3. • In the design of columns, cross-sectional area is
selected such that
- allowable stress is not exceeded
all
A
P
- deformation falls within specifications
spec
AE
PL
• After these design calculations, may discover
that the column is unstable under loading and
that it suddenly becomes sharply curved or
buckles.
Stability of Structures
4. • Long slender members subjected to axial
compressive force are called columns.
• The lateral deflection that occurs is
called buckling.
• The maximum axial load a column can
support when it is on the verge of
buckling is called the critical load, Pcr.
Stability of Structures
5. • Consider model with two rods and torsional
spring. After a small perturbation,
moment
ing
destabiliz
2
sin
2
moment
restoring
2
L
P
L
P
K
• Column is stable (tends to return to
aligned orientation) if
L
K
P
P
K
L
P
cr
4
2
2
Stability of Structures
6. • Assume that a load P is applied. After a
perturbation, the system settles to a new
equilibrium configuration at a finite
deflection angle.
sin
4
2
sin
2
cr
P
P
K
PL
K
L
P
• Noting that sinθ < θ, the assumed
configuration is only possible if P > Pcr.
Stability of Structures
7. • Summing moments, M = Py, equation of
bending becomes
• General solution is
• Since y = 0 at x = 0, then B = 0.
Since y = 0 at x = L, then
0
2
2
y
EI
P
dx
y
d
x
EI
P
cos
B
x
EI
P
sin
A
y
0
L
EI
P
sin
A
Euler’s Formula for Pin-Ended Beams
8. • Disregarding trivial solution for A = 0, we get
• Which is satisfied if
or
,...
3
,
2
,
1
2
2
2
n
L
EI
n
P
n
L
EI
P
0
sin
L
EI
P
Euler’s Formula for Pin-Ended Beams
9. • Smallest value of P is obtained for n = 1, so critical load for column
is
• This load is also referred to
as the Euler load. The
corresponding buckled
shape is defined by
• A represents maximum
deflection, ymax, which occurs
at midpoint of the column.
2
2
L
EI
Pcr
L
x
sin
A
y
Euler’s Formula for Pin-Ended Beams
10. • A column will buckle about the principal axis of the x-
section having the least moment of inertia (weakest axis).
• For example, the meter stick shown will
buckle about the a-a axis and not
the b-b axis.
• Thus, circular tubes made excellent
columns, and square tube or those
shapes having Ix ≈ Iy are selected
for columns.
Euler’s Formula for Pin-Ended Beams
11. • Buckling equation for a pin-supported long slender column,
Pcr = critical or maximum axial load on column just before it
begins to buckle.
E = modulus of elasticity of material
I = Least modulus of inertia for column’s x-sectional area.
L = unsupported length of pinned-end columns.
2
2
L
EI
Pcr
Euler’s Formula for Pin-Ended Beams
12. • The value of stress corresponding to the critical load,
A
L
EI
A
P
L
EI
P
cr
cr
cr
cr
2
2
2
2
Slenderness Ratio
13. Expressing I = Ar2 where A is x-sectional area of column and
r is the radius of gyration of x-sectional area.
cr = critical stress, an average stress in column just before
the column buckles.
E = modulus of elasticity of material
L = unsupported length of pinned-end columns.
r = smallest radius of gyration of column
2
2
r
L
E
cr
Slenderness Ratio
14. • The geometric ratio L/r in the equation is known as the
slenderness ratio.
• It is a measure of the column’s flexibility and will be used to
classify columns as long, intermediate or short.
Slenderness Ratio
15. IMPORTANT
• Columns are long slender members that are subjected to
axial loads.
• Critical load is the maximum axial load that a column can
support when it is on the verge of buckling.
• This loading represents a case of neutral equilibrium.
Ideal Column with Pin Supports
16. IMPORTANT
• An ideal column is initially perfectly straight, made of
homogeneous material, and the load is applied through the
centroid of the x-section.
• A pin-connected column will buckle about the principal axis
of the x-section having the least moment of intertia.
• The slenderness ratio L/r, where r is the smallest radius of
gyration of x-section. Buckling will occur about the axis
where this ratio gives the greatest value.
• Preceding analysis is limited to centric loadings.
Ideal Column with Pin Supports
20. The A-36 steel W20046 member shown 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.
Example
21. From calculation, column’s x-sectional area and
moments of inertia are A = 5890 mm2,
Ix = 45.5106 mm4,and Iy = 15.3106 mm4.
By inspection, buckling will occur about the y-y axis.
Applying buckling equation, we have
kN
6
.
1887
m
4
mm
1000
/
m
1
mm
10
3
.
15
kN/m
10
200
2
4
4
4
2
6
2
2
2
L
EI
Pcr
Example
22. When fully loaded, average compressive stress in column is
Since this stress exceeds yield stress (250 N/mm2), the load P
is determined from simple compression:
kN
5
.
1472
mm
5890
N/mm
250 2
2
P
P
2
2
N/mm
5
.
320
mm
5890
N/kN
1000
kN
6
.
1887
A
Pcr
cr
Example
32. Exercise 1
Determine
(a) The crtical load for the steel strut
(b) The dimension d for which the aluminium strut will have same critical load
(c) Express the weight of the aluminium alloy strut as a percent of the weight of
the steel strut.
Refer to Figure No
10.12 Page 621
Steel
E = 200 GPa
Ρ = 7860 kg/m3
33. Exercise 2
A compression member of 1.5 m effective length consists of a solid 30 mm
diameter brass rod. In order to reduce the weight of the member by 25%, the
solid rod is replaced by a hollow rod of the cross section shown. Determine
(a) the percent reduction in the critical load
(b) The value of the critical load for the hollow rod. Use E = 70 GPa
Refer to Figure No
10.13 Page 621
15
mm
30 mm
30 mm