6 Examples

7/30/2019 6 Examples

1/77

Design ofSteelStructures

Prof. S.R.Satish Kumar andProf. A.R.Santha Kumar

Problem 1

Design a hand operated overhead crane, which is provided

in a shed, whose details are:

Capacity of crane = 50 kN

Longitudinal spacing of

column = 6m Center to center

distance of gantry girder =

12m Wheel spacing = 3m

Edge distance = 1m Weight of

crane girder = 40 kN Weight of

trolley car = 10 kN

Design by allowable stress method as per IS: 800 - 1984

To find wheel load (refer fig.1):


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Indian Institute of Technology Madras


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RA = 20 + 60 (11 / 12) = 75 kN

Wheel load = RA / 2 = 37.5 kN To find maximum

BM in gantry girder (refer fig.2):

RA = 46.88 kN

RB = 28.12 kN Max. BM = 28.12 x 2.25 = 63.27 kN-m Adding 10% for impact,

M1 = 1.1 x 63.27 = 69.60 kN-m Max. BM due to self - weight of girder and rail

taking total weight as 1.2 kN/m

M2 = = 5ABfm 8

Therefore Total BM, M = 75 kN-m

To find maximum shear force (refer fig.3):

SF = RA = 59.85 kN.

To find lateral loads:

This is given by 2.5% of (lateral load / number of wheel = 0.025 x 60 / 2 kN = 0.75kN

Therefore Max BM due to lateral load by proportion is given by, ML = (63.27 /37.5) x 0.75 = 1.27 kN-m

Design of section

Approximate section modulus Zc required, (M / a bc) = 75 x 106 / 119 = 636 x103mm3 [forX= 120, D / T = 25].

Since, the beam is subjected to lateral loads also, higher section is selected.

For, ISMB 450 @ 710.2 N/m,

Zx = 1350.7 cm3 , T = 17.3 mm,


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sign of Steel Structures Prof. S.R.Satish Kumar and Prof..Santha Kumar

t = 9.4mm, Iyy = 111.2cm3 ry

= 30.1mm, bf = 150mm

To find allowable stresses, T/t =

17.4 / 9.4 = 1.85 < 2 D/T = 450 /

17.4 = 25.86 ~ 26 L/ry = 6000 / 30.1

= 199.3 ~ 200

Therefore allowable bending compression about major axis is, sbc' x = 77.6

N/mm2Actual stress in compression side, sb' x = M / Z = 55.5 N / mm2.

The bending moment about Y-axis is transmitted only to the top of flange

and the flange is treated as rectangular section. The allowable stress is sbc. y =

165 MPa (i.e. 0.66 fy).

Zy of the flange = 111.2 / 2 = 55.6 cm3 Therefore sby = My / Zy = 1.27 x 10

6 / 55.6 x

103 = 22.84 MPa The admissible design criteria is = (sbx / sbcx) + ( sb y / sbcy) =

(55.5 / 77.6) + (22.84 / 165) = 0.715 + 0.138 = 0.854 < 1. Hence , the design is

safe. So, ISMB 450 is suitable Check for shear: Design shear stress, xx V / (Dt) =

59.85 x 103 / 450 x 9.4 = 14.15 MPa. This is less than xa (0.4fy). Hence, the design

is o.k.

Check for deflection and longitudinal bending can be done as usual.

an Institute of Technology Madras


5/77


Design by limit state method as per IS: 800 draft code

For ISMB 450, properties are given below:

T = 17.4mm, t = 9.4mm, b = 150mm, ry = 30.1mm, Zp = 1533.33 cm3, Zc =

1350.7 cm3, Shape factor = 1.15, Izz = 30390.8 cm4, H1 = d = 379.2mm

Section classification:

Flange criteria: b / T = 75 / 17.4 = 4.31 < 9.4

No local buckling. Therefore OK

Web criteria: d / tw = 379.2 / 9.4 =

40.34 < 83.9 No local buckling.

Therefore OK

Section plastic.

Shear capacity:

= (250 x 450 x 9.4)/ (fi x 1.1) = 555043

N ^ 555 kN

Fv / Fvd = (59.85 x 1.5) / 555 = 0.1634 < 0.6

Check for torsional buckling (CI.8.2.2):

tf / tw


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7/77

sign of Steeluctures

Prof. S.R.Satish Kumar and Prof.A.R.Santha Kumar

i+-L

2

0

KL

h

1.27? x 450 x 2 x 10' x

B34 x

10* 2

x

60002

x 1 + I

20

60

00

30

.1

45

0

17

4

= 246 x 10

m

pV; = 1.248

(Pb = 1 for plassection)

Therefore fyu =

[1 + O,LT (^LT

+X2u

=

0

.

5

[

1


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+ 0.21

(1.248 -

0.2) +

1.2482] =

1.389

i

&T + - 7210J

1/1/999 = 0.5

^LT fy (0.5 x 2

13.7MPa

ThereforeM

d=

b.z

p.fbd

= 1 x 1533.3

x 103 x 113.7

=

1

.

7

4

3

4

x

1

0

8

=

1

7

4

.

3

4

k

N

-

m

F

a

c

t

o

=


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t,

M

f

=

7

5

x

1

.

5

=

1

1

2

.

5

k

N

-

m

F

a

c

t

o

r

e

d

l

a

t

e


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r

a

l

m

o

m

e

n

t,

M

fL

=

1

.

2

7

x

1

.

5

=

1

.

9

1

k

N

-

m

Indian Institute of Technology Madr


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Lateral BM capacity = MdL

-21 fS!h.-:ru h'sr/^far} f^

0 1.10

! 1

^

1.15 x 250

2

1 111

= 14,53 x 106Nor 14.53kN-m

For Safety,

M M

M * M.

= (112.5 / 135.9) + (1.91 / 14.53) = 0.96 Mdz (=47.15

T-m ) Therefore Mndz = Mdz = 47.15T-m

For, n


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Therefore (My / Mndy)a1 + (Mz / Mndz)a2 = (1.44 / 6.88)1 + (12.60 / 47.15)2 =0.281 = 1, therefore a1 = 5 x 0.066 = 0.33 = 1

a,2 = 2 (As per Table 9.1)}

Alternatively,

(N / Nd ) + (Mz / Mdz) + (My / Mdy) = { (16.5 x 103 x 1.10) / (110.7 x 2500) +(12.60 x 105 x 1.10) / (302.9 x 2500) = 0.54


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Mcr= [(Pi_T7r2Elyh) / {2(KL)

2}] [1 + 1/20 {(KL / ry) / (h / tf)}2]0'5

=[(1.20 x 7r2x 2 x 105 x 1369.8 x 104 x 500) / (2 x 42502)] [1 + 1/20

{(120.70) / (500 / 17.2)}2]05

= 6.129 x 108 N-mm

Xa = JiAZJrtM,,)

= ^(1.0 x 2074.70 x 103

x 250 / 6.129 x 108

= 0,92

4a = 0.5 [1 + aa. {Xa- 0.2) + V]

= 0.5 [1 + 0.21(0.92 - 0.2) + 0.922]

= 0.999

x* = i / [ 4 + {Ara

-vn= 0.72 < 1.0fu = XJTVY* = -72* 250 ' 1-1 = 168.3 Mftl

J^* = A-^.-/M

= 1.0 x 2074.7 x 1638 x 10 J

= 33.98r-

[ &r= 1.20, &= 1.0,^ = 0.21]

(c) Determination of Mdy (Clause 9.3.2.2)

Mdy = Pb Zpy. fy / ym

= 1.0 x 302.90 x 2500 / 1.1 x 10"5 = 6.88T"m

(d) Determination of Cz (Clause 9.3.2.2) From Table - 9.2,

^=9.15/12.6=0.726

.-. 0^ = 1.8- 0.7x0.726 = 1.292

/.M* = 4(^m - 4) + (Zm -ZJ/Za < 0.9



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= 0.237 (2 x 1.292 - 4) + 0.1469 =

-0.188 For torsional buckling,

fin- = 01S^r- 0.15 < 0.90

Since, /?Mz = /?Afflr^

:.(!. =0.15x1.292x1.359-0.15 = 0.113

Since jutis larger of ^ and^-,^ =0.113

= 1-0.113x16.5/249.65 = 0,993


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-

1

Thebeam(ISMB400

)inFig.1isdes

ignedconsidering it is fullyrestrainedlaterally.

1. WSM (clause 6IS:800 - 1984):

Bending

moment,M=18.75Tm

For ISMB 45

s bc(cai) = 1

*105 / 1350.

= 1388.17 kg /cm2 Fv (

= 26.25T ), the section is safe against shear. For checking of

deflection : a = ( 35000 * 3003 ) / ( 48 * 2.1 * 106 * 45218.3 )

= 0.207 cm = 2.07 mm = L / 1446

Hence O.K.



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Example - 3

Thebeam ,ISMB500 asshowninFig.3is tobedesignedconsideringnorestraintalong the

spanagainst

lateralbuckling .

) : i ) In WSM (clause 6.2, 6.2.2, 6and 6.2.4 and 6.2.4A of IS : 8001984

Bending moment, M = 2.1 * 62

9.45Tm

For, ISMB500 :

L = 600 cm r y = 3.52 cm

T = 17.2 mm Z = 1808.7 c


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0T(DL)+1.1

Xt**

**f+t

+*v+r6.0

6

.3

Fig.3

The

refore,

Y=26.

5*105

/ (600/3.5

2)2=91.2

X

=

9

1.21

+ ( 1 / 20 ){(

1.72 ) / (3.52

50 )}2]

=

1

5

0

.

4

0

a

n

d

f

C

b

=

k

i

(

X

+


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k

2

y

)

C

2

/

C

i

=

X

(

s

i

n

c

e

C

2

=

C

1

,

k

i

=

1

a

n

d

k

2

=

0

)

N

o

w

,

Obc(perm) = (0fcb-fy ) / {fcb" + f

= 746 Kcm2.


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ian Institute ofchnology Madras


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and abc(cai) = 9.45 * 105 / 1808.7

= 522.5 Kg / cm2. Therefore , percentage strength attained is

( 522.5 / 746 ) = 0.7 or 70 %. ii ) In LSM (clause 8.2 of draft IS : 800 ) : For

MB 500 :

D = 500 mm T = 17.2 mm

B = 180 mm t = 10.2 mm

Zp = 2025.74 cm3 ryy = 3.52 cm

Hi = d = 424.1 mm

lzz = 45218.3 cm4 , lyy = 1369.8 cm4 iii) Classification of

section ( ref . Table 3.1 of the code ):

b / T = 90 / 17.2 = 5.2 < 9.4 , hence O.K.

d /1 = 424.1 / 10.2 = 41.6 < 83.90 ( O.K)

Therefore , the section is Plastic . iv ) Check for

torsional buckling ( clause 8.2.2 ) :

tf / tw for ISMB 500 = 17.2 / 10.2 = 1.69 -2.0 Therefore ,

PLT = 1-20 ,for plastic and compact aestions. Mcr =

Elastic critical moment given as :

Mcr = { ( PLT TT2 Ely ) / (KL)2 }[ 1 + ( 1 / 20 ) {KL / ry ) / ( h / tf ) }2 ]05 ( h/ 2 )

( ref . clause 8.2.2.1 ) = (1.20 7r2 * 2 *

106 * 1369.8 / 6002 )[ 1 + ( / 20 ){ (600 / 3.52) / (50 / 1.72)}2 ]05 (50 / 2)

= 371556.3 Kg-cm .

Now ,Xn= * ( Pb Zp .fy / Mcr ) = 1.1675

( since Pb = 1.0 for plastic and compact sections )

Therefore ,



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4>LT = 0.5 [ 1 + CXLT ( ^LT - 0.2 ) +XLT2]

= 0.5 [ 1 + 0.2 ( 1.1675 - 0.2 ) + 1.16752 ] =

1.283 ( CXLT = 0.21 for rolled section ).

Therefore XLT = 1 / [ 4>LT + { (!>LT2 - ^LT2 }0.5 ]

= 1 / [ 1.283 + { 1.2832 - 1.16752 }0.5 ] = 0.55 Md =

1.0 * 0.55 * 2025.74 * 2500 / 1.10 * 10-5

= 25.32Tm Now actual moment is obtained

as follows: Factored load = 1.0 * 1.5 + 1.1 * 1.5

= 3.15TmFactored moment = 14.175Tm < Md

Percentage strength attained is (14.75 / 25.32 ) = 0.56 or 56 % which is less

than 70 % in case of WSM.

Hence , the beam is safe both in LSM and WSM design but percentage

strength attained is comparatively less in LSM for the same section

Problem:

The girder showed in Fig. E1 is fuly restrained against lateral buckling throughtoutits sapan. The span is 36 m and carries two concentrated loads as show in Fig. E1.Design a plate girder.

Yield stress of steel, fy = 250 N/mm2

Material factor for steel, ym = 1.15

Dead Load factor, yfd = 1.50

Imposed load factor, yfl = 1.50


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d

:

U

n

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o

a

d

,

w

d

=

1

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k

N

/

m

C

o

n

c

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t

r

a

t

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l

o

a

d

,

W

i

d

=

1

8

0

K

N

Concentrated load

W2d

= 180 KN

Live load:

Uniformly distribut

load, W = 35 kN


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Co

nc

ent

rat

ed

loa

d,

Wn

=

40

0

kN

Co

nc

ent

rat

ed

loa

d,

W2|

=

40

0

kN

Fa

cto

red

Lo

ads

w' = Wd * Yfd + wi

= 18 * 1.5 + 35 * 1

79.5 kN/m

Wi = Wid *yfd + W

*yf| = 180 * 1.5 + 4

1.5 = 870 kN

W2 = W2d * yfd + * yfi = 180 * 1.5 + 41.5 = 870 kN



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2.0 Bending momand shear force

Be

UDL effectw----dolG

Concentrated load effect

Total

The desigh shearforces and bendingmoments are showFig. E2.

3.0 Initial sizing oplate girder

Depth of the plate

girder:

Th

e

recomm

ended

span/dep

th ratio

for

simply

supporte

d girder

varies

between

12 for

short

span and

wo for


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s

i

d

e

r depth

of the

girder as

2600

mm.

1 = 36000=13S

d 2600

Depth of 2600 mm

acceptable.

(For

drawing

the

bending

moment

and

shear

force

diagram

s,

factored

loads

are

conside

red)


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Fig.EBendmomand

sheaforcediag

8

7

0

k

X

9000mm

?

9

.

5

k

X/

m

SH

OO

O

mm

I^^iS

cV"WkX

WH

Omm

36001 mm


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N

-

i

n

Fi

Be

g

mo

an

sh

fo

dia

ms

Flange:

P

y

=

2


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5

0

/

1

.

1

5

=

2

1

7

.

4

N

/

m

m

2

S

i

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g

l

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f

l

a

n

g

e

a

r

e

a

,

* d^ 2600*

By thumb

rule, the

flange

width is

assumed

as 0.3

times the

depth of

the

section.

Try 780 X

50 mm,

giving an

area =

39000

mm2


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Web:

Minimum web thickness for plate girder in builidings usually varies between 10

mm to 20 mm. Here, thickness is assumed as 16mm. Hence, web size is 2600 X

16 mm

4.0 Section classification

Flange:

7.6

T 50 Hence, Flange is COMPACT SECTION.

b:

2600 ,-.------= 162.5 > bis

250

nce, the web is checked for shear buckling.

Checks

Check for serviceability:

d 2600

250 Web is adequate for serviceability.


B;t=780ll6=3322

0.4mm


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eck for flange buckling in to web:

uming stiffener spacing, a > 1.5 d

i i2600

X234

Since, t (=16 mm) > 8.2 mm, the web is adequate to avoid flange buckling

into the web.

Check for moment carrying capacity of the flanges:

The moment is assumed to be resisted by flanges alone and the web

resists shear only. Distance between centroid fo flanges, hs = d + T = 2600 + 50 =

2650 mm

Af = B * T = 780 * 50 = 39000 mm2

Mc = Pyf * Af * hs = 217.4 * 39000 * 2650 * 10-6 = 222468.3 kN-m

> 20709 kN-m Hence, the section in

adequate for carrying moment and web is designed for shear.

6.0 Web design

The stiffeners are spaced as shown in Fig.E5. Three different spacing

values 2500, 3250 and 3600 mm are adopted for trail as shown in fig. E5.

End panel (AB) design:

d=2600 mmt = 16 mm

Maximum shear stress in the panel is

7.4'

mtf


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52/77


/ flll *1[L --=i =---------------= 55.3 N/mm

A 2600*16

a 2500

!-*5-i5

Calcualtion of critical stress,

q((wfcen a/d < 1) = [0 75+ 1/ (a/d)3] [l000/(d/t)]2 = [o.75+

l/(O.96)2l[l0OO/(162.5)V

= 69.5 N/mm2

Slenderness parameter,

K = [0.6(f^)/qe]

= [0.6(250/1.15)/69.5] =

1.37 > 1.25

Hence, Critical shear strength (qcr= qe) = 69.5 N/mm2

Since, fv < qcr (55.3 < 69.5)

Tension field action need not be utilised for design.

Checks for the end panel AB:

End panel AB should also be checked as a beam (Spanning between the

flanges of the girder) capable of resisting a shear force Rtf and a moment Mtfdue

to anochor forces. (In the following calculations boundary stiffeners are omitted for

simplicity)


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Check for shearcapacity of the enpanel:

tin ~ U. /. 0.6P

*a

*

2

2

A

.

-

t.

a

-

1

6

*

2

5

00

-

4

0

0

0

m

m3


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P

O.eP^A

, -0

6*(250/

1.15)*4

0000/1

000-

5217

kN

Since,

Rtf < P

, the

end

panel

can

cany

the

shear

force.

Checkformoment

capacityofendpanelAB:

Had

10

4636*2600 3=--------------*10_3 = 1205kN-m

10

,--*-12502 2

I = taJ = *16*25003 =20S3a:l07 mm*12 12

1 ?0Ri*in7

Ma = -py = *f250/1.15)*l(3623kN-m

q y 1250

Since,M^< M

Th

e

en

d

pa

nel

ca

n

car

ry

the

be

ndi

ng

mo

me

nt.

Design of panel B

P

a

n

(1205 < 3623)


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a

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t

i

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d

=

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6


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m

m

;

t

=

1

6

m

m

ian Institute of

chnology Madras


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_FVE _2102.3*103 '

~d7 2600*16

a _

3250 1) = [l. 0 + 0

(a/d )21 [l 000

t)]*

= [l.O + 0.75 /

(1.25)*] [1000/

(162.5)f

Slendemesspa

er, \

Hence, Critical

shear stength= [0.6(25

15)/56 0]

= 2.33 > 1=

1.25


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0

m

(qI

I =

qe)

1.5qCT _

1.5*56.0

=525

V'+C'-

25)2

yt =

(P^2

-

3q,a

_

V)5

" A"

(21

7.42

- 3

*56

02 +

52

52 )

* -

52.

5 =149

*

+|

l*U

=

=5

6.

0+

-

:

Fl2

5 +Jl +

(l2

5)3

B

2.1

N

/m

mI

e

,

(82.

1>50.

5)

BCissafe

againstsh

earbucklin

g.


Hi


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7.0 Design of stiffeners Load

bearing stiffener at A:

Design should be made for compression force due to bearing and moment.

Design force due to bearing, Fb = 2301 kN

Force (Fm) due to moment Mtf, is

M^=12O5,103=482kNa s

2500

Total compression = Fc = Fb + Fm = 2301 + 482 = 2783 kN

Area of Stiffener in contact with the flange, A:

Area (A) should be greater than

Q8F 10241 mm2 Bearing

check is ok.

Check for outstand:

Outstand from face of web should not be greater than

20t,e

150y


61/77

an Institute of Technology Madras Outstand bs = 250mm < 20 tss( =20*25*1.0=500) bs =250 mm < 13

t(s(- 13.7*25*1.0 = 342.5)


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Hence, outstand

criteria is satisfied

Check stiffener fo

buckling:

The effective stiffe

section is shown in

E3

b*

=

2

7

0

m

m

16 mmt'---^t--r-'i

f"Tb, -

270

mm

Stiffener

Fig. E3 End beating stiffener

-~&:~.ib

i.


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Thebucklingresistancedue

toweb isneglectedhereforthe

sakeofsimplicity.

= LJ JJU-_*25*163 =35807*10* mm4

^12

12A. = Effective

area = 270 *25 *2 = 13500 mma

07*1

3500

= 162.8

mm

F

l

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t

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f

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r

,

t

h

e

n

I

e

=

0

.

7

1

=

0

.

7

*

2

6

0


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0

=

1

8

2

0



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^ 162.8 For fT = 250 N/mm1

and A=11.2 of = 250N/mm2

from table(30 of chapter on axially compressed columns

Buckling resistance of stiffener is

Pc = acA/Tm - 250*1350/1.15 - 2935fcN

Since Fc < Pc (2783 < 2935)

Therefore, stiffener provided is safe against buckling.

Check stiffener A as a bearing stiffener:

Local capacity of the web:

Assume, stiff bearing length b1=0

n2 = 2.5 * 50 * 2 = 250 BS 5950 : Part -1, Clause 4.5.3

Pcrip = (b1 + n2) tPyw

= (0 + 250) * 16 * (250/1.15) * 10-3 = 870 kN

Bearing stiffener is designed for FA

FA = Fx = Pcrip = 2783 - 870 = 1931 kN

Bearning capacity of stiffener alone

PA = Pys * A = (50/1.15) * 13500/1000 = 2935 kN

Since, FA < PA (1931 < 2935)

The designed stiffener is OK in bearing. Stiffener

A - Adopt 2 flats 270 mm X 25 mm thick



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sign of intermediate stiffener at B:

fener at B is the most critical one and will be chosen for the design. Minimum Stiffness

1, > 0.75dt3 fora > cK/2s 0.75dt3 r pr

\ >------jFor a < dV2a

dV2=V2*2600 = 3677mm

a/2 (3250 < 3677)

Conservatively ' t ' is taken as actual web thickness and minimum ' a ' is used.

-1022*10* mm4

a3

32502

intermediate stiffener of 2 flats 120 mm X 14 mm

14*2563 14*163(I) . = -11-^-= 1957*10* mm4

\ s'St umifA ip IT

1 5dV1^ > j.the section stab sfied mini mum stiffen ess requirement a

eck for outstand:

stand of the sti ffen er < 13.7 i%s 13.7tss 13.7*14 "1.0 = 192 mm

stand = 120 mm (120 < 192)

nce, outstand criteria is satisfied.

ckling check:

fener force, Fq = V - Vs Where V = Total shear force Vs = Vcrof

web a / d = 3600 / 2600 = 1.38


d3t3 1 5*26003*163


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d /1 = 2600 / 16 162.5

Elastic critical stress

(when afa> 1)- ft 0 75/(a/d)a] TlOOO/(d

= 1.0 + 0.75/

(1.38) [1000/

(162.5)1

Slendemess parameter, X

Hence, Criticalshear strength =

[0.6(250/l.l5

/52.8f =

2.47> 1.25

= 52N/mm= qe)

Vcr= qCrdt = 52.8 *2600 * 16 * 10"3 = kN

Buckling resistanof intermediatestiffener at B:


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70/77


1979

*104

13600

It =0.7*2600 = 1820

X = i = l^ = 48.0 * 38 1

For fy = 250 N/mm2 and *=48.0 From table3 of chapter on axially compressed

umns, ac =213.2 N/mm1

ckling resistance = (213.2/1.15) * 13600 * 10-3 = 2521 kN

ear force at B, VB = 2301-{(2301 - 1585.5)*(2500/9000)] = 2102 kN

fener force, Fq = [21201 - 2196] < 0

< Buckling resistance

nce, intermediate stiffener is adequate

ermediate stiffener at B - Adopt 2 flats 120 mm X 14 mm

ermediate stiffener at E (Stiffener subjected to external load): Minimum stiffness calculation:

a = 3600 d^ = 3677

a < d^ (3600 < 3677)

I >lJdV = 1.3-2600^163 =833,1Q4

a2 3600*

intermediate stiffener 2 flats 100 mm X 12 mm thick


i_ * ,i( niA1j(r1 64(1*11^ 14*16*14-2563 + ^lil-^-il = 1979*10* mm4

12 12 12

240*14+6-10*16 = 13600 mmJ

*


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\ ^f?

l

P'nce,

ckling Check:

VV

-vfV

vs " VB = qCTdt =52.8*2600*16*10-* = 2196 kN

Fq is negatiue and Fq - Fx =0

Ms = 0

Fx = 870 kN

ckling resistance of load carrying stiffener at D: (Calculation is similar to stiffener at B)

20 tw =20*16 = 320 mm

,. - 1*12*216' 2 -H_1ii . ,029-10-

^ 12 12 12A = 200*12 + 640*16 =12640 mm2

.1

2 =28 5

12640

I, =0.7*2600 = 1820

X = = l = 63.9 t 28.5

Forfy = 250 N/mm3 and A=63 9

m table3 of chapter on axially comperssed columns, a, = 180 N/mm

3

ckling resistance, Px (180/1.15) * 2640 * 10-3 = 1978 kN. Fx/Px = 870/1978 = 0.44 S33 *10+

9*10*


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Hence, Stiffener aOK against bucklin

S

t

i

f

f

e

n

e

r

a

t

D

-

A

d

o

p

t


73/77

f

l

a

t

s

1

0

0

m

m

X

1

2

m

m

t

h

i

c

k

W

e

b

c

h

e

c

k

b

e

t

w

e

e

n

s

t

i

f

f

e

n

e

r

s

:


74/77

f

< P

fed=w1 /t =79.

5 /16=4.97N/mm2

When

compressionflangeisrestraine

dagainstrotationrelativeto

theweb

2.75+ .

(a

/d

)

3.79*

20

00

026

4

0

6

(a

n

y

= 28.7

N/mm3200000

2600j \J

Since,

fed^ed

[4.97

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