Project 2

Q1). 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.

Answer:

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 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.

INTRODUCTION:

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

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

Web buckling:

• 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 certain 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 = 1 in 12

Width, b = 30 m

X/(b/2) = 1/12

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

X = 1.25 m

GEOMETRY / MODEL CREATION:

PROPERTIES:

3D VEIW OBJECT:

Material: STEEL

BETA Angle 90 degree for wind / gable columns

Materials:

SPECIFICATION:

• Member Tension
• End MY MZ
• Start MY MZ

SUPPORT:

• Pinned

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

SEISMIC DEFINATION:

Zone Parameters:

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

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 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.

Step - 1

Type 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 ratios:

KLx/rx = 9000/166.1 =54.1

KLy/ry = 9000/51.6 =174.41

Non-dimensional Effective Slenderness ratio

λ">λ = 1.96

Step - 3

To calculate φ:

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 = 6 m

Dead load = 18KN/m

Imposed load = 40KN/m

support bearing = 100mm

yield strength = 250N/mm2

Step - 1

Design load calculation:

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

Factored Bending moment = 391.5 KN/m

Step - 2

Section modulus Required:

Z reqd = 1722600mm^3

= 1722.60cm^3

Step - 3

Section classification:

Flange criterion B/2T = 5

Web criterion (D/2T)t =32.61

ɛ=1

B/2T<9.4

(D/2T)t<83.9

The section is plastic.

ISMB-250

A = 475.5mm2

D = 250mm

B = 125mm

t = 6.9 mm

T = 12.5 mm

Izz = 5131.6cm^4

Iyy = 344.5 cm^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 = 1044.491 KN/m

<391.50

Hence section ISMB 250 IS Adequate for flexure.

Step - 4

Shear resistance of the cross section:

Shear capacity Vc = 0.6fyAv/γm

Av = 1725mm^2

Vc = 235.23kn

V= 261kn

V/Vc = 1.11

>0.6

Step - 5

Check for web buckling:

Slenderness ratio of web = 2.5d/t

d = 225mm

Design 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 ISMB250 R = 13mm

tf + R =25.5mm

Dispersion length n^2 = 63.75mm

Pcrip = 245625N

=245.63kn

<261

Hence section adequate for web crippling.

Step - 7

Check for serviceability deflectio:

Design load = 58kn/m

Deflection = 95.36

Allowable deflection = 30mm

Hence serviceability is satisfied

DESIGN OF BASE PLATE:

Step -1

Strength of concrete, Fcu = 40 N/mm2

Yield strength of steel, fy = 250 N/mm2

Material factor, γm">γmγmFA−6XMzBL2= 1.1 KN

Factoresl oad = 1500 KN

Step - 2

Steel column section:

Thickness of flange, T = 12.7 mm

Step - 3

Area required:

Bearing strength of concrete = 0.4fcu = 0.4 x 40 = 16 N/mm2

= (1500x1000) /( 16)

= 93750 mm2

Let size of plate , Bplate = 350 mm

Dplate = 350 mm

Area of plate= 122500 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

LXσ1σ1+σ2

ts= 7.7 mm < 12.7 mm

Size of Base plate 350mm x 350mm x 12mm

d) DESIGN OF PEDESTAL:

Grade of concrete = 40 N/mm2

Load = 123 KN

Moment = 116 KNm

Horizontal shear = 13 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:

Step -1

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

Step - 2

Base plate Details

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

Thickness of plate = 16 mm

Yield stress of plate = 330 N/mm2

Step - 3

Anchor bolt design

Area of the plate = 157500 mm2

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

= 10.6 N/mm2 < 16 Hence OK

Minimum pressure =F/A-6 X Mz/BL^2

= -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

Step - 4

Shear Check

Factored shear force = Vsb = 1.63 KN

Vd,sb = 81290.9 N

= 81 KN

Factored Vd,sb = 65.03 KN

Step - 5

Tensile Check

Factored tensile force in bolt, Tb = 91.1 KN

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

Td,b = 98.65 KN

Step - 6

Combined Unity Check

Vsb/Vdb = 0.025

Tb/Tdb = 0.92

Unity check = 0.85 < 1, Hence OK

Step - 7

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:

Step -1

Span of the purlin = 7 m

spacing of purlin = 1.5 m

No:of sag rods = 1

slope of the roof = 4.76 = 5 degree

Step - 2:

Dead load:

Weight of sheeting = 5kg/m2

self weight of purlin = 4.22 kg/m2

Additional load = 10% = 0.42 kg/m^2

Total Dead load = 0.096 kn/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 = 1949.9N/m^2 = 1.94 KN/m^2

Length of building, L = 50 m

Breadth of building, W = 30 m

Height of building, H = 10.5 m

Height of eaves, = 9 m

h/w = 0.34

L/w = 1.67

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

Step - 3

Design Load Calculation:

spacing of purlin 1.5 m

slope of roof = 5degree

Total dead load = 0.096KN/M^2

DL Normal component = 0.144 KN/m

DL Tangential component = 0.013 KN/m

Total Live load = 0.75Kn/m^2

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 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 l = 20mm.

Internal bending radius = 3mm

Area = 8.07 cm2

Zxx = 47.72cm^3

Zyy = 10.22cm^3

lxx = 477.18cm^4

lyy = 61.34cm^4

Weight of purlin = 6.335 kg/m

Check for basic properties:

t = 2.30mm

w = 49.40mm

Fy = 2400kg/cm^2

W/t = 21.48

<60

OK

Check for stress:

Span major axis = 7m

Span minor axis = 2m

BM coefficient for mxx = 10

BM 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

lxx = 477.18cm^4

δ = 34.29mm

Allowable deflection δ = Span/180

= 33.33mm

>34.29

Not Ok.

Foundations using STAAD Foundation.

RESULTS:

PROPERTIES:

• First you must Analyze the file.
• Select Analyze from the Staad.Pro top menu bar.
• Select the Analysis option. After this, click on Run Analysis at the bottom of the small window dialog box.
• After the analysis of the file is completed, click on the Done button.
• Next, we go to the Post Processing mode to view the forces and results graphically.
• To enter into the Post Processing mode, select Mode from the top menu bar and select Post Processing.
• To view the Bending Moment Diagrams, select the Beam page from the left side. From the top menu bar, choose Results.
• Under the Beam Results tab, you will see the options Bending, Shear, Axial, Displacement and Stresses.
• Click on the Annotate button. Then click on the close button.
• Here are a group of of other options to view BMD, SFDs
• To view the bending moment diagram for individual members you may go to the Beam > Graphs page on the left and click on any member and the BMD, SFD and axial force diagrams for the member would be displayed on the right.
• Alternately you may also double click on any member to bring up the query box and go to the Shear Bending tab to see the bending moment or shear force diagrams for that member.

Q2). Design a simply supported gantry girder to carry electric overhead travelling crane

Answer:

Given that:

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

Properties of selected section:

ISMB [email protected] KN/m

A 15600MM^2

h 600mm

b 210mm

tf 20.3mm

tw 12mm

Izz 91800x10^4mm^4

Iyy 2650x10^4mm^4

R 20mm

ISMB [email protected] KN/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

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

y" = 388.93mm

lzz = 1348.13x10^6mm^4

Zez = 3466240mm^3

For compression flange about y-y axis

l = 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 =fyZp/1.1

= 1002.34knm

Mdz = 1.2 Zefy/1.1

= 945.34knm

For top flange

Mdy = 245.11knm

Mdy = 228.64knm

If the above which ever 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

ly=17850 x 10^4mm^4

A=21980mm^2

ry = 90.12mm

fcr,b = 417.43kn/m^2

λ=0.744

φ=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

Step - 12

Check for web buckling:

b = 175mm

n1 = 232.6mm

d = 519.4mm

t = 12mm

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

lzz = 1348.13 X 10^5mm^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.