Aircraft Design Day5

8/14/2019 Aircraft Design Day5

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DAY 5



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 =σ STRESS



ONE DIMENSIONAL STRESS

• Engineering stress / Nominal stress

– The simplest definition of stress, σ = F / A,

where A is the initial cross-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 σ ε σ )1( +=• Relation between Engineering & Nominal stress



TYPES OF STRESSES

TENSILE

BENDING

COMPRESSIVE

SHEAR

TORSION

http://www.allstar.fiu.edu/aero/pic1-2e.htm

http://www.allstar.fiu.edu/aero/pic1-2d.htm

http://www.allstar.fiu.edu/aero/pic1-2c.htm

http://www.allstar.fiu.edu/aero/pic1-2b.htm

http://www.allstar.fiu.edu/aero/pic1-2a.htm



SHEAR STRESS

TORSION

12

12

B Aτ z τ z

dz

dx

τ

z d

z d

y τ

z d

z d

y

τ xdxdy

τ xdxdy

( ) ( )dz dxdydx dzdy xz τ=τ

D

C

xz τ=τ

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



TWO DIMENSIONAL STRESS• Plane stress

• Principal stress

yσ

xσ xσ

yσ

xyτ

xyτ

yxτ

yxτ

2

2,122

xy

y x y xτ

σ σ σ σ σ +

−±

+=



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

=

zz zy zx

yz yy yx

xz xy xx

ij

σ τ τ

τ σ τ

τ τ σ

σ

iJ,X

ijτ

jI,JX

IJS = X – Deformation gradient



• Stress invariants of the Cauchy stress

• Characteristic equation of 3D principal stress is

•Invariants in terms of principal stress

3D PRINCIPAL STRESS

z y x I σ σ σ ++=1222

2 zx yz xy x z z y y x I τ τ τ σ σ σ σ σ σ −−−++=222

3 2 xy z zx y yz x zx yz xy z y x I τ σ τ σ τ σ τ τ τ σ σ σ −−−+=

032

2

1

3 =−+− I I I σ σ σ

3211 σ σ σ ++= I

1332212 σ σ σ σ σ σ ++= I

3213

σ σ σ = I



VON-MISES STRESS

• Based on distortional energy

( ) ( ) ( )2

2

13

2

32

2

21 σ σ σ σ σ σ σ

−+−+−=v

( ) ( ) ( ) ( )222222

62

1

zx yz xy x z z y y xvτ τ τ σ σ σ σ σ σ σ +++−+−+−=





VOLUMETRIC STRAIN

• Volumetric strain

0

0

V

V V −=υ

z y x ε ε ε υ ++=



TWO DIMENSIONAL STRAIN

• Plane strain

• Principal strain

y

ε

xε xε

yε

xyγ xyγ

yxγ

yxγ

+

−±

+=

2222,1

xy y x y x γ ε ε ε ε ε



3D STRAIN

=

zz

zy zx

yz

yy

yx

xz xy

xx

ij

ε γ γ

γ ε

γ

γ γ ε

ε

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

δ

( ) FF

ij

Ekj

1-

ki

1

2

1 −−=ij

δ

Green Lagrangian Strain tensor

Almansi Strain tensor



STRESS-STRAIN CURVE

Mild steel

Thermoplastic

Copper

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



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



TERMINOLOGY• SHEAR FORCE

– A shear force in structural mechanics is an exampleof an internal force that is induced in a restrainedstructural element when external forces are applied

• BENDING MOMENT – A bending moment in structural mechanics is an

example of an internal moment that is induced in arestrained structural element when external forcesare applied

• CONTRAFLEXURE – Location, where no bending takes place in a beam



TYPES OF BEAMS

• CANTILEVER BEAM

• SIMPLY SUPPORTED BEAM

• FIXED-FIXED BEAM• OVER HANGING BEAM

• CONTINUOUS BEAM



BEAMS (Contd…)

• STATICALLY DETERMINATE• STATICALLY INDETERMINATE

C

BA

D



BEAM

•TYPES OF BENDINGHoggingSagging



SHEAR FORCE & BENDING

MOMENTBEAM

BENDING MOMENT

SHEAR FORCE FREE BODY DIAGRAM

P

P/2

PL/8P

LP/2

P/2

P/2

PL/8

PL/8 PL/8



SHEAR FORCE & BENDING

MOMENTBEAM

BENDING MOMENT

SHEAR FORCE FREE BODY DIAGRAM

P

5P/16

P

L11P/16

11P/16

5P/16

3PL/8

3PL/8

RELATI N BETWEEN



RELATI N BETWEEN

BM, SHEAR & LOAD

M+dMM V+dV

w

VO

dx

dx

dM V

dxwdxVdxdM M M

M O

=

=

−++−

⇒=∑ 02

)(0

Taking moments about O

Force equilibrium gives ( )

dx

dV w

dxwdV V V

=

=++− 0*



BEAM THEORY

• ASSUMPTIONS – MATERIAL IS HOMOGENOUS

– MATERIAL IS ISOTROPIC

– THE BEAM IS SYMMETRICAL

– THE TRANSVERSE PLANE SECTION

REMAIN PLANE AND NORMAL TO THE

LONGITUDIONAL FIBRES AFTER BENDING

(NEUTRAL PLANE REMAINS SAME

AFTER BENDING)



BENDING STRESSo

R MM

ba

c edf

From similar triangles edf & cod

(1)... (

cd

ef

lengthOriginal

lengthinChangestrain ==)ε

(2)... co

de

cd

ef =

...(3) (

R

y

cd

ef strain ==)ε

Hooks law

...(5) R

E

y

f

=

...(4) (

( E

or f strain

σ ε

)) =

(6)... dA

R

Eyarea stressdF == *

(7)... dA

== R

Ey ydF ydM *

(8)... dAy 2

=

= ∫ R

EI

R

E M

(9)... R

E

I

M =

From (3) & (4) From (5) & (9)

(9)... y

f

R

E

I

M

==



FINITE ELEMENTS

• TRUSS / BAR / LINK ELEMENT• BEAM ELEMENT



3D BEAM ELEMENT



3D BEAM ELEMENT

k

s

l V

……… (1)

k

sz

l q

k k k

k

tz

l q

k k k k

l q

k k

k

sy

l q

k k k

k

ty

l q

k k k k

l q

k k

k

sx

l q

k

k k

k

tx

l q

k

k k k

l q

k

k

V hb s

V hat

z ht sr z

V hb s

V hat

yht sr y

V hb s

V hat

xht sr x

∑∑∑

∑∑∑

∑∑∑

===

===

===

++=

++=

++=

111

111

111

22),,(

22),,(

22

),,(

t

t

t

Cartesian coordinate of any point in the element

Cartesian coordinate of any nodal point k

Cross sectional dimensions of the beam at nodal point k

Components of unit vector in direction t at nodal point k

Components of unit vector in direction s at nodal point k

We call and the normal vectors or director vectors at

nodal point k

===

==

k

sz

l k

sy

l k

sx

l

k

tz

l k

ty

l k

tx

l k k

k

l

k

l

k

l

l l l

V V V

V V V ba

z y x z y x

,,

,, ,

,,,,

k

t

l V

k

t

l V

k

s

l V

k

s

l V

k

s

l V



3D BEAM ELEMENT

The displacement components are

From (1) & (2) we get

……… (2)

……… (3)

z z t sr z y yt sr y x xt sr x

01

01

01

),,(),,(),,(

−=−=−=

k

sz

q

k k k

k

tz

q

k k k k

q

k k

k

sy

q

k

k k

k

ty

q

k

k k k

q

k

k

k

sx

q

k k k

k

tx

q

k k k k

q

k k

V hb s

V hat

z ht sr z

V hb s

V hat

yht sr y

V hb s

V hat

xht sr x

∑∑∑

∑∑∑

∑∑∑

===

===

===

++=

++=

++=

111

111

111

22),,(

22

),,(

22),,(

k

s

k

s

k

s

k

t

k

t

k

t

V V V

V V V 01

01

−=

−=



3D BEAM ELEMENT

[ ]

[ ]

[ ]

∂∂

=

∂

∂

∂∂

∂∂

∑=

k

z

k

y

k

x

k

q

k

k

i

k

i

k

ik

k

i

k

i

k

ik

k

i

k

i

k

i

r

u

) g ( ) g ( ) g ( h

) g ( ) g ( ) g ( h

(g)(g)(g)r

h

t

u

s

u

r

u

θ

θ

θ

1

321

321

321

1

ˆˆˆ1

1

∑=

=

q

k k k

u B1

ˆ

η ξ

η ς

η η

γ γ ε where [ ]

k z

k y

k xk k k k wvuu θ θ θ =ˆ

Strain displacement relation

and the matrices Bk ,k=1,…..,q, together constitute the matrix B,

[ ]q B B B ....

1=



3D BEAM ELEMENT

Jacobian Transformation

( )

=

0

0

0

2ˆ

00

00

00

-

-

-

k

sx

k

sy

k

sx

k

sz

k

sy

k

sz

k k

V V

V V

V V

b g

ξ ∂

∂=

∂

∂ −1 J

x

( )

=

0

0

0

200

00

00

-

-

-

k

tx

k

ty

k

tx

k

tz

k

ty

k

tz

k k

V V

V V

V V a

g

( ) ( ) ( ) k ij

k

ij

k

ij g t g s g += ˆ



3D BEAM ELEMENTStrain displacement relation

Where

Stiffness

Load

ξ d l

DB BT ∫ −

=1

1

2

2K

T

K

T f d l

N B f +=

∫ −ξ

1

1

2

2

∂∂

∂∂

∂∂

=

∂∂

∂∂

∂∂

∑=

−

−

−

k

z

k

y

k

x

k

q

k

k

i

k

i

k

i

k

k

i

k

i

k

i

k

k

i

k

i

k

i

k u

)(G )(G )(Gr

h J

)(G )(G )(Gr

h J

)(G )(G )(Gr h J

z

u

y

u

xu

θ

θ

θ

1

333

1

31

222

1

21

111

1

11

321

321

321

( ) ( )[ ] ( ) ( )[ ]k

k

min

k

min

k k

min

k

in

h g J g J r

h g J Gm 1

3

1

2

1

1

ˆ −−− ++∂

∂=

STIFFNESS MATRIX



STIFFNESS MATRIX

−

−

−

−

−

−

−

=

L

EI

L

EI

L

EI

L

EI L

EI

L

EI

L

EI

L

EI L

GJ L

GJ

L

EI

L

EI

L

EI

L

EI L

EI

L

EI

L

EI

L

EI L

AE

L

AE L

EI

L

EI

L

EI

L

EI L

EI

L

EI

L

EI

L

EI L

GJ

L

GJ L

EI

L

EI

L

EI

L

EI L

EI

L

EI

L

EI

L

EI L

AE

L

AE

K e

400

600

200

600

04

006

002

006

0

0000000000

600

1200

600

1200

06

0012

006

0012

0

0000000000

2

00

6

00

4

00

6

00

02

006

004

006

0

0000000000

600

1200

600

120

06

00126

0012

0

0000000000

22

22

2323

2323

22

22

2323

2323

-

--

---

THREE MOMENT EQUATION







THREE MOMENT EQUATION(Developed by clapeyron)

Continuity condition

Equating the above equations

Using second moment-area theorem

R

C R

L

C L

L Ltantan ∆=∆

R R

R R

L L

L L

R

R

R

C

R

R

L

L

L

L

L

EI L

A x

EI L

A xM

EI

LM

EI

L

EI

LM

EI

L 662 −−=+

++

++=∆ L L L LC L L L

L

C LLM L LM L A x

EL 21

31

21

321

tan

++=∆

R R R RC R R R

R

C RLM L LM L A x

EL 2

1

3

1

2

1

3

21tan



THREE MOMENT THEOREM

+=+++

2

22

1

11

22116)(2

L

x A

L

x A LM L LM LM C B A

+=

+

++

22

22

11

11

2

2

2

2

1

1

1

1 62 I L

x A

I L

x A

I

LM

I

L

I

LM

I

LM

C B A

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Aircraft Design Day5