11-Introduction to Axially Compression Members (Steel Structural Design & Prof. Shehab Mourad)

1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Introduction to Axially Compression Members
Compression members can be found in:
1- Columns
2- Top chord of simply supported trusses
3- Bracing members
4- Parts of members such as the compression flange of rolled beams
Compression members can have three different modes of failure
.
1- Flexural buckling (Euler buckling –
Global buckling), it depends on the
slenderness ratio (KL/r) and end
conditions.
L
(a)
P
P
(b)
Pcr
Pcr
What is buckling?
Buckling occurs when a
straight column subjected to
axial compression suddenly
undergoes bending as
shown in the Figure 1(b).
Buckling is identified as a
failure limit-state for
columns

2- Local buckling occurs when some part of cross section are so thin that can
buckle locally in compression before flexural buckling occur.

3- Torsional buckling may occur in columns having certain cross sections,
where the shear center does not coincide with the center of gravity. These
columns failed by twisting or combination of torsion and flexural buckling
yo
c.g.
s.c.
s.c
c.g.
yo
xo
yo
s.c

Euler Formula
For a pined-pined column with length (L) and cross section (A) and has a
moment of Inertia (I) with a modulus of Elastically (E), The buckling load is
given by Euler formula (PE)
And the buckling stress (critical Stress) , Fcr is given by:
In general, for different end conditions, the buckling length will be equal to KL
where K, depends on the end conditions and Fcr , is given by
Fcr = π2
E , where KL/r is called slenderness ratio of the column
(KL/r)2
Fcr = PE = π2
E (I/A) = π2
E
A L2
(L/r)2
PE = π2
E I
L2

Column Formulas ( Short, Intermediate and long column)
(Φ = 0.85 for compression members)
Ag = gross area of cross section even there are holes for bolts
(elastic buckling) , long columns (for λc > 1.5)
Fcr = 0.877 π2
E = 0.877 Fy where λc = KL/r √ Fy / E
(KL/r)2
λc
2
π
( Inelastic buckling), short and intermediate columns
For λc ≤ 1.5 Fcr = (0.658 ) Fy
Φ Pn = 0.85 Ag Fcr
λc
2
x
y
For wide flange sections,
x is the major axis and y
is the minor axis. Major
axis means axis of
greater moment of
inertia (Ix > Iy)

Local buckling of the cross section elements
Local buckling of flanges and webs of column cross section can occur before
the member reaches its buckling load. LRFD specifications provide limiting
values for the width-thickness ratios of the individual elements of
compression members as follows;

Limiting width-thickness ratio for compression elements, local buckling will
occur if those limits are exceeded
1- Unstiffened elements:
- Single angle and double angles
λ r = b/t = 0.45 √ E / Fy
- Flanges of W-shapes and channels;
λ r = b/t = 0.56 √ E / Fy in pure compression
- Stem of tees;
λ r = d/t = 0.75 √ E / Fy
2- Stiffened elements:
- Flanges of rectangular box and hollow structural sections
λ r = b/t = 1.40 √ E / Fy
- All other elements uniformly compressed and supported along two edges
as web of W-shapes
λ r = b/t = 1.49√ E / Fy
Effect of local buckling in the compression strength
If the limits of b/t > λ r , a reduction factor Q is applied to the yield strength,
where ;
Q = Qs . Qa
Qa: reduction factor for stiffened elements as webs
Qs : reduction factor for unstiffened elements as angle legs and W-shape
flanges

Note that ;
Qs = 1.0 if the cross section is composed of only stiffened elements as
hollow sections
Qa = 1.0 if the cross section is composed of only unstiffened elements
as angles and Tees
For sections that are composed of both stiffened and unstiffened
elements such as W-shapes and channels, Q = Qs . Qa
For Unstiffened compressed elements
a) For single angle
0.45√ E / Fy < b/t < 0.91 √ E / Fy
b/t ≥ 0.91 √ E / Fy
b) For flanges of W-shapes and channels
0.56√ E / Fy < b/t < 1.03 √ E / Fy
b/t ≥ 1.03 √ E / Fy
c) For stems in tees
0.75√ E / Fy < d /t < 1.03 √ E / Fy
b/t ≥ 1.03 √ E / Fy
Qs = 0.534 E
Fy (b/t)2
Qs = 1.34 – 0.76 (b/t) √ Fy / E
Qs = 1.415 – 0.74 (b/t) √ Fy / E
Qs = 0.69 E
Fy (b/t)2
Qs = 1.908 – 1.22 (d/t) √ Fy / E
Qs = 0.69 E
Fy (d/t)2

For Stiffened Compression elements
If h/tw ≥ 1.49 √ E / Fy
be = 1.91 * tw * 













−
f
E
thf
E
w
*
)/(
34.0
1* < hw
f : computed elastic compression stress in the stiffened element based on
design properties
Qa =
A
effectiveArea.Σ
=
[ ]
A
tbA w**
−
Factored Strength of Column
If Qc*λ < 1.5 Fcr = Q (0.658) * Fy
If Qc*λ > 1.5 Fcr = 0.877 * Fy
λc2
where λc = KL/r √ Fy / E
π
be/2
be/2
b*
hw
Q λc2
Φ Pn = 0.85 Ag Fcr

11-Introduction to Axially Compression Members (Steel Structural Design & Prof. Shehab Mourad)

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