L18 Plate Girder

7/31/2019 L18 Plate Girder

1/21

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Timber and Steel DesignTimber and Steel Design

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

LectureLecture 1188 BuiltBuilt--up Beamsup Beams


2/21

Plate Girder Cross SectionsPlate Girder Cross Sections

(a) Welded

(b) Riveted without Stiffeners

Flange plate

Flange angle

Stiffener angle

Web plate

Filler plate

(rivets not shown)

(c) Riveted with Stiffeners

Typical components of riveted plate girder

Cover plates

Flange

angles

Flange angles

Intermediate

stiffener

angles

Web

Flange angles

Fillerplate

Endstiffener

anglesWeb

(a) Cross-section (b) Elevation at end of span


3/21

Typical components of welded plate girder

Flange plates

Intermediate

stiffener

plate

Web

Bearing

stiffenerplates

Webplate

(a) Cross-section (b) Elevation at end of span

Cover-plated Beams

d

tp

tp

2

22

reqd s

dI I A

= +

22 ( / 2)

/ 2reqd s

s

A dS S

d

S Ad

= +

= +


4/21

1 7 . 1 6 0 . 1 , 4 0 0 ../ .2 W 6 0 0 1 3 7

12 t/m

7 m

W600137 (I= 103,000 .4 S= 3,530 .3)

=180 ../

M = (12.18)(7)2/8 = 74.6 -

Sreqd = (74.6105)/1,400 = 5,329 .3

< Sreqd NOT ENOUGH

58.2 cm 59.0 cm59.8 cm 5,329 .2 OK

W600 137 PL8 450 ..

8 ..

Sreqd = Ss +A d

5,329 = 3,530 +A (59.0)

A = 30.5 .2

PL 8 450 .. (A = 36 .2)


5/21


6/21

231 2

2 212

2 2

ffw

reqd

f f

thAt h

Sh h

t t

+ = +

+ +

:

fv= 4 3 , 0 0 0 / ( 1 . 5 1 5 0 ) = 1 9 1 ../ .2

Fv= 0 .4 2 ,5 0 0 = 1 ,0 0 0 ../ .2 > 191 ../ .2

Since h/tw< 260 (AISC) and web shear stress < max. value

Stiffeners are not needed

:

Sreqd = 215 x 105 / 1,650 = 13,030.3

OK

( )

2

3 146 21 2(1.5) 1462 21213,030

146 1462 2

2 2

fA + = +

+ +

tf= 2.

Af == 53.7.2 ( 22 8 .)

1 .5 1 4 6 . 22 8 .


7/21

Riveted or bolted

WT

WT

PL

Welded Welded box girder

PL girder (may be for

full depth of story)

PLATE GIRDERSPLATE GIRDERS


8/21


9/21

Allowable Shearing StressAllowable Shearing Stress

yv

y

v FCF

F 40.089.2

=where

8.0when/

585,1

or8.0when)/(

000,165,32 >=+=


10/21

If properly spaced and proportioned intermediate stiffeners are used,

tension field action can be accounted for, and the allowable shearing

stress may be increased to:

yv

v

y

v Fha

CC

FF 40.0

)/(115.1

1

89.2 2

+

+=

provided that Cv 1.0

(AISC Eq.1.10-2)

If intermediate stiffeners are not used, use Fvfrom (AISC Eq.1.10-1) since

there will be no tension field.

Intermediate StiffenerIntermediate Stiffener

AISC also requires that h/tw< 260 if intermediate stiffeners are not used.

If intermediate stiffeners are deemed to obtain Fv from (AISC Eq.1.10-2), the

clear spacing amust be such that

a h3and/

2602

h

a

thh

a

w


11/21

The minimum cross-sectional area of an intermediate stiffener

wv

st thDYha

ha

h

aCA

+

=

2

2

)/(1

)/(

2

1

whereAst = total cross-sectional area of the stiffener

D = 1.0 for stiffeners in pairs (angles or plates)

= 1.8 for single-angle stiffeners

= 2.4 for single-plate stiffeners

stiffenerof

webof

y

y

F

FY =

The minimum moment of inertia:

4

50

=

hIst

A

A

B

B

B - BIntermediate stiffener

Bearing StiffenersBearing Stiffeners

A - A

Bearing stiffener

required when the web has insufficient strength of web yielding,

web crippling, or sidesway web buckling.

Stiffener

PLs

Corners are cut to avoid

flange-to-web weld.Width outside of fillets

Chamfer

Filler PLs

Stiffener L


12/21

Column Stability Criterion Bearing stiffeners are designed as the columnconsisted of the stiffener plus web portion.

End stiffener

Flange

t

End of

girder Web

w

12 tw

tw

Intermediate stiffener

25 tw

x

0 < x< 1.25 cmapproximately to

edge of flange

Bearing stiffener

cross section

Portion of web to support load: 12twat girder ends

25twat interior concentrated loads

Effective length of bearing stiffener columns: KL = 0.75 h

Example 9.1 Investigate the plate girder shown in the figure below. The uniform load

of 6 t/m includes the weight of the girder. The compression flange has lateral support

at the ends and at the points of application of the concentrated loads. There are no

intermediate stiffeners, and A36 steel is used throughout.

6 t/m (includes girder wt.)

18 ton18 ton18 ton

A B C DE

3.5 m 3.5 m 3.5 m 3.5 m

14 m

15 cm

Detail at A & E Detail at B, C, DTypical section

PL 2.5x40 cm

PL 2.5x40 cm

150 cmPL1.5x145 cm


13/21

SolutionSolution

Allowable bending stress:

y

40 cm

2.5 cm

145/6

= 24 cm

Compute the radius of gyration rT:

4

33

cm340,13

12

)5.1(24

12

)40(5.2

=

+=yI

2cm136)5.1(24)40(5.2 =+=A

cm90.9136

340,13===

A

Ir yT

4.3590.9

)100(5.3 ==TrL

==< 6.53

500,2)1(10173,710173,7

33

y

b

FC

2kg/cm500,1)500,2(60.060.0 === yb FF

Local stability of web:

==


14/21

Bending stress:

Flange: PL 2.5 cm x 40 cm

Web: PL 1.5 cm x 145 cm

Moment of inertia of the girder cross section:

4

23

cm890,468,1

2

5.2145)40)(5.2(2

12

)145(5.1=

++=I

25

kg/cm394,1890,468,1

)2/150(10273=

==

I

Mcfb

2kg/cm500,1=< bF OK

Shear stress:2

3

kg/cm317)5.1(145

1069=

==

w

vth

Vf

41.2145

350

==h

a

a= 3.5 m h= 145 cm

Allowable shear stress:

41.2145

350==

h

a

=

=

< 23.7

5.1/145

260

/

26022

wth3 1, 03.6)41.2(

434.5

)/(

434.5

22=+=+=

hakv

817.0)5.1/145(500,2

)03.6(000,165,3

)/(

000,165,322

===wy

vv

thF

kC > 0.8 NG

805.0500,2

03.6

5.1/145

585,1

/

585,1===

y

v

w

vF

k

thC > 0.8 OK

[ ]ksc000,140.0ksc753)41.2(115.1

805.01805.0

89.2

500,22

==

+

+= yv FF

fv = 317 ksc < Fv OK


15/21

Plate Girder DesignPlate Girder Design

1. Select the overall depth:

Well proportioned girder will have a depth of 1/10 1/12 span length

2. Select a trial web size:

Web depth h = Overall depth 2 flange thickness

Web thickness twcan be found by satisfying local stability requirements:

yw Ft

h 360,6 To prevent web buckling due to flexural compression

and either

)160,1(000,984+

yyw FFt

h To prevent vertical web buckling

or

yw Ft

h 670,16 If transverse stiffeners spacing 1.5h

3. Estimate the flange size:- Estimate girder weight

- Compute maximum moment

h

Af

Af

tw

The required flange area can be estimated from:

2

3

flangeweb

22

12

1

Let

+

+=

hAht

III

fw

hAht

h

hA

h

htS f

wfw +=+=62/

)2/(2

2/

12/ 223

Equating required section modulus to S:

0.60f

y

MA

F hAssume Fb= 0.60Fy:

hAht

SF

Mf

w

b

+==6

2

6

ht

hF

MA w

b

f =

yf

f

Ft

b 795

2

Select bf

& tf

from


16/21

1 7 . 3 1 5 2 .2 /

3 2 1 / 3

A3 6

32 ton 32 ton2.5 ton/m

5 m 5 m 5 m50.75t 50.75t

50.75t

50.75t

38.25t

6.25t

38.25t

6.25t

222.5t-m 222.5t-m230.3t-m

= 300 ../ ,

6,360 6,360164

1,500w b

h

t F= = =

( )

984,000 984,000325

2,500(2,500 1,160)1,160w yf yf

h

t F F = =

++

142w

h

t=

: = ( 1 / 1 0 ) ( 1 5 0 0 ) = 1 5 0 .

= 1 5 0 8 = 1 4 2 .

h/tw

tw= 142/164 = 0.866.

1 .5

tw= 142/325 = 0.437 .

tw= 0.8.

1 1 4 2 . ( Aw= 1 4 2 .2 )


17/21

:

= ( 2 3 4 0 + 1 1 4 2 ) ( 7 , 8 5 0 ) / 1 0 02

= 2 9 9 .8 7 3 0 0 ../ OK

Use 3 x 40 cm2cm108)142(5.1

)100(3.230 ==hF

MA

b

f

67.6)3(2

40

2==

f

f

t

b

==< 9.15

500,2

795795

yFOK

:

142 cm

1 cm

3 cm

148 cm

3 cm

40 cm

No reduction in allowablebending stress Fb

I = (1/12)(1.0)(142)3+(2)(3x40)(72.5)2

= 1,500,107.4

S = 1,500,107/74 = 20,272.3

fb

= (230.3x105)/20,272

= 1,136 ../ .2

==


18/21

: ( )

: V= 50.75 tons

ksc357)1(142

)000,1(75.50===

w

vth

Vf

Tension field actionAISC Eq. (10.1-1)

A B

BFlange force

Stiffener force

Tension field force

32 ton 32 ton

2.5 ton/m

5 m 5 m 5 m

50.75t 50.75t

38.25t

6.25t

yv

y

v FCF

F 40.089.2

=

a/h= 500/142 = 3.52

a/h> 1.0

( )2 2

4.00 4.005.34 5.34 5.66

3.52/vk

a h= + = + =

( )

2 2

3,165,000 3,165,000(5.66)0.355

(2,500)(142)/

vv

y w

kC

F h t

= = =

2,500(0.355)

2.89 2.89

307 ksc 357 ksc

y

v v

FF C= =

= < NG

USE intermediate stiffeners

: a/h= 500/142 = 3.52


19/21

a/h= 250/142 = 1.76

2.5 V= 50.75 2.5(2.5) = 44.5

fv= (44.5)(1,000)/(142) = 313.4../ .2

.4 Fv= 357 ksc h/tw= 142,

:

32 ton

2.5 m

50.75t

38.25t

6.25t

CL

2.5 t/m

a/h= 1.81 a= 1.81(142) = 257 . (2.5)

No tension

field action

( )2 2

4.00 4.005.34 5.34 6.63

1.76/vk

a h= + = + =

( )2 2

3,165,000 3,165,000(6.63)0.416

(2,500)(142)/

vv

y w

kC

F h t= = =

2,500(0.416) 360 ksc 313.4 ksc

2.89 2.89

y

v v

FF C= = = > OK

No need for additional intermediated stiffeners

fv= (38.25)(1,000)/(142) = 269./ .2

Fb= (0.825 0.375(269/360))(2,500)

= 1,362./ .2 > 269./ .2 OK

32 ton

5 m

50.75t

38.25t

6.25t

CL2.5 t/m

:

22

2

1 0.416 1.761.76 (142)(1.0) 9.53 cm

2 1 1.76stA

= =

+

0.8 x 10. ( Ast= 16.2 )

OK

0.8 cm

10 cm

1 cm

Web

=


20/21

4 4

4142Min 65.05 cm50 50

st

hI

= = = Moment of inertia:

0 .81 21 3 6 .

= 142 0.6 6(1) = 135.4.

:

3 cm

0.6 cm

k=3.6 cm32,000

( 5 ) 1.0(0 5 3.6)

1,778 ksc 0.66 1,650 ksc

w

y

R

t N k

F

=+ +

= > =

Check web yielding for interior loads

Bearing stiffeners are required

( )

44

23

2

cm05.65cm617

stiffeners25.05)10(8.012

)10(8.0

provided

>=

++=

+= AdIIst

c

c 4tw to 6tw

:

fa = / Aeff = 50.75(1,000)/48

= 1,057./ .2 < 1,457./ .2 OK

88.748

979,2

=

I= ( 1 / 1 2 ) ( 1 .2 ) ( 3 1 ) 3 = 2 ,9 7 9 .4

Aeff= ( 2 ) ( 1 5 ) ( 1 .2 ) + ( 1 2 ) ( 1 ) = 4 8 .2

r= .KL = ( 0 .7 5 ) ( 1 4 2 ) = 1 0 6 .5 .

KL/r= 1 0 6 .5 / 7 .8 8 = 1 3 .5 2

Fa = 1 ,4 5 7 ./ .2 .1

Web

12 mm

Aeffshown shaded15 cm

1 cm

15 cm

31 cm

12tw = 12 cm

2 PL 1 .2 1 5 .

2 PL 1.2 x 15 x 140.


21/21

2.5 m 2.5 m 2.5 m 2.5 m 2.5 m 2.5 m

5.0 m 5.0 m 5.0 m

2 PL 0.8 x 10 cm2 PL 1.2 x 15 cm 2 PL 1.2 x 15 cm

FINAL DESIGN

Documents

L18 Plate Girder