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8/13/2019 Seminar03 MPE Truss
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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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