Columns
Department of Civil Engineering - Fall 2016 1 | MOS-II
Stability of Structures
• The selection of structural elements is based on three
characteristics: strength, stiffness, and stability.
• The procedures of stress and deformation analyses in a
state of stable equilibrium were discussed already.
• Structure is not necessarily stable even though the stress is
in the allowable range.
• Some members, may be subjected to compressive
loadings, and if these members are long and slender the
loading may be large enough to cause the member to
deflect laterally or sidesway.
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Columns and Buckling
Long slender members subjected to an axial compressive
force are called columns, and the lateral deflection that
occurs is called buckling. Quite often the buckling of a
column can lead to a sudden and dramatic failure of a
structure or mechanism, and as a result, special attention
must be given to the design of columns so that they can
safely support their intended loadings without buckling.
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Critical Load
The maximum axial load that
a column can support when it
is on the verge of buckling is
called the critical load, 𝑃𝑐𝑟 .
Any additional loading will
cause the column to buckle
and therefore deflect laterally
as shown in Figure.
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Critical Load
In order to better
understand the nature of
this instability, consider a
two-bar mechanism
consisting of weightless
bars that are rigid and pin
connected as shown in
figure.
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Critical Load
When the bars are in the
vertical position, the spring,
having a stiffness k, is
unstretched, and a small
vertical force P is applied at
the top of one of the bars.
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Critical Load
We can upset this equilibrium
position by displacing the pin
at A by a small amount Δ as
shown in figure. The spring
will produce a restoring force
𝐹 = 𝑘∆, while the applied
load P develops two
horizontal components, 𝑃𝑥 =
𝑃𝑡𝑎𝑛𝜃
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Critical Load
These horizontal forces tend to
push the pin (and the bars)
further out of equilibrium.
𝐿
Since 𝜃 is small, ∆≈ 𝜃 and
2
𝑡𝑎𝑛𝜃 ≈ 𝜃. Thus the restoring
spring force becomes 𝐹=
𝐿
𝑘𝜃 2 , and the disturbing force
is 2𝑃𝑥 = 2𝑃𝜃.
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Critical Load
If the restoring force is greater than the disturbing force,
𝑘𝐿
then 𝑃 < and the system is said to be in stable
4
equilibrium as the force developed in the spring is adequate
to restore the bars back to their vertical position.
If the restoring force is smaller than the disturbing force,
𝑘𝐿
then 𝑃 > and the system is said to be in unstable
4
equilibrium as the the system will tend to move out of
equilibrium and not be restored to its original position.
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Critical Load
If the restoring force is equal to the disturbing force, then
𝑘𝐿
𝑃= . This load is referred as the critical load, 𝑃𝑐𝑟 . This
4
load represents a case of the mechanism being in neutral
equilibrium. Since 𝑃𝑐𝑟 is independent of the (small)
displacement of the bars, any slight disturbance given to the
mechanism will not cause it to move further out of
equilibrium, nor will it be restored to its original position.
Instead, the bars will remain in the deflected position.
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Stability of Equilibrium
Concept of stable, unstable and neutral equilibrium is easy
to visualize by considering a ball on a surface.
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Euler Load for Columns with
Pinned Ends
• Consider the ideal perfectly
straight column with pinned
supports at both ends as shown
in figure.
• The least force at which a
buckled mode is possible is the
critical or Euler buckling load.
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Euler Load for Columns with
Pinned Ends
Assume the compressed column is displaced as shown in the
figure. In this position, the bending moment according to the
beam sign convention is −𝑃𝑣.
By substituting this value of moment into the deflection
equation, the differential equation becomes:
𝑑2𝑣/𝑑𝑥2 = 𝑀/𝐸𝐼 = −(𝑃/𝐸𝐼)𝑣
Letting 2 = 𝑃/𝐸𝐼 , 𝑑2𝑣/𝑑𝑥2 + 2 𝑣 = 0
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Euler Load for Columns
with Pinned Ends
This equation is of the same form as the one for simple
harmonic motion, and its solution is
𝑣 = 𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥
Boundary conditions
𝑣(0) = 0, 𝑣(𝐿) = 0
Then, 𝐵 = 0
Also, for the non-zero solution, 𝐿 = 𝑛
1
Since, 2 = 𝑃/𝐸𝐼, (𝑃/𝐸𝐼)2 × 𝐿 = 𝑛
Thus, 𝑃 = (𝑛22𝐸𝐼)/(𝐿2)
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Euler Load for Columns
with Pinned Ends
These 𝑃𝑛’𝑠 are eigenvalues for this problem.
However, since in stability problems, only the least value
of 𝑃𝑛 is of importance, 𝑛 must be taken as unity, and the
critical or Euler Load 𝑃𝑐𝑟 for an initially perfectly straight
elastic column with pinned ends becomes
𝑃𝑐𝑟 = (2𝐸𝐼)/(𝐿2)
According to the previous equation of the deflection
equation, at the critical load, the equation of the buckled
elastic curve is
𝑣 = 𝐴𝑠𝑖𝑛𝑥
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Euler Load for Columns
with Pinned Ends
First, second, and third modes
of this deflection equation are
also shown in the figure.
However, higher modes have
no meaning because the
primary mode is most critical.
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Euler Loads for Columns with
Different End Restraints
The procedure to find out the Euler buckling loads for
columns with different end restraints is very similar to
the previously discussed pinned ends restraints
conditions. Only difference is different boundary
conditions.
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Euler Loads for Columns with
Different End Restraints
Columns fixed at one end and
pinned at the other
Here the effect of unknown end
moment 𝑀𝑜 and the reactions must
be considered in setting up the
differential equation for the elastic
curve at the critical load:
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Euler Loads for Columns with
Different End Restraints
𝑀 = −𝑃𝑣 + 𝑀𝑜 − (𝑀𝑜 /𝐿)𝑥
𝑑2𝑣/𝑑𝑥2 = 𝑀/𝐸𝐼
= [−𝑃𝑣 + 𝑀𝑜 (1 − 𝑥/𝐿)]/𝐸𝐼
Letting, 2 = 𝑃/𝐸𝐼, and transposing,
𝑑2𝑣/𝑑𝑥2 + 2𝑣 = (2 𝑀𝑜 /𝑃)(1 − 𝑥/𝐿)
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Euler Loads for Columns with
Different End Restraints
The homogeneous solution of
this differential equation is the
same as that given previously.
The particular solution, due to the
non-zero right side, is given by
dividing the term on that side by
2.
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Euler Loads for Columns with
Different End Restraints
The complete solution then becomes
𝑣
= 𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥 + (𝑀𝑜 /𝑃)(1 − 𝑥/𝐿)
Boundary conditions,
𝑣(0) = 0, 𝑣(𝐿) = 0, 𝑣´(0) = 0
𝑀𝑜
𝑣 0 = 0 = 𝐵 +
𝑃
𝑣(𝐿) = 0 = 𝐴𝑠𝑖𝑛 𝐿 + 𝐵𝑐𝑜𝑠 𝐿
𝑣´(0) = 0 = 𝐴 − 𝑀𝑜 /𝑃𝐿
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Euler Loads for Columns with
Different End Restraints
Solving these equations simultaneously,
one obtains 𝐿 = tan 𝐿
The smallest 𝐿 which satisfies the
above equation is L = 4.493.
Then one obtains the critical load
𝑃𝑐𝑟 = (20.19𝐸𝐼)/𝐿2
= (2.052𝐸𝐼)/𝐿2
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Euler Loads for Columns with
Different End Restraints
Columns fixed on both ends
By similar procedure, it can be shown that
𝑃𝑐𝑟 = (42𝐸𝐼)/𝐿2
Cantilever Columns
By similar procedure, it can be shown that
𝑃𝑐𝑟 = (2𝐸𝐼)/(4𝐿2)
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Euler Loads for Columns with
Different End Restraints
Effective Column Length
The previously derived equations are similar, and can be
expressed in one equation using the effective column length
(𝐿𝑒):
𝑃𝑐𝑟 = (2𝐸𝐼)/𝐾𝐿2 = (2 𝐸𝐼)/(𝐿𝑒2)
where, 𝐾 = 1 for pinned - pinned end conditions
= 0.7 for fixed - pinned end conditions
= 0.5 for fixed - fixed end conditions
= 2 for free standing columns
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Euler Loads for Columns with
Different End Restraints
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Limitations of the Euler Formulas
Elastic modulus 𝐸 was used in the derivation of the
Euler formulas for columns. Therefore all the reasoning
presented earlier is applicable while the material
behavior remains linearly elastic.
By this limitation, the previously derived buckling
equation can be rewritten in a different form using the
relationship 𝐼𝑚𝑖𝑛 = 𝐴𝑟𝑚𝑖𝑛 2.
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Limitations of the Euler Formulas
𝑃𝑐𝑟 = (2𝐸𝐼)/(𝐿𝑒2) = (2𝐸𝐴𝑟2)/(𝐿𝑒2) 𝑜𝑟
𝑐𝑟 = 𝑃𝑐𝑟/𝐴 = (𝜋 2 𝐸)/(𝐿𝑒 /𝑟)2
where, 𝐿𝑒/𝑟 of the column length to the least radius of gyration
is called the column slenderness ratio.
A graphical interpretation of the previous equation is shown on
next slide. Since the previous equation is based on the elastic
behavior of a material, 𝑐𝑟 , determined by this equation cannot
exceed the proportional limit of a material.
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Limitations of the Euler Formulas
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Limitations of the Euler Formulas
Therefore, the hyperbolas shown in the figure are drawn
dashed beyond the individual material’s proportional limit,
and these portion of the curves cannot be used.
Also, note that a precise definition of a long column is now
possible with the aid of this figure.
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Example 1
Find the shortest length 𝐿 for a steel column with pinned
ends having a cross-sectional area of 60 × 100 𝑚𝑚, for
which the elastic Euler formula applies.
Let E = 200 GPa and assume the proportional limit to be
250 MPa.
Solution:
𝐼𝑚𝑖𝑛 = 100 × 603/12 = 1.8 × 106 𝑚𝑚4
𝑟𝑚𝑖𝑛 = 𝐼𝑚𝑖𝑛/𝐴 = 1.8 × 106/60 × 100 = 3 × 10 𝑚𝑚
Because of end conditions,
𝐿𝑒 = 𝐿, 𝑐𝑟 = (2𝐸)/(𝐿/𝑟)2
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Example 1
Solve for slenderness ratio (𝐿/𝑟) at the proportional
limit.
(𝐿/𝑟)2 = (2𝐸)/ 𝑐𝑟
= (2 × 200 × 103)/250 = 8002
(𝐿/𝑟) = 88.9 𝑎𝑛𝑑
𝐿 = 88.9 × 3 × 10 = 1540 𝑚𝑚
Therefore, shortest length 𝐿 which does not cause the
buckling of the column is 1.54 𝑚.
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Example 2
The A-36 steel member W8 × 31 shown in
figure is to be used as a pin-connected
column. Determine the largest axial load it
can support before it either begins to buckle
or the steel yields.
Column’s cross-sectional area and moments
of inertia are
𝐴 = 9.13 𝑖𝑛2, 𝐼𝑥 = 110 𝑖𝑛4, 𝐼𝑦 = 37.1 𝑖𝑛4
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Example 2
By inspection, buckling will occur about the y–y axis.
Since this stress exceeds the yield stress (36 ksi), the
load P is determined from simple compression:
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Example 3
A W6 × 15 steel column is 24 ft long
and is fixed at its ends as shown in
figure. Its load-carrying capacity is
increased by bracing it about the y–y
(weak) axis using struts that are
assumed to be pin connected to its
mid-height. Determine the load it
can support so that the column does
not buckle nor the material exceed
the yield stress.
Take 𝐸𝑠𝑡 = 29 × 103 ksi and 𝜎𝑌 =
60 ksi.
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Example 3
The buckling behavior of the column will be different
about the x–x and y–y axes due to the bracing. The buckled
shape for each of these cases is shown in figures below.
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Example 3
The effective length for buckling about the x–x axis is
𝐾𝐿 𝑥 = 0.5 × 24 𝑓𝑡. = 12 𝑓𝑡. = 144 𝑖𝑛.
The effective length for buckling about the y–y axis is
24
𝐾𝐿 𝑦 = 0.7 × 𝑓𝑡. = 8.40 𝑓𝑡. = 100.8 𝑖𝑛.
2
Column’s cross-sectional area and moments of inertia are
𝐴 = 4.43 𝑖𝑛2,
𝐼𝑥 = 29.1 𝑖𝑛4,
𝐼𝑦 = 9.32 𝑖𝑛4
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Example 3
By comparison, buckling will occur about the y–y axis.
Since this stress is less than the yield stress, buckling will
occur before the material yields. Therefore, 𝑃𝑐𝑟 =
263 𝑘𝑖𝑝𝑠
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Example 3
It can be seen that buckling will always occur about the
column axis having the largest slenderness ratio, since a
large slenderness ratio will give a small critical stress.
Thus, using the data for the radius of gyration from the
table in Appendix B, we have
Hence, y–y axis buckling will occur, which is the same
conclusion reached by earlier calculations.
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Example 4
The aluminum column is fixed at its
bottom and is braced at its top by
cables so as to prevent movement at
the top along the x axis as shown in
figure. If it is assumed to be fixed at
its base, determine the largest
allowable load P that can be applied.
Use a factor of safety for buckling of
F.S. = 3.0. Take 𝐸𝑎𝑙 = 70 𝐺𝑃𝑎,
𝜎𝑌 = 215 𝑀𝑃𝑎,
𝐴 = 7.5 × 10−3 𝑚2 , 𝐼𝑥 = 61.3 × 10−6 𝑚4 ,
𝐼𝑦 = 23.2 × 10−6 𝑚4
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Example 4
Buckling about the x and y
axes is shown in figures.
For x–x axis buckling, K = 2,
𝐾𝐿 𝑥 = 2 × 5 = 10 𝑚.
For y–y axis buckling, K =
0.7,
𝐾𝐿 𝑦 = 0.7 × 5 = 3.5 𝑚.
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Example 4
the critical loads for each case are
Column will buckle about the x–x axis. The allowable load
is therefore
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Eccentric Loads and the Secant
Formula
Since no column is perfectly straight nor are the applied forces
perfectly concentric, behavior of real columns may be studied
with some imperfections or possible misalignment of the
applied loads. Also, there are situations where load is applied
at an eccentricity deliberately. Therefore, an eccentrically
loaded column can be studied and its capacity determined on
the basis of an allowable elastic stress. This capacity will not
be the ultimate capacity of the column.
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Eccentric Loads and the Secant
Formula
To analyze the behavior of an
eccentrically loaded column, consider the
column shown in figure.
Differential equation for the elastic curve
is the same as for concentrically loaded
column
𝑑2𝑣/𝑑𝑥2 = 𝑀/𝐸𝐼 = −(𝑃/𝐸𝐼)𝑣
Again letting 2 = 𝑃/𝐸𝐼, the general
solution is as before
𝑣 = 𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥
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Eccentric Loads and the Secant
Formula
Boundary conditions
𝐿
𝑣(0) = 𝑒, 𝑣(𝐿) = 𝑒, 𝑣´ = 0
2
Using the boundary conditions,
𝐿
𝑠𝑖𝑛 2
𝐵 = 𝑒, 𝐴 = 𝑒
𝐿
𝑐𝑜𝑠 2
𝐿
𝑠𝑖𝑛 2
𝑣=𝑒 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
𝐿
𝑐𝑜𝑠
2
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Eccentric Loads and the Secant
Formula
Maximum deflection will occur at 𝐿/2
𝐿
𝐿 𝑠𝑖𝑛2 2 𝐿 𝐿
𝑣 = 𝑣𝑚𝑎𝑥 = 𝑒 + 𝑐𝑜𝑠 = 𝑒 𝑠𝑒𝑐
2 𝐿 2 2
𝑐𝑜𝑠 2
2
Largest bending moment 𝑀 will be equal to 𝑃𝑣𝑚𝑎𝑥
Maximum compressive stress will be
𝑃 𝑀𝑐 𝑃 𝑃𝑣𝑚𝑎𝑥 𝑐
𝜎𝑚𝑎𝑥 = + = +
𝐴 𝐼 𝐴 𝐴𝑟 2
𝑃 𝑒𝑐 𝐿
= 1 + 2 𝑠𝑒𝑐
𝐴 𝑟 2
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Eccentric Loads and the Secant
Formula
But = 𝑃/𝐸𝐼 = 𝑃/𝐸𝐴𝑟 2 , hence,
𝑃 𝑒𝑐 𝐿 𝑃
𝜎𝑚𝑎𝑥 = 1 + 2 𝑠𝑒𝑐
𝐴 𝑟 𝑟 4𝐸𝐴
This equation is known as the secant formula for the
columns.
In the secant formula, the radius of gyration r may not be
the minimum, since it is obtained from the value of I
associated with the axis around which bending occurs.
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Eccentric Loads and the Secant
Formula
It can be noticed that there is a nonlinear relationship
between the load and the stress. Hence, the principle of
superposition does not apply, and therefore the loads
have to be added before the stress is determined.
Furthermore, due to this nonlinear relationship, any
factor of safety used for design purposes applies to the
load and not to the stress.
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Eccentric Loads and the Secant
Formula
For an allowable force 𝑃𝑎 acting on the column, 𝑃 must be
substituted by 𝑛𝑃𝑎 in secant formula, where 𝑛 is the factor
of safety, and 𝜎𝑚𝑎𝑥 must be set at 𝜎𝑌 , yield point of the
material
𝑛𝑃𝑎 𝑒𝑐 𝐿 𝑛𝑃𝑎
𝜎𝑚𝑎𝑥 = 𝜎𝑌 = 1 + 2 𝑠𝑒𝑐
𝐴 𝑟 𝑟 4𝐸𝐴
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Eccentric Loads and the Secant
Formula
Application of secant formula is cumbersome as it involves
trail and error procedure. Alternatively, it can be studied
graphically, as shown in figure (next slide). From the
figure, it can be noted that load eccentricity has greater
effect on short columns and negligible effect on very
slender columns. These types of graphs form a suitable aid
in practical design. Secant formula covers whole range of
column lengths but the greatest handicap in using it is that
some eccentricity is to be assumed even for supposedly
straight columns.
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Eccentric Loads and the Secant
Formula
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Example 5
A W8 × 40 A-36 steel column,
shown in figure, is fixed at its
base and braced at the top so that
it is fixed from displacement, yet
free to rotate about the y–y axis.
Also, it can sway to the side in
the y–z plane. Determine the
maximum eccentric load the
column can support before it
either begins to buckle or the
steel yields.
𝐴 = 11.7 𝑖𝑛2, 𝐼𝑥 = 146 𝑖𝑛4, 𝐼𝑦 = 49.1𝑖𝑛4
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Example 5
From the support conditions it is seen that about the y–y axis
the column behaves as if it were pinned at its top and fixed at
the bottom and subjected to an axial load P. About the 𝑥– 𝑥
axis the column is free at the top and fixed at the bottom, and
it is subjected to both an axial load P and moment M = 9 × P
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Example 5
𝑦– 𝑦 Axis Buckling,
K = 0.7, 𝐾𝐿 𝑦 = 0.7 × 12 = 8.4 𝑓𝑡. = 100.8 in.
𝑥– 𝑥 Axis Yielding,
K = 2, 𝐾𝐿 𝑥 = 2 × 12 = 24 𝑓𝑡. = 288 in.
Using A = 11.7 in2, c = 8.25 in./2 = 4.125 in., rx = 3.53 in. and
applying secant formula
𝑃𝑥 𝑒𝑐 𝐿 𝑃𝑥
𝜎𝑌 = 1 + 2 𝑠𝑒𝑐
𝐴 𝑟𝑥 𝑟𝑥 4𝐸𝐴
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Example 5
Substituting the data and simplifying yields
421.2 = 𝑃𝑥 1 + 2.979 sec (0.0700𝑃𝑥 )
Solving for 𝑃𝑥 by trial and error, noting that the argument for
the secant is in radians, we get
𝑃𝑥 = 88.4 𝑘𝑖𝑝
Since this value is less than 𝑃𝑐𝑟 𝑦 = 1383 kip, failure will
occur about the x–x axis.
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27
