Title: Chapter 11 Energy Method
1Chapter 11 Energy Method
-- Utilize the Energy Method to solve
engineering mechanics problems. -- Set aside
the Equations of quilibrium
21. Introduction
The relations between forces and deformation
Stress -- Ch 1
Fundamental concept of
Strain Ch 2
Strain Energy Ch 11
We will learn
- Modulus of Toughness
- Modulus of resilience
- 3. Castigliano Theorem
311.2 Strain Energy
(11.1)
(11.2)
4If the material response is elastic
(11.3)
511.3 Strain-Energy Density
(11.4)
Modulus of Toughness Toughness area
under the ?-? curve.
6(11.5)
(11.6)
(11.7)
(11.8)
Modulus of Resilience
711.4 Elastic strain Energy for Normal Stresses
(11.9)
8(11.9)
(11.11)
(11.12)
9(11.13)
1011.5 Elastic Strain Energy for Shearing
Stresses
(11.18)
(11.19)
(11.20)
(11.21)
11Strain Energy in Torsion
(11.19)
(11.21)
(11.22)
1211.6 Strain Energy for a General State of Stress
(11.25)
From Eq. (2.38)
(2.38)
(11.26)
13(11.26)
If the principal stresses are used
(11.27)
Where ?a, ?b, ?c the principal stresses
14(11.28)
Where uv the part of energy leading to volume
change hydrostatic stress ud
deviatoric energy the part of energy leading
to shape change.
Defining Mean Stress or Average Stress
(11.29)
15And set
(11.30)
where
mean stress
deviatoric stress
Combining Eqs. (11-29) and (11.30)
(11.31)
16(11.31)
-- They only change the shape, but do not lead to
the change of the volume of the material.
The dilatation, ?V/V, caused by the state of
stress can be obtained, via Eq. (2.31)
as
or e 0
17The uv can be obtained by substituting
into Eq. (11.27)
to obtain
By means of Eq. (11.29) we have
(11.32)
18The distortion or deviatoric energy can be
obtained as
After simplification
Recalling Eq. (2.43)
(2.43)
or
19Hence, the previous equation takes a new form
(11.33)
For 2-D cases,
Eq. (11.33) reduces to
(11.34)
For uniaxial tension, i.e. 1-D cases, ?b 0 and
?a ?y
20Substituting this equation to the previous
equation, it leads to
(7.26)
Expanding the same operation to a 3-D case, one
can have
(11.35)
Replacing lt by , it follows
(11.36)
This is a circular cylinder of radius
21The 2- D Yield Locus
(7.26)
22The 3- D Yield Locus
(11.35)
2311.7 Impact Loading
K. E. of the ball K. E.
The strain energy in the bar
(11.37)
Assuming 1. No heat dissipation 2. The
ball sticks with the rod after impact.
(11.38)
24If the stress is uniform within the rod
Therefore, ?m can be determined as
(11.39)
2511.8 Design for Impact Loads
Case A For a Uniform-Diameter Rod
(11.45a)
Case B For a Multiple-Section Rod
(11.45b)
26Case C For a Circular Cantilever Rod
(11.44)
However,
Hence,
Eq. (11.44) can be reduced to
(11.45c)
27Since
We conclude that --- in order to develop
lower ?m in the rod, the rod should have 1.
Lower E 2. Larger V 3. Uniform ?m
2811.9 Work and Energy under a Single Load
Case A For an Uniaxial Load
(11.2)
(11.3)
Case B For a Cantilever Beam
(11.46)
29Case C For a Beam in Bending
(11.47)
(11.48)
Case D For a Beam in Torsion
(11.49)
3011.10 Deflection under a Single Load by the
Work-Energy Method
Case A For an Uniaxial Load
(11.3)
where x1 deflection due to P1
Case B For a Beam in Bending
(11.47)
where ?1 deflection due to single moment M1
3111.11 Work and Energy under Several Loads
Deflection due to P1
(11.54)
Deflection due to P2
(11.54)
?ij influence coefficients
The combining effect of P1 and P2
(11.54)
(11.54)
32Calculating the Work Done by P1 and P2
Case I Assuming P1 is applied first ---
At Point 1, the work done by P1 is
(11.58)
At Point 2, the work done by P1 is zero.
P2 is applied next ---
At Point 2, the work done by P2 is
(11.59)
33At Point 1, the work done by P1 due to additional
defection caused by P2 is
(11.60)
(11.58) (11.59) (11.60), the total strain
energy is
(11.61)
34Case II Assuming P2 is applied first ---
(11.62)
Equating Eqs. (11.61) and (11.62) leads to
-- Maxwell Reciprocal Theorem
3511.12 Castiglianos Theorem
(11.61)
(11.63)
(11.64)
Or, in general
Castiglianos Theorem
(11.65)
(11.66)
36General Formulation for Castiglianos Theorem
For multiple loading, P1, P2, ., Pn the
deflection of the point of application of Pi can
be expressed as
(11.66)
The total strain energy of the structure is
(11.67)
Differentiating U w.r.to Pj
37Since ?ij ?ji, the above equation becomes
(11.65)
-- for concentrated loads
-- for moment loads
-- for torsion
3811.13 Deflections by Castiglianos Theorem
Total strain energy of a beam subjected to bending
But, the differentiation can be applied prior to
integration.
(11.17)
Deflection at point Pj
(11.70)
Total strain energy of a truss member
(11.71)
Deflection at point Pj
(11.72)
39If no load is applied to a point, where we desire
to obtain a deflection
-- Apply a dummy (fictitious) load Q at that
point, determine
Castiglianos Theoem
(11.76)
Then, set Q 0.
4011.14 Statically Indeterminate Structures
Structure indeterminate to the 1st degree
Procedures 1. Assume one support as
redundant 2. Replace it with an unknown force
3. y ?U/?RA 0 ? solving for RA
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