Challenge

Week 5 Challenge

MMallela Sri Sai Satya Ashish12 Jun 2021

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

Ans)

Aim: To generate manual wind loading in the design report based IS code as per the following input which are mentioned in the question.

Tools Required: Tekla model, Dimensions, Microsoft Excel Worksheet

Procedure:

Firstly open the Tekla software and open the designed model for which you desired to apply wind load.
Now take the dimensions of the model such as height, lenght,width etc.
Now place the values in the Microsoft Excel Worksheet and type the required content.
Multiply the factors as in when required.


Wind load calculations as per IS: 875 P3-1987

Basic parameters

	Basic wind speed Vb=		39	m/s		Terrain category=		2
	Probable design life=		50	years		class of structure =		A

	Greater horizontal dimension, l =			24	m	Ratio for wall and roof co-efficients:
	Lesser horizontal dimension, w =			18	m
	Eaves level height =			14	m	wall =	l/w =	1.33
	Width of frame-1, grid A and B			7	m		h/w =	0.78
						Roof =	h/w =	0.78


		Topography factor, k3=			1
		Risk co-efficient, k1=			1	(for 50 years)

			Height(m)=			upto 10
	Terrain factor & str.size factor k2=					1
	Important factor for cyclonic, k4=					1.15
	Design wind speed , Vz= Vbk1k2k3k4=					44.85
	Design wind pressure (N/m)		(N/m^2), Pz= 0.6*vz^2=			1206.9
	Design wind speed(kN/m^2), Pz =					1.207

			Roof angle, α =		16	degree
		Internal pressure coefficient, α =			0.5	(upto 20% opening) ()		(Cl.6.2.3.3 IS 875 P3:1987)
		Internal pressure coefficient, α =			-0.5	(upto 20% opening) ()


	Wall loading calculations :
	Wind direction : +Y
	Wind load along Y-Direction			(Face A):		_+ Cpi	_- Cpi

		Height at different level =				upto 10	upto 10
		External pressure coefficient, Cpe =				0.7	0.7
		Internal pressure coefficient, Cpi =				0.5	-0.5
		Net pressure coefficient, Cp =				0.2	1.2
		Design wind pressure for wall =				1.207	1.207
		Loading to be applied on the wall =				0.241	1.448
		factor ( loading * 1.05) =				0.253	1.52



	Wind load along Y-Direction			(face B):		_+Cpi	_-Cpi

		Height at different level =				upto 10	upto 10
		External pressure coefficient, Cpe =				-0.25	-0.25
		Internal pressure coefficient, Cpi =				0.5	-0.5
		Net pressure coefficient, Cp =				-0.75	0.25
		Design wind pressure for wall =				1.207	1.207
		Loading to be applied on the wall =				-0.905	0.302
		factor ( loading * 1.05) =				-0.95	0.32



	Wind load along Y-Direction			(Face C & D)		_+Cpi	_-Cpi

		Height at different level =				upto 10	upto 10
		External pressure coefficient, Cpe =				-0.6	-0.6
		Internal pressure coefficient, Cpi =				0.5	-0.5
		Net pressure coefficient, Cp =				-1.1	-0.1
		Design wind pressure for wall =				1.207	1.207
		Loading to be applied on the wall =				-1.328	-0.12
		factor ( loading * 1.05) =				-1.4	-0.13



	Roof loading calculations
	Wind direction -Y					_+Cpi		_-Cpi
					h/w = 0.78		h/w = 0.78
	External pressure coefficient, Cpe =				-0.88	-0.4	-0.88	-0.4
	Internal pressure coefficient, Cpi =				0.5	0.5	-0.5	-0.5
	Net pressure coefficient, Cp =				-1.38	-0.9	-0.38	0.1
	Design wind pressure for roof, Pz =				1.207	1.207	1.207	1.207



	Wind load along X-direction ( Roof EG):					_+Cpi	_-Cpi

		Height at different level =				upto 10	upto 10
		External pressure coefficient, Cpe =				-0.76	-0.76
		Internal pressure coefficient, Cpi =				0.5	-0.5
		Net pressure coefficient, Cp =				-1.26	-0.26
		Design wind pressure for wall =				1.207	1.207
		Loading to be applied on the wall =				-1.52	-0.314
		factor ( loading * 1.05) =				-1.596	-0.33



	Wind load along X- direction ( Roof FH) :					_+Cpi	_-Cpi

		Height at different level =				upto 10	upto 10
		External pressure coefficient, Cpe =				0	0
		Internal pressure coefficient, Cpi =				0.5	-0.5
		Net pressure coefficient, Cp =				-0.5	0.5
		Design wind pressure for wall =				1.207	1.207
		Loading to be applied on the wall =				-0.6	0.6
		factor ( loading * 1.05) =				-0.63	0.63

Result: Generation of wind load calculations of a model manually is done successfully.

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

Ans)

Aim: To apply the loadings on the model.

Tools Required: Loads, Load cases, Area load, Properties, Direction, Load value.

Procedure:

Now open the model that was previously designed for last week challenge.

Now select the load cases option, a dialogue box pops out and add in the load cases of our choice and naming the load as wind load and etc., as in below picture.

Now in the below bar of the workspace assign a particular load to the frames that are to be assigned as per requirement.

Repeat the loads for the Faces of A,B,C,D and roof accordingly.

Result: Hence, applying the loadings on the model is done successfully.

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

Ans)

Aim: To create a separate model and generate wind loading using wind wizard.

Tools Required: Levels, Manage Properties, New, Members, Concrete Column, Grids, Columns, Manage Properties, New, Members, steel beam, Beams, Properties tab, Load, Wind Wizard

Procedure:

Firstly on the workspace create a model of your choice.

Now create the grids and then assigning the columns, beams as per the requirement for the building.

Now select the wind wizard option and you will be directed to a dialogue box on the workspace.

Now select next and finish because all the values of the structure are already taken by the software itself and there is no values that you have to change.

Click ok and the work is done.

Result: Hence, creation of a separate model and generating of wind loading using wind wizard is done successfully.

Week 5 Challenge

Ship work like this
in your own course.

Talk to a career counsellor

Ship work like thisin your own course.

Talk to a career counsellor

Ship work like this
in your own course.