CE 591 – Advanced Structural Steel Design

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CE 591 – Advanced Structural Steel Design
CE 591
Advanced Structural Steel Design
Fall 2013
Lecture 7: Plate Girders; Design
 Rules of Thumb
 Flange-to-web weld
 Design Aids
 Design Example
Proportioning the Section

Goals

Satisfy limit states



Strength
Serviceability
Minimum cost


assume cost proportional to
weight of steel
but remember that least-weight
may not provide the most cost
effective design!
Rules of Thumb

Span-Depth Ratio

Salmon & Johnson, Steel Structures, 4th ed.
L
 10 to 12
d

bf
Rules of Thumb, MSC 2000
L
 15
d
L
d
Modern Steel Construction, 2000
“Design It Like You Are Going to Build It”
Karl Frank
2013 NASCC Educator Session: Bridge Design for the
Classroom
media.aisc.org/NASCC2013/ES1.mp4

Recommendations for composite plate
girders for bridges
“Design It Like You Are Going to Build It”
Karl Frank
2013 NASCC Educator Session: Bridge Design for the
Classroom
media.aisc.org/NASCC2013/ES1.mp4

Recommendations for composite plate
girders for bridges
AASHTO Cross Sectional Limits
Similar to AISC
eq. F13-2
Not stable if
outside limit
Rules of Thumb – Flange Width
bf
bf
 0.2 to 0.3
d
d
“shallow
section”
“deep
section”
Optimum Depth (another option)

Based on minimizing weight (i.e. gross
cross-sectional area), supposing no depth
restriction
f is average stress on flange (i.e. Fcr)
 bw is an assumed constant h/t
 bw of 320 for ‘optimum’ proportion A36
 C1 is a factor to account for reducing
flange size at regions of lower moment
 C2 is factor to account for reducing web
thickness at regions of reduced shear

Optimum Depth, cont’d
h3

3MC1bw
f (3C 2  C1)
Suppose C1= C2= 1 (i.e. no section
reduction in regions of lower stress)
3Mbw
h3
2f
Rules of Thumb – Flange Area
bf
Mu/f
Af
C
Af
d h
Aw
T

M Aw

fh 6
Average
stress on
flange
Sx / h
Rules of Thumb – Plate dimensions

Plate widths


2” increments
Stiffener spacing

3” multiples
Rules of Thumb – Plate thickness
Increments
(inches)
1/16
Range
(inches)
t ≤ 9/16
1/8
5/8 ≤ t ≤ 1-1/2
1/4
t > 1-1/2
Rules of Thumb – Flange Plates, p. 1

Based on minimum volume (weight) and Af
equation used earlier
Mmax
L/2
L
Groove weld
Af1
Af
Af 1 1

Af 2
Rules of Thumb – Flange Plates, p.2
Mmax
L/3
L
Af 1 5

Af 9
Groove weld

Af1
Af
Unless save 200 – 300 lbs of
material, added cost of weld
makes flange plate
transition uneconomical
(Salmon et al., Steel Structures, 5th ed.)
Other flange plate recommendations

If concerned about LTB, keep bf /2tf at
about the lp value in maximum moment
regions


Could then reduce flange thickness in low
moment regions (reduce thickness instead of
width)
No LTB? Reduce width if desired



“slight advantage in fatigue strength”
Transition slope should be less than 1 in 2-1/2
for either width or thickness change
1 in 4 to 1 in 12 recommended for change in
width
2013 NASCC Educator Session: Bridge Design for the Classroom
media.aisc.org/NASCC2013/ES1.mp4
Weld of flange to web

X
X
Must provide for factored
horizontal shear flow
Vu Q
shear flow 
Ix
(kips/in)
1st moment
of area
of flange
about
neutral axis
Weld of flange to web, p.2

DOTs typically require SAW
for these welds


X
X
“more thermally efficient”
More uniform weld cross section
and strength


no stops/starts and other
irregularities that concentrate
stress
SMAW uses stick electrodes of
limited length and diameter
Rules of Thumb – Web

“Reasonable range” for web stress
fVn
Aw


 12 to 16 ksi
< 9 ksi? May be able to use thinner web
Practical minimum web thickness (tw)

5/16”
Design Aids – Shear, Stiffeners

Tables 3-16a, 3-17a (without TFA)

36 ksi and 50 ksi steel

Starting on AISC Manual p. 3-152

Tables 3-16b, 3-17b (with TFA)

fvVn/Aw graphed as a function of h/tw and a/h
NOTE: here, Aw = d tw (AISC G2)
---- means
exceeded
“practical limit”
on
stiffener spacing
a  260 

 
h  h tw 
a/h > 3.0
kv = 5.0
2
Corresponds approximately
to limit for vertical
flange buckling
Plate Girder Design Example
Consider a simply supported plate girder that carries the factored
uniform load and two concentrated loads as shown. Design a
doubly symmetric, non-hybrid girder with A36 steel. Assume
that lateral support is provided at the ends and at the
concentrated loads. 1.2D + 1.6L load combination controls.
Factored loads shown. L/360 deflection limit (total service load).
150 k
150 k
5 k/ft
A
24'
330 k
B
24'
C
Lateral Support
24'
D
Lateral
Support
Design Example, p. 2
150 k
150 k
5 k/ft
A
24'
B
330 k
24'
C
24'
D
Lateral
Support
210
60
Shear
(kip)
60
210
330
Moment
(ft-kip)
6480
6480
6840
Design Example, p. 3a

Sizing the Section, try ‘Rule of Thumb’
L
 10 to 12
d


Try h = 72”
Try tw = 5/16”
(72 ft )(12 in / ft)
 72 in
12
h
72"

 230
tw 5 / 16"
Design Example, p. 3b

Sizing the Section, try formula for h
3Mbw
3( Mu / f ) bw
h3
3
2f
2 RpgFcr
3(6840kipft / 0.9)(12in / ft )(320)
h3
2(1.0)(36ksi)
 107in
Design Example, p. 3c
h 107"

 342
tw 5 / 16"

Try h = 107”
Try tw = 5/16”

AISC F13.2 for a/h ≤ 1.5

29000ksi
h
 341
   12.0
36ksi
 tw  max


a/h > 1.5 limit will be even smaller …
Also, web substantially heavier than for h=72”
with practical minimum web thickness
 So, try h=72”
Design Example, p. 4

Slender Web?
E
lr  5.70
Fy
Table B4.1b (Case 15)

29000ksi
lr  5.70
 162
36ksi
h
 230  lr
tw
Check shear stress (recommended)
fVn
Vu
330kips

 14.7
 12 to 16 ksi 
5
A
w
(
72
"
)(
")
Aw
16
Design Example, p. 5

Double check AISC Limitations, h/tw = 230
h
E
 12.0
tw
Fyf
a
for  1.5
h
29000 ksi
 12.0
 341
36 ksi
h 0.40 E
a

for  1.5
tw
Fy
h
0.40(29000ksi)

 322
36ksi
(F13-3)
(F13-4)
√ OK, Limitations
satisfied
Design Example, p. 6

Estimate Flange Size
Mu
Aw
M Aw
Af 



fh
6 fb RPG Fcr h 6

Assume RPG = 1.0, Fcr = Fy
6840 ft  kip(12in / ft ) 72" (5 / 16" )
2
Af 

 31.4in
(0.9)(1.0)(36ksi)(72" )
6
Design Example, p. 7


bf
Possible flange dimensions
tf (in)
bf (in)
Af
1.375
24
33.0
8.72
1.25
26
32.5
10.4
1.125
28
31.5
12.4
(in2)
bf /2tf
tf
Use FLB compactness limit to help
choose size (optional) – Table B4.1b
E
29000
l p  0.38
 0.38
 10.8
Fyf
36

Flange width rule
of thumb
bf
26"

 0.34
d (72"2(1.25" ))
> 0.3
Design Example, p. 8a

LTB – Flexural Capacity; Lb = 24 ft
1
1
5 3
3
(
26
)(
1
.
25
)

(
12
)(
)
bf =26”
12
16  7.1 in
rt  12
5
(26)(1.25)  (12)( )
tf =1.25”
16
hc/6
tw
=5/16”
=72”/6
=12”
Lp  1.1rt
(F4-7)
E
Fy
Lr   rt
E
0.7 Fy
(F5-5)
29000ksi
Lp  1.1(7.1in )
 222in  18.5 ft
36ksi
29000ksi
Lr   (7.1in )
 757in  63.1 ft
0.7(36ksi)
150 k
150 k
5 k/ft
Design Example, p. 8b
Lp = 18.5 ft < Lb = 24 ft < Lr =24'63.1 ft 24'
B
A
Lb  Lp
Fcr  Cb[ Fy  (0.3Fy )(
)]  Fy
Lr  Lp
330 k
~5% difference in
Mu within Lb –
assume constant
moment
24'
210
60
Shear
(kip)

C
Lateral
Support
D
(F5-3)
60
210
330
Moment
(ft-kip)
6480
6480
6840
24  18.5
Fcr  (1.0)[36  (0.3(36))(
)]  36ksi
63.1  18.5
Fcr  34.5ksi
Design Example, p. 9

Calculate Section Modulus, etc.
1 5
72 1.25 2
1
( )(723 )  2(26)(1.25)( 
)  2( )(26)(1.253 )
2
2
12
S x  12 16
72
(  1.25)
2
 2602 in 3
hctw
aw 
bfctc
(F4-12)
(72" )(5 / 16" )
aw 
 0.69
(26" )(1.25" )
<10
√ OK
26”
1.25”
72”
5/16”
Design Example, p. 10

Calculate Flexural Capacity
aw
hc
E
Rpg  1 
(  5.7
)  1.0
1200  300aw tw
Fy
(F5-6)
0.69
72"
29000ksi
 1
(
 5.7
)  0.97
1200  300(0.69) 5 / 16"
36ksi
fM n  fb S xc Rpg Fcr
(F5-2) and AISC F1
 0.9(2602in 3 )(0.97)(34.5ksi)  78368kip  in  6531kip  ft
< Mu = 6840 kip-ft N.G.
Design Example, p. 11
28”
TRY 28” x 1.5” flange
 Recalculate properties:

1.5”
Af  42 in 2 bf  28"  9.33  lp  10.8
rt  7.74in
2t f
2(1.5" )
compact wrt FLB
Lp  20.1 ft  Lb  24  Lr  68.7 ft
Fcr  35.8ksi
S x  3285in 3
aw  0.535
72”
5/16”
Design Example, p. 12

Recalculate Flexural Capacity, cont’d.
0.535
72"
29000ksi
Rpg  1 
(
 5.7
)  0.973
1200  300(0.535) 5 / 16"
36ksi
fM n  fb S xc R pg Fcr
 0.9(3285in )(0.973)(35.8ksi)
 102985 kip  in  8582 kip  ft
3
Design Example, p. 13

Check against Mu including self-weight
5
2
Area  2(28in )(1.5in )  (72in )( in )  106in
16
106in 2
Weight 
(490 pcf )  361 lb / ft
2
in
144
ft 2
361lb / ft
(72 ft )2
1000lbs / kip
M u  6840kip  ft  1.2
 7120kip  ft
8
< 8582 kip-ft
fM n  M u

flexural capacity of section is adequate
Design Example, p. 14

Check Deflection Limits

Estimate service loads
150 k
150 k
5 k/ft
a
A
24'
B
C
24'
24'
D Lateral
Support
330 k
wu  5k / ft  1.2(0.210
361k / ft )  5.43k / ft
wShear
service
(kip)
60
 5.43 / 1.5  3.62k / ft
Pservice  Pu / 1.5  150kips / 1.560 100kips
210
Moment
330
Design Example, p. 15

Deflection limits, cont’d.
I x  2(
1
72" 1.5" 2 1 5
)(28" )(1.5"3 )  (2)(28" )(1.5" )(

)  ( " )(72"3 )  123,183in 4
12
2
2
12 16
4
5wservice L4 Pservicea
5
(
3
.
62
k
/
ft
)(
72
ft
)(1728in 3 / ft 3)
2
2


(3L  4a ) 

4
384 EI
24 EI
384(29,000ksi)(123,183in )
100kips(24 ft )(1728in 3 / ft 3)
2
2
(
3
(
72
ft
)

4
(
24
ft
))  0.61  0.64  1.25in
4
24(29,000ksi)(123,183in )
 max 
L
72 ft (12in / ft )

 2.40in
360
360
> 1.25 in
√ OK
Design Example, p. 16
150 k
150 k

5 k/ft
A
24'
B
330 k
C
24'
24'
Lateral
Support
210
60
Shear
(kip)
60
210
Moment
(ft-kip)
D
330
Low shear demand –
Region BC
h
 230  260
tw
260 is limit for unstiffened girders
(F13.2); stiffeners not required
unless needed for capacity
h
kv E
5.0(29000ksi)
 230  1.37
 1.37
 87.0
tw
Fyw
36ksi
6480
6480
6840
1.51kv E
Cv 
h 2
( ) Fy
tw
1.51(5)( 29000ksi)

 0.12
2
(230 )(36ksi)
(G2-5)
Design Example, p. 17

Shear Capacity – Region BC
fVn  0.9(0.6 Aw FyCv )
(G2-1) and AISC G1
5
 0.9(0.6)(75")( ")(36ksi)(0.12)  54.6kips
16
2


a  260 

h 
h
 tw 
2
260



  1.28
 230 
a  1.28h  1.28(72" )  92.2"
<Vu = 60 kips
w/o self-weight N.G.
a = 90”
a/h=1.25
Design Example, p. 18

Shear Capacity – Region BC
5
kv  5 
( a / h) 2
5
kv  5 
 8.2
2
(1.25)
kv E
8.2(29000ksi)
h
1.37
 1.37
 111   230
Fy
36ksi
tw
1.51kv E
Cv 
h 2
( ) Fyw
tw
(G2-5)
1.51(8.2)(29000ksi)

 0.19
2
(230 )(36ksi)
Design Example, p. 19

Shear Capacity – Region BC
fVn  0.9(0.6 Aw Fy )(Cv 
1  Cv
1.15 1  (a / h)2
)
(G3-2)
5
1  0.19
 0.9(0.6)(75")( ")(36ksi)(0.19 
)
16
1.15 1  1.252
 86  200  286 kips
>>Vu = 60 kips √ OK
By inspection, adequate for Vu
including self-weight
24 ft (12in / ft )
 3.2 panels
90in / panel
24 ft (12in / ft )
a 72"
a
 72in; 
 1.0
4spaces
h 72"
Small adjustment needed later since ‘a’ is clear distance between stiffeners
Design Example, p. 20
150 k
150 k

5 k/ft
A
24'
330 k
Shear
(kip)
B
24'
C
D
Lateral
Support
Design End Panels first
NO TFA Permitted
210
60
60
210
Moment
(ft-kip)
24'
Shear Capacity –
Regions AB and CD
330
361lb / ft
(72 ft )
1000lbs / kip
6480 kips  1.2 6480
Vu  330
 346 kips
6840
2
346kips
Required
 14.8 ksi
Stress
(75")(5 /16")
Design Example, p. 21

Shear Capacity, End
Panels

Use Table 3-16a (No
TFA) for estimate
Will need a/h < 0.5
fVn  Vu  346kips
Requires:
Cv >0.791
kv > 34.4
a/h < 0.41
a < 29.5”
Design Example, p. 22

Shear Capacity, End Panel
Try a = 27”; a/h = 0.375
5
5
kv  5 
 5
 40.6
2
2
( a / h)
(0.375)
1.37
kv E
40.6(29000ksi)
h
 1.37
 247   230
Fy
36ksi
tw
1.10
kv E
40.6(29000ksi)
h
 1.10
 199   230
Fy
36ksi
tw
1.10 kvE Fy 1.10 (40.6)(29000ksi) (36ksi)
Cv 

 0.86
h tw
230
5
Vn  0.9(0.6 Aw Fy Cv )  0.9(0.6)(75")( ")(36ksi)(0.86)  392 kips
16
√ OK
Design Example, p. 23
150 k
150 k

5 k/ft
A
B
24'
330 k
24'
24'
D
Lateral
Support
210
60
Shear
(kip)
60
210
~27”
Moment
(ft-kip)
C
330
Shear Capacity –
Regions AB and CD
After End Panels
TFA Permitted?
Check AISC G3.1
(a) and (b) satisfied;
(c) and (d) ??
2 Aw
2(72" )(5 16" )

 0.54 < 2.5
( Afc  Aft ) 2(28" )(1.5" )
6480
TFA OK!
6480
6840
h
h 72"
 
 2.6
bfc bft 28"
< 6.0
TFA OK!
Design Example, p. 24
150 k
150 k

5 k/ft
A
B
24'
330 k
24'
C
24'
D
Lateral
Support
210
60
Shear
(kip)
60
210
~27”
330
Moment
(ft-kip)
Vu  334kips
Including self-weight
6480
Required
Stress
6480
6840
334kips
 14.2 ksi
(75")(5 /16")
Shear Capacity –
Regions AB and CD
After End Panels
TFA Permitted
Design
Example, p. 25

Shear Capacity, after
End Panels

Use Table 3-16b (with
TFA) for estimate

Based on required
stress, try a/h = 0.80?
Design Example, p. 26

Shear Capacity, after End Panel, with TFA
Try a = 56”; a/h = 0.78
1.37
5
5
kv  5 
 5
 13.2
2
2
(a / h)
(0.78)
kv E
13.2(29000ksi)
h
 1.37
 141   230
Fy
36ksi
tw
1.51kv E 1.51(13.2)( 29000ksi)
Cv 

 0.304
2
h 2
(230 )(36ksi)
( ) Fy
tw
1  Cv
Vn  0.9(0.6 Aw Fy )(Cv 
)
2
1.15 1  (a / h)
5
1  0.304
 0.9(0.6)(75)( )(36)(0.304 
)  356kips >334 kips
2
16
1.15 1  0.78
√ OK
Design Example, p. 27
150 k
150 k

5 k/ft
A
B
24'
330 k
C
24'
D
After first 2 panels;
TFA permitted
Lateral
Support
210
60
Shear
(kip)
27”
24'
Shear Capacity –
Regions AB and CD
60
210
56”
330
Moment
(ft-kip)
Vu  309kips
Including self-weight
6480
Required
Stress
6480
6840
309kips
 13.2 ksi
(75")(5 /16")

Based on Table 3-16b,
repeat a/h= 0.78
Design Example, p. 28
27”
Repeat process for next panel(s); determine stiffener
layout (another layout might be more efficient)
CL
sym.
56” 56” 73”
76”
4 @ 72”
Note: ‘a’ dimension used for stiffener spacing; therefore,
actual ‘a’ (clear distance) will be smaller
(May also modify to get multiples of 3” for spacing)
Design Example, p.29

Size flange-to-web
weld
h tf
Q  Af (  )
2 2
72" 1.5"
 28" (1.5" )(

)  1544 in 3
2
2
X
X
Vu Q
shearflow 
Ix
346kips(1544in 3)

 4.34kips / in
4
123,183in
Design Example, p. 30

Flange-to-web welds,
cont’d

AISC Table J2.4 minimum
size


3/16” fillet for 5/16” plate
(thinner part joined)
Assume Submerged Arc
Weld (SAW)


Try w=1/4”
Use matching weld
electrode, 70ksi
X
X
Design Example, p. 31

Flange-to-web weld, cont’d.
Weld Metal (AISC J2)
2
fRn  f (0.6FEXX ) Aw  0.75(2)(0.6)(70ksi)
(0.25" )  11.1kips / in
2
Base Metal – Shear Yield (AISC J4.2)
fRn  f (0.6Fy ) Ag  1.0(0.6)(36ksi)(0.3125" )  6.75kips / in
Base Metal – Shear Rupture (AISC J4.2)
CONTROLS
fRn  f (0.6Fu ) Anv  0.75(0.6)(58ksi)(0.3125" )  8.2kips / in
>4.34 kips/in
√ OK
Design Example, p. 32

Intermediate Transverse Stiffeners

Assume single-plate A36 stiffeners

Design stiffener between end panel and first panel with
TFA
End panel a/h = 0.375; adjacent panel (TFA) a/h = 0.78
Design Example, p.33
28”
28" 5 16"
bst 
 13.8"
2
1.5”
Try bst = 8”
E
29000
b t st  0.56
 0.56
 15.9
Fyst
36
 15.9
tst
tst  0.503"
8"
Try tst = 9/16”
(G3-3)
bst
72”
5/16”
Design Example, p. 34

Intermediate Transverse Stiffeners, cont’d. (adequate
stiffness for web buckling; AISC G2.2)
End panel a/h = 0.375; adjacent panel (TFA) a/h = 0.78
2.5
j
 2  0.5
2
( a / h)
I st  btw j
3
2.5

 2  15.7
2
0.375
5 3
 (27" )( " ) (15.7)  12.9in 4
16
for adjacent panel a/h = 0.78, Ist = 3.61 in4
Check 8” x 9/16”
0.5625" (8" )3
Ist 
 96in 4
3
√ OK
Design Example, p. 35

Intermediate Transverse Stiffeners, cont’d.

Adequate stiffness for TFA
 Vu  Vc1  (G3-4)
Ist  Ist1  ( Ist 2  Ist1) 

V
c
2

V
c
1


h 4 st1.3  Fyw 1.5 7241.01.3  36 1.5
4
Ist 2 

 29.4in




40  E 
40  29000 
(G3-5)
 346  138 
Ist  12.9  (29.4  12.9) 

356

138


Check other stiffeners
√ OK
96.0in 4  28.6in 4
Design Example, p. 36

Size welds for stiffeners
f nv  0.045h

3
36ksi3
 0.045(72" )
 4.1kips / in
E
29000ksi
Fyw
Try minimum weld size for 5/16” plate
(thinner plate) AISC Table J2.4



w=3/16”
Assume SMAW
Use matching electrode, 70 ksi
Design Example, p. 37

Stiffener welds, cont’d.
Weld Metal (AISC J2)
fRn  f (0.6FEXX ) Aw  0.75(2)(0.6)(70ksi)(0.707)(0.1875" )  8.35kips / in
Base Metal – Shear Yield (AISC J4.2)
CONTROLS
fRn  f (0.6Fy ) Ag  1.0(0.6)(36ksi)(0.5625" )  12.2kips / in
Base Metal – Shear Rupture (AISC J4.2)
fRn  f (0.6Fu ) Anv  0.75(0.6)(58ksi)(0.5625" )  14.7kips / in
or
fRn  f (0.6Fu ) Anv  0.75(0.6)(58)(2)(0.1875" )  9.79kips / in
>4.1 kips/in
√ OK
Design Example, p. 38

Stiffener welds, cont’d.
3/16”
Use nominal weld size (Table
J2.4) to connect to compression
flange (to prevent uplift of
flange – single stiffener)
3/16”
5
 4tw  4( " )  1.25"
16
USE 1.5”
5
 6tw  6( " )  1.88"
16
Design Example, p. 39

Bearing Stiffeners
Check LWY, LWC, etc.
(given bearing length of
support)
Design bearing stiffener
as needed
Typically make full
depth; check
capacity of
intermediate
transverse
stiffeners for
BEARING
Design Example, p. 40

Bearing Stiffeners



Will design / check for
homework
Use pairs of stiffeners (as
shown to left)
Use full depth

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