Chapter 4

Energy Methods • vectorial methods force/displacement, stress/strain equilibrium equations • energy methods deformation work, strain energy variation methods (unit load method, stationary potential energy method) Terminology • displacements: linear (translational) and rotational • forces: linear and rotational (bending and twisting moments)

{ } T, , , , ,xa ya za xa ya zaa F F F M M M⎢ ⎥= ⎣ ⎦F

{ } T, , , , ,a a a xa ya zaa u v w⎢ ⎥Δ = θ θ θ⎣ ⎦

Fyax

zy

uaθxa

va

θza

waFxaMxa

MyaFza

Mza

θya

a

Degrees of Freedom

va

vb

θa

θb

Mza

Fxa

Fya

ua

ub

Mzb

Fxb

Fyb

The overall behavior of a member (element) of a structure is defined by the displacements of the structure's joints (nodes) under the action of forces applied at the joints.

Strains and stresses within individual members can be calculated separately.

The movement of each joint can be described by three translational and three rotational displacement components.

Further reduction of the degrees of freedom (DOF) is possible by neglecting certain deformations, i.e., the DOF is not unique in approximations.

Indeterminacy

physical indeterminacy is always zero

static indeterminacy is the number of “elasticity” equations required for analysis in addition to the equations of equilibrium

kinematic indeterminacy is the number of displacements that are not predetermined

SI UR EE= −

UR R KR= −

KI KR ND= −

SI Static Indeterminacy

UR number of Unknown Reaction forces

EE number of Equilibrium Equations

R number of Reaction forces

KR number of Known Reaction forces

KI Kinematic Indeterminacy

ND Negligible Displacements

Examples of Indeterminacy (Bending Only)

loading force

unknown reaction force (UR)

kinematically indeterminate displacement (KI)

static indeterminacy: 3 - 3 = 0, kinematic indeterminacy: 3 - 1 = 2

static indeterminacy: 4 - 3 = 1, kinematic indeterminacy: 5 - 2 = 3

static indeterminacy: 6 - 3 = 3 kinematic indeterminacy: 0 – 0 = 0

static indeterminacy: 6 - 3 = 3, kinematic indeterminacy: 6 – 3 = 3

Work and Energy

Δ

F

dWF1

Δ1

W

dΔ

W*

Δ

F

F1

Δ1dΔ

F = kΔ

1

0W F d U

Δ= Δ =∫ ,

1

0* *

FW dF U= Δ =∫ , 1 1* *W W U U F+ = + = Δ

linear behavior

2 2

1 1 1 11 1 1*2 2 2

W W F k Fk

= = Δ = Δ =

vectorial form

1 1 ( )2 2 x y zW F u F v F w= ⋅ + +F Δ =

structure with n DOFs

{ } T1 2 1 2

1 1 , , , ,2 2 n nW F F F= = Δ Δ ⋅⋅ ⋅Δ ⋅ ⋅ ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦FΔ

[ ]{ }12

W U= =⎢ ⎥⎣ ⎦ KΔ Δ

free DOFs

{ } { }1 12 2f f f ff fW U⎢ ⎥ ⎢ ⎥ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦ ⎣ ⎦F KΔ Δ Δ

Method of Virtual Displacement/Work Virtual Displacement

imaginary, very small change in the configuration relative to the equilibrium condition

neither external loads nor internal forces are altered by such a virtual displacement

Total Potential

0( )loadU WΠ = − +Π or loadd dU dWΠ = −

( )internal loadd dW dWΠ = − + or d dVΠ = −

internaldU dW= −

internal loaddV dW dW= +

Π total potential of the system

U strain energy of the deformed body

Wload work done by the external loads

internalW work done by the internal forces

V virtual work done by external loads and internal forces during a virtual displacement

Stable Equilibrium

Fourier inequality 0dV ≤

Minimum potential 0dΠ ≥

Normalized Lattice Distance

Pote

ntia

l Ene

rgy

[a. u

.]

0 1 2

typical

parabolic

potential well

Reciprocity

flexibility and stiffness matrices are symmetric

local

[ ] [ ]T=d d or ij jid d=

[ ] [ ]T=k k or ij jik k=

global

[ ] [ ]T=D D or ij jiD D=

[ ] [ ]T=K K or ij jiK K=

1

2

F1

F2

Δ1

Δ2

1 11 12 1

2 21 22 2

d d Fd d F

Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ⎩ ⎭ ⎩ ⎭⎣ ⎦

1

2

F1

Δ11

Δ21

Δ12

Δ22

1

2

F2

11 11 12 1

21 21 22 0d d Fd d

Δ⎧ ⎫ ⎡ ⎤ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ ⎩ ⎭⎩ ⎭ ⎣ ⎦

12 11 12

22 21 22 2

0d dd d F

Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ⎩ ⎭ ⎩ ⎭⎣ ⎦

1,2 11 1 12 1 22 21 12 2

W F F F= Δ + Δ + Δ

2,1 22 2 21 2 11 11 12 2

W F F F= Δ + Δ + Δ

1,2 2,1W W=

12 1 21 2F FΔ = Δ

12 21d d=

Maxwell's Reciprocal Theorem

andij ji ij jid d k k= =

Strain Energy of Beams

oV

U U dV= ∫

2 2 2

2 2 2

1 [ 2 ( )]21 ( )

2

o x y z x y y z x z

xy yz xz

UE

G

= σ + σ + σ − ν σ σ + σ σ + σ σ

+ τ + τ + τ

• straight (approximately) prismatic bar • bisymmetrical cross section

symmetrical bisymmetrical

Cx

y

S

x

y

C, S

cross sectional area A dx dy= ∫∫

first moments xQ y dx dy= ∫∫ and yQ x dx dy= ∫∫

second moments 2xI y dx dy= ∫∫ , 2

yI x dx dy= ∫∫ , and xyI x y dx dy= ∫∫

centroid 0x yQ Q= =

symmetrical 0xyI = ( x xMθ ∝ and y yMθ ∝ )

shear center ( ) 0z zy zxM x y dx dy= τ − τ =∫∫ in shear

extension

2 2 2

20 0 02 22

L L L

A A

P PU dAdx dAdx dxE EAEA

σ= = =∫ ∫∫ ∫ ∫∫ ∫

torsion

2 2 22

20 0 02 22

L L L

A A

T TU dAdx r dAdx dxG GJGJ

τ= = =∫ ∫∫ ∫ ∫∫ ∫

2

0 2

L TU dxGK

= ∫

bending

2 2 22

20 0 02 22

L L L

A A

M MU dAdx y dAdx dxE EIEI

σ= = =∫ ∫∫ ∫ ∫∫ ∫

shear-bending

2 2 2 2 2

2 2 2 20 0 02 2 2

L L L

A A A

V Q V A QU dAdx dAdx dAdxG G GAI t I t

τ= = =∫ ∫∫ ∫ ∫∫ ∫ ∫∫

2

0 2

L VU k dxGA

= ∫

2

2 2A Qk dy dzI t

= ∫∫

combined loading

2 22 22 2

0 2 2 2 2 2 2

L y yz zy z

y z

M VM VP TU k k dxEA GK EI EI GA GA

⎧ ⎫⎪ ⎪= + + + + +∫ ⎨ ⎬⎪ ⎪⎩ ⎭

Shear Stress Distribution in Bending

τ

τ = 0

τ = 0

VV

x

z

y

z z + dz

yτyz C

σz

A'

t

σzτzy

equilibrium in the z-direction

' '( , ') ' ( , ') ' ( ) ( ) 0z z zy

A Az dz y dA z y dA y t y dzσ + − σ − τ =∫∫ ∫∫

( )( , ') 'x

zx

M zz y yI

σ =

'

( ) ( ) ' ' ( ) ( ) 0x xzy

x A

M z dz M z y dA y t y dzI

+ −− τ =∫∫

'( ) ' '

AQ y y dA= ∫∫

( ) ( ) ( )( )

( )x x

zyM z dz M z Q yy

dz I t y+ −

τ =

xdMV

dz=

( )( )( )xy

V Q yyI t y

τ =

Transverse Shear Factor

2

2 2A Qk dy dzI t

= ∫∫

rectangular cross section

h/2

z

y

y( /2 + )/2h y

b

ht

Qh/2h/2

z

y

y( /2 + )/2h y( /2 + )/2h y

b

ht

Q

A hb= 3

12h bI =

22( ) ( ) ( ) ( )

2 2 2 2 4 2h h b h h bQ y b y y y y= − = − + = −

t b=

2 2 2/ 22 2 2 2

2 2 5 5/ 2

36 36( ) ( )4 4

h

h

A Q h hk dy dz y dy dz y dyI t h b h −

= = − = −∫∫ ∫∫ ∫

/ 24 2 2 4 2 3 5/ 2 4

5 50 0

72 72( )16 2 16 6 5

hh h h y h y h y yk y dyh h

⎡ ⎤= − + = − +⎢ ⎥∫

⎢ ⎥⎣ ⎦

65

k =

Unit Load Method

A

ΔA

loadingforces

PA

, , 0A AV

W W U U dV= = = ∫

2 2 2

2 2 2

1 [ 2 ( )]21 ( )

2

o x y z x y y z x z

xy yz xz

UE

G

= σ + σ + σ − ν σ σ + σ σ + σ σ

+ τ + τ + τ

A

x x xσ = σ + σ , Ay y yσ = σ + σ , ...

A A

o o o oU U U U= + +

1 [ 2 ( )]

1 ( )

A A A A A A Ao x x y y z z x y y z z x

A A Axy xy yz yz zx zx

UE

G

= σ σ + σ σ + σ σ − ν σ σ + σ σ + σ σ

+ τ τ + τ τ + τ τ

A A

o A AV

U U dV P= = Δ∫

For a unit load (PA = 1):

A AA o A

VD U U dV= = = Δ∫

General case of combined tension torsion, bending, and shear:

0

L y y y y yz z z z z

y z

M m k V vN n T t M m k V vD dxE A G K E I E I G A G A

⎛ ⎞= + + + + +⎜ ⎟∫ ⎜ ⎟

⎝ ⎠

Example 1:

x

q0

L

y, v

P = 1A

x

q0

L

y, v

P = 1AP = 1A

202

q xM = , m x=

2 2 2

0 0 0 0

( )2 2 2

L L L LM m M m M mU dx dx dx dxEI EI EI EI+

= = + +∫ ∫ ∫ ∫

4

30 0

0 02 8

L LAA

q q LM mv U dx x dxE I E I E I

= = = =∫ ∫

Example 2:

1P =B

x

q0

L

y, v

1P =B 1P =BP =B

x

q0

L

y, v

20

0 2q xM q L x= − , m x= −

42 3 4

0 0 0

0 0 0

5( )2 3 8 24

LL LB

q q q LM m x L x xv dx L x x dxE I E I E I E I

⎡ ⎤= = − − = − − = −⎢ ⎥∫ ∫

⎢ ⎥⎣ ⎦

Example 3:

45°B1P =B

x

a

y, v

aP

A C

D

k

Nn

from A to B:

2P xM = ,

2xm = − ,

2PN = , 1

2n = −

for the whole system:

32

0 02

2 2 2 6

a aB

N n M m P P P P av dx x dxk E I k E I k E I

= + = − − = − −∫ ∫

Curved Members

circular ring in the x-y plane

zx

y

R

Fx

Cx

Fz

CzFy

Cy

VR

MR

VzMz

T

N

φz

x

y

R

Fx

Cx

Fz

CzFy

Cy

VR

MR

VzMz

T

N

φ

Equilibrium Equations:

sin cosx yN F F= φ − φ

cos sinR x yV F F= − φ − φ

z zV F= −

sin cos sinR z x yM F R C C= φ − φ − φ

sin (1 cos )z x y zM F R F R C= − φ − − φ −

(1 cos ) sin cosz x yT F R C C= − − φ + φ − φ

Slender Uniform Half-Ring

P, 1

A

B

1•R

1•R

φ

P

R

A

B

C1

P, 1

A

B

1•R

1•R

φ

P

R

A

B

C1

0 0

L LR R

CM m T tD d d

E I G K= +∫ ∫

/ 2 / 2

0 0

R RC

M m T tD R d R dE I G K

π π= φ + φ∫ ∫

Load effect:

sinRM P R= − φ

(1 cos )T P R= − φ

Unit load effect:

sin ( )cos ( )sin cosRm R R R R= − φ − φ + φ = − φ1 1i i

(1 cos ) ( )sin ( ) cos (1 sin )t R R R R= − φ + φ + φ = + φ1 1i i

Displacement at C:

3 3/ 2 / 2

0 0sin cos (1 cos ) (1 sin )C

P R P RD d dE I G K

π π= φ φ φ + − φ + φ φ∫ ∫

3 312 2CP R P RD

E I G Kπ −

= +

Castigliano's Theorems

Castigliano's first theorem

0loadd dU dWΠ = − =

kk

U P∂=

∂Δ, k

k

U M∂=

∂θ

Castigliano's second theorem

*k

k

UP

∂= Δ

∂,

*k

k

UM

∂= θ

∂

Example:

x

q0

L

y, v

PA

x

q0

L

y, v

PAPA

2202

0 0

( )( ) 2*2 2

L L Aq x P xM mU dx dx

EI EI

++= =∫ ∫

2

0

0

0

( )* 2

A

L AA

A

P

q x P x xUv dxP EI

=

+∂= = ∫

∂

4

30 0

02 8

LA

q q Lv x dxE I E I

= =∫

Finite Element Method

va

vb

θa

θb

Mza

Fxa

Fya

ua

ub

Mzb

Fxb

Fyb

The overall behavior of a member (element) of a structure is defined by the displacements of the structure's joints (nodes) under the action of forces applied at the joints.

Strains and stresses within individual members can be calculated separately.

Structural analysis:

1. Basic mechanics; fundamental constitutive, compatibility, and equilibrium equations

2. Finite element mechanics; exact or approximate solutions of the

fundamental equations for an element 3. Equation formulation; establishment of the governing algebraic

equations for the structure 4. Equation solution; computational methods and algorithms 5. Solution interpretation; presentation of the results for design or

analysis

Example: Beam in Bending

va

vbθa

L

θb

Vb

Va

MaMb

1. Basic mechanics

x,u

y,v

V(x+dx)V(x)

M(x) M(x+dx)

compatibility relationship: y ′′ε = − v

constitutive relationship:

Eσ = ε

fundamental bending equation:

''M E I v=

equilibrium equations:

' '''M E I v V= = −

' '''' 0V E I v= − =

general solution

2 30 1 2 3v a a x a x a x= + + +

boundary conditions

0(0)av v a= =

1'(0)a v aθ = =

2 3

0 1 2 3( )bv v L a a L a L a L= = + + +

21 2 3'( ) 2 3b v L a a L a Lθ = = + +

specific solution

0 aa v=

1 aa = θ

2 2 23 3 2 1

a b a ba v vL LL L

= − + − θ − θ

3 3 3 2 22 2 1 1

a b a ba v vL L L L

= − + θ + θ

2 3

0 1 2 3v a a x a x a x= + + +

internal displacement

2 3 2 3 2 3 2 3

2 3 2 2 3 23 2 2 3 2(1 ) ( ) ( ) ( )a a b b

x x x x x x x xv v x vL LL L L L L L

= − + + − + θ + − + − + θ

2 3( ) (2 6 )M x E I a a x= +

3( ) 6V x E I a x= −

local stiffness matrix

2 2

3

2 2

12 6 12 6

6 4 6 212 6 12 6

6 2 6 4

a a

a a

b b

b b

L LV vM L L L LE IV vL LLM L L L L

−⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪θ−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥− − −⎪ ⎪ ⎪ ⎪⎢ ⎥

⎪ ⎪ ⎪ ⎪θ⎢ ⎥⎩ ⎭ ⎩ ⎭−⎣ ⎦

bending strain energy

2

0 2

L MU dxEI

= ∫

2

2 22 3 ( ) 3b a b aa a b b a b

E I v v v vUL L L

⎡ ⎤− −⎛ ⎞= θ + θ θ + θ − θ + θ +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Example

P

a b

1 2 3

U WΠ = −

1 1 3 3 0v v= θ = = θ =

2 22 22 2 2 22 2 2 2 22 2

2 23 3 3 3E I v v E I v v P va a b ba b

⎡ ⎤ ⎡ ⎤Π = θ − θ + + θ + θ + −⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

20

v∂Π

=∂

and 2

0∂Π=

∂θ

2 23 3 2 21 1 1 112 6E I v E I Pb a b a

⎛ ⎞ ⎛ ⎞+ + − θ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 22 21 1 1 16 4 0E I v E I

b ab a⎛ ⎞ ⎛ ⎞− + + θ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

3 3

2 33 ( )P a bv

E I a b= −

+

2 2

2 3( )

2 ( )P a b a b

E I a b−

θ =+