AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmet Unsymmetric ric Columns Columns
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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 5.1
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmet Unsymmetric ric Columns Columns
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There’s always a solution in steel!
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Copyright © 2016 American Institute of Steel Construction 5.2
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmet Unsymmetric ric Columns Columns
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There’s always a solution in steel!
Course Description Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns March 1, 2016 This session examines single and double angle and single and double WT compression members as well as cruciform and I-shapes that behave in a similar fashion. Limit states of flexural buckling and torsional or flexural-torsional flexural-torsional buckling are explained for members with and without without slender elements. Several examples will be presented.
There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction 5.3
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmet Unsymmetric ric Columns Columns
Learning Objectives • Gain Gain an under understa stand ndin ing g of AIS AISC C Specification provisions that govern design of symmetric, singly symmetric and unsymmet unsymmetric ric columns columns • Underst Understand and and and apply apply the provis provisions ions for for compre compressi ssion on members with slender elements • Calcula Calculate te compress compression ion strengt strength h of members members based based on the the limit states of flexural buckling and torsional or flexuraltorsional buckling • Analyze the interaction interaction of compression compression and and flexural flexural loads on a single angle member There’s always a solution in steel!
Steel Design 2: Selected Topics Topics based on AISC 360-10 Specification for Structural Steel Buildings Lesson 5 – Symmetric, Singly Singly Symmetric and Unsymmetric Columns 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 5.4
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmet Unsymmetric ric Columns Columns
Steel Design 2: Selected Topics based on AISC 360-10 Specification for Structural Steel Buildings Night School 10 Lesson 5 Symmetric, Singly Symmetric and Unsymmetric Columns There’s always a solution in steel!
5.9
Lesson Lesson 5 – Comp Compres ressio sion n • Compre Compressi ssion on membe members rs are cove covered red in in Chapter E • This chapter chapter is organized organized differen differently tly from from Chapter F for flexural members • Single Single angle angles s are the only only shape shape treat treated ed with with their own section, E5 • All other other shapes shapes are are treate treated d throug through h the the same set of provisions
There’s always a solution in steel!
5.10
Copyright © 2016 American Institute of Steel Construction 5.5
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Lesson 5 – Compression • The limit state of flexural buckling is covered in E3 • The limit states of torsional buckling and flexural-torsional buckling are covered in E4 • Section E7 addresses compression members with slender elements and refers back to E3 and E4 when necessary There’s always a solution in steel!
5.11
E3. Flexural Buckling • For nonslender element members in uniform compression Pn = Fcr Ag
When
KL r
E
≤ 4.71
E3-1
F y
F y
or
F e
≤ 2.25
E3-2
F F Fcr = 0.658 F y y e
There’s always a solution in steel!
5.12
Copyright © 2016 American Institute of Steel Construction 5.6
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E3. Flexural Buckling • For nonslender element members in uniform compression When
KL r
> 4.71
E F y
F y > 2.25 or F e
E3-3
Fcr = 0.877 F e
and F e =
π2 E
KL r
E3-4
2
There’s always a solution in steel!
5.13
E3. Flexural Buckling F F Fcr = 0.658 F y
E3-2
Fcr = 0.877F e
E3-3
y
F y = 36 ksi
e
Eq. E3-2
Eq. E3-3
Inelastic Buckling
Elastic Buckling
F e =
π2 E
KL r
2
E3-4
Flexural buckling is the same, regardless of the cross section shape. There’s always a solution in steel!
5.14
Copyright © 2016 American Institute of Steel Construction 5.7
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E4. Torsional and Flexural-Torsional Buckling • Applies to singly symmetric, unsymmetric, and some doubly symmetric members – Double angle and tee-shaped members • Determine F cr directly through Section E4(a)
– All other cases, determine F e through Section E4 and go back to Section E3 to determine F cr
There’s always a solution in steel!
5.15
E4. Torsional and Flexural-Torsional Buckling • E4.(b)(i) Doubly symmetric members π2 EC 1 w Fe = + GJ 2 K L ( z ) I x + I y
E4-4
• E4.(b)(ii) Singly symmetric members 4 Fey Fez H Fey + Fez F e = 1 − 1 − 2 2 H Fey + F ez ) (
There’s always a solution in steel!
E4-5
5.16
Copyright © 2016 American Institute of Steel Construction 5.8
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E4. Torsional and Flexural-Torsional Buckling • E4.(b)(iii) Unsymmetric members 2
2
x y ( Fe − Fex ) ( Fe − Fey ) ( Fe − Fez ) − F ( Fe − Fey ) o − Fe2 ( Fe − F ex ) o = 0 ro r o 2 e
E4-6
F e is the lowest root of this cubic equation
where F ex =
π2 E
K x L r x
2
F ey =
E4-7
π 2 E
K y L r y
2
E4-8
π2 EC 1 w + GJ Fez = 2 2 ( K z L ) A g r o
There’s always a solution in steel!
E4-9
5.17
E4. Torsional and Flexural-Torsional Buckling and H = 1 −
xo2 + yo2
E4-10
2
r o
ro2 = xo2 + yo2 +
I x + I y
E4-11
A g
r o = polar radius of gyration about the shear center xo , yo = coordinates of the shear center with respect to the centroid For a double symmetric member xo = yo = 0. There’s always a solution in steel!
5.18
Copyright © 2016 American Institute of Steel Construction 5.9
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E4. Torsional and Flexural-Torsional Buckling • E4.(a) Double angle and tee-shaped members 4 Fcry Fcrz H Fcry + Fcrz F cr = 1 − 1 − 2 H 2 ( Fcry + F crz )
E4-2
Fcry = F cr from Eq. E3-2 or E3-3 for flexural buckling about
y-axis of symmetry F crz =
GJ A gr o2
E4-3
Note that this is Eq. E4-9 with C w taken as zero
There’s always a solution in steel!
5.19
E5. Single Angle Members • This section gives important direction for all single angles. – The nominal compressive strength of single angle members shall be determined in accordance with Section E3 or E7. This means just consider flexural buckling when b/t ≤ 20. – It also addresses the special cases described where eccentricity can be neglected. • Uses a modified effective slenderness ratio There’s always a solution in steel!
5.20
Copyright © 2016 American Institute of Steel Construction 5.10
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E5. Single Angle Members • But, when b/t >20 use Section E4 also – This means we must address flexural-torsional buckling for those cases when b/t >20 – However, all hot rolled angles, although they may be slender element members, have a leg slenderness less than or equal to 20. – This is fortunate since using E4 requires that we solve the cubic equation for unequal-leg angle compression members, Eq. E4-6 There’s always a solution in steel!
5.21
E7. Slender Element Members • There are 7 categories of slender elements, 4 for unstiffened elements and 3 for stiffened elements. – E7.1(a) projections from rolled columns (Table B4.1a Case 1) – E7.1(b) projections from built-up columns (Table B4.1a Case 2) – E7.1(c) single angles (Table B4.1a Case 3) – E7.1(d) stems of tees (Table B4.1a Case 4) – E7.2(a) stiffened elements except HSS (Table B4.1a Cases 5 & 8) – E7.2(b) rectangular HSS (Table B4.1a Case 6 & 7) – E7.2(c) round HSS (Table B4.1a Case 9) There’s always a solution in steel!
5.22
Copyright © 2016 American Institute of Steel Construction 5.11
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
E7. Slender Element Members • E7.1(c) For single angles When
b t
≤ 0.45
E F y
Q s = 1.0
when
b t
> 0.91
Q s =
when 0.45 E7-10
E
<
F y
b t
≤ 0.91
E F y
b F y E7-11 Q s = 1.34 − 0.76 t E
E F y = 36 ksi
F y
0.53 E
b F y t
2
5x¼the most slender leg angle
Eq. E7-10
E7-12
Eq. E7-12 b t = 12.8
b t = 25.8
There’s always a solution in steel!
5.23
E7. Slender Element Members • Section E7 provides its own equations for flexural buckling but they are the same as those from Section E3 with F y replaced with QF y. For
For
KL r KL r
≤ 4.71
> 4.71
E QF y E QF y
QF F Fcr = 0.658 QF y y
;
;
e
E7-2 (E3-2)
Fcr = 0.877 F e
E7-3 (E3-3)
There’s always a solution in steel!
5.24
Copyright © 2016 American Institute of Steel Construction 5.12
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 1 • Determine the compressive strength of an L8x4x7/16 A36 loaded at its centroid x
w
z
B
C
y
α
P
y
z
w and z represent the major and minor principal axes. x and y represent the major and minor geometric axes.
A g = 5.11 in.2 r z = 0.867 in. A
x w
Although this is an unlikely loading condition, we will address it first.
There’s always a solution in steel!
5.25
Example 1 • Determine the leg slenderness b t
=
8 7 16
= 18.3 > 0.45
E F y
= 12.8
• Therefore the angle is slender and we must use Section E7.1(c) which specifically addresses single angles • We already know that we do not need to address flexural torsional buckling, Eq. E4-6 There’s always a solution in steel!
5.26
Copyright © 2016 American Institute of Steel Construction 5.13
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 1 • Our angle falls in the region of Eq. E7-11 when 0.45
E F y
= 12.8 <
b t
= 18.3 ≤ 0.91
E F y
b F y Q s = 1.34 − 0.76 t E Q s = 1.34 − 0.76 (18.3 )
= 25.8 E7-11
36 29,000
Q = 0.850 can also be found in Manual Table 1-7
= 0.850
The 4 in. leg is not slender. Had it been slender, it would have been conservative to use Q s for the most slender leg. Therefore Q = Q s
There’s always a solution in steel!
5.27
Example 1 • For flexural buckling the division between E 29,000 inelastic and elastic KLr = 4.71 QF = 4.71 = 145 0.85(36 ) y
KL r z
Determine the critical stress for this column with an effective length KL = 5 ft, buckling about the z -axis,
= 69.2
Q = 1.0 Q = 0.85
KL r
= 133.7
KL r
= 145
There’s always a solution in steel!
KL r z
=
5.0 (12 ) 0.867
= 69.2
5.28
Copyright © 2016 American Institute of Steel Construction 5.14
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 1 • For this 5 ft compression member With
QF y F e
KL r z
=
=
5.0 (12 ) 0.867
0.85( 36 ) 59.8
= 69.2,
π2 ( 29, 000 ) π2 E F e = = = 59.8 ksi 2 2 KL 69.2 ( )
E3-4
r z
= 0.511 ≤ 2.25
QF 0.85( 36 ) F Fcr = 0.658 QF y = 0.658 59.8 0.85 ( 36 ) = 24.7 ksi
E7-2
Pn = Fcr Ag = 24.7 ( 5.11) = 126 kips
E7-1
y
e
There’s always a solution in steel!
5.29
Example 2 • A more realistic location for loading might be at the edge of the leg as shown. x
w B
4.0 in.
P
y
C
α
z
A
z
x w
y This loading arrangement will result in bending moment about both principal axes. Thus, we will need to determine moment strength as we did in Lesson 1 Example 5.
There’s always a solution in steel!
5.30
Copyright © 2016 American Institute of Steel Construction 5.15
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Section properties for L8x4x7/16 x
w B
4.0 in.
z
P
Manual Table 1-7
C
y
α
y
w
S wA = 11.0 in.3
t = 7
S wB = 14.6 in.3
in.
y = 2.81 in.
Shapes Database A
d = 4.00 in. 16 x = 0.829 in.
z
x
b = 8.00 in.
Shapes Database I w = 36.4 in.4
A = 5.11 in.2
w A = 2.39 in.
z A = 3.33 in.
w B = 1.53 in.
z B = 2.50 in.
4
I x = 34.2 in.
4
I y = 6.03 in.
wC = 0.758 in. z = 5.17 in. C
S wC = 7.04 in.3 S zA = 1.61 in.3 S zB = 2.51 in.3 S zC = 5.09 in.3
I z = 3.84 in.4 0.867 in. r z = t an α = 0.268
There’s always a solution in steel!
5.31
Example 2 • The determination of flexural strength of the angle follows the procedure discussed in Lesson 1. • F10.1 Yielding M nw = 1.5 ( 36 )( 7.04 ) = 380 in.-kips M nz = 1.5( 36)(1.61) = 86.9 in.-kips
F10-1
• F10.3 Leg local buckling (different than for compression) 0.54
E F y
= 0.54
29, 000 36
= 15.3 <
b t
=
8 29, 000 E = 18.3 > 0.91 = 0.91 = 25.8 7 36 F y 16
Therefore noncompact There’s always a solution in steel!
5.32
Copyright © 2016 American Institute of Steel Construction 5.16
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • F10.3 Leg local buckling – point C is in compression for bending about the w -axis b F y M nw = Fy S wC 2.43 − 1.72 t E
F10-7
8 36 = 36 ( 7.04) 2.43 −1.72 7 29,000 16 = 335 in.-kips There’s always a solution in steel!
5.33
Example 2 • F10.2 Lateral-torsional buckling – Does not apply to z -axis bending
• F10.2(ii) for bending about the major principal axis of an unequal-leg angle
2
L t 4.9 EI z Cb β2w + 0.052 b + βw M e = 2 Lb
r z
=
4.9 ( 29,000)( 3.84)(1.0 )
( 5.0(12 ) )
2
F10-5
2 5.0 (12 ) ( 7 ) 16 + ( −5.48 ) = 505 in.-kips ( −5.48) + 0.052 ( 0.867 ) 2
βw is negative for shear center in tension for bending about major principal axis, w
There’s always a solution in steel!
5.34
Copyright © 2016 American Institute of Steel Construction 5.17
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • F10.2 Lateral-torsional buckling M y = 36 ( 7.04 ) = 253 in.-kips < M e = 505 in.-kips
= 1.92 − 1.17
M nw = 1.92 − 1.17
M yw
M yw ≤ 1.5 M yw M ew 253 ( 253 ) 505
F10-3
= 276 in.-kips < 1.5 ( 253 ) = 380 in.-kips There’s always a solution in steel!
5.35
Example 2 • Thus, for each flexural limit state and each axis, the nominal moment strength, M n, is Limit State
-axis bending (in.-kips)
w
-axis bending (in.-kips)
z
Yielding
380
86.9
Leg local buckling
335
NA
Lateral-torsional Buckling
276
NA
• For compression, from Example 1 P n = 126 kips There’s always a solution in steel!
5.36
Copyright © 2016 American Institute of Steel Construction 5.18
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Determine applied moments w
x
1.19 in.
B
4.0 in.
P
α
y
z C
e z
ew
x = 0.829
Principal axes
y
z
A y = 2.81
w x
w
= Pew
z
= Pez
Geometric axes
We need the principal axis moments so either determine the eccentricities or convert the geometric axis moments to principal axis moments
M x = P (1.19 ) M y = P ( 0.829 )
There’s always a solution in steel!
5.37
Example 2 • Determine applied moments w
x
Resultant Moments
B
z
P
y
z
= 1.36 P
C
x
y
w
α
z
= 0.492 P
y
M w = M x cos α + M y sin α
= 1.19 P cos (15.03 ) + 0.829 P sin (15.03 ) = 1.36 P
A
w x
M z = − M x sin α + M y cos α
= −1.19 P sin (15.03 ) + 0.829 P cos (15.03 ) = 0.492 P
There’s always a solution in steel!
5.38
Copyright © 2016 American Institute of Steel Construction 5.19
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Chapter C requires that we consider all second-order effects. • We are permitted to use the approximate approach in Appendix 8 • Since this single angle column is very flexible about the z -axis in particular, we can not ignore these second-order effects.
There’s always a solution in steel!
5.39
Example 2 • Appendix 8 approximate second-order effects – with no sidesway
– where
B1w =
= B1w M w
rz
= B1z M z
C m α P r
1−
B1 z =
rw
P e1w
and A-8-3
C m α P r
1−
A-8-1
P e1w = P e1z =
π2 EI w
( KL )
2
π2 EI z
( KL )
A-8-5
2
P e1z
There’s always a solution in steel!
5.40
Copyright © 2016 American Institute of Steel Construction 5.20
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Second-order effects C m = 1 for uniform moment along length
α = 1.0 for LRFD α = 1.6 for ASD
P e1w = P e1z =
π2 EI w
( KL )
2
π2 EI z
( KL )
2
= =
π2 ( 29, 000 )( 36.4)
( 60 )
2
π2 ( 29, 000 )( 3.84)
( 60 )
2
= 2890 kips = 305 kips
There’s always a solution in steel!
5.41
Example 2 • Second-order effects for LRFD • If the applied load is P u = 75 kips B1w =
C m = α P r
1−
B1 z =
P e1w
C m = α P r
1−
P e1z
1.0 = 1.03 75
1−
2890
1.0 = 1.33 75
1−
305
There’s always a solution in steel!
5.42
Copyright © 2016 American Institute of Steel Construction 5.21
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Required moment strength for LRFD M uw = 1.03 (1.36 ( 75) ) = 105 in.-kips M uz = 1.33 ( 0.492 ( 75) ) = 49.1 in.-kips
• H2. Interaction equation f ra
+
Fca
f rbw Fcbw
+
f rbz F cbz
≤ 1.0
H2-1
Remember from Lesson 1 that we do not need to use stresses as long as we properly address signs. There’s always a solution in steel!
5.43
Example 2 • Interaction at points A, B, and C for LRFD Pra Pca
−
M rbw M cbw
−
M rbz M cbz A
=
75.0 0.9(126 )
−
105 0.9 ( 276 )
−
49.1 0.9 (86.9)
A
= +0.661 − 0.423 − 0.628 A = 0.390 ≤ 1.0 Pra Pca
−
M rbw M cbw
+
M rbz
M cbz B
=
75.0 0.9 (126 )
−
105 0.9 ( 276 )
+
The heel, point B, is the critical location
49.1 0.9 ( 86.9)
B
= +0.661 − 0.423 + 0.628 B = 0.866 ≤ 1.0 Pra Pca
+
M rbw M cbw
−
M rbz
M cbz
C
=
75.0 0.9 (126 )
+
105 0.9 ( 276 )
−
49.1 0.9 ( 86.9)
C
= +0.661+ 0.423 − 0.628 C = 0.456 ≤ 1.0
There’s always a solution in steel!
5.44
Copyright © 2016 American Institute of Steel Construction 5.22
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Second-order effects for ASD • If the applied load is P a = 50 kips B1w =
B1 z =
C m 1.0 = = 1.03 α P r 1.6 (50.0 ) 1− 1− P e1w 2890 C m = α P r
1−
P e1z
1.0 1−
1.6 (50.0 )
= 1.36
305
There’s always a solution in steel!
5.45
Example 2 • Required moment strength for ASD M aw = 1.03 (1.36 ( 50 ) ) = 70.0 in.-kips M az = 1.36 ( 0.492 ( 50 ) ) = 33.5 in.-kips
• H2. Interaction equation f ra Fca
+
f rbw Fcbw
+
f rbz F cbz
≤ 1.0
H2-1
Remember from Lesson 1 that we do not need to use stresses as long as we properly address signs. There’s always a solution in steel!
5.46
Copyright © 2016 American Institute of Steel Construction 5.23
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 2 • Interaction at points A, B, and C for ASD Pra Pca
−
M rbw M cbw
−
M rbz M cbz A
=
50.0
−
70.0
−
33.5
(126 1.67) ( 276 1.67) (86.9 1.67) A
= +0.663 − 0.424 − 0.644 A = 0.405 ≤ 1.0 Pra Pca
−
M rbw M cbw
+
M rbz
M cbz B
=
50.0
−
70.0
+
33.5
(126 1.67) ( 276 1.67) ( 86.9 1.67) B
The heel, point B, is the critical location
= +0.663 − 0.424 + 0.644 B = 0.883 ≤ 1.0 Pra Pca
+
M rbw M cbw
−
M rbz
M cbz C
=
50.0
+
70.0
−
33.5
(126 1.67) ( 276 1.67) ( 86.9 1.67) C
= +0.663 + 0.424 − 0.644 C = 0.443 ≤ 1.0
There’s always a solution in steel!
5.47
E4. Torsional and Flexural-Torsional Buckling • This section applies to “certain doubly symmetric members” – Cruciform – All doubly symmetric members when the torsional unbraced length exceeds the lateral unbraced length
There’s always a solution in steel!
5.48
Copyright © 2016 American Institute of Steel Construction 5.24
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 3 • Determine the compressive strength of the cruciform column shown when the effective length for lateral buckling and torsional buckling are each 10.0 ft, F y = 50 ksi The cruciform has essentially no warping strength, thus ignore C w.
Section properties 2
A = 7.75 in.
0.5 in. 8 in.
I x = I y = 21.4 in.4 r x = r y = 1.66 in. J =
8 in.
bt 3 3
=
8 ( 0.5) 3
3
+
7.5 ( 0.5) 3
3
= 0.646 in.4
There’s always a solution in steel!
5.49
Example 3 • Check element slenderness Table B4.1a, Case 3 b t
=
3.75 0.5
E
= 7.5 < 0.45
0.5 in. 8 in.
F y
= 0.45
29, 000 50
= 10.8
So the member does not have slender elements
8 in.
There’s always a solution in steel!
5.50
Copyright © 2016 American Institute of Steel Construction 5.25
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 3 • E3. Flexural buckling KL r y
=
10.0 (12 )
F e =
1.66
π2 E
KL r
2
E
= 72.3 < 4.71
=
F F Fcr = 0.658
y e
F y
π2 ( 29,000)
( 72.3)
2
= 4.71
29,000 50
= 113
= 54.8
E3-4
50 54.8 F y = 0.658 (50 ) = 34.1 ksi
There’s always a solution in steel!
E3-2
5.51
Example 3 • E4. Torsional buckling – E4.(b)(i) for doubly symmetric members – Use Eq. E3-2 or E3-3 with F e determined through Eq. E4-4 π2 EC 1 1 w Fe = GJ + = [GJ ] 2 I x + I y ( K z L ) I x + I y =
E4-4
(11, 200 )( 0.646 ) ( 21.4 + 21.4)
= 169 ksi There’s always a solution in steel!
5.52
Copyright © 2016 American Institute of Steel Construction 5.26
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 3 • E4. Torsional buckling – For torsional buckling we don’t have KL/r , therefore it requires that we use the F y/ F e limit F y F e
=
50 169
= 0.296 < 2.25
– So use
F F Fcr = 0.658 F y 50 = 0.658169 ( 50 ) = 44.2 ksi y e
E3-2
There’s always a solution in steel!
5.53
Example 3 • The controlling limit state is flexural buckling so P n = 7.75 ( 34.1) = 264 kips
• Note that for a cruciform, the torsional strength is independent of the torsional effective length since we ignore warping • For torsional buckling to control, this column must be shorter than 5.7 ft There’s always a solution in steel!
5.54
Copyright © 2016 American Institute of Steel Construction 5.27
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 • Determine the strength of a T-shaped column using the same plates as used in Example 3. Section properties A = 7.75 in.2
8 in.
I x = 48.6 in.
The user note says for tees omit C w when computing F ez and take xo as 0.
4
I y = 21.4 in.4
0.5 in. . n i
= 2.50 in. r x
8 in.
r y = 1.66 in.
1 8 . 5
J =
3
bt 3
=
8 ( 0.5 )
3
3
+
7.5 ( 0.5) 3
3
= 0.646 in.4
There’s always a solution in steel!
5.55
Example 4 • Additional properties xo = 0 yo = 8.0 − 5.81 − 0.5 2 = 1.94 in. (distance from centroid to shear center) ro2 = xo2 + yo2 +
I x + I y A g
2
= 0 + (1.94 ) +
48.6 + 21.4 7.75
= 12.8
E4-11
r o = 3.58 (polar radius of gyration) H = 1 −
2 2 xo + yo
r o2
2
=1−
0 + 1.94 12.8
= 0.706
There’s always a solution in steel!
E4-10
5.56
Copyright © 2016 American Institute of Steel Construction 5.28
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 • Check flange and stem slenderness, Table B4.1a Flange: Case 1 b f 2t f
8.0
=
2 ( 0.50 )
E
= 8.0 < 0.56
F y
= 13.5
Nonslender
Stem: Case 4 d t
=
8.0 0.5
= 16 < 0.75
E F y
= 18.1
Nonslender
There’s always a solution in steel!
5.57
Example 4 • E3. Flexural buckling KL r y
=
F e =
10.0 (12 ) 1.66
π2 E
KL r
2
E
= 72.3 < 4.71
=
F F Fcr = 0.658
y e
F y
π2 ( 29,000)
( 72.3)
2
= 4.71
= 54.8
29,000 50
= 113
E3-4
50 E3-2 F y = 0.658 54.8 (50 ) = 34.1 ksi y-axis controls so no change
here from the cruciform There’s always a solution in steel!
5.58
Copyright © 2016 American Institute of Steel Construction 5.29
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 • E4. Flexural-torsional buckling – E4.(a) for double angle and tee shaped compression members – Calculate F cr directly with 4 Fcry FcrzH Fcry + Fcrz F cr = 1 − 1 − 2 2 H ( Fcry + F crz )
E4-2
There’s always a solution in steel!
5.59
Example 4 • where determined) = 34.1 ksi Fcry = F cr (already F crz =
GJ A g r o2
=
11,200 (0.646 ) 7.75 (3.58 )
2
There’s always a solution in steel!
= 72.8 ksi
E4-3
5.60
Copyright © 2016 American Institute of Steel Construction 5.30
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 • Therefore, the critical stress is 4 Fcry Fcrz H Fcry + Fcrz F cr = 1 − 1 − 2 2 H ( Fcry + F crz ) 34.1 + 72.8 4 ( 34.1)( 72.8)( 0.706 ) 1 1 = − − 2 2 (0.706 ) 34.1 72.8 + ( ) = 28.6 ksi < F cr = 34.1 ksi
E4-2
so P n = 28.6 ( 7.75) = 222 kips
E4-1
There’s always a solution in steel!
5.61
Example 4 • E4. Flexural -torsional buckling • We could also use the general provisions – E4.(b)(ii) for singly symmetric members – Use Eq. E3-2 or E3-3 with F e determined through Eq. E4-5 4 Fey Fez H Fey + Fez F e = 1 − 1 − 2 2 H Fey + F ez ) ( There’s always a solution in steel!
E4-5
5.62
Copyright © 2016 American Institute of Steel Construction 5.31
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 thus
F ey =
π2 E
K y L r y
2
=
π2 ( 29,000 )
( 72.3)
2
= 54.8
User note said to ignore C w which makes this equation the same as E4-3
π2 EC 1 GJ w Fez = GJ + = 2 2 Ag r o2 ( K z L ) A g ro
=
11,200 (0.646 ) 7.75 (3.58 )
The same as we had already determined
2
E4-9
= 72.8 ksi
There’s always a solution in steel!
5.63
Example 4 • Therefore, the elastic buckling stress is 4 Fey Fez H Fey + Fez F e = E4-5 1 − 1 − 2 2 H + F F ( ey ez ) 54.8 + 72.8 4 ( 54.8)( 72.8)( 0.706 ) = 2 2 (0.706 ) 1 − 1 − + 54.8 72.8 ( ) = 40.2 ksi
There’s always a solution in steel!
5.64
Copyright © 2016 American Institute of Steel Construction 5.32
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 4 • E4. Flexural-torsional buckling – For flexural-torsional buckling we again use the F y/ F e limit F y F e
=
50 40.2
= 1.24 < 2.25
– So use
F F Fcr = 0.658 F y 50 = 0.658 40.2 ( 50 ) = 29.7 ksi < F cr = 34.1 ksi y e
E3-2
There’s always a solution in steel!
5.65
Example 4 • The controlling limit state is flexuraltorsional buckling so P n
=
(
7.75 29.7
)
=
230 kips
• It is seen that determining the strength according to E4.(a), the special provisions for double angles and tees, is a bit conservative and not all that simpler. E4.(a): Pn = 222 kips and E4.(b)(ii): P n = 230 kips There’s always a solution in steel!
5.66
Copyright © 2016 American Institute of Steel Construction 5.33
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 5 • Using the same area, make an I-shape and determine its strength. 8 in.
User note indicates that for I-shapes
Section properties . n i 5 7 . 7 =
2
A = 7.75 in.
0.25 in. 8 in.
0.5 in.
C w =
I x = 77.7 in.4 I y = 21.4 in.4
o
h
=
= 3.17 in. r x
I y ho2 4
21.4 ( 7.75) 4
2
= 321 in.6
r y = 1.66 in.
Note that J has decreased but we now have a C w.
J =
bt 3 3
=
2 ( 8)( 0.25)
3
3
+
7.5 ( 0.5) 3
3
= 0.396 in.4
There’s always a solution in steel!
5.67
Example 5 • Check flange and web slenderness, Table B4.1a Flange: Case 2 b f 2t f
=
8.0 2 ( 0.25 )
= 16 > 0.64
kc = 4
h t w = 4
kc E F y
= 0.64
0.76( 29,000 ) 50
= 13.4
Slender
7.75 0.5 = 1.02 > 0.76
Web: Case 5 h tw
=
7.5 0.5
= 15 < 1.49
E F y
= 35.9
There’s always a solution in steel!
Nonslender
5.68
Copyright © 2016 American Institute of Steel Construction 5.34
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 5 • For the slender unstiffened flange of a built-up member use E7.1(b) 0.64
0.76 ( 29, 000 )
= 13.4 <
50
b f 2t f
= 16 < 1.17
kc E F y
= 1.17
0.76 ( 29, 000 ) 50
= 24.6
therefore b F y Q s = 1.415 − 0.65 t Ek c = 1.415 − 0.65 (16 )
E7-8
50 0.76 ( 29,000 )
= 0.920
There’s always a solution in steel!
5.69
Example 5 • E3. Flexural buckling KL r y
=
F e =
10.0 (12 ) 1.66
π2 E
KL r
2
= 72.3 < 4.71
=
E
QF y
π2 ( 29,000)
( 72.3)
QF Fcr = Q 0.658 F
y
e
2
= 4.71
29,000 0.92 ( 50 )
= 118
= 54.8
0.92(50 ) F y = 0.92 0.658 54.8 (50 ) = 32.4 ksi
There’s always a solution in steel!
E3-4
E7-2
5.70
Copyright © 2016 American Institute of Steel Construction 5.35
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 5 • E4. Torsional buckling – E4.(b)(i) for doubly symmetric members – Use Eq. E3-2 or E3-3 with F e determined through Eq. E4-4 π2 EC
Fe =
+ GJ
w 2
1 E4-4
( K z L ) I x + I y π2 ( 29000 )( 321) 1 11200 0.396 = + ( ) 2 (10 (12 ) ) ( 77.7 + 21.4) = 64.4 + 44.8 = 109 ksi
There’s always a solution in steel!
5.71
Example 5 • E4. Torsional buckling – For torsional buckling we don’t have KL/r , therefore it requires that we use the F y/ F e limit QF y 0.92( 50 ) F e
=
109
= 0.422 < 2.25
– So use
QF E7-2 F Fcr = Q 0.658 F y 0.92(50 ) = 0.92 0.658 109 ( 50 ) = 38.9 ksi > F cr = 32.4 ksi y
e
There’s always a solution in steel!
5.72
Copyright © 2016 American Institute of Steel Construction 5.36
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 5 • The controlling limit state is flexural buckling so P n = 7.75 ( 32.4 ) = 251 kips
• Comparison of results Limit State Flexural buckling Flexuraltorsional buckling Strength
F e F cr
F e F cr
P n
Ex. 3
Ex. 4
Ex. 5
54.8
54.8
54.8
34.1
34.1
32.4
169
40.2
109
44.2
29.7
38.9
264
230
251
There’s always a solution in steel!
5.73
Example 6 • To better understand when torsional buckling can control for doubly symmetric I-shapes, consider this column at L= 30 ft, braced at the third points for flexural buckling about the x- and y-axes but not braced torsionally at these points KL r y
=
10 (12 ) 1.66
= 72.3, Fe =
π2 ( 29, 000 )
( 72.3)
2
0.92(50 ) = 54.8, F cr = 0.92 0.658 54.8 50 = 32.4 ksi
Flexural buckling is unchanged from Example 5 There’s always a solution in steel!
5.74
Copyright © 2016 American Institute of Steel Construction 5.37
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 6 • For torsional buckling, with the unbraced length K z L = 30 ft π2 EC 1 w Fe = GJ + E4-4 2 I I + K L ( z ) x y π2 ( 29000 )( 321) 1 11200 0.396 = + ( ) 2 ( 30 (12 ) ) ( 77.7 + 21.4) = 51.9 ksi
There’s always a solution in steel!
5.75
Example 6 • Since F e for torsional buckling is less than F e for flexural buckling, torsional buckling controls and QF y F e
=
0.92(50 ) 51.9
= 0.886 < 2.25
Which results in
QF P n = 7.75 ( 31.8) = 246 kips F E7-2 Fcr = Q 0.658 F y 0.92( 50) = 0.92 0.658 51.9 ( 50 ) = 31.8 ksi y
e
There’s always a solution in steel!
5.76
Copyright © 2016 American Institute of Steel Construction 5.38
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Compression • In the 6 examples we have considered so far, we have treated, single angle, cruciform, tee, and I-shape • For flexural buckling, all shapes are treated the same • For torsional or flexural-torsional buckling, shapes are broken into 4 classifications • For slender elements, there are 7 categories
There’s always a solution in steel!
5.77
Compression • To consider some additional possibilities, we will look at a double angle and a double WT • We will find that we are using the same provisions already considered • For the built-up members, we will assume, for this Lesson, that they are sufficiently attached to each other so that they do work together as one • Requirements for that connection will be discussed in Lesson 6 There’s always a solution in steel!
5.78
Copyright © 2016 American Institute of Steel Construction 5.39
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 7 • Determine the nominal compressive strength of 2-L5 x 3 x ¼ LLBB A36 angles used as the top chord of a truss. The angles are attached with welds as needed to insure they work as one. 10 ft
There’s always a solution in steel!
5.79
Example 7 • Determine the nominal compressive strength of 2-L5 x 3 x ¼ LLBB A36 angles used as the top chord of a truss. Single Angle Table 1-7
3/8 in.
A g = 1.94 in.2
I x = 5.09 in.4
r x = 1.62 in.
I y = 1.41 in.4
r y = 0.853 in. r z = 0.652 in.
x
x
J = 0.0438 in.4
x = 0.648 in.
Q = 0.804 Shape has a slender leg
There’s always a solution in steel!
y 5.80
Copyright © 2016 American Institute of Steel Construction 5.40
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 7 • Combined properties Table 1-15 A g = 3.88 in.2
H = 0.638 in. r o = 2.52 in.
4
I x = 2 ( 5.09 ) = 10.2 in. r x = 1.62 in.
3/8 in.
2 4 I y = 2 1.41 + 1.94 ( 0.648 + 3 ) = 5.53 in. 16
r y =
5.53 3.88
= 1.19 in.
x
x = 0.648 in.
J = 2 ( 0.0438) = 0.0876 in.4
y There’s always a solution in steel!
5.81
Example 7 • E3. Flexural buckling KL r y
=
10.0 (12 ) 1.19
F e =
QF F Fcr = Q 0.658
y
e
= 101 < 4.71
π2 E
KL r
2
=
E
QF y
= 4.71
π2 ( 29,000 )
(101)
2
29,000 0.804 ( 36 )
= 149
= 28.1
0.804(36 ) F y = 0.804 0.658 28.1 ( 36 ) = 18.8 ksi
There’s always a solution in steel!
E3-4
E7-2
5.82
Copyright © 2016 American Institute of Steel Construction 5.41
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 7 • E4. Flexural-torsional buckling – E4.(a) for double angle and tee shaped compression members – Calculate F cr directly with 4 Fcry FcrzH Fcry + Fcrz F cr = 1 − 1 − 2 2 H ( Fcry + F crz )
E4-2
There’s always a solution in steel!
5.83
Example 7 • where determined) = 18.8 ksi Fcry = F cr (already F crz =
GJ A g r o2
=
11, 200 ( 0.0876 ) 3.88 ( 2.52 )
2
= 39.8 ksi
E4-3
Note that there is no impact from the slender element in F crz
There’s always a solution in steel!
5.84
Copyright © 2016 American Institute of Steel Construction 5.42
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 7 • Therefore , the critical stress is 4 Fcry Fcrz H Fcry + Fcrz F cr = 1 − 1 − 2 2 H ( Fcry + F crz ) 18.8 + 39.8 4 (18.8) 39.8 (0.638 ) 1 1 = − − 2 2 (0.638 ) 18.8 39.8 + ( ) = 15.3 ksi < F cr = 18.8 ksi
E4-2
so P n
=
(
15.3 3.88
)
=
59.4 kips
There’s always a solution in steel!
E4-1
5.85
Double Angles • Leg slenderness for double angles is a function of spacing, gap (3) or no gap (1). F y = 36 ksi Case 1, Q s = 0.894
Double angles in contact, E7.1(a)
Single angles and double angles with a gap, E7.1(c)
Case 3, Q s = 0.804
5x¼
There’s always a solution in steel!
5.86
Copyright © 2016 American Institute of Steel Construction 5.43
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Double WT • Since WT’s are made form W-shapes, the flange will not be slender since the Wshape flange is not slender (Same Table B4.1a Case 1) • The stem is often slender according to Case 4 in Table B4.1a • Unlike for double angles, combining WT’s does not alter the element slenderness check
There’s always a solution in steel!
5.87
Double WT • Stem slenderness, E7.1(d) F y = 50 ksi
Eq. E7-13
Eq. E7-15 0.75
E F y
1.03
E F y
There’s always a solution in steel!
5.88
Copyright © 2016 American Institute of Steel Construction 5.44
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Double WT • For torsional buckling use the provisions for a doubly symmetric member, E4(b)(i) • As for the cruciform, ignore the warping contribution, C w • The controlling axis for flexural buckling is not always the y-axis. • As we did with double angles, we will assume that they are connected sufficiently to be considered a single member There’s always a solution in steel!
5.89
Example 8 • Determine the nominal compressive strength of a member composed of 2-WT6 x 29 A992 shapes with an effective length Single WT, of 20 ft. 2-WT calculated y
y = 1.03 in.
Manual Table 1-8 2 A = 8.52 in. I x = 19.1 in.4
Gap = 3/8 in.
4 I y = 53.5 in.
x
A = 2 ( 8 .5 2) = 1 7. 0 in.2 2 I x = 2 19.1 + 8.52 ( 1.03+ 3 16)
= 63.5 in.
= 1.50 in. r x r y = 2.51 in. 4
4
r x =
63.5 17.0
= 1.93 in.
J = 1.05 in.
I y = 2 ( 5 3. 5) = 1 07 in.4
Q s = 1.00
J = 2 (1 .0 5) = 2 .1 0 in.4
There’s always a solution in steel!
5.90
Copyright © 2016 American Institute of Steel Construction 5.45
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 8 • E3. Flexural buckling KL r x
F e =
20.0 (12 )
=
1.93
π2 E
KL r
2
=
= 124 > 4.71
π2 ( 29,000 )
(124 )
2
E F y
= 4.71
29,000 50
= 113
= 18.6 ksi
E3-4
Fcr = 0.877 F e = 0.877 ( 18.6) = 16.3 ksi
E3-3
There’s always a solution in steel!
5.91
Example 8 • E4. Torsional buckling – E4.(b)(i) for doubly symmetric members – Use Eq. E3-2 or E3-3 with F e determined through Eq. E4-4 π2 EC 1 1 w Fe = GJ + = [GJ ] 2 I x + I y ( K z L ) I x + I y =
E4-4
(11, 200 )( 2.10) ( 63.5 + 107 )
= 138 ksi There’s always a solution in steel!
5.92
Copyright © 2016 American Institute of Steel Construction 5.46
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 8 • E4. Torsional buckling – For torsional buckling we don’t have KL/r , therefore it requires that we use the F y/ F e limit F y F e
=
50 138
= 0.362 < 2.25
– So use
F F Fcr = 0.658 F y 50 = 0.658138 ( 50 ) = 43.0 ksi y e
There’s always a solution in steel!
E3-2
5.93
Example 8 • The controlling limit state is flexural buckling so P n = 17.0 (16.3) = 277 kips
• Since flexural buckling about the x-axis is critical, the connectors will be in shear and will be critical. We have assumed that this connection is adequate.
There’s always a solution in steel!
5.94
Copyright © 2016 American Institute of Steel Construction 5.47
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 8 • What would have changed if we had included the contribution of warping? Single WT C w = 2.08 in.4 Manual Table 1-8 C w = 2 ( 2.08) = 4.16 in.4
2-WT
π2 EC
Fe =
w 2
+ GJ
1
E4-4
( K z L ) I x + I y π2 ( 29,000)( 4.16) 1 = + 11, 200 2.10 ( )( ) 2 ( 63.5 + 107 ) ( 20 (12 )) =
20.6 + 23,520
( 63.5 + 107 )
= 138 ksi
No change
There’s always a solution in steel!
5.95
Example 9 • Now consider 2-WT8 x 33.5 A992 shapes with an effective length of 10 ft. y y = 1.56 in.
Single WT, Manual Table 1-8 2 A = 9.81 in. I x = 48.6 in.4
Gap = 3/8 in.
x
I y = 59.5 in.4
2-WT calculated A = 2 ( 9 .8 1) = 1 9. 6 in.2 2 I x = 2 48.6 + 9.81( 1.56 + 3 16)
= 157 in.
r x = 2.22 in. r y = 2.46 in.
4
r x =
157 19.6
= 2.83 in.
J = 1.19 in.4
I y = 2 ( 5 9. 5) = 1 19 in.4
Q s = 0.859
J = 2 (1 .1 9) = 2 .3 8 in.4
There’s always a solution in steel!
5.96
Copyright © 2016 American Institute of Steel Construction 5.48
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 9 • E3. Flexural buckling KL r y
F e =
=
10.0 (12 ) 2.46
π2 E
KL r
2
=
E
= 48.8 < 4.71
F y
π2 ( 29,000 )
( 48.8)
QF Fcr = Q 0.658 F y
e
2
= 4.71
29,000 50
= 113
= 120 ksi
E3-4
0.859( 50) F y = 0.859 0.658 120 (50 ) = 37.0 ksi
There’s always a solution in steel!
E3-2
5.97
Example 9 • E4. Torsional buckling – E4.(b)(i) for doubly symmetric members – Use Eq. E3-2 or E3-3 with F e determined through Eq. E4-4 π2 EC 1 1 w Fe = GJ + = [GJ ] 2 I x + I y ( K z L ) I x + I y =
E4-4
(11, 200 )( 2.38) (157 + 119 )
= 96.6 ksi There’s always a solution in steel!
5.98
Copyright © 2016 American Institute of Steel Construction 5.49
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Example 9 • E4. Torsional buckling – For torsional buckling we don’t have KL/r , therefore it requires that we use the F y/ F e limit QF y 0.859 ( 50 ) F e
– So use
=
= 0.445 < 2.25
96.6
QF F Fcr = Q 0.658 F y 0.859( 50) = 0.859 0.658 96.6 (50 ) = 35.7 ksi y
e
E3-2
There’s always a solution in steel!
5.99
Example 9 • The controlling limit state is torsional buckling so P n = 19.6 ( 35.7 ) = 670 kips
Note: There are 273 different WT-shapes. Of these the x -axis will control for 72 and the y -axis will control for 201 when the gap is 3/8 in.
There’s always a solution in steel!
5.100
Copyright © 2016 American Institute of Steel Construction 5.50
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
Summary • We have considered 6 different shape compression members, single angle, double angle, WT, double WT, cruciform, and I-shape • Four different slender elements were addressed • Flexural buckling and torsional or flexuraltorsional buckling were addressed • Combined axial and bending for an unsymmetric single angle was treated. • All built up members were assumed to act as a single member There’s always a solution in steel!
5.101
Lesson 6 • The built-up members treated in this lesson will again be looked at to assess the assumption of action as a single member • Additional built-up shapes will be considered such as multiple angles and double channels • Connections between elements will be addressed as lacing, battens, and tie plates
There’s always a solution in steel!
5.102
Copyright © 2016 American Institute of Steel Construction 5.51
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
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5.103
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There’s always a solution in steel!
Copyright © 2016 American Institute of Steel Construction 5.52
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
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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 5.53
AISC Night School March 1, 2016
Steel Design 2: Selected Topics Session 5: Symmetric, Singly Symmetric and Unsymmetric Columns
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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 5.54