Mechanics of structures module4

Mechanics of structuresMechanics of structures
Stresses in 2D plane,
ColumnsColumns
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1

Module IV
Transformation of stresses and strains (two dimensional case only)
Module IV
Transformation of stresses and strains (two-dimensional case only) -
equations of transformation - principal stresses - mohr's circles of
stress and strain - strain rosettes - compound stresses - superposition
d it li it tiand its limitations –
Eccentrically loaded members - columns - theory of columns -
buckling theory - Euler's formula - effect of end conditions -
eccentric loads and secant formula
2

Analysis of plane stress and plane strainy p p
cosnA A θ=
P
L
AA
θ
P
A
σ = cos
n
n
A
A
θ
=
AAn
A
2
2cos cos
cosn
P P
A A
θ θ
σ σ θ= = =
cosP θ
θ
nA A
sin sin cos sin2P Pθ θ θ σ θ
τ
P
θ
sinP θAn
θ
3
2
n
nA A
τ = = =

2 2
σ σ τ+
( ) ( )
2 22
cos sin cos
R n nσ σ τ
σ θ σ θ θ
= +
= +
Resultant stress on inclined plane
2 2
cos cos sin
cos
σ θ θ θ
σ θ
= +
=
0 2 cos sin 0nd
d
σ
σ θ θ
θ
= ⇒ =For maximum normal stress,
0
sin2 0 0σ θ θ= ⇒ =
0
When 0
P
θ σ σ= = =
0 2 0ndτ
θFor maximum shear stress
,maxWhen 0 , n
A
θ σ σ= = =
0 0
0 cos2 0
2 90 45
n
d
σ θ
θ
θ θ
= ⇒ =
⇒ = ⇒ =
For maximum shear stress,
4
0
,maxWhen 45 ,
2 2
n
P
A
σ
θ τ= = =

nσ σ
nτ σ
Rσ 2 2
n nσ τ+
σ
,maxnτ
σ
5

Element under pure shear
τ
B
θ σ
nτ
τ
θ
τ
θ nσ
ττ θ
cos sinBC AC ABσ τ θ τ θ= − −
A C
ττ
. .cos . .sin
sin cos cos sin
sin2
n
n
BC AC ABσ τ θ τ θ
σ τ θ θ τ θ θ
σ τ θ
= − −
= − −
. . .sin . .cosn BC AC AB
AC AB
τ τ θ τ θ= − +
sin2nσ τ θ= −
2 2
. .sin . .cos
cos sin
n
n
AC AB
BC BC
τ τ θ τ θ
τ τ θ τ θ
= − +
= −
6
cos2
n
nτ τ θ=

Element under biaxial normal stress
B
nτ
yσ
θ nσ
n
σ
xσθxσ
A C
xσ
yσ
sin cosBC AB ACτ σ θ σ θ= −cos sinBC AB ACσ σ θ σ θ= +
yσ
. . .sin . .cos
. .sin . .cos
n x y
n x y
BC AB AC
AB AC
BC BC
τ σ θ σ θ
τ σ θ σ θ
=
= −
. .cos . .sin
. .cos . .sin
n x y
n x y
BC AB AC
AB AC
BC BC
σ σ θ σ θ
σ σ θ σ θ
= +
= +
( )
cos sin sin cos
sin2
n x y
BC BC
τ σ θ θ σ θ θ
θ
= −2 2
cos sin
y
n x y
BC BC
σ σ θ σ θ= +
7
( ) 2
n x yτ σ σ= −

Element under a general two-dimensional stress
xyτ
xyτ
B
nτ
τ
σ
θ θ nσ
xσ
xyτ
xσ
xyτ
xyτ
A C
x
xyτ
. .cos . .sin . .cos . .sinn x y xy xyBC AB AC AC ABσ σ θ σ θ τ θ τ θ= + − −
yσ
xy
yσ
. .cos . .sin . .cos . .sinn x y xy xy
AB AC AC AB
BC BC BC BC
σ σ θ σ θ τ θ τ θ= + − −
2 2
cos sin sin2n x y xyσ σ θ σ θ τ θ= + −
+
8
cos2 sin2
2 2
x y x y
n xy
σ σ σ σ
σ θ τ θ
+ −
= + −

i iBC AB AC AB ACθ θ θ θ. . .sin . .cos . .cos . .sin
sin cos cos sin
n x y xy xyBC AB AC AB AC
AB AC AB AC
τ σ θ σ θ τ θ τ θ
= − + −
= − + −
2 2
. .sin . .cos . .cos . .sin
cos sin sin cos cos sin
n x y xy xy
n x y xy xy
BC BC BC BC
τ σ θ θ σ θ θ τ θ τ θ
= − + −
= − + −n x y xy xy
sin2 cos2
2
x y
n xy
σ σ
τ θ τ θ
−
= +
2
n xy
Note: In the above derivations,
Sign convention for θ: Anticlockwise angle is +ve.g g
Sign convention for shear stress: With respect to a point inside
the element, clockwise shear stress is +ve.
9
the element, clockwise shear stress is +ve.

σ σ σ σ+ −
cos2 sin2
2 2
x y x y
n xy
σ σ σ σ
σ θ τ θ
+
= + −
σ σ−
Th b ti i th l d t ti l ( h ) t
sin2 cos2
2
x y
n xy
σ σ
τ θ τ θ= +
The above equations give the normal and tangential (shear) stresses on
any plane inclined at θ with the vertical.
To find maximum/minimum value of normal stress
σ∂
( )0nσ
θ
∂
=
∂
( )sin2 2 cos2 0x y xyσ σ θ τ θ⇒ − − − =
2
tan2 xyτ
θ
−
⇒
( )
tan2 y
x y
θ
σ σ
⇒ =
−
10

2
σ σ−⎛ ⎞ sin2 xyτ
θ
−
=
xyτ−
( )
2
2
x y
xy
σ σ
τ
⎛ ⎞
+⎜ ⎟
⎝ ⎠
2θ
2
2
sin2
2
y
x y
xy
θ
σ σ
τ
=
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
( )
2
x yσ σ−
2
cos2 x yσ σ
θ
−
=
2
2
2
2
x y
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
cos2 sin2
2 2
x y x y
n xy
σ σ σ σ
σ θ τ θ
+ −
= + −
2
2
max
2 2
x y x y
xy
σ σ σ σ
σ τ
+ −⎛ ⎞
⇒ = + +⎜ ⎟
⎝ ⎠⎝ ⎠
2
2
i
x y x yσ σ σ σ
σ τ
+ −⎛ ⎞
= − +⎜ ⎟
11
min
2 2
xyσ τ+⎜ ⎟
⎝ ⎠

222
2
max,min 1,3
2 2
x y x y
xy
σ σ σ σ
σ σ τ
+ −⎛ ⎞
= = ± +⎜ ⎟
⎝ ⎠
( )
2
tan2 xy
x y
τ
θ
σ σ
−
=
−
2
2
max,min 1,3
2 2
x y x y
xy
σ σ σ σ
σ σ τ
+ −⎛ ⎞
= = ± +⎜ ⎟
⎝ ⎠
( )
2
tan2 xy
x y
τ
θ
σ σ
−
=
−
Principal stresses Principal planes
σ σ−
Maximum shear stress
We know, sin2 cos2
2
x y
n xy
σ σ
τ θ τ θ= +
F i h t 0nτ∂
For maximum shear stress, 0n
θ
=
∂
( )cos2 2 sin2 0x y xyσ σ θ τ θ⇒ − − =( )x y xy
( )tan2
x yσ σ
θ
−
⇒ =
12
tan2
2 xy
θ
τ
⇒

2
( )
2
For max normal stress, tan2 xy
n
x y
τ
θ
σ σ
−
=
−
( )For max shear stress, tan2
2
x y
s
σ σ
θ
τ
−
=
1
tan2θ
−
=
2 xyτ tan2 nθ
( )0
cot 2 tan 2 90n nθ θ= − = ±( )n n
0
2 2 90s nθ θ= ±s n
0
45s nθ θ= ±
Hence, planes of maximum shear stress are at 450 to the principal
planes
13
p

To get maximum shear stress
( )
2
2
2
x y
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
2
2
cos2 xy
s
x y
τ
θ
σ σ
τ
=
−⎛ ⎞
+⎜ ⎟
g
τ
( )
2
x yσ σ− 2
y⎜ ⎟
⎝ ⎠
2 sθ
2
y
xyτ+⎜ ⎟
⎝ ⎠
sin2 x yσ σ
θ
−
=xyτ 2
2
sin2
2
2
s
x y
xy
θ
σ σ
τ
=
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
,max 2 2
2 2
2
2
x y x y xy
n xy
x y x y
σ σ σ σ τ
τ τ
σ σ σ σ
τ τ
− −
= +
− −⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
2
σ σ−⎛ ⎞
2
2 2
xy xyτ τ+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
σ σ⎛ ⎞2
,max
2
x y
n xy
σ σ
τ τ
⎛ ⎞
= +⎜ ⎟
⎝ ⎠
2
,max,min
2
x y
n xy
σ σ
τ τ
−⎛ ⎞
= ± +⎜ ⎟
⎝ ⎠
14

2
⎛ ⎞ 2
1,3
2
We kno
2
w, x y x y
xy
σ σ σ σ
σ τ
+ −⎛ ⎞
= ± +⎜ ⎟
⎝ ⎠
2
2
1 3 ,max2 2
2
x y
xy n
σ σ
σ σ τ τ
−⎛ ⎞
− = + =⎜ ⎟
⎝ ⎠
1 3
,max
2
n
σ σ
τ
−
=
2
To get normal stress on planes of maximum shear stress
2 2
2 2
2 2
x y x y xy x y
n xy
x y x y
σ σ σ σ τ σ σ
σ τ
σ σ σ σ
+ − −
= + −
− −⎛ ⎞ ⎛ ⎞2 2
2
2 2
x y x y
xy xy
σ σ σ σ
τ τ
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
x yσ σ+
th l f h t
15
2
x y
nσ = , on the planes of max shear stressaverageσ=

Problem: Find the principal stresses (including principal planes) and
maximum shear stress (including its plane)
2
60 N
2
80 N mm
Principal stresses
2
120 N mm
2
60 N mm
2
120 N mm
2
2
max,min 1,3
2 2
x y x y
xy
σ σ σ σ
σ σ τ
+ −⎛ ⎞
= = ± +⎜ ⎟
⎝ ⎠
Principal stresses
2
80 N mm
2
60 N mm
⎝ ⎠
2
2
max min 1 3
120 80 120 80
60σ σ
− − − +⎛ ⎞
= = ± +⎜ ⎟
⎝ ⎠
max,min 1,3 60
2 2
σ σ ± +⎜ ⎟
⎝ ⎠
100 63 24σ σ= = − ±max,min 1,3 100 63.24σ σ= = ±
2
max 1 163.24 N mmσ σ∴ = = −max 1
2
min 3and 36.75 N mmσ σ= = −
16

2
tan2 xyτ
θ
−
=Principal planes
2 60− ×
= 3=
( )
tan2
x y
θ
σ σ
=
−
Principal planes
( )120 80
=
− +
3=
( )1
2 tan 3 71 57θ −
= = 35 78θ∴ =( )2 tan 3 71.57θ = = 35.78θ∴ =
1 35.78θ∴ =1
3 35.78 90 125.78θ = + =
xyτ
xyτ
35 78
xσ
xyτ 3σ
1σ
35.78
yσ
xyτ
17

1 3
,max
2
n
σ σ
τ
−
=
163.25 36.75
2
− +
= 2
63.25 N mm= −
2 2
Planes of maximum shear stress
( )tan2
x yσ σ
θ
−
=
120 80− +
=tan2
2
s
xy
θ
τ
=
2 60
=
×
xyτ
xyτ 1 120 80
2 tθ − − +⎛ ⎞
⎜ ⎟
18.43= −σ
xy
9.22
1
2 tan
2 60
θ
⎛ ⎞
= ⎜ ⎟
×⎝ ⎠
18.43xσ
xyτ
xyτ ,maxnτ
9.22θ = −
18
yσ
xy

cos2 sin2x y x yσ σ σ σ
σ θ τ θ
+ −
= +
Mohr’s circle
cos2 sin2
2 2
n xyσ θ τ θ= + −
cos2 sin2x y x yσ σ σ σ
σ θ τ θ
+ −
= i
sin2 cos2
2
x y
n xy
σ σ
τ θ τ θ
−
= +
cos2 sin2
2 2
n xyσ θ τ θ− = − i
ii2
n xy
Squaring and adding the above equations,
( ) ( )
2 2
22
2 2
x y x y
n n xy
σ σ σ σ
σ τ τ
+ −⎛ ⎞ ⎛ ⎞
− + = +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠2 2⎝ ⎠ ⎝ ⎠
( ) ( )
2 2 2
0n av n Rσ σ τ− + − =
This is equation of a circle with centre and radius ( )
2
2
2
x y
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
( ),0avσ
19
Let us draw this circle!

Mohr’s circle
xyτ
( ),y xyσ τ
xσ
xyτ
xy
xyτ
2
2
2
y x
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
yσ
xyτ
σy
τxyτ
2
y xσ σ+
y
σx σσ3 σ1
α
2
y xσ σ−
τxy
( ),x xyσ τ− 1 2
tan 2xy
y x
τ
α θ
σ σ
− −
= =
−
20: Principal stressesσ1, σ3
measured clockwiseα

Mohr’s circle
xyτ
xσ
xyτ
y
xyτ
( ),x xyσ τ
yσ
xyτ
τ
2
2
2
y x
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠
σyσx
τxyτ
σ3
αx
σ
3
σ1
2
y xσ σ+
2
y xσ σ−
τxy
2
1 2
tan 2xy
y x
τ
α θ
σ σ
−
= =
−( ),y xyσ τ−
21
y
measured anticlockwiseα

•Mohr’s circle is a graphical representation of the state of stress in an•Mohr s circle is a graphical representation of the state of stress in an
element.
E i h i l h l d h•Every point on the circle represents the normal and shear stress on a
plane.
•While x-coordinate of a point on the circle represents the normal
stress on a plane, y-coordinate represents the shear stress on that plane.
•Procedure for construction of Mohr’s circle
22

Maximum shear stress from Mohr’s circle
xyτ
( ),y xyσ τ
xσ
xyτ
xy
xyτ
Max shear
2
2
2
y x
xy
σ σ
τ
−⎛ ⎞
+⎜ ⎟
⎝ ⎠ yσ
xyτ
Max shear
stress
1 3
max
2
n
σ σ
τ
−
=
σy
τxyτ
,max
2
n
y
σx σσ3 σ1
2θ
τxy
( ),x xyσ τ−
23

Principal planes from Mohr’s circle
xyτ( ),y xyσ τ
xy
xyτ
θ
σ
τxyτ
σ
σy
σx
τxy
σσ3 σ1
( )
2θ
3σ1σ
xσ
xyτ( ),x xyσ τ−
σ
yσ
xyτ
3σ
1σ
3σ
24
3
1σ

2
60 N
2
80 N mmProblem: Find principal stresses, principal
2
120 N mm
2
60 N mm
2
120 N mm
planes and max shear stress analytically. Draw
Mohr’s circle and verify graphically.
2
80 N mm
2
60 N mm
( )( )120,60−
τ60 18.44
2
71 56
θ
80120
60
σ σ3σ1
3σ
1σ
35.78
71.56=
τ
9.22
60 1
,maxnτ
25
( )80, 60− −

Problems: Find principal stresses, principal planes and max shear stress
analytically. Draw Mohr’s circle and verify graphically.analytically. Draw Mohr s circle and verify graphically.
1510
80
5015
15
15
50
50
10
10
10
50
50
30
15
30
15
80
10
10
15
50
5
20
20
20
50
5
50
20
26

Transformation of strains
σ σ σ σ+ − sin2 cos2x yσ σ
τ θ τ θ
−
+cos2 sin2
2 2
x y x y
n xy
σ σ σ σ
σ θ τ θ
+
= + −
sin2 cos2
2
y
n xyτ θ τ θ= +
The above equations, which give the normal and tangential (shear) stresses on
any plane inclined at θ with the vertical, are called stress transformation
equations.equations.
Similarly strain transformation equations can be derived as follows:
2 2
cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + −
cos2 sin2
2 2 2
x y x y xy
n
ε ε ε ε γ
ε θ θ
+ −
= + −OR,
27
sin2 cos2
2 2 2
x y xyn
ε ε γγ
θ θ
−
= +and,

Principal strains:
Planes on which
i i l t i tPrincipal strains: principal strains act:
2 2
max min 1 3
x y x y xyε ε ε ε γ
ε ε
+ −⎛ ⎞ ⎛ ⎞
= = ± +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ( )
tan2 xyγ
β
−
=
max,min 1,3
2 2 2⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ( )x yε ε−
Strain Rosettes
M t f l t i i i l• Measurement of normal strains is simple.
• Strain gages are placed as a cluster, along several gage lines through
a pointa point
• This arrangement of strain gages is called a strain rosette
• If three measurements are taken at a rosette (in three directions), the
information is sufficient to get the complete state of plane strain at a
point
p

0
3 90θ =
1 2 3, ,θ θ θε ε ε are measured from strain gages
0
2 45θ =
1 2 3, ,θ θ θε ε ε are measured from strain gages
0 45 90ε ε ε45 degree rosette:
1 0θ =
0 45 90, ,ε ε ε45 degree rosette:
0 60 120, ,ε ε ε60 degree rosette:
Rectangular strain rosette
From strain transformation equations,
g
(45 degree rosette)2 2
Hence, for a 45 degree rosette, 0 0 0x xε ε ε= + − =
45 0.5 0.5 0.5x y xyε ε ε γ= + −y y
90 0 0y yε ε ε= + − =
From the above, we can get , ,x y xyε ε γ

Problem: Using a 60 degree rosette, the following strains are obtained at
i t D t i t i t d i i l t ia point. Determine strain components and principal strains.
0 60 12040 , 980 , 330ε μ ε μ ε μ= = =
, ,x y xyε ε γ
We have, 2 2
2 2
0 cos 0 sin 0 sin cos0x y xyε ε ε γ θ∴ = + −
i.e., 40 0 0 40x xε ε= + − ⇒ =
0 x y xyγ
60 980 0.25 0.75 0.433x y xyε ε ε γ⇒ = + −
120 330 0.25 0.75 0.433x y xyε ε ε γ⇒ = + +
40 , 860 , 750x y xyε μ ε μ γ μ= = = −
y y
From the above, we can get
Principal strains and their planes can be obtained from:
2 2
ε ε ε ε γ+ −⎛ ⎞ ⎛ ⎞ tan2 xyγ
β
−
=
max,min 1,3
2 2 2
x y x y xyε ε ε ε γ
ε ε
+ ⎛ ⎞ ⎛ ⎞
= = ± +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )
tan2
x y
β
ε ε
=
−

Theory of columnsy
Compression member: A structural member loaded in compression
Column: A vertical compression member
Strut: An inclined compression member – as in roof trusses
Stanchion: A compression member made of rolled steel section
Classification of columns based on mode of failure
Short columns: Failure by crushing under axial
compression
L ( l d ) l F il b l l b di
Intermediate (medium length) columns: Failure by
Long (slender) columns: Failure by lateral bending
(buckling)
31
Intermediate (medium length) columns: Failure by
combination of buckling and crushing

Equilibrium:qu b u :
Stable, neutral, unstable
32

L d th t b i d b th b b f f il
Critical load
Load that can be carried by the member before failure
Least load that causes elastic instability
depends on
dimensions of the member end conditions
modulus of elasticity
Slenderness ratio: Ratio of length to the least radius of gyration.
Buckling tendency varies with slenderness ratio
modulus of elasticity
Buckling tendency varies with slenderness ratio.
33

Euler’s theory – Leonhard Euler (1757)y ( )
2
d y
Both ends hinged PEI
M
R
=
2
d y
EI P
2
d y
2
d y
EI M
dx
⇒ = A2
y
EI Py
dx
⇒ = −
2
0
d y
EI Py
dx
+ =
2
2
0
d y P
y+ =2
y
dx EI
P P⎛ ⎞ ⎛ ⎞
Solution for the above differential equation is:
y
XX
l1 2cos sin
P P
y C x C x
EI EI
⎛ ⎞ ⎛ ⎞
= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
When 0, 0x y= = 1 0C⇒ =
B
x
When 0, 0x y 1
When , 0x l y= = 20 sin
P
C l
EI
⎛ ⎞
⇒ = ⎜ ⎟
⎝ ⎠
P
B⎝ ⎠
sin 0 0, ,2 ,3 ,4 ...
P P
l l
EI EI
π π π π
⎛ ⎞
= ⇒ =⎜ ⎟
⎝ ⎠
34
P
,where 0,1,2,3,4...
P
l n n
EI
π= =

2 2
2
n EI
P
l
π
=
2 sin
P
y C x
EI
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
2 sin
n x
C
L
π⎛ ⎞
= ⎜ ⎟
⎝ ⎠
The least practical
2
EI
P
π
The least practical
value for P is:
Critical load2cr
EI
P
l
π
=
sin
x
y C
π⎛ ⎞
= ⎜ ⎟The corresponding mode shape is:
35
2 siny C
L
= ⎜ ⎟
⎝ ⎠
The corresponding mode shape is:

Assumptions in Euler’s theoryAssumptions in Euler s theory
• Material is homogeneous and isotropic
• Axis of column is perfectly straight when unloaded
• Line of thrust coincides exactly with the unstrained axis of the column
• Column fails by buckling alone
• Flexural rigidity EI is uniform
S lf h f l l d• Self weight of column is neglected
• Stresses are within elastic limit
36

Euler’s theory
P
y
2
d y
EI M
One end fixed and the other end free
( )
2
d y
EI P yδ⇒ = −
P
A
δ
2
y
EI M
dx
= ( )2
EI P y
dx
δ⇒ =
2
d y
EI P Pδ
2
d y P P
y
δ
+ = y2
y
EI Py P
dx
δ+ = 2
y
dx EI EI
+ = y
lSolution for the above differential equation is:
1 2cos sin
P P
y C x C x
EI EI
δ
⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
xEI EI⎝ ⎠ ⎝ ⎠
When 0, 0x y= = 1C δ⇒ = −
x
B
37

dy P P P P⎛ ⎞ ⎛ ⎞
dy P
1 2sin cos
dy P P P P
C x C x
dx EI EI EI EI
⎛ ⎞ ⎛ ⎞
= − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
When 0, 0
dy
x
dx
= = 2 20 0 0
P
C C
EI
⇒ = = ⇒ =
When x l y δ= = cos
P
lδ δ δ
⎛ ⎞
⇒ +⎜ ⎟
3 5
cos 0
P P
l l
π π π⎛ ⎞
⇒⎜ ⎟
When ,x l y δ= = cos l
EI
δ δ δ⇒ = − +⎜ ⎟
⎝ ⎠
cos 0 , , ...
2 2 2
l l
EI EI
= ⇒ =⎜ ⎟
⎝ ⎠
P
l
π
Th l t ti l l i
2
P
l
EI
π
=
2
EI
P
π 2
EI
P
π
The least practical value is:
2el l= Effective length
2
4
EI
P
l
π
= 2
e
EI
P
l
π
=
P⎛ ⎞ ⎛ ⎞
e Effective length
38
cos 1 cos
2 e
P x
y x
EI l
π
δ δ δ
⎛ ⎞ ⎛ ⎞
= − + = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠

Euler’s theory Py
( )
2
d y
EI M P H l
One end fixed and the other hinged
H
A
( )2
y
EI M Py H l x
dx
= = − + −
( )
2
d y
EI P H l y( )2
y
EI Py H l x
dx
+ = −
( )cos sin
P P H
C C l
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
y
l
( )1 2cos siny C x C x l x
EI EI P
= + + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
x
⎛ ⎞ ⎛ ⎞
x
1 2sin cos
dy P P P P H
C x C x
dx EI EI EI EI P
⎛ ⎞ ⎛ ⎞
= − + −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
When 0 0x y= = 1 10
H H
C l C l⇒ = + ⇒ = −
B
M
39
When 0, 0x y 1 1
P P
P
M

When 0, 0
dy
x
dx
= = 2 20
P H H EI
C C
EI P P P
⇒ = − ⇒ =
When 0x l y= = 0 cos sin
H P H EI P
l l l
⎛ ⎞ ⎛ ⎞
= − +⎜ ⎟ ⎜ ⎟
tan
P P
l l
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
When , 0x l y= = 0 cos sinl l l
P EI P P EI
+⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
tan l l
EI EI
=⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
P
2
20.25 2EI EI
P
π
4.5 radians
P
l
EI
= 2 2
P
l l
= ≈
2
EI
P
π l
l2
e
EI
P
l
π
=
2
el =
40

Euler’s theory
P
y
2
d y
EI M M P
Both ends fixed A M0
02
y
EI M M Py
dx
= = −
2
d y
EI P M
y2
0d y P M
02
y
EI Py M
dx
+ =
0P P M⎛ ⎞ ⎛ ⎞
y
l
0
2
y
y
dx EI EI
+ =
0
1 2cos sin
P P M
y C x C x
EI EI P
⎛ ⎞ ⎛ ⎞
= + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
x
1 2sin cos
dy P P P P
C x C x
dx EI EI EI EI
⎛ ⎞ ⎛ ⎞
= − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
x
When 0 0x y= =
0 0
1 10
M M
C C⇒ = + ⇒ = −
B
M
41
When 0, 0x y 1 1
P P
P
M0

When 0, 0
dy
x
dx
= = 2 20 0
P
C C
EI
⇒ = ⇒ =
When 0x l y= =
0 0
0 cos
M P M
l
P EI P
⎛ ⎞
= − +⎜ ⎟ 0
1 cos 0
M P
l
⎡ ⎤⎛ ⎞
⇒ − =⎢ ⎥⎜ ⎟
cos 1
P
l
⎛ ⎞
⎜ ⎟
When , 0x l y= = P EI P
⎜ ⎟
⎝ ⎠ P EI
⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
0 2 4 6
P
l π π π⇒ =cos 1l
EI
=⎜ ⎟
⎝ ⎠
2
4 EI
P
π
0,2 ,4 ,6 ...l
EI
π π π⇒ =
2
P
l π
2
P
l
=
2
EI l
2l
EI
π=
2
2
e
EI
P
l
π
=
2
e
l
l =
42

Effective lengthEnd conditions
Both ends hinged
g
l
One end fixed and the
other end free 2l
B th d fi d
One end fixed and the
other hinged 2l
Both ends fixed 2l
43

Limitations of Euler’s theoryLimitations of Euler s theory
• Applicable to ideal cases only. There may be imperfections in the
l th l d t tl th h th t id f th
2
column, the load may not pass exactly through the centroid of the
column section
• Direct stress is not taken into account
2
2E
e
EI
P
l
π
=• Strength of the material is not taken into account
2
2
E
E
e
P EI
A Al
π
σ = =
2 2 2
EAr Eπ π
22E
e e
EAr E
Al l
r
π π
σ⇒ = =
⎛ ⎞
⎜ ⎟
⎝ ⎠
44

σEσE
l /
Validity limits of Euler’s formula
le /r
45
Critical stress for mild steel with E=2x105 MPa

2
l Eπe
E
l E
OD
r
π
σ
= =
5 2
2 10 NEL t
2
250 N mmPLσ =Stress at limit of proportionality
5 2
2 10 N mmE = ×Let
( )2 5
2 10
89
250
el
r
π ×
∴ = =
250r
i.e., Euler’s theory is applicable for 89el
> for mild steel
r
46

Rankine’s theoryy
2
2Euler
EI
P
π
=
c cP Aσ=For short compression members,
For long columns, 2Euler
el
g ,
Rankine proposed a general empirical formula:
1 1 1
Rankine c EulerP P P
= + For a short compression member, PE is very large.
P P∴ ≈Rankine cP P∴ ≈
For long columns, 1/PEuler is very large.
2
2
1 el
P EIπ
=
Rankine EulerP P≈
EulerP EIπ
2
1 1 1
R ki
EIP A πσ
= + 2
I Ar=
47
2
Rankine c
e
EIP A
l
πσ

1 1 1
2
2 r
Eπ σ
⎛ ⎞
+⎜ ⎟
2
2
1 1 1
Rankine cP A r
EA
l
σ
π
= +
⎛ ⎞
⎜ ⎟
⎝ ⎠
2
2
c
e
Rankine
c
E
lA
P r
E
l
π σ
σ π
+⎜ ⎟
⎝ ⎠=
⎛ ⎞
⎜ ⎟
⎝ ⎠
2 2
c c
Rankine
A A
P
l l
σ σ
σ
= =
⎛ ⎞ ⎛ ⎞
c
a
σ
=
el⎝ ⎠ c
el⎜ ⎟
⎝ ⎠
2
1 1c e el l
a
E r r
σ
π
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2
a
Eπ
2 2
Crushing loadc
Rankine
A
P
l l
σ
= =
⎛ ⎞ ⎛ ⎞
2
1 el
a
r
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠1 1e el l
a a
r r
⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
r⎝ ⎠
Factor that accounts
for bucklingg
48

Find the length of the column for which Rankine’s and Euler’s formulae
give the same buckling load:
Rankine EulerP P=
2
A EIσ π 1/2
2 2
⎛ ⎞2 2
1
c
ee
A EI
ll
a
r
σ π
=
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
1/2
2 2
2e
c
Er
l
Ea
π
σ π
⎛ ⎞
= ⎜ ⎟
−⎝ ⎠
49

Problem: Compare the buckling (crippling) loads given by Rankine’sProblem: Compare the buckling (crippling) loads given by Rankine s
and Euler’s formulae for a tubular strut hinged at both ends, 6 m long
having outer diameter 15 cm and thickness 2 cm. Given,
4 2 2 1
8 10 N mm , 567 N mm ,
1600
cE aσ= × = =
For what length of the column does the Euler’s formula cease to apply?For what length of the column does the Euler s formula cease to apply?
2
2Euler
EI
P
l
π
=
( )4 4 4
150 110 17663604.69 mm
64
I
π
= − =
el
6 m 6000 mmel l= = =( )64
e
2
2
387406.2 NEuler
EI
P
l
π
= = 387.4 kN=
2Euler
el
( )2 2 2π
50
( )2 2 2
150 110 8168.141 mm
4
A
π
= − =

c A
P
σ
2
1
c
Rankine
e
P
l
a
r
σ
=
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
2 4
17663604.69 mmI Ar= = 17663604.69
46.503 mm
8168.141
I
r
A
= = =
567 8168.141
406097 78 NP
×
406 098 kN=2
406097.78 N
1 6000
1
1600 46.503
RankineP = =
⎛ ⎞
+ ⎜ ⎟
⎝ ⎠
406.098 kN=
2
2E
Eπ
σ = ( )2 4
8 10
37 317el π ×
∴ = =
2
EP EIπ
σ = =
To find the length of the column above which Euler’s formula is applicable
2E
el
r
σ
⎛ ⎞
⎜ ⎟
⎝ ⎠
37.317
567r
∴ = =2E
eA Al
σ = =
46 503 37 317l 1735 34 mm 1 735 m
51
46.503 37.317el∴ = × 1735.34 mm 1.735 m= =

Long column under eccentric loading P
σ =For short columns
P
A
σ =
M P e=
For short columns,
e
.M P e
Z Z
σ = =
.M P e=
.P P e
y
A I
σ = ±
Z Z A I
. .
1
P P e y P ey⎛ ⎞
± +⎜ ⎟2 2
1
y y
A Ar A r
σ ⎛ ⎞
= ± = +⎜ ⎟
⎝ ⎠
Aσ
2
1
A
P
ey
r
σ
=
⎛ ⎞
+⎜ ⎟
⎝ ⎠
2
A
P
al ey
σ
=
⎛ ⎞⎛ ⎞
For long columns,
52
2 2
1 1eal ey
r r
⎛ ⎞⎛ ⎞
+ +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠

Secant formula P
A
e
2
2
0
d y
EI Py⇒ + =
2
2
d y
EI Py= −
Both ends hinged
2
y
dx
2
2
0
d y P
y
d EI
+ =
2
y
dx
y
l
2
dx EI
P P⎛ ⎞ ⎛ ⎞
y
x
1 2cos sin
P P
y C x C x
EI EI
⎛ ⎞ ⎛ ⎞
= +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
When 0,x y e= = 1C e⇒ =
B
When 0,x y e 1C e⇒
2sin cos
dy P P P P
e x C x
dx EI EI EI EI
⎛ ⎞ ⎛ ⎞
= − +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠dx EI EI EI EI⎝ ⎠ ⎝ ⎠
When 0
l dy
x = =
sin
2
l P
EI
C e
⎛ ⎞
⎜ ⎟
⎝ ⎠=
53
When , 0
2
x
dx
= = 2
cos
2
C e
l P
EI
=
⎛ ⎞
⎜ ⎟
⎝ ⎠

l P⎡ ⎤⎛ ⎞
sin
2
cos sin
cos
l P
EIP P
y e x x
EI EIl P
⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟
⎛ ⎞ ⎛ ⎞⎝ ⎠⎢ ⎥= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠⎢ ⎥⎜ ⎟
2
sin
2
l P
EI
⎡ ⎤⎛ ⎞
⎢ ⎥⎜ ⎟
⎛ ⎞
cos
2 EI
⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
l P⎛ ⎞
max
2
When , cos
2 2
cos
2
EIl l P
x y y e
EI l P
EI
⎢ ⎥⎜ ⎟
⎛ ⎞ ⎝ ⎠⎢ ⎥= = = +⎜ ⎟⎢ ⎥⎛ ⎞⎝ ⎠⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
max .sec
2
l P
y e
EI
⎛ ⎞
⇒ = ⎜ ⎟
⎝ ⎠
max max . .sec
2
l P
M Py P e
EI
⎛ ⎞
= = ⎜ ⎟
⎝ ⎠2 EI⎝ ⎠
P My P Pey l P P ey l P⎡ ⎤⎛ ⎞ ⎛ ⎞
max 2
sec 1 sec
2 2
c c cP My P Pey l P P ey l P
A I A I EI A r EI
σ
⎡ ⎤⎛ ⎞ ⎛ ⎞
= + = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
54

Secant formula P
δ
One end fixed and the other end free
A
δ e
( )
2
2
d y
EI P e yδ= + −
( )
2
2
d y P P
y e
d EI EI
δ+ = +
y
( )2
y
dx
( )2
dx EI EI
P P⎛ ⎞ ⎛ ⎞
l
( )1 2cos sin
P P
y C x C x e
EI EI
δ
⎛ ⎞ ⎛ ⎞
= + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
When 0, 0x y= = ( )1C eδ⇒ = − +
x
When 0, 0x y ( )1
( ) 2sin cos
dy P P P P
e x C x
dx EI EI EI EI
δ
⎛ ⎞ ⎛ ⎞
= − + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
B
dx EI EI EI EI⎝ ⎠ ⎝ ⎠
When 0 0
dy
x = = 0C⇒ =
55
When 0, 0x
dx
= = 2 0C⇒ =

When x l y δ= = ( ) ( )cos
P
e l eδ δ δ
⎛ ⎞
⇒ = + + +⎜ ⎟When ,x l y δ= = ( ) ( )cose l e
EI
δ δ δ⇒ = − + + +⎜ ⎟
⎝ ⎠
( ) 1
P
lδ δ
⎡ ⎤⎛ ⎞
⇒ + ⎢ ⎥⎜ ⎟( ) 1 cose l
EI
δ δ⇒ = + −⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦
( )cos
P
e l eδ
⎛ ⎞
⇒ + ⎜ ⎟( )cose l e
EI
δ⇒ + =⎜ ⎟
⎝ ⎠
( ) sec
P
e e lδ
⎛ ⎞
⇒ + = ⎜ ⎟
( )
P
M P P lδ
⎛ ⎞
+ ⎜ ⎟
( ) .sece e l
EI
δ⇒ + = ⎜ ⎟
⎝ ⎠
( )max . .secM P e P e l
EI
δ= + = ⎜ ⎟
⎝ ⎠
⎡ ⎤⎛ ⎞ ⎛ ⎞
max 2
sec 1 secc c cP My P Pey P P ey P
l l
A I A I EI A r EI
σ
⎡ ⎤⎛ ⎞ ⎛ ⎞
= + = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
56

max . .sec
2
el P
M P e
EI
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
In general,
2 EI⎝ ⎠
max 2
1 sec
2
c eP ey l P
A r EI
σ
⎡ ⎤⎛ ⎞
= +⎢ ⎥⎜ ⎟
⎢ ⎥⎝ ⎠⎣ ⎦2A r EI⎢ ⎥⎝ ⎠⎣ ⎦
For short compression members (no buckling), max .M P e=
For long columns (with buckling) el P
M P
⎛ ⎞
⎜ ⎟For long columns (with buckling),
max . .sec
2
e
M P e
EI
= ⎜ ⎟
⎝ ⎠
57

Problem: A hollow mild steel column with internal diameter 80 mm
and external diameter 100 mm is 2.4 m long, hinged at both ends,
carries a load of 60 kN at an eccentricity of 16mm from the geometrical
axis. Calculate the maximum and minimum stresses in the column.
Also find the maximum eccentricity so that no tension is induced in the
section. 5 2
2 10 N mmE = ×
( )4 4 4
100 80 2898119 mmI
π
= − = ( )2 2 2
100 80 2827 4 mmA
π
= − =( )100 80 2898119 mm
64
I
2400 mml l
( )100 80 2827.4 mm
4
A
2898119
32 015 mm
I
r 2400 mmel l= =32.015 mm
2827.4
r
A
= = =
58

sec el P
M P e
⎛ ⎞
= ⎜ ⎟
3
3 2400 60 10
60 10 16 secM
⎛ ⎞×
× × × ⎜ ⎟ 0 96 kN
max . .sec
2
e
M P e
EI
= ⎜ ⎟
⎝ ⎠
max 5 6
60 10 16 sec
2 2 10 2.898 10
M = × × × ⎜ ⎟
⎜ ⎟× × ×⎝ ⎠
0.96 kNm=
3 6
⎧max
max
min
cP M y
A I
σ = ±
3 6
max
min
37.78 MPa60 10 0.96 10 50
4.69 MPa2827.4 2898119
σ
⎧× × ×
= ± = ⎨
⎩
To find the maximum eccentricity so that no tension is induced
in the section
max
0cP M y
A I
− =
.
.sec 0
2
e
c
P P l
y
I
e P
A EI
⎛ ⎞
− =⎜ ⎟
⎝ ⎠
20.5 mme =
59

SummarySummary
Transformation of stresses and strains (two dimensional case only)Transformation of stresses and strains (two-dimensional case only) -
equations of transformation - principal stresses - mohr's circles of
stress and strain - strain rosettes - compound stresses - superposition
d it li it tiand its limitations –
Eccentrically loaded members - columns - theory of columns -
buckling theory - Euler's formula - effect of end conditions -
eccentric loads and secant formula
60

"A teacher is one who makes himself
progressively unnecessary "progressively unnecessary."
Thomas Carruthers
61

Mechanics of structures module4

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