Using Contact Force to Determine Adhesive Bonding Characteristics

[smirnov.da]

Can I use the contact force to obtain a force between two surfaces of the parts to determine the characteristics of the adhesive joint? If the modulus of elasticity is the same, then the force of the sum contact force(Fy) per slice on one surfase part  is comparable to the applied force. If the modules are different, the sum contact force(Fy) per slice is different from the applied force. The example shows two plates with different modulus of elasticity (200,000 MPa and 25,000 MPa), between them a bonded contact. If I make a sliding constants in a around, then I get half of the given force on each part. I think this is due to Hooke's law, but I can't find an explanation yet. Haw determinet correct forse for determinet the characteristics of the adhesive joint?

[John_Holtz]

Hi @smirnov.da 

The sum of the contact force is independent of the modulus of elasticity of the parts. If the results do not confirm that, then there is probably a warning in the analysis that indicates the results are inaccurate.

[smirnov.da]

Hello, John.  Thanks for the answer. There is a warning message. The calculation is not correct?

The results of the contact force values in the elasticity modulus part 25 GPa are shown below.

One node has a value of -615. If you remove this value, the total Y force of the part is equal to the total Y force of the other part.
But if there is a large assembly, then how not to determine the correct contact force?

I also made another calculation. There he is.

The contact force values are shown below. They differ in places in the same values ​ ​ along the X axis.

Warning calculation does not have.

I use Nastran Inventor 2022.

[smirnov.da]

Applied load 500 N.

[smirnov.da]

Show the correct Y contact force data for the last calculation. They differ from the applied load and SPS force.

[John_Holtz]

Hi @smirnov.da 

Please attach the model. (Compress the assembly file .iam and part files .ipt and attach the .zip or .rar to the forum post.)

The only think I see is that there are at least 54 nodes in contact (based on the G3051 warnings), but you are summing the contact force from only 14 nodes. Perhaps there are 27 nodes on one face and 27 on the other face, but that still leaves 13 nodes in contact that you are not including in your sum. If the contact force at the other nodes really zero?

[smirnov.da]

Hi, John. 

I send two examples.

[John_Holtz · Accepted solution]

Hi @smirnov.da 

Do not use "Continuous Meshing" with solid parts.

Continuous Meshing only matches the mesh at some nodes around the perimeter of the contact when using solids. (By matching, I mean the same node number is used on both parts.) If a node is matched, there is no contact at that node. Since there is no contact, there is no contact force.

In other words, your two parts were connected by a matched mesh at some locations and by contact at other locations. Some of the load was transmitted through the matched mesh, and some of the load was transmitted through the bonded contact. Therefore, the summation of the contact force was only a portion of the total load transmitted.

When not using continuous meshing, the entire load is transmitted through the contact, and the sum of the contact forces is as expected. (These results are for assembly2.iam.)

	Result:	[202] T1 CONTACT FORCE	[203] T2 CONTACT FORCE	[204] T3 CONTACT FORCE
Subcase	Node
[1] SUBCASE 1	1	-0.36	-49.61	0.22
[1] SUBCASE 1	2	0.19	-49.36	-0.07
[1] SUBCASE 1	46	0.11	-51.13	0.20
[1] SUBCASE 1	47	-0.07	-33.10	-0.22
[1] SUBCASE 1	48	0.02	-27.00	-0.05
[1] SUBCASE 1	49	0.24	-39.15	-0.12
[1] SUBCASE 1	61	0.22	-51.23	-0.10
[1] SUBCASE 1	62	0.05	-33.66	0.04
[1] SUBCASE 1	63	0.09	-27.67	0.03
[1] SUBCASE 1	64	-0.54	-39.47	-0.10
[1] SUBCASE 1	100	-0.32	-12.78	0.21
[1] SUBCASE 1	101	0.15	-35.79	0.00
[1] SUBCASE 1	103	0.04	-36.54	-0.06
[1] SUBCASE 1	104	0.17	-13.42	0.04
[1] SUBCASE 1	Sum	0.00	-499.91	0.00

Let us know if you have any questions. 

[smirnov.da · Accepted solution]

Hello, John. Thanks for the answer! I think that I will have questions, but on other topics.