AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
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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 4.1
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
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There’s always a solution in steel!
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Copyright © 2016 American Institute of Steel Construction 4.2
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
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There’s always a solution in steel!
Course Description Session 4: Plate Girders February 23, 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 4.3
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Learning Objectives • Gain an understanding of AISC Specification provisions that apply to the design of plate girders • Identify which provisions apply for different cases of singly or doubly symmetric shapes with compact, noncompact, or slender flanges and webs • Determine flexural strength associated with the limit states of yielding, lateral-torsional buckling, and flange local buckling • Calculate the shear strength of a plate girder including the effects of stiffeners and tension field action There’s always a solution in steel!
Steel Design 2: Selected Topics based on AISC 360-10 Specification for Structural Steel Buildings Lesson 4 – Plate Girders 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 4.4
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Steel Design 2: Selected Topics based on AISC 360-10 Specification for Structural Steel Buildings
Night School 10 Lesson 4 Plate Girders
There’s always a solution in steel!
4.9
Lesson 4 – Plate Girders • Plate girders as a term has not been used in AISC 360 since the 2005 Specification • The previous ASD and LRFD Specifications had a separate chapter, Chapter G, that dealt with them by that name • Currently the provisions are found in Chapter F for bending and Chapter G for shear There’s always a solution in steel!
4.10
Copyright © 2016 American Institute of Steel Construction 4.5
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Lesson 4 – Plate Girders • A member made (built-up) from plates in the form of a singly or doubly symmetric Ishape is what we will be referring to as plate girders Compression y1 1
y1
Tension S xc = S xt
S xc < S xt
Sxc > S xt
There’s always a solution in steel!
4.11
Plate Girders • For bending, plate girders are a part of – F2. Doubly symmetric and compact – F3. Doubly symmetric with compact web and noncompact or slender flanges – F4. Doubly symmetric or singly symmetric with compact or noncompact webs – F5. Doubly symmetric or singly symmetric with slender webs
There’s always a solution in steel!
4.12
Copyright © 2016 American Institute of Steel Construction 4.6
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders All rolled W-shapes
Doubly Symmetric Flange compact Web compact Flange noncompact, slender Web compact Flange all Web noncompact Flange all Web slender
Singly Symmetric
compact compact noncompact, slender F3 compact all F4 compact, noncompact all F5 slender F2
NA NA F4 F5
Plate girders could fall into any of these categories
There’s always a solution in steel!
4.13
Plate Girders • For shear, plate girders are a part of – G2. Stiffened or unstiffened webs without tension field action – G3. Stiffened webs with tension field action Rolled W-shapes do not benefit from stiffeners Plate girders may benefit from stiffeners and may benefit from tension field action There’s always a solution in steel!
4.14
Copyright © 2016 American Institute of Steel Construction 4.7
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • For proportioning of plate girders, F13; – Singly symmetric I-shaped members 0.1 ≤
I yc I y
≤ 0.9
F13-2
– I -shaped members with slender webs When
a h
≤ 1.5
h E = 12.0 F y tw max
When
a h
> 1.5
h 0.40 E = F y tw max
For unstiffened girders h t w ≤ 260 There’s always a solution in steel!
4.15
Plate Girders • For our purposes, we will not address those plate girders that fall within the provisions that also cover W-shapes, that is F2 and F3. • We will first look at bending of doubly symmetric plate girders • Then we will look at bending of singly symmetric plate girders • We will conclude by looking at the shear provisions with and without tension field action There’s always a solution in steel!
4.16
Copyright © 2016 American Institute of Steel Construction 4.8
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • F4. for doubly and singly symmetric with noncompact web – The limit state of web local buckling does not lead to a specific nominal strength – Rather, web local buckling modifies the strength determined for the other limit states; yielding , flange local buckling and lateraltorsional buckling , through the use of the web plastification factor, R pc. 4.17
There’s always a solution in steel!
Plate Girders • F4.1 Compression flange yielding n
= R pc M yc = R pc FyS xc
F4-1
• F4.4 Tension flange yielding n
= R pt M yt = R pt FyS xt
F4-15
• F4.3 Compression flange local buckling λ − λ pf λ rf − λ pf
Noncompact M n = Rpc M yc − ( R pc M yc − FL S xc ) Slender M n =
0.9 Ekc S xc
λ 2
F4-13
F4-14
There’s always a solution in steel!
4.18
Copyright © 2016 American Institute of Steel Construction 4.9
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • F4.1 Compression flange yielding n
= R pc M yc = R pc FyS xc
F4-1
For a doubly symmetric shape, these are all equal since S xc = S xt
• F4.4 Tension flange yielding n
= R pt M yt = R pt FyS xt
F4-15
• F4.3 Compression flange local buckling λ − λ pf λ rf − λ pf
Noncompact M n = Rpc M yc − ( R pc M yc − FL S xc ) Slender M n =
0.9 Ekc S xc
F4-13
F4-14
λ 2
There’s always a solution in steel!
4.19
Plate Girders • Look at the web plastification factor, R pc When
hc t w
R pc =
When
≤ λ pw
M p M yc
hc t w
R pc =
=
Z FyZ Fy S
=
Z
Rectangle = 1.5
= Shape Factor W-shapes = 1.1-1.3 S As you add thickness to the flanges, the shape could approach a rectangle, thus the shape factor will approach 1.5.
S
> λ pw p yc
M p λ − λ pw − − 1 M yc λ rw − λ pw
M p ≤ M yc
For doubly symmetric we will not be concerned with this limit p
There’s always a solution in steel!
= Fy Z x ≤ 1.6 Fy S xc 4.20
Copyright © 2016 American Institute of Steel Construction 4.10
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders When I yc/ I y > 0.23
M p M y
If I yc I y ≤ 0.23
M p M y
then
I yc/ I y range for W-shapes 0.49-0.51
Noncompactweb
= 1.6
= 1.0
Compact web
R pc = 1
λ p
λ r
For doubly symmetric, this limit can only be exceeded if the compression flange contributes less than 0.426 I y of the web
There’s always a solution in steel!
4.21
Plate Girders • F4.2 Lateral-torsional buckling When L p < Lb ≤ Lr
Lb − Lp Lr − Lp
M n = Cb Rpc M yc − ( Rpc M yc − FL Sxc )
When Lb > Lr M n =
Cb π2 ES xc
Lb r t
2
1 + 0.078
J Lb
For rolled W-shapes, F L = 0.7 F y
≤ Rpc M yc
F4-2
2
≤ Rpc M yc S xc ho r t
F4-3, F4-5
For rolled W-shapes, r t = r ts
There’s always a solution in steel!
4.22
Copyright © 2016 American Institute of Steel Construction 4.11
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • Determine the nominal moment strength of the given doubly symmetric plate girder, A36 steel. Section Properties I x = 30,600 in.4 I y = 2560 in.4 t f =0.875 in.
S x = S xc = S xt = 1230 in.3 h =hc= 48 in.
Z x = 1330 in.3
t w =0.375 in.
d = 49.75 in. b f = 26.0 in. t w = 0.375 in.
b f = 26.0 in.
t f = 0.875 in.
There’s always a solution in steel!
4.23
Example 1 • Check flange slenderness, Table B4.1b Case 11 b f 2t f
λ p = 0.38
E F y
= 0.38
k c = S xt S xc
29,000 36
4 h t w
=
26 2 ( 0.875)
= 10.8
=
= 14.9
λ r = 0.95
4 48 0.375
kc E FL
= 0.354
= 1.0 ≥ 0.7 therefore F L = 0.7F y
There’s always a solution in steel!
= 0.95
k c ( 29,000) F L
=?
(but no less than 0.35 nor more than 0.76) F4-6a
4.24
Copyright © 2016 American Institute of Steel Construction 4.12
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • Check flange slenderness, Table B4.1b Case 11 kc E
λ r = 0.95
F L
= 0.95
λ p = 10.8 <
b f 2t f
0.354 ( 29,000 ) 0.7 (36 )
= 14.9 < λ r = 19.2
• Check web slenderness, h t w
λ p = 3.76
E F y
= 3.76
29, 000 36
=
= 19.2
48 0.375
= 107
Flange noncompact
Table B4.1b Case 15 0.40 E h t = F = 322 F13-4 max y
= 128
λ r = 5.70
E F y
= 5.70
29, 000 36
= 162
Web - noncompact
There’s always a solution in steel!
4.25
Example 1 • Since the web is noncompact, we must use Section F4. • The web plastification factor impacts all limit states, so first determine R pc. I yc I y M p M yc
3
= =
0.875( 26.0) 12 1282
=
2560 Fy Z Fy S xc
=
1330 1230
2560
= 0.50 > 0.23
= 1.08
There’s always a solution in steel!
4.26
Copyright © 2016 American Institute of Steel Construction 4.13
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • So R pc becomes
In this equation, this inequality will ALWAYS be satisfied
M p M p λ − λ pw M p R pc = − − 1 ≤ M M λ − λ M yc pw yc yc rw
F4-9b
128 − 107 = 1.08 − (1.08 − 1) = 1.05 ≤ 1.08 162 107 −
There’s always a solution in steel!
4.27
Example 1 • F4.1 Compression flange yielding M n = R pc M yc = 1.05 ( 36 )(1230 ) = 46,500 in.-kips
F4-1
• F4.3 Compression flange local buckling – We found that the flange was noncompact λ − λ pf F4-13 λ rf − λ pf 14.9 − 10.8 = 46, 500 − ( 46, 500 − 0.7 ( 36 )(1230 ) ) = 38, 900 in.-kips 19.2 − 10.8
M n = R pc M yc − ( Rpc M yc − FL S xc )
There’s always a solution in steel!
4.28
Copyright © 2016 American Institute of Steel Construction 4.14
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • F4.4 Tension flange yielding Since S xt ≥ S xc this limit state does not apply
• F4.2 Lateral-torsional buckling – Additional section properties J =
r t =
bt 3 3
3
=
3 b fc
ho
12
d
3
2 ( 26 )( 0.875 ) + 48 ( 0.375 )
1 6
h hod 2
+ aw
= 12.5 in.4 r t can be approximated as the radius of gyration of the compression flange plus 1/6 the web
F4-11
There’s always a solution in steel!
4.29
Example 1 • F4.2 Lateral-torsional buckling – Additional section properties aw =
hc t w b fc t fc
=
48 (0.375 ) 26 (0.875 )
= 0.791
r t =
h = 48 in. ho = 48 + 0.875 = 48.875 in.
b fc
F4-11
h 1 h2 12 o + aw d 6 hod 26
=
2 ( 48.875 ) 1 (48 ) + ( 0.791) ( 49.75 ) 6 (48.875 )(49.75 )
12
d = 48 + 2 ( 0.875) = 49.75 in.
= 7.13
There’s always a solution in steel!
4.30
Copyright © 2016 American Institute of Steel Construction 4.15
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • F4.2 Lateral-torsional buckling L p = 1.1r t
E
Lr = 1.95r t F L
E F y
= 1.1( 7.13 )
29,000 36
2
J F L + + 6.76 S xc ho E Sxc ho J
= 223 in. 18.6 ft
29, 000 12.5 = 1.95 ( 7.13) 0.7 ( 36 ) 1230 ( 48.875 ) + = 796 in. 66.3 ft
F4-7
F4-8
2
2
0.7 (36 ) 12.5 + 6.76 29, 000 1230 ( 48.875 )
There’s always a solution in steel!
2
4.31
Example 1 • F4.2 Lateral-torsional buckling When L p < Lb ≤ Lr
Lb − Lp M n = Cb Rpc M yc − ( Rpc M yc − FL Sxc ) ≤ Rpc M yc Lr − Lp
F4-2
L − 18.6 = 1.0 46,500 − ( 46,500 − 0.7( 36)( 1230) ) b 66.3 − 18.6 = 46, 500 − 325 ( Lb − 18.6) (in.-kips)
There’s always a solution in steel!
4.32
Copyright © 2016 American Institute of Steel Construction 4.16
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 1 • F4.2 Lateral-torsional buckling When Lb > Lr M n =
=
=
Cb π2 ES xc
Lb r t 1.0π
2
2
1 + 0.078
1.24 ×108 L
2
≤ Rpc M yc
F4-3, F4-5
S xc ho r t
12 Lb 2 12.5 1 + 0.078 1230( 48.875) 7.13
( 29,000 )(1230 ) 12 Lb 7.13
2 b
J Lb
2
−
1 + 4.59 ×10 5 L2b
(in.-kips)
There’s always a solution in steel!
4.33
Example 1 • Nominal strength
Compression flange local buckling
R pc M yc = 3880 ft-kips
M n = 3240 ft-kips M r = 0.7 FyS xc = 2580 ft-kips
L p
L p′
Lr
There’s always a solution in steel!
4.34
Copyright © 2016 American Institute of Steel Construction 4.17
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • For singly symmetric girders, nothing will change in our approach • However,
hc ≠ h S xc ≠ S xt
• So we must be careful when h and hc or S xc and S xt are called for There’s always a solution in steel!
4.35
Example 2 • Determine the nominal moment strength of the given singly symmetric plate girder, A36 steel. Section Properties b fc = 20.0 in. y
=
4 I x = 32,200 in. d = 50.125 in. I = 2120 in.4 b fc = 20.0 in.
25.78 in.
hc
t fc =1.25 in.
y
2
h = 48 in.
t w =0.375 in.
S xc = 1320 in.3 S xt = 1250 in.3
t ft =0.875 in.
3 Z x = 1380 in.
t fc = 1.25 in. b ft = 26.0 in. t ft = 0.875 in. t w = 0.375 in.
b ft = 26.0 in.
There’s always a solution in steel!
4.36
Copyright © 2016 American Institute of Steel Construction 4.18
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • Check flange slenderness b fc 2t fc
λ p = 0.38
E F y
S xc
=
h t w 1250 1320
=
20 2 (1.25)
= 10.8
36
4
k c = S xt
29,000
= 0.38
=
4 48 0.375
= 8.0 kc E
λ r = 0.95
FL
= 0.95
k c ( 29,000) F L
=?
(but no less than 0.35 nor more than 0.76)
= 0.354
= 0.947 ≥ 0.7 therefore F L = 0.7F y
F4-6a
There’s always a solution in steel!
4.37
Example 2 • Check flange slenderness λ r = 0.95 b f 2t f
kc E F L
= 0.95
0.354 ( 29,000 ) 0.7 (36 )
= 8.0 < λ p = 10.8 < λ r = 19.2
• Check web slenderness hc t w
λ p = 3.76
E F y
= 3.76
= 19.2
29, 000 36
=
46.2 0.375
= 107
hc = 2 ( 48.0 + 0.875 − 25.78 ) = 46.2 in. Less slender than Example 1
= 123
λ r = 5.70
There’s always a solution in steel!
Flange - compact
E F y
= 5.70
29, 000 36
Web - noncompact
= 162
4.38
Copyright © 2016 American Institute of Steel Construction 4.19
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • Since the web is noncompact, we again should use Section F4. • The web plastification factor impacts all limit states, so first determine R pc. I yc I y M p M yc
3
= =
1.25 ( 20.0) 12 2120 Fy Z Fy S xc
=
1380 1320
=
833 2120
= 1.05
= 0.39 > 0.23 0.1 ≤
I yc I y
≤ 0.9 F13-2
There’s always a solution in steel!
4.39
Example 2 • So R pc becomes M p M p λ − λ pw R pc = − − 1 M yc M yc λ rw − λ pw
Remember from Example 1, this inequality will ALWAYS be satisfied
M p ≤ M yc 123 − 107 = 1.05 − (1.05 − 1) = 1.04 ≤ 1.05 162 107 −
There’s always a solution in steel!
F4-9b
4.40
Copyright © 2016 American Institute of Steel Construction 4.20
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • F4.1 Compression flange yielding M n = R pc M yc = 1.04 ( 36 )(1320 ) = 49, 400 in.-kips
F4-1
• F4.3 Compression flange local buckling – We found that the flange was compact so this limit state does not apply
There’s always a solution in steel!
4.41
Example 2 • F4.4 Tension flange yielding Since S xt < S xc this limit state applies M p M yt
and
=
FyZ FyS xt
=
1380 1250
= 1.10
M p M p λ − λ pw M p R pt = − − 1 ≤ λ − λ M yt M yt rw pw M yt
123 − 107 = 1.10 − (1.10 − 1) = 1.07 ≤ 1.10 162 − 107
There’s always a solution in steel!
F4-16b
Note that R pc and R pt are now different 4.42
Copyright © 2016 American Institute of Steel Construction 4.21
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • F4.4 Tension flange yielding M n = Rpt Fy S xt = 1.07 ( 36 )(1250 ) = 48, 200 in.-kips F4-15
There’s always a solution in steel!
4.43
Example 2 • F4.2 Lateral-torsional buckling – Additional section properties J = r t =
bt 3 3
3
=
3
3
20 (1.25 ) + ( 26 )(0.875 ) + 48 (0.375 ) 3 b fc
2 h 1 h 12 o + aw d 6 hod
F4-11
There’s always a solution in steel!
= 19.7 in.4
r t can be approximated as the radius of gyration of the compression flange plus 1/6 the web
4.44
Copyright © 2016 American Institute of Steel Construction 4.22
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • F4.2 Lateral-torsional buckling – Additional section properties aw =
hc t w b fc t fc
=
46.2 (0.375) 20 (1.25)
b fc
r t =
= 0.693
h = 48 in. ho = 48 + 1.25 2 + 0.875 2 = 49.1 in.
h 1 h2 12 o + aw d 6 hod
F4-11
20
=
2 ( 49.1) 1 ( 48 ) 12 + ( 0.693 ) ( 50.1) 6 ( 49.1 )(50.1 )
d = 48 + 1.25 + 0.875 = 50.1 in.
= 5.53
There’s always a solution in steel!
4.45
Example 2 • F4.2 Lateral-torsional buckling L p = 1.1r t
E
Lr = 1.95r t F L
E F y
= 1.1 ( 5.53 )
29,000 36
2
= 173 in. 14.4 ft
J F L + + 6.76 S xc ho S h E xc o J
29, 000 19.7 = 1.95 ( 5.53) 0.7 ( 36 ) 1320 ( 49.1) + = 631 in. 52.6 ft
There’s always a solution in steel!
F4-7
F4-8
2
2
19.7 0.7 (36 ) + 6.76 29, 000 1320 ( 49.1)
2
4.46
Copyright © 2016 American Institute of Steel Construction 4.23
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • F4.2 Lateral-torsional buckling When L p < Lb ≤ Lr
Lb − Lp ≤ Rpc M yc Lr − Lp L − 14.4 = 1.0 49,400 − ( 49, 400 − 0.7 ( 36)( 1320) ) b 52.6 − 14.4
M n = Cb Rpc M yc − ( Rpc M yc − FL Sxc )
= 49, 400 − 422 ( Lb − 14.4)
F4-2
(in.-kips)
There’s always a solution in steel!
4.47
Example 2 • F4.2 Lateral-torsional buckling When Lb > Lr M n =
=
=
Cb π2 ES xc
Lb r t 1.0π
2
2
1 + 0.078
8.02 ×107 2 b
2
≤ Rpc M yc
F4-3, F4-5
S xc ho r t
( 29,000 )(1320 ) 12 Lb 5.53
L
J Lb
2
19.7 12 Lb 2 1 + 0.078 1320( 49.1) 5.53
1 + 1.12 ×10 −4 L2b
There’s always a solution in steel!
(in.-kips) 4.48
Copyright © 2016 American Institute of Steel Construction 4.24
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 2 • Nominal strength
Tension flange yielding
R pc M yc = 4120 ft-kips
M n = 4020 ft-kips
M r = 0.7 FyS xc = 2770 ft-kips
L p L p′
Lr
There’s always a solution in steel!
4.49
Example 3 • Reverse the flanges of the shape in Example 2 so that the smaller width flange is in tension. Section Properties b fc = 26.0 in.
t fc =0.875 in.
4 I x = 32,200 in. d = 50.125 in. I = 2120 in.4 b fc = 26.0 in. y
hc 2
t w =0.375 in.
d = 48 in.
S xc = 1250 in.3 S xt = 1320 in.3
t ft =1.25 in.
3 Z x = 1380 in.
t fc = 0.875 in. b ft = 20.0 in. t ft = 1.25 in. t w = 0.375 in.
b ft = 20.0 in.
There’s always a solution in steel!
4.50
Copyright © 2016 American Institute of Steel Construction 4.25
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • Check flange slenderness b fc 2t fc
λ p = 0.38
E F y
= 0.38
S xc
=
36
4
k c = S xt
29,000
h t w 1320 1250
=
=
26 2 ( 0.875)
= 10.8
Note change from Example 2
= 14.9 kc E
λ r = 0.95
4 48 0.375
FL
= 0.95
k c ( 29,000) F L
=?
(but no less than 0.35 nor more than 0.76)
= 0.354
= 1.06 ≥ 0.7 therefore F L = 0.7F y
F4-6a
Note change from Example 2
There’s always a solution in steel!
4.51
Example 3 • Check flange slenderness λ r = 0.95
kc E F L
= 0.95
λ p = 10.8 <
b f 2t f
0.354 ( 29,000 ) 0.7 (36 )
= 14.9 < λ r = 19.2
• Check web slenderness hc t w
λ p = 3.76
E F y
= 3.76
29, 000 36
=
= 19.2
49.8 0.375
= 107
hc = 2 ( 25.78 − 0.875 ) = 49.8 in. Change from Example 2
= 133
λ r = 5.70
There’s always a solution in steel!
Flange noncompact
E F y
= 5.70
29, 000 36
Web - noncompact
= 162
4.52
Copyright © 2016 American Institute of Steel Construction 4.26
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • Since the web is noncompact, we again will use Section F4. • The web plastification factor impacts all limit states, so first determine R pc. I yc I y M p M yc
3
= =
0.875( 26.0) 12 1280
=
2120 Fy Z Fy S xc
=
1380 1250
2120
= 1.10
= 0.604 > 0.23 0.1 ≤
I yc I y
≤ 0.9 F13-2
There’s always a solution in steel!
4.53
Example 3 • So R pc becomes M p M p λ − λ pw R pc = − − 1 M yc M yc λ rw − λ pw
Remember this inequality will ALWAYS be satisfied
M F4-9b ≤ M 133 − 107 = 1.10 − (1.10 − 1) = 1.05 ≤ 1.10 162 107 −
There’s always a solution in steel!
p
yc
4.54
Copyright © 2016 American Institute of Steel Construction 4.27
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • F4.1 Compression flange yielding M n = R pc M yc = 1.05 ( 36 )(1250 ) = 47,300 in.-kips
F4-1
• F4.3 Compression flange local buckling – We found that the flange was noncompact λ − λ pf F4-13 λ rf − λ pf 14.9 − 10.8 = 47, 300 − ( 47, 300 − 0.7 ( 36 )(1250 ) ) = 39, 600 in.-kips 19.2 − 10.8
M n = Rpc M yc − ( Rpc M yc − FL S xc )
There’s always a solution in steel!
4.55
Example 3 • F4.4 Tension flange yielding Since S xt ≥ S xc this limit state does not apply
• F4.2 Lateral-torsional buckling – Additional section properties J = r t =
3
3
bt 3
=
3
3
20 (1.25 ) + ( 26 )( 0.875 ) + 48 (0.375 ) 3
= 19.7 in.4
b fc 2 h 1 h 12 o + aw d 6 hod
r t can be approximated as the radius of gyration of the compression flange plus 1/6 the web
There’s always a solution in steel!
4.56
Copyright © 2016 American Institute of Steel Construction 4.28
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • F4.2 Lateral-torsional buckling – Additional section properties aw =
hc t w b fc t fc
=
49.8(0.375 ) 26 (0.875 )
r t =
= 0.821
h = 48 in.
=
ho = 48 + 1.25 2 + 0.875 2 = 49.1 in. d = 48 + 1.25 + 0.875 = 50.1 in.
b fc
h 1 h2 12 o + aw d 6 hod
F4-11
26 2 ( 49.1) 1 ( 48 ) 12 + ( 0.821) ( 50.1) 6 ( 49.1 )(50.1 )
= 7.13
There’s always a solution in steel!
4.57
Example 3 • F4.2 Lateral-torsional buckling L p = 1.1r t
E F y
= 1.1( 7.13 )
29,000 36 2
= 223 in. 18.6 ft
J F L + + Lr = 1.95r t + 6.76 F L S xc ho S h E xc o E
J
29, 000 19.7 = 1.95 ( 7.13) 0.7 ( 36 ) 1250 ( 49.1) + = 816 in. 68.0 ft
There’s always a solution in steel!
F4-7
F4-8
2
2
19.7 0.7 (36 ) + 6.76 29, 000 1250 ( 49.1)
2
4.58
Copyright © 2016 American Institute of Steel Construction 4.29
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • F4.2 Lateral-torsional buckling When L p < Lb ≤ Lr
Lb − Lp ≤ Rpc M yc Lr − Lp L − 18.6 = 1.0 47,300 − ( 47,300 − 0.7 ( 36)( 1250) ) b 68.0 − 18.6
M n = Cb Rpc M yc − ( Rpc M yc − FL Sxc )
F4-2
= 47, 300 − 320 ( Lb − 18.6) (in.-kips)
There’s always a solution in steel!
4.59
Example 3 • F4.2 Lateral-torsional buckling When Lb > Lr M n =
=
=
Cb π2 ES xc
Lb r t 1.0π
2
2
1 + 0.078
1.26 ×108 L
2
≤ Rpc M yc
F4-3, F4-5
S xc ho r t
( 29,000 )(1250 ) 12 Lb 7.13
2 b
J Lb
2
19.7 12 Lb 2 1 + 0.078 1250( 49.1) 7.13
1 + 7.09 ×10−5 L2b
There’s always a solution in steel!
(in.-kips) 4.60
Copyright © 2016 American Institute of Steel Construction 4.30
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 3 • Nominal strength
Compression flange local buckling
R pc M yc = 3940 ft-kips M n = 3300 ft-kips M r = 0.7 Fy S xc = 2630 ft-kips
L p
L p′
Lr
There’s always a solution in steel!
4.61
Example 3 • Nominal strength
Ex 2
Ex 3
Ex 1
There’s always a solution in steel!
4.62
Copyright © 2016 American Institute of Steel Construction 4.31
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • F5. for doubly and singly symmetric with slender web – As with F4, the limit state of web local buckling does not lead to a specific nominal strength – Rather, web local buckling modifies the strength determined for the other limit states; yielding , flange local buckling and lateraltorsional buckling , through the use of the bending strength reduction factor, R pg . 4.63
There’s always a solution in steel!
Plate Girders • F5.1 Compression flange yielding n
= R pg Fy S xc
F5-1
• F5.4 Tension flange yielding, S xt < S xc n
= Fy S xt
F5-10
• F5.3 Compression flange local buckling λ − λ pf λ rf − λ pf
Noncompact Fcr = Fy − ( 0.3 F y ) n
= R pg Fcr S xc
F5-7
Slender F cr = There’s always a solution in steel!
0.9 Ek c
b f 2t f
2
F5-8
F5-9
4.64
Copyright © 2016 American Institute of Steel Construction 4.32
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • Look at the bending strength reduction factor, R pg R pg = 1 −
aw =
h E c − 5.7 ≤ 1.0 1, 200 + 300aw t w F y aw
hc t w b fc t fc
≤ 10.0
F5-6
F4-12 plus the limit to 10
There’s always a solution in steel!
4.65
Plate Girders
aw = 0.79 aw = 2 aw = 5 aw = 10
λ pw
λ rw
There’s always a solution in steel!
4.66
Copyright © 2016 American Institute of Steel Construction 4.33
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders • F5.2 Lateral-torsional buckling n
= R pg Fcr S xc
F5-2
When L p < Lb ≤ Lr
Lb − L p Fcr = Cb Fy − ( 0.3Fy ) Lr − Lp
≤ F y
F5-3
When Lb > Lr 2
F cr =
Cb π E
Lb r t
2
F5-4
4.67
There’s always a solution in steel!
Plate Girders • F5.2 Lateral-torsional buckling L p = 1.1r t
E
F4-7
F y r t =
Lr = πr t
b fc
ho 1 h2 12 + aw d 6 hod
F4-11
F5-5
0.7 F y
There’s always a solution in steel!
4.68
Copyright © 2016 American Institute of Steel Construction 4.34
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Plate Girders F4
F2, F3 Z S
F5
t c a p m o c n o N
= 1.6
R pg aw = 0.79
R pc
aw = 10
Compact
Slender
λ pw
λ rw
A user note in Section F4 says that Section F5 may conservatively be used for shapes that fall under the provisions of Section F4 There’s always a solution in steel!
4.69
Example 4 • Reconsider the plate girder from Example 1 using Section F5 Section Properties I x = 30,600 in.4 I y = 2560 in.4 t f =0.875 in.
S x = 1230 in.3 h = 48 in.
t w =0.375 in.
Z x = 1330 in.3 d = 49.75 in. b f = 26.0 in.
b f = 26.0 in.
There’s always a solution in steel!
t w = 0.375 in. t f = 0.875 in. 4.70
Copyright © 2016 American Institute of Steel Construction 4.35
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 4 • Flange and web slenderness are the same as already calculated in Example 1 λ p = 10.8 ≤ λ p = 107 ≤
b f 2t f h t w
= 14.9 < λ r = 19.2 = 128 < λ r = 162
• Since the web is noncompact, we could use Section F4 as we did in Example 1 but we are permitted to conservatively use Section F5 There’s always a solution in steel!
4.71
Example 4 • The bending strength reduction factor impacts other limit states, so first determine R pg . From Example 1 aw = 0.791 hc E F5-6 − 5.7 ≤ 1.0 1, 200 + 300aw t w F y 48.0 0.791 29, 000 5.7 = 1− − 1, 200 + 300 ( 0.791) 0.375 36 = 1.02 > 1.0 therefore R pg = 1 When using F5 in place of
R pg = 1 −
aw
F4, R pg will always be 1.0 There’s always a solution in steel!
4.72
Copyright © 2016 American Institute of Steel Construction 4.36
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 4 • F5.1 Compression flange yielding M n = R pg M yc = 1.0 ( 36 )(1230 ) = 44,300 in.-kips
F5-1
• F5.3 Compression flange local buckling – We found that the flange was noncompact λ − λ pf λ rf − λ pf 14.9 − 10.8 = 36 − 0.3 ( 36 ) 19.2 − 10.8 = 30.7 ksi
Fcr = Fy − ( 0.3F y )
M n = R pg Fcr S xc
F5-8
F5-7
= 1.0 ( 30.7)(1230) = 37,800 in.-kips
There’s always a solution in steel!
4.73
Example 4 • F5.4 Tension flange yielding Since S xt ≥ S xc this limit state does not apply
• F5.2 Lateral-torsional buckling r t = 7.13 L p = 1.1r t Lr = πr t
F4-11
E F y
= 1.1( 7.13)
E 0.7 F y
= π ( 7.13)
29,000 36
= 223 in. 18.6 ft
29,000 0.7 ( 36 )
There’s always a solution in steel!
= 760 in. 63.3 ft
F4-7
F5-5
4.74
Copyright © 2016 American Institute of Steel Construction 4.37
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 4 • F5.2 Lateral-torsional buckling n
= R pg Fcr S xc
F5-2
When L p < Lb ≤ Lr
Lb − Lp ≤ F y Lr − Lp L − 18.6 = 1.0 36 − ( 0.3 ( 36) ) b 63.3 − 18.6
Fcr = Cb Fy − ( 0.3Fy )
= 36 − 0.24 ( Lb − 18.6)
F5-3
(ksi)
There’s always a solution in steel!
4.75
Example 4 • F5.2 Lateral-torsional buckling When Lb > Lr Fcr =
=
=
Cb π2 E
≤ F y 2 Lb r t 1.0π2 ( 29,000 )
12 Lb 7.13 5 1.01× 10 L2b
There’s always a solution in steel!
F5-4
2
(ksi) 4.76
Copyright © 2016 American Institute of Steel Construction 4.38
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 4 From this example, it does appear that using F5 in place of F4 is conservative. Is the simplicity worth it?
Section F4 Example 1
Section F5 Example 4
There’s always a solution in steel!
4.77
Example 5 • Determine the nominal strength of a plate girder with a slender web. This is Example 1 with a thin web Section Properties I x = 29,500 in.4 I y = 2560 in.4 t f =0.875 in.
S x = 1190 in.3 h = 48 in.
t w =0.25 in.
Z x = 1260 in.3 d = 49.75 in. b f = 26.0 in.
b f = 26.0 in.
There’s always a solution in steel!
t w = 0.250 in. t f = 0.875 in. 4.78
Copyright © 2016 American Institute of Steel Construction 4.39
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 5 • Check flange slenderness b f 2t f
λ p = 0.38
E F y
= 0.38
29,000
2 ( 0.875)
= 10.8
36
k c =
26
=
4 h t w
= 14.9
λ r = 0.95 4
=
48 0.25
kc E FL
= 0.95
= 0.289
k c ( 29,000) F L
=?
A change from Example 1
(but no less than 0.35 nor more than 0.76)
k c = 0.35
There’s always a solution in steel!
4.79
Example 5 • Check flange slenderness λ r = 0.95
kc E F L
= 0.95
λ p = 10.8 <
b f 2t f
0.35(29,000 ) 0.7 (36 )
= 19.1
= 14.9 < λ r = 19.1
Flange noncompact
• Check web slenderness λ p = 107 ≤
h t w
0.40 E h = 322 F13-4 = F y t max
=
48 0.25
= 192 > λ r = 162
Therefore the web is slender and we must use Section F5
There’s always a solution in steel!
4.80
Copyright © 2016 American Institute of Steel Construction 4.40
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 5 • The bending strength reduction factor, R pg . aw =
hc t w b fc t fc
=
48 ( 0.25) 26 (0.875)
= 0.527 < 10
F4-12
h E F5-6 c − 5.7 ≤ 1.0 1, 200 + 300aw tw F y 48.0 0.527 29, 000 = 1− − 5.7 1, 200 + 300 ( 0.527 ) 0.250 36 = 0.988 < 1.0 therefore R pg = 0.988
R pg = 1 −
aw
There’s always a solution in steel!
4.81
Example 5 • F5.1 Compression flange yielding M n = R pg M yc = 0.988( 36)(1190) = 42,300 in.-kips
F5-1
• F5.3 Compression flange local buckling – We found that the flange was noncompact λ − λ pf λ rf − λ pf 14.9 − 10.8 = 36 − 0.3 ( 36 ) 19.1 − 10.8 = 30.7 ksi
Fcr = Fy − ( 0.3F y )
F5-8
There’s always a solution in steel!
M n = R pg Fcr S xc
F5-7
= 0.988 ( 30.7)(1190) = 36,100 in.-kips
4.82
Copyright © 2016 American Institute of Steel Construction 4.41
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 5 • F5.4 Tension flange yielding Since S xt ≥ S xc this limit state does not apply
• F5.2 Lateral-torsional buckling – Additional section properties r t =
aw = 0.527 h = 48 in. ho = 48 + 0.875 = 48.875 in.
b fc
26
=
d = 48 + 2 ( 0.875) = 49.75 in.
F4-11
2 h 1 h 12 o + aw d 6 hod
2 ( 48.875 ) 1 ( 48 ) + ( 0.527 ) ( 49.75 ) 6 ( 48.875)( 49.75 )
12
= 7.27
There’s always a solution in steel!
4.83
Example 5 • F5.2 Lateral-torsional buckling L p = 1.1r t Lr = πr t
E F y
= 1.1( 7.27 )
E 0.7 F y
29,000
= π ( 7.27 )
36
= 227 in. 18.9 ft
29,000 0.7 ( 36 )
There’s always a solution in steel!
= 775 in. 64.6ft
F4-7
F5-5
4.84
Copyright © 2016 American Institute of Steel Construction 4.42
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 5 • F5.2 Lateral-torsional buckling n
= R pg Fcr S xc = 0.988Fcr S xc
F5-2
When L p < Lb ≤ Lr
Lb − Lp ≤ F y Lr − Lp L − 18.9 = 1.0 36 − ( 0.3 ( 36) ) b 64.6 − 18.9
Fcr = Cb Fy − ( 0.3Fy )
F5-3
= 36 − 0.236 ( Lb − 18.9) (ksi) There’s always a solution in steel!
4.85
Example 5 • F5.2 Lateral-torsional buckling When Lb > Lr Fcr =
=
=
Cb π2 E
≤ F y 2 Lb r t 1.0π2 ( 29,000 )
12 Lb 7.27 5 1.05 × 10 L2b
There’s always a solution in steel!
F5-4
2
(ksi) 4.86
Copyright © 2016 American Institute of Steel Construction 4.43
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 5 Flange local buckling
R pg M yc = 3530 ft-kips
M n = 3010 ft-kips
M r = Rpg ( 0.7 Fy S xc ) = 2470 ft-kips
L p
L p′
Lr
There’s always a solution in steel!
4.87
Example 5 • Look at the impact of reducing the web thickness
Example 1: 0.375 in. web
Example 5: 0.250 in. web
There’s always a solution in steel!
4.88
Copyright © 2016 American Institute of Steel Construction 4.44
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Compare F4 and F5 • Compression flange yielding M n = R pc FyS xc
F4-1
M n = R pg FyS xc
F5-1
• Compression flange local buckling Noncompact
λ − λ pw λ rw − λ pw λ − λ pw M n = R pg M yc − ( R pg M yc − R pg ( 0.7 Fy S xc ) ) λ − λ pw rw M n = R pc M yc − ( R pc M yc − FL S xc )
F4-13
F5-8 modified
There’s always a solution in steel!
4.89
Compare F4 and F5 • Compression flange local buckling Slender M n =
M n =
0.9 Ekc S xc
b f 2t f
F4-14
2
R pg ( 0.9 Ekc S xc )
b f 2t f
2
There’s always a solution in steel!
F5-9 modified
4.90
Copyright © 2016 American Institute of Steel Construction 4.45
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Compare F4 and F5 • Tension flange yielding n
= Rpt FyS xt
M n = Fy S xt
F4-15 F5-10
• Lateral-torsional buckling When L p < Lb ≤ Lr
Lb − Lp M n = Cb Rpc M yc − ( Rpc M yc − FL S xc ) Lr − Lp
≤ Rpc M yc Lb − Lp ≤ Rpg M yc M n = Cb Rpg M yc − ( Rpg M yc − Rp g ( 0.3 FyS xc ) ) Lr − Lp There’s always a solution in steel!
F4-2
F5-3 modified
4.91
Compare F4 and F5 • Lateral-torsional buckling When Lb > Lr 2
M n =
M n =
Cb π ES xc 2
1 + 0.078
Lb r t 2 R pg Cb π ES xc Lb r t
2
J Lb
2
≤ Rpc M yc
F4-3, F4-5
S xc ho r t
≤ Rpg M yc
There’s always a solution in steel!
F5-2, F5-4
4.92
Copyright © 2016 American Institute of Steel Construction 4.46
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Shear • Chapter G addresses shear strength • For all shapes the limit states to be considered are shear yielding and shear buckling • Tension field action will also be considered but it is not a separate limit state • G2. Members with unstiffened or stiffened webs
Vn = 0.6 Fy AwCv
G2-1
There’s always a solution in steel!
4.93
Shear • The web shear coefficient is a function of web slenderness. For h tw ≤ 1.10 kv E Fy C v = 1.0
G2-3
For 1.10 kv E Fy < h t w ≤ 1.37 kv E Fy C v =
1.10 kv E F y
G2-4
h t w
For h t w > 1.37 kv E Fy C v =
1.51kv E
( h tw )
There’s always a solution in steel!
2
G2-5
F y 4.94
Copyright © 2016 American Institute of Steel Construction 4.47
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Shear • G3. For tensio tension n field field action action h tw ≤ 1.10 kv E Fy Vn = 0.6 Fy Aw
G3-1
From the previous slide, this is C v = 1. Here it means you can not benefit from tension field action.
h tw > 1.10 kv E Fy
1 − C v Vn = 0.6 Fy Aw Cv + 2 1.15 1.15 1 + ( a h )
This is the tension field action benefit
G3-2
There’s always a solution in steel!
4.95
Shear • Web plate plate shear shear buckling buckling coefficient coefficient,, k v – For webs without transverse stiffeners k v = 5 – With transverse stiffeners k v = 5 +
5
( a h)
2
G2-6
260 = 5 when a h > 3 or a h > ( h t w )
2
– Where a = clear distance between stiffeners There’s always a solution in steel!
4.96
Copyright © 2016 American Institute of Steel Construction 4.48
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Shear = 5, that is either no stiffeners Coefficient, C v, for k v = 5 or stiffeners spaced greater than a/h = 3.
F y = 50 ksi
λ pwv
λ rwv
There’s always a solution in steel!
4.97
Shear • With With stiffene stiffeners rs but withou withoutt tension tension field field action action
F y = 50 ksi a h = 0.5 a h = 1 a h ≥ 3
h
There’s always a solution in steel!
a
4.98
Copyright © 2016 American Institute of Steel Construction 4.49
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Shear • With stiffener stiffeners s and and tension tension field action action F y = 50 ksi
a h = 0.5 a h = 1 a h ≥ 3
h a
There’s always a solution in steel!
4.99
Example 6 • Determine Determine the shear shear strength strength of the the thin thin web plate girder from Example 5 Aw = 12.0 in.2 h = 48.0 in.
t f =0.875 in. h = 48 in.
t w = 0.250 in. h t w = 192
t w =0.25 in.
For k v = 5.0 b f = 26.0 in.
λ pwv = 1.10 kv E F y = 69.8 λ rwv = 1.37 kv E F y = 86.9
There’s always a solution in steel!
4.100
Copyright © 2016 American Institute of Steel Construction 4.50
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Since h t w = 192 > λ rwv = 86.9
the web is slender for shear and C v =
1.51kv E
(h
2
tw ) F y
=
1.51( 5)( 29,000) 2
(192) ( 36)
= 0.165
G2-5
and the shear strength is Vn = 0.6 Fy AwCv = 0.6 ( 36)(12.0)( 0.165) = 42.8 kips There’s always a solution in steel!
G2-1
4.101
Example 6 • Assume this girder is on a 120 ft span with bracing every 20 ft. From Example 5 with Lb = 20 ft, the strength is controlled by flange local buckling, M n = 3010 ft-kips • For LRFD φ M n = 0.9 ( 3010 ) = 2710 ft-kips and for uniform load wuM =
2710( 8)
(120 )
2
= 1.51 kip/ft
There’s always a solution in steel!
4.102
Copyright © 2016 American Institute of Steel Construction 4.51
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Based on shear strength φV n = 0.9 ( 42.8 ) = 38.5 kips and for uniform load wuV =
38.5( 2 ) 120
= 0.642 kips/ft
• So the strength of this beam is controlled by shear since wuV = 0.642 < wuM = 1.51 There’s always a solution in steel!
4.103
Example 6 • So increase the shear strength by adding stiffeners with a/h < 3 so that k v will be greater than 5 • Required shear strength V u =
1.51(120 ) 2
= 90.6 kips
so that V n =
V u
φ
=
90.6 0.9
= 101 kips
There’s always a solution in steel!
4.104
Copyright © 2016 American Institute of Steel Construction 4.52
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Determine the minimum C v to provide the needed strength Since Vn = 0.6 Fy AwCv
G2-1
then C v =
V n 0.6 F y Aw
=
101 0.6 ( 36 )(12 )
= 0.390
There’s always a solution in steel!
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Example 6 • To increase C v, we must put in sufficient stiffeners to increase k v. • For the slenderness of the web that we have C v =
1.51kv E 2
( h tw ) F y
=
1.51( k v )( 29, 000) 2
(192) ( 36)
= 0.390
G2-5
so that k v =
0.390 (192)
2
( 36) = 11.8 1.51( 29,000)
There’s always a solution in steel!
4.106
Copyright © 2016 American Institute of Steel Construction 4.53
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Thus, since k v = 5 +
5
( a h)
2
= 11.8
G2-6
• a/h must be 5
a h=
k v − 5
5
=
11.8 − 5
= 0.857
so that a (max) = 0.857 ( 48 ) = 41.1 in.
There’s always a solution in steel!
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Example 6 • Therefore add stiffeners at 40 in. a h = 40 48 = 0.833 k v = 5 +
5
( 0.833)
2
= 12.2
G2-6
For k v = 12.2
λ pwv = 1.10 kv E F y = 109 λ rwv = 1.37 kv E F y = 136 There’s always a solution in steel!
The web will still buckle elastically
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Copyright © 2016 American Institute of Steel Construction 4.54
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Determine the strength of this stiffened web C v =
1.51kv E 2
( h tw ) F y
=
1.51(12.2)( 29, 000) 2
(192 ) ( 36)
= 0.403
G2-5
and Vn = 0.6 Fy AwCv = 0.6 ( 36)(12.0)( 0.403) = 104 kips wuV =
0.9 (104)( 2 ) 120
G2-1
= 1.56 kips/ft > wuM = 1.51 kips/ft
There’s always a solution in steel!
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Example 6 • This solution requires stiffeners every 40 in. but we saw that even for a/h = 3 tension field action can increase strength. • So check to see if with a/h = 3 tension field action will give enough additional strength so shear will not control.
There’s always a solution in steel!
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Copyright © 2016 American Institute of Steel Construction 4.55
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • G3 limits use of tension field action. May not use it if any of the following occur AISC 360-10 limitations
Our example
end panels
not in end panels
a h > 3.0 or 260 ( h tw ) 2 Aw
(A
fc
+ A ft ) > 2.5
h b fc or h b ft > 6.0
2
2
a h = 3.0 and 260 (192 ) = 1.83 2 Aw
(A
fc
+ Aft ) = 0.527
h b fc or h b ft = 1.85
This limit is to aid in fabrication. It will be removed from AISC 360-16 So we will ignore it here There’s always a solution in steel!
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Example 6 • Determine strength with tension field action a = 3 ( 48) = 144 in. 12.0 ft k v = 5 C v = 0.165 as found earlier
2 1.15 1 + ( a h ) 1 − 0.165 = 0.6 ( 36 )(12.0 ) 0.165 + 2 1.15 1+ ( 3.0 )
Vn = 0.6Fy Aw Cv +
1 − C v
There’s always a solution in steel!
G3-2
= 102 kips 4.112
Copyright © 2016 American Institute of Steel Construction 4.56
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Based on shear strength φV n = 0.9 (102 ) = 91.8 kips and wuV =
91.8( 2 ) 120
= 1.53 kips/ft
• So the strength of this beam is fairly well matched between shear and bending wuV
=
1.53
kips/ft
≈
wuM
=
1.51
kips/ft
There’s always a solution in steel!
4.113
Example 6 • This appears to be the most efficient placement of stiffeners since any further apart will not be effective as stiffeners • Since we can not use tension field action in the end panel, put the first stiffener at 40 in. which was determined earlier without tension field action to be sufficient
There’s always a solution in steel!
4.114
Copyright © 2016 American Institute of Steel Construction 4.57
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Example 6 • Final design using tension field action Mid-span
40 in.
144 in.
144 in.
248 in.
in. 144
shear strength envelope 93.6 k
91.8 k
V u = 90.6 k
V u = 35.8 k
35.8 k
316 in.
a h ≥ 3 ok
There’s always a solution in steel!
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Example 6 • Stiffeners must be sized according to G2.2 or G3.3, depending on use of tension field action or not. • This will be left to the student to review.
There’s always a solution in steel!
4.116
Copyright © 2016 American Institute of Steel Construction 4.58
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Summary • We have determined the flexural strength of doubly and singly symmetric girders with compact webs • We have also looked at how girder strength changed as we altered the flange size • We have treated doubly symmetric girders with slender webs and noted how singly symmetric girders would be treated • We also addressed shear strength of plate girders There’s always a solution in steel!
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Lesson 5 • The next lesson will begin our treatment of compression members • We will look at symmetric, singly symmetric and unsymmetric members • This will include single and double angles • It will also include tees and double tees • We will also look at cruciform and Ishaped members that behave similarly There’s always a solution in steel!
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Copyright © 2016 American Institute of Steel Construction 4.59
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
Thank You
American Institute of Steel Construction One East Wacker Drive Chicago, IL 60601 There’s always a solution in steel!
4.119
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 4.60
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
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 4.61
AISC Night School February 23, 2016
Steel Design 2: Selected Topics Session 4: Plate Girders
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 4.62