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Challenge

Project 2

VVarshita C28 Apr 2023
DESIGN

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
  2. Create structural model using STAAD Pro Connect Edition
  3. Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
  4. Design using MS – Excel
    1. Steel column
    2. Rafter
    3. Base plate
    4. Pedestal
    5. Z- purlin
  1. Design Foundations using STAAD Foundation.
  2. Attach necessary sketches and drawings wherever required.
  3. All stipulations, assumptions and design parameter must adhere to Indian Standards.

ANS :

Roof slope calculation

Roof slope = 1 in 12

x/(b/2) = 1/12

b = 30 m

x/(30/2) = 1/12

x = 15/12 = 1.25 m

Total height = 9 + 1.25 = 10.25 m

Calculation of Design Loads

Dead Load Calculation

  1. Dead Load – Rafters
  • Self weight – FOS = 1.15
  • UDL on Rafter (Main frame)
  • Weight for roof sheeting = 5Kg/m2
  • Weight for sag rods and braces = 5 Kg/m2
  • Collateral load = 10 Kg/m2
  • Total load = 20 Kg/m2 = 0.2 KN/m2
  • UDL on Rafters
  • Spacing = 6 m
  • UDL = 0.2 x 6 = 1.2 m
  • UDL on Rafters from Purlin
  • Choose suitable purlin section for weight of purlin (IS-811)
  • Say Z =270 x 75 x 20 x 2.55 = 8.77 Kg/m
  • (No. of purlin x Length of purlin x Weight of purlin) / Length of rafter
  • (13 x 6 x 8.77) / 15 = 0.456 KN/m
  • UDL on Main frame = 1.2 + 0.456 = 656 KN/m
  • UDL on Gable frame = 1.656 / 2 = 828 KN/m
  1. Dead Load – Columns
  • Load on Column (Main frame)
  • Weight of side wall sheeting = 5 Kg/m2
  • Weight of sag rods and braces = 5 Kg/m2
  • Load from sag rods and braces = 10 Kg/m2 = 0.1 KN/m2 = 0.1 x 6 = 0.6 KN/m
  • Weight of grits (270 x 75 x 20 x 2.55) = 8.77 Kg/m
  • Weight of column = (6 x 6 x 8.77)/6 = 52.62 Kg/m = 0.53 KN/m
  • Total Load = 0.6 + 0.53 = 13 KN/m
  • Load on Gable frame column = 1.13 / 2 = 565 KN/m
  • Load on wind column
  • Load from sheet and sag rods = 0.1 x 6 = 0.6 KN/m
  • Weight of grits ( 230 x 75 x 20 x 2) = 6.32 Kg/m
  • Load on wind column = (6 x 6.32 x 6)/6 = 37.92 Kg/m = 0.38 KN/m
  • Total load = 0.6 + 0.38 = 98 KN/m
  1. Dead Load – Eave strut and Gutter
  • Point load
  • Weight on Eave strut = CS 270 x 75 x 20 x 3.15 = 10.7 Kg/m
  • Weight of Eave gutter = 0.25 x 0.25 x 0.001 = [0.25 + 0.25 + 0.25] x 0.001 x 7850 = 5.89 Kg/m
  • Load due to Eave strut and gutter = 10.7 + 5.89 = 16.59 kg/m
  • Load due to Eave strut and gutter (Main frame) = 16.59 x 6 = 99.54 kg/m = 995 KN
  • Load due to Eave strut and gutter (Gable frame) = 0.995/2 = 497 KN

Live Load calculation

  • Live load on roof = 0.75 KN/m2 (IS 875-II)
  • UDL on Rafter
  • Main Frame rafter = 0.75 x 6 = 5 KN/m
  • Gable frame rafter = 4.5/2 = 25 KN/m
  • Eave gutter (250 x 250 x 1 mm)
  • Load due to weight of water in gutter
  • Main frame = [0.25 x 0.25 x 9 x 6] = 375 KN
  • Gable frame = 3.375/2 = 687 KN

Earthquake load

  • Base shear, Vb = Ah x W
  • Zone, Z = 0.16
  • Response reduction factor, R = 4
  • Importance factor, I = 1.2
  • Damping Ratio = 5%
  • Soil Category = Medium

Wind Load Calculations

Basic wind speed, Vb = 50 m/s for Chennai

Probability factor, K1 = 1 (T.1)

Terrain factor, K2 = 1 (T.2, Cat.2)

Topography factor, K3 = 1 (Cl.6.3.3.1)

Importance factor, K4 = 1.15 (Industrial Structure)

Design Wind speed, Vz = 50 x 1 x 1 x 1 x 1.15 = 57.5 m/s

Wind Pressure, Pz = 0.6 Vz^2 = 0.6 x 57.5^2 = 1.984 KN/m2

Tributary area = 6 x 9 = 54 m2

From table 4 (Clause 7.2.2) for Ka Value(Area Averaging factor)

25

0.9

54

x

100

0.8

X = 0.9 + (((54-25)/(100-25))*((0.9)-(0.8))) = 0.861

After interpolation,

Area Averaging factor, Ka = 0.861 (Clause 7.2.2)

Wind directionality factor, Kd = 1 (Clause 7.2.1)

Combination factor, Kc = 0.9 (Clause 7.3.3.13)

Design Wind Pressure, Pd = Kd x Ka x Kc x Pz = 1 x 0.861 x 0.9 x 1.984 = 1.537 KN/m2

WL(0+CPI)

WL(0-CPI)

WL(90+CPI)

WL(90-CPI)

Design of 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

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

λ">λ =2.5X(174)2π2X2X105">√2.5X(174)2π2X2X1052.5X(174)2π2X2X105= 3.83

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

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

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">Wl28Wl28= 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">γm= 1.1 KN

Factoresl Load = 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

Area = (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 < T =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 =10.6 N/mm2 < 16 Hence OK

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 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 /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= 1.4 N/mm2

Anchor length required = Tb(3.14*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

  1. 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

ANS :

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

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 whichever 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

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.

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