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Challenge

Project 2

VVishnu Kumar10 Oct 2023

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
  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.


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

 

 

AIM:

To design a Warehouse Building located in Chennai using STAAD Pro Connect Edition. 

INTRODUCTION:

STAAD Pro is a popular structural analysis application known for analysis , diverse application of use, interoperability, and time-saving capabilities. It helps structural engineer perform 3d structural analysis and design for both steel and concrete structures.

PROCEDURE:

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.

 

Given data:

Width = 30m

Length  = 50m

Eave Height = 9m

Bay spacing = 6m

Soil type = Medium

Safe bearing capacity = 200 kN/m2

Roof slope = 1 in 12

Roof slope calculation:

Roof slope = 1 in 12

Width, b = 30m

X / (b/2) = 1/12

X / 15 = 1/12

X = 15 / 12

X = 1.25m

Total height = 9 + 1.25 = 10.25m

Load Calculation:

Dead load calculation:

Dead load - Rafter:

Self weight, FOS = 1.15

UDL on Rafter (Main frame):

Weight of roof sheeting = 5 kg/m2

Weight of sag rods and braces = 5 kg/m2

Collateral load = 10 kg/m2

Total load = 20 kg/m2 = 0.2 kN/m2

Spacing = 6m

UDL = 0.2 x 6 = 1.2 kN/m

UDL on Rafter 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 pulins x length of purlin x weight of purlin)/(length of rafter)

(8 x 6 x 8.77)/15 = 28.064 kg/m = 0.281 kN/m

UDL on Main frame = 1.2 + 0.281 = 1.481 kN/m

UDL on gable frame = 1.481/2 = 0.741 kN/m

Dead load - Column:

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 grit (270 x 75 x 20 x 2.55) = 8.77 kg/m

(8 x 6 x 8.77)/6 = 70.16 kg/m =0.701 kN/m

Total load = 0.6 + 0.701 = 1.301 kN/m

Load on column (Gable frame):

Total load = 1.301/2 = 0.651 kN/m

Load on Wind column:

Load from sheeet 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

=> (6 x 6.32 x 6)/6 = 37.92 kg/m = 0.38 kN/m

Total load = 0.6 + 0.38 = 0.98 kN/m

Dead load - Eave strut & Gutter:

Point load:

Weight of Eave strut = CS 270 x 75 x 20 x 3.15 = 10/7 kg/m

Weight of Eave gutter = (0.25 + 0.25 + 0.25) x 0.001 x 7850 = 5.89 kg/m

Load due to eave strut & gutter = 10.7 + 5.89 = 16.59 kg/m

Main frame = 16.59 x 6 = 99.54 kg/m = 0.995 kN

Gable frame = 0.497 kN

 

Live load calculation: 

From (IS 875 Part II),

Live load on roof = 0.75 kN/m2

UDL on Rafter:

Main frame rafter = 0.75 x 6 = 4.5 kN/m

Gable frame rafter = 0.5 x 4.5 = 2.25 kN/m

Eave gutter (250 x 250 x 1mm)

Load due to weight of water in gutter:

Main frame = (0.25 x 0.25 x 10 x 6) = 3.75 kN

Gable frame = 0.5 x 3.75 = 1.875 kN

 

Seismic load :(IS 1893:2016)

Location - Chennai

Zone, Z = 0.16

Reduction factor, R = 4

Importance factor, I = 1.2

Damping ratio = 5%

Soil category = Medium 

 

Wind load calculation:(IS 875 Part III)

Design Wind Speed    Vz = Vb k1 k2 k3 k4 m/s
Basic Wind Speed     Vb = 50 m/s
Probability factor     k1 = 1  
Terrain factor , Category   for k2 = 1  
Topography factor   k3 = 1  
Importance factor     k4 = 1.15  
Design Wind Pressure   Pz = 0.6 Vz2  
               
    Design wind speed Vz = 57.5 m/s
    Design wind pressure Pz = 1.984 kN/m2
               
Wind directionality factor   kd = 1  
Area averaging factor   ka      
      Tributary area = 54.00 sq.m
        ka = 0.86  
Combination factor   kc = 0.90  
Design wind pressure   Pd = kd x ka x kc x Pz
        Pd = 1.535 kN/m2

 

Wind pressure coefficient: 

  Wind load on members, F = (Cpe - Cpi) * A * pd
Cl 7.3.2.2 Internal pressure coefficient, Cpi = ±0.5  
Table 5 & 6 External pressure coefficient, Cpe      
Length of the building, L = 50 m
  Width of the building, w = 30 m
  Height of column, h = 9 m
  Ratio L/w = 1.67  
    h/w = 0.30  
  Roof angle  1 in 12 = 4.76 degrees
      = 5 degrees
  Design wind pressure, pd = 1.535 kN/m2
  For internal columns and internal rafters   = 6.00 m
  For gable column and gable rafters   = 5.00 m
  For internal wind columns   = 6.00 m
  For gable wind columns   = 5.00 m
Cpe values - Roof        
Table 6 0O 90O
  EF GH EG FH
  -0.9 -0.4 -0.8 -0.4
Cpe values - Walls          
Table 5 Degree A B C D
  0O 0.7 -0.25 -0.6 -0.6
  90O -0.5 -0.5 0.7 -0.1

Wind load Rafter:

Angle Cpe Cpi (+) Cpi (-) (Cpe-Cpi) for "+" (Cpe-Cpi) for "-" pd for "+" pd for "-"
0O -0.9 0.2 -0.2 -1.1 -0.7 -10.13 -6.45
-0.4 0.2 -0.2 -0.6 -0.2 -5.53 -1.84
90O -0.8 0.2 -0.2 -1 -0.6 -9.21 -5.53
-0.4 0.2 -0.2 -0.6 -0.2 -5.53 -1.84

Wind load Column:

Angle Cpe Cpi (+) Cpi (-) (Cpe-Cpi) for "+" (Cpe-Cpi) for "-" pd for "+" pd for "-"
0O 0.7 0.2 -0.2 0.5 0.9 4.61 8.29
-0.25 0.2 -0.2 -0.45 -0.05 -4.14 -0.46
-0.6 0.2 -0.2 -0.8 -0.4 -7.37 -3.68
-0.6 0.2 -0.2 -0.8 -0.4 -7.37 -3.68
90O -0.5 0.2 -0.2 -0.7 -0.3 -6.45 -2.76
-0.5 0.2 -0.2 -0.7 -0.3 -6.45 -2.76
0.7 0.2 -0.2 0.5 0.9 4.61 8.29
-0.1 0.2 -0.2 -0.3 0.1 -2.76 0.92

Structural model using STAAD Pro:

1. Create skeleton model:

2. Assign the section to the structure:

3. Enter specification:

  • Assign tension member
  • Release moment

4. Assign the Support:

5. Load case details:

6. Dead load and Live load:

7. Seismic definition:

  • Generate seismic parameter

  • Member weight and Joint weight

  • EL +X and EL +Z

8. Wind Pressure Coefficient:

  • WL(0+CPI)

  • WL(0-CPI)

  • WL(90+CPI)

  • WL(90-CPI)

9. Define Parameters:

10. Define Commands:

11. Run Analysis:

A.Design of Steel Column:

Height of column h = 9 m
Yield strength fy = 250 N/mm2
Elastic modulus E = 200000 N/mm2
Material factor γm = 1.1  
Section classification:  

Properties of selected section

ISHB 400
A 10466 mm2
h 400 mm
B 250 mm
t 10.6 mm
T 12.7 mm
rx 166.1 mm
ry 51.6 mm
Zp   cm3

d = 374.6

Flange criterion b/T = 9.84
Web criterion d/t = 35.34
  Ɛ = 1.00
  b/T < 10.5
  d/t < 42

The section is classified as COMPACT

Effective Length:              
T7.5 As both ends pin-jointed        
    KLx = 9.00 m
    KLy = 9.00 m
    KLz = 9.00 m
           
Slenderness ratio:              
    KLx/rx = 54.18  
    KLy/ry = 174.42  
Non-dimensional effective slenderness ratio:    
 
 
Cl 7.1.2.1 λ =    
         
    = 1.964  
To calculate φ:  
  φ = 0.5[1+α(λ-0.2)+λ2]
  h/bf = 1.60
    > 1.2
  T = 12.7 mm
T 7.1 so, buckling class = b
  α = 0.34
  φ = 2.73
Calculation of X:      

 
 
      χ =      
               
               
        χ = 0.216 N/mm2
Calculation of fcd:    
        fcd = χfy/γm
          = 49.17
Factored Axial Load:    
        Pd = Afcd
        = 514.614 kN

B.Design of Rafter:

Span of I beam = 3 m
Dead Load = 18 kN/m
Imposed Load = 40 kN/m
Length of bearing = 100 mm
Yield strength = 250 N/mm2
Design load calculation:        
  Factored Load = 87.00 kN/m
  Factored bending moment = 97.88 kNm
Section modulus - Required:        
  Z reqd = 430650.0 mm3
    = 430.65 cm3
Section classification:

Properties of selected section

ISMB 250
A 475.5 mm2
D 250 mm
B 125 mm
t 6.9 mm
T 12.5 mm
Izz 5131.6 cm4
Iyy 344.5 cm4
Zp 459.76 cm3
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 classified as PLASTIC

Moment of resistance of the cross section:

Moment Md = (Zpxfy)/γm  
      Md = 104.491 kNm
    > 97.88  

Hence section ISMB 250 is adequate for flexure

Shear resistance of the cross section:

Shear capacity Vc = 0.6fyAv/γm
  Av = 1725 mm2
  Vc = 235.23 kN
Factored shear V = 130.50 kN
  V/Vc = 0.55
    < 0.6

Check for Web Buckling:

Slenderness ratio of web   = 2.5d/t  
      d = 225  
        = 70.33  
Design compressive stress fc = 203 N/mm2
      Pw = 315158  
        = 315.16 kN
        > 130.50  

Hence web is safe against shear buckling

Check for web crippling at support:

Root radius of ISMB 250 R = 13 mm
      tf  + R = 25.5 mm
Dispersion length n2 = 63.75 mm
      Pcrip = 245625 N
        = 245.63 kN
        > 130.50  

Hence section adequate for web crippling 

Check for serviceability - Deflection:

Design load   = 58 kN/m
Deflection   δ = 5.96 mm
Allowable deflection   = 15 mm

Hence serviceability is satisfied

C.Design of Base plate:

Strength of concrete fcu = 40 N/mm2
Yield strength of steel fy = 250 N/mm2
Material factor γm = 1.1 kN
Factored load   = 1500 kN

Steel column section:

Thickness of flange  T = 12.7 mm

Area required:

Bearing strength of concrete   = 0.4 fcu  
        = 16 N/mm2
Area required     = 93750 mm2
             
Let size of plate   Bplate = 450 mm
      Dplate = 300 mm
Area of plate     = 135000 mm2
Let projection on each side a = b  = 25 mm
      w = 11.11 N/mm2
Cl 7.4.3.1 Thickness of slab base ts = 7.3 mm
          < 12.70 mm

Base plate of size 450mm x 300mm x 16mm

D.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

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 = (F/A) + 6(Mz/BL^2)

                                         = 10.6 N/mm2 < 16>

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

                                         = -9.04 N/mm2 < 16>

 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

E.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

Design Foundations using STAAD Foundation.

  • Create job title then click design option in created job title tab

--------------------------------------------------------------------------------

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

ANSWER:

Given data:

Span of gantry girder = 7 m
Span of crane girder = 9 m
Crane capacity = 250 kN
Self weight (trolley,hook,motor) = 40 kN
Self weight of crane girder = 100 kN
Minimum hook approach = 1 m
Distance between wheels = 3 m
Self weight of rails = 0.2 kN/m

 

Step 1: Maximum moment due to vertical load

Weight of trolley + Weight of load lifted = 290 kN
Self weight of crane girder = 100 kN

For maximum reaction on girder, the moving load should be as close to the gantry as possible

 

Reaction at A RA = 501 kN
Load on gantry girder from each wheel = 251 kN
Factored wheel load   = 375.833 kN
      RD = 281.875 kN
Maximum moment ME = 634.22 kNm
let self weight of girder   = 2 kN/m
Dead load (self weight + rails)   = 2.2 kN/m
Factored DL   = 3.30 kN/m
M due to DL   = 20.21 kNm
M due to impact load   = 158.55 kNm
Factored moment due to all vertical loads        
M = 812.99 kNm

Step 2: Maximum moment due to lateral load

Hzl. Force transferred to rails   = 29 kN
Horizontal force on each wheel = 7.25 kN
Factored horizontal force   = 10.875 kN
Maximum moment   = 18.35 kNm
Vertical shear due to wheel loads = 563.75 kN
impact   = 140.94 kN
self weight   = 9.90 kN
Total shear   = 714.59 kN
Lateral shear due to surge   = 20.68 kN

Step 3: Preliminary section determination

Minimum economic depth D = L/12  
    = 583.333 mm
Width of compression flange b = L/40 to L/30
    = 175 233
Required plastic modulus Zp = 1.4 x M / fy
    = 4552721  
    = 4552.72 x103 mm3

Let us try ISMB 600 with ISMC 400 on compression flange

Step 4: Properties of selected section

ISMB 600 @ 1.21 kN/m
A 15600 mm2
h 600 mm
b 210 mm
tf 20.3 mm
tw 12 mm
Izz 91800 x104mm4
Iyy 2650 x104mm4
R 20 mm
ISMC 400 @ 0.49 kN/m
A 6380 mm2
h 400 mm
b 100 mm
tf 15.3 mm
tw 8.8 mm
Izz 15200 x104mm4
Iyy 508 x104mm4
Cyy 2.42 mm

Step 5: Determination of section modulus

Let the distance of N.A from the tension range by y'

y' = 388.93 mm
Izz = 1348.13 x106mm4
Zez = 3466240 mm3

For compression flange about y-y axis

I = 16766.7 x104mm4
Zey = 838333 mm3
 Total area of section = 21980 mm2

Let plastic NA be at a distance Yp from tension flange. 

Yp = 580.88 mm
Zpz = 4410317 mm3
For top flange    Zpy = 1078488 mm3

Step 6: Section classification  

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 7: Check for local moment capacity

Local moment capacity for bending in vertical plane

    Mdz = fyZp / 1.1  
      = 1002.34 kNm
      = 1.2 Zefy / 1.1
      = 945.34 kNm
    Mdz = 945.34 kNm
For top flange        
    Mdy = 245.11 kNm
      = 228.636 kNm
    Mdy = 228.64 kNm

Step 8: Check for combined local capacity

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

                               0.947<1>

Step 9: Check for buckling resistance

      Md = βb Zp fbd
For plastic section βb = 1
LLT = 7000 mm
E = 200000 N/mm2
hf = 597.5 mm
Iy = 17850 x104mm4
A  = 21980 mm2
ry = 90.12 mm

fcr,b = 417.43 N/mm2
λ = 0.774
ϕ = 0.95
X = 0.97 N/mm2
fbd = 166.591  
       
Mdz = 734.72 kNm
Mdy = 202.84 kNm
Hence 1.20 > 1

Hence section is unsafe against torsional buckling

Step 10: Check for shear

Vz = 714.59 kN  
Shear capacity = Av*fyw / √3 *1.1
  = 944755 N  
  = 944.755 kN  
  > 714.59  
Now = 566.853 kN  
    low shear  

Step 11: Check for web buckling

b = 175 mm
n1 = 232.6 mm
d = 519.4 mm
t = 12 mm
λ = 104.746  
Fcd = 128.6 N/mm2
Buckling resistance = 629008 kN
  > 714.59 kN

Step 12: Check for deflection at working load

Vertical deflection            
  Serviceability vertical wheel load = 251 kN

Deflection at mid span

a = 2000 mm
Izz = 1348.13 x106mm4
δ = 25.32 mm

Horizontal deflection

I = 16766.65 x104mm4
δ = 4.88 mm
L/750 = 9.33 mm

Unsafe in vertical direction.

Safe in vertical direction

RESULT:

  • Hence the structural model is created using STAAD Pro connect edition
  • Prepare DL, LL, WL and EQL load calculations as per IS 875 standards
  • Design a Steel column, Rafter, Base plate, Pedestal, Z-purlin as per condition
  • Design Foundations are prepared using STAAD Foundation
  • Design a gantry grider calculation using MS-Excel
  • All stipulations, assumptions and design parameter are followed in Indian Standards.
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