cee380 elementary structures ii focus - uw courses...
TRANSCRIPT
CEE380
Elementary Structures II
Columns-1
columns-1.ppt 2
Focus
• Overview of column behavior: short,
long, intermediate
• Euler buckling formula for long columns
• Concept of slenderness ratio
• How to determine axial design strength
using tables from AISC
columns-1.ppt 3
Mathematical Models
• Short or stub: Failure when the
compressive strength of the material is
exceeded
• Long: Failure by buckling
• Intermediate: Combination of strength
and stiffness
columns-1.ppt 4
Design Strategies
• Short
– Keep compressive stress of P/A below a
specified value
• Long
– Avoid buckling behavior
columns-1.ppt 5
Euler Critical Buckling Load
�
PCR
=!
2EI
KL( )2
K = ratio of effective column length to actual unbraced length
L = unbraced length [in]
columns-1.ppt 6
Derivation of KL
Use equilibrium to
solve for the
buckling load using
this FBD of a fixed-
pinned column
columns-1.ppt 7
�
M0! = "Py + Rx" EI # # y = 0
# # y + PEI( )y = R
EI( )x
y = A sin PEI
x$
% &
'
( ) + B cos P
EIx
$
% &
'
( ) +
Rx
P
columns-1.ppt 8
Use boundary conditions to evaluate constants and reaction
at x = 0, y = 0 ! B = 0at x = L, y = 0 :
0 = Asin PEI( )L"
#$%&'+RL
P
A =(RL
P sin PEI( )L"
#$%&'
at x + L, )y = 0 and since B = 0:
0 = PEI( )Acos P
EI( )L"#$
%&'+R
P
A =(R
P PEI( )cos P
EI( )L"#$
%&'
columns-1.ppt 9
Set the two equations for “A” equal to each
other.
�
PEI
L = tan PEI
L!
" #
$
% &
PEI
L ' 0; i.e., it can't be negative
4.493 = tan(4.493)
The only solution is in the third quadrant.
( PEI
L = 4.493
P = 4.493( )2 EI
L2( )
Rewriting, factoring out ) 2 ,we obtain
PCR
=) 2EI
0.7L( )2
where K = 0.7 * effective length factor
columns-1.ppt 10
K values from the steel manual: This table
is in your handout for columns.
columns-1.ppt 11
Design
Design Equation:
Pu ! "cPn
"cPn = "cAgFcr with "c = 0.85
Slenderness Ratio:
KL
r
KL = effective length
r=minimum radius of gyration
columns-1.ppt 12
AISC design strategies
columns-1.ppt 13
columns-1.ppt 14
columns-1.ppt 15
Examples of using the columns tables
Using the columns tables:
Fy = 50ksi : Find !cPn for the following members.
W12 x 106, KL = 20 ' :!cPn = 858k
W14 x 109, KL = 14 ' :!cPn = 1170k
W14 x 53, KL = 21' :!cPn = ?
KL = 20 ',!cPn = 213kKL = 21',!cPn = ?KL = 22 ',!cPn = 176k
20 " 21
20 " 22=
213" ?
213"176so ? = 194.5k
columns-1.ppt 16
Example using Table 3-50
GIVEN:W12x87, L=30ft, Simply-Supported ("pinned") ends
FIND: !cPn
SOLUTION:K = 1.0 from Table C-C2.1
KL = 30 ft
W12x87, radius of gyration ry = 3.07 inches from W Tables; Area A = 25.6 in2
KL
r=
30(12)
3.07= 117.3
Using Table 3-50
KL = 117, !cFCR = 15.6ksiKL = 118, !cFCR = 15.3ksi
so !cFCR = 15.5ksi
and !cPn = 25.6in2 (15.5ksi) = 396.8k
columns-1.ppt 17
Select the lightest available W 12 sections for the following columns:
a) Pu = 500k, L = 14 ft, pinned end supports
K = 1.0 for pinned-pinned (Table C-C2.1); KL = 14, Fy = 50 ksi, W12x58 is OK, !cPn=521k
b) Pu = 360 k, L = 12 ', fixed end supports
K = 0.65 for fixed-fixed (Table C-C2.1); KL = 7.8 '; round up to 8', Fy = 50 ksi,
W12x40,!cPn = 416k
c)Pu = 700k, L = 16.5 ft, fixed at bottom, pinned on top
K = 0.80 (Table C-C2.1); KL = 13.2', Fy = 50 ksi,
KL = 13 : W12x72,!cPn = 740k
KL = 14 : W12x72,!cPn = 717ksoKL = 13.2,!cPn = 735.2k
Examples using the Columns Tables
columns-1.ppt 18
Design Example
Enter the column load table for W shapes at an effective length of KL=16 ft.
Select W14 x 99, good for 1040 kips > 1000 kips.
rx/ry = 1.66
Equivalent effective length for X-X axis:
31/1.66 = 18.6 ft
Since 18.6 > 16 ft, X-X axis controls.
Re-enter table for effective length of 19 ft to satisfy axial load of 1000 kips, select
W 14 x 109.
Design the lightest W shape of Fy=50 ksi steel to support a
factored concentric load [i.e. Pu] of 1000 kips.
The effective length with respect to its minor axis is 16 ft.
The effective length with respect to the major axis is 31 ft.
columns-1.ppt 19
Summary
• More example problems with complex end conditions provided on hand-written
sheets.
• Overview of column behavior: short, long, intermediate
• Euler buckling formula
• Concept of slenderness ratio
• How to determine axial design strength using tables from AISC