steel designing project

8/13/2019 Steel Designing Project

http://slidepdf.com/reader/full/steel-designing-project 1/84

[Year]

S MAITRA

[Company name]

[Date]

[Document title]



1

We are supposed to design an Industrial truss in this particular project. It

would be of a specified kind. Here are some given data according to which I am supposeddesign the particular truss. They are given below →

General data:

Height of the truss = (3.6+.2y) m [where y = my position in the group]

Length of the truss = (15+N) m [where N = my group no]

Hence my general data is

Height of the truss = (3.6+.2×2) m = 4 m

Length of the truss = (15+6) m = 21 m

STEEL DESIGNING PROJECT

•

TEACHER - B.K.R SIR• STUDENT - SUBHAJIT MAITRA



2

From the assumption of the angle shown in the fig. we have,

( )

( )=

Now,

From we have,

()

[Where, L=AF] …………. (1)

Again,

From we have,

()

[Where, L=AF] …………. (2)

Here, AB =√

= 5.62.

Now,

()()

()()

()

()

() () () ()

()

()

()

()

CALCULATION OF ANGLE:-



3

We have,

()

Dead Load due to concr ete asbestos sheeting: -Unit load=.13KN

Effective area of point B, C, D, E, F, G, H

=2.81×3× cos(20.85°)=7.88 m2

Effective area of point E

=7.88 m2

Effective area of point A, I

=1.71×3× cos(20.85°)=4.8 m2

Hence,

Dead load due to sheeting →

@ Point B, C, D, E, F, G, H = 7.88×.13=1.03KN

@ Point B, C, D, E, F, G, H = 7.88×.13=1.03KN

@ Point B, C, D, E, F, G, H = 4.8×.13=.624KN

Dead Load due to purl in: -We are using MC-100

Unit load=.098KN

Hence,

Dead load due to purlin →

@ All points except E = .098×3=.294KN

@ Point E= .294×2=.588KN

Dead Load due to pur li n: -Total length of tie=21m

DEAD LOAD CALCULATION



4

Total length of rafter=2×11.24m=22.48m

()

() 73

5.73×2=11.46m

( )

Sin48.88°=

KE=6.33m

()

BJ=2.81×tan20.35°=1.05m

CK=5.62tan20.35°=2.1m

Hence total length of truss

= (4.2+4.2+12.66+21+22.48+11.46+9.54) m=85.54m

Total weight

= (2×85.54×.08) =13.686m

This total load will be divided to each hinge of the rafter. Now as we can see from the fig of

the truss that hinge point A & I will bear half the load than the other points. Because the

effective area of the other points is double than the effective area of the points A & I.

Hence, this total load will be distributed to the 8 points of the rafter.



5

Hence total Dead Load →

@ node A & I =(.867+.294+.624)×1.05=1.87KN

@ node B ,C ,D ,F ,G & I =(1.03+.294+1.733)×1.05=3.21KN

@ node E =(1.03+.588+1.733)×1.05=3.52KN

From IS-875-2007 we have,

[Where θ = angle of inclination of the roof]

Hence,

The total Live Load →

@ nodes A & I = 0.533 =2.558 KN/m2

@ all other nodes = 0.533 =5.1168 KN/m2

LIVE LOAD CALCULATION

The formula• The live load intensity on the roof

• =0.75-0.02(-10)KN/m2

The result

• The live load intensity on the roof

• =0.75-0.02(20.85-10)KN/m2=0.533KN/m2



6

My roll no = 27

Hence my city will be Bangalore.

So my basic wind speed according to initial consideration V b = 33 m/s

Design wind speed

Here,

k 1=1

The total height = (4+4.5) =8.5m

Maximum dimension = 21mAccording to our consideration our building is at category-4 and class-2.

k 2=.076

Now, considering the upwind sped angle to be less than 3° we get,

k 3=1

Wind load calculation

Vz= Vbk1k2k3

Vz= Vbk1k2k3=3310.761=25.1m/s



7

As our opening is less than less than 5%(<5%)

We will take C pi=0.2

Now,

w=21 m

h=4.5m

Hence,

WIND ANGLE

=0°

WIND ANGLE

=90°

EF GH EG FH

20° -0.4 -0.4 -0.7 -0.6

30° 0 -0.4 -0.7 -0.6

20.85° -0.366 -0.4 -0.7 -0.6

C pi=0.2 WIND ANGLE

=0°

WIND ANGLE

=90°

C pe - C pi

EF GH EG FH

-0.2-0.366

=-0.566

-0.2-0.4

=-0.6

-0.7-0.2

=-0.9

--0.6-0.2

=-0.8

C pi=-0.2 WIND ANGLE

=0°

WIND ANGLE

=90°

C pe - C pi

EF GH EG FH

0.2-0.366

=-0.166

0.2-0.4

=-0.2

-0.7+0.2

=-0.5

--0.6+0.2

=-0.4

Now, from the IS-875-1984 we have,

Wind load F= (C pe - C pi)Apa



8

FOR WIND ANGLE θ=0°

Point Influence

Area (m2

)

Wind angle,

Θ=0°

pa Fz

C pi=0.2 C pi=

-0.2

0.387 C pi=0.2 C pi=

-0.2

A 4.8 -0.566 -0.166 0.387 -1.03 -0.3

B 7.88 -0.566 -0.166 0.387 -1.68 -0.5

C 7.88 -0.566 -0.166 0.387 -1.68 -0.5

D 7.88 -0.566 -0.166 0.387 -1.68 -0.5

E 3.94 -0.566 -0.166 0.387 -0.84 -0.25

E 3.94 -0.6 -0.2 0.387 -0.9 -0.3

F 7.88 -0.6 -0.2 0.387 -1.79 -0.6

G 7.88 -0.6 -0.2 0.387 -1.79 -0.6

H 7.88 -0.6 -0.2 0.387 -1.79 -0.6

I 4.8 -0.6 -0.2 0.387 -1.1 -0.36


FOR EG

Point Influence

Area (m2)

Wind angle,

Θ=0°

pa Fz

C pi=0.2 C pi=

-0.2

0.387 C pi=0.2 C pi=

-0.2

A 4.8 -0.9 -0.5 0.387 -1.63 -0.91



9

B 7.88 -0.9 -0.5 0.387 -2.68 -1.5

C 7.88 -0.9 -0.5 0.387 -2.68 -1.5

D 7.88 -0.9 -0.5 0.387 -2.68 -1.5

E 3.94 -0.9 -0.5 0.387 -1.38 -0.75

E 3.94 -0.9 -0.5 0.387 -1.38 -0.75

F 7.88 -0.9 -0.5 0.387 -2.68 -1.5

G 7.88 -0.9 -0.5 0.387 -2.68 -1.5

H 7.88 -0.9 -0.5 0.387 -2.68 -1.5

I 4.8 -0.9 -0.5 0.387 -1.63 -0.91


FOR FH

Point Influence

Area (m2)

Wind angle,

Θ=0°

pa Fz

C pi=0.2 C pi=

-0.2

0.387 C pi=0.2 C pi=

-0.2

A 4.8 -0.8 -0.4 0.387 -1.45 -0.72

B 7.88 -0.8 -0.4 0.387 -2.38 -1.19

C 7.88 -0.8 -0.4 0.387 -2.38 -1.19

D 7.88 -0.8 -0.4 0.387 -2.38 -1.19

E 3.94 -0.8 -0.4 0.387 -1.29 -0.6

E 3.94 -0.8 -0.4 0.387 -1.29 -0.6

F 7.88 -0.8 -0.4 0.387 -2.38 -1.19

G 7.88 -0.8 -0.4 0.387 -2.38 -1.19

H 7.88 -0.8 -0.4 0.387 -2.38 -1.19

I 4.8 -0.8 -0.4 0.387 -1.45 -0.72



10



11



12

DEAD LOAD ANALYSIS

Plan length of AI=21m

Hence, plan length of AB=

2.625m

Vertical length of AB = 1m

Now we will evaluate & find out the member loads considering each points.

As the structure is having symmetrical loading, then →

R A=R I=

KN

Consideri ng Poin t A: -

V=0,

FAB FAJ=-R A+1.87

LOAD ANALYSIS



13

FAB FAJ= -11.39

H=0,

FAB+ FAJ= 0

FAB FAJ= 0

FAB=-32.745KN

FAJ=30.647KN

Consideri ng Poin t B: -

V=0,

FBC FBJ= FAB

FBC FBJ= 8.445

H=0,



14

FBC + FBJ = FAB

FBC + FBJ= 30.616

FBC=31.6KN

FBJ=3 KN

Consideri ng Poin t J:-

V=0,

FJC FJK = FJA FJB

FJC FJK =

H=0,

FJC+ FJK = FJB FJB

FJC+ FJK =

FJC=4.318KN

FJK =26.328 KN

Taking a section 11’

M=0,

FKL ( )

FKL=17.47KN

Consideri ng Poin t K: -

V=0,



15

FKN FKC = FJK

FKN FKC =

H=0,

FKN + FKC= FJK FKL

FKN + FKC=

FKN=10.26KN

FKC=8.91 KN

Considering Point C:-

V=0,

FCD FCN= FCK FCB FCJ

FCD FCN=

H=0,



16

FCD+ FCN= FCB FCJ FCK

FCD+ FCN=

FKN=38.32KN

FKC=12.713 KN

Consideri ng Poin t D: -

V=0,

FDE FDN= FDC

FDE FDN=

H=0,

FDE + FDN = FDC

FDE + FDN =

FKN=KN

FKC=3 KN

Consideri ng Poin t N: -

V=0,

FDE FDN= FDC

FDE FDN=



17

LIVE LOAD ANALYSIS

H=0,

FDE

+ FDN

= FDC

FDE + FDN =

FKN=KN

FKC=3 KN

In this case also,



2.625m




18



R A=R I=

KN


V=0,

FAB FAJ= -R A+1.87

FAB FAJ=

H=0,

FAB+ FAJ= 0

FAB FAJ= 0

FAB=51.53KN

FAJ=48.22KN


V=0,

FBC FBJ= FAB

FBC FBJ= 13.22



19

H=0,

FBC + FBJ = FAB FBC + FBJ= 48.06

FBC=49.68KN

FBJ=4.78 KN


V=0,

FJC FJK = FJA FJB

FJC FJK =

H=0,

FJC+ FJK = FJB FJB

FJC+ FJK =



20

FBC=6.72KN

FBJ=52.8 KN


M=0,

FKL ( )

FKL=27.22KN


V=0,

FKN FKC = FJK

FKN FKC =

H=0,

FKN + FKC= FJK FKL

FKN + FKC=

FKN=47.72KN

FKC=42.19 KN


V=0,

FCD FCN= FCK FCB FCJ

FCD FCN=



22

H=0,

FDE + FDN = FDC

FDE + FDN =

FKN=KN

FKC=4.78 KN


V=0,

FDE FDN= FDC

F NE=KN



23

WIND LOAD ANALYSIS

WL θ=90° EG Cpi=0.2

In this case also, Plan length of AI=21m


2.625m




R A=R I=

KN


V=0,

FAB FAJ= R A1.63

H=0,



24

FAB+ FAJ= 1.63

FAB=25.22KN

FAJ=23KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN

FBJ= KN


V=0,

FJC FJK = FJA FJB



25

H=0,

FJC+ FJK = FJB FJA

FBC=KN

FBJ= KN


M=0,

FKL

(

)

FKL=11.3KN


V=0,

FKN FKC = FJK



26

H=0,

FKN + FKC= FJK FKL

FKN=33.44KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ

H=0,

FCD+FCN= FCB FCJ FCK

FKN=70.1KN



27

FKC=97.82 KN


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC

FKN=

FKC= KN




28

WIND LOAD ANALYSIS

WL θ=90° EG C pi = -0.2

[Cite your source here.]

V=0,

FDE FDN= F NC F NK

F NE=KN

In this case also,




29


2.625m




R A=R I=

KN


V=0,

FAB FAJ= R A0.91

H=0,

FAB+ FAJ= 0.91

FAB=14.1KN

FAJ=12.84KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN



30

FBJ= KN


V=0,

FJC FJK = FJA FJB

H=0,

FJC+ FJK = FJB FJA

FBC=KN

FBJ= KN


M=0,



31

FKL( )

FKL=6..3KN


V=0,

FKN FKC = FJK

H=0,

FKN + FKC= FJK FKL

FKN=KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ



32

H=0,


FKN=KN

FKC= KN


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC



33

WIND LOAD ANALYSIS

WL θ=90° FH C pi = 0.2


FKN=KN

FKC= KN


V=0,

FDE FDN= F NC F NK

F NE=KN



34

In this case also,



2.625m




R A=R I=

KN


V=0,

FAB FAJ= R A1.45

H=0,



35

FAB+ FAJ= 1.45

FAB=22.37KN

FAJ=20.39KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN

FBJ=2.38 KN


V=0,

FJC FJK = FJA FJB



36

H=0,

FJC+ FJK = FJB FJA

FJC=KN

FJK = KN


M=0,

FKL ( )

FKL=8.1KN


V=0,

FKN FKC = FJK

H=0,

FKN + FKC= FJK FKL

FKN=KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ



37

H=0,


FKN=KN

FKC= KN


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC

FDE=KN



38

WIND LOAD ANALYSIS

WL θ=90° FH C pi = - 0.2


FDN= KN


V=0,

FDE FDN= F NC F NK

F NE=KN



39

In this case also,



2.625m




R A=R I=

KN


V=0,

FAB FAJ= R A0.72

H=0,

FAB+ FAJ= 0.72



40

FAB=11.28KN

FAJ=10.29KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN

FBJ=1.2 KN


V=0,

FJC FJK = FJA FJB



41

H=0,

FJC+ FJK = FJB FJA

FJC=KN

FJK = KN


M=0,

FKL( )

FKL=5.1KN


V=0,

FKN FKC = FJK



42

H=0,

FKN + FKC= FJK FKL

FKN=KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ

H=0,




43

FCD=KN

FCN= KN


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC

FDE=KN

FDN= KN




44

WIND LOAD ANALYSIS

Θ=0°


V=0,

FDE FDN= F NC F NK

F NE=KN

In this case also,



2.625m



Here, we get R A=6.785 KN

R B=6.985 KN


V=0,

FAB FAJ= R A1.03



45

H=0,

FAB+ FAJ= 1.03

FAB=15.76KN

FAJ=13.9KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN

FBJ= KN



46


V=0,

FJC FJK = FJA FJB

H=0,

FJC+ FJK = FJB FJA

FJC=KN

FJK =

KN


M=0,

FKL ( )

FKL=12.4KN


V=0,

FKN FKC = FJK



47

H=0,

FKN + FKC= FJK FKL

FKN=KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ

H=0,


FCD=KN



48

FCN= KN


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC

FDE=

FDN= KN




49

WIND LOAD ANALYSIS

Θ=0°


V=0,

FDE FDN= F NC F NK

F NE=KN

In this case also,



2.625m



Here, we get R A=2.03 KN

R B=2.18 KN


V=0,

FAB FAJ= R A0.3



50

H=0,

FAB+ FAJ= 0.3 FAB=5.79KN

FAJ=4.89KN


V=0,

FBC FBJ=FAB

H=0,

FBC + FBJ = FAB

FBC=KN

FBJ= KN




52

H=0,

FKN + FKC= FJK FKL

FKN=KN

FKC= KN


V=0,

FCD FCN=FCK FCB FCJ

H=0,


FCD=KN

FCN= KN



53


V=0,

FDE FDN= FDC

H=0,

FDE + FDN = FDC

FDE=

FDN= KN




54

V=0,

FDE FDN= F NC F NK

F NE=KN

factored compressive load =335.175 KN

The angle belongs to buckling class ‘c’(from IS -800-2007,table 10). f cd varies from 227 to 24.3 MPa

depending on ⁄ .

Assuming f cd =100 N/mm2

Ac=

mm

2

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

DESIGNING OF MEMBERS

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



55

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 110×110×12 @ 19.7 kg/m. The angles are joined

by fillet welding.

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

λ c=2 2

1 2 3vvk k k

2810

21.5 1.461 88.812

250

l

r vv

vv E

1 2 110 110

2 2 12 0.1031 88.812

250

b b

t

E

For hinge support we have ,k 1=0.70, k 2=0.60, k 3=5.

2 20.70 0.60 1.46 5 0.103 1.425e

Again for fixed support we have, k 1=0.20, k 2=0.35, k 3=20.

2 20.20 0.35 1.46 20 0.103 1.076e

It is given in the IS-800-2007 that

ϕ = 0.5[1+0.49(1.251-0.2) +1.2512] = 1.539

• DESIGN FOR COMPRESSIVEFORCE

IS code specification

•For partial restraint, the λe can be interpolated between the λe results for fixed & hingedcases.

the interpolated result

•λe=1.251



56

2

2 2 0.5 N

2501.1

93.341.539 [1.539 1.251 ]

/mmcd f

Pc = cd c f A = 93.33 × 2 × 21.1×100 = 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,

COMPRESSION

ELEMENT

RATIO CONDITION RESULT

Double angle with

components separated,

axial compression

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.

According to IS-800-2007 table no-2

(The limiting width to thickness ratio )

clauses (3.7.2 & 3.7.4)



57

Hence our assumed section is ok.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f

Maximum size of welding (a) = 8−1.5 = 6.5 mm

Minimum size of welding (a) = 5 mm

So, taking weld size of 6 mm

Then t =0.7×a = 0.7×4 = 4.2 mm

Now,

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

2×L×t×189.371= 335.175×103

L = 335.175 1000

210.72 4.2 189.371

mm

So we have to provide 225 mm length of welding.

.

Factored tensile load = 37.48×1.5=56.22 KN

DESIGN OF WELDED JOINTS

• DESIGN FOR TENSILESTRENGTH

TENSION

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 × f y/γm0

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



58

25.1 250 100570.454

1.1

g y

dg

mo

dg

A f T

T KN

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f

Maximum size of welding (a) = 8−1.5 = 6.5 mm Minimum size of welding (a) = 5 mm


Then t =0.7×a = 0.7×6 = 4.2 mm

Now,


2×L×t×189.371= 56.22×103

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.

Here bs = 110 mm, and Lc = 175 mm, W = 110 mm, t = 12 mm, f y =250N/mm2 , and f u = 410 N/mm

2

0.7≰ β =1.19

β = 1.19

0.9 (110 12) 12 410 1.19 (110 12) 12 250 665.2091.25 1.1

dnT KN


Design strength due to Rupture of Critical Section

Tdn = 0.9Anc× f u/γm1+β×Ago× f y/γm0 Where, β = 1.4−0.076× (w/t)×(f y/f u)× (bs/Lc)



59

Design tensile strength=570.454×2=1140.908KN > 56.22KN


.

DESIGN OF MEMBER aj mi


• TENSION DESIGN

Compressive load on AJ and MI member

• = 23 KN

Tensile load on the AJ and MI member

• =133.305 KN



60

factored compressive load =23×1.5=34.5 KN


depending on ⁄ .


Ac=

mm

2

Since we will be using double angle so area of each angle = 345/2 = 172.5


thickness 8mm and the dimension of the angle is 70×70×6 @ 6.3 kg/m. The angles are joined byfillet welding


λ c=2 2

1 2 3vvk k k

2865

13.6 2.371 88.812

250

l

r vv

vv E

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






•λe=1.811



61

ϕ = 0.5[1+0.49(1.811-0.2) +1.8112] = 2.534

2

2 2 0.5 N

2501.1

52.8542.534 [2.534 1.811 ]

/mmcd

f

Pc = cd c f A = 52.854 × 2 × 8.06×100 = 85.2 KN > 34.5KN







Now,0.825 2865

168.53 18013.6

KL

r

Hence ok.

Again we know that,

COMPRESSION

ELEMENT


Double angle with


axial compression





clauses (3.7.2 & 3.7.4)



62

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×4 = 2.8 mm

Now,


4×L×t×189.371= 28.8×103

L = 28.8 1000

38.02

4 2.8 189.371

mm

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

Factored tensile load = 133.305 KN



TENSION



63

8.06 100 250183.181

1.1

g y

dg

mo

dg

A f T

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×4 = 2.8 mm

Now,


4×L×t×189.371= 133.305×103

L = 133.305 1000

62.854 2.8 189.371

mm









64


2

0.7≰ β = .89 ≤ (f uγm0/f yγm1)=1.44

β = 0.89

0.9 (70 6) 6 410 0.89 (70 6) 6 250191.02

1.25 1.1dn

T KN

Design tensile strength=183.181×2=366.362KN > 133.305KN



Tdn = 0.9Anc× f u/γm1+β×Ago× f y/γm0

Where, β = 1.4−0.076× (w/t)×(f y/f u)× (bs/Lc)

DESIGN OF MEM BER kl


• TENSION DESIGN



65

.

factored compressive load =11.3×1.5=16.95 KN


depending on ⁄ .


Ac=

mm

2

Since we will be using double angle so area of each angle = 169.5/2 = 84.75


thickness 8mm and the dimension of the angle is 200×200×20 @ 4.5 kg/m. The angles are joined by

fillet welding


λ c=2 2

1 2 3vvk k k

9542

39.3 2.7341 88.812

250

l

r vv

vv E

Compressive load on KL member• = 11.3 KN

Tensile load on the KL member

• =65.16 KN




66

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


ϕ = 0.5[1+0.49(1.811-0.2) +1.8112] = 2.451

22 2 0.5 N

250

1.1 54.822.451 [2.451 1.77 ]

/mmcd f

Pc = cd c f A = 54.82 × 2 × 76.4×100 = 837.649 KN > 16.95KN










•λe=1.77





67

Now,0.825 9542

194.24 25039.3

KL

r

Hence ok.

Again we know that,

COMPRESSION

ELEMENT


Double angle with


axial compression

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×6 = 4.2 mm

Now,


4×L×t×189.371= 28.8×103

L = 16.95 1000

3.894 4.2 189.371

mm

According to IS-800-2007 table no-2(The limiting width to thickness ratio )

clauses (3.7.2 & 3.7.4)




68

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


76.4 100 250 1736.361.1

g y

dg

mo

dg

A f T

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×6 = 4.2 mm

Now,



TENSION








69

4×L×t×189.371= 65.16×103

L = 65.16 1000

40.962 4.2 189.371

mm


Here bs = 200 mm, and Lc = 350 mm, W = 200 mm, t = 20 mm, f y =250N/mm2 , and f u = 410 N/mm2

0.7≰ β = 1.135 ≤ (f uγm0/f yγm1)

β = 1.135

0.9 (200 20) 20 410 1.135 (200 20) 20 2501991.35

1.25 1.1dn

T KN

Design tensile strength=1736.36×2=3472.72KN > > 65.16KN







70

.

factored compressive load =103.631 KN


depending on ⁄ .


Ac=

mm

2

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



fillet welding


λ c=2 2

1 2 3vvk k k

DESIGN OF MEMBER ck lg


• TENSION DESIGN

Compressive load on CK and LG member

• = 103.631 KN

Tensile load on the CK and LG member

• =8.97 KN




71

2084

13.5 1.781 88.812

250

l

r vv

vv E

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


ϕ = 0.5[1+0.49(1.426-0.2) +1.4262] = 1.817

2

2 2 0.5 N

2501.1

77.221.817 [1.817 1.426 ]

/mmcd f

Pc = cd c f A = 77.22 × 2 ×10.6×100 = 163.766 KN > 103.631KN









•λe=1.426



72


Now,0.825 2084

123.49 18013.5

KL

r

Hence ok.

Again we know that,

COMPRESSION

ELEMENT


Double angle with


axial compression

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×6 = 4.2 mm

Now,





clauses (3.7.2 & 3.7.4)




73


2×L×t×189.371= 103.631×103

L = 103.631 1000

65.172 4.2 189.371

mm


Factored tensile load = 8.97×1.5=13.455 KN

10.6 100 250240.9

1.1

g y

dg

mo

dg

A f T

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




TENSION








74


Then t =0.7×a = 0.7×4 = 2.8 mm

Now,

Shearing area at the throat × design shear strength of the weld = design load4×L×t×189.371= 13.455×10

3

L = 13.455 1000

6.3434 2.8 189.371

mm



2

0.7≰ β =1.021 ≤ (f uγm0/f yγm1)

β = 1.021

0.9 (70 8) 8 410 1.021 (70 8) 8 250261.513

1.25 1.1dn

T KN

Design tensile strength=240.9×2=481.8KN > 13.455 KN







75

.

factored compressive load =19.82×1.5=29.73 KN


depending on ⁄ .


Ac=

mm

2

Since we will be using double angle so area of each angle = 345/2 = 172.5



fillet welding


DESIGN OF MEMBER CN GO


• TENSION DESIGN

Compressive load on CN and OG member• = 19.82 KN

Tensile load on the CN and OG member

• =275.925 KN




77


Now,0.825 2865

168.53 18013.6

KL

r

Hence ok.

Again we know that,

COMPRESSION

ELEMENT


Double angle with

components separated,axial compression

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f




Then t =0.7×a = 0.7×4 = 2.8 mm





clauses (3.7.2 & 3.7.4)




78

Now,


2×L×t×189.371= 19.82×103

L = 19.83 1000 18.682 2.8 189.371

mm



8.06 100 250183.181

1.1

g y

dg

mo

dg

A f T

T KN

.

From IS 800:2007 clause 10.5.7.1.1

2

1

410

189.3713 1.25

N/mm3

uwd

m

f f


• DESIGN FOR TENSILE

STRENGTH

TENSION








79



Then t =0.7×a = 0.7×4 = 2.8 mm

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

2×L×t×189.371= 275.925×103

L = 275.925 1000

230.22 2.8 189.371

mm



2

0.7≰ β =1.21 ≤ (f uγm0/f yγm1)

β = 1.21

0.9 (70 8) 8 410 1.21 (70 8) 8 250282.819

1.25 1.1dn

T KN

Design tensile strength=183.181×2=366.362 > 275.925







80

.

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 θ = 9.44 Tension

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

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

1.5(DL+WL 3) 1.5{(0.588+1.03) × cos θ – 2.68} = − 1.75 Compression

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

1.5(DL+WL 5) 1.5{(0.588+1.03) × cos θ – 2.38} = 1.30 Compression

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

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





DESIGN OF PURLIN

LO D COMBIN TION



81

1.2(DL+LL+WL 6) 1.2[{(0.588+1.03+5.12} × cos θ 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 θ = 3.59 Tension

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






1.2(DL+LL+WL 1) 1.2×{(0.588+1.03)+5.12} ×sin θ = 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

KN – m

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

1 WL 1 (9.54) 31.07

10 8 10 8

KN-m


According to IS 808: 1989,

DESIGN



82

Try with ISMC 100,

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

Z pz = 43.83cm3

According to IS 800: 2007, Now, b/ tf = 50/7.7 =6.493 < 9.4ϵ

And d/tw = (100-2×7.7)/5 = 16.92 < 42ϵ

So the section is plastic.

Now, Z py =31220 50

152502 2 2 2

a bmm

Design strength under corresponding moment

Mndz = (β b × Z pz × f y)/ γm0 and Mndy = (β b × Z py × f y)/ γm0

For plastic section β b = 1,

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

= 9.9613×106 N-mm = 9.9613 KN-m

Mdy = (1 ×15250 × 250) / 1.1

= 3.465×106 N-mm = 3.465 KN-m

Now,1 2

1 y z

ndy ndz

M M

M M

1 20.8619 0.3064

0.2497 13.465 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 35.762

384 384 2 10 192 10

WLmm

EI

Whereas allowable deflection is 3000 20150

CHECK FOR DEFLECTION



So it is also safe from deflection.

steel designing project

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