lecture-20_ generalized forces and generalized displacements_ strain energy of the rod in general...
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V. Demenko Mechanics of Materials 2012
1
LECTURE 20 Generalized Forces and Generalized Displacements.
Strain Energy of the Rod in General Case of Loading
Generalized force nF is a force factor in general (F, q, M,.....).
Generalized displacement nδ is a geometrical parameter on which the
generalized force nF does work. If, for example, nF is understood to be an
external moment M , then nδ represents an angular displacement at the moment
application point in the direction of the moment.
1 Displacements of a Rod under Arbitrary Loading
1.1 The work done by external force
(a) The work done by a concentrated force:
Fig. 1
In general case the elementary work dA is equal to F⋅ dΔ ,i.e.
Δ⋅= dFdA . (1)
By summing these quantities we obtain
∫Δ
Δ=kFdA
0. (2)
If F = cΔ (within elasticity limitations (Fig. 2)), then
V. Demenko Mechanics of Materials 2012
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Fig. 2
2222
2
0
2
0
Δ=
ΔΔ=
Δ=
Δ=ΔΔ=
ΔΔ
∫FcccdcA . (3)
The work done by concentrated force is equal to the area of the triangle OAB.
(b) The work done by a couple of forces:
Fig. 3
The displacements of the points A and 1A produced by the couple of forces are
Θ=Δ2h .
The work done by a couple of forces is
242
22 Θ
=Θ
=Δ
=FhFhFA .
But MhF =⋅ , then
2Θ
=MA . (4)
V. Demenko Mechanics of Materials 2012
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1.2 The work done by internal forces. Potential energy of strain (strain
energy)
It is evident, that internal forces really describe the stresses acting at any
point of elastic solid (see chapter 2 in Lecture 4).
The work done by internal forces is equal to potential energy of strain
stored in the element (within elasticity limitations): U = A.
(a) Determine the potential energy of strain in tension (compression)
σ2 = σ3 = 0.
Fig. 5
It is known (see Eq. 34 in Lecture 9) that
( )[ ]32312123
22
210 2
2σσσσσσμσσσ ++−++==
EdVdVUdU .
In this case, σ1 = σ, σ2 = σ3 = 0.
As a result, dVE
dU2
2σ= . (5)
Strain energy density E
U2
20
σ= . (6)
(b) Determine the potential energy of strain in pure shear
Fig. 6
V. Demenko Mechanics of Materials 2012
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[ ] ( ) =+=⋅−−+== μτττμττ 12
2)(221 222
0 EdVdV
EdVUdU
( )( ) dV
GGE
EdV
21212
2
22 τμ
μ
τ=
⎭⎬⎫
⎩⎨⎧
=+
=
+
= , (7)
In this case, G
U2
20
τ= . (8)
It should be observed here that within the elementary volume of solid its stress
state is homogeneous.
(c) Determine the strain energy of a rod in general case of loading:
Fig.7
dxdAdV ⋅= – elementary (unit) volume
To determine potential energy, let’s isolate an elementary portion of length dx
from a rod. In general, six internal force factors occur at each cross section:
three moments and three forces ),, and ,,( yzxzyx QQNMMM .The potential
energy of the element may be regarded as the sum of work done by each of the
six force factors acting separately
( ) ( )++== yx QdUNdUdUdA ( ) ( ) ( )++++ yxz MdUMdUQdU ( )zMdU+ . (9)
V. Demenko Mechanics of Materials 2012
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As it is known
dxdAUdVUdUdVdUU
A⎟⎟⎠
⎞⎜⎜⎝
⎛==→= ∫ 000 . (10)
Then
(1) =⎭⎬⎫
⎩⎨⎧ ==⎟
⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡= ∫∫ A
NdxdAE
dxdANUNdU x
AAxx σσ
2)()(
20
dxEA
NdxEA
dAN x
A
x22
2
2
2== ∫ . (11)
(2) =⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡= ∫∫ )(2
)()(*2
0 zbISQ
dxdAG
dxdAQUQdUy
yz
AAzz ττ
dxGA
QKdxdAzb
S
IA
GAQdxdA
zbGI
SQ zz
A
y
y
z
A y
yz2)]([2)]([2
2
2
2*
2
2
22
2*2=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛∫∫ . (12)
(3) By analogy: dxGA
QKQdU y
yy 2)(
2= . (13)
(4) =⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
==⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
= ∫∫ρ
ρττI
MdxdAG
dxdAMUMdU x
AAxx 2)()(
20
dxGIMIdAdxdA
GIMdxdA
GIM x
AA
x
A
xρ
ρρρ
ρρρ222
222
2
2
2
22=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
==⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛= ∫∫∫ .
( ) dxGIMMdU x
xρ
=2
2. (14)
(5) =⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
==⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎟⎠
⎞⎜⎜⎝
⎛= ∫∫
y
y
AAyy I
zMdxdA
EdxdAMUMdU σσ
2)()(
20
Kz
V. Demenko Mechanics of Materials 2012
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dxEI
MdxdA
EI
zM
y
y
A y
y22
2
2
22=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛= ∫ . (15)
(6) By analogy, ( ) dxEI
MMdUz
zz 2
2= . (16)
Now expression (9) becomes
zz
y
y
pxz
zy
yx
EIdxM
EIdxM
GIdxM
GAdxQK
GAdxQ
KEA
dxNdU222222
222222+++++= . (17)
In order to obtain the potential energy of the whole rod this expression
must be integrated over the length of the rod
dxEI
MdxEIM
dxGIMdx
GAQKdx
GAQK
dxEA
NUl z
z
l y
y
l
x
l
zz
l
yy
l
x ∫∫∫∫∫∫ +++++=222222
222222
ρ. (18)
All the terms in expression (18) are not always equivalent. In the great
majority of systems encountered in practice with components acting in bending
or torsion, the last three terms in expression (18) are less appreciable than the
first three. In other words, the energy due to tension and shear is considerably
less then the energy due to bending and torsion.