cd and fg member

8/13/2019 CD and Fg Member

1/24

factored compressive load =335.175 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa

depending on .Assuming fcd=100 N/mm

2

Ac=

mm2

Since we will be using double angle on the same side of the gusset plate.

so area of each angle = 3351.75/2 = 1675.875 As this is the double angle rafter so the section must be joined on a single side of a gusset plate of

thickness 8mm and the dimension of the angle is 11011012 @ 19.7 kg/m. The angles are joinedby fillet welding.

As per IS 800:2007 SECTION 7 clause no. 7.5.1.2and 7.1.2.1 and table 5, 7, 10, 12

c=2 2

1 2 3vvk k k

2810

21.5 1.461 88.812

250

l

rvv

vvE

DESIGN OF MEMBER cd fG COMPRESSION DESIGN

TENSION DESIGN

Compressive load on CD and FG member

= 335.175 KN

Tensile load on the CD and FG member

=37.48 KN

DESIGN FOR COMPRESSIVE FORCE


2/24

1 2 110 110

2 2 12 0.1031 88.812

250

b b

t

E

For hinge support we have ,k1=0.70, k2=0.60, k3=5.2 20.70 0.60 1.46 5 0.103 1.425

e

Again for fixed support we have, k1=0.20, k2=0.35, k3=20.

2 20.20 0.35 1.46 20 0.103 1.076e

It is given in the IS-800-2007 that

= 0.5[1+0.49(1.251-0.2) +1.251

2] = 1.539

2

2 2 0.5 N

2501.1

93.341.539 [1.539 1.251 ]

/mmcdf

Pc = cd cf A = 93.33 2 21.1100 = 393.852 KN > 335.175KN

Hence our design is safe.Here we are assuming that all the joints are partially hinged.

So for hinged joint K = 1

Again for fixed joint K = 0.625

So for our case K = 0.825

Hence K =0.825 is alright.

Now,0.825 2810

159.88 18014.5

KL

r

Hence ok.

Again we know that,

IS code specification

For partial restraint, the e can be interpolated between the eresults for fixed & hingedcases.

the interpolated result

e=1.251

According to IS-800-2007

In case for bolted, riveted & welded trusses the effective length,KL,of compression members shallbe taken as 0.7 to 1 times of distance between centres of commections depending upon degree ofend restraint provided.


3/24

COMPRESSIONELEMENT

RATIO CONDITION RESULT

Double angle withcomponents separated,axial compression

Hence our assumed section is ok.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.371

3 1.25

N/mm

3

uwd

m

ff

Maximum size of welding (a) = 81.5 = 6.5 mm

Minimum size of welding (a) = 5 mm

So, taking weld size of 6 mm

Then t =0.7a = 0.74 = 4.2 mm

Now,

Shearing area at the throat design shear strength of the weld = design load

2Lt189.371= 335.175103

L = 335.175 1000

210.72 4.2 189.371

mm

So we have to provide 225 mm length of welding.

.

Factored tensile load = 37.481.5=56.22 KN

According to IS-800-2007 table no-2

(The limiting width to thickness ratio )

clauses (3.7.2 & 3.7.4)

DESIGN OF WELDED JOINTS

DESIGN FOR TENSILESTRENGTH

TENSION


4/24

25.1 250 100570.454

1.1

g y

dg

mo

dg

A fT

T KN

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.76 = 4.2 mm

Now,Shearing area at the throat design shear strength of the weld = design load

2Lt189.371= 56.22103

L = 56.22 1000

35.342 4.2 189.371

mm

So as we have provided 225 mm length of welding , so it is ok.

Herebs= 110 mm, and Lc= 175 mm, W = 110 mm, t = 12 mm, fy =250N/mm2, and fu= 410 N/mm

2

0.7 =1.19

= 1.190.9 (110 12) 12 410 1.19 (110 12) 12 250

665.2091.25 1.1dnT KN

Design tensile strength=570.4542=1140.908KN > 56.22KN

As per IS 800: 2007 SECTION 6 clause 6.1, 6.2 and 6.3.3

Design strength due to Yielding of Gross Section,

Tdg = Ag fy/m0

Hence we can can find out the trial area for the design form this.


Design strength due to Rupture of Critical Section

Tdn = 0.9Anc fu/m1+Ago fy/m0

Where, = 1.40.076 (w/t)(fy/fu) (bs/Lc)


5/24

Hence our design is safe.

.

factored compressive load =231.5=34.5 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa


2

DESIGN OF MEMBER aj mi COMPRESSION DESIGN

TENSION DESIGN

Compressive load on AJ and MI member

= 23 KN

Tensile load on the AJ and MI member

=133.305 KN



6/24

Ac=

mm2

Since we will be using double angle so area of each angle = 345/2 = 172.5 As this is the double angle rafter so the section must

be joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is

70706 @ 6.3 kg/m. The angles are joined byfillet welding


c=2 2

1 2 3vvk k k

2865

13.6 2.371 88.812

250

l

rvv

vvE

1 2 70 70

2 2 6 0.1311 88.812

250

b b

t

E

For hinge support we have ,k1=0.70, k2=0.60, k3=5.

2 20.70 0.60 2.37 5 0.131 2.038e


2 20.20 0.35 2.37 20 0.131 1.584e


= 0.5[1+0.49(1.811-0.2) +1.811

2] = 2.534

2

2 2 0.5 N

2501.1

52.8542.534 [2.534 1.811 ]

/mmcdf

Pc = cd cf A = 52.854 2 8.06100 = 85.2 KN > 34.5KN


Here we are assuming that all the joints are partially hinged.







e=1.811


7/24


Now,0.825 2865

168.53 18013.6

KL

r

Hence ok.

Again we know that,

COMPRESSIONELEMENT



From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.74 = 2.8 mm

Now,


4Lt189.371= 28.8103

L = 28.8 1000

38.024 2.8 189.371

mm

So we have to provide 40 mm length of welding.(as per IS Code).





clauses (3.7.2 & 3.7.4)



8/24

Factored tensile load = 133.305 KN

8.06 100 250183.181

1.1

g y

dg

mo

dg

A fT

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.74 = 2.8 mm

Now,


4Lt189.371= 133.305103

L =133.305 1000

62.854 2.8 189.371

mm



TENSION



Tdg = Ag fy/m0




9/24


2

0.7 = .89 (fum0/fym1)=1.44

= 0.890.9 (70 6) 6 410 0.89 (70 6) 6 250

191.021.25 1.1

dnT KN

Design tensile strength=183.1812=366.362KN > 133.305KNHence our design is safe.




DESIGN OF MEMBER kl COMPRESSION DESIGNTENSION DESIGN


10/24

.

factored compressive load =11.31.5=16.95 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa


2

Ac=

mm2

Since we will be using double angle so area of each angle = 169.5/2 = 84.75 As this is the double angle rafter so the section mustbe joined on a single side of a gusset plate of

thickness 8mm and the dimension of the angle is

20020020 @ 4.5 kg/m. The angles are joined

by fillet welding


c=2 2

1 2 3vvk k k

9542

39.3 2.7341 88.812

250

l

rvv

vvE

1 2 200 200

2 2 20 0.1121 88.812

250

b b

t

E


2 20.70 0.60 2.734 5 0.112 1.789e


2 20.20 0.35 2.734 20 0.112 1.751e

Compressive load on KL member

= 11.3 KN

Tensile load on the KL member

=65.16 KN



11/24


= 0.5[1+0.49(1.811-0.2) +1.811

2] = 2.451

2

2 2 0.5 N

2501.1

54.822.451 [2.451 1.77 ]

/mmcdf

Pc = cd cf A = 54.82 2 76.4100 = 837.649 KN > 16.95KN

Hence our design is safe.Here we are assuming that all the joints are partially hinged.





Now,0.825 9542

194.24 25039.3

KL

r

Hence ok.

Again we know that,

COMPRESSIONELEMENT


Double angle with

components separated,axial compression




e=1.77





clauses (3.7.2 & 3.7.4)


12/24

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.76 = 4.2 mm

Now,


4Lt189.371= 28.8103

L = 16.95 1000

3.894 4.2 189.371

mm

So we have to provide 40 mm length of welding.(as per IS Code).

Factored tensile load = 65.16 KN

76.4 100 2501736.36

1.1

g y

dg

mo

dg

A fT

T KN

.



TENSION



Tdg = Ag fy/m0



13/24

From IS 800:2007 clause 10.5.7.1.1

2

1

410 189.3713 1.25 N/mm3

uwd

mff




Then t =0.7a = 0.76 = 4.2 mm

Now,


4Lt189.371= 65.16103

L = 65.16 1000

40.962 4.2 189.371

mm



2

0.7 = 1.135 (fum0/fym1)

= 1.1350.9 (200 20) 20 410 1.135 (200 20) 20 250

1991.351.25 1.1

dnT KN

Design tensile strength=1736.362=3472.72KN > > 65.16KNHence our design is safe.






14/24

.

factored compressive load =103.631 KNThe anglebelongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa


2

Ac=

mm2

Since we will be using double angle so area of each angle = 1036.31/2 = 518.155

As this is the double angle rafter so the section must

be joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is



c=2 2

1 2 3vvk k k

DESIGN OF MEMBER ck lg COMPRESSION DESIGN

TENSION DESIGN

Compressive load on CK and LG member = 103.631 KN

Tensile load on the CK and LG member

=8.97 KN



15/24

2084

13.5 1.781 88.812

250

l

rvv

vvE

1 2 710 70

2 2 8 0.0981 88.812

250

b b

t

E


2 20.70 0.60 1.78 5 0.098 1.627e


2 20.20 0.35 1.78 20 0.098 1.225e


= 0.5[1+0.49(1.426-0.2) +1.426

2] = 1.817

2

2 2 0.5 N

2501.1

77.221.817 [1.817 1.426 ]

/mmcdf

Pc = cd cf A = 77.22 2 10.6100 = 163.766 KN > 103.631KN







Now,0.825 2084

123.49 18013.5

KL

r

Hence ok.

Again we know that,




e=1.426




16/24

COMPRESSIONELEMENT



From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.76 = 4.2 mm

Now,


2Lt189.371= 103.631103

L = 103.631 1000

65.172 4.2 189.371

mm


Factored tensile load = 8.971.5=13.455KN



clauses (3.7.2 & 3.7.4)



TENSION


17/24

10.6 100 250240.9

1.1

g y

dg

mo

dg

A fT

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.74 = 2.8 mmNow,


4Lt189.371= 13.455103

L = 13.455 1000

6.3434 2.8 189.371

mm



2

0.7 =1.021 (fum0/fym1)

= 1.0210.9 (70 8) 8 410 1.021 (70 8) 8 250 261.513

1.25 1.1dn

T KN



Tdg = Ag fy/m0







18/24

Design tensile strength=240.92=481.8KN > 13.455 KN


.

factored compressive load =19.821.5=29.73 KNThe angle belongs to buckling class c(from IS -800-2007,table 10). fcd varies from 227 to 24.3 MPa

depending on .

Assuming fcd=100 N/mm2

DESIGN OF MEMBER CN GO COMPRESSION DESIGN

TENSION DESIGN

Compressive load on CN and OG member

= 19.82 KN

Tensile load on the CN and OG member

=275.925 KN



19/24

Ac=

mm2

Since we will be using double angle so area of each angle = 345/2 = 172.5 As this is the double angle rafter so the section mustbe joined on a single side of a gusset plate ofthickness 8mm and the dimension of the angle is



c=2 2

1 2 3vvk k k

2865

13.6 2.371 88.812

250

l

rvv

vvE

1 2 70 70

2 2 6 0.1311 88.812

250

b b

t

E


2 20.70 0.60 2.37 5 0.131 2.038e


2 20.20 0.35 2.37 20 0.131 1.584e


= 0.5[1+0.49(1.811-0.2) +1.811

2] = 2.534

2

2 2 0.5 N

2501.1

52.8542.534 [2.534 1.811 ]

/mmcdf

Pc = cd cf A = 52.854 2 8.06100 = 85.2 KN > 29.732KN









e=1.811


20/24


Now,0.825 2865

168.53 18013.6

KL

r

Hence ok.

Again we know that,

COMPRESSIONELEMENT



From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.74 = 2.8 mm

Now,


2Lt189.371= 19.82103

L = 19.83 1000

18.682 2.8 189.371

mm






clauses (3.7.2 & 3.7.4)



21/24

Factored tensile load = 275.925KN

8.06 100 250183.181

1.1

g y

dg

mo

dg

A fT

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

ff




Then t =0.7a = 0.74 = 2.8 mm

Now,


2Lt189.371= 275.925103

L =275.925 1000

230.22 2.8 189.371

mm



TENSION



Tdg = Ag fy/m0




22/24


2

0.7 =1.21 (fum0/fym1)

= 1.210.9 (70 8) 8 410 1.21 (70 8) 8 250

282.8191.25 1.1

dnT KN

Design tensile strength=183.1812=366.362 > 275.925


.

Here we are supposed to find out the maximum load at minor & major axis. To do so we

need to do the load combinations for various condition. They are explained below

LOAD COMBINATIONS

Load Types Major axis MAX. LOADING IN KN Type

1.5(DL+LL) 1.5{(0.588+1.03)+5.12} cos = 9.44 Tension

1.5(DL+WL 1) 1.5{(0.588+1.03) cos 1.68} = 0.25 Compression

1.5(DL+WL 2) 1.5{(0.588+1.03) cos 0.6} = 1.35 Tension





DESIGN OF PURLIN

LO D COMBIN TION


23/24

1.5(DL+WL 4) 1.5{(0.588+1.03) cos 1.5} = 0.018 Tension


1.5(DL+WL 6) 1.5{(0.588+1.03) cos 1.19} =0.255 Compression

1.2(DL+LL+WL 1) 1.2[{(0.588+1.03)+5.12} cos 1.68] = 5.26 Tension





1.2(DL+LL+WL 6) 1.2[{(0.588+1.03+5.12} cos 1.19] = 6.13 Tension

Load Types Minor axis MAX. LOADING IN KN Type

1.5(DL+LL) 1.5{(0.588+1.03)+5.12} sin = 3.59 Tension

1.5(DL+WL 1) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.5(DL+WL 2) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.5(DL+WL 3) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.5(DL+WL 4) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.5(DL+WL 5) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.5(DL+WL 6) 1.5{(0.588+1.03) sin } = 0.864 Tension

1.2(DL+LL+WL 1) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension

1.2(DL+LL+WL 2) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension1.2(DL+LL+WL 3) 1.2{(0.588+1.03)+5.12} sin = 2.88 Tension




Maximum design load on plane of major axis = 9.44 KN Maximum design load on plane of minor axis = 3.59 KN

Let assume length of purlin = 3 m

Now factor applied moment about the minor axis, Mz=2 2

1 WL 1 (3.59) 30.41

10 8 10 8

KNm

Now factor applied moment about the major axis, My=2 2

1 WL 1 (9.54) 31.07

10 8 10 8

KN-m

Assuming fcd =100 N/mm2

DESIGN


24/24

According to IS 808: 1989,

Try with ISMC 100,

Area (a) = 1220 mm2, b = 50 mm, tw = 5 mm, tf = 7.7 mm,

Zpz = 43.83cm3

According to IS 800: 2007,

Now, b/ tf= 50/7.7 =6.493 < 9.4

And d/tw= (100-27.7)/5 = 16.92 < 42

So the section is plastic.

Now, Zpy=31220 50

152502 2 2 2

a bmm

Design strength under corresponding moment

Mndz= (b Zpz fy)/ m0

and Mndy = (b Zpy fy)/ m0

For plastic section b= 1,

So, Mdz = (1 43.83 103 250) / 1.1

= 9.9613106N-mm = 9.9613 KN-m

Mdy= (1 15250 250) / 1.1

= 3.465106N-mm = 3.465 KN-m

Now,

1 2

1y z

ndy ndz

M M

M M

1 2

0.8619 0.30640.2497 1

3.465 9.9613

So our design is safe

CHECK FOR DEFLECTION

DEEFLECTION OF THE PURLI N:-

=

4

4

5 4

9.445 3000

5 1.5 3 5.762384 384 2 10 192 10

WL mmEI

Whereas allowable deflection is3000

20150

So it is also safe from deflection.

CHECK FOR DEFLECTION

cd and fg member

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