Transcript Columns:

Columns:
Are compression members which are subjected to concentric axial
compressive forces. These are to be found in trusses and as a lateral
bracing members in frame building. Short columns are sometimes
referred to as to as “struts” or “stanchions”.
P
Beam-Columns:
Are members subjected to combined axial compressive
and bending stresses; These are found in single storey of
multi-storey framed structures. These are treated
independently in this course (chap. 12 in your text book).
P
Columns Theory:
Stocky columns (short) fail by yielding of the material at
the cross section, but most columns fail by “buckling” at
loads for less then yielding forces.
P
P
(a)
(b)
C-1
For “slender” columns, Euler (1759) predicted the critical buckling load (Pcr)
– also known as Euler Buckling Load as:
 2 EI
Pcr  2            (C  1)
L
where:
E = Young Modulus of Elasticity.
I = Minor moment of Inertia.
L = Unbraced length of column.
Derivation of Euler Buckling Load:
y
y
2
d y
M


dx 2
EI
Pcr
y"
y0
EI
Solution of this differential equation:
y = A cos (cx) + B sin (cx)
Pcr
Pcr
x
x
L
where:
Pcr
c
EI
, A and B are constants.
C-2
From boundary conditions:
y = 0 @ x = 0, and
y = 0 @ x = L, we get (A = 0) and (B sin cL = 0)
if B ≠ 0, then cL = n
where n = 0, 1, 2, 3 ………
 cL = 
Note:
Pcr
L  
EI
 2EI
Pcr  2
---- Euler Buckling Critical Load
L
 2E Ar 2

L2
where: r = minor radius of gyration
 2EA

L / r 2
Fcr 
Pcr
π E

             (C  2)
2
Ag
L
r
2
The critical buckling load
is a function of the section
properties (A, L, r) and the
modulus of elasticity for
material, and is not a
function of the strength or
grade of the material.
 
C-3
Example C-1
Find the critical buckling load for W 12 x 50, supported in a pinned-pinned
condition, and has an over-all length of 20 feet?
Solution:
Fcr 
 2E
L r 2
rmin = ry = 1.96 inch (properties of section).
Fcr 
 2  29000


2012 2
1.96
 19 ksi
Pcr = Fcr A = 19.1 x 14.7 = 280.8 kips
Note:
The steel grade is not a factor affecting buckling.
C-4
For short (stocky) columns; Equation (C-2) gives high values for
(Fcr), sometimes
greater then proportional limit, Engessor (1889)
proposed to use (Et) instead of (E) in Euler formula:
Pcr 
 2 Et I
2
L
         (C  3)
where:
Et = Tangent Modulus of Elasticity
Et < E
When (Fcr) exceeds (FPR), this is called
“Inelastic Buckling”, constantly variable
(Et) need to be used to predict (Fcr)
in the inelastic zone.
Shanley (1947), resolved this inconsistency.
C-5
Depending on (L/r) value the column buckling
strength was presented as shown by Shanley.
Residual Stresses:Due to uneven cooling of hot-rolled sections,
residual stresses develop as seen here.
The presence of “residual stresses” in almost all
hot-rolled sections further complicates the issue
of elastic buckling and leads towards inelastic
buckling.
C-6
The Euler buckling formula (C-1) is based on:
1 – Perfectly straight column. (no crookedness).
2 – Load is concentric (no eccentricity).
3 – Column is pinned on both ends.
The
Previous
conditions
are
where:
K = Effective length factor.
(Kl) = Effective length.
(Kl/r) = Effective slenderness ratio.
very
difficult to achieve in a realistic building
condition, especially the free rotation of pinned
ends. Thus an “effective slenderness factor” is
introduced
to
account
for
various
end
conditions:
Thus:
Fcr 
 2E
 r
KL
2
, or
Fcr 
see commentary
(C – C2.2) (page 16.1-240)
π 2 Et
 r
Kl
2
   C  4
C-7
AISC (Chapter E) of LRFD code stipulates:
Pu (factored load)  c Pn
where:
Pu = Sum of factored loads on column.
c = Resistance factor for compression = 0.90
Pn = Nominal compressive strength = Fcr Ag
Fcr = Critical buckling Stress. (E3 of LFRD)
a) for
Kl
 4.71
r
E
Fy
or
F
e
 0.44Fy 
 Fy 


F
Fcr  0.658  e  Fy


b) for
Kl
 4.71
r
E
Fy
or
F
e
Fcr  0.877Fe
where:
Fe 
π 2E
 KLr 2
E - 3.2 
 0.44Fy 
E - 3.3 
E - 3.4 
C-8
The above two equations of the LRFD code can be
illustrated as below:
Kl Fy
where:
λc 
rπ E
* The code further stipulates that an upper value for
for column should not exceed (200).
* For higher slenderness ratio,
Equation (E-3.3) controls and
(Fy) has no effect on (Fcr).
C-9
Example C-2
Determine the design compressive strength (cPn) of W 14x74 with an
untraced length of (20 ft), both ends are pinned, (A-36) steel is used?
Solution:
Kl =1 x 20 x 12 = 240 in
Rmin = ry = 2.48
240
 Kl 

 96.77  200 (0k)
 
r
2.48
 max.
π 2 x2900
Fe 

 30.56 ksi
2
2
(96.77)
 Kl 
 
r 
Fy


Fe
Fcr  0.658  Fy  (0.658)1.178 x 36


 0.611 36  21.99 ksi
c Pn = 0.9 x Fcr x Ag = 0.9 x (21.99) x 21.8
= 433.44 kips (Answer)
π 2E
0.44 Fy = 0.44 x 36 =15.84 ksi
Fe ≥ 0.44 Fy  Equ. E-3.2
(controls)
 Also from (table 4-22) LFRD Page 4-320
c Fcr = 19.75 ksi (by interpolation)
c Pn = c Fcr Ag = 430.55 kips
(much faster)
C-10
For must profiles used as column, the buckling of thin elements in the section may
proceed the ever-all bucking of the member as a whole, this is called local bucking.
To prevent local bucking from accruing prior to total buckling. AISC provides upper
limits on width to thickness ratios (known as b/t ratio) as shown here.
See AISC (B4)
(Page 16.1-14)
See also:
Part 1 on properties
of various sections.
C-11
Depending on their ( b/t ) ratios (referred to as ) ,
sections are classified as:
a) Compact sections are those with flanges fully welded
(connected) to their web and their:
  p
(AISC B4)
b) Non compact Sections:
p    r
(B4)
c) Slender Section:
 > r
(B4)
Certain strength reduction factors (Q) are introduced for slender
members. (AISC E7). This part is not required as most section
selected are compact.
C-12
Example C-3
Determine the design
compressive strength
(c Pn) for W 12 x 65
column shown below,
(Fy = 50 ksi)?
K xL x 1 x 24 x 12

 54.55
rx
5.28
K yL y
ry

1 x 8 x 12
 31.79
5.28
π 2E
Solution:
A) By direct LRFD
From properties:
Ag =19.1 in2
rx = 5.28 in
ry = 3.02 in
π 2 x29000
Fe 

 96.2 ksi
2
2
(54.55)
 Kl 
 
r 
Fe  0.44 Fy (  22 ksi)
Equ. (E 3.2)
50


96.2
Fcr  0.658  Fy


 0.8045 x 50  40.225 ksi
c Pn = 0.9 x Fcr Ag = 0.9 x 40.225 x 19.1 = 691.5 kips
C-13
B) From Table (4.22) LRFD
Evaluate
 Kl 

 r max
= 
= 54.55
Enter table 4.22 (page 4 – 318 LRFD)
cFc = 36.235 ksi (by interpolation)
Pn = Fc x Ag = 692.0 kips
C) From (Table 4.1 LRFD)
K xL x 1x24
(KL)y 

 13.7 ft
rx
1.75
ry
Enter table (4.1 ) page 4.17 LFRD with (KL)y = 13.7
Pn = 691.3 kips (by interpolation).
C-14
Design with Columns Load Table (4) LFRD:A) Design with Column Load Table (4) LFRD:
The selection of an economical rolled shape to resist a given
compressive load is simple with the aid of the column load tables.
Enter the table with the effective length and move horizontally until
you find the desired design strength (or something slightly larger). In
some cases, Usually the category of shape (W, WT, etc.) will have
been decided upon in advance. Often the overall nominal
dimensions will also be known because of architectural or other
requirements. As pointed out earlier, all tabulated values correspond
to a slenderness ratio of 200 or less. The tabulated unsymmetrical
shapes – the structural tees and the single and double-angles –
require special consideration and are covered later.
C-15
EXAMPLE C - 4
A compression member is subjected to service loads of 165 kips dead load
and 535 kips live load. The member is 26 feet long and pinned in each end.
Use (A572 – Gr 50) steel and select a W14 shape.
SOLUTION
Calculate the factored load:
Pu = 1.2D + 1.6L = 1.2(165) + 1.6(535) = 1054 kips
 Required design strength cPn = 1054 kips
From the column load table for KL = 26 ft, a W14  176
has design strength of 1150 kips.
ANSWER
Use a W14  145, But practically W14  132 is OK.
C-16
EXAMPLE C - 5
Select the lightest W-shape that can resists a factored compressive load Pu of
190 kips. The effective length is 24 feet. Use ASTM A572 Grade 50 steel.
SOLUTION
The appropriate strategy here is to fined the lightest shape for each nominal
size and then choose the lightest overall. The choices are as follows.
W4, W5 and W6:
None of the tabulated shape will work.
W8:
W8  58, cPn = 194 kips
W10:
W10  49, cPn = 239 kips
W12:
W12  53, cPn = 247 kips
W14:
W14  61, cPn = 276 kips
Note that the load capacity is not proportional to the weight (or crosssectional area). Although the W8  58 has the smallest design strength of
the four choices, it is the second heaviest.
ANSWER Use a W10  49.
C-17
B) Design for sections not from Column Load Tables:
For shapes not in the column load tables, a trial-and-error approach must be used.
The general procedure is to assume a shape and then compute its design strength. If
the strength is too small (unsafe) or too large (uneconomical), another trial must be
made. A systematic approach to making the trial selection is as follows.
1) Assume a value for the critical buckling stress Fcr. Examination of AISC Equations
E3-2 and E3-3 shows that the theoretically maximum value of Fcr is the yield stress Fy.
2) From the requirement that cPn  Pu, let
cAgFcr  Pu and A g  φPFu
c cr
3) Select a shape that satisfies this area requirement.
4) Compute Fcr and cPn for the trial shape.
5) Revise if necessary. If the design strength is very close to the required value,
the next tabulated size can be tried. Otherwise, repeat the entire procedure,
using the value of Fcr found for the current trial shape as a value for Step 1
6) Check local stability (check width-thickness ratios). Revise if necessary.
C-18
Example C-6
Select a W18 shape of A36 steel that can resist a factored load of 1054 kips.
The effective length KL is 26 feet.
Solution:
Try Fcr = 24 ksi (two-thirds of Fy):
Required Ag 
Pu
1054

 48.8 in 2
c Fcr 0.9( 24)
Try W18 x 192:
Ag = 56.4 in2 > 48.8in2
KL 26(12)

 111.8  200
rmin
2.79
(OK)
π 2E
π 2 x29000
Fe 

 22.9 ksi
2
2
111.8 
 Kl 
 
r 
Fe  0.44 Fy (15.84)  LRFD Equ. E . 3.2
C-19
Fy
36




22.9
Fcr  0.658 Fe  Fy  0.658  x36  0.532 x 36




 18.64 ksi
cPn  0.9 A gFcr  0.9 x 56.4 x 18.64  943kips  1054k
(N.G.)
Try Fcr  18.64 ksi (the value just computed for the W18 x 192) :
Required A g 
Pu
1054

 62.83 in2
c Fcr 0.9(18.64)
Try W18 x 234 :
A g  68.8 in2 .  62.83 in 2
KL 26(12)

 109.5  200
rmin
2.85
(OK)
C-20
Fe 
π 2E
Klr 2
π 2  29000

 23.87ksi
2
109.5 
Fe  0.44Fy
 Use LFRD (Equ.E - 3.2)
Fy
36



23.87 
Fe
Fcr  0.658  Fy  0.658
 x 36  0.532 x 36  19.15 ksi




φ cPn  0.9 A gFcr  0.9 x 68.8 x 19.15  1185kips  1054k
(OK)
This shape is not in the column load tables, so the width - thickness
ratios must be cheacked :
bf
95
 2.8 
 15.8
(OK)
2t f
36
h
253
 13.8 
 42.2
(OK)
tw
36
Answer
Use a W18 x 234
C-21