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8/14/2019 Aircraft Design Day5
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DAY 5
8/14/2019 Aircraft Design Day5
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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
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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
8/14/2019 Aircraft Design Day5
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TYPES OF STRESSES
TENSILE
BENDING
COMPRESSIVE
SHEAR
TORSION
8/14/2019 Aircraft Design Day5
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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
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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 +
−±
+=
8/14/2019 Aircraft Design Day5
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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
8/14/2019 Aircraft Design Day5
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• 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
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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 +++−+−+−=
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8/14/2019 Aircraft Design Day5
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VOLUMETRIC STRAIN
• Volumetric strain
0
0
V
V V −=υ
z y x ε ε ε υ ++=
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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
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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
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STRESS-STRAIN CURVE
Mild steel
Thermoplastic
Copper
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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
8/14/2019 Aircraft Design Day5
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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
8/14/2019 Aircraft Design Day5
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TYPES OF BEAMS
• CANTILEVER BEAM
• SIMPLY SUPPORTED BEAM
• FIXED-FIXED BEAM• OVER HANGING BEAM
• CONTINUOUS BEAM
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BEAMS (Contd…)
• STATICALLY DETERMINATE• STATICALLY INDETERMINATE
C
BA
D
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BEAM
•TYPES OF BENDINGHoggingSagging
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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
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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
8/14/2019 Aircraft Design Day5
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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*
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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)
8/14/2019 Aircraft Design Day5
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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
==
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FINITE ELEMENTS
• TRUSS / BAR / LINK ELEMENT• BEAM ELEMENT
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3D BEAM ELEMENT
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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
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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
−=
−=
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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=
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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
8/14/2019 Aircraft Design Day5
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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
8/14/2019 Aircraft Design Day5
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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
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THREE MOMENT EQUATION
THREE MOMENT EQUATION
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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
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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