Project 2
1. Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:
Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875
- Prepare a Design Basis Report for the project
- Create structural model using STAAD Pro Connect Edition
- Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
- Design using MS – Excel
- Steel column
- Rafter
- Base plate
- Pedestal
- Z- purlin
- Design Foundations using STAAD Foundation.
- Attach necessary sketches and drawings wherever required.
- All stipulations, assumptions and design parameter must adhere to Indian Standards.
- Design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. The specification must be as follows:
Assume suitable sections for structural elements. Follow IS800:2007, IS1893 and IS 875
Width = 30m
Length=50m
Eave height =9m
Bay spacing=6m
Soil type=medium
Safe bearing capacity=200KN/m^2
Roof slope = 1 in 12
- Prepare a Design Basis Report for the project
- Create structural model using STAAD Pro Connect Edition
- Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
- Design using MS – Excel
- Steel column
- Rafter
- Base plate
- Pedestal
- Z- purlin
- Design Foundations using STAAD Foundation.
- Attach necessary sketches and drawings wherever required.
- All stipulations, assumptions and design parameter must adhere to Indian Standards.
Solution:
ROOF SLOPE CALCULATION:
Roof Slope = 1 in 12
Width, b = 30 m
X / (b/2) = 1/12
X = (30/2) / 12 = 15/12
X = 1.25 m
DESIGN OF RAFTER:
Solution:
Span of Rafter = 6 m
Dead load = 18KN/m
Imposed load = 40KN/m
support bearing = 100mm
yield strength = 250N/mm2
Design load calculation:
Factored load 1.5(DL+LL) =(18*1.5)+(7.875*1.5)
=87 KN/m
Factored Bending moment = (87*6*6)/8=65.25KN/m
Section modulus Required:
Z reqd = (65.25*10^6*1.1)/250=287.1cm3
Section classification:
ISMB-200
A = 323.3 mm2
D = 200mm
B = 100mm
t = 5.7 mm
T = 10.8 mm
Ixx = 2235.4 cm4
Iyy = 150 cm4
Zp = 375.35 cm3
DESIGN OF STEEL COLUMN:
Flange thickness = T = 12.7 mm.
Overall height of Column ISMB400 = h = 400 mm.
Clear depth between flanges = d = 400 – (12.7 x 2)
= 374.6 mm.
Thickness of web = t = 10.6mm.
Flange width = 2b = bf = 250 mm.
Hence, half Flange Width = b = 125 mm.
Self –weight = w = 0.822 kN/m.
Area of cross-section = A = 10466 mm2.
Radius of gyration about x = rx = 166.1 mm.
Radius of gyration about y = ry = 51.6 mm.
Type of section:
b/T = 125/12.7= 9.8 < 10.5
d/t =374.6/10.6 =35.3 < 42
(Table 3.1 of IS: 800)
Hence, cross-section can be classified as “COMPACT”.
Effective Sectional Area, Ae = 10,466 mm2
(Since there is no hole, (Clause 7.3.2 of IS: 800)
no reduction has been considered)
Effective Length:
As, both ends are pin-jointed effective length, KLx = KLy = 3m
Slenderness ratios:
KLx/rx = 9000/166.1 =54.1
KLy/ry = 9000/51.6 =174.41
DESIGN OF BASE PLATE:
Strength of concrete, Fcu = 40 N/mm2
Yield strength of steel, fy = 250 N/mm2
Material factor, = 1.1 KN
Factoresl oad = 1500 KN
Steel column section:
Thickness of flange, T = 12.7 mm
Area required:
Bearing strength of concrete = 0.4fcu = 0.4 x 40 = 16 N/mm2
= (1500x1000) /( 16)
= 93750 mm2
Let size of plate , Bplate = 450 mm
Dplate = 300 mm
Area of plate= 135000 mm2
projection on each side = a=b =25mm
W = (1500x1000) /( 450 x 300)
= 11.11 N/mm2
Therefore, Thickness of Base Plate, clause 7.4.3.1
ts= 7.3 mm < 12.7 mm
Size of Base plate 450 x 350 x 16 mm
DESIGN OF PEDESTAL:
Grade of concrete = 40 N/mm2
Load = 200 KN
Moment = 120 KN
Horizontal shear = 20 KN
Yield strength = 250 N/mm2
Length of base plate = 450 mm
Width of base plate = 350 mm
C/C distance of bolt in group-Z = 300 mm
C/C distance of bolt in group-X = 180 mm
Bearing strength of concrete Fc = 16 N/mm2
Depth of Column = 300 mm
Width of Column = 250 mm
Anchor Bolt Details
Dia of anchor bolts =24 mm
No:of anchor bolts in each side = 4
Total no:of anchor bolts, n = 8
Gross area of the bolt ;Asb' = 452.16 mm2
Net area of bolt 'Anb' = 352 mm2
Ultimate tensile strength of bolt 'fub' = 400 N/mm2
Fyb (anchor bolts) = 240 N/mm2
Base plate Details
Ultimate tensile strength of plate 'fu' = 490 N/mm2
Thickness of plate = 16 mm
Yield stress of plate = 330 N/mm2
Anchor bolt design
Area of the plate = 157500 mm2
Stress, Maximum pressure , =F/A+6*Mz/BL^2
=10.6 N/mm^2<16
It is ok
Stress, Manimum pressure , =F/A-6*Mz/BL^2
=-9.04 N/mm^2<16
a =L/2-C/3 = 144.04 mm
e = (L-Ld) / 2 =75 mm
y = (L - C/3-e) = 294.04 mm
Tension in anchor bolt along the length of plate, FT = (Mz - Fa)/Y
= 364.38
Tension per bolt = 91 KN
Shear per bolt = 1.63 KN
Shear Check
Factored shear force = Vsb = 1.63 KN
Vd,sb = 81290.9 N
= 81 KN
Factored = 65.03 KN
Tensile Check
Factored tensile force in bolt, Tb = 91.1 KN
Tensile strength of bolt Ts,b = Tn,b /[Math Processing Error]���= 126 KN
Td,b = 98.65 KN
Combined Unity Check
Vsb/Vdb = 0.025
Tb/Tdb = 0.92
Unity check = 0.85 < 1, Hence OK
DESIGN OF Z-PURLIN:
Span of the purlin = 6 m
spacing of purlin = 1.5 m
No:of sag rods = 1
slope of the roof = 4.76 = 5 degree
Dead load:
Weight of sheeting = 6kg/m2
self weight of purlin = 4.22 kg/m2
Additional load = 10% = 0.42 kg/m2
Total Dead load = 0.106 kg/m2
Live load:
Live load on roof = 75 kg/m2
Wind Load:
Basic wind speed = 50 m/s
Terrain caterogy = 2
Building class = B
K1 = 1
K2 = 1
K3 = 1
Design wind speed, Vz = 57 m/s
Design wind pressure, Pz = 1949.9 m/s = 1.94 KN
Length of building, L = 50 m
Breadth of building, W = 30 m
Height of building, H = 10.25 m
Height of eaves, = 9 m
h/w = 0.34
L/w = 1.667
External Pressure Co-eeficient
Maximum downward =Cpe = -0.4
Maximum upward =Cpe = -0.7
Internal Pressure Co-eeficient
Maximum positive =Cpi = 0.5
Maximum negative =Cpi = -0.5
For Maximum upward wind force
Max upward Cpe = - 0.7
Cpi = - 0.5
Cpe+Cpi = -1.2
Pz = 1.95
Wind Pressure for Purlin Design = - 2.339 KN/m2
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 = 0.195 KN/m2
Design Load Calculation
spacing of purlin 1.5 m
slope of roof = 5degree
Total dead load = 0.096
DL Normal component = 0.144 KN/m
DL Tangential component = 0.013 KN/m
Total Live load = 0.75
LL Normal component = 1.121 KN/m
LL Tangential component = 0.098 KN/m
Total Wind load = -2.339
WL Normal component = - 3.496 KN/m
WL2 WL load = 0.195 KN/m
WL Normal component = 0.291 KN/m
Maximum Normal component = DL+LL = 1.265 KN/m
Purlin section
Selected section Z200 x 6 x 2.3
Area = 8.07 cm2
Weight of purlin = 6.335 kg/m
2. Design a simply supported gantry girder to carry electric overhead travelling crane
Given:
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
Maximum moment due to vertical load
weight of trollley + crane capacity = 40 + 250 = 290 kN
Self-weight of crane girder excluding trolley = 100 kN
For maximum reaction on gantry girder, the moving load should be as close the gantry girder as possible.
Ra=290X8/9+100/2=307.77KN
This load is transferred to gantry girder, through two wheels, the wheel base being 3m.
So load on gantry girder from each wheel = 307.77/2 = 158.88 KN
Factored load = 158.88 x 1.5 =230.83 KN
Maximum moments due to moving loads occur under a wheel when the c.g of wheel load &
The wheel are equidistant from the ccentre of girder, as shown below.
RD= 230.83 x (3-1.5-0.75) + 230.83 x (3+0.75) = 148.40 KN
7
Max. moment, Me = 148.40 x 2.25 = 333.90 KN-m
Moment due to impact = 0.25 x 333.90 = 83.47 KN-m
Assume self weight of girder = 2KN-m
Dead load due to self weight + rails = 2+0.2= 2.2 KN-m
Factored DL = 2.2 x 1.5 = 3.3 KN-m
Moment due to DL = = 20.21 KN-m
Factored moment due to all vertical loads,
Mz = 333.90 + 83.47 + 20.21 = 437.58 KN-m
Maximum moment due to Lateral force:
Horizontal force transferred to rails = 10% of weight of trolley plus load lifted = (10/100) x (250+40) = 29 KN
This is distributed over 4 wheels.
So, horizontal force on each wheel = 29/4 = 7.25 KN
Factored horizontal force on each wheel = 1.5 x 7.25 = 10.875 KN
For maximum moment in gantry girder the position of loads is same as earlier except that it is horizontal, hence byproportioning we get.
My = (10.875/230.83) x 333.90 = 15.73 KN-m.
Shear force:
For maximum shear force on the girder, the trailing wheel should be just on the girde
Priliminary Section:
Minimum economic depth, L/12 = 7000/12 = 583.33 mm
Width of the compression flange may be taken as (1/40) to (1/30) of span
So, flange width can be taken, L/40 = 7000/40 = 175 mm to L/30 = 7000/30 = 233.33 mm.
Required, Zp = 1.4 x M/fy = 1.4X437.58X10^6/250 = 2450 x 103103mm3
Let us try a ISMB 550 with ISMC 250 on compression flange.