* energy = work = (magnitude of force)(translation of force in the direction of force) cf. (stress)(strain)
* unit of energy(work): Nm (= J, Joule), kgcm, lb-ft, etc ----- scalar quantity
* Two approaches in mechanics
 Vector mechanics [Newtonian mechanics] – Equilibrium of forces(vector quantity)
 Scalar mechanics [Lagrangian mechanics] – Conservation of energy(scalar quantity)
10.1 Introduction
* internal work = (internal forces; stresses) (deformations; strains)  elastic strain energy
* strain energy
 Axial force: U A  12 F  
L
0
F2
dx
2 EA
 Bending moment: U M  12 M  
L
0
 Shearing force: U S  Sv  k 
1
2
 Torque: U T  12 T  
L
0
L
0
T2
dx
2GJ
M2
dx
2 EI
S2
dx
2GA
10.2 Strain-Energy Density
* strain-energy density = amount of strain energy per unit volume
* Calculation of strain-energy density 
known stresses, strains, and stress-strain relations
* stress-strain relations
   ( )
- general case:    ( ) or
- linear elastic case:   E or
  /E
* uni-axial strain-energy density in the differential volume dV  dxdydz [ Fig (a)]
  x   x ( x ) 
x
x
0
0
dU x    x (dydz )d x (dx)  dV   x d x
x
dU x
   x d x
 strain-energy density: u x 
0
dV
  x   x ( x ) 
=== area under the  x   x curve
x
dU x*  dV   x d x
0
 complementary strain-energy density: u*x 
x
dU x*
   x d x
0
dV
=== area above the  x   x curve
 u x  u *x   x x
1
1 2  x2
 linear elastic stress-strain relations [  x  E x ]: u x  u   x x  E x 
2
2
2E
*
x
* strain-energy density due to shearing in the differential volume dV  dxdydz [ Fig (b)]
- internal force =  xy (dxdz) ;
deformation =  xy dy
 xy
 strain-energy density: uxy    xy d xy
0
 xy
 complementary strain-energy density: u*xy    xy d xy
0
 uxy  u*xy   xy xy
2
1
1
 linear elastic stress-strain relations [  xy  G xy ]: u xy  u *xy   xy xy  G xy2  xy
2
2
2G
* strain-energy density for general stress 3-D stress state
1
u  ( x x   y y   z z   xy xy   yz yz   zx zx )
2
 generalized Hooke’s law
1

1 2


u( , ) 
 x2   y2   z2    x y   y z   z x  
 xy   yz2   zx2 
2E
E
2G
1 2
or, u( ,  )  e  2G  x2   y2   z2   G  xy2   yz2   zx2 
2

* relations
u ( , )
 x
 x
u ( ,  )
 x
 x

u ( , )
  xy
 xy
u ( ,  )
  xy
 xy


* strain-energy density (volume integral) strain energy of a deformable body
U   udV   udxdydz
V
[<Example 10.2.1>] Elastic strain energy of a axially-loaded bar
Length = L, cross-sectional area = A, concentric axial load = P, stress:  x 
E,
A
L
P
A
P
L 1 P
2 
P2 L
U   udV    x dV  
( Adx) 


V
V 2E
0 2E
2 EA
 A


2
[<Example 10.2.2>] Elastic strain energy in cantilever beam with rectangular cross-section
* strain energy due to bending and shearing
My
- bending: M   Px ;  x 
I
- shearing: S  P
;  xy  
P
2I

1
4
h2  y2 
Bending
 x2
M 2 y 2 P2 x2 y 2

2 E 2 EI 2
2 EI 2
- volume integration ( dV  dxdydz  dxdA )
- strain-energy density: u x 
U M   u x dV 
V
P2
2 EI 2

P
 x y dV  2EI 
2
2
2
2
V
A
L
y 2 dA x 2 dx 
0
P2
2 EI

L
0
x 2 dx 
P 2 L3
6EI
Shearing
- strain-energy density: u xy 
 xy2

2G
- volume integration ( dV  Lbdy ) )
U S   u xy dV 
V
 U
2
P2 1 2

h  y2 
2 4
8GI
P 2 Lb h2 1 2
P 2 Lb h 5
3P 2 L 6 P 2 L
2 2


(
)
h

y
dy

U


A

bh
S
4
8GI 2  h2
8GI 2 30
5GA 5 2GA
P 2 L3 3P 2 L

6 EI
5GA
Note ---- shape factor for rectangle (Table 7.5.1):
6/5
[<Example 10.2.3>] Elastic strain energy due to shear: shape factors for various sections
Length = L, cross-sectional area = A, moment of inertia = I, shearing force = S, stress:  xy 
U S   u xy dV  
V
 xy2
V
2G
2
dV 
1  SQ 
1

 dV 

V
2G  Ib 
2GI 2
A
I2
2
Q 2
V  b  S dV
2
2
2
 S2
L A
L kS
1 L Q 2
Q 
dx  
dx
  S dAdx  0  2 A   dA
0 2GA
2GI 2 0 A  b 
b
 I
 2GA
( dV  dAdx )
- shape factor: k 
2
Q
A  b  dA

A
bh 3
b  h2
 rectangle(bh): Q    y 2  , I 
, A  bh , dA  bdy  k  2 2
I b
12
2 4

 circle(r): Q 
2
3
r
2
 y2  2 , I 
 k
SQ
Ib
3
A
I2
r 4
4
, A  r 2 , b  2r 2  y 2  2 ,
5
2 2
10
2 2


r

y
dy 
r 9
9
r
1
dA  bdy
2
b2  h 2
6
2
 h2 4  4  y  bdy  5
h
2
10.3 Principle of Mechanical Energy Conservation
* work done by a force: {P = force, moment, or torque}  { = displacement, rotation, or twisting}

 general force-displacement relations [Fig. 10.3.1a]: W   Pd
0
 a constant force P and displacement due to other source [Fig. 10.3.1b]: W  P
 a static force and linear elastic displacement [Fig. 10.3.1c]: W  12 P
* conservation law of mechanical energy (neglecting energy dissipation)
 general case: (work) = (kinetic energy; T  12 mv 2 ) + (strain energy; U ) : W  T  U
 Statics ( T  0 ) : W  U
 deformable body: WE  WI  U
(external work done by applied forces) = (internal work done by stresses)  (strain energy)
- Assume static loads( WE  12 P ), no acceleration( T  0 ), and elastic deformations(no dissipation).
[<Example 10.3.1>] Impact load and elastic deformations
* assumptions for idealization
- conservation of energy
- no rebound
- proportionality of Pi   i relations
Ps W

k
k
* maximum dynamic deflection for equivalent impact
P
load:  i  i
k

P
* impact factor: i  i  i
 s Ps
* static deflection for static load:  s 
* (external work done by falling body W = Ps) = (strain energy of spring) : Ps (h   i )  12 Pi i
( Pi 
i
2
Ps ) Ps ( h   i )  12 i Ps , or,  i2  2 s i  2h s  0
s
s

2h 

  i   s 1  1 


s


* impact factor: i  1  1 
---- maximum deflection of the spring due to the impact load W
2h
s
h   s very
high
 i  2h /  s ]
sudden load
- extreme cases [ h  0 

 i  2 ;
* discussion and application
Structure & loading
Deflection due to static load Equivalent spring constant
PL
EA
s  s
k
A bar under axial loading
EA
L
3
48EA
PL
k
Simple beam and a concentrated center load
s  s
L3
48EI
3EA
Ps L3
k 3
Cantilever beam and a concentrated free-end load
s 
L
3EI
TL
GJ
s  s
k
Twisting of circular shaft
GJ
L
[<Example 10.3.1>] Vertical bar on a simple beam
*
 sD
static
deflection
Wl WL3
  sl   sC 

Ea 48EI
* max deflection of D due to impact:
2h
i  1 1
 sD
 iD  i sD
of
D:
10.4 Method of Real Work
* Deflection of a structure may be evaluated using the conservation law of energy, WE  WI  U .
* external work done by applied load = real work  method of real work to find deflections
 deflection of free end (= point of load application) using the method of real work
-  = deflection of free end: WE  12 P
- WI  U 
P 2 L3 3P 2 L

6 EI
5GA
- WE  U


PL3 6PL

3EI 5GA
* limitations
- only one concentrated load
- deflection at the point of load application
[<Example 10.4.1>] Simple beam and a concentrated load at center-span (Case 7 of Table 8.5.1)
-  = deflection at center-span: WE  12 P
- WI  U  
L
0
- WE  U
2
L
M2
P 2 L3
2 ( Px / 2)
dx  2 
dx 
0
2 EI
2 EI
96 EI


PL3
48EI
[<Example 10.4.2>] Cantilever beam and a moment load at free end (Case 1 of Table 8.5.1)
-  = deflection angle of free end: WE  12 M 0
- WI  U  
L
0
- WE  U
2
L M
M2
M 2L
0
dx  
dx  0
0 2 EI
2 EI
2 EI


M 0L
EI
10.5 Method of Virtual Work : Unit-Load Method
* method of real work --- only for load-directional deflection of the loading point
* method of virtual work: applicable for all kinds of structure, for any deflection of any point
virtual force at the point of deflection = unit load  unit-load method
* real work vs. virtual work
- real work = real force + real displacement
- virtual work = virtual force + real displacement, or real force + virtual displacement
10.5.1 Virtual Work
* virtual work: work done by virtual forces in equilibrium
along the real and kinematically admissible displacements
* principle of virtual work - conservation law of energy
 (external virtual work) = (internal virtual work): WE  WI
- rigid body: WI  0   WE  0
- elastic deformable body: WE  WI  U
- generally for inelastic behaviour, deformation due to temperature change and support settlement
* displacement of a point A in the direction of P  1 (virtual force)
Fig. 10.5.1(a): (1) Apply a virtual
load P  1 at the point (A) and
direction of desired displacement.
(2) Compute the virtual internal
forces [ f ] which are in
equilibrium with the virtual load
P  1 .
Fig. 10.5.1(b): (1) Apply real
loads P1 , P2 , P3 ,  , and compute
internal deformations [ dL ].
(2)  = real displacement of the
point (A) in the direction of the
virtual load P  1 .
 application of principle of virtual work
- external virtual work done by the virtual load P  1 along the real displacement : WE  P  1 
- internal virtual work done by the virtual forces [ f ] along the real deformations: WI   fdL
Real displacements
* virtual work equation: 1    f dL
Virtual forces
 virtual unit moment M   1 : 1    f dL
- f = virtual internal forces due to the virtual external moment M   1
10.5.2 Linear Elastic Systems
* internal forces and deformations
Virtual internal forces [ f ] Real internal forces Real deformation [ dL ]
Fdx
d 
f
F
Axial
EA
Mdx
d 
m
M
Bending
EI
kSdx
dv 
s
S
Shearing
GA
Tdx
d 
t
T
Torsion
GJ
* virtual work equation for the displacement of linear elastic system
fF
mM
ksS
tT
dx   
dx   
dx   
dx
 (1   ) or (1   )   
EA
EI
GA
GJ
 truss:   
f FL
EA
 beam:  or    
mM
dx
EI
* procedure of unit-load method (method of virtual work)
(1) real member forces due to real applied loads: (F, M, S, T)
(2) virtual member forces due to virtual unit load: (f, m, s, t)
(3) application of virtual work equation to get displacement(rotation)
[<Example 10.5.1>] Simple beam with uniform load  center-span deflection, support rotation
M  12 ( wLx  wx 2 ) ,
0 x L
0  x  12 L
 1x,
m  1 2
1
 2 ( L  x) , 2 L  x  L
m  1
x
,
L
center-span deflection
L
L mM
L 1
5wL4
2 1
2
2
1
1
1
1


C  
dx  
( 2 x )  2 ( wLx  wx )dx  L
2 ( L  x )  2 ( wLx  wx ) dx 
0 EI
0 EI
384 EI
2 EI
support rotation
L mM
L 1 
x 1
wL3
2
A  
dx  
1    2 ( wLx  wx )dx 
0 EI
0 EI
24 EI
 L
0 x L
[<Example 10.5.2>] Cable-supported beam with uniform load  center-span deflection
- support reaction, tension of cable, bending moment of beam
RA  12 wL ;
f 
F  12 wL
w
( Lx  x 2 ) , 0 x L
2
- deflection due to cable elongation and beam bending
fF
mM
C   
dx   
dx
EA
EI
1 h  1  wL 
1 L2 1
1

dx

( 2 x )  12 w( Lx  x 2 )dx 






EA 0  2  2 
EI 0
EI
wLh
5wL4


4 EA 384 EI
0  x  12 L
 1x,
m  1 2
1
 2 ( L  x) , 2 L  x  L
M 

L
L
2
1
2
( L  x )   12 w( Lx  x 2 )dx
[<Example 10.5.3>] Space structure [ E = 200 GPa, G = 80 GPa,
Member
CB
BA
M
-30x
-50x
m
-x
-x
T
0
60
t
0
2
A mM
B tT
A tT
mM
dx  
dx  
dx  
dx
C EI
B EI
C GJ
B GJ
4
1  2
1

(  x )( 30 x )dx   (  x )( 50 x )dx  

 GJ
0
EI  0
C  
1
2
I = 100106 mm4, J = 180106 mm4]
Integration range
02
04
B
 260dx
4
0
= 0.0093 + 0.0333 = 0.0427 m
[<Example 10.5.4>] Deflection of truss – vertical displacement of joint C
Member
AB
AC
BC
BD
BE
CE
DE
DF
EF
C  
L
10
8
6
8
10
8
6
10
8
F
-37.5
+60.0
+30.0
-50.0
-12.5
+60.0
+37.5
-62.5
+50.0
f
-10/9
+8/9
+1
-4/9
-5/9
+8/9
+1/3
-5/9
+4/9
f FL
3750/9
3830/9
180
1600/9
625/9
3840/9
225/3
3125/9
1600/9
f FL
20675 / 9

 0.023 m
EA (200  106 )(500  106 )
10.6 Reciprocal Theorem
* linear elastic structures
- Fig. (c): total work done by static load P1 & P2
W  12 P11  P2  2 
- Fig. (a): deflection due to static load P1 only
11  11P1 , 21   21P1
 ij = deflection of point i due to a unit load at j
= influence coefficient
- Fig. (b): deflection due to static load P2 only
12  12 P2 , 22   22 P2
- superposition
 1  11 12   P1 
 
   , or    P
 2   21  22   P2 
* load-deflection relations : apply P1 first and then P2
 work done by P1 & P2 :
 W12  W1  W2 
1
2
W1  12 P111  P112  12 11P12  12 P1P2 , W2  12 P2  22  12  22 P22

P  212 P1 P2   22 P22
2
11 1

* load-deflection relations : apply P2 first and then P1 [change the roles of 1 and 2]
 work done by P2 & P1 :
 W21  W2  W1 
1
2
W2  12 P2  22  P2  21  12  22 P22   21P2 P1 , W1  12 P111  12 11P12

P  2 21P2 P1   22 P22
2
11 1

* Total work should be the same: W  W12  W21
 12   21 [Maxwell’s reciprocal theorem]
- linear elastic behavior
- neither temperature change nor support settlement
 applicable to moment load and rotation
10.7 Castigliano’s Second Theorem
- strain energy
U  12 11P12  212 P1 P2   22 P22


- partial differentiation
U
 11P1  12 P2  1
P1
U
  21P1   22 P2   2
P2
- generally,
U
: Castigliano’s 2nd theorem
j 
Pj
-- linear elastic behavior
-- neither temperature change nor support settlement
 applicable to moment load and rotation
* moment load and deflection angle:  j 
torque and twisting angle:  j 
U
M j
U
T j
* general formula of the Castigliano’s 2nd theorem
F  F 
M  M 
k S  S 
T
 j  
dx   
dx   
dx   






EA  Pj 
EI  Pj 
GA  Pj 
GJ
 basically same as the unit-load method
 T 

dx
 P 
 j
[<Example 10.7.1>] Displacement of the point with a load
Cantilever beam with a concentrated load at free end – deflection of free end [E8.3.1]
- moment: M   P( L  x )
M
 ( L  x )  Same as the m in the unit-load method!!!!
- partial derivative:
P
M  M 
P L
PL3
2
 B   
( L  x ) dx 

dx 
EI  P 
EI 0
3EI
[<Example 10.7.2>] Displacement of the point without a load – dummy load method
(1) Apply a dummy load ( Pd or M d ) at the point of interest, whose magnitude is zero.
(2) Calculate the member forces in terms of dummy load.
(3) Compute the partial derivatives and substitute in the general formula of the Castigliano’s 2nd theorem.
(4) Put the dummy load as zero and evaluate the integral to get the displacement.
Fig. (a): a dummy load PA for the calculation of  A
- moment in terms of dummy load: M   PA x  12 wx 2
M
  x  Same as the m in the unit-load method!!!!
- partial derivative w.r.t. dummy load:
PA
- Put PA  0 ( or, M   12 wx 2 ), and then use the Castigliano’s 2nd theorem:
A  
M  M 
1

dx 
EI  PA 
EI
 
L
0
1
2
wx 2 (  x )dx 
wL4
8EI
Fig. (b): a dummy load M A for the calculation of  A
- moment in terms of dummy load: M   M A  12 wx 2
M
 1  Same as the m in the unit-load method!!!!
- partial derivative w.r.t. dummy load:
M A
- Put M A  0 ( or, M   12 wx 2 ), and then use the Castigliano’s 2nd theorem:
A  
M  M 
1

dx 
EI  M A 
EI
 
L
0
1
2
wx 2 ( 1)dx 
wL3
6EI
10.8 Method of Least Work : Statically Indeterminate Structures
* method of consistent deformation + Castigliano’s 2nd theorem  method of least work
* procedure of method of least work to get redundant forces of SIS
 primary structure and redundant forces: R1 , R2 ,  , Rn
 member forces as functions of redundant forces: F  F ( R1 , R2 ,  , Rn ) , M  M ( R1 , R2 ,  , Rn ) , …
 strain energy of primary structure as a function of redundant forces: U  U ( R1 , R2 ,  , Rn ) ,
 Castigliano’s 2nd theorem for consistent deformation [ principle of least work]:
U
 0 ; i  1,, n [system of linear equations]
Ri
 Solve system of linear equations to determine the redundant forces.
 U

* 
 0 ; i  1,, n   necessary conditions for the minimum of U  U ( R1 , R2 ,  , Rn )
 Ri

 method of least work
[<Example 10.8.1>] Three-bar truss
- redundant force: R = F2 = reaction at the support B
- member forces: F1  F3  ( P  R) / 2 , F2  R
- strain energy: U  
F 2L
L  ( P  R)2 R 2 

 
2 EA EA 
2
2
- Castigliano’s 2nd theorem:
- solution: R 
2P

1 2
U
 2( P  R)
0
R
R
2
F1  F3 
P
,
2 2
F2 
2P
2 2
[<Example 10.8.2>] Propped beam with linearly-distributed load
- redundant force: R = reaction at the support A
- bending moment: M  Rx 
wx 3
, 0 x L
6L
- strain energy due to bending: U  
L
0
- Castigliano’s 2nd theorem:
- solution: R 
U
0
R
M2
dx
2 EI

L
0
L 1 
M M
wx3 
 Rx 
( x )dx
dx  
0 EI
EI R
6L 

wL
10
wL
4wL
wL2
 RA 
, RB 
, MB 
10
10
15
3
2
2 
wL  x   x  
wL 
x 
M 
3

5
V

1

5
,
  
     

3 0   L   L  
10 
 L  
 You may use double-integration method to get the deflection and slope [9.6.2(4)].
[<Example 10.8.3>] Fixed beam with linearly-distributed load
- redundant forces: RA , M A = reactions at the support A
wx3
, 0 x  L
6L
- bending moment: M  M A  RA x 
- strain energy due to bending: U  
L
0
M2
dx
2 EI
- principle of least work:
L M M
L 1 
U
wx3 
0 
 M A  RA x 
( x )dx
dx  
0 EI R
0 EI
R A
6L 

A
U
0
M A
- solution: R A 
M 

L
0
3
wL ,
20
wL2
60
L 1 
M M
wx3 
 M A  RA x 
(1)dx
dx  
0 EI
EI M A
6
L


MA  
wL2
30

RB 
7
wL2
wL , M B 
20
20
3
2

wL 
x
x 
x 

2

9

10
V

3

10
,
 
  
  


20 
 L
 L  
 L  
