Transcript Lecture 9 – Compression Members Problems

N.W.F.P. University of Engineering and
Technology Peshawar
Lecture 09: Compression Members
By: Prof Dr. Akhtar Naeem Khan
[email protected]
1
Effective length of columns
in frames
 Rotation of the ends of the columns in
building frames is usually limited by the
beams connecting to them, while
compression members in trusses may
have restricted end rotations because of
other members connecting at the joints.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
2
Effective length of columns
in frames
 KL is called effective length of column and K
effective length factor.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
3
Effective length of columns
in frames
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
4
Effective length of columns
in frames
 So far, we have looked at the buckling strength of
individual columns. These columns had various
boundary conditions at the ends, but they were
not connected to other members with moment
(fix) connections.
 The effective length factor K for the buckling of an
individual column can be obtained for the
appropriate end conditions from Table C-C2.1 of
the AISC Manual
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
5
Effective length of columns
in frames
 However, when these individual columns are part
of a frame, their ends are connected to other
members (beams etc.).
 These frames are sometimes braced and
sometimes un braced.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
6
Effective length of columns
in frames
 A Braced frame is one in which a sideway (joint
translation) is prevented by means of bracing,
shear walls, or lateral support from adjoining
structure.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
7
Effective length of columns
in frames
 A Un Braced does not have any bracing and
must depend on stiffness of its own members and
rotational rigidity of joints between frame
members to prevent lateral buckling.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
8
Effective length of columns
in frames
Conclusions
 Effective length coefficient increases with
decreasing stiffness of the beam and becomes
unity with zero stiffness.
 Critical loads for a column depends on:
 Its stiffness relative to that of beams framing
into it and
 Presence or absence of restraint to lateral
displacement of its ends.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
9
Effective length of columns
in frames
Braced and Un-braced Frames
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
10
Effective length of columns
in frames
Braced and Un-braced Frames
 Similarly you can analyze multi bay, multistory
frames.
 Assumptions

Subjected to vertical loads only

All columns become unstable simultaneously

All joint rotations at floor are equal

Restraining moment distributed in proportion to
stiffness.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
11
Effective length of columns
in frames
Method of Analysis
 First, you have to determine whether the
column is part of a braced frame or an
unbraced (moment resisting) frame.
 Then, you have to determine the relative
rigidity factor G for both ends of the column
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
12
Effective length of columns
in frames
Method of Analysis
 G is defined as the ratio of the summation of the
rigidity (EI/L) of all columns coming together at an
end to the summation of the rigidity (EI/L) of all
beams coming together at the same end.
 G=
E Ic
L
c
E Ib
L
b
 It must be calculated for both ends of the column.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
13
Effective length of columns
in frames
Method of Analysis
 Then, you can determine the effective length
factor K for the column using the calculated value
of G at both ends, i.e., GA and GB and the
appropriate alignment chart.
 There are two alignment charts provided by the
AISC manual.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
14
Effective length of columns
in frames
Method of Analysis
 One is for columns in braced (side sway
inhibited) frames. See Figure C-C2.2a on page
16.1-191 of the AISC manual. 0 < K ≤ 1
 The second is for columns in un-braced (side
sway uninhibited) frames. See Figure C-C2.2b on
page 16.1-192 of the AISC manual. 1 < K ≤ ∞
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
15
Effective length of columns
in frames
Alignment Chart
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
16
Effective length of columns
in frames
Method of Analysis: Inelastic Case
 G is a measure of the relative flexural rigidity of
the columns (EIc/Lc) with respect to the beams
(EIb/Lb)
 However, if column buckling were to occur in the
inelastic range (lc < 1.5), then the flexural rigidity
of the column will be reduced because Ic will be
the moment of inertia of only the elastic core of
the entire cross-section.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
17
Effective length of columns
in frames
Method of Analysis: Inelastic Case
rc = 10 ksi
rt = 5 ksi
Yielded zone
rt = 5 ksi
Elastic core, I c
rc = 10 ksi
rt = 5 ksi
(a) Initial state – residual stress
CE-409: Lecture 09
(b) Partially y ielded state at buckling
Prof. Dr Akhtar Naeem Khan
18
Effective length of columns
in frames
Method of Analysis: Inelastic Case
 The beams will have greater flexural
rigidity when compared with the
reduced rigidity (EIc) of the inelastic
columns. As a result, the beams will be
able to restrain the columns better,
which is good for column design.
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
19
Effective length of columns
in frames
Method of Analysis: Inelastic Case
The ratio Fcr/ Fe is called Stiffness reduction factor
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
20
Effective length of columns
in frames
Method of Analysis: Inelastic Case
GInelastic 
 E I / L 
 E I / L 
t
Cols
 GElastic  
Beams
Et

E
Fcr, Inealstic

Fcr, ealstic
lc 2

 lc 2 
Fcr, inelastic
(0.658 ) Fy
lc 2
 0.658

 
Fcr, elastic
 0.877
 0.877 
Fy
 lc 2 

The ratio Fcr/ Fe is called Stiffness reduction factor
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
21
Procedure for Column Design
1. Design Load
2. Assume Fcr

ØPn = ØAg Fcr = Pu

Find Ag

Select a section
3. Find
K L Fy
4. Find lc =
r E
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
22
Procedure for Column Design
For lc ≤ 1.5
For lc > 1.5
Fcr = 0.658
l2c
Fcr =
 Fy
 0.877
 2 
 l c 
Fy
5. Fcr Calculated > Fcr Assumed
6. ØP > Pu…………………. Check
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
23
Procedure for Column Design
Using Design Aids
 LRFD mannual contains variety of
Design aids, helpful in making original
trial section.
1. Design load
2. Find section for Corresponding P & KL
using table 4-21
K L
x
3. Calculate an equivalent (KL)eq =
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
x
rx / ry
24
Procedure for Column Design
Using Design Aids
4. Use the calculated (KL)eq value to find
(ØcPn) the column strength.
1.
Select section and its properties
5. Find lc
6. Find Fcr
7. Find ØP
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
25
Problem 4-11-1
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
26
Problem 4-11-1
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
27
Problem 4-11-1
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
28
Example Problem 01 ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
29
Example Problem 01 ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
30
Example Problem 01 ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
31
Example Problem 01 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
32
Example Problem 01 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
33
Example Problem 01 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
34
Example Problem 01 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
35
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
36
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
37
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
38
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
39
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
40
Example Problem 02 LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
41
Column Bases
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
42
Column Bases
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
43
Example ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
44
Example ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
45
Example ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
46
Example ASD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
47
Example LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
48
Example LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
49
Example LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
50
Example LRFD
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan
51
THANKS
CE-409: Lecture 09
Prof. Dr Akhtar Naeem Khan