1. Energy Methods
(Chapter 14)
1. Internal Strain Energy Stored and External Work done
2. Conservation of Energy
3. Impact Loading
4. Principle of Virtual Work
5. Castigliano’s Theorem
External Work done
Due to an Axial Load on a Bar
Consider a bar, of length L and cross-sectional area A, to be subjected to an end
axial load P. Let the deformation of end B be ∆1. When the bar is deformed by axial load,
it tends to store energy internally throughout its volume. The externally applied load P,
acting on the bar, does work on the bar dependent on the displacement ∆1 at its end B,
where the load is applied. Let this external work done by the load be designated as ue.
Drawing the force-deformation diagram of the bar, as it is loaded by P.
Applied force P
A
P
F
B
∆1
P End displacement
∆1
∆ d∆
1

2. ∆1
ue = ∫ Fd∆ = External work done
0
Since the force versus the end displacement relationship is linear, F at any displacement ∆
can be represented by
F = k ∆, where k = a constant of proportionality
∆1
∆1 ∆1  k∆2  k∆2 P∆1
u e = ∫ Fd∆ = ∫ k∆d∆ = 
 2  =

1
= , since P = k∆1 (A)
0 0
 0 2 2
The external work done on the bar by P increases from zero to the maximum as the load
P increases from 0 to P (in a linear manner). Therefore the total work done can be
represented by the average magnitude of externally applied force (viz., P/2), multiplied
by the total displacement ∆1 (as given by equation (A)).
Let an additional load P′ be applied to the bar after the load P has caused an end
extension of ∆1 at B. Considering the deformation of the end B of the bar due to the
application of an additional load P′ at B, let the additional deformation of the bar be equal
to ∆′.
Applied force
Area = P′∆1
A F I E
Area = ½ P′∆′
L P′
G H
J
P Area = P∆′
B
∆1
P
∆′ O C End displacement
D
P′ ∆1
2

3. ∆′
Area = ½ P∆1
1
The total external work done u e = ( P + P ′)(∆1 + ∆′)
2
1 1 1 1
= P∆1 + P∆′ + P ′∆1 + P ′∆′
2 2 2 2
Incremental work done
Incremental work
done on the bar on the bar if the load P′ is
when load P was applied to the bar
applied at B, resulting in a
initially displacement ∆′
Considering ∆S OEF and OED, Area of Figure GHIF = Area of Figure CDJH
i.e., P ′∆1 = P∆ ′
1 1 1
∴u e = P∆1 + ( P∆′ + P ′∆1 ) + P ′∆′
2 2 2
1 1 1
= P∆1 + ( P∆′ + P∆′) + P ′∆′
2 2 2
1 1
= P∆1 + P∆′ + P ′∆′
2 2
Additional work
done by P as the
bar deforms by
an additional ∆′
Hence when a bar (having a load P acting on it) is subjected to an additional load P′, then
the work done by (the already acting) P due to the incremental deformation ∆′ (caused by
P′) is equal to P∆′. This is similar, to a suddenly applied load P creating an instantaneous
deformation ∆, producing an external work of P∆′.
3

4. Work done due to an end moment
θ
M1
A
B
Let a moment M be applied to end B of the beam AB. Let the rotation at end B be θ1 due
to M. Since M and θ gradually increase from zero to θ1 (following earlier formulations for
an axially loaded bar),
Moment
M1 θ1
U e = external work = ∫ Mdθ
0
1
= M 1θ1
2
θ
θ1
________________________________________________________________________
Work done due to the externally applied torque T1
T1
4

5. Torque
T1
φ1 φ
φ
u e = ∫ Tdφ
1
0
1
= T1φ1
2
Internal Energy Stored (or Internal Work Done)
Due to an end axial force
The internal strain energy stored in the material is dependent on the amount of stresses
and strains created within the volume of the structure.
z
σz σ
Complementary
σz strain energy
Stress
dz
dy dx
y Strain
dz
dx energy
σz
dy Strain ε
x
w is the σz
displacement P
Z
5

6. The internal strain energy Ui stored within the body is given by
U i = ∫ du i = ∫[(average force on element)] ×(distance moved)
1 
= ∫  σ z dxdy ( ε z d z )
V 
2 
1
= ∫σ z ε z ( dxdydz ) dw
2V = εz
dz
1
=
2V∫σ z ε z dv dw = ε z dz
1 σz 2
σz
= ∫ 2 E dV since ε z = E
2V
Due to Shear Stresses and Strains
z
H G
γzydz
τzy
E F
τyz
γzy
D C y
dz
dx
A dy B
x
U i = ∫ du i = ∫  τ zy dydx (γ zy dz )
1 
 2 
Average force on distance moved
top face, i.e.,
EFGH
[Force on other faces do not do any work since motion of face ABCD is zero]
6

7. 1
2∫
= τ xy γ zy dxdydz
1
= ∫τ xy γ zy dv
2v
1 τ xy τ xy
2
2∫ G
= dv since γ xy =
v
G
Due to a bending moment
M
M
x y
2
 My 
 
σ 2
 I  dxdydz
Ui = ∫ dV = ∫
v
2E V
2E
L
M2 “I” - can be constant or varying
=∫ dx ∫ y 2 dA
0 2 EI 2 A
L L
M 2I M 2 dx
=∫ dx = ∫
0 2 EI 2 0
2 EI
Due to an axial force
N 2
N 
 
σ 2
A
Ui = ∫ dV = ∫   dV
V
2E 2E
N
7

8. 2 2 2
N dV N ( Adx ) N dx
∫ 2 = ∫ 2 =∫
V L L
2A E 2A E 2 AE
Due to a transverse shear force
y
V
z
x
2
τ2 1  VQ 
Ui = ∫ dV = ∫   dAdx
V
2G V
2G  It 
V2  Q2 
L
=∫ 2 ∫ 2
 dA dx

0 2GI  A t 
V  A Q 
L 2 2
=∫  2 ∫ 2 dA dx
I 
0
2GA  A t 
L 2
f V dx
=∫ s , where f s = form factor for shear.
0
2GA
Since shear varies depending on shape of cross - section
[See pages 11 and 12 for additional details concerning the form factor fs]
Due to a torsional moment
T
L
T
8

9. τ2 1  Tρ 
Ui = ∫ dV = ∫  dV
V
2G V
2G  J 
(∫ ρ dA)dx
L 2
T
=∫ 2
0 2GJ 2
L
T 2J T2
=∫ dx = ∫ dx
0 2GJ 2 2GJ
Due to Three dimensional Stresses and Strains
σz
σ3
τyz
τxz
τyx
τxy σy σ2
σx
Fig. 14-5 σ1
Multi-axial Stresses: The previous development may be expanded to determine the
strain energy in a body when it is subjected to a general state of stress, Figure shown
above. The strain energies associated with each of the normal and shear stress
components can be obtained from Eqs. I and II. Since energy is a scalar, the strain energy
in the body is therefore
1 1 1 1 1 1 
U i = ∫  σ x ε x + σ y ε y + σ z ε z + τ xy γ xy + τ yz γ yz + τ xz γ xz  (I)
V 
2 2 2 2 2 2 
The strains can be eliminated by using the generalized form of Hook’s law given by
Equations given in Chapter 10. After substituting and combining terms, we have
9

10. ν
 1
Ui = ∫  (2 2
)
σ x + σ x + σ z2 − (σ xσ y + σ y σ z + σ xσ z ) +
1 2
(
τ xy + τ yz + τ xz
2 2
) dV

V 
2E E 2G 
(II)
where,
εx =
1
E
[ ]
σ x − ν (σ y + σ z )
1
[
ε y = σ y − ν (σ z + σ x )
E
]
[
ε z = σ z − ν (σ x + σ y )
E
1
]
τ xy
γ xy =
G
τ yz
γ yz =
G
τ
γ xz ( = γ zx ) = xz
G
E
G=
2(1 + ν )
If only the principal stresses σ1 , σ 2 , σ 3 act on the element, as shown in the earlier figure,
this equation reduces to a simpler form, namely,
ν
 1
Ui = ∫  ( ) 
σ 12 + σ 2 + σ 32 − ( σ 1σ 2 + σ 2σ 3 + σ 1σ 3 ) dV
2
(14 –14)
V 
2E E 
Recall that we used this equation in Sec. 10.7 as a basis for developing the maximum-
distortion-energy theory.
________________________________________________________________________
Using the principle of conservation of energy
Internal strain energy stored in the structure due to the applied load =
External work done by the applied load.
10

11. _____________________________________________________________
Appendix to: Effect of Transverse Shear Forces
1  V 2 ( x) Q 2 ( x, y ) 
L
Ur = ∫
2 0 G ( x) I 2 ( x) ∫ t 2 ( y )
dAdx (1)
A 
To simplify this expression for Ur , let us define a new cross-sectional properties fs, called
the form factor for shear. Let
A( x ) Q 2 ( x, y )
I 2 ( x) ∫ t 2 ( y )
f s ( x) ≡ dA (2)
A
(The form factor is a dimensionless number that depends only on the shape of the cross
section, so it rarely actually varies with x). Combining Eqs. 1 and 2 we get the following
expression for the strain energy due to shear in bending:
L
1 f sV 2 dx
Ur = ∫ (3)
2 0 GA
The form factor for shear must be evaluated for each shape of cross section. For
example, for a rectangular cross section of width b and height h, the expression
h 
2
b 2
Q = 
 −y 

2  4 
was obtained in example Problem 6.14 (Chapter 6). Therefore, from Eq. 2 we get
2
bh 1 h/2 b  h 2 2 
 6
2 ∫ h/2 2   4 − y  bdy = 5
fs = 
1 3
− b 2   (4)
 bh 
 12 
11

12. The form factor for other cross-sectional shapes is determined in a similar manner.
Several of these are listed in Table A, given below. The approximation for an I-section or
box section is based on assuming that the shear force is uniformly distributed over the
depth of the web(s).
Table A: Form Factor fs for shear
Section fs
Rectangle 6/5
Circle 10/9
Thin tube 2
I-section or box section ≈A/Aweb
Impact Problems Using Energy Methods
WHAT ARE IMPACT FORCES?
Suddenly applied forces that act for a short duration of time
- Collision of an automobile with a guard rail
- Collision of a pile hammer with the pile
- Dropping of a weight on to a floor
W δ
h
t
δst
Plastic impact
δmax
12

13. Loaded member vibrates till equilibrium is established.
Assumptions:
1. At impact, all kinetic energy of striking mass is entirely transferred to the structure. It
is transferred as strain energy within the deformable body.
1 1 W  2
U = mvi2 =  v
2 2 g



This means that the striking mass should not bounce off the structure and retain
some of its kinetic energy.
2. No energy is lost in the form of heat, sound or permanent deformation of the striking
mass.
Axial Impact of an Elastic Rod
δ
m vi
L
vi = velocity of impact
2 2 2
σ σ σ V
x x x
U = ∫ dV = ∫ dV =
i vol vol
2E 2E 2E
1 2
U = mv
e i
2
Equating Ui = Ue
13

14. 2
1
σ Vol 1 2
x
= mv ,
2 E 2 i
2 2
mv E mv E
i i
∴σ = =
x
Vol AL
δ Eδ
ε = ,σ = Eε = ,
x x x
L L
2 2
mv E mv L
L L
i i
∴δ = σ = =
x
E E AL EA
Impact Response of an elastic spring
W (m = W/g)
h
Velocity = vi (just
before impact)
δmax δmax W
W
Static deflection of spring = = δ st
k
k = spring constant = load per unit deformation
δmax = maximum deflection of spring due to impact = δ
Fe = maximum force in spring during impact = kδ max = kδ
14

15. 1  1  1
U e = U i → W (h + δ ) =  Fe  × ( δ ) =  kδ  × δ = kδ 2
2  2  2
1 2W 2Wh
i.e., kδ 2 − Wδ − Wh = 0 → δ 2 − δ− =0
2 k k
i.e., δ 2 − ( 2δ st )δ − 2hδ st = 0
 2h 
∴ δ = δ st ± δ st + 2hδ st = δ st 1 ± 1 +
2
 - -I
 δ st 
If we use the velocity at impact as a parameter, just before impact
1 1W 2
Wh = mvi2 = vi
2 2 g
vi2
∴h = (II)
2g
Substituting in Eqn. (I),
 v2 
δ = δ st 1 + 1 + i  (III)

 gδ st 

Impact Bending of a Beam
W
h
δ
U e =Ui
1
W (h + δ ) = Pimpact ×δ
2
For a central load,
15

16. (P
impact )L
3
δ =
48EI
48EIδ
∴ Pimpact =
L3
1  48EIδ 
∴W (h + δ ) = ×   ×δ
2  L3 
24EIδ 2
=
L3
24 EIδ 2
− Wδ − Wh = 0
L3
 WL3   WL3 
i.e., δ 2 −  δ − 
 24 EI   24 EI h = 0

   
WL3
Let δ st =
48EI
∴δ 2 − 2δ st δ − 2δ st h = 0
∴δ = δ st ± δ st + 2hδ st
2
  h 
= δ st 1 ± 1 + 2
δ 


  st 
To find the impact bending stress,
 Pimpact L  c
M impact c
σ max = = 4 
I
I  
 48EIδ  Lc 
= 3  
 L  4 I 
Virtual Work Method for Deflections (or Deformations)
Work-energy method, of equating the external work to internal strain energy, has the
disadvantage that normally only the deflection (or deformation) caused by a single force
16

17. can be obtained. The method of virtual work provides a general procedure to determine
the deflections and slopes (or rotations) at any point in the structure (which can be a truss,
a beam or frame) subjected a number of loadings.
To develop the virtual work method in a general manner, let us consider a body or
a structure of arbitrary shape (later this body will be made to represent a specific truss,
beam or frame) shown in the figure below.
(b) Real Forces
(acting on the
body) P2
(a) Virtual Forces (Internal virtual
force)
P1
u
∆L u
L L
P P
u O u O
A A
B
(magnitude = 1)
P3
∆
∆ = Deformation at A, along AB, caused by the loads P1, P2 and P3.
Let us assume that we want to determine the deflection ∆ of a point A, along the line AB,
caused by a number of actual (or real) forces P1, P2 and P3 acting on the body, as shown in
Figure (b). To find ∆ at A, along AB, due to the applied loads (P 1, P2 and P3), using the
virtual work method, the following procedure could be used.
(Internal virtual
force)
u
L
P
u O
A
B
1 17
(unit virtual
force)

18. Figure (a)
Step 1: Place a virtual force (here we use a unit virtual force) on the body at point A in
the same direction AB, along which the deflection is to be found. The term virtual force
is used to indicate that the force is an imaginary one and does not exist as part of the real
forces. This unit force, however, causes internal virtual forces throughout the body. A
typical virtual force (acting on a representative element of the body) is shown in Figure
(a).
P2
P1
dL u
L
P
u O
A
B
1
P3 (unit virtual
force)
∆
Figure b
Step 2: Next place the real forces, P1, P2 and P3 on the body [Figure (b)]. These forces
cause the point A to deform by an amount ∆ along the line AB, while the representative
element, of length L, now deforms by an amount dL. As these deformations occur within
the body, the external unit virtual force (already acting on the body before P 1, P2 and P3
are applied) moves through the displacement ∆; similarly the internal virtual force u
18

19. acting on the element (before P1, P2 and P3 are applied) moves through the displacement
dL. These forces, moving through displacements ∆ and dL, do work.
Step 3: The external virtual unit force, moving through displacement ∆, performs
external virtual work given as (1) times (∆), on the body. Similarly, the internal virtual
force u, moving through displacement dL, performs internal virtual work given as (u)
times (dL). Since the external virtual work is equal to the internal virtual work done on all
elements making up the body, we express the virtual work equation as:
Virtual forces
1 ×∆ = ∑ u ×dL)
( (A)
Real
deformations
The summation sign, in Eqn. (A), indicates that all the internal virtual work in the whole
body must be included. Eqn. (A) gives the deflection ∆ along the line of action of unit
virtual force. A positive value for ∆ indicates that the deflection is in the same direction
as the unit force.
In writing down Eqn. (A), one has to remember that the full values of the virtual
forces (unit force at A, and all the internal forces, u i) were already acting on the body
when the real forces were applied (viz. P1, P2 and P3). Therefore, no one-half appears in
any term of Eqn. (A).
In a similar manner, the rotation (or slope) at a point in a body can be determined by
applying a virtual unit moment or couple (instead of a unit force) at the point where the
rotation is desired (see Figure below).
19

20. P2
(Internal virtual
force)
P1
u
dL u
L L
P P
u O u O
A A
B
1
(Virtual unit P3
moment)
θ
(a) Virtual unit moment applied (b) Real forces P1, P2 and P3 applied
(Develop virtual force u, within (Virtual unit moment rotates through an
the body) angle θ)
Virtual unit
moment Virtual internal forces
(1) × (θ ) = ∑( u m ) × ( dL ) (B)
Real deformation
Real slope
Specific Structures
Trusses
(i) Subjected to applied external loads only
20

21. If ui represents the internal forces developed in the members, due to an applied
unit load (at the point where the deformation is to obtained) in the required
direction, then Eqn. (A) can be expressed as
u i Pi Li
(1) × ( ∆) = ∑ (C)
Ai E i
(ii) For trusses subjected to a temperature change (causing internal forces)
The incremental deformation caused in member due to a temperature rise is dL,
where
dL = α( ∆T ) L
Also
n
(1) × ( ∆ ) = ∑ u iα i ( ∆Ti ) Li (D)
i =1
Temperature
change
(iii) Trusses with Fabrication Errors
n
(1) × ( ∆) = ∑ u i ∆Li (E)
i =1
where
∆L = difference in length of the member from its intended length, caused
by a fabrication error.
Beams
For loads acting on a beam subjected to bending moments alone, the deformation
∆, at a given point along a given direction is given by
21

22. (1) × ( ∆) = ∫ mMdx (F)
EI
where m is the bending moment in the member when a unit load is applied on the
structure at the specified point in the specified direction. For a general loading on the
beam, generating axial, shear, bending and torsional forces/moments in the beam
f s vV
(1) × ( ∆) = ∫ nN dx + ∫ mM dx + ∫ dx + ∫
tT
dx (G)
AE EI GA GJ
where n is the axial force generated in the beam when a unit load is applied on the beam
in the required direction; similarly m, v and t are the bending moment, shear force and
torsional moment generated under the applied unit load.
Consider a truss subjected to loads F1, F2 and F3
D
C
B
A
1
Unit virtual load is applied in the direction in which the deflection is required, say at B in
the vertical direction. Let uAB, uBC, uCA and uCD be the internal forces generated when
the unit load is applied at B.
D F1
C
B F2
A
F3
22

23. Let PAB, PBC, PCA and PCD be the internal forces generated in the truss members due to the
given loads F1, F2 and F3 acting on the beam. Then the vertical deflection at B is obtained
as,
n
u i Pi Li
∆v B = ∑ (H)
i =1 Ai Ei
Considering a Beam Subjected to Bending Loads P1, P2 and P3
A B C
L L/2
Let us say that it is required to find the vertical deflection at C due to the given loads.
(i) Apply a unit vertical load (virtual) at C in the vertical direction and find the
moment m in the beam.
(ii) Then apply the given loads on the beam (say P 1, P2 and P3) and compute the
bending moments M in the beam. Then the deflection ∆v at C is obtained
P1
P2 P3
A C
B
L L/2
mMdx
∆vC = ∫ (I)
EI
23

24. Castigliano’s Theorem
(Based on the strain energy stored in a body)
Consider a beam AB subjected to loads P1 and P2, acting at points B1 and B2 ,
respectively.
P1 P2
B1 B2
v2
v1
P1 =
B2
B1
v21
v11
+
P1 P2
v22
v12
v1 = v11 + v12
v 2 = v 21 + v 22
If f 11 P = v11 ,
1
where f11 = deflection at B1 due to a unit load at B1
and f 21 P = v 21 with f21 = deflection at B2 due to a unit load at B1
1
and
v 22 = f 22 P2 , with f22 = deflection at B2 due to a unit load at B2 &
v12 = f 12 P2 , with f12 = deflection at B1 due to a unit load at B2.
24

25. Then
v1 = v11 + v12
= f 11 P + f 12 P2
1 (I)
Similarly,
v 2 = v 21 + v 22
= f 21 P + f 22 P2
1 (II)
Considering the work done = Ui
1 1
= P v11 + P2 v 22 + P v12
1 1
2 2
1 1
= P × f 11 P + P2 × f 22 P2 + P × f 12 P2
1 1 1
2 2
1 1
= f 11 P 2 + f 22 P22 + f 12 P P2
1 1 (III)
2 2
Now we reverse the order the application of loads P 1 and P2, viz., applying P2 at B2 first
and then applying P1 at B1,
P1 P2
B1 B2
v2
v1
= P2
B1 B2
v22
v12
25
+
v11 P1 P2 v21

26. v1 = v12 + v11 = f 12 P2 + f 11 P1
Similarly,
v 2 = v 22 + v 21 = f 22 P2 + f 21 P1
Ui =
1 1
= P2 v 22 + P2 v 21 + P v11
1
2 2
1 1
= P2 × f 22 P2 + P2 × f 21 P + P × f 11 P
1 1 1
2 2
1 1
= f 22 P22 + f 21 P P2 + f 11 P 2
1 1 (IV)
2 2
Considering equation (III) and (IV), and equating them, it can be shown that
1 1
Ui = f 11 P 2 + f 22 P22 + f 12 P P2
1 1
2 2
1 1
= f 22 P22 + f 21 P P2 + f 11 P 2
1 1
2 2
f 12 = f 21 This is called Betti – Maxwell’s reciprocal theorem
1
B1 B2
f21
1
26
B1 B2
f12 f22

27. Deflection at B2 due to a unit load at P1 is equal to the deflection at B1 due to a unit load
at P2.
From Eqn. (III)
∂U i
= f 11 P + f 12 P2 = v1
1
∂P 1
From Eqn. (IV)
∂U i
= f 22 P2 + f 21 P1 = v 2
∂P2
This is Castigliano’s first theorem.
Similarly the energy Ui can be express in terms of spring stiffnesses k 11, k12 (or k21), & k22
and deflections v1 and v2; then it can be shown that
∂U i
=P 1
∂v1
∂U i
= P2
∂v 2
This is Castigliano’s second theorem. When rotations are to be determined,
∂v
θi =
∂M i
27

28. Deflection at B2 due to a unit load at P1 is equal to the deflection at B1 due to a unit load
at P2.
From Eqn. (III)
∂U i
= f 11 P + f 12 P2 = v1
1
∂P 1
From Eqn. (IV)
∂U i
= f 22 P2 + f 21 P1 = v 2
∂P2
This is Castigliano’s first theorem.
Similarly the energy Ui can be express in terms of spring stiffnesses k 11, k12 (or k21), & k22
and deflections v1 and v2; then it can be shown that
∂U i
=P 1
∂v1
∂U i
= P2
∂v 2
This is Castigliano’s second theorem. When rotations are to be determined,
∂v
θi =
∂M i
27

29. Deflection at B2 due to a unit load at P1 is equal to the deflection at B1 due to a unit load
at P2.
From Eqn. (III)
∂U i
= f 11 P + f 12 P2 = v1
1
∂P 1
From Eqn. (IV)
∂U i
= f 22 P2 + f 21 P1 = v 2
∂P2
This is Castigliano’s first theorem.
Similarly the energy Ui can be express in terms of spring stiffnesses k 11, k12 (or k21), & k22
and deflections v1 and v2; then it can be shown that
∂U i
=P 1
∂v1
∂U i
= P2
∂v 2
This is Castigliano’s second theorem. When rotations are to be determined,
∂v
θi =
∂M i
27

30. Deflection at B2 due to a unit load at P1 is equal to the deflection at B1 due to a unit load
at P2.
From Eqn. (III)
∂U i
= f 11 P + f 12 P2 = v1
1
∂P 1
From Eqn. (IV)
∂U i
= f 22 P2 + f 21 P1 = v 2
∂P2
This is Castigliano’s first theorem.
Similarly the energy Ui can be express in terms of spring stiffnesses k 11, k12 (or k21), & k22
and deflections v1 and v2; then it can be shown that
∂U i
=P 1
∂v1
∂U i
= P2
∂v 2
This is Castigliano’s second theorem. When rotations are to be determined,
∂v
θi =
∂M i
27