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8/12/2019 Members in Compression - V
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Compression Members
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COLUMN STABILITY
A.Flexural Buckling
Elastic Buckling
Inelastic Buckling
Yielding
B. Local BucklingSection E7 pp 16.1-39
and B4 pp 16.1-14
C. Lateral Torsional Buckling
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AISC Requirements
CHAPTER E pp 16.1-32
Nominal Compressive Strength
gcrn AFP
AISC Eqtn E3-1
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AISC Requirements
LRFD
ncu PP
loadsfactoredofSumuP
strengthecompressivdesignncP
0.90ncompressioforfactorresistance c
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In Summary
877.0
44.0or
71.4658.0
otherwiseF
FF
F
E
r
KLifF
F
e
ye
y
yF
F
cr
e
y
200
r
KL
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In Summary - Definition of Fe
Elastic Buckling Stress corresponding to the controlling mode of
failure (flexural, torsional or flexural torsional)F
e:
Theory of Elastic Stability (Timoshenko & Gere 1961)
Flexural Buckling Torsional Buckling
2-axis of symmetry
Flexural Torsional
Buckling
1 axis of symmetry
Flexural Torsional
Buckling
No axis of symmetry
22
/ rKLEFe
AISC EqtnE4-4
AISC EqtnE4-5
AISC EqtnE4-6
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Column Design Tables
Assumption : Strength Governed by Flexural Buckling
Check Local Buckling
Column Design Tables
Design strength of selected shapes for effective length KL
Table 4-1 to 4-2, (pp 4-10 to 4-316)
Critical Stress for Slenderness KL/r
table 4.22 pp (4-318 to 4-322)
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Design of Members in Compression
Selection of an economical shape: Find lightest shape
Usually category is defined beforehand, e.g. W, WT etc
Usually overall nominal dimensions defined in advance
because of architectural and other requirements.
USE OF COLUMN LOAD TABLES
IF NOT APPLICABLE - TRIAL AND ERROR
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EXAMPLE I COLUMN LOAD TABLES
A compression member is subjected to service loads pf 165 dead and 535
kips live. The member is 26 feet long and pinned at each end
LRFD
Calculate factored load
kips054,1)535(6.1)165(2.16.12.1 LDPu
Required Design Strength
kips054,1ncP
Enter Column Tables with KL=(1)(26)=26 ft
uPXW kips230,1:strengthdesign14514 OK
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EXAMPLE I COLUMN LOAD TABLES
A compression member is subjected to service loads pf 165 dead and 535
kips live. The member is 26 feet long and pinned at each end
ASD
Calculate factored load
kips700)535()165( LDPa
Required Allowable Strength
kips700c
nP
Enter Column Tables with KL=(1)(26)=26 ft
aPXW kips702:strengthdesign13214 OK
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EXAMPLE Ii COLUMN LOAD TABLES
Select the lightest W-shape that can resist a service dead load of 62.5
kips and a service live load of 125 kips. The effective length is 24 feet.
Use ASTM A992 steel
LRFD
Calculate factored load and required strength
nu PLDP ckips275)125(6.1)5.62(2.16.12.1
Enter Column Tables with KL=(1)(24)=24 ft
No Footnote: No need to
check for local buckling
kips275withW8No:8 ncPW
kips282,5410:10 c nPXWW
kips293,5810:12 c nPXWW
kips293,6114:14 c nPXWW
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IF COLUMNS NOT APPLICABLE
1. Assume a value for Fcr
ycr FF
2. Determine required areaLRFD
crc
ugugcrc
F
PAPAF
ASD
cr
ag
g
acr
F
PA
A
PF
6.06.0
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IF COLUMNS NOT APPLICABLE
3 Select a shape that satisfies area requirement
4 Compute Fcrfor the trial shape
5 Revise if necessary
If available strength too close to required value try next tabulated value
Else repeat 1-4 using Fcrof trial shape
6 Check local stability and revise if necessary
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Calculate factored load and required strength
kips600)300(6.1)100(2.16.12.1 LDPu
Try ksi333
2 ycr FF
Required Area 2in2.20339.0
600
crc
ugF
PA
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x71
2.20in8.20
2
gA
Slenderness 2005.18370.1
1226
min
r
KLOK
OKAvailable Area
Eulers Stress ksi5.85.183
)000,29(
/2
2
2
2
rKL
E
Fe
11350
000,2971.471.4
yF
EElastic Buckling
Slenderness Limit
5.183
min
r
KL
ELASTIC BUCKLING
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress ksi455.75.8877.0877.0 ecr FF
NG
Design Strength
kips140)8.20)(455.7(9.0 gcrcnc AFP kips600
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Required Area
2in3.33209.0
600
crc
ug
F
PA
Assume NEW
Critical Stress
2
455.733ksi20crF
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x119
3.33in1.35
2
gA
Slenderness 2000.11669.2
1226
min
r
KLOK
OKAvailable Area
Eulers Stress ksi27.210.116
)000,29(
/2
2
2
2
rKL
E
Fe
11350
000,2971.471.4
yF
EElastic Buckling
Slenderness Limit
116
min
r
KL
ELASTIC BUCKLING
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress ksi65.1827.21877.0877.0 ecr FF
Design Strength
kips589)1.35)(65.18(9.0 gcrcnc AFP kips600 NG
This is very close, try next larger size
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Try W 18x130
2
in2.38gA
Slenderness 2006.11570.2
1226
min
r
KLOK
Available Area
Eulers Stress ksi42.216.115
)000,29(
/2
2
2
2
rKL
E
Fe
11350
000,2971.471.4
yF
EElastic Buckling
Slenderness Limit
6.115
min
r
KL
ELASTIC BUCKLING
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Example
Select a W18 shape of A992 steel that can resist a service dead load of
100 kips amd a service live load of 300 kips. Effective length KL=26 ft
Critical Stress ksi79.1842.21877.0877.0 ecr FF
OK
Design Strength
kips646)2.38)(79.18(9.0 gcrcnc AFP kips600
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More on Effective Length Factor
A
B
IgLg IgLg
IcLc
IcLc
Assumptions
All columns under
consideration reach bucklingSimultaneously
All joints are rigid
Consider members lying in the
plane of buckling
All members have constant A
Elastic Behavior
Define:
gg
cc
ALI
LIG
gg
cc
BLI
LIG
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Effective Length Factor-Alingnment Charts
Use alignment charts (Structural Stability Research Council SSRC)
AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241
Connections to foundations
(a) Hinge
G is infinite - Use G=10
(b) Fixed
G=0 - Use G=1.0
Assumption of Elastic Behavior is violated whenInelastic Flexural Buckling min
71.4r
KL
F
E
y
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Example
gg
cc
LI
LIG
Joint A
94.018/183020/1350
12/107012/833 AG
A
B
W12x120
C
W12x120
W24x68
W24x68W24x55
W24x55
12
15
12W12x96
20 18
Joint B
95.018/183020/1350
15/107012/1070
BG
Joint C
0.10CGPinned EndSway Uninhibited
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Example
AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241
COLUMN AB
94.0AG
95.0BG
3.1xK
COLUMN BC
0.10cG
95.0BG
85.1xK
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More on Effective Length
Assumptions
All columns underconsideration reach bucklingSimultaneously
All joints are rigid
Consider members lying in theplane of buckling
All members have constant A
Elastic Behavior Violated
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Example
Compute Stiffness Reduction Factor per LRFD for an axial compressive stress of25 ksi and Fy=50 ksi
ksi25
g
u
A
P
y
FF
gc
uinelasticcr
F.A
PF ey6580ksi27.78
9.0
25)(
ksi61.35506580ksi27.78 50 eF F. e
23.3161.35877.08770)(
eelasticcr F.F
890023.31
28.27
)(
)(.
F
F
elasticcr
inelasticcr
a
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