2.001 - MECHANICS AND MATERIALS I
Lecture #26
12/11/2006
Prof. Carol Livermore
Energy Methods
3 Basic Ingredients of Mechanics:
1. Equilibrium
2. Constitutive Relations
σ − . . . Stress-Strain
F − δ Force-Deformation
3. Compatibility
Dealing with geometric considerations
Castigliano’s Theorem
Start with work:
F · ds
W =
Conservative Forces:
EX: Gravity
1
u = mgh ⇒ Potential Energy, Path Independent
h
W = F · ds =
mgdz = mgh
0
Note: Elastic systems are conservative.
Plastic deformation is not conservative.
EX: Springs
Energy = 1/2kδ 2
W =
δ
δ
P dδ =
kδdδ =
0
0
Stored Energy = Work Put In
U =W
What if the spring were stretched a little further?
2
1 2
kδ
2
dU = P dδ
dU
P =
dδ
Complementary Energy (U ∗ )
∗
U =
δdP
Note: U = U ∗ due to linearity
U=
U ∗ if this is non-linear
What if the load were stretched a little further?
3
dU ∗ = δdP
dU ∗
δ=
dP
Recall, for linear elastic material
δ=
dU
dP
Castigliano’s Theorem:
Express complementary energy in terms of the loads
Add up all the complementary energy stroed in all the pieces
Find displacement of given point from derivative of U ∗ with respect to P at
that point in that direction
Example: Linear Spring
U∗ =
δdP =
k2 δ2
1
P
P2
dP =
=
= kδ 2
k
2k
2k
2
Example: Springs in series
F1 = F2 = P
U=
Ui =
i
F12
F2
P2
P2
+ 2 =
+
2k2
2k1
2k2
2k1
Use Castigliano’s Theorem
δ=
P
P
U∗
=
+
dP
k1
k2
4
Find δ1 using Castigliano’s Theorem
Put a fictitous Pf to coinside with δ1
Fx = 0
−F1 + Pf + P = 0 ⇒ F1 = Pf + P
−F2 + P = 0 ⇒ F2 = P
U=
F2
(Pf + P )2
P2
F12
+ 2 =
+
2k1
2k2
2k1
2k2
δ1 =
∂U
Pf + P
=
∂Pf
k1
Recall Pf =0, so:
δ1 =
δ=
P
k1
dU
Pf + P
P
=
+
dP
k1
k2
Pf = 0
So:
δ=
P
P
+
k2
k1
Note: This was done without doing compatibility explicitly.
Example Truss
5
U=
Ui =
i
2
FAB
F2
+ BC
2kAB
2kBC
Equilibrium of B
Fx = 0
FAB − FBC sin θ = 0
Fy = 0
−P − FBC cos θ = 0
P = −FBC cos θ
−P
cos θ
= P tan θ
FBC =
FAB
U=
(P tan θ)2
+
2kAB
−uB
y =
P
cos θ
2
1
2kBC
dU
P tan2 θ
P
=
+
dP
kAB
cos2 θkBC
Strain Energy Density
W =
F dδ
u=
σd
This is useful in beam bending.
6
Stored Energy in A Beaam (σxx , xx )
Total Energy Stored
U=
dV
V
0
σxx dxx
Recall:
For a beam:
σxx = Exx
For one material:
σxx =
−M y
I
Beam:
U
=
dV
σxx dxx
σxx
σxx dσxx
=
dV
E
V
0
2
σxx
=
dV
2E
V
V
For one material:
U=
V
M (x)2 2
y dV
2EI 2
So:
U=
1
2EI
M (x)2 dx For special case of one material beam.
a
7
Example:
M (x) = −P (L − x)
L
P 2 L3
1
U=
P 2 (L − x)2 dx =
2EI 0
6EI
δtip =
dU
P L3
=
dP
3EI
8