Week 10 Challenge

1.

AIMS & INTRODUCTION:

Total number of purlins and girts are required to be calculated-

Inputs for given building :

Rafter span = 50 m

Bay spacing = 8 m

Roof slope = 1 in 10

Height of column = 12 m

Brick wall height = 2 m

Purlin spacing = 1.2 m

Girt spacing = 1.5 m

PROCEDURE IN DETAILS-

In the print screen of the model various given data have been indicated on the basis of which manual solution of the challenge has been initiated and should not be treated for calculating no of Purlins & Girt.

Calculation of numbers of purlins-

Half of the span of Rafter = 50/2= 25 m

Height of the Truss= 2.5 m ( As per slope 1 in 10)

Inclined length of Rafter `

Length of Rafter :

[Math Processing Error]252+2.52 = 25.124`m

Number of purlins = ( Inclined length of rafter/spacing of purlins) +1

= (25.124/1.2)+1 = 20.936 +1 = 22

Total purlins on entire span of Truss = 22x2 = 44

Calculation of numbers od Girts-

Bay length = 8 m ( This data is not required for the solution of given problem)

Column height = 12 m

Brick height = 2 m where no Girts are not required to be installed

Remaining height of columns for grit installation = 12 - 2 = 10 m

Girt spacing = 1.5 m

Number of Girts on one side =( 10/1.5) +1= 8

Total nos of Girts on both side = 8x2 =16

RESULT -It is concluded that As per above calculation Total nos of Purlins on both side of slope are 44 and total nos of Grit on both side are 16.

3.

asic wind speed [Math Processing Error]��= 50 m/s ( For Chennai as per IS 475 part-3 , Appendix A , clause 5.2 )

STEP-1:

Design wind speed [Math Processing Error]��=���1�2�3�4

where K1 = Risk/Probability factor ( IS 875 Part-III, Page-11, Table-1)

Considering general building and structure and [Math Processing Error]��=50 m/s

K1= 1 from Table 1 clause 5.3.1 for design life of structue 50 years & and for given Basic wind speed of 50 m/s

K2 = Terrain height & height factor ( IS 875 Part-III, Page-12, Table-2) =0.91

K3 = Topography factor = 1.0 ( Assuming upwind slope < [Math Processing Error]30) (clause 5.3.3)

K4 = Imporance factor for cyclonic region =1.15 Assuming Industrial structure ( clause 5.3.4)

Therefore Vz = 1x0.91x1x1.15x50 =52.33 m/s

STEP-2 :

Wind pressur Pz = 0.6 [Math Processing Error](��)2

where [Math Processing Error]�� = wind pressure in N/[Math Processing Error]�2 at height z

Therefore [Math Processing Error]��= 0.6x[Math Processing Error](52.33)2= 1643.06 N/ [Math Processing Error]�2= 1.64 KN/[Math Processing Error]�2

STEP-3:

Design wind pressure [Math Processing Error]��=��������

where [Math Processing Error]�� = wind directionality factor 1.0 for cyclone affected region

[Math Processing Error]��= Area averaging factor =0.9 for tributary area < [Math Processing Error]25�2

(Tributary area = 6x2.5 = 15 m^2)

K_c= combination factor =0.8

Therefore P_d = 1x0.9x0.8x1.64 = 1.181 KN/[Math Processing Error]�2

STEP-4 :

Wind load acting in a direction normal to the individual structural element or cladding unit F = [Math Processing Error]{���-���}���

where [Math Processing Error]��� = internal pr coefficient = 0.5 for opening of 15% of wall area i.e. medium permeability ( clause 6.2.2 of IS :475, part 3)

External pressure coefficient:

l= 10 m ; b = 5 m ; h = 6 m

[Math Processing Error]ℎ�=65=1.2

[Math Processing Error]��=105=2

Thus as per table 5 of clause 6.2.3.1 of IS:475, Part 3

[Math Processing Error]12<ℎ�≤32 and [Math Processing Error]