Energy Methods
• We have seen how to find the strain in axial loaded bar,
deflection of points on a beam etc.
• Same can be done using energy methods
• We have to develop a relation between work done by
externally applied forces and strain energy of the
deforming body
Work done
• A uniform rod is subjected to a slowly increasing load
• The elementary work done by the load P as the rod
elongates by a small dx is
dU P dx elementary work
which is equal to the area of width dx under
the load-deformation diagram
Work done
• The total work done by the load for a deformation x1,
x1
U P dx total work strain energy
0
which results in an increase of strain energy in the rod.
• For linear elastic deformation
x1
U kx dx 12 kx12 12 P1x1
0
k= ??
Strain energy
• When loads are applied to a body, they will deform the
material. Provided no energy is lost in the form of heat,
the external work done by the loads will be converted
into internal work called strain energy.
• This energy, which is always positive, is stored in the
body and is caused by the action of either normal or
shear stress.
Strain energy density
• To remove the size effects due to specimen geometry, we
calculate the energy density
x1
U P dx
V
A L
0
1
u x d strain energy density
0
• The total strain energy density resulting from the deformation is equal to the
area under the curve up to 1.
• As the material is unloaded, the stress returns to zero but there is a
permanent deformation. Only the strain energy represented by the
triangular area is recovered.
• Remainder of the energy spent in deforming the material is dissipated as heat.
Strain energy density
• The strain energy density resulting from
setting 1 R is the modulus of toughness.
• The energy per unit volume required to cause
the material to rupture is related to its
ductility as well as its ultimate strength.
• If the stress remains within the proportional
limit,
1
E12 12
u E1 d x
2 2E
0
• The strain energy density resulting from
setting 1 Y is the modulus of resilience.
Y2
uY modulus of resilience
2E
Elastic Strain Energy for Normal Stresses
• In an element with a nonuniform stress
distribution,
U dU
u lim U u dV total strain energy
V 0 V dV
• For values of u < uY , i.e., below the proportional
limit,
x2
U dV elastic strain energy
2E
• Under axial loading, x P A dV A dx
L
P2
U dx
2 AE
0
• For a rod of uniform cross-section,
P2L
U
2 AE
Elastic Strain Energy for Normal Stresses
• For a beam subjected to a bending load,
x2 M 2 y2
U dV 2
dV
2E 2 EI
• Setting dV = dA dx,
M 2 2
L L
M 2 y2
My U dA dx 2
y dA dx
x 2
I 0 A 2 EI 0 2 EI A
L
M2
dx
2 EI
0
• For an end-loaded cantilever beam,
M Px
L
P2 x2 P 2 L3
U dx
2 EI 6 EI
0
Strain Energy For Shearing Stresses
• For a material subjected to plane shearing
stresses,
xy
u xy d xy
0
• For values of xy within the proportional limit,
2
2 xy
u 12 G xy 12 xy xy
2G
• The total strain energy is found from
U u dV
2
xy
dV
2G
Strain Energy For Shearing Stresses
• For a shaft subjected to a torsional load,
2
xy T 2 2
U dV 2
dV
2G 2GJ
• Setting dV = dA dx,
T 2 2
L L
T 2 2
T U dA dx 2
dA dx
xy 2GJ 2
2GJ
J 0A 0 A
L
T2
dx
2GJ
0
• In the case of a uniform shaft,
T 2L
U
2GJ
Work and Energy Under a Single Load
• Strain energy may be found from the work of other types
of single concentrated loads
• Transverse load • Bending couple • Torsional couple
y1 1 1
U P dy 12 P1 y1 U M d 12 M11 U T d 12 T11
0 0 0
3 2 3
1 P P1L P1 L 1 M M1L M12 L 2
1 T T1L T1 L
2 1 EI
2 1 JG 2 JG
2 1 3EI 6 EI 2 EI
Strain Energy for a General State of Stress
• Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,
u 12 x x y y z z xy xy yz yz zx zx
• With respect to the principal axes for an elastic, isotropic body,
u
1 2
2E
a b2 c2 2 a b b c c a
uv ud
1 2v
uv a b c 2 due to volume change
6E
ud
1
12G
a b 2 b c 2 c a 2 due to distortion
• Basis for the maximum distortion energy failure criteria,
Y2
ud ud Y for a tensile test specimen
6G
Conservation of energy
• All energy methods used in mechanics are based on a balance
of energy, often referred to as the conservation of energy
• Only mechanical energy will be considered in the energy
balance
– The energy developed by heat, chemical reactions, and
electromagnetic effects will be neglected
• if a loading is applied slowly to a body, then physically the
external loads tend to deform the body so that the loads do
external work Ue as they are displaced
• This external work on the body is transformed into internal
work Ui or strain energy which is stored in the body
Conservation of energy
• When the loads are removed, the strain energy restores
the body back to its original undeformed position,
provided the material’s elastic limit is not exceeded
• The conservation of energy for the body can therefore be
stated mathematically as
• This equation can be applied to determine the
displacement of a point on a deformable member or
structure
Example
• Vertical displacement of truss at point P
External work done by P:
If N is the axial force developed in truss,
then strain energy stored,
Using conservation of energy:
Example
• Vertical displacement at load point P
External work done by P:
Strain energy due to internal bending moment M:
Using conservation of energy:
Impact loading
• So far we have considered all loadings to be applied to a
body in a gradual manner
– when they reach a maximum value the body remains static
• Some loading are dynamic – they vary with time – such
as collision of objects
• This is called an impact loading
– impact occurs when one object strikes another, such that large
forces are developed between the objects during a very short
period of time
– If we assume no energy is lost during impact, due to heat, sound
or localized plastic deformations, then we can study the
mechanics of impact using the conservation of energy
Block and spring system
• When the block is released from rest, it falls a distance h,
striking the spring and compressing it a distance
If we neglect the mass of the spring and assume that the
spring responds elastically, conservation of energy requires
that the energy of the falling block be transformed into
stored (strain) energy in the spring.
External work done by the block’s weight:
Force in the spring F:
Internal work done:
Block and spring system
• According to conservation of energy,
If the weight W is supported statically by the spring, then static displacement
Block and spring system
• Determine the maximum displacement of the end of the spring if
the block is sliding on a smooth horizontal surface with a known
velocity v just before it collides with the spring
In this case the block’s kinetic energy will be
transformed into stored energy in the spring.
Maximum stress
• Once max is determined, the maximum dynamic force can
then be calculated from
• If we consider Pmax to be an equivalent load which produces
same strain energy, then we can calculate maximum stress
using statics
• The ratio of the equivalent static load Pmax to the static load
Pst = W is called the impact factor, n
This factor represents the magnification of
a statically applied load so that it can be
treated dynamically.
Maximum stress
• To determine the maximum stress m
- Assume that the kinetic energy is
transferred entirely to the structure,
U m 12 mv 02
- Assume that the stress-strain
diagram obtained from a static test
is also valid under impact loading.
• Consider a rod which is hit at its
• Maximum value of the strain energy,
end with a body of mass m moving
with a velocity v0. m2
Um dV
2E
• Rod deforms under impact. Stresses • For the case of a uniform rod,
reach a maximum value m and then
disappear. 2U m E mv02 E
m
V V
Example
• Find the static load Pm which produces
the same strain energy as the impact.
Pm2 L 2 Pm2 L 2 5 Pm2 L
Um
AE 4 AE 16 AE
16 U m AE
Pm
5 L
SOLUTION: • Evaluate the maximum stress resulting
• Due to the change in diameter, from the static load Pm
the normal stress distribution is Pm
m
nonuniform. A
16 U m E
U m 12 mv02
5 AL
2 2
m m V
dV 8 mv02 E
2E 2E
5 AL
