm Av = 1725mm^2 Vc = 235.23kN V = 261kN V/Vc = 1.11 .> 0.6 Step :- 5 Check for web buckling Slenderness ratio of web = 2.5d/t d = 225mm = 70.33mm Design compressive stress fc = 203N/mm^2 Pw = 315.16kN > 261 Hence web is safe against shear buckling Step :- 6 Check for web crippling at support Root radius of ISMB250 R = 13mm tf + R = 25.5mm Dispersion length n^2 = 63.75mm Pcrip = 245625N = 245.63kN < 261 Hence section adequate for web crippling Step :- 7 Check for serviceability Deflection Design load = 58kN/m Deflection δ = 95.36mm Allowable deflection = 30mm Hence serviceability is satisfied 1. c) Base plate :- Step :- 1 Strength of concrete fcu = 40 N/mm^2 Yield strength of steel fy = 250 N/mm^2 Material factor γm = .kN Factored load = 1500 kN Step :- 2 Steel column section Thickness of flange T = 12.7 mm Step :- 3 Area required Bearing strength of concrete = 0.4fcu = 16 N/mm^2 Area required = 93750 mm^2 Let size of plate Bplate = 350 mm Dplate = 350 mm Area of plate = 122500 mm^2 Let projection on each side a = b = 25 mm w = 12.24 N/mm^2 Thickness of slab base ts = 7.7 mm < 12.70 mm So use base plate of size 350mmx350mmx12mm. 1. d) Pedestal :- Grade of concrete = 40N/mm^2 Load = 123kN Moment = 116kNm Horizontal shear = 13kN Yield strength = 250N/mm^2 Length of base plate L = 450mm Width of the base plate B = 350mm C/C distance of (Ld) of bolt group in Z = 300mm C/C distance of (Ld) of bolt group in X = 180mm Bearing strength of concrete fc = 16N/mm^2 Depth of column D = 300mm Width of column W = 250mm Step :- 1 Anchor Bolt Details Dia of anchor bolt = 24mm No of anchor bolt on each side = 4 Total number of anchor bolt,n = 8 Gross area of the bolt 'Asb' = 452.16mm^2 Net area of the bolt 'Anb' = 352mm^2 Ultimate tensile strength of the bolt fub = 400N/mm^2 Fyb(Anchor bolt) = 240N/mm^2 Step :- 2 Base plate details Ultimate tensile strength of a plate,fu = 490N/mm^2 Thickness of plate = 16mm Yield stress of plate = 330N/mm^2 Step :- 3 Anchor bolt design Area of the plate = 157500mm^2 Stress Max pressure ζ1 = F/A+6*Mz/BL^2 = 10.60N/mm^2 < 16 OK Min pressure ζ2 = F/A-6*Mz/BL^2 = -9.04N/mm^2 < 16 C = L*ζ1/(ζ1+ζ2) = 242.89mm a = L/2-C/3 = 144.04mm e = (L-Ld)/2 = 75mm y = (L-C/3-e) = 294.04mm Tension in anchor bolts along the length of plate FT = (Mz-Fxa)/y = 364.38 Tension per bolt = 91.10kN shear per bolt,V = 1.63kN Step :- 4 Shear check Factored shear force Vsb = 1.63kN Vd,sd = 81290.9N = 21.291kN Factored Vd,sb = 65.03kN Step :- 5 Tensile check Factored tensile force in bolt Tb = 91.10kN Tensile strength of bolt Td,b = Tnb/γmb Tn,b = 0.9fubAn = 126720N = 126.72kN Tn,b = fybAsb(γmb/γmo) = 123316N = 123.316kN Lesser of two =123.32kN So, Td,b = 98.65kN Step :- 6 Combined unity check (Vsb/Vdb)^2+(Tb/Tdb)^2 Vsb/Vdb = 0.025 Tb/Tdb = 0.92 Unity check = 0.85 < 1 Hence OK Step :- 7 Anchor bolt length Bond strength in tension bd = 1.4N/mm^2 Anchor length required = Tb/(3.14*d*ηbd) = 863.44mm Let anchor bolt bolt length = 900mm 1. e) Purlin :- Step :- 1 Span of the purlin = 7m Spacing of the purlin = 1.5m Number of sag rods = 1 Slope of the roof = 5deg Step :- 2 Dead loads Weight of sheeting = 5kg/m^2 Self weight of purlin = 4.22kg/m^2 Additional load = 10% = 0.42kg/m^2 Total dead load = 0.096kN/m^2 Live loads Live load on roof = 75kg/m^2 = 0.75kN/m^2 Wind loads Basic wind spped = 50m/s Terrain category = 2 Building class = B k1 = 1 k2 = 1 k3 = 1 Design wind speed Vz = 57m/s Design wind pressure Pz = 1949.40N/m^2 = 1.95kN/m^2 Length of the building I = 50m Breath of the building w = 30m Height of the building h = 10.5m Height of the building at eaves = 9m h/w = 0.35 I/w = 1.67 External pressure coefficient Maximum downward Cpe = -0.4 Maximum upward Cpe = -0.7 Internal pressure coefficient Maximum positive Cpi = 0.5 Maximum negative Cpi = -0.5 For maximum upward wind force Maximum upward Cpe = -0.7 Cpi = -0.5 Cpe + Cpi = -1.2 Pz = 1.95 Wind pressure for purlin design = -2.339kN/m^2 For maximum upward wind force Max upward Cpe = -0.4 Cpi = 0.5 Cpe + Cpi = 0.1 Pz = 1.95 Wind pressure for purlin design = 0.195kN/m^2 Step :- 3 Design load calculation Spacing of purlin = 1.5m Slope of roof = 5deg Total dead load = 0.096kN/m^2 DL normal component = 0.144kN/m DL tangential component = 0.013kN/m Total live load = 0.75kN/m^2 LL normal component = 1.121kN/m LL tangential component = 0.098kN/m WL1 wind load = -2.339kN/m^2 WL normal component = -3.496kN/m WL2 wind load = 0.195kN/m^2 WL normal component = 0.291kN/m Summary of loads DL+LL DL+WL1 DL+LL+WL2 Normal load 1.265 -3.351 1.556 Tangential load 0.111 0.013 0.111 DL+LL 0.75(DL+WL1) 0.75(DL+LL+WL2) Normal load 1.265 -2.514 1.167 Tangential load 0.111 0.009 0.083 Maximum normal component = 1.265kN/m Step :- 4 Purlin section Section selected = Zx200x6x2.3 Yield stress of material = 2400kg/cm^2 Flange width b = 60mm Depth of section d = 200mm Thickness t = 2.30mm Length of Lip lip_I = 20mm Internal bending radius = 3mm Area = 8.07cm^2 Zxx = 47.72cm^3 Zyy = 10.22cm^3 Ixx = 477.18cm^4 Iyy = 61.34cm^4 Weight of purlin = 6.335kg/m =4.223kg/m^2 Check for basic properties t = 2.30mm w = 49.40mm Fy = 2400kg/cm^2 w/t = 21.48 < 60 OK Check for stress Span major axis = 7m Span minor axis = 2m BM coefficient for Mxx = 10 BM coefficient for Myy = 10 DL+LL DL+WL1 DL+LL+WL2 Normal load 6.198 16.422 7.625 Tangential load 0.060 0.007 0.060 Mxx/Zxx 129.878 344.133 159.789 Myy/Zyy 5.895 0.672 5.895 (Mxx/Zxx)+(Myy/Zyy) 135.773 344.805 165.684 Maximum compressive stress = 344.80N/mm^2 1435/sq.root of f = 24.44 > 21.48 Hence section OK Step :- 5 Combined Bending And Shear stresses in web h/t = 84.96 DL+LL DL+WL1 DL+LL+WL2 Fbw 506.539 673.697 637.697 Fv 735.223 977.847 977.847 Actual stress fbw 129.878 344.133 159.789 fv 5.895 0.672 5.895 fbw/Fbw 0.256 0.511 0.237 fv/Fv 0.008 0.001 0.006 Sqrt of sum of squares 0.257 0.511 0.237 Maximum combined stresses = 0.511 < 1 OK Step :- 6 Check for deflection Number of spans = 3 Deflection = (3/384)(wl^4/EI) w = 3.351kN/m Span = 6m E = 207400N/mm^2 Ixx = 477.18cm^4 δ = 34.29mm Allowable deflection δ = Span/180 = 33.33mm > 34.29 Not OK 2) Design a simply supported gantry girder to carry electric overhead travelling crane Step :- 1 Span of gantry girder = 7 m Span of crane girder = 9 m Crane capacity = 250 kN Self-weight of trolley, hook, electric motor etc. = 40 kN Self-weight of crane girder excluding trolley = 100 kN Minimum hook approach = 1.0 m Distance between wheels = 3 m Self-weight of rails = 0.2 kN/m Step :- 2 Maximum moment due to vertical load Weight of trolley + weight of load lifted = 290kN Self weight of crane girder = 100kN For maximum reaction on girder, the moving load should be as close to the gantry as possible Reaction at A RA = 501kN Reaction at B RB = 281.875kN Load on gantry girder from each wheel = 251kN Factored wheel load = 375.833kN Maximum moment ME = 634.21kNm Letr self weight of girder = 2kN/m Dead load = 2.2kN/m Factored DL = 3.3kN/m M due to DL = 20.21kNm M due to impact load = 158.55kNm Factored moment due to all vertical loads M = 812.99kNm Step :- 3 Maximum moment due to lateral load Horizontal force transferred to rails = 29kN Horizontal force on each wheel = 7.25kN Factored horizontal force = 10.875kN Maximum moment = 18.35kNm Vertical shear due to wheel loads = 563.75kN Impact = 140.94kN Self weight = 9.90kN Total shear = 714.59kN Lateral shear due to surge = 20.68kN Step :- 4 Preliminary section determination , ,, Minimum economic depth D = L/12 = 583.333mm Width of compression flange b = L/40 to L/30 = 175 to 233 Required plastic modulus Zp = 1.4M/fy = 4552721 = 4552.72x10^3 Let us try ISMB600 with ISMC400 on compression flange. Step :- 5 Properties of selected section ISMB [email protected]/m A 15600mm^2 h 600mm b 210mm tf 20.3mm tw 12mm Izz 91800x10^4mm^4 Iyy 2650x10^4mm^4 R 20mm ISMC [email protected]/m A 6380mm^2 h 400mm b 100mm tf 15.3mm tw 8.8mm Izz 15200x10^4mm^4 Iyy 508x10^4mm^4 Cyy 2.42mm Step :- 6 Let the distance of N>A> from, t,he tensio,n range by y' y' = 388.93mm Izz = 1348.13x10^6mm^4 Zez = 3466240mm^3 For compression flange about y-y axis I = 16766.7x10^4mm^4 Zey = 838333mm^3 Total area of section = 21980mm^2 Let plastic N>A> be at a distance Yp = 580.88mm Zpz = 4410317mm^3 For top flange Zpy = 1078488mm^3 Step :- 7 Section classfication For ISMB b/t = 4.90 < 9.4 d/t = 43.28 < 84 For ISMC b/t = 5.96 < 9.4 Hence section is plastic Step :- 8 Check for local moment capacity Local moment capacity for bending in vertical plane Mdz = fyZp/1.1 = 1002.34kNm Mdz = 1.2Zefy/1.1 = 945.34kNm For top flange Mdy = 245.11kNm Mdy = 228.64kNm In the above which ever the minimum value is taken Mdz = 945.34kNm Mdy = 228.64kNm Step :- 9 Check for combined local capacity (812.99/945.34)+(18.35/228.64)<1 0.947<1 Step :- 10 Check for buckling resistance Md = βbZpfbd βb = 1 Lt = 7000mm E = 200000N/mm^2 hf = 597.5mm Iy = 17850x10^4mm^4 A = 21980mm^2 ry = 90.12mm fcr,b = 417.43N/mm^2 λ = 0.744 θ = 0.95 = 0.97N/mm^2 fbd = 166.591 Mdz = 734.72kNm Mdy = 202.84kNm Hence 1.20>1 Hence section is unsafe against torsional buckling Step :- 11 Check for shear Vz = 714.59kN Shear capacity = Avfyw/(3^0.5x1.1) = 944755N = 944.755kN > 714.59 Now = 566.853kN Low shear Step :- 12 Check for web buckling b = 175mm n1 = 232.6mm d = 519.4mm t = 12mm = 104.746 Fcd = 128.6N/mm^2 Buckling resistance = 629008kN > 714.59kN Step :- 13 Check for deflection at working load Vertical deflection Serviceability vertical wheel load = 251kN Deflection at mid span a = 2000mm Izz = 1348.13x10^5mm^4 δ = 25.32mm Horizontal deflection I = 16766.65x10^4mm^4 δ = 4.88mm L/750 = 9.33mm Unsafe in vertical direction. Safe in lateral direction.