Project 2

Questions & Answers

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.

Roof slope calculation:

Roof slope = 1 in 12

Width, b = 30m

x / (b/2) = 1/12

x = (30/2) / 12 = 15/12

x = 1.25 m

Geometry / Model creation:

Load case defination:

1. EQ+X

* IS 1893 Load-x 1

2. EQ+Z

* IS 1893 Load-z 1

3. DEAD LOAD

* Self wegth Y-1.15

* Main frame refter 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 refter 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

DESIGN OF STEEL COLUMN:

Solution:

Flange thickness = T = 12.7 mm.

Overall height of column ISMB400 = h = 400mm.

Clear depth between flanges = d =400 - (12.7 x 2 ) = 374.6 mm.

Thickness of web = t = 10.6 mm.

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 sectinal 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-joined effective length, KLx = KLy = 3m

Slenderness ratios:

KLx/rx = 9000/166.1 = 54.1

KLy/ry = 9000/51.6 = 174.41

Non-dimentional Effective slenderness ratio, λ">3.3x7²/8.

λ">λ = √ƒy/ƒ∞

Value of ø " >øts = √2.5w (a2 - 0.3b2) ϒm0fy > tf from equation:

DESIGN OF RAFTER:

Solution:

Span of rafter = 6m.

Dead load = 18 KN/m

Imposed load = 40 KN/m

Support bearing = 100mm

yield strength = 250N/mm2

Design load calculation:

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

Factores Bending moment = W128" > 10 Wl²/8 = 65.25 KN/m

Section modulus Required:

Z reqd = (65.25 x x 1.1 )/ 250 = 287100 mm3 = 287.1 cm³

Section classification:

ISMB-200

A = 323.3 MM²

D = 200 MM

B = 100 mm

t = 5.7 mm

T = 10.8 mm

lxx = 2235.4 cm4

lyy = 150 cm4

zρ = 375.35 cm³

Moment of resistance of the cross-section:

Md = βbZfy/ϒm0

= (1x375.35x250)/(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, ϒm">ϒm F/A-6XMz/BL² = 1.1 KN

Factore load = 1500 KN

Steel Column section:

Thickness of flange, T = 12.7 mm

Area of required:

Bearing strength of concrete = 0.4fcu = 0.4x40 = 16 N/mm2

= (1500x1000)/(16)

= 93750 mm2

Let size of plate, Bplate = 450mm

Dplate = 300mm

Area of plate = 135000 mm2

Projection on each side = a=b= 25mm

W = (1500x1000) / (450x300)

= 11.11 N/mm2

Therefore, Thickness of base plate, clause 7.43.1

LXσ1/σ1+σ2

ts = 7.3 mm < 12.7mm

Size of Base plate 450 x 350 x 16 mm

DESIGN OF PEDESTAL:

Grade of concrete = 40 N/mm2

Load = 200 KN

Moment = 120KN

Horizontal shear = 20 KN

Yield strength = 250 N/mm2

Length of base plate = 450 mm

Width of base plate in group - z = 300 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

Achor 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' = 352 mm2

Ultimate tension strength of bolt 'fub' = 400 N/mm2

Fyb (anchor bolts) = 240 mm2

Base plate Details

Ultimate tension strength of plate 'fu' = 490 N/mm2

Thickness of plate = 16 mm

Yield stress of plate = 330 N/mm2

Anchor bolt desgn

Area of the plate = 157500 mm2

Stress, Maximum pressure = F/A + 6XMz /BL²

= 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 tensile factore in bolt, Tb = 91.1 KN

Tensile strength of bolt Ts,b/γmb = 126 KN

Td,b = 98.65 KN

Tensile Check

Factored tensile force in bolt, Tb = 91.1 KN

Tensile strength of bolt Ts,b = Tn,b/ymb = 126 KN

Td,b = 98.65 KN

Combined Unit 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 = τb(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:

Weigth of sheeting = 6kg/m2

self weigth of purlin = 4.22 kg/m2

Additional load = 10% = 0.42 kg/m2

Total Dead load = 0.106 kg/m2

Live load:

Love 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 bulding, H = 10.25 m

Height of eaves, = 9m

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

Total Live Load = 0.75

LL Normal component = 0.121 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

Weigth 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

Solution:

Maximum moment due to vertical load

weigth of trollley + crane capacity = 40 + 250 = 290 kN

Self-weigth of crane grider excluding trolley = 100 kN

For maximum reaction on granty, the moving load should ba as close the granty

grider as possible.

This load is transferred to granty grider, through two wheel base being 3m.

So load on granty grider from each wheel = 307.77/2 = 158.88 KN

Factored load = 158.88 x 1.5 = 230.83 KN

Maximum moment due to moving loads occur under a wheel when the c.g of wheel load & The wheel are equidistant

from the centre of girder, as show below.

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 weigth of girder = 2KN-m

Dead load due to self weigth + 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 weigth of trolley plus load lifted = (10/100) x (250+40) = 29KN

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 granty girder the position of load is same as earlier expert that it is horizonntal, hence byproportioning we get.

My = (10.875/230.83) x 333.90 = 15.73 KN-m.

Priliminary section:

Minimum economic depth, L/12 = 7000/12 = 583.33 mm

Width of the compressive 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 = 1.4x 437.58x10/250 = 2450x10³mm3

Let us try a ISMB 550 with ISMC 250 on compressive flange.