Project 2

Width 30m

Length50m

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

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 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 

 

 

 

Earthquake load.

LOAD CASE DEFINATION:

1. EQ+X

IS 1893 Load-X 1

2. EQ+Z

IS 1893 Load-Z 1

 

3. DEAD LOAD

Self weight Y -1.15

Main frame rafter GY -1.55 KN/m

Gable frame rafter GY -0.775 KN/m

Main frame column GY -1.2 KN/m

Gable frame column GY - 0.6 KN/m

Gable / wind column GY -1.04 KN/m

Wind frame point load FY - 1.14 KN/m

Gable frame point load FY -0.57 KN/m

4. LIVE LOAD

Main frame rafter GY -4.85 KN/m

Gable frame rafter GY - 2.425 KN/m

Wind frame point load FY - 4.15 KN/m

Gable frame point load FY - 2.06 KN/m 

Member Weight: 

 

Assign to Edit list from followed by the Dead Load

 

Self weight Y -1.15

Main frame rafter GY -1.55 KN/m

Gable frame rafter GY -0.775 KN/m

Main frame column GY -1.2 KN/m

Gable frame column GY - 0.6 KN/m

Gable / wind column GY -1.04 KN/m

Joint Weight: 

 

Assign to Edit list from followed by the Dead Load (point load) 

 

Wind frame point load FY - 1.14 KN/m

Gable frame point load FY -0.57 KN/m 

 

 

Manual calculation of DD, LL, WL

DESIGN OF STEEL COLUMN:

Solution: 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

Non-dimensional Effective Slenderness ratio, λ">λ3.3X728:

λ">λ =2.5X(174)2π2X2X105">√2.5X(174)2π2X2X105= 3.83

Value of ϕ">ϕts=√2.5w(a2−0.3b2)γm0fy>tf from equation: 

 

 Hence, α">α = 0.34 for buckling class ‘b’ will be considered.

Hence, ϕ">ϕ = 0.5 x [1+0.34 x (0.654-0.2)+0.6542] = 0.791

Calculation of x from equation x = 0.809

 

Calculation of fcd from the following equation:

= 0.809 x 250/1.1 = 183.86 N/mm2

Factored axial load in kN.

pd = Ag fcd= 10466 x 183.86/1000 = 1924.28 kN.

 

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:

Factores load 1.5(DL+LL) =87 KN/m

Factores Bending moment = Wl28">106Wl28">Wl28= 65.25 KN/m

Section modulus Required:

Z reqd = (65.25 x x1.1) /250 =287100 mm3 = 287.1 cm3

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

Moment of resistence of the cross-section:

= (1 x 375.35 x 250) / ( 1.1)

Md = 85.306 > 65.25 KN/m

DESIGN OF BASE PLATE:

Strength of concrete, Fcu = 40 N/mm2

Yield strength of steel, fy = 250 N/mm2

Material factor, γm">γmF/A)−(6XMz/BL^2)= 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

pojection 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

 

  

 Minimum pressure = -9.04 N/mm2 < 16 Hence OK

Centroid = = 242.89 mm

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 

 = 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 /γmb">γmb= 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

Anchor Bolt Length

Bond strength in tension, τbd">τbd= 1.4 N/mm2

Anchor length required = Tb(3.14*τbd">τbd) = 863.44 mm

Let Anchor Bolt Length = 900 mm 

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