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

Answer the question no. 1

Aim: Aim is to Generate manual wind loading in the design report based IS code as per the following input

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

Procedure: We have to follow the following steps

Step-1

• We are calculating the load combination.

Step-2

• Wind Direction +y :
• Wind direction +y
• Wind Load Along y Direction FAce A Cpi+ Cpi-

Height at diffrent level	upto 10	upto10
External pressure coeff Cpe=	0.7	0.7
Intenal pressure coeff Cpe=	0.5	-0.5
Net Pressure coeff Cpe-Cpi=	0.2	1.2
Design wind pressure for wwall Pz	1.207	1.207
Loading to be applied on the wall=	0.241	1.448
K2 factor from table 2= 1.05	0.2537	1.521

Wind Load Along y Direction FAce B Cpi+ Cpi-

Height at diffrent level	upto 10	upto10
External pressure coeff Cpe=	-0.25	-0.25
Intenal pressure coeff Cpe=	0.5	-0.5
Net Pressure coeff Cpe-Cpi=	-0.75	0.25
Design wind pressure for wwall Pz	1.207	1.207
Loading to be applied on the wall=	-0.905	0.302
K2 factor from table 2= 1.05	-1.394	-0.127

Result: i Completly write the answer of the question and also attached the working screenshhots

Answer the question no. 2

Aim: Aim is to Based on the above calculation apply the loadings on the model

Procedure: We have to follow the following steps

Step-1

• Open the tekla software. and open week 4 file.
• Selectt load tab, in that select loadcases and add new load cases.

Step-2

• Apply the load Wind load = 0.253kn/m^2
• Apply load frame 5 = wind = -0.951kn/m^2

Step-3

• Apply load on fram 1 wind = 1.52 kn/m^2
• Apply load on faces 5 = wind = 0.317 kn/m^2
• Apply load on face face C winf =-1.394kn/m^2
• Apply load on face D = wind = cpi o.127 kn^2

Step-4

• Now we will apply load of -y
• All loading will be similar just the direction og wind is changed
• Frame 5 = wind = -0.253kn/m^2
• Apply load on Frame 1 -wind = 0.951kn/m^2
• Apply load on frame 5 = wind = 1.521kn/m^2
• Apply load on Frame 1 wind =-0.317kn/m^2
• Apply Load on frame D = wind = 1.394 kn/m^2 on glovbal X axis

Step-5

• Now we will applying wind load on roofs and take angle as 25 degree
• From IS 875 part 3 table 3 clause 6.2.3.2
• and apply roof load
• Frame 2 = wind = -1.204kn/m^2

Step-6

• Apply load on frame 2 wind = 0.063kn/m^2
• Roof XY
• Apply load on Frame 1 = wind = o kn/m^2
• Roof PQ
• Apply load on frame on frame 1 = wind = 1.204kn/m^2Apply load on frame 1 wind cpi= 0.063 kn/m^2 on roof PR

Result: i Completly write the answer of the question and also attached the working screenshhots

Answer the question no. 3

Aim: Aim is to On a separate model generate wind loading using wind wizard

Procedure: We have to follow the following steps

Step-1

• We have to open the file
• delete all load cases and load load.
• select wind wizard set default.

Step-2

Check all details s of the building like as wind load speed and height of the building

Step-3

• Select terrain cateogory
• Put distance 10
• Click next then select topography and then next
• After that it shows resultant wind pressure.

Step-4

• All the loads are applied and click on validation to check for errors.

Result: i Completly write the answer of the question and also attached the working screenshhots