- This document discusses a teaching schedule for a course on the failure of slender and stocky columns. It covers topics like column stability, unsymmetric bending, and complex stress/strain over 11 weeks. - The key learning outcomes are to derive the Euler critical load for slender pinned-pinned columns under compression and to predict the failure mode of short and slender columns. - The document motivates the importance of considering both the stiffness and strength of materials, and how the slenderness of a column affects its failure mode in compression.

1. Failure
of
slender
and
stocky
columns
Dr
Alessandro
Palmeri
<A.Palmeri@lboro.ac.uk>

2. Teaching
schedule
Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff
1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- ---
2 Shear centres A P Basic Concepts J E-R Shear Centre A P
3 Principle of Virtual
forces
J E-R Indeterminate Structures J E-R Virtual Forces J E-R
4 The Compatibility
Method
J E-R Examples J E-R Virtual Forces J E-R
5 Examples J E-R Moment Distribution -
Basics
J E-R Comp. Method J E-R
6 The Hardy Cross
Method
J E-R Fixed End Moments J E-R Comp. Method J E-R
7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R
8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R
9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P
10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric
Bending
A P
11 Complex Stress/Strain A P Complex Stress/Strain A P Complex
Stress/Strain
A P
Christmas
Holiday
12 Revision
13
14 Exams
15
2

3. Mo@va@ons
(1/5)
• Load-­‐carrying
structures
may
fail
in
a
variety
of
ways,
depending
upon:
– Type
of
structure
(truss,
frame,
…)
– Condi@ons
of
support
(pinned,
fixed,
…)
– Loads
applied
(sta@c,
dynamic,
…)
– Materials
used
(briQle,
duc@le,
…)
• Failures
are
prevented
by
designing
structures
so
that
maximum
stresses
(strength
criterion)
and
maximum
displacements
(s,ffness
criterion)
remain
within
admissible
limits
3

4. Mo@va@ons
(2/5)
4
• For
the
fans
of
The
Big
Bang
Theory:
– Sheldon
and
Howard
have
got
this
seriously
wrong!
• You
can’t
use
the
Young’s
modulus
to
quan@fy
the
strength
of
material,
but
its
s,ffness!

5. Mo@va@ons
(3/5)
• S@ffness
and
strength
of
materials
– In
the
stress-­‐strain
curve
for
a
duc@le
material
(e.g.
steel),
the
Young’s
modulus
E
defines
the
s@ffness,
while
the
yield
stress
σy
represents
the
strength
5

6. Mo@va@ons
(4/5)
• S@ffness
criterion:
“Slender
Column”
6
• Strength
criterion:
“Short
Column”

7. Mo@va@ons
(5/5)
7
Coventry
Cathedral
ç
Slender
column
Detail
of
the
support
è

8. Learning
Outcomes
• When
we
have
completed
this
unit
(2
lectures
+
1
tutorial),
you
should
be
able
to:
– Derive
the
Euler’s
cri@cal
load
for
slender
pinned-­‐
pinned
columns
in
compression
– Predict
the
mode
of
failure
for
both
short
and
slender
columns
in
compression
8

9. Further
reading
• R
C
Hibbeler,
“Mechanics
of
Materials”,
8th
Ed,
Pren@ce
Hall
–
Chapter
13
on
“Buckling
of
Column”
• T
H
G
Megson,
“Structural
and
Stress
Analysis”,
2nd
Ed,
Elsevier
–
Chapter
21
on
“Structural
Instability”
(eBook)
9

10. Short
and
Slender
Struts
10
• Increasing
the
length
of
a
strut
reduces
its
buckling
load
– For
instance,
a
matchs,ck
is
reasonably
strong
in
compression
(lek),
but
a
longer
s,ck,
with
the
same
cross
sec@on
and
the
same
material,
would
be
weaker
and
buckles
in
compression
(right)
– The
slenderness
of
a
strut
plays
an
important
role
in
its
mode
of
failure
in
compression

11. Buckling,
i.e.
Lateral
Instability
(1/2)
11
• That
is,
if
a
column
is
rela@vely
slender,
it
may
deflect
laterally
when
subjected
to
a
compressive
force
P
(Fig
(a))
and
fail
by
bending
(Fig
(b)),
rather
than
failing
by
direct
compression
of
the
material

12. Buckling,
i.e.
Lateral
Instability
(2/2)
12
• Pcrit
is
the
so-­‐called
cri,cal
buckling
load
– If
the
axial
load
P
is
less
than
Pcrit,
bending
is
caused
by
lateral
loads
only
– If
P
is
greater
than
Pcrit,
the
ruler
bends
even
without
lateral
loads

13. Euler’s
Cri@cal
Load
for
Pinned-­‐Pinned
Slender
Columns
• One
of
the
Learning
Outcomes
of
this
Unit
is
for
you
to
become
able
to
mathema@cally
derive
(and
remember
as
well)
the
expression
of
the
cri@cal
load
Pcrit
for
pinned-­‐pinned
slender
column
• Pcrit=PE
is
oken
called
Euler’s
buckling
load
– Aker
the
Swiss
mathema@cian
Leonhard
Euler
(1707-­‐1783)
!!
Pcrit
=
π2
EImin
L2
13

14. Mathema@cal
Deriva@on:
Bending
Equa@on
• What’s
the
equa@on
ruling
the
beam’s
downward
deflec@on,
uz(x),
for
a
given
bending
moment
diagram,
My(x)?
• We
used
this
second-­‐order
differen@al
equa@on
in
part
A
to
calculate
the
beam’s
deflec@on
under
transverse
loads…
• where,
as
usual:
– E=
Young’s
modulus
– Iyy=
Second
moment
of
area
about
the
horizontal
neutral
axis
14
EIyy
d2
uz (x)
dx2
= −My (x)

15. Mathema@cal
Deriva@on:
Sign
Conven@on
• Do
you
remember
from
last
year?
15

16. Mathema@cal
Deriva@on:
P-­‐Delta
(1/2)
• What’s
the
bending
moment
My
in
this
circumstance?
• We
don’t
have
transverse
loads
this
@me
– the
column
is
subjected
to
the
axial
load
P
only
• How
can
we
have
a
bending
moment?
• In
order
to
derive
the
expression
of
the
Euler’s
buckling
load,
we
need
to
assume
that
– a
disturbance/imperfec@on
exists
in
the
column,
– therefore
the
buckling
occurs
– and
My
can
be
consistently
evaluated
by
using
the
equilibrium
equa@ons
in
the
deformed
shape
16

17. Mathema@cal
Deriva@on:
P-­‐Delta
(1/2)
17
Deformed
shape
Equilibrium
condi@on
P
P
My=
P
uz
EIyy
z
uz

18. Mathema@cal
Deriva@on:
Buckling
Equa@on
(1/2)
• Knowing
the
bending
moment
My
in
the
deformed
shape:
• we
can
subs@tute
it
within
the
deflec@on
equa@on:
• This
equa@on
can
be
rewriQen
as:
• Where
α
is
a
posi@ve
quan@ty,
given
by:
18
My = Puz
EIyy
d2
uz
dx2
= −Puz
d2
uz
dx2
+α2
uz = 0
α =
P
EIyy

19. Mathema@cal
Deriva@on:
Buckling
Equa@on
(2/2)
• What
do
we
do
in
order
to
solve
an
ordinary
differen@al
equa@on?
• First,
we
find
the
general
solu,on,
which
contains
as
many
integra@on
constants
as
the
order
of
the
differen@al
equa@on
(two,
in
this
case)
19
uz
= C1
cos(α x) + C2
sin(α x)

20. Mathema@cal
Deriva@on:
Boundary
Condi@ons
(1/2)
• Second,
we
apply
the
boundary
condi,ons
(BCs)
to
get
the
values
of
the
integra@ons
constants
for
the
par@cular
case
– For
two
unknown
constants,
C1
and
C2,
two
BCs
are
needed!
• For
a
pinned-­‐pinned
column,
the
BCs
read:
• uz=0
@
x=0
(i.e.
the
transverse
transla@on
is
prevented
at
the
lek-­‐hand
side
end)
• uz=0
@
x=L
(i.e.
the
transverse
transla@on
is
prevented
at
the
right-­‐hand
side
end
as
well)
20
EIyy
z

21. Mathema@cal
Deriva@on:
Boundary
Condi@ons
(2/2)
21
• The
applica@on
of
the
first
BC
is
quite
straighrorward
uz
(x) = C1
cos(α x) + C2
sin(α x)
uz
= 0 @ x = 0 1 20 1 0C C⇒ = × + ×
1 0C⇒ =
General
solu5on
Boundary
condi5on

22. Mathema@cal
Deriva@on:
Non-­‐Trivial
Solu@on
(1/3)
22
• The
second
BC
does
require
more
effort
• Trivial
solu@on:
– It
would
follow
y=0
for
any
value
of
the
abscissa
x
– No
transverse
displacements
would
occur
(straight
column)
– This
solu@on
is
therefore
unacceptable
• Non-­‐trivial
solu@on:
uz
(x) = C1
cos(α x) + C2
sin(α x)
uz
= 0 @ x = L 20 sin( )C Lα⇒ =
sin( ) 0Lα⇒ = L nα π⇒ =
n
n
L
π
α α⇒ = =
2 0C⇒ =

23. Mathema@cal
Deriva@on:
Non-­‐Trivial
Solu@on
(2/3)
23
• Recalling
now
the
expression
of
the
parameter
a,
one
obtains:
• The
associated
modes
of
instability,
for
n=
1,
2,
3,
…,
are
sinusoidal
func@ons,
having
a
total
number
n
of
peaks
and
valleys
P
EIyy
=
nπ
L
⇒
P
EIyy
=
n2
π2
L2
⇒ Pn
= n2
π2
EIyy
L2
n= 3
n= 2
n= 1

24. Mathema@cal
Deriva@on:
Non-­‐Trivial
Solu@on
(3/3)
24
• Larger
values
of
the
buckling
load
are
associated
to
more
complicated
modes
of
instability
• Theore@cally,
these
modes
could
be
achieved
if
roller
supports
are
applied
at
the
points
of
contraflexure
• However,
in
prac@ce,
the
lower
value
P1
is
never
exceeded

25. Mathema@cal
Deriva@on:
Euler’s
Buckling
Load
• The
actual
cri,cal
load,
i.e.
the
so-­‐called
Euler’s
buckling
load,
is
the
“engineering
solu@on”,
which
is
the
minimum
among
the
mathema@cal
solu@ons
P1,
P2,
P3,
…,
and
is
obtained
for
n=1
• Moreover,
in
order
to
be
truly
the
minimum,
you
must
use
the
minimum
value
of
the
second
moment
of
area,
which
might
not
be
Iyy
• The
laQer
expression
is
very
important
in
Structural
Engineering
– You
are
requested
to
remember
it
– You
must
be
able
to
derive
this
expression
as
well
25
Pcrit
= PE
= P1
=
π2
EIyy
L2
2
min
E 2
EI
P
L
π
=

26. FEM-­‐Computed
Modes
of
Instability
26
• Euler’s
buckling
load
(PE=
P1=
181
kN)
• Higher
buckling
load
in
the
orthogonal
direc@on
• (P4=
3,518
kN)
XY X
Z
Y
Z
XY X
Z
Y
Z
4
min 330 cmI = = 4
max 6572 cmI
Horizontal
sway
Ver0cal
deflec0on

27. Effects
of
the
Boundary
Condi@ons
(1/2)
27
The
more
the
column’s
ends
are
restrained,
the
higher
is
the
buckling
load
Similar
sinusoidal
shapes
are
observed
for
different
BCs

28. Effects
of
the
Boundary
Condi@ons
(2/2)
28
(a) Pinned-­‐pinned
(b) Can@levered
(c) Fixed-­‐fixed
(d) Propped
L0
is
the
distance
between
two
consecu@ve
crosses
of
the
horizontal
axis

29. Effec@ve
Length
• It
is
useful
to
introduce
the
concept
of
equivalent
length,
Le=k
L
as
the
length
of
a
pinned-­‐pinned
column
having
the
same
Euler’s
cri@cal
load
• We
therefore
must
know
the
value
of
the
coefficient
k
for
different
BCs
29
k=2
k=1
k=0.7
k=0.5
Can5levered
Pinned-­‐
pinned
Propped
Fixed-­‐
fixed
2
min
E 2
e
EI
P
L
π
=

30. Stocky
Columns
(1/2)
• If
we
divide
the
Euler’s
cri@cal
load
PE
by
the
cross
sec@onal
area
A,
we
get
the
so-­‐called
Euler’s
cri@cal
stress
σcrit:
• This
is
the
maximum
normal
stress
which
is
allowable
to
prevent
buckling
instability,
and
is
inversely
propor@onal
to
the
square
of
the
equivalent
length
Le
• If
we
introduce
the
parameter
ρmin
as
the
minimum
radius
of
gyra@on
of
the
cross
sec@on,
and
then
the
slenderness
ra@o
λ=Le/ρmin,
the
above
equa@on
can
be
rewriQen
as:
30
σcrit
=
π2
E Imin
A( )
Le
2
=
π2
E ρmin
2
Le
2
=
π2
E
Le
ρmin( )
2
=
π2
λ2
Eρmin
=
Imin
A
2
min
crit 2
e
EP EI
A AL
π
σ = =

31. Stocky
Columns
(2/2)
31
• The
s@ffer
the
material,
i.e.
the
larger
the
Young’s
modulus
E,
the
higher
is
σcrit:
• The
larger
the
slenderness
ra@o
λ,
the
lower
is
σcrit,
i.e.
very
slender
columns
will
have
very
low
values
of
σcrit
• Conversely,
stocky
columns,
with
a
small
slenderness
ra@o
λ,
will
not
experience
the
buckling
failure,
as
the
yielding
of
the
material
is
likely
to
happen
first:
2
crit 2
E
π
σ
λ
=
crit y Material’s yield stressfσ > =

32. Strength
and
S@ffness
Criteria
(1/2)
• “Strength”
criterion
• Stocky
columns
tend
to
fail
because
the
elas@c
limit
of
the
material
is
reached
• The
safety
checks
is:
• “S,ffness”
criterion
• Slender
columns
tend
to
fail
because
the
elas@c
configura@on
is
unstable
• The
safety
check
is:
32
2
min
E 2
e
EI
P P
L
π
< =y yP P f A< =
Both
must
be
sa0sfied

33. Strength
and
S@ffness
Criteria
(2/2)
• For
briQle
materials
such
as
concrete,
the
yielding
stress
fy
is
replaced
with
the
crushing
stress
fc
• The
safety
check
then
reads:
33
2
min
E 2
e
EI
P P
L
π
< =P < Pc
= fc
A
Both
must
be
sa0sfied
fc

34. Strength
and
S@ffness
Criteria
34
• The
Rankine’s
failure
load
PR
combines
these
two
different
criteria,
therefore
taking
into
account
both
material
and
geometrical
nonlineari@es
• PR=Py
for
λ=0
• PR
approaches
PE
as
λ
goes
to
infinity
0 100 200 300 400
0.0
0.5
1.0
1.5
2.0
l
PêPy
Py
PE
PR
PR
=
Py
PE
Py
+ PE
λ
P/Py
Rankine
(1820-­‐1872)
was
a
Scoush
civil
engineer,
physicist
and
mathema@cian

35. Ul@mate
Normal
Stress
35
• …
experimentally
derived
(dots)
for
wide-­‐flange
steel
columns
• …
as
a
func@on
of
the
slenderness
λ=
k
L/ρmin
λ

36. Key
Learning
Points
1. Columns
in
compression
may
fail
because
– Insufficient
bending
s@ffness:
è
Lateral
buckling
– Insufficient
axial
capacity:
è
Yielding/Crushing
2. Euler’s
buckling
load
PE
depends
on:
– Minimum
second
moment
of
area,
Imin
– Length
of
the
column,
L
– Boundary
condi@ons
3. Interac@on
between
lateral
buckling
and
axial
capacity
can
be
taken
into
account
through
the
(approximate)
Rankine’s
formula
36
è
Effec@ve
length,
Le
PR
=
Py
PE
Py
+ PE
PE
=
π2
EImin
Le
2