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3. Seminar Energy Methods, FEM Class of 2013

Topic Potential of Internal Energy 30.10.2013

A c c e s s

1 MPE: Planar truss system

Wanted: internal potential energy depending on nodal displacements v1(k) and v2(k) of nodesk= 1 . . . 5 in terms ofi = f

v1(k) , v2(k)

for linear elastic material and small deformations

1 2

34 5

x41

x12

x31

x43 x35

x52x32

x ,v2 2

x ,v1 1

E, A = const.L = L (for all truss members)ik

remember last seminar: i=VwidV

1.1 internal potential for a truss member

definition of truss member: stress/strain distribution over cross section:

i

k

x,uE, A, L

u(i)

u(k)

u, , = const x x

x

specific stress condition for each truss member (solution step 1)

x= 0 y =z = 0 trussesonly axial forces

xz =xy= 0 (no torsion and transversal loading)yz = 0 (no shearing/slipping in longitudinal direction)

stress-strain relationship (solution step 2)

x= E x (1)internal potential for each truss member (in local coordinates):

i=

V

x0

xd xd V =E

V

x0

xd xd V =1

2E

V

2xd V (2)

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Appendix B: consideration of large deformations

strain-displacement relationship

large deformations (2nd order theory, linear and quadratic part of Taylor-serie):

x =d u

d x+1

2du (x)

x2 =u+12 u2 (22)

inserting (22) in (2)

i=1

2EA

L

u+

1

2u

2

2d x (23)

ansatz-functions for displacement field

u(x) =L

L (24)

inserted in potential

=1

2EA

L2

L +

L3

L2 +

1

4

L4

L3

(25)

(for each truss member)

undeformed initial and deformed situation as before but in case of large deformations:

cos = 1sin =

forLfollows (similar as before)

L = L (1 cos) + v1 cos+ v2 sin (26)

determination of using eq. (9):

cos(+ ) = x1(k) x1(i) +v1(k) v1(i)

L+ L

sin(+ ) = x2(k) x2(i) +v2(k) v2(i)

L+ Ltan(+ ) =

L sin+v2(k) v2(i)L cos+v1(k) v1(i)

recalling:

tan(a+b) = tan a+ tan b

1 tan a tan b

can be determined

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