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 EngineeringGovt. 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 ti and 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

LAA

θ

PA

σ = cos

n

nAAθ

=AAn

A

22cos cos cosn

P PA Aθ θσ σ θ= = =

cosP θθ

nA A

sin sin cos sin2P Pθ θ θ σ θτ

Pθ

sinP θAn

θ


32n

nA Aτ = = =

2 2σ σ τ+

( ) ( )2 22cos sin cos

R n nσ σ τ

σ θ σ θ θ

= +

= +

Resultant stress on inclined plane

2 2cos cos sincos

σ θ θ θσ θ

= +=

0 2 cos sin 0nddσ σ θ θθ= ⇒ =For maximum normal stress,

0sin2 0 0σ θ θ= ⇒ =

0When 0 Pθ σ σ= = =

0 2 0ndτ θFor maximum shear stress

,maxWhen 0 , n Aθ σ σ= = =

0 0

0 cos2 0

2 90 45

n

dσ θ

θθ θ

= ⇒ =

⇒ = ⇒ =

For maximum shear stress,


40

,maxWhen 45 ,2 2n

PA

σθ τ= = =

nσ σ

nτσ

Rσ 2 2n nσ τ+

σ,maxnτ

σ


5

Element under pure shearτ

B

θ σnτ

τ

θτ

θ nσττ θ

cos sinBC AC ABσ τ θ τ θ= − −

A Cττ

. .cos . .sinsin cos cos sinsin 2

n

n

BC AC ABσ τ θ τ θσ τ θ θ τ θ θσ τ θ

= − −= − −

. . .sin . .cosn BC AC ABAC AB

τ τ θ τ θ= − +sin 2nσ τ θ= −

2 2

. .sin . .cos

cos sin

n

n

AC ABBC BC

τ τ θ τ θ

τ τ θ τ θ

= − +

= −


6cos2

n

nτ τ θ=

Element under biaxial normal stress

Bnτ

yσ

θ nσn

σxσθ

xσ

A C

xσ

yσsin 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 ACAB ACBC BC

τ σ θ σ θ

τ σ θ σ θ

=

= −

. .cos . .sin

. .cos . .sin

n x y

n x y

BC AB ACAB ACBC BC

σ σ θ σ θ

σ σ θ σ θ

= +

= +

( )

cos sin sin cos

sin 2n x y

BC BCτ σ θ θ σ θ θ

θ

= −2 2cos sin

y

n x y

BC BCσ σ θ σ θ= +


7

( ) 2n x yτ σ σ= −

Element under a general two-dimensional stress

xyτxyτ

Bnττ

σ

θ θ nσ

xσ

xyτ

xσ

xyτ

xyτA C

x

xyτ

. .cos . .sin . .cos . .sinn x y xy xyBC AB AC AC ABσ σ θ σ θ τ θ τ θ= + − −

yσxy

yσ

. .cos . .sin . .cos . .sinn x y xy xyAB AC AC ABBC BC BC BC

σ σ θ σ θ τ θ τ θ= + − −

2 2cos sin sin 2n x y xyσ σ θ σ θ τ θ= + −

+


8cos2 sin 2

2 2x y x y

n xy

σ σ σ σσ 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

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 BCτ σ θ σ θ τ θ τ θ

τ σ θ θ σ θ θ τ θ τ θ

= − + −

= − + −n x y xy xy

sin 2 cos22

x yn xy

σ στ θ τ θ

−= +

2n 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 sin 2

2 2x y x y

n xy


+= + −

σ σ−

Th b ti i th l d t ti l ( h ) t

sin 2 cos22

x yn xy

σ στ 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∂ ( )0nσθ

∂=

∂( )sin 2 2 cos2 0x y xyσ σ θ τ θ⇒ − − − =

2tan 2 xyτ

θ−

⇒ ( )tan 2 y

x y

θσ σ

⇒ =−


10

2σ σ−⎛ ⎞ sin 2 xyτθ

−=

xyτ−

( )

2

2x y

xy

σ στ⎛ ⎞

+⎜ ⎟⎝ ⎠

2θ

22

sin 2

2

y

x yxy

θσ σ

τ

=−⎛ ⎞

+⎜ ⎟⎝ ⎠

( )2

x yσ σ−

2cos2 x yσ σ

θ−

=2

222

x yxy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠

cos2 sin 22 2

x y x yn xy


+ −= + −

22

max 2 2x y x y

xy

σ σ σ σσ τ

+ −⎛ ⎞⇒ = + +⎜ ⎟

⎝ ⎠⎝ ⎠

22

ix y x yσ σ σ σ

σ τ+ −⎛ ⎞

= − +⎜ ⎟


11

min 2 2 xyσ τ+⎜ ⎟⎝ ⎠

2222

max,min 1,3 2 2x y x y

xy

σ σ σ σσ σ τ

+ −⎛ ⎞= = ± +⎜ ⎟

⎝ ⎠ ( )2

tan 2 xy

x y

τθ

σ σ−

=−

22

max,min 1,3 2 2x y x y

xy

σ σ σ σσ σ τ

+ −⎛ ⎞= = ± +⎜ ⎟

⎝ ⎠ ( )2

tan 2 xy

x y

τθ

σ σ−

=−

Principal stresses Principal planes

σ σ−

Maximum shear stress

We know, sin 2 cos22

x yn xy

σ στ θ τ θ= +

F i h t 0nτ∂For maximum shear stress, 0n

θ=

∂

( )cos2 2 sin 2 0x y xyσ σ θ τ θ⇒ − − =( )x y xy

( )tan 2 x yσ σ

θ−

⇒ =


12

tan 22 xy

θτ

⇒

2

( )2

For max normal stress, tan 2 xyn

x y

τθ

σ σ−

=−

( )For max shear stress, tan 2

2x y

s

σ σθ

τ−

= 1tan 2θ−

=2 xyτ tan 2 nθ

( )0cot 2 tan 2 90n nθ θ= − = ±( )n n

02 2 90s nθ θ= ±s n

045s nθ θ= ±

Hence, planes of maximum shear stress are at 450 to the principal planes


13

p

To get maximum shear stress

( )2

2

2x y

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠

22

cos2 xys

x y

τθ

σ στ

=−⎛ ⎞

+⎜ ⎟

g

τ

( )2

x yσ σ− 2 y⎜ ⎟⎝ ⎠

2 sθ2

yxyτ+⎜ ⎟

⎝ ⎠

sin 2 x yσ σθ

−=xyτ

22

sin 2

22

s

x yxy

θσ σ

τ

=−⎛ ⎞

+⎜ ⎟⎝ ⎠

,max 2 22 2

22

x y x y xyn xy

x y x y

σ σ σ σ ττ τ

σ σ σ στ τ

− −= +

− −⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟

2σ σ−⎛ ⎞

22 2xy xyτ τ+ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

2σ σ⎛ ⎞2,max 2

x yn xy

σ στ τ⎛ ⎞

= +⎜ ⎟⎝ ⎠

2,max,min 2

x yn xy

σ στ τ

−⎛ ⎞= ± +⎜ ⎟

⎝ ⎠


14

2⎛ ⎞ 2

1,3 2

We kno2

w, x y x yxy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

22

1 3 ,max2 22

x yxy n

σ σσ σ τ τ

−⎛ ⎞− = + =⎜ ⎟

⎝ ⎠

1 3,max 2n

σ στ −=

2

To get normal stress on planes of maximum shear stress

2 22 2

2 2x y x y xy x y

n xy

x y x y

σ σ σ σ τ σ σσ τ

σ σ σ σ

+ − −= + −

− −⎛ ⎞ ⎛ ⎞2 222 2

x y x yxy xy

σ σ σ στ τ⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x yσ σ+ th l f h t


152x y

nσ = , on the planes of max shear stressaverageσ=

Problem: Find the principal stresses (including principal planes) and maximum shear stress (including its plane)

260 N

280 N mmPrincipal stresses

2120 N mm

260 N mm

2120 N mm2

2max,min 1,3 2 2

x y x yxy

σ σ σ σσ σ τ

+ −⎛ ⎞= = ± +⎜ ⎟

⎝ ⎠

Principal stresses

280 N mm

260 N mm

⎝ ⎠

22

max min 1 3120 80 120 80 60σ σ − − − +⎛ ⎞= = ± +⎜ ⎟

⎝ ⎠max,min 1,3 60

2 2σ σ ± +⎜ ⎟

⎝ ⎠

100 63 24σ σ= = − ±max,min 1,3 100 63.24σ σ= = ±

2max 1 163.24 N mmσ σ∴ = = −max 1

2min 3and 36.75 N mmσ σ= = −


16

2tan 2 xyτ

θ−

=Principal planes 2 60− ×= 3=( )tan 2

x y

θσ σ

=−

Principal planes( )120 80

=− +

3=

( )12 tan 3 71 57θ −= = 35 78θ∴ =( )2 tan 3 71.57θ = = 35.78θ∴ =

1 35.78θ∴ =1

3 35.78 90 125.78θ = + =

xyτxyτ

35 78

xσxyτ 3σ1σ

35.78

yσxyτ


17


1 3,max 2n

σ στ −=


163.25 36.752

− += 263.25 N mm= −

2 2

Planes of maximum shear stress

( )tan 2 x yσ σ

θ−

=120 80− +

=tan 22s

xy

θτ

=2 60

=×

xyτxyτ 1 120 802 tθ − − +⎛ ⎞

⎜ ⎟

18.43= −σ

xy

9.22

12 tan2 60

θ ⎛ ⎞= ⎜ ⎟×⎝ ⎠

18.43xσ

xyτ

xyτ ,maxnτ

9.22θ = −


18yσ

xy

cos2 sin 2x y x yσ σ σ σσ θ τ θ

+ −= +

Mohr’s circle

cos2 sin 22 2n xyσ θ τ θ= + −

cos2 sin 2x y x yσ σ σ σσ θ τ θ

+ −= i

sin 2 cos22

x yn xy

σ στ θ τ θ

−= +

cos2 sin 22 2n xyσ θ τ θ− = − i

ii2n xy

Squaring and adding the above equations,

( ) ( )2 2

22

2 2x y x y

n n xy

σ σ σ σσ τ τ

+ −⎛ ⎞ ⎛ ⎞− + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠2 2⎝ ⎠ ⎝ ⎠

( ) ( )2 2 20n av n Rσ σ τ− + − =

This is equation of a circle with centre and radius ( )2

2

2x y

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠( ),0avσ


19Let us draw this circle!

Mohr’s circlexyτ

( ),y xyσ τxσxyτ

xyxyτ

22

2y x

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠yσ

xyτ

σy

τxyτ

2y xσ σ+

yσx σσ3 σ1α

2y xσ σ−

τxy

( ),x xyσ τ− 1 2tan 2xy

y x

τα θ

σ σ− −

= =−


20: Principal stresses σ1, σ3measured clockwise α

Mohr’s circlexyτ

xσxyτ

y

xyτ

( ),x xyσ τ

yσxyτ

τ

22

2y x

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠

σyσx

τxyτ

σ3α

x

σ3

σ1

2y xσ σ+

2y xσ σ−

τxy

2

1 2tan 2xy

y x

τα θ

σ σ−= =

−( ),y xyσ τ−


21

y

measured anticlockwise α

•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 circlexyτ

( ),y xyσ τxσxyτ

xyxyτ

Max shear 2

2

2y x

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠ yσxyτ

Max shear stress

1 3max 2n

σ στ −=

σy

τxyτ ,max 2n

yσx σσ3 σ12θ

τxy

( ),x xyσ τ−


23

Principal planes from Mohr’s circle

xyτ( ),y xyσ τxy

xyτθ

στxyτ

σ

σyσx

τxy

σσ3 σ1

( )

2θ3σ1σ

xσxyτ( ),x xyσ τ−

σ

yσxyτ

3σ1σ

3σ


24

3

1σ

260 N

280 N mmProblem: Find principal stresses, principal

2120 N mm

260 N mm

2120 N mm

planes and max shear stress analytically. Draw Mohr’s circle and verify graphically.

280 N mm

260 N mm

( )( )120,60−

τ60 18.442

71 56θ

80120

60

σ σ3σ1

3σ1σ

35.7871.56=

τ

9.22

60 1,maxnτ

Dept. of CE, GCE Kannur Dr.RajeshKN25

( )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.

151080

5015

1515

5050

10

1010

5050

3015

30

15

80

1010

15

50

5

2020

20

50

5

5020


26

Transformation of strains

σ σ σ σ+ − sin 2 cos2x yσ στ θ τ θ

−+cos2 sin 2

2 2x y x y

n xy


+= + − sin 2 cos2

2y

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 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + −

cos2 sin 22 2 2

x y x y xyn

ε ε ε ε γε θ θ

+ −= + −OR,

Dept. of CE, GCE Kannur Dr.RajeshKN27

sin 2 cos22 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 3x y x y xyε ε ε ε γ

ε ε+ −⎛ ⎞ ⎛ ⎞

= = ± +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ( )tan 2 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

03 90θ =

1 2 3, ,θ θ θε ε ε are measured from strain gages0

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 rosetteFrom strain transformation equations,

g(45 degree rosette)2 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + −

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 2cos sin sin cosn x y xyε ε θ ε θ γ θ θ= + −

2 20 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 2ε ε ε ε γ+ −⎛ ⎞ ⎛ ⎞ tan 2 xyγβ

−=


max,min 1,3 2 2 2x y x y xyε ε ε ε γ

ε ε+ ⎛ ⎞ ⎛ ⎞

= = ± +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ( )tan 2

x y

βε ε

=−

Theory of columnsyCompression 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 compressionL ( 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 ( )

2d y

Both ends hinged PEI MR

=2d yEI P

2d y2

d yEI Mdx

⇒ = A2

yEI Pydx

⇒ = −2 0d yEI Py

dx+ =

2

2 0d y P y+ =2 ydx EI

P P⎛ ⎞ ⎛ ⎞Solution for the above differential equation is:

yXX

l1 2cos sinP Py C x C xEI EI

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠When 0, 0x y= = 1 0C⇒ =

B

xWhen 0, 0x y 1

When , 0x l y= = 20 sin PC lEI

⎛ ⎞⇒ = ⎜ ⎟

⎝ ⎠

P

B⎝ ⎠

sin 0 0, ,2 ,3 ,4 ...P Pl lEI EI

π π π π⎛ ⎞

= ⇒ =⎜ ⎟⎝ ⎠


34

P, where 0,1,2,3,4...Pl n n

EIπ= =

2 2

2

n EIPlπ

=

2 sin Py C xEI

⎛ ⎞= ⎜ ⎟

⎝ ⎠

2 sin n xCLπ⎛ ⎞= ⎜ ⎟

⎝ ⎠

The least practical

2EIP π

The least practical value for P is:

Critical load2crEIP

lπ

=

sin xy C π⎛ ⎞= ⎜ ⎟The corresponding mode shape is:


35

2 siny CL

= ⎜ ⎟⎝ ⎠

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 theoryP

y

2d yEI M

One end fixed and the other end free

( )2d yEI P yδ⇒ = −

P

Aδ

2

yEI Mdx

= ( )2EI P ydx

δ⇒ =

2d yEI P Pδ2d y P Py δ

+ = y2

yEI Py Pdx

δ+ = 2 ydx EI EI

+ = y

lSolution for the above differential equation is:

1 2cos sinP Py C x C xEI EI

δ⎛ ⎞ ⎛ ⎞

= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ xEI EI⎝ ⎠ ⎝ ⎠

When 0, 0x y= = 1C δ⇒ = −

x

B


37

dy P P P P⎛ ⎞ ⎛ ⎞

dy P

1 2sin cosdy P P P PC x C xdx EI EI EI EI

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

When 0, 0dyxdx

= =2 20 0 0PC C

EI⇒ = = ⇒ =

When x l y δ= = cos Plδ δ δ⎛ ⎞

⇒ +⎜ ⎟

3 5cos 0P Pl l π π π⎛ ⎞⇒⎜ ⎟

When ,x l y δ= = cos lEI

δ δ δ⇒ = − +⎜ ⎟⎝ ⎠

cos 0 , , ...2 2 2

l lEI EI

= ⇒ =⎜ ⎟⎝ ⎠

Pl πTh l t ti l l i2

PlEI

π=

2EIP π 2EIP π

The least practical value is:

2el l= Effective length24EIPl

π= 2

e

EIPl

π=

P⎛ ⎞ ⎛ ⎞

e Effective length


38cos 1 cos

2 e

P xy xEI l

πδ δ δ⎛ ⎞ ⎛ ⎞

= − + = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Euler’s theory Py

( )2d yEI M P H l

One end fixed and the other hinged HA

( )2

yEI M Py H l xdx

= = − + −

( )2d yEI P H l y( )2

yEI Py H l xdx

+ = −

( )cos sinP P HC C l⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟

y

l( )1 2cos siny C x C x l x

EI EI P= + + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

x⎛ ⎞ ⎛ ⎞

x1 2sin cosdy P P P P HC x C x

dx EI EI EI EI P⎛ ⎞ ⎛ ⎞

= − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

When 0 0x y= = 1 10 H HC l C l⇒ = + ⇒ = −

B

M


39

When 0, 0x y 1 1P PP

M

When 0, 0dyxdx

= = 2 20 P H H EIC CEI P P P

⇒ = − ⇒ =

When 0x l y= = 0 cos sinH P H EI Pl l l⎛ ⎞ ⎛ ⎞

= − +⎜ ⎟ ⎜ ⎟

tan P Pl l⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

When , 0x l y= = 0 cos sinl l lP EI P P EI

+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

tan l lEI EI

=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P220.25 2EI EIP π

4.5 radiansPlEI

= 2 2Pl l

= ≈

2EIP π ll2

e

EIPl

π=

2el =


40

Euler’s theory Py

2d yEI M M P

Both ends fixed A M0

02

yEI M M Pydx

= = −

2d yEI P M y20d y P M

02

yEI Py Mdx

+ =

0P P M⎛ ⎞ ⎛ ⎞

y

l

02

y ydx EI EI

+ =

01 2cos sinP P My C x C x

EI EI P⎛ ⎞ ⎛ ⎞

= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞x

1 2sin cosdy P P P PC x C xdx EI EI EI EI

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

x

When 0 0x y= = 0 01 10 M MC C⇒ = + ⇒ = −

B

M


41

When 0, 0x y 1 1P P PM0

When 0, 0dyxdx

= = 2 20 0PC CEI

⇒ = ⇒ =

When 0x l y= =0 00 cosM P Ml

P EI P⎛ ⎞

= − +⎜ ⎟ 0 1 cos 0M Pl⎡ ⎤⎛ ⎞

⇒ − =⎢ ⎥⎜ ⎟

cos 1Pl⎛ ⎞⎜ ⎟

When , 0x l y= = P EI P⎜ ⎟⎝ ⎠ P EI⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

0 2 4 6Pl π π π⇒ =cos 1lEI

=⎜ ⎟⎝ ⎠

24 EIP π

0,2 ,4 ,6 ...lEI

π π π⇒ =

2Pl π2P

l=

2EI l

2lEI

π=

2

2e

EIPl

π= 2e

ll =


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

2Ee

EIPl

π=• Strength of the material is not taken into account

2

2E

Ee

P EIA Al

πσ = =

2 2 2EAr Eπ π22E

e e

EAr EAl l

r

π πσ⇒ = =⎛ ⎞⎜ ⎟⎝ ⎠


44

σ Eσ E

l /

Validity limits of Euler’s formula

le /r


45Critical stress for mild steel with E=2x105 MPa

2l Eπe

E

l EODr

πσ

= =

5 22 10 NEL t2250 N m mPLσ =Stress at limit of proportionality

5 22 10 N m mE = ×Let

( )2 52 1089

250elr

π ×∴ = =

250r

i.e., Euler’s theory is applicable for 89el > for mild steelr


46

Rankine’s theoryy

2

2EulerEIP π

=

c cP Aσ=For short compression members,

For long columns,2Euler

elg ,

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 elP EIπ

=

Rankine EulerP P≈

EulerP EIπ

21 1 1

R ki EIP A πσ= + 2I Ar=


472

Rankine c

e

EIP Al

πσ

1 1 1

22 rEπ σ

⎛ ⎞+⎜ ⎟

22

1 1 1

Rankine cP A rEAl

σπ

= +⎛ ⎞⎜ ⎟⎝ ⎠

22

ce

Rankinec

ElA

P rEl

π σ

σ π

+⎜ ⎟⎝ ⎠=

⎛ ⎞⎜ ⎟⎝ ⎠

2 2c c

RankineA AP

l lσ σσ

= =⎛ ⎞ ⎛ ⎞

ca σ=

el⎝ ⎠ cel

⎜ ⎟⎝ ⎠

21 1c e el laE r r

σπ

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2aEπ

2 2Crushing loadc

RankineAPl l

σ= =

⎛ ⎞ ⎛ ⎞

2

1 elar

⎛ ⎞+ ⎜ ⎟⎝ ⎠1 1e el la 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=

2A EIσ π 1/22 2⎛ ⎞2 2

1

c

ee

A EIlla

r

σ π=

⎛ ⎞+ ⎜ ⎟⎝ ⎠

1/22 2

2ec

ErlEa

πσ π⎛ ⎞

= ⎜ ⎟−⎝ ⎠


49

Problem: Compare the buckling (crippling) loads given by Rankine’s Problem: 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 18 10 N mm , 567 N mm ,1600cE 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

2EulerEIP

lπ

=

( )4 4 4150 110 17663604.69 mm64

I π= − =

el

6 m 6000 mmel l= = =( )64 e

2

2 387406.2 NEulerEIP

lπ

= = 387.4 kN=2Euler

el

( )2 2 2π


50

( )2 2 2150 110 8168.141 mm4

A π= − =

c AP σ2

1

cRankine

e

Plar

σ=

⎛ ⎞+ ⎜ ⎟⎝ ⎠

2 417663604.69 mmI Ar= = 17663604.69 46.503 mm8168.141

IrA

= = =

567 8168.141 406097 78 NP × 406 098 kN=2 406097.78 N1 60001

1600 46.503

RankineP = =⎛ ⎞+ ⎜ ⎟⎝ ⎠

406.098 kN=

2

2EEπσ = ( )2 48 10

37 317el π ×∴ = =

2EP EIπσ = =

To find the length of the column above which Euler’s formula is applicable

2Eelr

σ⎛ ⎞⎜ ⎟⎝ ⎠

37.317567r

∴ = =2E

eA Alσ = =

46 503 37 317l 1735 34 m m 1 735 m


51

46.503 37.317el∴ = × 1735.34 m m 1.735 m= =

Long column under eccentric loading Pσ =For short columns

PA

σ =

M P e=

For short columns,

e.M P e

Z Zσ = =

.M P e=

.P P e yA I

σ = ±Z Z A I

. . 1P P e y P ey⎛ ⎞± +⎜ ⎟2 21y yA Ar A r

σ ⎛ ⎞= ± = +⎜ ⎟⎝ ⎠

Aσ

21

APeyr

σ=⎛ ⎞+⎜ ⎟⎝ ⎠

2

APal eyσ

=⎛ ⎞⎛ ⎞

For long columns,


522 21 1eal ey

r r⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Secant formula P

A e2

2 0d yEI Py⇒ + =2

2

d yEI Py= −

Both ends hinged

2 ydx

2

2 0d y P yd EI

+ =

2 ydx

yl

2dx EI


y

x

1 2cos sinP Py C x C xEI EI

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠When 0,x y e= = 1C e⇒ =

B

When 0,x y e 1C e⇒

2sin cosdy P P P Pe x C xdx EI EI EI EI

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠dx EI EI EI EI⎝ ⎠ ⎝ ⎠

When 0l dyx = =sin

2l P

EIC e

⎛ ⎞⎜ ⎟⎝ ⎠=


53

When , 02

xdx

= = 2

cos2

C el P

EI

=⎛ ⎞⎜ ⎟⎝ ⎠

l P⎡ ⎤⎛ ⎞sin

2cos sin

cos

l PEIP Py e x x

EI EIl P

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎛ ⎞ ⎛ ⎞⎝ ⎠⎢ ⎥= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎛ ⎞⎝ ⎠ ⎝ ⎠⎢ ⎥⎜ ⎟

2sin2l P

EI⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟⎛ ⎞

cos2 EI

⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

l P⎛ ⎞max

2When , cos

2 2cos

2

EIl l Px y y eEI l P

EI

⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥= = = +⎜ ⎟⎢ ⎥⎛ ⎞⎝ ⎠⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

max .sec2l Py e

EI⎛ ⎞

⇒ = ⎜ ⎟⎝ ⎠

max max . .sec2l PM Py P e

EI⎛ ⎞

= = ⎜ ⎟⎝ ⎠2 EI⎝ ⎠

P My P Pey l P P ey l P⎡ ⎤⎛ ⎞ ⎛ ⎞max 2sec 1 sec

2 2c c cP My P Pey l P P ey l P

A I A I EI A r EIσ

⎡ ⎤⎛ ⎞ ⎛ ⎞= + = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


54

Secant formula Pδ

One end fixed and the other end free Aδ e

( )2

2

d yEI P e yδ= + −

( )2

2

d y P Py ed EI EI

δ+ = +y

( )2 ydx

( )2dx EI EI


l

( )1 2cos sinP Py C x C x eEI EI

δ⎛ ⎞ ⎛ ⎞

= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

When 0, 0x y= = ( )1C eδ⇒ = − +

xWhen 0, 0x y ( )1

( ) 2sin cosdy P P P Pe x C xdx EI EI EI EI

δ⎛ ⎞ ⎛ ⎞

= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Bdx EI EI EI EI⎝ ⎠ ⎝ ⎠

When 0 0dyx = = 0C⇒ =


55

When 0, 0xdx

= = 2 0C⇒ =

When x l y δ= = ( ) ( )cos Pe l eδ δ δ⎛ ⎞

⇒ = + + +⎜ ⎟When ,x l y δ= = ( ) ( )cose l eEI

δ δ δ⇒ = − + + +⎜ ⎟⎝ ⎠

( ) 1 Plδ δ⎡ ⎤⎛ ⎞

⇒ + ⎢ ⎥⎜ ⎟( ) 1 cose lEI

δ δ⇒ = + −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

( )cos Pe l eδ⎛ ⎞

⇒ + ⎜ ⎟( )cose l eEI

δ⇒ + =⎜ ⎟⎝ ⎠

( ) sec Pe e lδ⎛ ⎞

⇒ + = ⎜ ⎟

( ) PM P P lδ⎛ ⎞

+ ⎜ ⎟

( ) .sece e lEI

δ⇒ + = ⎜ ⎟⎝ ⎠

( )max . .secM P e P e lEI

δ= + = ⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞max 2sec 1 secc c cP My P Pey P P ey Pl l

A I A I EI A r EIσ

⎡ ⎤⎛ ⎞ ⎛ ⎞= + = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


56

max . .sec2el PM P e

EI⎛ ⎞

= ⎜ ⎟⎝ ⎠

In general,

2 EI⎝ ⎠

max 21 sec2

c eP ey l PA r EI

σ⎡ ⎤⎛ ⎞

= +⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦2A r EI⎢ ⎥⎝ ⎠⎣ ⎦

For short compression members (no buckling), max .M P e=

For long columns (with buckling) el PM P⎛ ⎞⎜ ⎟For long columns (with buckling),

max . .sec2eM P e

EI= ⎜ ⎟

⎝ ⎠


57

Problem: A hollow mild steel column with internal diameter 80 mmand 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 geometricalaxis. Calculate the maximum and minimum stresses in the column.Also find the maximum eccentricity so that no tension is induced in thesection. 5 22 10 N mmE = ×

( )4 4 4100 80 2898119 mmI π= − = ( )2 2 2100 80 2827 4 mmA π

= − =( )100 80 2898119 mm64

I

2400 mml l

( )100 80 2827.4 mm4

A

2898119 32 015 mmIr 2400 mmel l= =32.015 mm2827.4

rA

= = =


58

sec el PM P e⎛ ⎞

= ⎜ ⎟

33 2400 60 1060 10 16 secM

⎛ ⎞×× × × ⎜ ⎟ 0 96 kN

max . .sec2eM P e

EI= ⎜ ⎟

⎝ ⎠

max 5 660 10 16 sec2 2 10 2.898 10

M = × × × ⎜ ⎟⎜ ⎟× × ×⎝ ⎠0.96 kNm=

3 6 ⎧max

maxmin

cP M yA I

σ = ±3 6

maxmin

37.78 MPa60 10 0.96 10 504.69 MPa2827.4 2898119

σ⎧× × ×

= ± = ⎨⎩

To find the maximum eccentricity so that no tension is induced in the section

max 0cP M yA I− =

. .sec 02e

cP P ly

Ie 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 ti and 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