ACD504 Session 01 BasAnalTech

M. S. Ramaiah University of Applied Sciences

1Faculty of Engineering & Technology

Session I

Basic Analytical Techniques

Session Speaker

Dr. Vinod K. Banthia



Session Objectives

At the end of this session students will be able to

Explain basic principles and concepts of Mechanics of

Materials

Classify, and analyse accordingly, structural components

based on their structural behaviour

Analyse structural components to assess their adequacy for

specific application without failure

Apply Theory of Elasticity approach for analysing structural

behaviour

Solve structural problems using Energy Methods



Structural Design

DesignTo create, fashion, execute, or construct according to plan - devise,

contrive

To conceive and plan out in the mind

To have as a purpose

To devise for a specific function or end

To indicate with a distinctive mark, sign, or name

To make a drawing, pattern, or sketch of

To draw the plans for

Intransitive verb

1: to conceive or execute a plan

2: to draw, lay out, or prepare a design

Structure: arrangement of elements for carrying loads

http://www.thefreedictionary.com/design

To devise for a specific function or end



Structural Design

Structures Loads/Forces

Interaction

Response of the Structure?

DeformationReaction to

the forces

(Stiffness) (Strength)

What makes a structure adequate?

What controls the response of the structure?



Forces



Forces/Loads

DesignWhat happens to solid bodies when subjected to various types of loading?

MechanicsMechanicsStudy of interaction between physical objects and the effect of

this interaction on the objects involved.

Effect of interaction between physical objects on the

objects involved

The source?

Relate

Load

Stress

Relate

Load

Displacement

(Strength of Materials, Mechanics of Deformable Bodies, Solid Mechanics, .)



Forces/Loads

N

S

vr

ir

N S



Types of Forces/Loads

Body forces: Act on each element volume of the body

Surface forces: Act on surface or area elements of the body

(Traction)

Idealisations



Forces/Loads Make bodies move

Prevent bodies from moving

Forces

Linear

Rotational

Motion

Linear

Rotational

Moment, Torque, Couple

Moment of Inertia

Angular acceleration

Rotation in the direction of

moment

M*q = Work/Energy

Force

Mass

Acceleration

Motion along the force

P*d = Work/Energy



Forces/Loads

Concept of force what causes things to move?

Measure of force/mass how to quantify?

Action/Reaction equilibrium

1 kg @ 1 m/s2

(derived unit)

)(mvdt

dF =

N

e

w

t

o

n

s

L

a

w

s

interaction

Force Laws

Magnitude of forces of interaction

Function of the properties of the body and its environment

Magnitude?

kx mg mg 2/ rGmM ( ) 420 /2/3 r



Equilibrium & Free Body Diagram

( ) = 0ii xmF &&r ( ) = 0ii IM &&

r ( )( ) = 0ii IM r

Free (isolate) the body of its interaction (Constraints)

Apply the forces at the interaction points (Applied/Constraint)

dAlemberts principle(accelerating to equivalent static)

Equivalent Force system

Inertia force through the c.g.,

Inertia torque through any point

St. Venants Principle

= 0iFr

(vector sum) = 0iMr

(moment/couple)

http://appliedmechanicsreviews.asmedigitalcolle

ction.asme.org/article.aspx?articleid=1399873




RPRPFx === 0 0

== 20 RFx +== PRRFy 310

P PR

P

E, L, I

P

R1R2 R3

a b

== PbLRM *10




P

R1 R1

P

gm1

gm2

T1

T2

T2

p

T1

w

a

R

M

R

M

P=wa

Equivalent Force System



http://www.engin.brown.edu/courses/en3/Notes/Statics/Staticequiv/Staticequiv.htm

Fn

rn

Mn

i

j

Fn

rn

Mn

i

j

AB

==

=3

1

4

1 n

B

n

n

A

n FF

====

+=+2

1

3

1

3

1

4

1 n

B

n

n

B

n

B

n

n

A

n

n

A

n

A

n MFrMFr

Moment about the same point




http://em-ntserver.unl.edu/negahban/em223/note10/note10.htm

http://www.engin.brown.edu/courses/en3/Notes/Statics/Staticequiv/Staticequiv.htm




Equivalent Force system Center of Mass

= gdmrmgr dmcgrr

=

i

ii

dm

dmxx

= xdmxMExtension to Systems of Particles

x

y

=

==n

i

i

n

i

ii

m

xm

x

1

1

x

y

g*dm

mg

cgrrdm

rr

= gdmmgTotal Force =

=

i

ii

dm

dmyy

=

==n

i

i

n

i

ii

m

ym

y

1

1



Stresses



Load Stress Deformation Stress

(Residual, Initial,Trapped, Locked in)

Micro Macro(micro-structure level

Independent of loads)

Influences material strength

Difficult to calculate

In the bulk of material(Load or Deformation)

Stresses



PP P

PP

P

P

M M

M

= P/A

As internal reaction/resistance to applied load

Stress



T1

T2

T3

P2

P1

P3

X

Y

Z

nr

P

A

Stress: A measure of internal force distribution (reaction of one part

of the body on the other) in a loaded body

A

PStress

A

0lim

plane specified aon force internal on theIntesity

=

Stress at a point or Stress on an area?

?lim0

AA

Stress

Internal

Resistance

Variable

Stress?



X

Y

Z

nr

P

X

Y

Z

nr

nP

X

Y

Z

nr

tP+=

Y

X

Znr P

Normal Shear

X

Y

Znr

nP

tPX

Y

Znr

txPtyP

nP

= =

Stress

Force Components Stress Components



Stress at a Point

X

Y

Z

nr

P

State of stress at a point



B = ?, B = ?

B

Magnitude, Direction and Orientation

Stress Tensor

Stress at a Point



n = Pn/A = P/A = Pt/A = 0

n = Pn/Asin = Pcos/Asin=Psincos/A

= Pt/Asin = P sin/Asin= Psin2/

P P

P

P

Pn

Pt

Stress at a Point



Notation for Stress

i j

Plane Dirrection

X

Y

Z

zzzx zy

xzxx xy

yz

xxyy

Sign Convention

(Sign for Plane)

*

(Sign for Direction)



State of Stress at a Point

All the components of stresses at a point

18

Equilibrium in Normal direction

18

Equilibrium in Tangential direction

xx

6

xyxx

yx

yy

yy

yx

yx

Sign Convention

Complimentary Shear

X

Y

Z

zzzx zy

xzxx xy

yz

xxyy



yy

yy

xxxx

xyxy

xyxy

cossin2cossin 22 xyyyxxnn ++=

yy

xx

xy

xy

AB

Cnnnt

( ) 22 cossincossin)( += xyxxyynt

( )

2tan5.0

cossin

cossin

)( 22=

=

xxyy

xyIf 0=nt

Transformation of Stresses

zxxzyzzyxyyxzzzyyyxxxn nnnnnnnnn 222222 +++++=



Pure shearNormal Stress

For any stress state at a point in a stressed body,

there exist three mutually perpendicular planes

which pass through the point and on which only

normal stresses, namely principal stresses, exist.

Normal Stress, Pure Shear and Principal Stress



yy

yy

xxxx

xyxy

xyxy

Principal Stresses

1

1

2

2

2

2

11

1

2

222

1 cossin +=

cossin)( 12 =

Transformation of Stresses



2

2

2,122

xy

yxyx

+

+=

( )

=

2tan2 1

yx

y

y

xx

xyxy

xyxy

2cos22

2121,

+=yx

2sin2

21 =xy

xx

y

y

xy

xy

yx

yx

x

y

Transformation of Stresses Transformation of Stresses Arbitrary conditionArbitrary condition



Strains



l

ll

l=

Strain

Uniform strain over the length

Real problems

Load on Structure Deformation Stress



dxudxx

uu /)(

+=dx

udx

x

uu

+x

u

=

y

xdx

dy

y

vy

=

x

ux

=

x

y

vu dxx

vv

+

dyy

uu

+

dxx

u

dyy

v

Shear Strain = =-x

v

y

uxy

+

=

Normal Strain: No distortion, Volume change

Shear Strain: Distortion, no volume change

Strain

Deformation behaviour in infinitesimal neighbourhood



x

y

xy

2sin2

2cos22

cossincossin 22'

xyyxyx

xyyxx

+

++

=

++=

2sin2

2cos22

cossinsincos 22'

xyyxyx

xyyxy

+

=

+=

2sin2

2sin2

)sin(cossincos2)( 22''

xyyx

xyyxyx

+

=

+=

n

yx

xy

= 11 tan2

Direction for

Principal Strain

22

21222

,

+

+

+= xyyxyx

Strain Transformation



Stress Strain Relationship



Load Stress

Load Deformation?

Strain Deformation

Stress Strain?

Uniqueness for structure?

Unique for material

Robert Hooke (1676)

ceiiinossssttuu

Ut tensio, sic vis (1678)

As the extension, so the force

Simeon Poisson

x

y yxlt ==AuxeticAuxetic MaterialsMaterials

http://home.um.edu.mt/auxetic/www/properties.htm

,

P()

E

1

Stress-Strain Relations



E=

Poissons Effect:

(Hookes Law)

lt =

[ ]zzyyxxxxE

=1 [ ]xxzzyyyy

E =

1 [ ]yyxxzzzzE

=1

xyxyG

1

= yzyzG

1

= zxzxG

1

=

EEE

zzyyxx

=== 321 ; ;x

y

zzz

xx

yyyy

zz

xx

321 =xx

x x E

xx

=

E

xzy

==

( )+=

12

EG

Stress-Strain Relationship



zyxzyxe ++=++= and Define

Ee

21=

Result

( )

KppE

e =

=213

(hydrostatic loading, Bulk Modulus)

( )( ) ( ) xxxGe

Ee

E

21211

+=+

++

=

( )( ) ( ) yyyGe

Ee

E

21211

+=+

++

=

( )( ) ( ) zzZGe

Ee

E

21211

+=+

++

=




+=

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

2

2100000

02

210000

002

21000

0001

0001

0001

)21)(1(

+

+

+

=

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

)1(200000

0)1(20000

00)1(2000

0001

0001

0001

1

Stiffness Representation

Flexibility Representation

Stress-Strain Relation



[ ]yyxxxx E += 21

Plane Stress:

zero be toassumed and , yzzx zz

x

y

x

y

yx

xy

yx

xy

[ ]yyxxxxE

=1

[ ]xxyyyyE

=1

[ ]xxyyyy E += 21

xyxyE

)1(2 +

=

[ ]yyxxzzE

+=

x

y




++

= yyxxxxE

121

Plane Strain:

zero be toassumed and , yzzx zz

++

= xxyyyyE

121

( )yyxxzz E

++

=

)21)(1(




Combined Stresses



What happens in these complex loading condition?

How to know the magnitudes of normal and shear stresses?

The real World



Combined StressesCombined Stresses

1. Find stress state for each loading

2. Sum up like stresses

3. Use Mohrs circle

Like Stresses: Normal (bending, axial)

Shear (torsion, shear)

I

Mc

A

Pn =J

Tr

A

Ps =

y

y

xx

xyxy

xyxy



M M

T T

b bMc/I Tr/J

y

y

xx

xyxy

xyxy

2

2

2,122

xy

yxyx

+

+=

( )

=

2tan2 1

yx

Combined Bending and Torsion



2

5

0

0

m

m

10000N

50 mm

500 mm

A B

( ) MPaaxial 1.5450*10000

2==

( )( ) MPabending 4.4076450*

25*500*100004

==

( ) MPaA 3.4021.54.407 =+=

( ) MPaB 5.4121.54.407 =+=

Ex. 1 Combined Bending and Torsion



30 mm

3

0

0

m

m

500 mm

Power = 35 kW

RPM = 500 rpm

$mmT 668416= $F 4340=

( ) $mmBM 5425004500*4340max ==

( ) MPatorsion 1.1261630*668416

3==

( ) MPabending 6.2043230*

5425003

==

MPacomb 367)2/6.204(1.1266.20422 =++=

MPacomb 4.162)2/6.204(1.12622 =+=

Example 2: Combined Normal and Shear Stress



Structural Behaviour

under Different

Loading Conditions



So many things So little time (for design)



Where do I begin?

http://www.daviddarling.info/images/electric_motor.jpg

http://www.easypedia.gr/el/ima

ges/shared/1/10/IC_engine.JPG http://boeingcockpit.com/image/c1f04.jpg http://image.absoluteastronomy.com/images/encyclopediaimages/w/wa/water_turbine.jpg

http://blog.mlive.com/chronicle/2008

/01/large_Wind_Turbine.jpg

What are we designing?

How?

Where do we start?



Whats the difference?

http://www.aerospace-

technology.com/projects/airbus_a380/im

ages/A380freighter_4.jpg http://edsphotoblog.com/wp-content/photos/800px/ba_airbus_a319_takeoff.j

pg

http://www.tourism-

india.in/delhi-

tourism/img/l-full.jpg www.nbtt.in/kolkata



Prismatic Member (Uniform Cross-section)

Loads/Weight act only along the axis of the element

Loads applied at the centroid of the cross-section

Applied Loads are constant

Material is homogeneous and isotropic

Poissons effect is neglected

Buckling effect is not considered

Cross-section remains plane

Application: Truss, Cables under tension

Components Subjected to Uniaxial Loading



P

P

A

,

L

E

P

=

P

Assume

P

AE

P

E==

1

L

L=

AE

PLL =

Deformation is uniform

Stresses and Strains are constant




Truss:

25.31

3)1020(6)10(9)1015(12

=

++++=

A

A

R

R

75.33

)1020()10()1015(

=

++++=

A

AB

R

RR

Method of joints

At joint A

31.25kN

FAC

FADo04.59)3/5(tan 1 ==

004.59cos

025.3104.59sin

=+

=+

ADAC

AC

FF

F

k$F

k$F

AD

AC

75.18

44.36

=

=




At joint D

FCD

FDEFAD

010

0

=

=

CD

ADDE

F

FF

10 kN

k$F

k$F

CD

DE

10

75.18

=

=

Method of sections

FDE

FDC

FAC

31.25 kN

k$F

FM

DE

DEC

75.18

053*25.310

=

===

k$F

FM

CD

CDA

10

03*1030

=

===




P

E, L, I

X,u(x,y)

Y,v(x)

M

P

q(x)

yvyx

vy

x

u==

=

=2

2

yvyx

xvyxu ==

=)(

),(

Eyx

vEyE =

==2

2

=== EIdxyEdxyM 2

=Mc/I

=Mc/2I=P/2a ``

Components Subjected to Bending Loading



P

E, L, I P

R1 R2

Shearing action of the cross-section (Vertical direction)

Shearing action of the cross-section (Horizontal direction)

Shearing Action




P

E, L, I

Compression

Stretch

M M

Rotation

Bending Action




Determination of Forces:

PM

q2(x)q1(x)

SFD

BMD

D

i

a

g

r

a

m

s

f

o

r

i

l

l

u

s

t

r

a

t

i

o

n

o

n

l

y

+ +




Drawing SFD and BMD

qdx

dV V

dx

dM

EI

M

dx

d

dx

dw w ====

Free Body Diagram




Drawing SFD and BMD




M

bdyy

c

IdAy

cdAy

c

yybdyM

c

c

c

c

c

c

max2maxmax)(

====I

Mc=max

I = Second Moment of Area

qdx

dVV

dx

dM== (Equilibrium)

EyI

M==


measure of resist to bending of a cross-section

I(distribution of material about bending axis)

Material farther Higher I



Longitudinal Shear Stress in Beam:

T1 T2

tI

yMaaTtdx

''*

===

It

yVa

dx

dM

It

ya ''==

bd

V3max = 2max 3

4

r

V

= [ ]222max )(

8bddDB

b

V+=

B

bdD

y




Example (Longitudinal shear):

y

12 mm

12 mm

2

2

0

m

m

180 mm

x

1

2

ycg

0

2.1621218012220

2261218011012220

=

=+

+=

cg

cg

x

mmy

X X

471066.2 mmI xx =A

B

C

Shear on Surface A = 0 (free surface)

Shear on Surface B (flange)

MPa432.01801066.2

8.63)12180(150007

=

=

Shear on Surface B (web)

MPa3.6121066.2

8.63)12180(150007

=

=

Shear at neutral axis (maximum shear)

MPa42.7121066.2

)9.28128.578.6312180(150007

=

+=



Shear Center:

Point of intersection of bending axis and plane of transverse section

Bending axis longitudinal axis through which the transverse bending load

must pass to ensure there is no twist of the section

If load passes through shear center, the section will not twist.

Load not passing through the shear center will cause twisting of the section

resulting in much higher longitudinal stress than predicted by bending

equation.




T

T

r

l

Angle of twist = (x)l/ =Pure TorsionlG / =Stress-Strain

dAlGT = )/(* Equilibrium

==

2

0 0

2 )/()/(

r

p lGIdAlGT

l

G

rJ

T ==

Prismatic member

Torque applied at the end

Small angle of twist

No change in r

Cross-section twists as rigid body

Components Subjected to Torsional Loading



Beams Torsionx

y

z

T

T

L

L

G

J

T

r

==

x

y

z

TT

tb )/(3/

3GbtTL =

)/(6 3btTy= )/(3 2max btT=

3/3btJ =

3)/(6 tbTy i=

3)(/3/ tbGTL i= is shape coefficientC(1.1), Z(1.17),

Angle(0.83), T(1.00)2

max )/(3 tbT i=




Beams Torsion

x

y

z

T

T

L

ds

0=

s

qShear flow is constant

around the section

A

Tq

2=

( ) ( )dstqGA = 21

= tdsAJ24

Shear flow pattern in

thin walled sections

t1

J. Powlowski, Vehicle Body Engineering, Business Books Limited, London




Beams Torsion

2211 22 qAqAT +=

In general,

nnqAqAqAT 2........22 2211 +++=

T

A1A2

q1 q2

t1 t2

t3

121 2 AGqq =+ 232 2 AGqq =

( )( )( )[ ]2213212213221231

21321231

2 AAlttAlttAltt

AAltAltT

+++

++=

( )( )( )[ ]2213212213221231

21312132

2 AAlttAlttAltt

AAltAltT

+++

++=

( )( )[ ]2213212213221231

2121211

2 AAlttAlttAltt

AltAltT

+++

=




22 rprtl =

=

t

pr

t

pr

l

l

2

2

dxrpdxtt = 22

=

t

pr

t

pr

t

t

Thin walled (r>5t)

22 rprtt =

t

prt

2=

t

prlt

42max =

=

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm

Components Subjected to Internal Pressure



Thick walled

( )22

22222/)

io

iooiooii

rrr

rpprrrprp

=

( )22

22222/)

io

iooiooii

trr

rpprrrprp

+=

22

22

io

ooiil

rr

rprp

=

[ ])(1 lrttE

+=

[ ])(1 lttrtE

++=

[ ])(1 trllE

+=

ltv += 2

Components Subjected to Internal Pressure



Rotating Disks

http://www.utm.edu/departments/engin/lemaster/Machine%20Design/Lecture%2016.pdf



++=

yR

y

ZaR

M 11

M M

daRy

y

aZ +=

1 Neutral Surface exists

Total deformation of the

fibers is proportional to

their distance from the

neutral surface (tension and

compression)

The strains of the fibers

are not proportional to its

distance from the neutral

surface

Bending of Curved Bars



c

c

b

R

ydy

+

=+

=+

=c

c

dyRy

y

cbdy

Ry

y

bcda

Ry

y

aZ

2

1

2

11

........11

4321

+

+

=

+=+

R

y

R

y

R

y

R

y

R

y

RRy

y

........7

1

5

1

3

1642

+

+

+

=R

y

R

y

R

yZ

Method 1

Method 2

+=

+=

c

c

c

c

dyRy

R

cdy

Ry

y

cZ 1

2

1

2

1

+

+=cR

cR

c

RZ elog

21

Bending of Curved Bars



,lT

lT

lP P

A,E

AE

PlTl = TE

A

PT ==

Temperature increase under constraints

Thermal gradient in a component

Phase change during solidification (locked in stresses)

Differential cooling rates (locked in stresses)

Compenents Subjected to Thermal Loading



P P

P P

=P/A =PL/(AE)

Neutral equilibrium

and are not proportional to the load(even though and are)

Elastic Buckling

s load increases increases abruptlyLoad is limited to critical load (Pcr)

Stress levels are lower than compressive yield

Axially loaded members

Buckling



Euler buckling

P

PAxially loaded member Homogeneous material

Pcr: Maximum load under which if the component when given a small

lateral deflection will come back to its equilibrium position.

P

Column with lateral deflection:

( ) PyMdxydEI ==22EIPyy ==+ 22 0

xBxAy cossin +=

For simply supported column

00

==== lxx

yy

Elastic buckling of ideal slender column



http://en.wikipedia.org/wiki/Buckling

http://physics.uwstout.edu/StatStr/Statics/Columns/cols61.htm

( )22

2

2

rl

EA

l

EIPcr

==

( )22

rl

EAP

e

cr

= le: effective length

le = kl

2ArI =(r: minimum radius

of gyration)

( )22

rl

E

e

cr

=

Elastic buckling of ideal slender column



Short columns: 60/0 cr)

Intermediate columns: 120/60



cr

(l/r)

( )22

rl

E

e

~ 100

?

Euler line

> c before P reaches Pcr Abrupt increase in at a

well defined load

cr

(l/r)

yc

yc/2

EulerJohnson

?

( )22

rl

E

e

( ) 2

2

1

=

rl

E

ey

ycr

yccr =

c

Johnsons Parabola

EEt( )2

2

rl

E

e

tcr

=

Tangent Modulus method

Intermediate columns/Inelastic buckling



Thin walled cylinder under axial or torsion or bending load Wide flanged thin walled beams Flat plates under in plane compression

Failure of functionality Redistribution of loads

Number of bulges Geometry, BC

( )2

2

2

112

==b

tEK

A

Pcrcr

( ) mttmbP crult max22 +=F.B.Seely and J.O.Smith, Advanced Mechanics of Materials, 2nd

Edition, McGraw-Hill Book Company.

Buckling of flat plates



2

21

=b

tEK

a/b=1 a/b=2 a/b=inf.

SS 7.75 5.43 4.40

Clampe

d

12.7 9.5 7.38

x x

y

y

+

+=

CCCCC x

yxy

SS

6122122

+

+=

CCCCC x

yxy

C

8431.2

3

4431.22

2

21

823.0

=b

tEC

( )5.12

2466.064.927.1

1H

l

tEHinged ++

=

T

T

l

( )5.12

2605.09.9639.2

1H

l

tEClamped ++

=

2

21tr

lH =

R.J.Roark and W.C.Young, Formulas for Stress and Strain, 5th

Edition,McGraw-Hill Book Company,

Buckling of flat plates



Material Behaviour



llAP / / ==

e

l

a

s

t

i

c

y

i

e

l

d

i

n

g

s

t

r

a

i

n

h

a

r

d

e

n

i

n

g

n

e

c

k

i

n

g

u

fy

Elastic limit

P

r

o

p

o

r

t

i

o

n

a

l

l

i

m

i

t

E

Material Behaviour



p

Proof Stress: Yield strength at a specified permanent set

Work Hardening: The phenomena of increased elastic limit after loading-unloading

Ductility: Total elongation due to plastic deformation

Toughness: Energy absorbed in the process of breaking

Tensile Strength: Highest stress the material can withstand before failure

Yield Strength: Stress at which plastic deformation starts

Bauschinger effect: Decrease in resistance of material to plastic deformation in

the direction opposite to the in which it was plastically deformed earlier

Reduction of Area: Measure of necking (reduction in the diameter) before breaking

Youngs Modulus/Secant Modulus/Tangent Modulus

Material Behaviour



General Stress-Strain Relationship



Different Kinds of Material Different Kinds of Material BehavioursBehaviours

IsotropicSame properties in all directions

2 material constants

OrthotropicProperties have two orthagonal planes of symmetry


Composites, rolled sheets

AnisotropicProperties have no planes of symmetry


Transversely Orthotropic

Same property in one plane and different normal to it




Failure of Material



y

y

x

xxy

xyxy

xy

FnMn?

Seely, F.B. and Smith, J.O, Advanced Mechanics of Materials, Second edition, John Wiley & Sons, Inc., New York

Material Behaviour and Failure



http://web.utk.edu/~prack/mse201/Chapter%208%20Failure.pdf

http://web.utk.edu/~prack/mse201/Chapter%208%20Failure.pdf



http://www-

outreach.phy.cam.ac.uk/physics_at_work/2005/exhibit/matsci.php

http://www.aloha.net/~icarus/




(Rankine Theory)

(Coulombs Theory)

(St. Vnenamts Theory)

Material Behaviour Theories of Failure




Material Behaviour Theories of Failure



1. Max. Principal Stress

2. Max. Shear Stress 2/)( minmaxmax =

max

3. Octahedral Stress 3)()()( 2132

32

2

21max ++=

4. von Mises Stress 2)()()( 2132

32

2

21max ++=

Theories of Failure



Bending and Torsion

Going Beyond SoM



Angle of twist = (x)l/ =Pure TorsionlG / =Stress-Strain

dAlGT = )/(* Equilibrium

==

2

0 0

2 )/()/(

r

p lGIdAlGT

l

G

rJ

T ==

Prismatic member

Torque applied at the end

Small angle of twist

No change in r

Cross-section twists as rigid body

No warping of cross-section

Torsion of Circular BarsT

T

r

l

http://www.transtutors.com/homework-help/mechanical-engineering/torsion/shafts-circular-section.aspx



http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect08.d/IAST.Lect08.pdf

Warping

Max. stress not farthest from centroid

JTt /max =

GJT /=

3ibtJ i =

Torsion of Non-Circular Sections



10tb

31=

3)/(6 tbTy i=

3)(/3/ tbGTL i= 2

max )/(3 tbT i=

http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect08.d/IAST.Lect08.pdf

xy

Torsion of Thin Rectangular Sections



Torsion of Solid Section

Torque is resisted by

the shear only on the

cross-section (xy=0)

Equilibrium Equation

Prandtls Stress Function

On the outside surface

Compatibility Condition

(assumption)

T.H.G. Megson, Aircraft Structures for Engineering Students, Elsevier, 4th Edition, 2000



Torsion of a Circular Bar

Using compatibility condition




Torsion of Narrow Rectangular Strip

Shape Factor

dz

dx

y

w =

dz

dxyw

=




St. Venants Warping FunctionAssumed mode of displacement

All sections rotate in a rigid manner

),( yxdz

dw

=

Warping function of x and y co-ordinate only

dxdyyyx

xxyz

GdxdyyxT zxzy

+

==

)(

dxdyyyx

xxy

J

+

=

= yxdz

dzx

+

= xydz

dzy

yu = xv =




Bending, Shear and Torsion

Of Thin-Walled beams



P

E, L, I

M M

Shearing Bending

+

+

qdx

dV V

dx

dM

EI

M

dx

d

dx

dw w ====


Assumptions

EyI

M==



Bending of Beam in two Mutually Perpendicular Directions

?21 +=

Resolution of BM



Bending of Beams of Arbitrary Cross-section

( ) ( )22

xyyyxx

xyxxy

xyyyxx

xyyyx

zIII

yIxIM

III

xIyIM

+

=

yy

y

xx

xz

I

xM

I

yM+=

yy

y

xx

xz

I

xMor

I

yM =

xyyyyx

xyxxxy

IMIM

IMIM

=tan

Neutral Axis

Through centroid


Section with symmetry 0or =yx MM



Bending Deflection of Beams of Arbitrary Cross-section


( )

=

y

x

xyyy

xxxy

xyyyxxM

M

II

II

IIIEv

u2

1

2

2

z

M

z

Sw xxy

=

=2

2

z

M

z

Sw

yy

x

=

=(q V M)

Deflection

EI

M

dx

yd=

2

2

qdx

dV V

dx

dM

EI

M

dx

d

dx

dw w ====

2

2

z

uu

=

2

2

z

vv

=



Approximations for Thin-walled Cross-section

Wall thickness is much smaller than cross-sectional dimensions

Stress is constant across thickness

Small contributions in I are neglected

0

1

2

3

4

5

6

7

8

9

10

0.0125 0.1375 0.2625 0.3875

0

0.5

1

1.5

2

2.5

3

3.5

4

Exact Approx.

% Diff



Bending Moment Variation and Shear Stress

qdx

dV V

dx

dM

EI

M

dx

d

dx

dw w ====

Shear Shear+Rotation



Shear in Open Section Beams

0=

+

zt

z

q z

tq =

Shear Flow

dstyIII

ISISdstx

III

ISISq

s

xyyyxx

xyxyyys

xyyyxx

xyyxxx

s

=

0

2

0

2

x

yS

z

M=

y

x Sz

M=

z

s0=

+

st

s

q s



Shear Centre

Shear

Centree

)/61(

3 2

hbh

be

+=

dstyI

Sq

s

xx

y

s =0

0=xS

+=h

bthI xx

61

12

3

12 61

6s

h

bh

Sq

y

lower

+

=

1

02

2 dsh

qeS

b

lowery =

h

b



Shear in Closed Section Beam

Load need not be on shear centre Shear and Torsion

Point at which shear flow is known is difficult/impossible to find

0,0,

0

2

0

2 sbs

s

xyyyxx

xyxyyys

xyyyxx

xyyxxx

s qqqdstyIII

ISISdstx

III

ISISq +=+

=

0,sq An unknown reference value



Shear in Closed Section Beam

dstyIII

ISISdstx

III

ISISq

s

xyyyxx

xyxyyys

xyyyxx

xyyxxx

b

=

0

2

0

2

Internal moment of this shear flow and external moment in equilibrium

+== pdsqdspqpqdsSS sbyx 0,00

+= 0,00 2 sbyx AqdspqSS Moment about intersection of Sx and Sy

+= 0,20 sb Aqdspq find



Twisting and Warp of Closed Section Beams

Shear loads not acting through shear center

Rotation of the cross-section

Axial displacement of the cross-section

+w

+

=z

v

s

wGtq ts

sincos vupvt ++=

= dsGtq

A

Ads

Gt

qww s

s

ss

0

0

0

= tdstdsww s0



Torsion of Closed Section Beams

0=

+

st

s

q z 0=

+

st

z

q s

Pure torque loading

0=

s

q0=

z

q

Cq =

Constant shear flow in the beam wall

Shear stress variation depends on variation of t

== AqpqdsT 2

Bredt-Batho Formula



Displacements in Closed Section Beams

+

=z

v

s

wGtq t

Pure torsion

(q constant)0=

z

q No direct stress

(z=0 constant)

02

2

=

z

vt 0sincos2

2

2

2

2

2

=++

dz

vd

dz

ud

dz

dp

BAz += DCzu += FEzv +=

= dsGtq

A

Ads

Gt

qww s

s

ss

0

0

0

=A

A

A

Tww OsOss

2

0

= Gtds

=s

OsGt

ds

0



Torsion of Open Sections Beams

Torsion of thin rectangular stripts >>

0 2 == znzsdz

dGn

max,dz

dGtzs

=

==t

dstJst

Jsec

33

3

1

3

max,J

tTzs =



Warping of the Cross-section

Waarping of a thin rectangular stripdz

dnswt

=

Waarping of a thin rectangular beamz

ps

w

z

v

s

wR

tzs

+

=

+

=

At mid-line of a section wall 0== zszs G

dspdz

dw

s

Rs = 0

Secondary

Small

Primary, constant over t

GJ

TA

dz

dAw RRs 22 ==



Structural Idealisation

1)(



Panel direct stress carrying boom + shear stress carrying skin

Idealisation



Theory of Elasticity



Mechanics of Materials

P

P

AP /=

M M

IMc /=

( )22

22222/)

rir

rpprrrprp

o

iooiooii

r

=

( )22

22222/)

rir

rpprrrprp

o

iooiooii

t

+=

T T

JTr /=



Mechanics of materials Theory of elasticity

Strain distribution is assumed

(simplified, unidirectional)

No simplifying assumption is

made about strain distribution

Hookess law for one component of

strain is used (=E)General statement of Hookess

law is used

Same material assumptions are used in both

Bodies are perfectly elastic

Body is homogenoeous

Body is isotropic (mostly)

Mechanics of Materials and Theory of Elasticity

Equilibrium in terms of applied

forces

Equilibrium in terms of internal

reactions

Compatibility Condition



Equilibrium ConditionEquilibrium Condition

External and Internal forces

P1

P2

P3

P4

P5

P6

X

Y

Z

nr

P

A

Over every small part of the body

Over the surface of the body

State of Stress in a body



Elasticity Equations of Equlibrium

= 0M = 0F

BC



dx

dydyx

dxy

( )( )dyxxx +

( )( )dxyyy +

dxyx

( )( )dyxxyxy + ( )( )dxyyxyx +

dyxy

( ) ( ) 00 =++ = XyxF xyxx ( ) ( ) 00 =++ = YyxF xyxyy

Differential equations of equilibrium:

Condition on rates of changes in stresses for equilibrium (internal)

Stresses must satisfy the force equilibrium on the boundary also

== yxxyM 0

2D Problem Equation of Equilibrium



( )( )( )x

udxudxxuux

=+= /

( )( )( )y

vdyvdyyvvy

=+= /

Assuming small strains

x

v

y

uxy

+

=

x

w

z

u

y

w

z

v

z

wzxyzz

+

=

+

=

= , ,

Components of Strains



x xE

xx

=

E

xzy

==

( )[ ]zyxxE

+=1

( )[ ]zxyyE

+=1

( )[ ]yxzzE

+=1

(Poissons effect)

Under the action of numerous forces, if the deformations are small and do not

affect the action of other forces, these deformations are neglected. Under this

condition, the resultant displacements can be obtained by superposition in the

form of linear functions of external forces.

xyxyG

1

=

yzyzG

1

=

zxzxG

1

=

( )+=

12

EG




zyxzyxe ++=++= and Define

Ee

21=

Result

( )

pE

e213

=(hydrostatic

loading)

( )( ) ( ) xxxGe

Ee

E

21211

+=+

++

=

( )( ) ( ) yyyGe

Ee

E

21211

+=+

++

=

( )( ) ( ) zzZGe

Ee

E

21211

+=+

++

=




Number of equations of equilibrium < Number of unknowns

Statically indeterminate problem

Use elastic deformation condition

x

ux

=

y

vy

=

x

v

y

uxy

+

=

u and v cannot be selected arbitrarily

yxxy

xyyx

=

+

2

2

2

2

2

u and v must satisfy the condition of compatibility

Strain Compatibility



( )yxyxxy

xyyxyx

+=

+

+

2

2

2

2

2

2

2

2

2

12

For general case

( ) ( )

+

+=+

+

y

Y

x

X

yxyx 12

2

2

2

For plane stress

( )( )

+

=+

+

y

Y

x

X

yxyx

1

12

2

2

2

For plane strain

Boundary conditions

x

y

xy

xy

X

Y

$

X

Yxyx

xyx

lmY

mlX

+=

+=

Compatibility Conditions in terms of Stresses



( ) ( ) 0=+ yx xyx ( ) ( ) 0=++ gyx yxy

For a two dimensional problems with weight being the only body force, the solution

to the problem is obtained by solving the following equations, along with satisfying

the prescribed boundary conditions

( ) 02

2

2

2

=+

+

yxyx

Define a function such that

, ,2

2

2

2

2

yxgy

xgy

yxyyx

=

=

=

satisfies the equilibrium equation, and will satisfy compatibility condition if

024

4

22

4

4

4

=

+

+

yyxx

2D Problems Stress Function



For a more general case of body forces, assuming these have potential

y

VY

x

VX

=

= ,

Define a function such that

, ,2

2

2

2

2

yxxV

yV xyyx

=

=

=

This leads to compatibility condition

( )

+

=

+

+

2

2

2

2

4

4

22

4

4

4

12y

V

x

V

yyxx

2D Problems Stress Function



2D Problem in Rectangular Coordinates

, ,2

2

2

2

2

yxxV

yV xyyx

=

=

=

With

body force

, ,2

2

2

2

2

yxxyxyyx

=

=

=

For a given polynomial expression for stress function

Corresponding variation of stresses can be found



222

22

22y

cxybx

a++=

222 , , bac xyyx ===

33232333

2*3222*3y

dxy

cyx

bx

a+++=

ycxbybxaydxc xyyx 333333 , , =+=+=

If all coefficients other than d3=0, pure bending If all coefficients other than a3,d3=0, pure bending

by normal stress on y = c If all coefficients are non zero, both normal and

shearing stresses on the sides of the plate.




34342243444

3*42*323*24*3xy

exy

dyx

cyx

bx

a++++=

244

2 ,0 , y

dxyd xyyx ===

2

44

2

4 ycxybxay ++=

244

24

22

2y

dxycx

bxy =

( )444 2 ace +=

If all coefficients other than d4=0




55453252354555

4*53*42*32*33*44*5y

fxy

eyx

dyx

cyx

bx

a+++++=

( ) ( ) 3552552535 23

132

3ydbxyacyxdx

cx +++=

352

5

2

5

3

53

yd

xycyxbxay +++=

( ) 355252535 323

1

3yacxydyxcx

bxy ++=




In the solutions presented, there is a correlations between the pattern of

variation of boundary loads and resulting stress variation.

A change in the pattern of variation of boundary loads will result in the change

of pattern of stress variation.

If the boundary loads are replaced by a

statically equivalent load system, the

effect on the pattern of stress variation

is localised to the region where such

change has been made.

Determination of Displacements Once stresses are known, use stress-strain relations to find strains. Use strain displacement relations

to find displacement components x

ux

=

y

vy

=

x

v

y

uxy

+

=

Displacements are obtained but not unique

St. Venants Principle



y = c faces are free of forces

Resultant of shear force on x=0 is P

( ) 24242

,0 , yd

bxyd xyyx ===

( )2

24

242

2 0

2 0

c

bdc

db

cyxy===

=

c

PbPdyy

c

bbPdy

c

c xy4

3

2 2

c

c-

2

2

22 ==

=+

====

2

2

31

4

3 ,0 ,

2

3

c

y

c

P

I

Pxyxy

c

Pxyyx

Example Bending of a Cantilever



( )222

, , ycIG

P

EI

Pxy

EI

Pxyxyyx ===

(x) fEI

Pxyvf(y)

EI

yPxu 1

22

2 ,

2+=+=

Substitute this in the third component, separate out functions of x and y,

apply boundary conditions (at y=0) to obtain

++=

GI

Pc

EI

Pl

GI

Py

EI

Py

EI

yPxu

22662

22332

EI

Pl

EI

xPl

EI

Px

EI

Pxyv

3262

3232

+++=

Example Bending of a Cantilever



qcyycyycyxy

======

,0 ,0

0 ,0 , ===+

+

+

c

c x

c

c x

c

c xyydydyqldy m

( )32 325 yyxdx = 33532 ydybay ++= xbxydxy 325 =

( )322

32 yyxI

qx =

( )3322

323 yyccI

qy +=

( )xycI

qxy

22

2=

To satisfy last condition, add another term d3y

( ) ( )52322

2322 ycyyxlI

qx +=

Example Bending of a Beam with UDL



r

01

=+

+

+

Rrrr

rrr

021

=++

+

Srrr

rr

Equilibrium equations

For the case when R=S=0, the stress function is defined by

2

2

2

11

+

=rrr

r 2

2

r

=

=

rr

r

1

Resulting in

01111

2

2

22

2

2

2

22

2

=

+

+

+

+

rrrrrrrr

r

ur

=

+=

v

rr

u 1

r

v

r

vu

rr

+

=

1

Strain components

2D Problem in Polar Coordinates



011

2

2

2

2

=

+

+

dr

d

rdr

d

dr

d

rdr

d

0112

32

2

23

3

4

4

=++dr

d

rdr

d

rdr

d

rdr

d

For axi-symmetric problems, equation to be solved is

DCrrBrrA +++= 22 loglog

CrBr

Ar 2)log21(2 +++=

0= r

CrBr

A2)log23(

2+++=

2D Problem in Polar Coordinates



Inner radius=a Outer radius=b

Internal pressure=pi External pressure=po

Cr

Ar 22 += Cr

A2

2+=

The solution is given by

Boundary conditions iarrp=

= obrr p==

ipCa

A=+ 2

2

opCb

A=+ 2

2

( )22

22

ab

ppbaA io

=

( )2222

2 ab

bpapC oi

=

Example Cylinder Under Internal Pressure



( )( )22

22

222

22 1

ab

bpap

rab

ppba oiior

+

=

( )( )22

22

222

22 1

ab

bpap

rab

ppba oiio

+

=

Radial displacement can be found using the relations

rr

uEE ==

Example Cylinder Under Internal Pressure



Boundary conditions

0==== brrarr

0=b

adr

0= r

Mrdrb

a=

02)log21( and 02)log21(22

=+++=+++ CbBb

ACaB

a

A

b

a

b

a

b

a

b

a

b

a

b

ar

rdr

rr

rrdr

rrdr

=

=

= 22

Mb

a=

Example Pure Bending of Curved Bars



DCrrBrrA +++= 22 loglogQ

( ) ( ) MabCaabbBa

bA =++ 2222 logloglog

( ) ( )[ ]aabbab$

MCab

$

MB

a

bba

$

MA loglog2 ,

4 ,log

4 22222222 +===

( )2

22222 log4

=a

bbaab$

++=

r

aa

b

rb

a

b

r

ba

$

Mr logloglog

4 222

22

+++= 2222

2

22

logloglog4

abr

aa

b

rb

a

b

r

ba

$

M

0= r

Example Pure Bending of Curved Bars



( )00

1 22 =+

=+

+

+

r

r

rR

rrr

rrrr

Using stress-strain relations and strain-displacement relations

322

2

22 1 r

Eu

r

ur

r

ur

=

+

( ) ( )

+= 32

2

18

1111

1r

rCCr

Eu

22

21 8

31r

rCCr

++=

22

21 8

311r

rCC

+=

Example Rotating Disk



Consider conditions on a circle with radius b (b>>a)

( )2

2sin ,

2

2cos1

SSbrrbrr

=+

===

2cos)(rf=

04141

22

2

22

2

=

+

+

r

f

dr

d

rdr

fd

rdr

d

rdr

d

( )( ) 2cos242 DrCBrAr +++=

Find stresses, apply boundary conditions to find integration constants

Example Plate with a Hole



2cos43

12

12 2

2

4

4

2

2

++

=

r

a

r

aS

r

aSr

2cos3

12

12 4

4

2

2

+

+=

r

aS

r

aS

+=

2

2

4

4 231

2 r

a

r

aSr

Example Plate with a Hole



Miscellaneous Applications



Energy Methods



Purpose of a structure to support loads

Response of the structure to loading deflection or deformation

Need a relationship between Load and Deflection

When extent of deformation/deflection is limited

For solving statically indeterminate problems

PP

S2S1l1 l2 KP =

P

Can the load-deflection

characteristics be used?

Energy Principles for P- Characteristics



P

Energy Methods

Strain Energy approach

P

PP

dP

P

S2S1l1 l2

21 UUUPdW P +=== (work=energy)

P

P

d

dUP

)(

=

If both horizontal and vertical forces are applied

H

Hd

dUP

=

V

Vd

dUP

=

= 111 deSU = 222 deSU(Ext.) (Int.)

P of Functionsde and de 21




P

dP

P

21 +=== dPW Pc

dP

dP

=

If both horizontal and vertical forces are applied

H

HdP

d =

V

VdP

d =

= 111 dSe = 222 dSe

(need to know

the forces)

Complementary Energy approach

For applicability

Force-deformation relationship is single valued

Force system is conservative

No unloading of components takes place in inelastic region




Example (Strain Energy): VB

A

B

C

HC

VB

HBHB

C`AC C`

CACABCBCABAB eSeSeSU2

1

2

1

2

1++=

CABCABHCVBHB el

AEe

l

AEe

l

AEU 222

2

1

2

1

2

1),,( ++=

HCAC

HCHBVBBC

VBHBAB

e

e

e

=

+=

=

*5.0*5.02/3

2/3*5.0

( )

=

==

= HCVBVB

BHCHB

HB

Bl

AEUV

l

AEUH

43

2

32

4 (2 equations

3 unknowns)



Example (Strain Energy) (contd):

Write an equation in terms of a dummy force at C

+=

= HCHBVBHC

Bl

AEUH

45

4

1

4

3

From these three equations, given the geometry and material property of the

bars, and the applied loads, the three deflections can be found.



Example (Complementary Energy): VB

A

B

C

HC

VB

HBHB

C`A C

C`

++=CABCAB S

CACAC

S

BCBC

S

ABAB dSedSedSe000

CABCABCABCAB SAE

lS

AE

lS

AE

lSSS 222

2

1

2

1

2

1),,( ++=

2/)32/(

3/

3/

BBAC

BBBC

BBAB

HVS

HVS

VHS

+=

=

=

+=

= BBB

VB HVAE

l

V 34

1

4

3

)32/(1

3/1

3/1

=

=

=

BCA

BBC

BAB

VS

VS

VS

2/1

1

1

=

=

=

BCA

BBC

BAB

HS

HS

HS

+=

= BBB

HB VHAE

l

H 34

1

4

9?=HC



Easier to useRequires more effort to get deformation

(solution of simultaneous equations

involving redundant variables)

Cumbersome for highly statically

indeterminate structures.

More suitable for statically

indeterminate structures where known

displacements are many.

Equations of equilibrium are used and

geometry of deformation is not used

Require use of geometry and

deformation, not of equilibrium eq.

Energy is expressed first in terms of

forces in the members and then in

terms of external loads

Energy is expressed first in terms of

deformation and then in terms of

deflection

Complementary EnergyStrain Energy

Both methods can be used for both linear and non-linear problems (restricted)Assumptions:

Rigid supports (W=0) Perfectly fitting joints

No buckling

Applicable beyond pin jointed structures

Applicable for forces and moments

Energy Methods



Castiglianos Theorem

In a structure subjected to external forces, with small deflections and linear

relation between loads and deflections, the deflection in the direction any one

force is equal to the partial derivative of internal strain energy with respect to

the force.

PP

U=

P

CE

SECE = SE

Energy Methods



Example (Castiglianos Theorem):

l/2 l/2P

x

( ) lxl for lxPxP

lx for xP

M



Minimum Energy

If a system consisting a structure and bodies that exert forces on it, is completely

isolated such that no energy is transferred in or out of it (rigid supports) total potential

energy of the system remains constant.

Total potential energy = Strain energy in the members + Energy of the load

U + E = constant

As work is done in deforming the body, the strain energy of the body will increase

0)(

=+

id

EU

Condition of max/min energy (kinematically admissible)

ii 0 0 Pd

E

d

EP

d

E

d

U

iiii

=

=

+=

+

(equilibrium)

Energy Methods



Example (Min. Potential Energy):

( ) ( )lxlBzxwzzu 2/cos2/)()( ==

( ) ( )lxlBzxwzz 2/sin2/)()( 2 ==)z(E)z( =

( ) ( ) ( ) ( )

==

===

dxlxlEIB

dAzdxlxlEB

dVwEzdVEdVU

2/sin2

2/2/sin

2

2/

2

1

2

1

2

1

2

4222

42

222

PBE =

lP

x

B

( )lxBxw 2/sin)( =

( ) ( ) PllBEIVU ==22

2/0

4

EI

PlB

044.3

3

=



Virtual Displacements

If a system of structure and bodies exerting loads is in equilibrium, increase in

its internal energy because of small virtual displacement is equal to the

work done on the system by the forces.

EU

EU

EEUU

=

=

=++

0

0

Expressions for strain energy

Axially loaded members )2/(2 EAlP

Members subjected to shear )2/(2 GAlV

Members subjected to bending )2/(2 EIM

Members subjected to torsion )2/(2 GJT

Energy Methods



Dummy Load Method

l/3 2l/3x l/3 2l/3

x

1

)2/()2/( 2wxwlxM = 2l/9

dxEI

Mmdx

EI

Mml

l

l

+=3/

2

3/

0

1

dxl

xxwxwlx

dxxwxwlx

EI

l

l

l

+

=

3/

23/

0

2

33

2

223

2

22

972243

2

324

444 wlwlwlEI =+=

Energy Methods



P Ql1 l

l

llxPQxQxM

1

1))(( 10 ++=

+

+=l

l

l

dxEI

xlxPQxdx

EI

xQU

1

1

2

))((

2

22

1

0

22

=

= )(2

)(3

11 21

213

1

3 llPl

llPEIQ

U

Example :

EI

lPdxx

EI

Pdx

EI

Pxdx

EI

MU

PxM

ll

62

1)(

2

1

2

1 32

0

22

0

22

====

=

EI

Pl

P

U

3

3

=

=



Session Summary

Basic principles and concepts of Mechanics of Materials

Classification of structural components based on their

structural behaviour

Analytical approaches for modelling different structural

behaviours

Modes of structural failure and assessment of design for

these

Theory of Elasticity approach for analysing structural

behaviour

Energy Methods to solve structural problems

In this session the topics below were covered

ACD504 Session 01 BasAnalTech

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Transcript of ACD504 Session 01 BasAnalTech