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Compression Members
COLUMN STABILITY
A. Flexural Buckling• Elastic Buckling• Inelastic Buckling• Yielding
B. Local Buckling – Section E7 pp 16.1-39 and B4 pp 16.1-14
C. Lateral Torsional Buckling
AISC Requirements
CHAPTER E pp 16.1-32
Nominal Compressive Strength
gcrn AFP
AISC Eqtn E3-1
AISC Requirements
LRFD
ncu PP
loads factored of Sum uP
strength ecompressiv design ncP
0.90 ncompressiofor factor resistance c
In Summary
877.0
44.0or
71.4 658.0
otherwiseF
FF
F
E
r
KLifF
F
e
ye
yy
F
F
cr
ey
200r
KL
In Summary - Definition of Fe
Elastic Buckling Stress corresponding to the controlling mode of failure (flexural, torsional or flexural torsional)
Fe:
Theory of Elastic Stability (Timoshenko & Gere 1961)
Flexural Buckling Torsional Buckling2-axis of symmetry
Flexural Torsional Buckling1 axis of symmetry
Flexural Torsional BucklingNo axis of symmetry
2
2
/ rKL
EFe
AISC EqtnE4-4
AISC EqtnE4-5
AISC EqtnE4-6
Column Design Tables
Assumption : Strength Governed by Flexural BucklingCheck Local Buckling
Column Design Tables
Design strength of selected shapes for effective length KLTable 4-1 to 4-2, (pp 4-10 to 4-316)
Critical Stress for Slenderness KL/rtable 4.22 pp (4-318 to 4-322)
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
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
kips 054,1)535(6.1)165(2.16.12.1 LDPu
Required Design Strength
kips 054,1ncP
Enter Column Tables with KL=(1)(26)=26 ft
uPXW kips 230,1 :strengthdesign 14514 OK
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
kips 700)535()165( LDPa
Required Allowable Strength
kips 700c
nP
Enter Column Tables with KL=(1)(26)=26 ft
aPXW kips 702 :strengthdesign 13214 OK
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 ckips 275)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
kips 275 with W8No:8 ncPφW
kips 282 ,5410 :10 c nPXWW
kips 293 ,5810 :12 c nPXWW
kips 293 ,6114 :14 c nPXWW
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
IF COLUMNS NOT APPLICABLE
3 Select a shape that satisfies area requirement
4 Compute Fcr for the trial shape
5 Revise if necessary• If available strength too close to required value try next tabulated value
• Else repeat 1-4 using Fcr of trial shape
6 Check local stability and revise if necessary
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
kips 600)300(6.1)100(2.16.12.1 LDPu
Try ksi 333
2 ycr FF
Required Area 2in 2.20
339.0
600
crc
ug F
PA
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
2.20in 8.20 2 gA
Slenderness 2005.18370.1
1226
min
r
KLOK
OKAvailable Area
Euler’s Stress
ksi 5.85.183
)000,29(
/ 2
2
2
2
rKL
EFe
11350
000,2971.471.4
yF
EElastic BucklingSlenderness Limit
5.183min
r
KL
ELASTIC BUCKLING
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 ksi 455.75.8877.0877.0 ecr FF
NG
Design Strength
kips 140)8.20)(455.7(9.0 gcrcnc AFP kips 600
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 2in 3.33
209.0
600
crc
ug F
PA
Assume NEW Critical Stress
2
455.733 ksi 20crF
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
3.33in 1.35 2 gA
Slenderness 2000.11669.2
1226
min
r
KLOK
OKAvailable Area
Euler’s Stress
ksi 27.210.116
)000,29(
/ 2
2
2
2
rKL
EFe
11350
000,2971.471.4
yF
EElastic BucklingSlenderness Limit
116min
r
KL
ELASTIC BUCKLING
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 ksi 65.1827.21877.0877.0 ecr FF
Design Strength
kips 589)1.35)(65.18(9.0 gcrcnc AFP kips 600 NG
This is very close, try next larger size
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
2in 2.38gA
Slenderness 2006.11570.2
1226
min
r
KLOK
Available Area
Euler’s Stress
ksi 42.216.115
)000,29(
/ 2
2
2
2
rKL
EFe
11350
000,2971.471.4
yF
EElastic BucklingSlenderness Limit
6.115min
r
KL
ELASTIC BUCKLING
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 ksi 79.1842.21877.0877.0 ecr FF
OK
Design Strength
kips 646)2.38)(79.18(9.0 gcrcnc AFP kips 600
More on Effective Length Factor
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 foundationsConnections 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
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’
12’W12x96
20’ 18’
Joint B
95.018/183020/1350
15/107012/1070
BG
Joint C
0.10CGPinned EndSway Uninhibited
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
More on Effective Length
Violated
Alingnment Charts & Inelastic Behavior
22
rKL
EFcr
elasticinelastic GE
E
LEI
LIEG t
gg
cct
elasticinelastic GG a
SRF: Table 4-21 AISC Manual pp 4-317
E
E
F
Ft
cr
cra
)elastic(
)inelastic(Stiffness Reduction Factor
22
rKL
EF t
cr
Elastic
Inelastic
Example
Compute Stiffness Reduction Factor per LRFD for an axial compressive stress of 25 ksi and Fy=50 ksi
ksi 25g
u
A
P
y
FF
gc
uinelasticcr F.
A
PF ey6580ksi 27.78
9.0
25)(
ksi 61.35506580ksi 27.78 50 eF F. e
23.3161.35877.08770)( eelasticcr F.F
890023.31
28.27
)(
)( .F
F
elasticcr
inelasticcra