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

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

A)

AIM: Design a warehouse Building located in chennai using STAAD Pro Connect Edition

1. Prepare a Design Basis Report for the project
2. Create structural model using STAAD Pro Connection Edition
3. Prepare DL,LL,WL and EQL Load calculations as per as IS 875 standards
4. Design using MS -Excel

1. Steel Column
2. Rafter
3. Base plate
4. Pedestal
5. Z-purlin

1. Design Foundation using STAAD Foundation.
2. Attach necessary sketches and drawings wherever required.
3. All stipulations, assumptions and design parameter must adhere to indian standards.

INTRODUCTION:

To create a model by using data and apply loads dead load live loads and wind loads

• Section 7 IS 800:2007
• BUckling class
• Effective length column
• Cross sectional shapes (Built up columns).

Web bucking:

• Web buckling occurs when intensity of compressive stress near the center of the section exceeds the critical buckling stress of web acting as a strut.
• at supports and under a concentrated load certian portion of the beam acts as a column to transfer the load.
• under this compressive force the web of the beam section may buckle.

Limit state of strength:

• Limit state strength are those associated with failures under the action of probable and most unfavorable load combinations on the structure using partial safety factor.
• The might endanger the safety of life and property.
• Checked by utilization ratio.

Limit state of serviceability:

• Limit state of serviceability are those associated with deformation and deflections which may adversely affective use of the structure.
• Checked by deflection limits.

PROCEDURE:

ROOF SLOPE CALCULATION:

Roof slope = 1in12

Width, b = 30m

X/(b/2) = 1/2

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

X = 1.25m

GEOMETRY/ MODEL CREATOR:

DEAD LOAD:

LIVE LOAD:

SEISIMIC LOAD:

WIND LOAD W(0+CPi):

W(0-CPi):

W(90+CPi):

W(90-CPi):

DESIGN OF STEEL COLUMN:

Solution:

Flange thickness t = 12.7mm

overall height of column ISMB400 = h = 400mm.

clear depth between flanges = d = 400 - (12.7 x 2)

=374.64mm

Thickneess of web = t = 10.6mm.

Flange width =2b = bf =250mm.

Hence half flange width = b =125mm.

slef- weight = w = 0.822 KN/m.

Area of cross- section =A = 10466mm2.

Radius of gyration about x = rx = 166.1mm.

Radius of gyratio anout Y = ry =51.6mm.

Step - 1

Types of section:

Properties of selected section

ISHB 400.

Flange criterion b/T = 125/12.7 = 9.8 <10.5

Web criterion d/t = 374.6/10.6 = 35.3<42

(Table 3.1 of IS : 800)

Hence,cross - section can be classified as "COMPACT".

step-2

Effective Length:

As, both ends are pin- jointed

Effective length ,KLx = 9m

KLy=9m

KLz=9m

Slenderness rations:

KLx/rx = 9000/166,1 =54.1

KLy/ry = 9000/51.6 = 174.41

Non-dimensional Effective slenderness ration

λ">λ =1.96

Step -3

To calculate :

h/bf = 1.60 >1.21

T=12.7mm

so buckling class = b = 0.34

φ = 0.5(1+(λ-0.2)+λ^2)

=2.76

Step - 4

Calculation of x from equation x :

= 0.214n/mm^2

Step - 5

Calculation of fcd:

fcd = Xfy/γm

=49.15

Step - 6

Factored axial load :

pd = Afcd

= 514.61 KN

DESIGN OF RAFTER :

Solution :

Span of Rafter = 6m

Dead load = 18KN/m

Imposed load = 40KN/m

support bearing = 100

yield strenght =250N/mm2

Step - 1

Design load calculation :

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

Factored Bending moment = 391.5 KN/m

Step - 2

Section modulus Required:

Z req =1722600mm^3

= 1722.60cm^3

Step -3

Section classification :

Flange criterion B/2T =5

Web criterion (D/2T)t = 32.61

∈ =1

B/2T <9.4

(D/2T)t<83.9

The section is Plastic.

ISMB - 250

A = 475.5mm^2

D = 250mm

B = 125mm

t = 6.9mm

T = 12.5mm

I_ZZ = 5131.6cm^4

I_yy = 344.5cm^4

Zp = 459.76 cm^3

Flange section B/2T =5

Web criterion (D - 2T)t = 32.61

B/2T <9.4

(D -2T ) /t< 83.9

The section is classified as plastic.

Step - 3

Moment of resistence of the cross- section:

Md = (Zp X fy) /γm)

Md = 10444.491KN/m

<391.50

Hence section ISMB 250 is Adequate for flexure.

Step -4

Shear resistance of the cross section :

Shear capacity Vc = 0.6fy Av/rm

AV = 1725 mm^2

Vc = 235.23KN

V = 261KN

V/Vc = 1.11

>0.6

Step - 5

Check for web bukling :

Slenderness ratio of web = 2.5d/t

d = 225mm

Desing 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 ISMB 250 R = 13mm

tf + R = 25.5mm

Dispersion length n^2 = 63.75mm

Pcrip = 245625N

=245.63 KN

<261

Hence section adequate for web crippling.

Step - 7

Check for serviceability deflection:

Design load = 58Kn/m

Deflection = 95.36

Allowable deflection = 30mm

Hence serviceability is satisfied

DESIGN OF BASEPLATE :

Step - 1

Strength of concrete, Fcu = 40N/mm2

Yield strengh of steel, fy = 250N/mm2

Material factor,γm">γm FA - 6XMzBL2 = 1.1KN

Factoresl Load = 1500 KN

Step -2

Steel column section :

Thickness of flange, T = 12.7mm

Step - 3

Area required :

Bearing strength of concrete = 0.4fcu = 0.4x 40 = 16N/mm^2

=(1500 x 1000)/(16)

=93750mm2

Let size of plate ,Bplate =350 mm

Dplate = 350mm

Area of plate = 122500 mm^2

Projection on each side = a= b= 25mm

W = (1500x 1000 ) /(450x300)

= 11.11 N/mm2

Therefore , Thickness of Base Plate , clause 7.4.3.1

ts = 7.7mm < 12.7mm

size of Base Plate 350mmx350mmx12mm

d) DESIGN PEDSTAL :

Grade of concrete = 40N/mm2

Load = 123KN

Moment = 116KNm

Horizontal shear = 13KN

Yield strength = 250N/mm2

Length of base plate = 350mm

Width of base plate = 350mm

C/C distance of bolt in group -z = 300mm

C/C distance of bolt in group -x = 180mm

Bearing strength of concrete Fc = 16N/mm2

Depth of column = 300mm

Width of column = 250mm

Anchor Bolt Details :

Step -1

Dia of anchor bolts =24mm

No:of anchor bolts in each side = 4

Total no :of anchor bolts, n = 8

Gross area of the bolt; Asb' = 452.16mm2

Net area of bolt 'Anb" = 352mm2

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

Fyb(anchor bolts) = 240N/mm2

Step-2

Base plate Details

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

Thickness of plate = 16mm

Yield stress of plate = 330N/mm2

Step -3

Anchor bolt desian

Area of the plate = 157500mm2

stress, Maximum pressure = F/A+6X Mz /BL^2

=10.6N/mm2 < 16 Hence Ok.

Centriod = 242.89

a =L/2 - C/3 = 144.04mm

e=(L-Ld)/2= 75mm

y = (L - C/3 -e) = 294.04mm

Tension in anchor bolt along the length of plate ,Ft = (Mz -Fa) /Y = 364.38

Tension per bolt = 91KN

Shear per bolt = 1.63KN

Step - 4

Shear check

Factored shear force = Vsb = 1.63KN

Step- 5

Tensile check

Factred tensile force in bolt ,Tb = 91.1KN

Tensile strength of bolt Td,b =Tn,b/γmb">γmb =126KN

Td,b = 98.65 KN

Step -6

Combined unity check

Vsb/Vdb = 0.025

Tb/Tbd = 0.92

Unit check = 0.85 < 1, Hence Ok

Step -7

Anchor Bolt Length

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

Anchor length require = Tb (3.14*tbd">tbd)

= 863.44mm

Let Anchor Bolt Length = 900mm

DESIGN OF Z -PURLIN

Step -1

Span of the purlin = 7m

Spacing of purlin =1.5m

No:of sag rods = 1

slope of the roof = 4.76= 5degree

Step -2

Dead load :

Weight of sheeting = 5Kg /m2

Self weight of purlin - 4.22 Kg/m^2

Additionl load = 10% = 0.42Kg/m^2

Total Dead load = 0.096 K/m2

Live Load :

Live load on roof = 75 kg/m2

Wind load :

Basic wind speed = 50m/s

Terrain caterogy =2

Building class = B

K1 =1

K2 = 1

K3 = 1

Design wind speed ,Vz =57m/s

Design wind pressure, Pz = 1949.9N.m2 = 1.94KN/m2

Length of building , L = 50

Breadth of building , w = 30m

Height of building , H = 10.5m

Height of eaves, =9m

h/w = 0.34

L/w = 1.67

External pressure co - efficient :

Maximum downward = cpe = -0.4

Maxiu upward = cpe = -0.7

Internal pressure co- efficient:

Maximum positive = Cpi = 0.5

Maximum negative = cpi = -0.5

For Maximu upward wind force :

Max upward Cpe = -0.7

CPi = -0.5

Cpe+Cpi = -1.2

Pz = 1.95

Wind pressure for purling 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 = 0.195KN/m2

Step -3

Design Load calculation:

Spacing of purlin 1.5m

Slope of roof = 5 degree

Total dead load = 0.096Kn/m^2

DL Normal component = 0.144KN/m

DL Tangential component = 0.144KN.m

Total Live load = 0.75KN/m2

LL Normal component = 1.12KN/m

LL Tangential component = 0.098KN/m

Total Wind load = -2.339

WL Noramal component = -3.498 KN/m

WL2 WL load = 0.195KN/m

Wl Normal component = 0.291KN/m

Maximum Noramal component = DL + LL = 1.265KN/m

Purlin section:

Select section Z X 200 x 6 x 2.3

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 = 47.72 cm^3

Zyy = 10.22cm^3

Ixx = 477.18cm^4

Iyy = 61.34 cm^4

Weight of purlin = 6.335 Kg/m

Check for basic properties :

t= 2.30 mm

w=49.40mm

Fy= 2400kg/cm^2

W/t = 21.48

<60

OK.

Check for stress :

Span major axis = 7m

Span minor axis = 2m

BIM cofficient for mxx = 10

BIM coefficient for myy = 10

Maximum compressive stress = 344.80N/mm^2

1435/Sq.root of f = 24.44 >21.48

Hence section OK.

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

d = 34.29mm

Allowable deflection d - span/180

= 33.33mm >34.29

Not OK.

Foundations using STAAD Foundation.

RESULTS:

• First analyze the file .
• Analysis Print

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

ANS)

Step -1

Span of gantry girder = 7m

span of crane girder = 9m

Crane capacity = 250KN

self -weight of trolley, hook , electric motor etc. 40 KN

Self -weight of crane grider excluding trolley = 100Kn

Minium hook approch = 1.0m

Distance between wheels = 3m

Slef - weight of rails = 0.2 KN/m

Step -2

Maximum moment due to vertical load :

Weigth of trolley +crane capacity = 40 +250 = 290Kn

Self - weight of crane grider excluding trolley = 100KN

For maxlimum reaction on ganty grider , the moving load should be as close the gantry girder as possible.

Reaction at RA = 50KN.

Reaction at RB = 281.875KN.

This load is transferred to ganty girder , through two wheels the wheel base beign 3m.

So load on gantry girder from each wheel = 251KN

factored wheel load = 375.833 KN

Maximum moments due to moving loads occur under a wheel when the c.g of wheel load

Max.moment due to moving loads occur under a wheel whe the c.g of wheel load

Max.Moment ,Me = 634.21KN-m

Momemnt due to impact = 0.25x 333.90 = 83.47-m

Assume self weight of girder = 2KN-m

Dead load due to self weigth +rails = 2+0.2 = 2.2 KN-m

Factored DL = 2.2x 1.5 = 33KN -m

Moment due to DL = 20221KN-m

M due to imact load = 158.55KNm

Factored moment due to all vertical loads Mz= 812.99KNm.

Step -3

Maximum moment due to lateral Load:

Horizontal force transfered to rails = 10% of weight of trolley plus load lifted = (10/100) x (250 +40) = 29KN

This is distributed over 4 wheels.

So, horizontal force on each wheel = 1.5x7.25 = 10.875Kn

For maximum moment in gantry girder the postion of loads is same as earlier except that it is horizontal , hence by proportioning we get,

My = 18.35KN- m

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

Shear force :

For maximum shear force on the grider , the trailling wheel should be just on the girder as shown below.

Vertical shear due to wheel loads = 230.83+(230.83 x 3)/ 7 = 329.75KN

Step- 4

Prilimary section determination :

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

Width of the compression flange may be taken as (L/40) to (L/30) of span

do, flange width can be taken , L/40 = 7000/40 = 175mm to L/30 =7000/30 = 233.33mm

Required plastic modulus Zp = 1.4M/fy

=4552721

=452.72 x 10^3

Let us try a ISMB 600 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 91800x 10^4mm ^4

Iyy 2650x10^4mm^4

R 20mm

ISMB [email protected]/m

A 6380MM^2

h 400mm

b 100mm

tf 15.3mm

tw 8.8mm

Izz 15200x 10 ^4mm^4

Iyy 508 x 10^4 mm^4

Cyy 2.42mm

Step-6

Let the distance of N>a> from the tension , n range by y

y" = 388.93mm

Izz= 1348.13 x 10^6mm^4

Zez = 3466240mm^3

For compression flange about y - y axis

I = 16766.7 x 10^4mm^4

Zey = 838333mm^3

Total area of section = 21980mm^2

Let plastic N>A>be at a distance

Yp = 580.88

Zpz = 4410317mm^3

For top flange Zpy = 10784888mm^3

Step -7

Section classification :

For ISMB b/t =4.90 <9.4

d/t = 43028 <84

For ISMC b/t = 5.96 <9.4

Hence section is plastic

Step - 9

Check for local moment capacity

Local moment capacity for bending in vertical plane

Mdz = fy zp/1.1

=1002.34Knm

Mdz = 1.2Zefy /1.1

=945.34KNm

For top flange

Mdy = 245.11 KNm

Mdy = 228.6KNm

check for combined local capacity:

(812.99/945.34)+(18.35/228.64)<1

0.947<1

Step -9

Check for buckling resistance:

Md = bzpfbd

Bb =1

Lt = 7000mm

E = 200000n/mm^2

hf = 597.5mm

Iy = 17850 x 10^4mm^4

A = 21980mm^2

ry = 90.12mm

fcr,b = 417.43 KN/m^2

y = 0.744

o = 0.95

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/(3^0.5 x 1.1)

=944755N

=944.755KN > 714.59

Now =566.853KN

Low shear

Check for web buckling:

b = 175mm

n1 =232.6mm

d = 519.4mm

t = 12mm

Fcd = 128.6N/mm^2

Bukling resistance = 629008KN > 714.59KN

Step -13

Check for deflection at working load :

Vertical deflection

Serviceabiity vertical wheel load = 251Kn

Deflection at mid span a = 2000mm

Izz = 1348.13 x 10^5mm^4

& = 25.32mm

Horizontal deflection

I = 16766.65 x 10^4mm^4

& = 4.88mm

L/750 = 9.33mm

Unsafe in vertical direction

safe in lateral direction.