1) The document discusses various buckling modes of columns including flexural, torsional-flexural, and torsional buckling. It provides examples of buckling in thin-walled tubes and prismatic members. 2) Euler buckling formulas are presented for columns with different end conditions, such as both ends pinned, one end fixed and one end pinned. The critical buckling load depends on the effective length which accounts for the end conditions. 3) Limitations of the Euler formulas and generalized formulas are discussed. The tangent modulus formula extends the elastic analysis to the inelastic range by using the tangent modulus.

1. Chapter 11
Stability of Equilibrium: Columns
Mechanics of Solids

2. • The selection of structural and machine elements is based on three
characteristics: strength, stiffness, and stability.
Examples of Instability
• If the wall thickness of tubular cross-section is thin, the plate-like
elements of such members can buckle locally, Fig. 1(a).
• At a sufficiently large axial load, the side walls tend to subdivide into a
sequence of alternating inward and outward buckles.
• As a consequence, the plates carry a smaller axial stress in the regions
of large amount of buckling displacement away from corners, Fig. 1(b).

3. • For such cases, it is customary to approximate the complex stress
distribution by a constant allowable stress acting over an effective
width 𝑤 next to the corners or stiffeners.
Fig. 1: Schematic of buckled thin-walled square tube

4. • Consider Fig. 2, where the emphasis is placed on the kind of buckling
that is possible in prismatic members.
Fig. 2: Column buckling modes: (a) pure flexural,
(b) and (c) torsional-flexural, and (d) pure torsional

5. • A plank of limited flexural but adequate torsional stiffness subjected
to an axial compressive force is shown to buckle in a bending mode,
Fig. 2(a).
• If the same plank is subjected to end moments, Fig. 2(b), in addition
to a flexural buckling mode, the cross sections also have a tendency to
twist.
• This is a torsion-bending mode of buckling, and the same kind of
buckling may occur for the eccentric force 𝑃, as shown in Fig. 2(c).
• A pure torsional buckling mode is illustrated in Fig. 2(d). This occurs
when the torsional stiffness of a member is small.
• A number of the open thin-walled sections in Fig. 3 are next examined
for their susceptibility to torsional buckling.

6. Fig. 3: Column sections exhibiting different buckling modes.
• Two section having biaxial symmetry, where centroids 𝐶 and shear
center 𝑆 coincide, are shown in Fig. 3(a).
• Compression members having such cross sections buckle either in
pure flexure, Fig. 2(a), or twist around 𝑆, Fig. 2(d).
• Flexural buckling would occur for the sections in Fig. 3(b) if the
smallest flexural stiffness around the major principal axis is less than
the torsional stiffness.

7. • Otherwise, simultaneous flexural and torsional buckling would
develop, with the member twisting around 𝑆.
• For the sections in Fig. 3(c), buckling always occur in the latter mode.
• It is assumed that the wall thicknesses of members are sufficiently
large to exclude the possibility of torsional or torsional-flexural
buckling.
• Consider Fig. 4, where two bars with pinned joints at the ends form a
very small angle with the horizontal. It is possible that applied force
𝑃 can reach a magnitude such that the deformed compressed bars
become horizontal.
• Then, on a slightly further increase in 𝑃, the bars snap-through a
new equilibrium position.

8. Fig. 4: Snap-through of compression bars Fig. 5: Spiral spatial twist-buckling of a slender shaft
• In Fig. 5, a slender circular bar is subjected to torque 𝑇. When
applied torque 𝑇 reaches a critical value, the bar snaps into a helical
spatial curve.

9. Criteria for Stability of Equilibrium
• Consider a rigid vertical bar with a torsional spring of stiffness 𝑘 at the
base, as shown in Fig. 6(a).
Fig. 6: Buckling behavior of a rigid bar

10. • The behavior of such a bar subjected to vertical force 𝑃 and
horizontal force 𝐹 is shown in Fig. 6(b) for a large and a small 𝐹.
• The system must be displaced a small (infinitesimal) amount
consistent with the boundary conditions. Then, if the restoring forces
are greater than the forces tending to upset the system, the system is
stable, and vice versa.
• The rigid bar shown in Fig. 6(a) can only rotate. For an assumed small
rotation angle θ, the restoring moment is 𝑘θ, with 𝐹 = 0, the
upsetting moment is 𝑃𝐿 sin θ ≈ 𝑃𝐿θ.
Stable System: 𝑘θ > 𝑃𝐿θ
Unstable system: 𝑘θ < 𝑃𝐿θ
Neutral system: 𝑘θ = 𝑃𝐿θ

11. • Critical or buckling load,
𝑃𝑐𝑟 = 𝑘 𝐿
• In the presence of horizontal force 𝐹, the 𝑃 − θ curves are as shown by
the dashed lines in Fig. 6(b) becoming asymptotic to the horizontal line
at 𝑃𝑐𝑟.
Fig. 7: (a) Stable, (b) unstable, and (c) neutral equilibrium
• A system is in a state of neutral equilibrium when it has at least two
neighboring equilibrium positions an infinitesimal distance apart.

12. • The horizontal line for 𝐹 = 0 shown in Fig. 6(b) is purely schematic for
defining 𝑃𝑐𝑟.
• Consider the rigid vertical bar shown in Fig. 6(a) and set 𝐹 = 0. The
equation of equilibrium,
𝑃𝐿 𝛿θ − 𝑘 𝛿θ = 0 or 𝑃𝐿 − 𝑘 𝛿θ = 0
• Two distinct solutions of above equation: (i) when 𝛿θ = 0, and (ii)
𝑃𝑐𝑟 = 𝑘 𝐿.
• Since at 𝑃𝑐𝑟, there are these two branches of the solution, such a
point is called the bifurcation (branch) point.
• By using the exact (nonlinear) differential equations for curvature, for
elastic columns, one can find equilibrium positions above 𝑃𝑐𝑟, Fig. 8

13. Fig. 8: Behavior of an ideal elastic column
• Increasing 𝑃𝑐𝑟 by a mere 1.5% causing a maximum sideways deflection
of 22% of the column length.
• Another illustration of the meaning of 𝑃𝑐𝑟 in relation to the behavior
of elastic and elastic-plastic columns based on nonlinear analyses is
shown in Fig. 9.

14. Fig. 9: Behavior of straight and initially curved columns where 𝑣0 𝑚𝑎𝑥 = ∆0
• Columns that are initially bowed into sinusoidal shapes with a
maximum center deflection of ∆0 are considered.

15. • Regardless of the magnitude of ∆0, critical load 𝑃𝑐𝑟 serves as an
asymptote for columns with a small amount of curvature, Fig. 9(b).
• A perfectly elastic initially straight long column with pinned ends,
upon buckling into approximately a complete circle, attains the
intolerable deflection of 0.4 of the column length.
• In elastic-plastic columns, Fig. 9(c), only a perfectly straight column
can reach 𝑃𝑐𝑟 and thereafter drop precipitously in its carrying
capacity.

16. Part A- Buckling Theory for Columns
• The moment of inertia is a maximum around one centroidal axis and
of the cross-sectional area a minimum around the other, Fig. 10.
Fig. 10: Flexural column buckling occurs in plane of major axis
• Consider the ideal perfectly straight column with pinned supports at
both ends, Fig. 11(a).

17. Fig. 11: Column pinned at both ends
𝑑2 𝑣
𝑑𝑥2 =
𝑀
𝐸𝐼
= −
𝑃
𝐸𝐼
𝑣
• Letting 𝜆2 = 𝑃 𝐸𝐼

18. • This is an equation of the same form as the one for simple harmonic
motion, and its solution is
𝑣 = 𝐴 sin λ𝑥 + 𝐵 cos λ𝑥
• Boundary conditions,
𝑣 0 = 0 and 𝑣 𝐿 = 0
Hence, 𝐵 = 0 and 𝑣 𝐿 = 0 = 𝐴 sin λ𝐿
𝜆𝐿 = 𝑛𝜋 or 𝑃 𝐸𝐼 𝐿 = 𝑛𝜋
𝑃𝑐𝑟 =
𝑛2 𝜋2 𝐸𝐼
𝐿2
• These 𝑃𝑛’s are the eigenvalues for this problem.
• Since in stability problems only the least value of 𝑃𝑛 is of importance,
𝑛 = 1.

19. • The critical or Euler load 𝑃𝑐𝑟 for an initially perfectly straight elastic
column with pinned ends becomes
𝑃𝑐𝑟 =
𝜋2 𝐸𝐼
𝐿2
• At the critical load, since 𝐵 = 0, the equation of the buckled elastic
curve is
𝑣 = 𝐴 sin λ𝑥
• This is the characteristic, or Eigen function of this problem.
• Since λ = 𝑛π 𝐿, 𝑛 can assume any integer value.
• For the fundamental case 𝑛 = 1, the elastic curve is a half-wave sine
curve. This shape and the modes corresponding to 𝑛 = 2 and 3 are
shown in Fig. 12.

20. Fig. 12: First three buckling modes for
a column pinned at both ends
Fig. 13: Column fixed at one end and
pinned at the other

21. Euler Load for Columns with Different End
Restraints
• Consider a column with one end fixed and the other pinned, Fig. 13.
• Differential equation for the elastic curve at the critical load:
𝑑2 𝑣
𝑑𝑥2 =
𝑀
𝐸𝐼
=
−𝑃𝑣+𝑀0 1− 𝑥 𝐿
𝐸𝐼
letting λ2 = 𝑃 𝐸𝐼 as before,
𝑑2 𝑣
𝑑𝑥2 + λ2 𝑣 =
λ2 𝑀0
𝑃
1 −
𝑥
𝐿
• The particular solution, due to the nonzero right side, is given by
dividing the term on that side by λ2.

22. • The complete solution,
𝑣 = 𝐴 sin λ𝑥 + 𝐵 cos λ𝑥 + 𝑀0 𝑃 1 − 𝑥 𝐿
• Kinematic boundary conditions,
𝑣 0 = 0 𝑣 𝐿 = 0 and 𝑣′ 0 = 0
• Using above boundary conditions, transcendental equation
λ𝐿 = tan λ𝐿
λ𝐿 = 4.493
• From which the corresponding least Eigen value or a critical load for a
column fixed at one end and pinned at the other is
𝑃𝑐𝑟 =
20.19𝐸𝐼
𝐿2 =
2.05π2 𝐸𝐼
𝐿2

23. Fig. 14: Effective lengths of columns with different restraints

24. • It can be shown that in the case of a column fixed at both ends, Fig.
14(d), the critical load is
𝑃𝑐𝑟 =
4𝜋2 𝐸𝐼
𝐿2
• The critical load for a free-standing column, Fig. 14(b), with a load at
the top is
𝑃𝑐𝑟 =
𝜋2 𝐸𝐼
4𝐿2
• In fundamental case, the effective column lengths can be used instead
of the actual column length. This length is the distance between the
inflection points on the elastic curves.
• The effective column length 𝐿 𝑒 for the fundamental case is 𝐿, for the
cases discussed it is 0.7𝐿, 0.5𝐿, and 2𝐿, respectively.

25. • For a general case, 𝐿 𝑒 = 𝐾𝐿, where 𝐾 is the effective length factor,
which depends on the end restraints.
• The axial force associated simultaneously with the bent and the
straight shape of the column is the critical buckling load. This occurs at
the bifurcation (branching) point of the solution.
• Elastic buckling load formulas do not depend on the strength of a
material, they determine the carrying capacity of columns.

26. Limitations of the Euler Formulas
• The reasoning is applicable while the material behavior remains
linearly elastic.
• By definition, 𝐼 = 𝐴𝑟2, where 𝐴 is the cross-sectional area, and 𝑟 is its
radius of gyration.
𝑃𝑐𝑟 =
π2 𝐸𝐼
𝐿2 =
π2 𝐸𝐴𝑟2
𝐿2
𝜎𝑐𝑟 =
𝑃𝑐𝑟
𝐴
=
π2 𝐸
𝐿 𝑟 2
• The critical stress 𝜎𝑐𝑟 is an average stress over the cross-sectional
area 𝐴 of a column at the critical load 𝑃𝑐𝑟.

27. • The ratio 𝐿 𝑟 of the column length to the least radius of gyration is
called the column slenderness ratio.
• A graphical interpretation of last equation is shown in Fig. 15. for each
material, 𝐸 is constant, and the resulting curve is hyperbola.
• The hyperbolas shown in Fig. 15 are drawn dashed beyond the
individual material’s proportional limit, and these portions of the
curves cannot be used.
• The column is said to be long if the elastic Euler formula applies.
• The beginning of the long-column range is shown for three materials
in Fig. 15.

28. Fig. 15: Variation of critical column stress with slenderness ratio for three different materials

29. Generalized Euler Buckling-Load Formulas
• A typical compression stress-strain diagram for a specimen that is
prevented from buckling is shown in Fig. 16(a). In the stress range
from 𝑂 to 𝐴, the material behaves elastically.
Fig. 16: (a) Compression stress-strain diagram, and (b) critical stress in column versus slenderness ratio

30. • This portion of the curve is shown as 𝑆𝑇 in Fig. 16(b).
• This curve does not represent the behavior of one column, but rather
the behavior of an infinite number of ideal columns of different
lengths.
• A column with an 𝐿 𝑟 ratio corresponding to point 𝑆 in Fig. 16(b) is
the shortest column of a given material and size that will buckle
elastically.
• If the stress level in the column has passed point 𝐴 and has reached
some point 𝐵 perhaps.
• At this point, the material stiffness is given instantaneously by the
tangent to the stress-strain curve, i.e., by the tangent modulus 𝐸𝑡, Fig.
16(a).

31. • The column remains stable if its new flexural rigidity 𝐸𝑡 𝐼 at 𝐵 is
sufficiently large, and it can carry a higher load.
• A column of ever “less stiff material” is acting under an increasing load.
• To make the elastic buckling formulas applicable in the inelastic range,
the generalized Euler buckling-load formula, or the tangent modulus
formula,
𝜎𝑐𝑟 =
π2 𝐸𝑡
𝐿 𝑟 2
• A plot representing this behavior for low and intermediate ratios of
𝐿 𝑟 is shown in Fig. 16(b) by the curve from 𝑅 to 𝑆.
• Columns having small 𝐿 𝑟 ratios exhibiting no buckling phenomena are
called short columns. They can carry very large loads.

32. • Plots of critical stress 𝜎𝑐𝑟 versus the slenderness ratio 𝐿 𝑟 for fixed-
ended columns and pin-ended ones are shown in Fig. 17.
• The carrying capacity for these two cases is in a ratio of 4 to 1 only for
columns having the slenderness ratio 𝐿 𝑟 1 or greater.
Fig. 17: Comparison of the behavior of columns with different end conditions

33. Eccentric Loads and the Secant Formula
• To analyze the behavior of an eccentrically loaded column, consider
the column shown in Fig. 18.
Fig. 18: Eccentrically loaded column

34. • If the origin of the coordinate axes is taken at the upper force 𝑃, the
bending moment at any section is −𝑃𝑣, and the differential equation
for the elastic curve is,
𝑑2 𝑣
𝑑𝑥2 =
𝑀
𝐸𝐼
= −
𝑃
𝐸𝐼
𝑣
by letting, λ = 𝑃 𝐸𝐼, the general solution,
𝑣 = 𝐴 sin λ𝑥 + 𝐵 cos λ𝑥
• Boundary conditions,
𝑣 0 = 𝑒 and 𝑣′ 𝐿 2 = 0
• The equation for the elastic curve is,
𝑣 = 𝑒
sinλ𝐿 2
cosλ𝐿 2
sin λ𝑥 + cos λ𝑥

35. • Because of symmetry, the elastic curve has a vertical tangent at the
midheight of the column.
• The maximum deflection occurs at 𝐿 2,
𝑣 𝐿 2 = 𝑣 𝑚𝑎𝑥 = 𝑒 sec
λ𝐿
2
• The largest bending moment = 𝑃𝑣 𝑚𝑎𝑥.
• The maximum compressive stress,
𝜎 𝑚𝑎𝑥 =
𝑃
𝐴
+
𝑀𝑐
𝐼
=
𝑃
𝐴
+
𝑃𝑣 𝑚𝑎𝑥 𝑐
𝐴𝑟2 =
𝑃
𝐴
1 +
𝑒𝑐
𝑟2 sec
λ𝐿
2
λ = 𝑃 𝐸𝐼 = 𝑃 𝐸𝐴𝑟2
𝜎 𝑚𝑎𝑥 =
𝑃
𝐴
1 +
𝑒𝑐
𝑟2 sec
𝐿
𝑟
𝑃
4𝐸𝐴

36. • This equation is known as the secant formula for columns, it is
applied to columns of any length, provided the maximum stress does
not exceed the elastic limit.
• The relation between 𝜎 𝑚𝑎𝑥 and 𝑃 is not linear, therefore the solutions
for maximum stresses in columns by different axial forces cannot be
superposed.
• For an allowable force 𝑃𝑎 on a column, where 𝑛 is the factor of safety,
𝑛𝑃𝑎 must be substituted for 𝑃,
𝜎 𝑚𝑎𝑥 = 𝜎 𝑦𝑝 =
𝑛𝑃 𝑎
𝐴
1 +
𝑒𝑐
𝑟2 sec
𝐿
𝑟
𝑛𝑃 𝑎
4𝐸𝐴
• Applications of these equations requires a trial-and-error procedure.
Alternatively, they can be studied graphically, Fig. 19.

37. Fig. 19: Results of analyses for different columns by the secant formula

38. • The large effect of the load eccentricity is on short columns and the
negligible one on very slender columns.
• The secant formula for short columns can be obtained when 𝐿 𝑟
approaches to zero. The value of the secant approaches unity.
𝜎 𝑚𝑎𝑥 =
𝑃
𝐴
+
𝑀𝑐
𝐼
=
𝑃
𝐴
+
𝑃𝑒𝑐
𝐴𝑟2

39. Beam Columns
• This is a special case of a member acted upon simultaneously by an
axial force and transverse forces or moments causing bending, known
as beam-columns.
• Consider the rigid bar shown in Fig. 20(a).
• This bar of length 𝐿 is initially held in a vertical position by a spring at
𝐴 having a torsional spring constant 𝑘.
• When vertical force 𝑃 and horizontal force 𝐹 are applied to the top of
the bar, it rotates and the equilibrium equation for deformed state,
𝑀𝐴 = 0 ↻ 𝑃𝐿 sin θ + 𝐹𝐿 cos θ − 𝑘θ = 0
𝑃 =
𝑘θ−𝐹𝐿 cos θ
𝐿 sin θ

40. Fig. 20: Rigid bar with one degree of freedom
• The qualitative feature of this result are shown in Fig. 20(b).

41. • As θ → π, provided the spring continues to function, a very large
force 𝑃 can be supported by the system.
• For a force 𝑃 applied in an upward direction, plotted downward in
the figure, angle θ decreases as 𝑃 increases.
• For small and moderately large deformations, sin θ ≈ θ, and cos θ ≈
1.
𝑃 =
𝑘θ−𝐹𝐿
𝐿θ
or θ =
𝐹𝐿
𝑘−𝑃𝐿
• As θ increases, the discrepancy between this linearized solution and
the exact one becomes very large and loses its physical meaning.

42. Alternative Differential Equations for Beam-
Columns
• Consider the beam-column element shown in Fig. 21.
Fig. 21: Beam-column element

43. • Small-deflection approximations:
𝑑𝑣 𝑑𝑥 = tan θ ≈ sin θ ≈ θ
cos θ ≈ 1 and 𝑑𝑠 ≈ 𝑑𝑥
• Two equilibrium equation,
𝐹𝑦 = 0 ↑ 𝑞 𝑑𝑥 + 𝑉 − 𝑉 + 𝑑𝑉 = 0
𝑀𝐴 = 0 ↻ 𝑀 − 𝑃 𝑑𝑣 + 𝑉 𝑑𝑥 + 𝑞 𝑑𝑥 𝑑𝑥 2 − 𝑀 + 𝑑𝑀 = 0
• Above equation yields,
𝑑𝑉
𝑑𝑥
= 𝑞
𝑉 =
𝑑𝑀
𝑑𝑥
+ 𝑃
𝑑𝑣
𝑑𝑥

44. • Using the usual beam curvature-moment relation 𝑑2 𝑣 𝑑𝑥2 = 𝑀 𝐸𝐼,
two alternative governing differential equations for beam columns:
𝑑2 𝑀
𝑑𝑥2 + λ2 𝑀 = 𝑞
𝑑4 𝑉
𝑑𝑥2 + λ2 𝑑2 𝑉
𝑑𝑥2 =
𝑞
𝐸𝐼
• In above equations, 𝐸𝐼 is assumed to be constant, and λ2 = 𝑃 𝐸𝐼.
𝑉 = 𝐸𝐼
𝑑3 𝑣
𝑑𝑥3 + 𝑃
𝑑𝑣
𝑑𝑥
• The homogeneous solution of 4th order equation and some of its
derivatives are

45. 𝑣 = 𝐶1 sin λ𝑥 + 𝐶2 cos λ𝑥 + 𝐶3 𝑥 + 𝐶4
𝑣′ = 𝐶1λ cos λ𝑥 − 𝐶2λ sin λ𝑥 + 𝐶3
𝑣′′ = −𝐶1λ2 sin λ𝑥 − 𝐶2λ2 cos λ𝑥
𝑣′′ = −𝐶1λ3 cos λ𝑥 + 𝐶2λ3 sin λ𝑥
• These equations are useful for expressing the boundary conditions in
evaluating constants 𝐶1, 𝐶2, 𝐶3, and 𝐶4.
• Solutions of homogeneous equations for particular boundary
conditions lead to critical buckling loads for elastic prismatic columns.