AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
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There’s always a solution in steel!
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There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
6.1
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
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There’s always a solution in steel!
AISC is a Registered Registered Provider with The American Institute Institute of Architects Continuing Education Systems (AIA/CES). Credit(s) earned on completion of this program will be reported to AIA/CES for AIA AIA members. Certificates of Completion for both AIA members and non AIA members are available available upon request. This program is registered with AIA/CES for continuing professional education. As such, it does not include content that that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner of handling, using, distributing, or dealing in any material or product. Ques Questi tion ons s rela relate ted d to spec specif ific ic mate materi rial als, s, meth method ods, s, and and serv servic ices es will will be addre address ssed ed at the the conc conclu lusi sion on of this this pres present entat atio ion. n. There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
6.2
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
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There’s always a solution in steel!
Course Description Session 6: Built-up Columns, Lacing, Battens, and Cover Plates March 8, 2016 This session focusses on flexural members built up from plates in the form of singly or doubly symmetric symmetric I-shaped sections. Bending and shear on girders girders built from plates will be studied. This includes girders that are compact as well as those with noncompact noncompact and slender flanges and webs. webs. Shear yielding and buckling along with tension field action action will be investigated. Several examples will be presented.
There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
6.3
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Learning Objectives • Gain Gain an under understa stand ndin ing g of AIS AISC C Specification provisions for the design of built-up columns • Underst Understand and the the influen influence ce of conne connector ctor spac spacing ing on slenderness and strength • Calcula Calculate te the stren strength gth of of built built up columns columns usin using g lacing, lacing, battens, and cover plates • Determi Determine ne the the require required d spacin spacing g of conn connecto ectors rs • Analyze Analyze the the value value of built-up built-up sectio sections ns versus versus similar similar W and and HSS shapes considering fabrication effort
There’s always a solution in steel!
Steel Design 2: Selected Topics Topics based on AISC 360-10 Specification for Structural Steel Buildings Lesson 6 – Built-up Columns, Lacing, Battens, and Cover Plates Presented by Louis F. Geschwindner, Ph.D., P.E. Emeritus Professor at Penn State University Former Vice-President at AISC
There’s always a solution in steel!
8
Copyright © 2016 American Institute of Steel Construction
6.4
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Steel Design 2: Selected Topics based on AISC 360-10 Specification for Structural Steel Buildings Night School 10 Lesson 6 Built-up Columns, Lacing, Battens, and Cover Plates There’s always a solution in steel!
6.9
Lesso Lesson n 6 – BuiltBuilt-up up Colum Columns ns • Columns Columns that are built-up built-up from multiple multiple shapes are considered • Shapes Shapes connecte connected d directly directly with bolts bolts or welds welds are discussed • Shapes Shapes connec connected ted with cover plates, plates, lacing lacing,, or battens are addressed • The influence influence of connect connector or spacin spacing g along along the the individual components on overall buckling will be treated There’s always a solution in steel!
6.10
Copyright © 2016 American Institute of Steel Construction
6.5
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • This section of the Chapter E on design of members for compression has two parts, – The first addresses strength through a modified slenderness ratio – The second gives the dimensional requirements
There’s always a solution in steel!
6.11
E6. Built-up Members • E6.1 Compressive Strength, applies to built-up members composed of two shapes – Interconnected by bolts or welds, or – With open sides interconnected by cover plates, lacing, or tie plates – The end connection shall be welded or connected by means of pretensioned bolts (designed to resist slip) There’s always a solution in steel!
6.12
Copyright © 2016 American Institute of Steel Construction
6.6
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • Strength – Use Sections E3, E4, or E7 as appropriate – If the buckling mode being investigated involves relative deformation that produces shear in the connectors use the modified slenderness ratio ( KL/r )m – The same built-up shape will have different strengths depending on the type of connectors; • (a) snug-tight bolts or • (b) welds and pretensioned bolts There’s always a solution in steel!
6.13
E6. Built-up Members • E6.1(a) for snug-tight bolted intermediate connectors KL r m KL r o
KL a + r o r i 2
=
2
E6-1
= the slenderness ratio of the built-up mem ber
acting as a unit buckling in the direction being considered
a = slenderness ratio of the individual component between connectors, a, r i based on the minimum radius of gyration of the component, r i
There’s always a solution in steel!
6.14
Copyright © 2016 American Institute of Steel Construction
6.7
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • E6.1(a) for snug-tight bolted intermediate connectors – The modified slenderness ratio will range from 1.41 times the larger slenderness ratio of the member or component down to equal to the larger slenderness ratio as the difference between these two increases. KL r m
=
(100)
2
KL r m
2
+ (100) = 141
=
(100 )
2
2
+ ( 20) = 102
There’s always a solution in steel!
6.15
E6. Built-up Members • E6.1(b) welded or pretensioned bolt connectors When
a r i
≤ 40
KL r m When
a r i
KL r o
=
E6-2a
> 40
KL r m
2
=
KL K i a r + r o i
There’s always a solution in steel!
2
K i = 0.50 for back-to-back angles E6-2b K i = 0.75 for back-to-back channels
K i = 0.86 for all other cases 6.16
Copyright © 2016 American Institute of Steel Construction
6.8
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • The effective slenderness ratio is used to account for the reduction in column strength due to shearing deformation in the connectors • Welds and slip-critical bolts will exhibit less shearing deformation than snug-tight bolts
There’s always a solution in steel!
6.17
E6. Built-up Members • AISC provisions in E6.1 indicate that for a/r i < 40 the effective length of the member will not be increased by the presence of the lacing and for a/r i > 40
KL r m
KL Ki a + r o r i 2
=
2
E6-2b
• Which can be reorganized as KL r m K ′ = KL r o There’s always a solution in steel!
2
=
K i a r i 1+ 2 KL r o
6.18
Copyright © 2016 American Institute of Steel Construction
6.9
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • This shows that the change in slenderness is greatest as the two contributing slenderness ratios approach each other. KL r m ′ K = KL r o
There’s always a solution in steel!
2
=
K ia r i 1+ 2 KL r o
6.19
E6. Built-up Members • Ziemian suggests, based on Bleich, that the worst case increase in effective length for latticed columns is a 10% increase which occurs for members with KL/r of 40 or less. This implies that a/r i is about 0.56 KL/r of the member, for latticed columns
There’s always a solution in steel!
6.20
Copyright © 2016 American Institute of Steel Construction
6.10
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members At
KL
= 40, a 10% increase r in slenderness yields a 1.7%
decrease in column strength
KL = 40 r
There’s always a solution in steel!
6.21
E6. Built-up Members • E6.2 Dimensional requirements – Individual component slenderness Ka ri
KL r governing
≤ 0.75
– End connections weld length ≥ maximum width of member or bolts spaced ≤ 4d b for length 1.5 times maximum width
There’s always a solution in steel!
6.22
Copyright © 2016 American Institute of Steel Construction
6.11
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • E6.2 Dimensional requirements – Open sides of compression members • Perforated cover plates • Lacing with tie plates
There’s always a solution in steel!
6.23
Example 1 • Consider the 2-L5 x 3 x ¼ LLBB A36 compression member from Lesson 5 Example 7 10 ft
3/8 in.
For the double angle member r x = 1.62 in.
x
x
r y = 1.19 in. y There’s always a solution in steel!
6.24
Copyright © 2016 American Institute of Steel Construction
6.12
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 1 • For flexural buckling, the y -axis controlled with 10 (12 ) KL
=
r y
1.19
= 101
• Overall member strength was limited by flexural-torsional buckling which was a function of the y -axis slenderness • Thus, the connectors will be in shear
There’s always a solution in steel!
6.25
Example 1 • With snug-tight bolts the only way to have the slenderness of the built-up member work as a single unit is if a/r i is less than 9 using KL r m
2
=
KL a + r o r i
=
1012 + 92 = 101
2
There’s always a solution in steel!
E6-1
6.26
Copyright © 2016 American Institute of Steel Construction
6.13
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 1 z
• For the individual angle ri = r z = 0.652 in.
thus a r i
=9=
a z
0.652
and
This does not appear to be a very workable solution
a = 5.87 in. There’s always a solution in steel!
6.27
Example 1 • With pretensioned bolts or welds, a
If
ri
≤ 40
KL r m
KL = 101 r o
=
E6-2a
thus a 0.652
≤ 40
so a ≤ 0.652 ( 40 ) = 26.1 in.
This might be a more acceptable solution, requiring 5 connectors There’s always a solution in steel!
6.28
Copyright © 2016 American Institute of Steel Construction
6.14
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 1 • However, a more reasonable design approach for the original problem would have been to decide on a realistic spacing of connectors and then proceed to determine the available strength. • If we had two intermediate connectors, for a 10 ft member a = 40 in. There’s always a solution in steel!
6.29
Example 1 • As a single unit
KL r o
• Between connectors
a r i
=
10 (12 ) 1.19
40
=
0.652
= 101
= 61.3 >
40
• Thus, KL r m
2
=
KL K i a + r o r i
2
There’s always a solution in steel!
=
0.5( 40) (101)o + 0.652 2
2
= 106
E6-2b
6.30
Copyright © 2016 American Institute of Steel Construction
6.15
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 1 • Check maximum slenderness between connectors a
40
=
r i
0.652
= 61.3 < 0.75 (106 ) = 79.5
• The overall impact on member strength, with KL/r changing form 101 to 106 is a reduction from P n = 59.4 kips to 57.4 kips as follows A 3.4 % reduction (See next 2 slides)
There’s always a solution in steel!
6.31
Example 1 • E3. Flexural-torsional buckling KL r y m
2
F e =
= 106
QF F Fcr = Q 0.658
y
e
F crz =
F y
π E
KL r
2
=
π
2
( 29,000 ) = 25.5 2 (106 )
0.804(36 ) = 0.804 0.658 25.5 ( 36 ) = 18.0 ksi
GJ A g r o2
=
11,200 ( 0.0876) 3.88( 2.52)
There’s always a solution in steel!
2
= 39.8
ksi
E3-4
E3-2
E4-3
5.32
Copyright © 2016 American Institute of Steel Construction
6.16
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 1 • Therefore , the critical stress is 4 Fcry Fcrz H 1 − 1 − 2 ( Fcry + F crz ) 18.0 + 39.8 4 (18.0)39.8 (0.638 ) 1 1 = − − 2 2 (0.638 ) 18.0 39.8 + ( )
Fcry + Fcrz F cr = 2 H
= 14.8
E4-2
ksi < F cr = 18.0 ksi
so P n
=
(
14.8 3.88
)
=
57.4 kips
E4-1
There’s always a solution in steel!
5.33
Example 2 • Consider the 2-WT6 x 29 A992 compression member from Lesson 5 Example 8 which had an effective length of 20 ft. The controlling slenderness y
y = 1.03 in.
was KL/r x = 124 For pretensioned connectors
Gap = 3/8 in.
a x
r i
=
a 1.50
≤ 40
a ≤ 1.50 ( 40) = 60 in.
This could be considered a reasonable spacing (3 @ 5 ft) There’s always a solution in steel!
6.34
Copyright © 2016 American Institute of Steel Construction
6.17
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • E6.2 Dimensional requirements – Open sides of compression members • Perforated cover plates – Continuous cover plates with access holes – Contributes to the strength of the member – Meet the width-to-thickness requirements of Table B4.1a Case 7 – Requirements for hole dimensions also given » Length/width of hole ≤ 2 » Clear distance between holes ≥ transverse distance between connectors » Hole minimum radius 1.5 in. There’s always a solution in steel!
6.35
E6. Built-up Members • E6.2 Dimensional requirements – Tie plates • End tie plates length between fasteners
≥
distance
• Intermediate tie plates ≥ ½ that distance • Tie plate thickness ≥ distance between fasteners /50 • Tie plate weld length ≥ 1/3 plate length • Tie plate bolts spaced ≤ 6 d b There’s always a solution in steel!
6.36
Copyright © 2016 American Institute of Steel Construction
6.18
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
E6. Built-up Members • E6.2 Dimensional requirements – Lacing • Spacing such that
a ri flange element
KL r governing
≤ 0.75
• Stress perpendicular to axis of member = 0.02 P c • Lacing slenderness
L ≤ 140 for single lacing r lacing ≤ 200
for double lacing
• For single lacing L = length between connections • For double lacing L = 0.7 length between connections There’s always a solution in steel!
6.37
E6. Built-up Members • E6.2 Dimensional requirements – Lacing α ≥ 45
attach at cross over
α ≥ 60
If spacing greater than 15 in. use double lacing or angles
Single Lacing There’s always a solution in steel!
Double Lacing 6.38
Copyright © 2016 American Institute of Steel Construction
6.19
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 3 • Determine the strength of a 25 ft laced column built-up from 4 - 6 x 6 x ¾ A36 angles 20.0 in.
. n i 0 . 0 2
Single Angle Properties, Table 1-7
1.77 in.
8.23 in. Gage line
13.0 in.
3.5 in.
A = 8.46 in.2
J = 1.61 in.4
I x = I y = 28.1 in.4
C w = 4.17 in.6
S x = 6.64 in.3
r o = 3.24 in.
r x = 1.82 in.
I z = 11.6 in.4
y = 1.77 in.
S z = 4.63 in.3
3 Z x = 11.9 in.
r z = 1.17 in.
y p = 0.705 in.
tan α = 1.00
Wt . = 28.7 lb/ft
Q s = 1.00( F y = 36 ksi)
There’s always a solution in steel!
6.39
Example 3 • Determine the strength of a 25 ft laced column built-up from 4 - 6 x 6 x ¾ A36 angles 20.0 in.
Built-up Member Properties
1.77 in.
A = 4 ( 8.46 ) = 33.8 in.2 . n i 0 . 0 2
8.23 in.
(
I x = I y = 4 28.1+ 8.46 (8.23 ) r x =
Gage line
13.0 in.
3.5 in.
2400 33.8
2
) = 4 (601) = 2400 in.
4
= 8.43 in.
If the member works as a single unit
There’s always a solution in steel!
KL r o
=
25 (12 ) 8.43
= 35.6
6.40
Copyright © 2016 American Institute of Steel Construction
6.20
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 3 • Determine the strength of a 25 ft laced column built-up from 4 - 6 x 6 x ¾ A36 angles 20.0 in.
. n i 0 . 0 2
y
=
1.77 in.
8.23 in. Gage line
13.0 in.
For built-up members with high slenderness, initial out-of-straightness is important For members with low slenderness, <40, shear effect is important and may increase slenderness by 10 %
3.5 in.
There’s always a solution in steel!
6.41
Example 3 • For our previous examples, it may not have been good to assume that the full slenderness ratio could be used. • For a laced column, the lacing provides very close spacing of connectors so that assumption may be more easily accommodated
There’s always a solution in steel!
6.42
Copyright © 2016 American Institute of Steel Construction
6.21
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 3 • Determine nominal strength – For
KL r o 2
F e =
π E
KL r
2
=
π
=
25 (12 ) 8.43
= 35.6
2
F y ( 29,000 ) = 226 and 2 F e ( 35.6)
36
=
226
= 0.159 < 2.25
so F F Fcr = 0.658
y e
F y
36
= 0.658 226
36 = 33.7 ksi
and P n = 33.7 ( 33.8) = 1140 kips
There’s always a solution in steel!
6.43
Example 3 • Lacing design Since the distance between gage lines in the angles is 13 in. < 15.0 in., single lacing may be used
α ≥ 60
Check angle slenderness between lacing a ri a ri
=
a r z
=
15.0 1.17
= 12.8 < 40
α≥
= 12.8 < 0.75( 35.6) = 26.7 ∴
KL r m
KL r o
=
60
. n i 0 . 5 1
E6-2a
So our assumption was good P n = 1140 kips
There’s always a solution in steel!
13.0 in.
13.0 in.
6.44
Copyright © 2016 American Institute of Steel Construction
6.22
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 3 • Lacing design Shear force in lacing
α≥
Fv = 0.02 P c = 0.02 (1140 ) = 22.8 kips
60
. n i 0 . 5 1
F v
Lacing on two sides F v (per side) = 11.4 kips
13.0 in.
Lacing bar force P bar =
11.4 cos30
= 13.2
For single lacing,
kips
L < 140 r lacing
There’s always a solution in steel!
6.45
Example 3 • Lacing design Try a ½ in. bar I y =
bt 3
12 A = bt bt 3 r y =
2 12 = t = bt 12
(1 2 ) 12
2
= 0.144
in.
Determine maximum permitted bar length L = 15.0 ≤ 140 ( 0.144 ) = 20.2 in. There’s always a solution in steel!
6.46
Copyright © 2016 American Institute of Steel Construction
6.23
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 3 • Lacing design Determine lacing strength 2
F e =
π E
KL r
2
=
π
2
( 29,000)
15.0 0.144
2
=
26.4 ksi
F y
=
F e
36 26.4
= 1.36 <
2.25
36 26.4 F cr = 0.658 36 = 20.3 ksi
1 Pn = 13.2 = 20.3 b; 2
b = 1.30 in.
Use a ½ x 1½ flat bar There’s always a solution in steel!
6.47
Example 3 • Tie plates End tie plate length ≥ 13.0 in. distance between fasteners t ≥
13.0 50
= 0.26
∴
use
3 8
plate each end
End tie plates, use 3 8
× 1 3× 1 '− 8 "
Attachment Weld length, l /3 = 4.4 in. or Min 3 bolts spaced ≤ 6d b There’s always a solution in steel!
. n i 0 . 3 1
20.0 in.
6.48
Copyright © 2016 American Institute of Steel Construction
6.24
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Replace the lacing from Example 3 with batten plates. (moment connection) Although there is nothing in AISC 36010 that specifically addresses battened columns, they are not prohibited.
. n i 0 . 0 3
As for Example 3, if the column works as a unit,
KL r o
. n i
= 35.6
There’s always a solution in steel!
0 . 0 3
13.0 in.
13.0 in.
6.49
Example 4 • Replace the lacing from Example 3 with batten plates. Between battens,
KL ≤ 0.75 ri r o a
. n i
= 0.75 ( 35.6 ) = 26.7
0 . 0 3
For r z = 1.17
a ≤ 26.7 (1.17) = 31.2 in.
. n i 0 . 0 3
Try a 30 in. spacing (the maximum)
30.0 = 25.6 r i 1.17 a
=
There’s always a solution in steel!
13.0 in.
13.0 in.
6.50
Copyright © 2016 American Institute of Steel Construction
6.25
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • F. Bleich, Buckling Strength of Metal Structures gives the following equation for slenderness ratio of battened columns (converted to the notation used in AISC 360-10) which accounts for the shear deformation in the battens KL r m
L r
=
2
a + 12 r i π
2
2
Bleich Eq. 351 Ziemian Eq. 3.38
There’s always a solution in steel!
6.51
Example 4 • Comparing to AISC Eq. E6-2b KL L = r Bleich r
2
a + 12 ri π
2
K i =
2
π
KL = r AISC
=
L r
2
2
K a + i ri
2
12
= 0.907
where AISC uses, for all but angles and channels,
K i = 0.86
The Bleich equation accounts for a bit more shear deformation than the AISC equation There’s always a solution in steel!
6.52
Copyright © 2016 American Institute of Steel Construction
6.26
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • So the column strength, taking into account the reduction in buckling strength due to shear deformation of the battens, is KL r m
L r
=
2
F e =
F F Fcr = 0.658
y e
π E
KL r
Fy
2
=
π
2
2
a + 12 r i π
2
2
=
( 35.6 )
( 29,000) = 158 ksi 2 ( 42.5)
36
= 0.658158 36 = 32.7
2
There’s always a solution in steel!
+
π
2
12 F y
=
F e
ksi
( 25.6 ) 36 158
2
= 42.5
Bleich Eq. 351 Ziemian Eq. 3.38
= 0.228 < 2.25
P n = 32.7 ( 33.8) = 1110 kips This is a reduction of 2.6% from the laced column 6.53
Example 4 • Batten design – Battens and their connections will be treated as vierendeel panels – The shearing force in the panel must be determined – Over the years, the required force has varied, based on the applicable code – We will use the recommendation by Bleich
There’s always a solution in steel!
6.54
Copyright © 2016 American Institute of Steel Construction
6.27
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Batten design – Assume all rigid connections between battens and chords – Assume hinge at mid point of batten – Assume hinge at mid height of chord between battens b
b
2
2
a 2
a
V max
V max
2
2
V b
2
There’s always a solution in steel!
V max
V max
2
2
6.55
Example 4 • Batten design
2010 E6.2 0.02 P = 22.2 k 2010 App 6 0.01 P = 11.1 k 2016 App 6 0.005 P = 5.55 k
– Maximum horizontal shear force in panel r L
Bleich Eq. 355
Vmax = 2π ( F y − Fcr ) A1 chord = 2π ( 36 − 32.7 )
( 2 (8.46 ) )
1 = 8.25 kips 42.5
b
b
2
2
– Shear force in battens By equilibrium about the chord
a 2
Vb = V max
V max
V max
2
2
a b
There’s always a solution in steel!
a 2
V b V max
V max
2
2
6.56
Copyright © 2016 American Institute of Steel Construction
6.28
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Batten design b = distance between centroids of chords = 20 − 2 (1.77 ) = 16.5 in. a = center to center of battens = 30 in.
Vb = V max
a b
30 = 15.0 kips 16.5
= 8.25
8.25 in. 8.25 in.
Since there are two planes of battens
V 1b =
15.0 2
= 7.5 kips 15.0 in.
16.5 = 61.9 in.-kips 2
8.25
8.25
2
2
M 1b = 7.5
15.0
15.0 in.
There’s always a solution in steel!
8.25
8.25
2
2
6.57
Example 4 • Batten design • Try a ½ x 4 in. batten Z =
0.5 ( 4.0 )
bh2
2 3
= = 2.0 in. 4 4 M p = 36 ( 2.0 ) = 72.0 in.-kips
S =
bh 2
=
6
0.5 ( 4.0 )
2
6
= 1.33
in.3
• F11.2 Lateral-torsional buckling Lb d t
2
=
16.5 ( 4.0 )
( 0.5 )
2
= 264 >
<
0.08 E
=
0.08 ( 29, 000 )
F y 1.9 E F y
36 =
1.9 (29000 ) 36
There’s always a solution in steel!
= 64.4
= 1530
6.58
Copyright © 2016 American Institute of Steel Construction
6.29
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • F11.2 for inelastic lateral-torsional buckling
Lb d F y M y 2 t E
M n = Cb 1.52 − 0.274
≤ M p
F11-2
16.5 ( 4.0 ) 36 ( 36 ) (1.33 ) = 68.5 ≤ M p = 1.0 1.52 − 0.274 ( 0.5 )2 29,000
= 72.0
in.-kips
Since the required batten strength is based on the nominal strength of the column, we will compare this nominal bending strength directly to the required bending strength
M 1b = 61.9 < M n = 68.5 in.-kips There’s always a solution in steel!
6.59
Example 4 • Check shear in batten F1v =
V b
=
bt
7.5 4.0 (0.5 )
= 3.75 < 0.6 F y = 0.6 ( 36 ) =
21.6 ksi
G2-1
• Check chord for local moment and axial V max 4
= 2.06
kips
M angle = 2.06 (15.0) = 30.9 in.-kips P angle =
1110 4
V b = 7.5 kips
+ 7.5 = 285 kips V max 4
There’s always a solution in steel!
= 2.06
kips
6.60
Copyright © 2016 American Institute of Steel Construction
6.30
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Strength of single angle in chord We are checking a local condition at the batten – chord junction Assume the column strength is based on axial yield ( KL = 0) and flexural plastic moment
P n = 36 (8.46 ) = 305 kips
E3-1
M p = 1.5 M y = 1.5 (36 )(6.64 ) = 359 in.-kips
F10-1
Check single angle for interaction at location of batten
Pr Pc
+
8 M r 9 M c
=
285
+
305
8 30.9 9 359
= 1.01 ≈ 1.0
H1-1a
There’s always a solution in steel!
6.61
Example 4 • If you are not willing to accept this slight overage, what can be done? – Since both moment and force are a direct function of P n = 1110 kips, reduce the nominal strength accordingly P n =
1.0 1.01
(1110 ) = 1100 kips
There’s always a solution in steel!
6.62
Copyright © 2016 American Institute of Steel Construction
6.31
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Moment connection of batten to angle V max 4
= 2.06
kips
Manual Table 7-7 addresses bolts at 3 in. spacing, which wont fit in our 4 in. batten
V b = 7.5 kips
V max 4
=
2.06 kips
As an approximation, these 4 bolts with the shear at an eccentricity of 8.25 in. have a strength of Rn = Cr n = 0.93 (17.9) = 16.6 kips > 7.5 kips
More detailed work on the batten to angle connection will need to be carried out. That may result in a wider batten to accommodate the required bolt group. There’s always a solution in steel!
6.63
Example 4 • End plates will be the same as for the latticed column End tie plates, use 3 8
× 1 3× 1 '− 8 "
Attachment Weld length, l /3 = 4.4 in. or Min 3 bolts spaced ≤ 6d b
. n i 0 . 3 1
20.0 in.
There’s always a solution in steel!
6.64
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 4 • Cautions – For a battened column to work as designed, the batten to chord connection must remain rigid – Shear deformation is critical and the effective length might be increased beyond what the Bleich equation gives – If there is any concern about the rigidity of the connections, treat the column as a spaced column and account only for the end plates There’s always a solution in steel!
6.65
Spaced Columns • Battened columns are not as commonly found as we might believe. • Many of what appear to be battened columns are actually spaced columns – The “battens” are not rigidly connected to the chords – End plates = stay plates still have an effect • Shorten the column length • Force reverse curvature bending There’s always a solution in steel!
6.66
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Cover Plates • E6.2 Dimensional Requirements – Open sides of compression members shall be provided with • Continuous cover plates • Perforated with access holes • Cover plates contribute to column strength – Width-to-thickness ratio conform to Table B4.1a – Length to width of hole shall not exceed 2 – Clear distance between holes greater or equal to distance between line of connectors – Holes have a minimum radius of 1 ½ in. There’s always a solution in steel!
6.67
Example 5 • Determine the strength of a built-up column composed of 4 - 4 x 4 x 5/16 angles with ½ in. cover plates perforated on two sides 20.0 in.
. n i 0 . 0 2
y = 1.11 in.
Single Angle Properties, Table 1-7
8.89 in. Gage line
10 in. 15.0 in.
2.5 in.
A = 2.40 in.2
J = 0.0832 in.4
I x = I y = 3.67 in.4
C w = 0.0963 in.6
S x = 1.27 in.3
r o = 2.21 in.
r x = 1.24 in.
I z = 1.46 in.4
y = 1.11 in.
S z = 0.936 in.3
Z x = 2.26 in.3
r z = 0.781 in.
y p = 0.300 in.
tan α = 1.00
Wt . = 8.2 lb/ft
Q s = 0.997( F y = 36 ksi)
There’s always a solution in steel!
6.68
Copyright © 2016 American Institute of Steel Construction
6.34
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 5 • Determine the strength of a built-up column composed of 4 - 4 x 4 x 5/16 angles with ½ in. cover plates perforated on two sides 20.0 in.
y = 1.11 in.
Built-up Member Properties
( 2 ) ( 20.0 ) + 2 ( 1 2 ) (10.0 ) = 39.6 in. = 4 ( 3.67 + 2.40 (8.89 ) )
A = 4 ( 2.40 ) + 2 1
. n i
8.89 in.
I x
0 . 0 2
2
2
Gage line
1 ( 20.0 )3 + 2 10 1 (10.25 )2 +2 2 2 12
( ( )
10 in.
)
4
15.0 in.
= 2490 in.
2.5 in.
r x =
2490 39.6
= 7.93 in.
There’s always a solution in steel!
6.69
Example 5 • Check slenderness of cover plates – E6.2 User note says to use Table B4.1a Case 7 b t
=
15.0 0.5
= 30.0 ≤ 1.40
E F y
= 1.40
29, 000 36
=
39.7
– The plate is not slender so our assumption is confirmed
There’s always a solution in steel!
6.70
Copyright © 2016 American Institute of Steel Construction
6.35
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 5 • As a single unit KL r o 2
F e =
π E
KL r
2
=
π
=
25 (12 ) 7.93
= 37.8
2
QF y ( 29, 000 ) = 200 and 2 F e ( 37.8 )
QF F Fcr = Q 0.658
y
e
F y
0.997(36 )
= 0.997 0.658
200
=
0.997 ( 36 ) 200
= 0.179 < 2.25
36 = 33.3 ksi
and
P n = 33.3( 39.6) = 1320 kips There’s always a solution in steel!
6.71
Example 5 • How much of the column strength can be attributed to the presence of the corner angles? For just the plates
( 2 ) ( 20.0 ) + 2( 1 2 ) (10.0 ) = 30.0 in.
A = 2 1
2
1 ( 20.0 )3 + 2 10 1 (10.25 ) 2 I x = 2 2 2 12
( ( )
)
4
= 1720 in.
r x =
1720 30.0
= 7.57 in.
There’s always a solution in steel!
6.72
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 5 As a single unit KL r o 2
F e =
π E
KL r
2
=
π
F Fcr = 0.658 F
25 (12 ) 7.57
= 39.6
2
y e
=
F y ( 29, 000 ) = 183 and 2 F e ( 39.6 )
F y
( 36 )
= 0.658 183
=
( 36 ) 183
= 0.197 < 2.25
36 = 33.2 ksi
and
P n = 33.2 ( 30.0) = 996 kips So the corner angles contributed 324 kips, about equivalent to their area contribution 6.73
There’s always a solution in steel!
Example 6 • Consider a box-shaped member composed of ½ in. plates w/o openings
( 2 ) ( 20.0 ) + 2 ( 1 2 ) (19.0 ) = 39.0 in.
A = 2 1
20.0 in.
. n i 0 . 0 2
1 0 . 0
i n .
2
1 ( 20.0 )3 + 2 19 1 ( 9.75 ) 2 I x = 2 2 2 12
( ( )
)
4
= 2470 in.
r x =
There’s always a solution in steel!
2470 39.0
= 7.96 in.
6.74
Copyright © 2016 American Institute of Steel Construction
6.37
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 6 • Member strength KL r o 2
F e =
π E
KL r
2
=
π
=
25 (12 ) 7.96
= 37.7
2
F Fcr = 0.658 F
y e
F y ( 29, 000 ) = 201 and 2 F e ( 37.7 )
F y
( 36 )
= 0.658 201
=
( 36 ) 201
= 0.179 < 2.25
36 = 33.4 ksi
and
P n = 33.4 ( 39.0) = 1300 kips There’s always a solution in steel!
6.75
Comparison of Results units
Latticed Ex.3
Battened Ex.4
F y
ksi
36
36
A
in.2
33.8
I
in.4
2400
r
in.
8.43
Approx. Wt
lb/ft
KL/r P n
kips
w/ Cover Box Ex.5 Ex.6
HSS 16x16x1/2
W14x132
46
50
36
36
33.8
39.6
3 9.0
28.3
38.8
2400
2490
2470
1130
548
8.43
7.93
7.96
6.31
3.76
172
141
169
133
103
132
35.6
42.5
37.8
37.7
47.5
79.8
1140
1100
1320
1 300
1120
1220
Question: Is all the fabricating effort really worth it?
There’s always a solution in steel!
6.76
Copyright © 2016 American Institute of Steel Construction
6.38
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Channel Columns • Another common shape for use in built-up columns is the double channel • A single channel has the same issues as a single angle with the shear center and centroid not at the same place. • Double channel compression members are doubly symmetric so this is not an issue There’s always a solution in steel!
6.77
Example 7 • Determine the strength of a back-to-back C10 x 15.3 double channel column with a length of 14 ft
Single Channel Properties, Table 1-5
Shear Center and center of gravity for combined channels
Gap = 3/8 in.
A = 4.48 in.2
Center of Gravity for single channel
x = 0.634 in.
I x = 67.3 in.
I y = 2.27 in.4
x = 0.634 in.
S x = 13.5 in.3
Z x = 15.9 in.3
S y = 1.15 in.3
Z y = 2.34 in.3
r x = 3.88 in.
J = 0.209 in.4
r y = 0.711 in.
C w = 45.5 in.6
b f = 2.60 in.
r o = 4.19 in.
t f = 0.436 in.
There’s always a solution in steel!
d = 10.0 in. t w = 0.240 in.
4
H = 0.884 6.78
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 7 • Determine the strength of a back-to-back C10 x 15.3 double channel column with a length of 14 ft Built-up Member Properties
Shear Center and center of gravity for combined channels
A = 2 ( 4.48) = 8.96 in.2
Gap = 3/8 in.
I y = 2 2.27 + 4.48 3 + 0.634 16
(
Center of Gravity for single channel
10.6
r y =
8.96
= 1.09 in.
)
2
= 10.6 in.4
(See also Table 1-16)
If the member works as a single unit KL r o
x = 0.634 in.
=
14 (12 ) 1.09
= 154
KL r x
=
14 (12 ) 3.88
There’s always a solution in steel!
= 43.3
6.79
Example 7 • Check web and flange slenderness – Flange Case 1 b f
=
t f
2.60 0.436
= 5.96 < 0.56
E F y
29, 000
= 0.56
36
= 15.9
– Web Case 5 h tw
= 35.3 < 1.49
E F y
= 1.49
AISC Shapes database
There’s always a solution in steel!
29,000 36
= 42.3
The channel is not a slender element member 6.80
Copyright © 2016 American Institute of Steel Construction
6.40
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 7 • Maximum spacing of connectors KL KL for r o r m
Use
a
a
=
r y
0.711
KL r m
≤ 0.75
= 0.75 (154) = 116
so a = 0.711(116) = 82.5 in.
Use two connectors at 56 in. There’s always a solution in steel!
6.81
Example 7 • Determine modified effective length a r i
=
56 0.711
= 78.8 > 40
Use Equation E6-2b KL r m
KL K ia + r o r i 2
=
2
=
0.75( 56 ) (154 ) + 0.711
There’s always a solution in steel!
2
2
= 165
6.82
Copyright © 2016 American Institute of Steel Construction
6.41
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 7 • Determine nominal strength – For
KL r = 165 m 2
F e =
π E
KL r
2
=
π
2
F y ( 29,000 ) = 10.5 and 2 F e (165)
36
=
10.5
= 3.43 > 2.25
so and
Fcr = 0 .8 77 F e = 0 .8 77 (10.5) = 9.21 ksi
P n = 9.21( 8.96) = 82.5 kips There’s always a solution in steel!
6.83
Example 8 • For the built-up member to be fully effective a
r i
≤ 40
• Which would require connectors at a ≤ 40r i = 40 ( 0.711) = 28.4 in.
• For a 14 ft column, use connectors at 24 in. Therefore a 24 =
r i There’s always a solution in steel!
0.711
= 33.8 ≤ 40
6.84
Copyright © 2016 American Institute of Steel Construction
6.42
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 8 • Determine nominal strength – For
KL r m 2
F e =
π E
KL r
2
=
π
2
KL = 154 r o
=
E6-2a
F y ( 29,000 ) = 12.1 and 2 F e (154)
36
=
12.1
= 2.98 > 2.25
so and
Fcr = 0.877 F e = 0.877 (12.1) = 10.6 ksi
P n = 10.6 ( 8.96) = 95.0 kips A 12.5 kip increase for 4 more bolts There’s always a solution in steel!
6.85
Example 9 • Space the channels so that the strength about the x - and y -axes is about the same • This requires that I x = I y Shear Center and center of gravity for combined channels
(assume a 10% increase in effective slenderness)
I y = 2 (1.1)( 67.3) = 148 = I y = 2 2.27 + 4.48x 2 x = 4.00 in. Center of Gravity x
for single channel
gap 2 x = 0.634 in.
There’s always a solution in steel!
= x − x = 4.00 − 0.634
gap = 2 ( 4.00 − 0.634 ) = 6.73 in.
6.86
Copyright © 2016 American Institute of Steel Construction
6.43
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 9 • Space the channels at 6.75 in. Built-up Member Properties A = 2 ( 4.48) = 8.96 in.2
6.75 in. x Center of Gravity for single channel
6.75 + 0.634 2
I y = 2 2.27 + 4.48 r y =
149
=
8.96
2
4 = 149 in.
4.08 in.
If the member works as a single unit
x = 0.634 in.
KL r o
=
14 (12 ) 4.08
KL r x
= 41.2
=
There’s always a solution in steel!
14 (12 ) 3.88
= 43.3
6.87
Example 9 • For this column to work as intended, we 8.95 in. need either lacing or battens. Gage = 1.5 in. • Use lacing at 60°
10.3 in.
For the channel between lacing a r i
=
10.3 0.711
= 14.5 < 40
So for y -axis, the member works as a unit b f = 2.60 in.
There’s always a solution in steel!
b f
6.75 in. b f 12.0 in.
6.88
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Example 9 • Thus, the x -axis is critical KL r y
KL r o
=
=
14 (12 ) 4.08
2
F e =
π E
KL r
F Fcr = 0.658 F
y e
2
=
F y
π
KL r x
= 41.2
=
14 (12 ) 3.88
= 43.3
2
F y ( 29,000 ) = 153 and 2 F e ( 43.3)
36
= 0.658153 36 = 32.6
ksi
=
36 153
= 0.235 < 2.25
and P n = 32.6( 8.96) = 292 kips
There’s always a solution in steel!
6.89
Example 9 • These two columns are equivalent 6.75 in.
x = 0.634 in.
3.83 in.
Center of Gravity for single channel
Center of Gravity for single channel
8.02 in.
7.77 in.
12.0 in.
9.04 in.
There’s always a solution in steel!
x = 0.634 in.
6.90
Copyright © 2016 American Institute of Steel Construction
6.45
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Comparison of Results units
2-bolt Ex. 7
6-bolt Ex. 8
6.5 in. spaced Ex. 9
HSS 9x9x1/4
W10x33
F y
ksi
36
36
36
46
50
A
in.2
8.96
8.96
8.96
8.03
9.71
I y
in.4
10.6
10.6
149
102
36.6
r
in.
1.09
1.09
4.08
3.56
1.94
Approx. Wt
lb/ft
30.6
30.6
30.6+
29.2
33
165
154
43.3
47.2
86.6
82.5
95.0
292
318
281
( KL/r )m P n
kips
Same Question: Is all the fabricating effort really worth it?
There’s always a solution in steel!
6.91
References • Bleich, F., (1952) Buckling Strength of Metal Structures, McGraw-Hill, NY • Ziemian, R., (ed.) (2010) Guide to Stability Design Criteria for Metal Structures, 6th Ed., John Wiley & Sons, Inc. Hoboken, NJ
There’s always a solution in steel!
6.92
Copyright © 2016 American Institute of Steel Construction
6.46
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Summary • Treated built-up compression members that were directly connected, latticed, battened and cover plated • Studied how the spacing of connectors influenced the overall strength of the member • Compared our strength calculations to the strength of similar HSS and W-shapes • Questioned the value of using built-up members other than those in direct contact
There’s always a solution in steel!
6.93
Lesson 7 • The next lesson will start with a brief treatment of effective length for prismatic members • It will then look at design of bracketed and stepped members with load introduced along the length • Tapered compression members will be considered and a simple starting point for design will be formulated There’s always a solution in steel!
6.94
Copyright © 2016 American Institute of Steel Construction
6.47
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Thank You
American Institute of Steel Construction One East Wacker Drive Chicago, IL 60601 There’s always a solution in steel!
6.95
Individual Webinar Registrants CEU/PDH Certificates Within 2 business days… • You will receive an email on how to report attendance from:
[email protected]. • Be on the lookout: Check your spam filter! Check your junk folder! • Completely fill out online form. Don’t forget to check the boxes next to each attendee’s name!
There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
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AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
Individual Webinar Registrants CEU/PDH Certificates Within 2 business days… • Reporting site (URL will be provided in the forthcoming email). • Username: Same as AISC website username. • Password: Same as AISC website password .
There’s always a solution in steel!
8-Session Registrants CEU/PDH Certificates One certificate will be issued at the conclusion of all 8 sessions.
There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
6.49
AISC Night School March 8, 2016
Steel Design 2: Selected Topics Session 6: Built-up Columns, Lacing, Battens, and Cover Plates
8-Session Registrants Quizzes Access to the quiz: Information for accessing the quiz will be emailed to you by Thursday. It will contain a link to access the quiz. EMAIL COMES FROM
[email protected] Quiz and Attendance records: Posted Wednesday mornings. www.aisc.org/nightschool - click on Current Course Details. Reasons for quiz: •EEU – must take all quizzes and final to receive EEU •CEUs/PDHS – If you watch a recorded session you must take quiz for CEUs/PDHs. •REINFORCEMENT – Reinforce what you learned tonight. Get more out of the course. NOTE: If you attend the live presentation, you do not have to take the quizzes to receive CEUs/PDHs.
There’s always a solution in steel!
8-Session Registrants Recording Access to the recording: Information for accessing the recording will be emailed to you by this Thursday. The recording will be available for two weeks. For 8-session registrants only. EMAIL COMES FROM
[email protected].
CEUs/PDHS – If you watch a recorded session you must take AND PASS the quiz for CEUs/PDHs.
There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction
6.50