Lecture 15: Revisiting bars and beams. - UVic.camech420/Lecture15.pdf · Lecture 15: Revisiting bars and beams. ! In Lecture 13 we stated that any structural/elastica problem could

MECH 420: Finite Element Applications

Lecture 15: Revisiting bars and beams.

n  §3.10 Potential Energy Approach to Derive Bar Element Equations. n  We place assumptions on the stresses developed inside the bar. n  The spatial stress and strain matrices become very ‘sparse’. n  We add (ad-hoc) nodal forces in the potential energy expression. n  The surface traction and body forces only exist in the x

direction. n  When we integrate over the volume (V(e)) and surface area (A(e))

of the element, the integrals reduce to simple 1-D integrals. n  This is one benefit of (reason for) the “finite element.” n  The other simplification comes from our definition of the

displacement field within the element.



An element of surface area S1, length L, and constant Young’s Modulus, E. The bar is subject to external body forces Xb, nodal loads fix, and surface traction Tx. The body

load and surface traction are forms of “distributed loads.”



n  Total potential energy is given by:

n  Consider the elastic potential energy stored internally:

n  Then consider the potential energy of the external loads.

dU = σ x dεx0

εx∫

"

#$$

%$$

&

'$$

($$

dV

U = dUVe∫∫∫ = Eεx dεx

0

εx∫

"

#$$

%$$

&

'$$

($$

Ve∫∫∫ dV =

12Ve

∫∫∫ Eεx2dV =

12

σ xεx dVVe∫∫∫

Ω = − uX b dV − uTx dSS1

∫∫ − fixi=1

2

∑Ve∫∫∫ dix

P Uπ = +Ω



n  When a body is at equilibrium (static or dynamic equilibrium), the total potential energy is at a local minimum.

n  The principle of virtual work just reflects the fact that if state variables are perturbed the rate of change of energy is 0.

Pπ

1a*1a

1 * *1( ,..., )

0P

Na aaπ∂

=∂

The problem is to calculate the values of a1, a2, …, aN that put the system in equilibrium.



n  Step 2: define the displacement field (or function).

n  The ‘correct’ state variable values will ensure a state of minimum potential energy.

u = Nd = 1− xL

xL

"

#$$

%

&''

d1x

d2x

()*

+*

,-*

.* ; N1 =1− x

L , N2 =

xL

∂π P∂ d{ }

= 0 or ∂π P∂d1x

= 0 , ∂π P∂d2x

= 0"#$

%$

&'$

($

T

u =uvw

!

"#

$#

%

&#

'#

; v = w = 0



n  Recall that:

n  Applying these definitions of strain, the elasticity matrix, and stress (which are now particular to our element type) we can form the potential energy expression…

εx = Su = SNd =∂∂x

"

#$

%

&' 1− x

LxL

"

#$$

%

&''

d1xd2x

)*+

,+

-.+

/+=

−1L

1L

"

#$$

%

&''

d1xd2x

)*+

,+

-.+

/+

B"# %&= SN =−1L

1L

"

#$$

%

&''

σ x = Dεx = DBdD"# %&= E

Strain

Stress (Constitutive Eqns)



n  Since we have set constant material and geometric properties:

ˆ ˆ ˆ ˆ2

T T TP

AL d B DB d d fπ = ⎡ ⎤ −⎣ ⎦

Note: the integrand expression is evaluated only at points on the element

surface.

1

2 1

ˆˆ ˆ ˆ

ˆx T T

x bx VeS

ff N T dS N X dV

f

⎧ ⎫⎪ ⎪= + +⎨ ⎬⎪ ⎪⎩ ⎭

∫∫ ∫∫∫

ˆ Sf ˆ Bf1 1

1 0

ˆˆ ˆ ˆ ˆ(1 )

where perimeter of the bar cross section.

Sx x x

L

S

xf N T dS P T dxL

P

= = −

≡

∫∫ ∫

Note the similarity to the work equivalence

treatment of distributed loads.

ˆ the distributed axial load we dealt with in our first visit to sec. 3.10.

xPT ≡



n  Enforcing the stationarity of :

n  Reading: §4.7. Potential Energy Approach to Derive Beam Element Equations (pg. # 218).

Pπ

{ }ˆ ˆ

1

ˆ ˆ ˆ

11 11ˆ

0

1 1

ˆ

111

T

T

P AL B DB d f

f kd

AL AELk B DB AL

d

L EL L

L

π= ⎡ ⎤ −⎣ ⎦

=

⎡ ⎤−⎢ ⎥ −⎡ ⎤⎡ ⎤= ⎡ ⎤ = − =⎢ ⎥⎣ ⎦ ⎢ ⎥⎢

=∂

⎥ −⎣ ⎦ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦



n  In Lecture 13 we stated that any structural/elastica problem could be reduced to Cauchy’s equations of motion.

n  When we set the displacement field by mixing discrete displacements using shape functions…

A(u) = 0

u = NaA(Na) ≈ 0 The method of weighted residuals states

that the original problem can not be exactly satisfied by the assumed displacement field.

The MWR seeks to solve another equation.



n  The error in the original equation is termed the residual.

n  The MWR looks to minimize the residual over a certain domain. n  The minimization is entirely ‘logical.’

n  The set of weighting functions used defines the type of criterion being applied. n  Subdomain collocation n  Galerkin’s method

( ) ( )A Na R a=

( )

( ) 0T

eVw R a dV⋅ =∫

w I=

w N=



n  In MECH 420 (and in most generalized FEM works) the Galerkin method is chosen – why?

n  Consider the variational principle of stationary potential energy.

δπ P = δuT A(u)

V (e)∫ ⋅dV = 0

u = Naδu = Nδa

δuT = δaT NT

0 = δaT NT A(Na)V (e)∫ ⋅dV

0 = NTR(a)V (e)∫ ⋅dV

If a natural variational principle exists, then the Galerkin

formulation will match it - guaranteed

A ‘variational principle’ a.k.a. an ‘energy functional’.



n  “Since Rayleigh-Ritz requires an energy functional, its use is limited to differential equations which have such functionals. Galerkin’s method does not need or use a functional and thus can be applied to equations where Rayleigh-Ritz can not.”

n  “Galerkin’s method can be applied literally to any differential equation, but when applied to a differential equation with an energy functional, it agrees exactly with the Rayleigh Ritz solution.”

n  “…the variational method is only applicable to those problems for which a variational priciple exists and has been found, a common situation only when the system of equations is linear and self-adjoint.”

n  “From a pragmatic standpoint the main shortcomings of variational formulations is that the variational methods they support provide no approximation scheme that can not be set-up more simply and quickly as one … or another version of MWR.”



n  The ‘design’ of a MWR solution relies on putting the original equation in a weak form.

n  We did this within the derivation of the potential energy expression.

n  The MWR gives the ‘user’ the freedom to integrate by parts as they ‘see fit.’

( ) ( ) ( )0 ( ) ( ) '( ) ( ) ( )T T T

e e eV V AN A Na dV B N A Na dV C N D Na dA= ⋅ ⋅ = − ⋅ ⋅ + ⋅ ⋅∫ ∫ ∫

Original Problem ‘Weak Form’ of the Problem

Integration by parts (or Green’s Theorum) A’ is a set of differential equations with lower order differentials than A.



n  The weighted residuals method provides a means to “sculpt” your finite element solution.

wiT A(u)

Ve∫ dV = 0

BT (wi ) !A (u)Ve∫ dV + CT (wi )D(

u)Ae∫ dA= 0

Boundary terms

Reduced order differentials on the displacements.

Higher order differentials on the weighting functions.



n  The principle of virtual work illustrates the relationship between variational and Galerkin formulations.

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Lecture 15: Revisiting bars and beams. - UVic.camech420/Lecture15.pdf · Lecture 15: Revisiting bars and beams. ! In Lecture 13 we stated that any structural/elastica problem could