Unit 5 - Compression Members

Compression Members
CE 422 Steel Design I
1
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”.
Beam-Columns:
P
Beam-columns are
members subjected to combined axial
compressive and bending stresses; These are found in single storey
of multi-storey framed structures.
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
(a)
2
P
(b)
Note that the ideal state is never realized in practice and some eccentricity
of load will be always present
3
For “slender” columns, Euler (1759) predicted the critical buckling load (Pcr)
– also known as Euler Buckling Load as:
Pcr 
where:
 2 EI
2
L
           (C  1)
E = Young Modulus of Elasticity.
I = Minor moment of Inertia.
L = Unbraced length of column.
4
Pcr
Pcr 
 2 EI
L2
 2 E Ar 2

L2
 2 EA

L / r 2
Fcr 
Note:
 2E
 
L
2
---- Euler Buckling Critical Load
where:
r = minor radius of gyration
Crippling Stress
r
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.
5
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
2
r
rmin = ry = 1.96 inch (properties of section).
Fcr 
 2  29000


2012 2
1.96
 19 ksi
Pcr = σcr A = 19.1 x 14.7 = 280.8 kips
Note:
The steel grade is not a factor affecting buckling,
also note
Fcr << Fy.
6
7
For short (stocky) columns; Equation gives high values for (σcr), sometimes
greater than proportional limit, Engessor (1889) proposed to use (Et) instead of
(E) in Euler formula:
Tangent modulus of elasticity is
 Et I
Pcr  2          (C  3)
L
2
The instantaneous rate of change of stress as a
function of strain. It is the slope at any point on a
stress-strain diagram.
where:
Et = Tangent Modulus of Elasticity
Et < E
When (Fcr) exceeds (Fpl), this is called
“Inelastic Buckling”, constantly variable
(Et) need to be used to predict (Fcr)
in the inelastic zone.
Shanley (1947), resolved this inconsistency.
8
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 hotrolled sections further complicates the issue of
elastic buckling and leads towards inelastic buckling.
9
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 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 
10
π 2E
 
KL
2
r
, or
Fcr 
π 2Et
 
Kl
2
r
   C  4 
where:
K = Effective length factor.
(Kl) = Effective length.
(Kl/r) = Effective slenderness ratio.
11
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:
12
Fe 
π 2E
 
KL 2
r
E - 3.2 
 0.44Fy 
E - 3.3 
E - 3.4 
The above two equations of the LRFD
code can be illustrated as below:
where:
Kl
λc 
rπ
Fy
E
 The code further stipulates
that an upper value for column
should not exceed (200).
 For higher slenderness ratio,
Equation (E-3.3) controls and
(Fy) has no effect on (Fcr).
13
14
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)
15
16
17
For most 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
occurring 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.
18
Upper limits on width to thickness ratios
Based on Table B4.1
19
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.
20
Example C-3
Determine the design
compressive strength
(c Pn) for W 12 x 65
column shown below,
(Fy = 50 ksi)?
K x L x 1 x 24 x 12

 54.55
rx
5.28 (controls)
K yLy
ry

1 x 8 x 12
 31.79
3.02
π 2E
Solution:
A) By direct LRFD
From properties:
Ag =19.1 in2
rx = 5.28 in
ry = 3.02 in
21
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
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).
22
23
24
Example C-4
P
Find the maximum load capacity
(Pn) of the W 14 x 53 (A-36)
C
B
25 ft.
15 ft.
Solution:
y-axis
C
10 ft.
column shown in figure ?
x-axis
P
A
Lx = 25 ft, kx = 0.8, rx = 5.89 in.
A
x
x
x
x
Section (AB) Ly = 15 ft, ky = 0.8, ry = 1.92 in.
Section (BC) Ly = 10 ft., ky = 1.0, ry = 1.92 in.
0.8  25  12
 Kl 

 41
 
5.98
 r x
max
 Kl 
 
 r y

Enter table (4-22) , Fc = 24.1 ksi
Column capacity Pn = Fcr Ag = 24.1 x 15.6 = 376 kips
0.8  15  12
 75
1.92
(controls)
25
26
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.
27
EXAMPLE C - 5
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  145
has design strength of 1230 kips.
ANSWER
Use a W14  145, But practically W14  132 is OK.
28
29
EXAMPLE C - 6
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 find 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:
W 8  58, cPn = 205 kips
W10:
W10  49, cPn = 254 kips
W12:
W12  53, cPn = 261 kips
W14:
W14  61, cPn = 293 kips
Note that the load capacity is not proportional to the weight (or cross-sectional
area). Although the W8  58 has the smallest design strength of the four choices,
it is the second heaviest.
ANSWER
Use a W10  49.
30
Example C-7
Select the lightest W – 10 section made of
A 572-Gr50 steel to resist a factored load
of (600 kips) ?
Solution:
Assume weak axis (y-y) controls buckling:
Enter design tables of AISC (Section 4) with KyLy = 9 ft.
Select W 10 x 54 (capacity = 625 k > 600 k
OK)
Check strong axis buckling strength:
Enter table for W10 x 54 with (KL)eq. = 10.53 ft.
Capacity = 595.8 kips (by interpolation) N.G.
Select W10 x 60 capacity = 698 kips for KyLy = 9 ft.
capacity = 666 kips for (KL)eq. = 10.5 ft.
31
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.
32
33
34
Example C-8
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
Fe 
35
π 2E

π 2 x29000
(OK)
 22.9 ksi
111.8 
 Kl 
 
r 
Fe  0.44 Fy (15.84)  LRFD Equ. E . 3.2
2
2
36
Fy
36




22.9
Fcr  0.658 Fe  Fy  0.658  x36  0.532 x 36




 18.64 ksi
φ cPn  0.9 A gFcr  0.9 x 56.4 x 18.64  943 kips  1054 k
(N.G.)
Try Fcr  18.64 ksi (the value just computed for the W18 x 192) :
Required A g 
Pu
1054

 62.83 in 2
φ cFcr 0.9(18.64)
Try W18 x 234 :
A g  68.8 in 2 .  62.83 in 2
KL 26(12)

 109.5  200
rmin
2.85
(OK)
37
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
Fcr  0.658 Fe  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  1185 kips  1054 k
(OK)
This shape is not in the column load tables, so the width - thickness
ratios must be cheacked :
bf
95
 2.8 
 15.8
2t f
36
h
253
 13.8 
 42.2
tw
36
Answer
Use a W18 x 234
(OK)
(OK)
38
The effective length factor (K) was introduced in page (C-7) for six ideal conditions,
these are not encountered in practical field conditions. LRFD commentary provides
both real conditions and standard ideal conditions (C-C2.2) (page 16.1-239 to 242)
Braced Frames:
Unbraced Frames:
No lateral movement is allowed
(0.5 < K < 1.0) (sideway prevented)
Lateral movement possible
(1.0 < K < 20.0) (sideway allowed)
a) Diagonal
bracing
b) Shear Walls
(masonry,
reinforcement concrete
or steel plate)
39
GA 
 Ic /L c
 Ig /L g
where
A is top of column
GB 
 Ic /L c
 Ig /L g
where
B is bottom of column
* For fixed footing G = 1.0
* For pinned support G = 10.0
40
Example C – 9:In the rigid frame shown below, Determine Kx for columns
(AB) & (BC). Knowing that all columns webs are in the plane.
W12 x 96
Solution:
Column (AB):
(A):
W24 x 55
A
12'
W24 x 68
B
W12 x 120
 Ic /L c
GA 
 Ig /L g
833/12  1070/12
1586


1350/20  1830/18 169.2
 0.94
W24 x 68
W24 x 55
W12 x 120
Joint
12'
15'
C
20'
20'
18'
41
42
For joint B,:G
ΣIc /L c 1070/12  1070/15 160.5


 0.95
ΣIg /L g
169.2
169.2
From the alignment chart for sideways uninhibited, with G A = 0.94 and GB = 0.95,
Kx = 1.3 for column AB.
Column (BC):
For joint B, as before,
G = 0.95
For joint C, at a pin connection the situation is analogous to that of a very
stiff column attached to infinitely flexible girders – that is, girders of zero
stiffness. The ratio of column stiffness to girder stiffness would therefore
be infinite for a perfectly frictionless hinge. This end condition is only be
approximated in practice, so the discussion accompanying the alignment
chart recommends that G be taken as 10.0.
From the alignment chart with GA = 0.95 and GB = 10.0, Kx = 1.85 for column BC.
43
When an axially loaded compression member becomes unstable overall (that is, not
locally unstable), it can buckle in one of three ways, as shown in figure.
1. Flexural buckling. We have considered
this type of buckling up to now. It is a
deflection caused by bending, or flexure,
about this axis corresponding to the
largest slenderness ratio (Figure a). This
is usually the minor principal axis – the
one with the smallest radius of gyration.
Compression members with any type of
cross-sectional configuration can fail in
this way.
44
2. Torsional buckling
This type of failure is caused by twisting about the
longitudinal axis of the member. It can occur only
with doubly symmetrical cross sections with very
slender cross-sectional elements (Figure b). Standard
hot-rolled shapes are not susceptible to torsional
buckling, but members built up from thin plate
elements may be and should be investigated. The
cruciform shape shown is particularly vulnerable to
this type of buckling. This shape can be fabricated
from plates as shown in the figure, or built up from
four angles placed back to back.
45
3. Flexural-torsional buckling. This type of failure is
caused by a combination of flexural buckling and
torsional buckling. The member bends and twists
simultaneously (Figure c). This type of failure can
occur only with unsymmetrical cross sections, both
those with one axis of symmetry – such as channels,
structural tees, double-angle shapes and equal-leg
single angles – and those with no axis of symmetry,
such as unequal-leg single angles.
The AISC Specification requires an
analysis of torsional or flexural-torsional
buckling when appropriate. Section E3 of
the Specification covers double-angle and
tee-shaped members, and Appendix E3
provides a more general approach that can
be used for any unsymmetrical shape.
46
When length exceeds requirements for a
single section, built-up compression
section are used as shown below:
The code provides details for built-up
section under LRFD EG.
47
Example C – 10 :- Calculate the capacity of the built-up column shown below.
Lx = Ly = 25 ft, Kx = 1.6, Ky = 1.0 Fy = 42 ksi ?
Solution:-
From table 4.22 page 4.321
cFcr = 18.3 ksi
Design Nominal Strength = cFcr Ag
=18.3 x 22.70
= 415.4 kips.
48
49