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

Sail Type               Medium

Safe Bearing Capacity  200KN/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 IS875 standards 

4.Design using MS-Excel

• Steel Column
• Rafter
• Base Plate
• Pedestal
• 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 Idian Standards.

Situation:

Roof slope Calculation:

Roof Slope =1 in 12

Width,b    =30m

x/(b/2)=1/12

x          =(30/2)/12=15/12

x          =1.25m

Geometry/Model Creation:

 

Properties:

3D View Object:

Material:STEEL

Beta angle 90 degree for wind/gable columns

Specification:

• Member Tension
• End MY MZ
• Start MY MZ

 

Support:

• Pinned

LOAD CASE DEFINATION:

1.EQ+x

• IS 1893 Load-x1

2.EQ+z

• IS 1893 Load-z1

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-1.2KN/m
• Gable frame column GY-0.6KN/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 loadFY-4.15 KN/m
• Gable frame point load FY-2.06 KN/m

SEISMIC DEFINATION:

Zone parameters:

Member weight:

Assign to Edit list from folllowed 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

LOAD CASE DEFINATION:

 5.Wind load(0+CPi)

6.Wind load(0-CPi)

7.Wind load(90+CPi)

8.Wind load(90-CPi)

 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-(12.7×2)

                      =374.6 mm

Thickness of web=t=10.6 mm

Flange width=2b=bf=250 mm.

Hence,half Flange Width =d=125 mm.

Self-weight=w=0.822 KN/m

Area of cross-section =A=10466 mm2.

Radious of gyration about x=rx=166.1 mm.

Radious 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">λ3.3X7^2/8:

=

λ">λ =

=√2.5X(174)2/π2X2X10^5

= 3.83

Value of ϕ">ϕts=√2.5w(a2−0.3b2)m0fy>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 RAFTER:

 Solution:

 Span of Rafter =6m

Dead load=18KN/m

Imposed load=40KN/m

Support bearing=100mm

Yield strength=250N/mm2

 

 Design load calculation:

 Factores load 1.5(DL+LL)=87KN/m

Factores Banding moment=

Wl28">10^6(Wl²/8)=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:

Md = βbZpfy/γm0

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

Md = 85.306 > 65.25 KN/m

 DESIGN BASE PLATE:

 Strength of concrete,Fcu=40N/mm2

Yield strength of steel ,fy=250N/mm2

Material factor, γm">γm F/A−(6XMz/BL²)

= 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 m

                      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

t s = √ 2.5 w ( a 2 − 0.3 b 2 ) γ m 0 f y > t f ( LXσ1/σ1+σ2)

ts= 7.3 mm < 12.7 mm

ts=7.3mm<12.7mm

Size of base plate 450×350×16mm 

 DESIGN OF PEDESTRIAL:

 Grade of concrete =40 N/mm2

Load=200 KN

moment=120KN

Horizontal shear=20KN

Yield strength=250 N/mm2

Length of base plate=450 mm

Width of base plate=350mm

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

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

Bearing strength 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'=400N/mm2

Fub(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)+(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 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=75kg/m2

Terrain  category=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,=9m

h/w=0.34

L/w=1.667

External pressure Co-efficient

 Maximum downward=Cpe=-0.4

Maximum upward=Cpe=-0.7

Internal pressure Co-efficient

 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=5 degree

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

Maximum Normal component=DL+LL

              =1.265 KN/m

Purlin Section

 Selected section 7200×6×2.3

Atrea=8.07 cm2

Weight of Purlin=6.335 Kg/m

Foundations using STAAD Foundation.

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

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.

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.

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 girder as shown below.

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

 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 = 1.4X(437.58X10^6)/250

 = 2450 x 10³ mm3

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