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[Xansys] [EXPLICIT] Tiebreak contact definition
 
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ilker.kazaz
User


Joined: 26 Jan 2011
Posts: 6

PostPosted: Wed Jul 02, 2014 6:01 am  Reply with quote

Hi members,

I am using Explicit Dynamics (LS-DYNA explicit) under component systems of
Workbench to model a masonry tower structure. A tiebreak contact with
tiebreak failure criterion that has normal and shear components is defined
between two parts standing on top of each other. Material is elastic for two
parts. Parts are meshed with solid elements ELFORM 13, "1 point tetrahedron
with nodal pressure".

The problem is that soon after the excitation the tiebreak develops and the
top part wanders around on the bottom part. I mean that the bond seems to be
weaker that it should be. I am suspicious about the contact definition.
The card entry in the k file defining the contact is given below. Units Pa.

I am asking whether the card is correct or should I seek some other source
of problem.



*CONTACT_AUTOMATIC_ONE_WAY_SURFACE_TO_SURFACE_TIEBREAK

$ 1SSID 2MSID 3SSTYP 4MSTYP 5SBOXID
6MBOXID 7SPR 8MPR

318 319 0 0
1 1

$ 1FS 2FD 3DC 4VC
5VDC 6PENCHK 7BT 8DT



$ 1SFS 2SFM 3SST 4MST
5SFST 6SFMT 7FSF 8VSF



$ 1OPTION 2NFLS 3SFLS 4PARAM 5ERATEN
6ERATES 7CT2CN

2 500000 1.5e+06


$ 1SOFT 2SOFSCL 3LCIDAB 4MAXPAR 5SBOPT
6DEPTH 7BSORT 8FRCFRQ

1



Best regards

Ţlker KAZAZ, Assoc.Prof.

Department of Civil Engineering

Erzurum Technical University

25050 Erzurum/Turkey



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jose.galan
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Joined: 21 Oct 2008
Posts: 140

PostPosted: Wed Jul 02, 2014 8:26 am  Reply with quote

Dear Mr. Kazaz,

"The tied contact options actually 'glue' the contact nodes (surfaces)
to the target surfaces."

"Tiebreak contact is identical to tied contact except that the contact
nodes (surfaces) are tied to the target surfaces only until a failure
criterion is reached. This is done by 'pinning' the contact nodes
(surfaces) to the target using a penalty stiffness; after the failure
criterion is exceeded, the contact nodes (surfaces) are allowed to slide
relative to, or separate from, the target surface. The tiebreak contact
options are typically used to represent spot-welded or bolted
connections."

The failure criteria that you have specified for your tiebreak is the
following:

(abs(sigma_n)/0.5 MPa)^2 + (abs(sigma_s)/1.5 MPa)^2 <=1

where sigma_n and sigma_s are the normal and shear stress, NFLS=0.5 MPa
and SFLS=1.5 MPa are the normal and shear failure stresses,
respectively.

The tiebreak can transmit both compression and traction.

I would like to highlight the following behaviour of tiebreak contact:
"after the failure criterion is exceeded, the contact surfaces are
allowed to slide relative to, or separate from, the target surface."
That means that, after failure, no friction between surfaces is
included. Perhaps you should consider wether the tiebreak contact
reproduces the physics of your problem. If the masonry blocks rest one
on top of the other, friction will develop in the contact, which is not
taken into account in the tiebreak.

The quotations are taken from the ansys manual (Mechanical APDL>Ansys
LS-DYNA User's Guide >6. Contact surfaces > 6.2 Contact Options).

Best regards,

Jose M. Galan

Contr. Engin. Dept.

Univ. Sevilla

Spain
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ilker.kazaz
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Joined: 26 Jan 2011
Posts: 6

PostPosted: Wed Jul 02, 2014 12:01 pm  Reply with quote

Quote:
(abs(sigma_n)/0.5 MPa)^2 + (abs(sigma_s)/1.5 MPa)^2 <=1

Quote:
where sigma_n and sigma_s are the normal and shear stress, NFLS=0.5 MPa and
SFLS=1.5 MPa are the normal and shear failure stresses, respectively.

Quote:
The tiebreak can transmit both compression and traction.

This is the definition of tiebreak contact in LS DYNA. OK

Quote:
I would like to highlight the following behaviour of tiebreak contact:
"after the failure criterion is exceeded, the contact surfaces are allowed
to slide relative to, or separate from, the target surface."

Quote:
That means that, after failure, no friction between surfaces is included.
Perhaps you should consider wether the tiebreak contact reproduces the
physics of your problem.
Quote:
If the masonry blocks rest one on top of the other, friction will develop
in the contact, which is not taken into account in the tiebreak.

Yes, this is exactly what happens with the current contact definition. But
as said in my previous mail, the problem is not the sliding after contact
failure, but the failure being premature.

I tried different options but the result is all the same. In Card 4, the
following parameters results in a contact definition with normal and shear
failure stresses criteria

$ 1OPTION 2NFLS 3SFLS 4PARAM 5ERATEN 6ERATES
7CT2CN
2 500000 1.5e+006

When failure criteria is reached, the body is free to slide. However using a
card 4 defined as below results in a tiebreak failure criterion that has
only a normal stress component:

abs(sigma_n)/0.5 MPa<=1

$ 1OPTION 2NFLS 3SFLS 4PARAM 5ERATEN 6ERATES
7CT2CN
4 500000 1.5e+006 1

This option adds friction to the contact before and after failure. For
OPTION=4, SFLS is a frictional stress limit if PARAM=1. This frictional
stress limit is independent of the normal force at the tie.

I tried all these options, increased the failure stress values in both
normal and transverse direction but the result is all the same: SLIDING.

The model is something similar to a 3 m cylindrical block with a diameter of
0.8 m standing on top of 12 m tall rectangular block with 2x2 m base
dimensions. No matter what value I try for failure stresses only sliding
develops, and its initiation is insensitive to the value of shear stress.
The expected behavior is the toppling of cylindrical block. This is why I
keep the shear stress higher when compared to normal stress.

May the solid element type has an effect on this problem?

Thank you.

Ţlker KAZAZ, Assoc.Prof.
Department of Civil Engineering
Erzurum Technical University
25050 Erzurum/Turkey




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jose.galan
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Joined: 21 Oct 2008
Posts: 140

PostPosted: Thu Jul 03, 2014 4:15 am  Reply with quote

Dear Mr. Kazaz,

in your post you do not mention anything about your loads. Could you
please describe them?

Have you checked the computed values of normal and shear stresses in
your contact zone? Are they similar to the expected values? What values
do they have right before the tiebreak fails? The failure may be caused
by an excessive value of compressive stresses, or an excessive value of
traction stresses, or an excessive value of shear stresses, or a
combination of both.

Best regards,

Jose M. Galan

Constr. Eng. Dept.

Univ. Sevilla

Spain

El 02/07/2014 21:01, ─░lker Kazaz escribi├│:

Quote:
Quote:
(abs(sigma_n)/0.5 MPa)^2 + (abs(sigma_s)/1.5 MPa)^2 <=1

Quote:
where sigma_n and sigma_s are the normal and shear stress, NFLS=0.5 MPa and

SFLS=1.5 MPa are the normal and shear failure stresses, respectively.

Quote:
The tiebreak can transmit both compression and traction.

This is the definition of tiebreak contact in LS DYNA. OK

Quote:
I would like to highlight the following behaviour of tiebreak contact: "after the failure criterion is exceeded, the contact surfaces are allowed

to slide relative to, or separate from, the target surface."

Quote:
That means that, after failure, no friction between surfaces is included.

Perhaps you should consider wether the tiebreak contact reproduces the
physics of your problem.

Quote:
If the masonry blocks rest one on top of the other, friction will develop

in the contact, which is not taken into account in the tiebreak.

Yes, this is exactly what happens with the current contact definition. But
as said in my previous mail, the problem is not the sliding after contact
failure, but the failure being premature.

I tried different options but the result is all the same. In Card 4, the
following parameters results in a contact definition with normal and shear
failure stresses criteria

$ 1OPTION 2NFLS 3SFLS 4PARAM 5ERATEN 6ERATES
7CT2CN
2 500000 1.5e+006

When failure criteria is reached, the body is free to slide. However using a
card 4 defined as below results in a tiebreak failure criterion that has
only a normal stress component:

abs(sigma_n)/0.5 MPa<=1

$ 1OPTION 2NFLS 3SFLS 4PARAM 5ERATEN 6ERATES
7CT2CN
4 500000 1.5e+006 1

This option adds friction to the contact before and after failure. For
OPTION=4, SFLS is a frictional stress limit if PARAM=1. This frictional
stress limit is independent of the normal force at the tie.

I tried all these options, increased the failure stress values in both
normal and transverse direction but the result is all the same: SLIDING.

The model is something similar to a 3 m cylindrical block with a diameter of
0.8 m standing on top of 12 m tall rectangular block with 2x2 m base
dimensions. No matter what value I try for failure stresses only sliding
develops, and its initiation is insensitive to the value of shear stress.
The expected behavior is the toppling of cylindrical block. This is why I
keep the shear stress higher when compared to normal stress.

May the solid element type has an effect on this problem?

Thank you.

─░lker KAZAZ, Assoc.Prof.
Department of Civil Engineering
Erzurum Technical University
25050 Erzurum/Turkey

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| The Online Community for users of ANSYS, Inc. Software |
| Hosted by PADT - www.padtinc.com |
| Send administrative requests to xansys-mod@tynecomp.co.uk |
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