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

• 2) Create structural model using STAAD Pro Connect Edition :-
• 
• 
• 
• 
• 
• 
• 
• 3) Prepare DL, LL, WL and EQL load calculations as per IS 875 standards :-
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 
• 

• Foundation

• 
  
• 

Designing using MS excel

• 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
• φ = 0.5(1+α(λ-0.2)+λ^2)
• φ = 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
• 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
• 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.
• 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 σ1 = F/A+6*Mz/BL^2
• = 10.60N/mm^2
• < 16
• OK
• Min pressure σ2 = F/A-6*Mz/BL^2
• = -9.04N/mm^2
• < 16
• C = L*σ1/(σ1+σ2)
• = 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*τbd)
• = 863.44mm
• Let anchor bolt bolt length = 900mm
• 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
• 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
• φ = 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.