Seminar04 MPE

8/13/2019 Seminar04 MPE

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4. Seminar Energy Methods, FEM Class of 2013

Topic Principal of Minimum of Potential Energy (MPE) 07.11.2013

A c c e s s

1 Principal of Minimum of Potential Energy (MPE)

total potential of the potential energy of an elastic body

= i + e (1)

i ... internal potential energy stored in the body/system

e ... potential energy due to external loads, e.g., body forces, traction forces, etc

= f (a) dependent on unknown parameters a,

e.g. in terms of deformations and rotations v with = f (v)

potential of internal energy

dwi

d

i = W i = V wi dV = V : d dV (2)

potential of external energy

e = W e

= V

p (x,y,z ) v (x,y,z ) d V (3)

p ... external loads (body forces, surface loads, line loads, single loads)

minimisation principal (extremum conditions)

necessary condition for extremum

= (i + e) = 0 (4)

condition for minimum

2 0 (5)

variation in direction of unknowns (.) = (.)

a a with unknown parameters a

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exact fullment of extremum conditionsEuler ian differential equations (e.g. one-dimensional)

f y

ddx

f y

= 0

lead to differential equilibrium conditionsa x 2 y + b x y + c y = (x)

e.g. for a beam

EI d4 wdx 4

= 0

e.g. for a plate

w = p

K

approximate fullment of extremum conditions ( Ritz -method)

ansatz-functions with unknowns

yN (x) = 0 (x) +n

i=1

a i i (x) (6)

yN (x) ... approximation/ansatz-function for displacement

0 (x) ... function for particular solution u = 0 at B (boundary)

i (x) ... homogeneous solution function for u = 0 at B

a i ... unknowns

requirements:

yN need to full kinematic boundary conditions (displacement boundary conditions) yN not necessary satisfy natural boundary conditions (traction boundary conditions)

solution steps

1. chose specic stress condition (e.g. plane stress, plane strain)2. set stress-strain relationship (e.g. Hooke s law)

3. set strain-displacement relationship (e.g. truss, beam, linear Bernoulli )

4. nd displacement eld over the body which leads to minimum potential of the totalsystem under consideration of boundary conditions

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2 Example

p(x)

S 0

s(x)

x, u

y, v

L

p 0I (x) = I 0 1

x3

2 L 3

s (x) = S 0

L p (x) = p0

xL

E = const.

wanted: maximum moment ( 2nd order theory)

simplication: negligence of torsion, lateral shear and axial strain energy

2.1 potential of internal energy

specic stress state (solution step 1)

x = 0 y = z = 0 bending (without axial strains)

xz = xy = 0 (no torsion and lateral shear)

yz = 0 (no shearing/slipping in longitudinal direction)

stress-strain relationship (solution step 2)Hookes law

x = E x

internal potential

i = V x

0

x d x d V = E V x

0

x d x d V = 12

E V 2x d V (7) strain-displacement relationship (solution step 3)

x = d u

d x (8)

kinematic hypothesis (Bernoulli)

y,v

u =u(y=0)0

u(y)=u - y0

= d v

d x

x = dd x

(u (x, y = 0) (x) y)

= d u

d x

d2 vd x 2

y = u v y

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internal potential under negligence of axial strains ( u 0)

i = 1

2E V ( v y)2 d V

12

E V v 2 y2 d V with A y2 dA = I zi =

12

E x I z (x) v (x) 2 d x (9)2.2 potential of external energy

axial loading transversal loading

x, u

y, v S(a) s(x)

P(a)p(x)

dx

single load W e = S (a) u (a) W e = P (a) v (a )

line load dW e = s (x) u (x) dx dW e = p (x) v (x) dx

W e = s (x) u (x) dx + C W e = p (x) v (x) dx + C total load W e = S (a) u (a) + L0 s (x) u (x) dx W e = P (a) v (a ) + L0 p (x) v (x) dx

potential of external energy (in general)

e = i

P (i) v (i) L

0

p (x) v (x) dx j

S ( j ) u ( j ) L

0

s (x) u (x) dx (10)

2nd order theory: relationship between u (x) and v (x) (equilibrium on deformed system)

dx

x,uy,v

dx

-du=v

du = dx dx cos

= dx

(1

cos

)

= dx 1 1 2

2! +

4

4!

small-angle approximation dx

2

2 = dx

v 2

2 (11)

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potential of external energy for example

e = L

0

p (x) v (x) dx + S 0 u (x = L) +L

0

s (x) u (x) dx (12)

with

u (x) = x

0

v (x) 22

dx + u0 u (x = 0) = 0 u 0 = 0

u (x = L) = L

0

v (x) 2

2 dx

nally

e = L

0

p (x) v (x) dx S 0 L

0

v (x) 2

2 dx

L

0

s (x) x

0

v (x) 2

2 dx dx (13)

2.3 ansatz-function

ansatz-function in form of power series with 4 unknowns

vN (x) = a 0 + a 1 x + a 2 x2 + a 3 x 3 =

3

j =0

a j x j (14)

boundary conditions

v (x = 0) = 0 v (x = L) = max

v (x = 0) = 0 v (x = 0) = 0

kinematic BC ( v, v ) need to be fullled

natural BC ( v , ...) not necessary satised

possible:

vN = a 2 x 2 constant moment (allowed, but no good approximation)

vN = a 3 x 3 linear moment distribution, but M (x = 0) = 0 (allowed)

vN = a (x 2 + x3 ) linear moment distribution and M (x = 0) = 0

vN = a 2 x 2 + a3 x 3 2 independent unknowns increasing effort

build derivations, e.g.

vN (x) = ax2

vN (x) = 2 axvN (x) = 2 a (15)

v 2 = . . . v 2 = . . . insert in and build integral

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2.4 MPE

total potential

= i + e

= 1

2E

L

0

I z (x) v (x)2

d x

L

0

p (x) v (x) dx S 0 L

0

v (x) 2

2 dx

L

0

s (x) x

0

v (x) 2

2 dx dx (16)

insertion of ansatz-function, loads and geometrical values

= 2EI 0 a 2 L0 1 x3

2 L 3d x

p0

L a

L

0x 3 dx 2S 0 a 2

L

0x 2 dx 2

S 0

L a 2

L

0x

0x 2 dx dx (17)

integration

= 2EI 0 a 2 x x4

8 L 3L

0

p0L

a x 4

4

L

0

2S 0 a 2 x 3

3

L

0

2S 0L

a 2L

0

x 3

3 dx

= 2 EI 0 a 2 L L8

p0L

a L 4

4 2S 0 a 2

L3

3 2

S 0L

a 2 x 4

12

L

0

= 2 EI 0 a2

78L p0 a

L3

4 2S 0 a2

L 3

3 2S 0L a

2

L 4

12 (18)

total potential depending on unknowns

= EI 0 74

L a 2 14 p0 L 3 a

56

S 0 L 3 a 2 (19)

extremum condition for determination of unknowns

= 0 d (a)

da a (one unknown)

(a 0 , a 1 , . . . , a n )

a j= 0 (n unknowns, system of equations)

for example

d (a )da

= 0 = EI 0 72

L a 14 p0 L 3

53

S 0 L 3 a (20)

determination of unknown a

a =14 p0 L

2

EI 0 72 53 S 0 L

2 (21)

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2.5 determination of moment

in general moment obtained by

M (x) = EI z (x) v (x)

cantilever beam: maximum moment M max = M (x = 0)

example

E = 2 107 kN m 2 L = 1 m I 0 = 10 4 m 4

S 0 = 10 kN p 0 = 10 kN m

v N x( ) a x2

0 0.2 0.4 0.6 0.8

1 10 4

2 10 4

3 10 4

4 10 4

v N

x( )

x

v N

L( ) 0.358 mm

N x( ) 2a x

0 0.2 0.4 0.6 0.8

2 10 4

4 10 4

6 10 4

8 10 4

N x( )

x

M x( ) E I x( ) 2 a

0 0.2 0.4 0.6 0.8

1.5 103

1 103

500

M x( )

x

M 0m( ) 1.432 kN m

M L( ) 0.716 kN m

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