Compression members are structural elements that are subjected only to axial
compressive forces; that is, the loads are applied along a longitudinal axis through the centroid
of the member cross section, and the stress can be taked as f=P/A, where f is considered to be
uniform over the entire cross section. This ideal state is never achieved in reality, however,
because some eccentricity of the load is inevitable. Bending will result, but it is usually regarded
as secondary.
There are several types of compression members, the Column being the best known.
Among the other types are the top chords of trusses and various bracing members. In addition,
many other members have compression flanges of rolled beams and built-up beam sections,
and members that are subjected simultaneously to bending and compressive loads.
Sections used for column
Classification of column:
Long Columns -The euler formula predicts very well the strength of long columns where the
axial buckling stress remains below the proportional limit. Such column will buckle elastically.
Intermediate Columns – For intermediate columns, some of the fibers will reach the yield stress
and some will not. The members will fail by both yielding and buckling, and their behavior is
said to be inelastic.
Short Columns – For short columns, the failure stress will equal the yield stress and no buckling
will occur.
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COLUMN FORMULAS (NSCP 2001 vs 2015) (ASD)
 allowable compressive strength (NSCP 2015 sec.505.3)
Pa = FaAg
Where:
Fcr = flexural buckling stress
Ag = gross cross sectional area of member
The allowable compressive stress Fa = 0.60Fcr
The flexural buckling stress Fcr, is determined as follows;
 allowable compressive strength (NSCP 2001 sec.505.3)
Pa = FaAg
Where:
Fa = allowable stress
Ag = gross cross sectional area of member
The allowable compressive stress Fa, is determined as follows;
When KL/r < Cc;
1−
=
2|
5 3
3+8∙
(
/ )
2
/
1 (
−8∙
/ )
Where:
=
2
When KL/r > Cc;
=
12
23( / )
EULER COLUMN BUCKLING THEORY
 Leonhard Euler, 1757 a swiss mathematician, who wrote a paper of great value
concerning the buckling of columns. When a column Is loaded to buckling load or Euler
load, column will fail by sudden buckling or bending.
EULER’S CRITICAL BUCKLING LOAD
=
(
)
EULER’S CRITICAL BUCKLING STRESS
=
=
(
/ )
EFFECTIVE LENGTH
 The effective length of a column is defined as the distance between points of zero
moment in the column, that is, the distance between its inflection points. Effective
length of column is referred to as KL, where K is the effective length factor.
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Table: Approximate Values of effective length factor, K
Two sets of K values are provided in the table, the theoretical values and the recommended
design values, based on the fact that perfectly pinned and fixed conditions are not possible.
SLENDERNESS RATIO
 The ratio of column’s effective length to its radius of gyration, and is a measure of
member’s slenderness.
=
Larger value between;
;
SRx =
SRx =
Where:
L = Laterally unbraced length of the member
r = radius of gyration = /
K = effective length factor
Critical Slenderness Ratio, Cc
 Slenderness ratio at which the Euler’s buckling stress reaches the proportional limit
→
4|
,
→
.
Hence,
=
The proportional limit of steel is approximately one-half of its yield strength.
≈
2
2
=
Therefore,
=
2
For members designed on the basis of compression, the slenderness ratio KL/r preferably
should not exceed 200.
SITUATION 1:
A wide flange section for a 4.5 m long column (hinged at both ends) has the following
properties:
A = 14500 mm2
E = 200 GPa
6
4
Ix = 1500 x 10 mm Iy = 170 x 106 mm4
1. Determine the Euler critical load of the column.
Ans. 16571.20 kN
Solution:
Since hinge at both ends, from table ; K=1
=(
)
=
=
5|
(
[( )(
)
)]
(
[( )(
)
)]
= 146216.4 kN
= 16571.20 kN (governs)
SITUATION 2:
A column 3m long and pinned at both ends carries an axial load of 190kN. The column is made
up of 2 angles of unequal legs with long legs back to back and separated by a gusset plate
whose thickness is 11mm. Use A36 steel with Fy = 248 Mpa and E = 200,000Mpa. Three
sections are being considered, as follows, with their respective properties relevant of this
problem (lengths units are in meters.).solve using code NSCP 2001 and NSCP 2015
Section
A(m2)
rx(m)
ry(m)
2L 125 x 75 x 12
2L 150 x 90 x 10
2L 150 x 90 x 12
0.00454
0.00463
0.00550
0.0390
0.0480
0.0500
0.0160
0.0195
0.0251
a. Which of the sections gives the largest allowable compressive stress?
Ans. 2L 150 x 90 x 12 (29.29 Mpa, 29.54 Mpa)
b. Which of the sections give the smallest allowable compressive stress?
Ans. 2L 125 x 75 x 12 (43.54 Mpa, 43.88 Mpa)
c. Which of the sections gives most economical (lightest) section for the given load?
Ans. 2L 150 x 90 x 10 (71.73 Mpa, 70.20 Mpa)
Solution: using NSCP 2001
a.
=
(
=
)
= 126.16
Since pinned at both ends, from table; K=1
For 2L 125 x 75 x 12:
=
Thus,
( )( )
= 187.50 > Cc
.
=
(
/ )
=
(
(
,
.
)
)
= 29.29 Mpa
For 2L 150 x 90 x 10:
=
Thus,
( )( )
= 153.80 > Cc
.
=
(
/ )
=
(
(
,
.
)
)
For 2L 150 x 90 x 12:
=
6|
( )( )
.
= 119.52 < Cc
= 43.54 Mpa
(
=
Thus,
/
∙
=
(
,
(
/ )
.
(
∙
)
)
(
(
=
/ )
.
.
.
.
∙
)
)
∙[
(
(
)
.
.
)
= 71.73 Mpa
]
= 43.54 Mpa
b. Based on calculation above, the section with smallest allowable compressive stress
is 2L 125 x 75 x 12.
c. Computing for most economical section
For 2L 125 x 75 x 12; Fa = 29.29 Mpa
[ P = Fa x A ]
P = 29.29(0.00454 x 1000²)
P = 132.98 kN
For 2L 150 x 90 x 10; Fa = 43.54 Mpa
[ P = Fa x A ]
P = 43.54 (0.00463 x 1000²)
P = 201.59 kN  governs since enough to carry 190 kN
For 2L 150 x 90 x 12:; Fa = 71.73 Mpa
[ P = Fa x A ]
P = 71.73 (0.00550 x 1000²)
P = 394.52 kN
Solution: using NSCP 2015
4.71
,
= 4.71
= 133.76
For 2L 125 x 75 x 12:
=
( )( )
.
Fcr
= 0.877 Fe
Fe
=
(
/ )
= 187.50 > 4.71
=
(
(
,
.
)
)
Thus, Fcr
= 0.877 (56.14) = 49.23 Mpa
Therefore;
Fa
= 0.60Fcr
= 0.60 (49.23) = 29.54 Mpa
For 2L 150 x 90 x 10:
=
Fcr
7|
( )( )
.
= 0.877 Fe
= 56.14 Mpa
= 153.85 > 4.71
Fe
=
(
=
/ )
(
(
,
)
)
.
Thus, Fcr
= 0.877 (83.39) = 73.13 Mpa
Therefore;
Fa
= 0.60Fcr
= 83.39 Mpa
= 0.60 (73.13) = 43.88 Mpa
For 2L 150 x 90 x 12:
=
( )( )
= 119.52 < 4.71
.
Fcr
= [0.658
Fe
=
(
] Fy
=
/ )
Thus, Fcr
= [0.658
Therefore;
Fa
= 0.60Fcr
.
(
(
,
.
] (248)
)
)
= 138.18 Mpa
= 117 Mpa
= 0.60 (117) = 70.20 Mpa
d. Based on calculation above, the section with smallest allowable compressive stress
is 2L 125 x 75 x 12.
e. Computing for most economical section
For 2L 125 x 75 x 12; Fa = 29.54 Mpa
[ P = Fa x A ]
P = 29.54(0.00454 x 1000²)
P = 134.11 kN
For 2L 150 x 90 x 10; Fa = 43.88 Mpa
[ P = Fa x A ]
P = 43.88 (0.00463 x 1000²)
P = 203.16 kN  governs since enough to carry 190 kN
For 2L 150 x 90 x 12:; Fa = 70.20 Mpa
[ P = Fa x A ]
P = 70.20 (0.00550 x 1000²)
P = 386.10 kN
SITUATION 3:
A W 250 x 73 is to be served as a pin-ended 12. Long column is braced at mid height with
respect to its weak axis. The proportional limit is 235 Mpa.
Properties of safety of 2.5
A = 9280 mm2
rx = 110
d = 253 mm
ry = 64.7
bf = 254 mm
Ix = 113 x 10⁶ mm⁴
tf = 14.2 mm
Iy = 38.8 x 10⁶ mm⁴
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Es = 200,000Mpa
a. Determine the slenderness ratio with respect to y-axis
Ans. 92.74
b. Determine the Euler’s buckling stress
Ans. 165.87 Mpa
c. Determine the Euler’s buckling load
Ans. 615.72 kN
Solution:
a.
Where :
K = 1 (pin-ended)
L = 6m (since braced at midheight)
Weak axis
Strong axis
Therefore :
=
b.
=(
( )(
))
.
= 92.74
/ )
; K = 1, Lx = 12m(since Lx is the strong axis-unbraced length)
=
( )(
)
.
= 109.09
governing slenderness (bigger value)
; K = 1, Ly = 6m(since Ly is the weak axis-braced at mid height)
=
( )(
)
.
= 92.74
Therefore;
=(
/ )
=
(
,
(
.
)
)
= 165.87 Mpa
c.
9|
( )
=(
( )
=(
)
)
=
=
(
,
)(
[( )(
(
,
)(
[( )(
⁶
⁴)
⁶
⁴)
)]
.
)]
= 1548.98 kN
= 2127.45 kN
govern
PRACTICE PROBLEMS:
 Provide solutions to the given situations below;
SITUATION 1:
A W 14 x 142 is used as a column having length of 9 m long. It is hinged at the upper end and
fixed at the lower end but there is a lateral bracing perpendicular to the minor axis of the W
section at the 5.4 m above the bottom support. It is assumed to be pinned connected at the
bracing point. Using A-36 steel Fy=248 MPa and the NSCP specifications. Es= 200 GPa.
Properties of W 14 x 142
A=26967.69 mm2
d=374.65 mm bf=393.70 mm
tf=27.00 mm tw= 17.27 mm
Ix= 695.11 x 106 mm4
Iy= 274.71 x 106 mm4
Sx=3719.18 x 103 mm3
Sy= 1396.18 x 103 mm3
rx=160.53 mm
ry= 100.84 mm
use NSCP 2001 & NSCP 2015
(Use the recommended k values)
a. Compute the critical slenderness ratio. Ans. 44.85
b. Compute the allowable axial stress. Ans. 129.48Mpa, 133.86Mpa
c. Compute the capacity of the column section. Ans. 3492kN, 3609.89kN
SITUATION 2:
Given properties of a steel column:
A = 8100 mm2
Fy = 345 MPa
x = 178 x 106 mm4
E = 200,000 MPa
6
4
y = 19 x 10 mm
Proportional limit = 310 MPa
The x – axis has an unbraced length of 8m which is pinned at the top and fixed at the bottom
with k = 0.70 to prevent sidesway. The y – axis has an unbraced length of 4 m due to the bracing
at the mid – height.
a. Determine the critical slenderness ratio.
Ans. 82.6
b. Compute the initial compressive load (buckling load) of the column.
Ans. 2344 kN
c. What is the minimum length of column that will not exceed the proportional limit?
Ans. 7.73m
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