Aircraft Design Day5

DAY 5

STRESS • Stress is a measure of force per unit area within a body. • It is a body's internal distribution of force per area that reacts to external applied loads.

P STRESSσ = A

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 • Relation between Engineering & Nominal stress

σ true = (1 + ε e )σ e

TYPES OF STRESSES

COMPRESSIVE

TENSILE

BENDING

SHEAR

TORSION

SHEAR STRESS dx 2 1

τ

z

A

2 1

dz

TORSION

D

τ zdzdy

z

τ zdzdy

Bτ

τ xdxdy

τ xdxdy

C Taking moment about CD, We get

τ z dzdy( dx ) = τ x dxdy( dz )

τz = τx 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

σy

τ yx

σx

τ xy

τ xy

σx

τ yx

σy

• Principal stress

σ 1, 2

σx +σ y σ x −σ y  2  + τ xy = ±  2 2  

THREE DIMENSIONAL STRESS • Cauchy stress – Force per unit area in the deformed geometry σ xx τ xy τ xz    σ ij = τ yx σ yy τ yz  τ τ σ  zz   zx zy

• Second Piola Kirchoff stress – Relates forces in the reference configuration to area in the reference configuration

S

IJ

= JX

τ X I, j ij J, i

X – Deformation gradient

3D PRINCIPAL STRESS • Stress invariants of the Cauchy stress I1 = σ x + σ y + σ z 2 2 2 I 2 = σ xσ y + σ yσ z + σ zσ x − τ xy − τ yz − τ zx I 3 = σ xσ yσ z + 2τ xyτ yzτ zx − σ τ − σ τ − σ τ 2 x yz

2 y zx

2 z xy

• Characteristic equation of 3D principal stress is

σ − I1σ + I 2σ − I 3 = 0 3

2

• Invariants in terms of principal stress I1 = σ 1 + σ 2 + σ 3 I 2 = σ 1σ 2 + σ 2σ 3 + σ 3σ 1 I 3 = σ 1σ 2σ 3

VON-MISES STRESS • Based on distortional energy σv =

(σ 1 − σ 2 )

2

+ (σ 2 − σ 3 ) + (σ 3 − σ 1 ) 2 2

2

1 2 2 2 (σ x − σ y ) + (σ y − σ z ) + ( σ z − σ x ) + 6(τ xy2 + τ yz2 + τ zx2 ) σv = 2

STRAIN • Strain is the geometrical expression of deformation caused by the action of stress on a physical body. Strain

δL ε= L

• Strain – displacement relations Normal Strain ∂u ε x= ∂x

∂v ε y= ∂y

∂w ε z= ∂z

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) ∂u ∂v ∂v ∂w γ xy= + γ yz= + ∂y ∂x ∂z ∂y

∂w ∂u γ zx= + ∂x ∂z

VOLUMETRIC STRAIN • Volumetric strain V − V0 υ= V0

υ =ε x+ ε y+ ε z

TWO DIMENSIONAL STRAIN • Plane strain

εx

γ xy

εy

γ yx

εy

γ xy

εx

γ yx

• Principal strain

ε x+ε y  ε x−ε y   γ xy   +   ε 1, 2= ±  2  2   2 

3D STRAIN

Strain tensor

 ε xx  γ yx ε ij=  2  γ zx  2

γ xy 2

ε yy γ zy 2

γ xz  2  γ yz  2   ε zz  

Green Lagrangian Strain tensor E = 1 ( F F − δ ) ij ij 2 ki kj 1  ∂ui ∂u j ∂uk ∂uk  =  + + 2  ∂x j ∂xi ∂xi ∂x j  Almansi Strain tensor

(

1 -1 −1 E = δ ij − Fki Fkj ij 2

)

STRESS-STRAIN CURVE

Mild steel

Thermoplastic

Copper

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 LONGEST DIMENSION

TERMINOLOGY • SHEAR FORCE – A shear force in structural mechanics is an example of an internal force that is induced in a restrained structural 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 a restrained structural element when external forces are 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

A

C

B

D

BEAM

•TYPES OF BENDING Hogging Sagging

SHEAR FORCE & BENDING MOMENT

BEAM

P

P PL/8

L

P/2 P/2 FREE BODY DIAGRAM

SHEAR FORCE P/2 P/2 BENDING MOMENT

PL/8

PL/8

PL/8

SHEAR FORCE & BENDING MOMENT

BEAM

P

P 3PL/8

L

11P/16 5P/16 FREE BODY DIAGRAM

SHEAR FORCE 11P/16 5P/16 BENDING MOMENT

3PL/8

RELATION BETWEEN BM, SHEAR & LOAD w

M

V+dV

V O

M+dM

dx Taking moments about O

∑M

O

=0⇒

 dx  M − ( M + dM ) + Vdx − wdx  = 0  2 dM V= dx Force equilibrium gives

V − (V + dV ) + w * dx = 0 dV w= dx

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 STRESS o

M

R c a

M d ef b

Change in length ef strain (ε ) = = Original length cd

... (1)

From similar triangles edf & cod ef de = ... (2) cd co ef y strain (ε ) = = ...(3) cd R

Hooks law

strain (ε ) =

f (or ) σ ...(4) E

From (3) & (4)

f E = ...(5) y R

Ey dF = stress * area = dA ... (6) R

 Ey  dM = y * dF = y dA ... (7)  R E  EI  M =   ∫ y 2dA =   ... (8)  R  R M E = ... (9) I R

From (5) & (9) M E f = = ... (9) I R y

FINITE ELEMENTS • TRUSS / BAR / LINK ELEMENT • BEAM ELEMENT

3D BEAM ELEMENT

3D BEAM ELEMENT q q t s t x(r , s, t ) =∑ hk l xk + ∑ ak hk lVtxk + ∑ bk hk lVsxk 2 qk =1 2 qk =1 k =1 q t s t l l k y (r , s, t ) =∑ hk yk + ∑ ak hk Vty + ∑ bk hk lVsyk ……… (1) 2 kq=1 2 kq=1 k =1 q t s t l l k z (r , s, t ) =∑ hk z k + ∑ ak hk Vtz + ∑ bk hk lVszk 2 k =1 2 k =1 k =1 q

x, y, z = Cartesian coordinate of any point in the element l l l xk , y k , z k = Cartesian coordinate of any nodal point k a k , bk = Cross sectional dimensions of the beam at nodal point k l V V Vtxk , l Vtyk , l Vtzk = Components of unit vector V in direction t at nodal point k l Vsxk , l Vsyk , l Vszk = Components of unit vector V in direction s at nodal point k l

l

l

l

l

k

s

k

t

l

l

k

s

k

s

l

k

We call Vt and lVsk the normal vectors or director vectors at nodal point k

3D BEAM ELEMENT The displacement components are

x ( r , s , t ) =1 x − 0 x y ( r , s , t ) =1 y − 0 y z ( r , s , t ) =1 z − 0 z

……… (2)

From (1) & (2) we get t q s q k x(r , s, t ) = ∑ hk xk + ∑ ak hkVtx + ∑ bk hkVsxk 2 qk =1 2 qk =1 k =1 q t s k y (r , s, t ) = ∑ hk y k + ∑ ak hkVty + ∑ bk hkVsyk 2 kq=1 2 kq=1 k =1 q t s z (r , s, t ) = ∑ hk z k + ∑ ak hkVtzk + ∑ bk hkVszk 2 k =1 2 k =1 k =1 q

Vt = Vt − Vt k 1 k 0 k Vs = Vs − Vs k

1

k

0

k

……… (3)

3D BEAM ELEMENT Strain displacement relation

ε η η q γ η ς  = ∑ Bk uˆ k γ  k =1  η ξ

where

uˆ k = [u k vk wk θ xk θ yk θ zk ]

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

B = [ B1 . . . . Bq ]

  ∂hr k k k   ∂u  [ ] 1 (g) (g) (g) 1i 2i 3i    ∂r uk   ∂r        θ k    q  ∂u  x k k k    ˆ ˆ ˆ [ ] = h 1 ( g ) ( g ) ( g )   ∑ k 1i 2i 3i  θ k  ∂ s k =1     y       k k k   ∂u  [ ] h 1 ( g ) ( g ) ( g ) 1i 2i 3i   k   k θz  ∂t     

3D BEAM ELEMENT ( gˆ )

k

(g)

k

 0 - 0 Vszk 0Vsyk    bk  0 k = Vsz 0 - 0 Vsxk   2  - 0 V k 0V k  0 sx  sy   0 - 0 Vtzk 0Vtyk    ak  0 k = Vtz 0 - 0 Vtxk   2   - 0 V k 0V k  0 tx  ty 

( g)

k ij

= s( gˆ ) ij + t ( g ) ij k

Jacobian Transformation

k

∂ −1 ∂ =J ∂x ∂ξ

3D BEAM ELEMENT Strain displacement relation  ∂u   −1 ∂hk  ∂x   J 11 ∂r     ∂u  q  ∂h   = ∑  J 21−1 k ∂r  ∂y  k =1      ∂u   J 31−1 ∂hk  ∂z   ∂r

 (G1 )ik1 (G 2 )ik1 (G3 )ik1  u k     k   θ x  k k k (G1 )i 2 (G 2 )i 2 (G3 )i 2     θ yk    (G1 )ik3 (G 2 )ik3 (G3 )ik3   k   θ z 

Where ( Gm ) k = [ J −1 ( g k ) ] ∂hk + [ J −1 ( gˆ k ) + J −1 ( g k ) ] h in n1 mi n2 mi n3 mi k

∂r

Stiffness

l K = ∫ B T DB 2 dξ 2 −1

Load

l2 f = ∫ B N dξ + f KT 2 −1

1

1

T

STIFFNESS MATRIX  AE  L   0   0    0   0    0 Ke =  AE −  L  0    0   0    0   0 

AE L

0

0

0

12 EI L3

0

0

12 EI L3

6 EI L2

0

0

6 EI L2

0

0

0

0

0

0

0

0

0

6 EI L2

0

0

0

0

6 EI L2

2 EI L

0

0

0

0

0

0

0

0 0

0 0

-

0

−

0

−

0

0

0

0

12 EI L3

0

0

12 EI L3

6 EI L2

0

0

GJ L

0

−

0

0

6 EI L2

GJ L

0

0

0

6 EI L2

4 EI L

0

0

4 EI L

0

0

0

0

0

12 EI L3

0

0

0

0

12 EI L3

6 EI L2

AE L

0

0

0

0

GJ L

6 EI L2

12 EI L3

12 EI L3

0

0

0

0

0

0

0

0

0

2 EI L

6 EI L2

GJ L

0

6 EI L2

4 EI L

0

0

0

-

0

0

6 EI L2

0

0

6 EI L2

2 EI L

0

0

0

-

-

-

−

−

0

−

-

6 EI L2

 0   0   6 EI  L2   0   0   2 EI  L   0   0   6 EI   L2  0    0  4 EI   L 

THREE MOMENT EQUATION

THREE MOMENT EQUATION (Developed by clapeyron) Continuity condition

∆ L tan C ∆ R tan C = LL LR

Using second moment-area theorem ∆ L tan C = ∆ R tan C

1 ELL

2 1 1 1   x A + L M L + L M L  L L L C L L L L 3 2 3 2  

1  2 1 1 1  =  xR AR + LR M C LR + LR M R LR  ELR  3 2 3 2 

Equating the above equations  LL LL LR  LR 6 xL AL 6 xR AR   M L + 2 + MC + MR = − −  EI L EI R LL EI L LR EI R  EI L EI R 

THREE MOMENT THEOREM  A1 x1 A2 x2   M A L1 + 2 M B ( L1 + L2 ) + M C L2 = 6 + L2   L1  L1   L1 L2   L2   A1 x1 A2 x2   M A   + 2 M B  +  + M C   = 6 +  I1   I1 I 2   I 2   L1 I1 L2 I 2 