Aircraft Design Day6

8/14/2019 Aircraft Design Day6

1/39

DAY 6


2/39

STRESS

Stress is a measure of force per unit areawithin a body. It is a body's internal distribution of force per

area that reacts to external applied loads.

A

P=STRESS


3/39

ONE DIMENSIONAL STRESS

Engineering stress / Nominal stress

The simplest definition of stress, = F/A,

where A is the initialcross-sectional area prior

to the application of the load

True stress True stress is an alternative definition in which

the initial area is replaced by the current area

eetrue )1( +=Relation between Engineering & true stress


4/39

TYPES OF STRESSES

TENSILE

BENDING

COMPRESSIVE

SHEAR

TORSION
http://www.allstar.fiu.edu/aero/pic1-2e.htmhttp://www.allstar.fiu.edu/aero/pic1-2d.htmhttp://www.allstar.fiu.edu/aero/pic1-2c.htmhttp://www.allstar.fiu.edu/aero/pic1-2b.htmhttp://www.allstar.fiu.edu/aero/pic1-2a.htm


5/39

SHEAR STRESS

TORSION

12

12

B A z z

dz

dx

zd

zd

y

zd

zd

y

xdxdy

xdxdy

( ) ( )dzdxdydxdzdy xz =

D

C

xz =

Taking moment about CD, We get

This implies that if there is a shear in one plane then there will be a shear in

the plane perpendicular to that


6/39

TWO DIMENSIONAL STRESS Plane stress

Principal stress

y

x x

y

xy

xy

yx

yx

2

2,122

xy

yxyx

+

+=


7/39

THREE DIMENSIONAL STRESS

Cauchy stress

Force per unit area in the deformed geometry

Second Piola Kirchoff stress

Relates forces in the reference configuration to

area in the reference configuration

=

zzzyzx

yzyyyx

xzxyxx

ij

iJ,XijjI,

JXIJS = X Deformation gradient


8/39

Stress invariants of the Cauchy stress

Characteristic equation of 3D principal stress is

Invariants in terms of principal stress

3D PRINCIPAL STRESS

zyxI ++=1 2222 zxyzxyxzzyyxI ++=

222

3 2 xyzzxyyzxzxyzxyzyxI +=

0322

1

3 =+ III

3211 ++=I

1332212 ++=I

3213

=I


9/39

VON-MISES STRESS

Based on distortional energy

( ) ( ) ( )2

2

13

2

32

2

21 ++

=v

( ) ( ) ( ) ( )222222 62

1

zxyzxyxzzyyxv +++++=


10/39

STRAIN Strain is the geometrical expression of deformation

caused by the action of stress on a physical body.

Strain displacement relations

Normal Strain

Shear strain (The angular change at any point

between two lines crossing this point in a body can

be measured as a shear (or shape) strain)

StrainL

L=

z

w

y

v

x

uzyx

==

=

z

u

x

w

y

w

z

v

x

v

y

uzxyzxy

+

=

+

=

+

=


11/39

VOLUMETRIC STRAIN

Volumetric strain

0

0

V

VV=

zyx ++=


12/39

TWO DIMENSIONAL STRAIN

Plane strain

Principal strain

y

xx

y

xyxy

yx

yx

+

+=

2222,1

xyyxyx


13/39

3D STRAIN

=

zz

zyzx

yz

yy

yx

xzxy

xx

ij

22

22

22Strain tensor

( )

+

+

=

=

j

k

i

k

i

j

j

i

ij

x

u

x

u

x

u

x

u

2

1

2

1

FFij

Ekjki

( )FFij

Ekj

1-

ki

1

2

1 =ij

Green Lagrangian Strain tensor

Almansi Strain tensor


14/39

STRESS-STRAIN CURVE

Mild steel

Thermoplastic

Copper
http://en.wikipedia.org/wiki/Image:Stress_v_strain_A36_2.png


15/39

BEAM

A STRUCTURAL MEMBER WHOSE THIRD DIMENSION IS

LARGE COMPARED TO THE OTHER TWO DIMENSIONS

AND SUBJECTED TO TRANSVERSE LOAD

A BEAM IS A STRUCTURAL MEMBER THAT CARRIES

LOAD PRIMARILY IN BENDING

A BEAM IS A BAR CAPABLE OF CARRYING LOADS IN

BENDING. THE LOADS ARE APPLIED IN THE

TRANSVERSE DIRECTION TO ITS LONGESTDIMENSION


16/39


17/39

TYPES OF BEAMS

CANTILEVER BEAM

SIMPLY SUPPORTED BEAM

FIXED-FIXED BEAM OVER HANGING BEAM

CONTINUOUS BEAM


18/39

BEAMS (Contd)

STATICALLY DETERMINATE STATICALLY INDETERMINATE

C

BA

D


19/39

BEAM

TYPES OF BENDINGHoggingSagging


20/39

DEFLECTION OF BEAMS

A loaded beam deflects by an amount that depends on several

factors including:

the magnitude and type of loading

the span of the beam

the material properties of the beam (Modulus of Elasticity)

the properties of the shape of the beam (Moment of Inertia)

the beam type (simple, cantilever, overhanging, continuous)


21/39

Deflections of beam can be calculated using

Double integration method

Moment area method

Castiglianos theoremStiffness method

Three moment theorem (Continuous beam)

DEFLECTION OF BEAMS

DOUBLE INTEGRATION


22/39

DOUBLE INTEGRATION

METHOD

(1)...REI

M 1=

(2)...1

d

2

2/32

21

+

=dx

dy

dxy

R

From Flexure formula

Radius of curvature

(3)...d

2

2

1

dx

y

R=

Ignoring higher order terms

From (1) & (3)(4)M...

dEI

2

=2dx

y

DOUBLE INTEGRATION METHOD


23/39

DOUBLE INTEGRATION METHOD

48EIy

3PL=

P

LRight of load

= xLP

dx

y

22

2dEI

Left of load

02

=C

3

2

22C

xLxP

dx

y+

=dEI

At x=L/2, dy/dx=08PL-

2

=3C

+=

228

22 xLxLP

dx

ydEI

4

322

648 CxLxxL

P +

+=EIy

24

PL-

3

=4

C

22Px

dx

y=

2dEI

1

2

4C

Px

dx

y+=

dEI

16PL

2

=2C

=

416

22xL

Pdx

ydEI

2

32

1216C

xxLP +

=EIy

48EIy

3PL

=

At x=0, y=0At x=L, y=0


24/39

First method

Second method

MOMENT AREA METHOD

EI

dxMd

*=

EI

dxM

xxd

*

.=


25/39

MOMENT AREA METHOD

L

P

PL/4

P/2

P/2

16

PL

4

PL

2

L2

12

=

Area of the moment diagram (1/2 L)

Taking moments about the end

48EIPL

16PL*2L*32

32

==EI1


26/39

Energy method derived by Italian engineer

Alberto Castigliano in 1879.

Allows the computation of a deflection at anypoint in a structure based on strain energy

The total work done is then:

U =F1 1+F2 2 F3 3+.Fn n

F1

F2

F3

Fn

CASTIGLIANOs THEOREM


27/39

Increase force Fn by an amount dF

This changes the state of deformation and

increases the total strain energy slightly:

Hence, the total strain energy after the increase

in the nth force is:

n

n

dFF

UdU

=

n

n

dFF

UU

+


(Contd )


28/39

Now suppose, the order of this process is reversed; i.e., Apply a small force dFn to this same body

and observe a deformation d n; then applythe

forces, Fi=1ton.

As these forces are being applied, dFn goes

through displacement n.(NotedFnisconstant)

and does work:dU = dFn n

Hence the total work done is:

U+ dFn

n


(Contd )


29/39

The end results are equal Since the body is linear elastic, all work is

recoverable, and the two systems are

identical and contain the same storedenergy:

n

n

nnnn

F

U

dFdFF

U

U

=

+=

+ U


(Contd )


30/39

The term force may be used in its most

fundamental sense and can refer forexample to a Moment, M, producing a

rotation, , in the body.M

n

nMU

=


(Contd )


31/39


32/39


L

P

PL/4

P/2

P/2

Strain energy

According to Castiglianos theorem

( )

( )dx

dxdxU

=

==2/

0

2

2/

0

2

2/

0

2

4

222

22

L

LL

EI

PxEI

Px

EI

M

]

48EI

6EI

dx

2EIL/2

0

3

3

2/

0

2

PL

xP

xP

P

U L

=

=

=

=

UNIT LOAD METHOD


33/39

UNIT LOAD METHOD

(VIRTUAL WORK METHOD)

dxx

=0 EI

MmQ

Deflection (Translation) at a point:

Rotation at a point:

==i

ii

ii

x

IE

hAdx

0 EI

mMQ


34/39

UNIT LOAD METHOD

L

Q=1

QL/4

Q/2

=i

ii

ii

IE

hAQ

Area of the moment diagram (1/2 L)

Unit load method

48EI

QL3

=

Q/2

A2

* *A1

d1d2

iih d2Q=

16

QL

4

QL

2

L2

12

==i

A

48

LQ

16

QL*

2

L*

3

2

2

Q

16

QL*

2

L*

3

2

2

QQ

32

22

=

+

=

DEFLECTIONS OF BEAMS


35/39




36/39


THREE MOMENT EQUATION


37/39




38/39

THREE MOMENT EQUATION(Developed by clapeyron)

Continuity condition

Equating the above equations

Using second moment-area theorem

R

CR

L

CL

LLtantan =

RR

RR

LL

LL

R

R

R

C

R

R

L

L

L

L

L

EIL

Ax

EIL

AxM

EI

LM

EI

L

EI

LM

EI

L 662 =+

++

++= LLLLCLLL

L

CLLMLLMLAx

EL 21

31

21

321

tan

++=

RRRRCRRR

R

CRLMLLMLAx

EL 2

1

3

1

2

1

3

21tan


39/39

THREE MOMENT THEOREM

+=+++

2

22

1

11

22116)(2

L

xA

L

xALMLLMLMCBA

+=

+

++

22

22

11

11

2

2

2

2

1

1

1

1 62IL

xA

IL

xA

I

LM

I

L

I

LM

I

LM

CBA

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Aircraft Design Day6