Energy and complementary energy
•In linear range U=C
Virtual displacements and generalized
coordinates
• System with n degrees of freedom under virtual
displacement
( x1 , x2 , … , xn ) ⇒ ( x1 + δ x1 , x2 + δ x2 , … , xn + δ xn )
• Generalized coordinates will typically be
distances or angles
• Generalized forces are defined as the
complementary quantity that when multiplied
by generalized displacement gives the
resulting work
δ W = Q1δ x1 + Q2δ x2 +
+ Qnδ xn
Conservative system
• Internal and external virtual work
δ W = δ We + δ Wi
δ Wi = −δ U =
∂U
∂U
δ x1 +
δ x2 +
∂x1
∂x2
δ We = P1δ x1 + P2δ x2 +
+
∂U
δ xn
∂xn
+ Pnδ xn
• Altogether
Q1δ x1 + Q2δ x2 +
−
Qi = Pi −
+ Qnδ xn = P1δ x1 + P2δ x2 +
∂U
∂U
δ x1 −
δ x2 +
∂x1
∂x2
∂U
∂xi
−
∂U
δ xn
∂xn
+ Pnδ xn
•
Principle of stationary potential
energy
In equilibrium δ W = 0 known as principle of
stationary potential energy
• Then
∂U
Pi =
∂xi
• Known as Castigliano’s first theorem (Carlo
Alberto Castigliano, 1847-1884)
Strain energy in Axial member
.
In an element with a nonuniform stress distribution
∆U dU
U = ∫ u dV = total strain energy
=
∆V→0 ∆V
dV
σ2x
P2
U = ∫ dV = ∫
dx since dV = A dx
2E
2AE
For a rod of uniform cross-section, under axial loading,
u = lim
σx = P A,
P2 L
dV = A dx → U =
2AE
Three-bar truss example
• Strains
• Strains
and stresses
v
εB =
L
•Strain energy
EA
nB =
v
L
(
⎛ E v + 3u
⎛ EA ⎞ L
U =⎜
v⎟
+⎜
4L
⎝ L ⎠ 2 AE ⎜
⎝
2
•Equilibrium
εA
)
2
v+
(
=
nA =
(
⎞
⎛ E v − 3u
L
(2
)
⎟
+⎜
⎟ 2 AE ⎜
4L
⎠
⎝
EA
∂U
= 0.75
u= p
∂u
L
)
(
3u
4L
) ε = (v −
E v + 3u
C
)
4L
3u
4L
nC =
)
(
E v − 3u
4L
⎞
EA (1.25v 2 + 0.75u 2 )
L
(2
)
⎟
=
⎟ 2 AE
2L
⎠
2
EA
∂U
= 1.25
v= p
∂v
L
)
Castigliano’s (second theorem)
• If an elastic system is supported so that rigidbody displacements of the system are
prevented, and if certain concentrated forces of
magnitude F1, F2,…, Fp act on the system, in
addition to distribted loads and thermal strains,
the displacement component qi of the point of
application of the force Fi is determined by the
equation
∂
C
• How do we calculate C?
q =
i
∂Fi
Beam in bending
.
σ 2x
My
M 2 y2
σx =
→ U = ∫ dV = ∫
dV dV = A dx
2
I
2E
2EI
L
L
L
M 2 y2
M2 ⎛ 2 ⎞
M2
U=∫ ∫
dA dx = ∫
y dA ⎟ dx = ∫
dx
2
2 ⎜∫
2EI
2EI ⎝ A
2EI
⎠
0 A
0
0
Simplest example
• What is the rotation of the end of a cantilever
beam under an end moment M
• Strain energy equal to complementary strain
energy
2
M L
C =U =
2 EI
• Rotation at tip
∂C ML
θ=
=
EI
∂M
• What do we do if we want displacement at tip?
Shear strain energy
γ xy
u=
.
∫τ
xy
dγ xy shear in plane stress
0
Within the proportional limit:
u = 12 Gγ 2xy = 12 τ xy γ xy =
τ2xy
2G
The total strain energy :U = ∫ u dV = ∫
τ2xy
2G
dV
For a shaft subjected to a torsional load,
U=∫
τ2xy
2G
dV
Tρ
T 2ρ2
Since τ xy =
dV dV = dA dx,
→U=∫
J
2GJ 2
L
L
L
T 2ρ 2
T2 ⎛ 2 ⎞
T2
U =∫∫
dA dx = ∫
dx
ρ dA ⎟ dx = ∫
2
2 ⎜∫
2GJ
2GJ
2GJ
⎝A
⎠
0A
0
0
T2L
Uniform shaft, U =
2GJ
Example
• The polar moment of inertia of a shaft is reduced by a
factor of two from root to tip as J=Jo/(1+x/L). Calculate
the percent reduction in weight and percent increase in
twist angle at the tip. Compare to uniform shaft of the
same volume.
• Area: The area is proportional to the square of the
radius while the moment of inertia is proportional to the
4th power. So the area at the tip is 70.7% of the area at
the root. Integrating, we find that the volume is reduced
by 17.2%.
• Twist angle: 1/J increases linearly to double at the tip,
so the twist angle increases by 50%.
• For uniform shaft, a reduction in volume of 17.2% will
reduce J to (1-0.172)2=0.686Jo, leading to a twist angle
increase by 1/0.686=1.46, only 46%.
Transverse shear
τ2
2
1 ⎛ VQ ⎞
U =∫
dV = ∫
⎜
⎟ dAdx
2G
2G ⎝ It ⎠
V
V
.
⎞
V2 ⎛ Q2
dA ⎟ dx
= ∫
2 ⎜∫ 2
2GI
t
0
⎝A
⎠
L
⎞
V 2 ⎛ A Q2
dA
= ∫
⎜ 2∫ 2
⎟ dx
2GA ⎝ I A t
0
⎠
L
L
kV 2 dx
,
=∫
2
GA
0
Cross Section
k
Rectangle
Solid Circular
1.20
1.33
Thin Walled Circular
2.00
I-Section, Box, Channel 1.0
where k = correction factor for shear
Reading assignment
Section 5.3-4: Question: What is “dummy” about “dummy load
method?”
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