l18 plate girder
TRANSCRIPT
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< Cover-plated Beams
< Built-up W Sections
< Plate Girder
< Stiffeners
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
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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
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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
= +
= +
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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)
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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 .
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Riveted or bolted
WT
WT
PL
Welded Welded box girder
PL girder (may be for
full depth of story)
PLATE GIRDERSPLATE GIRDERS
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Allowable Shearing StressAllowable Shearing Stress
yv
y
v FCF
F 40.089.2
=where
8.0when/
585,1
or8.0when)/(
000,165,32 >=+=
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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
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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
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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
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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:
==
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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
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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
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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 )
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:
= ( 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
==
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: ( )
: 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
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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
=
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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.
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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