To calculate wind load for industrial steel structures and to apply it by using TSD

1.) Generate manual wind loading in the design report based IS code as per the following input 

• Basic wind speed = 39m/s
• Terrain category 2

 

AIM : 

    To Generate manual wind loading in the design report based IS code as per the following input

INTRODUCTION : 

• Wind loading is one of primary horizontal loads acting on bridges, and its appropriate consideration is necessary to satisfy the design requirements.
• The dynamic wind effects are also important particularly for long-span bridges, which may induce significant vibrations not only in along wind direction but also in vertical and torsional directions, and they have to be avoided.
• In this chapter, wind effects on bridges are overviewed, and a typical procedure for wind resistant design of a long-span bridge is described.
• Design wind speeds and wind loads in codes, some examples of field measurements of full-scale bridges, and research results on stay cable vibrations are introduced.

 

PROCEDURE : 

Given Data :

• Basic wind speed = 39m/s
• Terrain category 2
• Total length of the building (l)  = 18m
• Total width of the building  (w) = 18m
• Total height of the building (h) = 12.5m
• Class of the structure = Class A
• Life of the structure = 50 Years
• l/w  = (18/18)    = 1
• h/w = (12.5/18) = 0.69

STEP :1

EXTERNAL PRESSURE CO-EFFICIENT (Cpe)

 

 

As per the IS 875 Part 3, Table 5

Building height ratio

Building height ratio = 1/2<h/w<3/2

                              = 0.5<0.69<1.5

Therefore, the building plan ratio = 1<l/w<3/2

= 1<1<1.5

Hence the plan we choose is given below

 

 

STEP :2

Finding the Factors (k1, k2, k3, k4)

k1:

From Table 1 for the basic wind speed for 39 m/s,

Risk Coeffiecent, K1 = 1.0

K2 :

From Table 2,

10m height = 1 for terrain category 2

15m height = 1.05

by using the interpolation method,

k2 = 1.02

  

K3:

Topography factor, k3 = 1 (from clause 6.3.3)

K4:

Importance factor K4 = 1.15 (from Clause 6.3.4)

 

STEP :3

Vz=Vb x k1 x k2 x k3 x k4

= 39x1x1.02x1x1.15

= 45.747 m/s

 

STEP :4

Pz = 0.6 Vz^2

= 0.6 x 45.747^2

= 1255.6728 n/sq.m

= 1.2556 kn/sq.m

 

STEP :5

Wind direction (up to roof level)

1.) Wind direction along y- direction (Face C & D):

•  

  	Cpe +	Cpi -
  Height of the building	12m	12m
  External pressure co-efficient Cpe	-0.6	-0.6
  Internal pressure co-efficient Cpe	0.5	-0.5
  Net pressure co-efficient Cp = Cpe - Cpi	-1.1	-0.1
  Design wind pressure, Pz	1.25	1.25
  Wind load on Wall (Cp x Cz) Kn/m^2	-1.375	-0.125

 

2.) Wind direction along X- direction (Face A):

•  

  	Cpe +	Cpi -
  Height of the building	12m	12m
  External pressure co-efficient Cpe	0.7	0.7
  Internal pressure co-efficient Cpe	0.5	-0.5
  Net pressure co-efficient Cp = Cpe - Cpi	0.2	1.2
  Design wind pressure, Pz	1.25	1.25
  Wind load on Wall (Cp x Cz) Kn/m^2	0.25	1.5

 

3.) Wind direction along X- direction (Face B):

•  

  	Cpe +	Cpi -
  Height of the building	12m	12m
  External pressure co-efficient Cpe	-0.25	-0.25
  Internal pressure co-efficient Cpe	0.5	-0.5
  Net pressure co-efficient Cp = Cpe - Cpi	-0.75	0.25
  Design wind pressure, Pz	1.25	1.25
  Wind load on Wall (Cp x Cz) Kn/m^2	-0.9375	0.3125

 

 

STEP :6

Roof Calculation:

W.k.t

h/w = 0.6

In order external coefficient from the table 6

Slope = Sin theta = opp/hyp

= 2.813/8.939

theta = 18.34

by using interpolation method

(18.34-10)/ (20-10) = (x+1.2)/(-1.4+1.2)

therefore, x =-1.36

Similarly
(18.34-10)/(20-10) = (x+0.4)/(-0.4+1.4)

X = -1.24

Hence, EF = -1.36

FG = -1.24

   

STEP :7

1.) Wind direction along Y- direction (Face EF):

•  

  	Cpe +	Cpi -
  Height of the building	12m	12m
  External pressure co-efficient Cpe	-1.36	-1.36
  Internal pressure co-efficient Cpe	0.5	-0.5
  Net pressure co-efficient Cp = Cpe - Cpi	-1.86	-0.86
  Design wind pressure, Pz	1.25	1.25
  Wind load on Wall (Cp x Cz) Kn/m^2	-2.325	-1.075

 

2.) Wind direction along y- direction (Face FG):

•  

  	Cpe +	Cpi -
  Height of the building	12m	12m
  External pressure co-efficient Cpe	-1.24	-1.24
  Internal pressure co-efficient Cpe	0.5	-0.5
  Net pressure co-efficient Cp = Cpe - Cpi	-1.74	-0.74
  Design wind pressure, Pz	1.25	1.25
  Wind load on Wall (Cp x Cz) Kn/m^2	-2.175	-0.925

 

RESULT : 

 The manual calculation of wind loading in the design report based IS code as been completed

 

2.) Based on the above calculation apply the loadings on the model 

 

AIM : 

    To Based on the above calculation apply the loadings on the model

INTRODUCTION : 

• Wind loading is one of primary horizontal loads acting on bridges, and its appropriate consideration is necessary to satisfy the design requirements.
• The dynamic wind effects are also important particularly for long-span bridges, which may induce significant vibrations not only in along wind direction but also in vertical and torsional directions, and they have to be avoided.
• In this chapter, wind effects on bridges are overviewed, and a typical procedure for wind resistant design of a long-span bridge is described.
• Design wind speeds and wind loads in codes, some examples of field measurements of full-scale bridges, and research results on stay cable vibrations are introduced.

 

PROCEDURE : 

• Open the tekla software
• Open the previous file we were working on
• First we want to create a load case by using load case option on under load panel option
• we want to create a 8 types of wind load 
• The wind loads are (Wind + Y + Cpi) , (Wind + Y - Cpi), (Wind + X + Cpi), (Wind + X - Cpi), (Wind - Y + Cpi) , (Wind - Y - Cpi), (Wind - X + Cpi), (Wind - X - Cpi),

 

 

• After completing the creating of wind load. Apply the load as per the calculated in the excel sheets above starting from the (wind + Y + Cpi) condition
• Go to the frame D apply the first wind load on the wall panel by using the area load
• Next go to the Frame A And apply the same load 
• Next apply the load -1.4 KN/m^2 on Frame 3 
• Next apply the load -1.4 KN/m^2 on Frame 7
• Next we want to applied the load on (Wind + Y - Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind + X + Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind + X - Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind - Y + Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind - Y - Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind - X + Cpi) by using the same above 4 frame 
• Next we want to applied the load on (Wind - X - Cpi), by using the same above 4 frame 

 

 

• Using the load we calculated on excel sheet 
• Next we want to applied the load on both left and right side of the roof by using (Wind - Y + Cpi)
• Next we want to applied the load on both left and right side of the roof by using (Wind - Y - Cpi)
• Next we want to applied the load on both left and right side of the roof by using (Wind + X + Cpi)
• Next we want to applied the load on both left and right side of the roof by using (Wind - X + Cpi)

 

 

RESULT : 

As per the question above calculation apply the loadings on the model as been completed 

 

3.) On a separate model generate wind loading using wind wizard 

 

AIM : 

    To Based on the above calculation apply the loadings on the model

INTRODUCTION : 

• Wind load is primarily horizontal load caused by the movement of air relative to earth.
• Wind load is required to be considered in structural design especially when the heath of the building exceeds two times the dimensions transverse to the exposed wind surface
• In this chapter, wind effects on bridges are overviewed, and a typical procedure for wind resistant design of a long-span bridge is described.

 

PROCEDURE : 

• Deleted all the previous wind load as been created in above
• Next go to the wind wizard option on under the load panel ,after wind wizard box as been appear 
• select the worst case data and unselect the (data for each direction and clade frame) --> Next 
• Give the wanted value on basic data box and click next
• Apply the fetch distance 10 and click next 
• And click next on topography box and results box
• And add the wind load on (load cases box) 
• Next go to the seismic load
• select the code spectra on (site specific spectra box) select next
• And given the basic information and select next
• Pick the next botton on structural irregularites
• select the rc steel concrete on fundamental period
• And select the steel building with OMRF on (Seismic force resisting system box) and pick next
• Again give the next botton and click finish

 

 

 

 

RESULT : 

 As per the question generate wind loading using wind wizard as been doned