attaching two materials of different modulus of elasticity

[Sjab]

Hi everybody,

I have the following problem:

I use two component epoxy to attach ceramic (270000 N/mm2) to a brick (14000 N/mm2). The ceramic piece is very thin and small with respect to the brick.

When the brick is subject to compressive stress, the brick will have a strain that follows Hooks low. What will be the strain in the ceramic piece?

I would assume that the epoxy has also its deformation and plays important role in transmitting the strain to the ceramic
I attach a simple drawing.

Any hint on how to make this calculation?

Thank you

 

[blakmax]

The LOADs will distribute in proportion to the modulus times thickness ratio if you are far enough away from the ends of the bond. Hence the load in the ceramic will be the total load P multipled by (Et)ceramic/((Et)ceramic + (Et)brick).

Once you have the load, you can work out the strain using Hooke's Law.

You just need to ensure you have sufficient overlap to enable the shear stress distribution to decay to near-zero in the centre of the joint for this calculation to be valid. (Do NOT use an average shear stress calculation. It is totally meaningless and has no relationship with the real world. Pity 78% of aircraft manufacturers use average shear stress to design aircraft joints.)

Regards

blakmax

 

[Sjab]

Hello Blackmax,

The ceramic piece is a thick film sensor.

The result of strain calculated this way is much bigger from the sensor reading.

I think the relationship you suggest is valid if the cross sectional areas of ceramic and brick is the same.

It seems that some geometrical factor is missing.

What do you think?

Jabir

 

[blakmax]

No, the load should be distributed between the two components in proportion to the proportion to the relative stiffness, and these terms already include thickness terms.

Let me be clear: Are the strain readings from the sensor higher or lower than those calculated?

 

[Sjab]

From the sensor reading, The strain in the sensor is 1/10th the calculated strain on the brick.

I think in this scenario the sensor has shear strain.

 

[blakmax]

I suggest that you have a very inefficient bond to the brick. Given the substanitaly higher modulus for the sensor and the comparative thinness of the sensor, it is not probable that shear through the sensor is your problem.

There will be several possible causes of inefficient bonds: inadequate overlap length, poor adhesive properties due to poor mixing of the adhesive, inadequate surface preparation, excessive service temperatures etc.

I'd need more details of all of the above issues to help you.

 

[Compositepro]

In the limit as ceramic and adhesive thickness approaches zero, the strain in the ceramic and the brick will be equal. This is the only condition where the analysis is simple. You say the ceramic is ten times stiffer than the brick. As the ceramic becomes thicker It will carry more load and strain less, creating complex stress patterns in the brick and ceramic. As the adhesive becomes thicker it will experience more shear strain and transfer less load between the adherands. Bondline end effects will become more pronounced. Only a finite element analysis will tell you what is going on.

 

[blakmax]

Sorry Composite Pro, but the strain in both adherends will be the same even for finite adhesive thicknesses. The non-uniform shear stresses in an adhesive bond have been known since the mid-1930s (Volkerson) and the behaviour of materials with different moduli, thickness and even thermal expansion coefficients were handled by Hart-Smith in the 1960s even for elatic-plastic behaviour in the adhesive. Load attraction effects for bonded doublers were handled by Rose in the 1980s. So there is no real need for any FEA.

Adhesive shear stresses peak at both ends of a bonded doubler, and if the overlap length is large enough, the shear stresses decay to zero in the middle of the joint. The reason they decay to zero is that the strain in both adherends is the same and will be a value as calculated above. There will be a strain ditribution in the ceramic from zero at the ends (no load in the adherends there) to the maximum once the shear stresses decay to zero.

Now if the overlap length is inadequate (or if the adhesive shear modulus is too low) and the shear stresses do not decay to zero, then your assertions may be correct.

Regards

blakmax

 

[Sjab]

Blackmax,

when you say strain distribution in the ceramic, I think you mean from the joint up, in the Y direction right?

I understand that the shear strain in the two joints should be the same. So it is the same as that of the brick.

What is the relation between F1 and F2 in the attached image?

the joint area on brick is A2 and the joint area on ceramic is A1. The cross section of ceramic is a1.

 

[blakmax]

Sjab

The strain in the ceramic is NOT uniform and it appears that your model assumes that it is. In reality where you have F1, there can be no force in the ceramic. The force will increase from zero at the end to a maximum towards the centre. That force is generated by adhesive shear transfer.

I will try to give a "canned" lesson in bonded joint mechanics. When your brick is loaded, a strain will develop in the brick. That causes a difference in displacements at the END of the ceramic between the brick and the ceramic (which at that point has no load so it is not strained). The relative DISPLACEMENT between the brick and the ceramic at the end of the ceramic will cause a shear strain in the adhesive. That shear strain causes a shear stress which will cause SOME load to be transferred from the brick into the ceramic. The ceramic, now being under load, will start to strain. I stress that this is not a uniform strain. The additional strain in the ceramic then actually reduces the displacement difference between the brick and the ceramic, so the shear strain in the adhesive (caused by the displacement difference) gradually reduces. So does the shear stress.

Eventually the adhesive transfers load until the strain in the ceramic matches the strain in the brick, and then there is no displacement difference accross the adhesive layer, so ther shear strains and shear stresses are zero. (There is a corresponding reduction in strain in the brick, but given the thickness that will not be significant.)

The same effect is occurring at the other end of the ceramic as load is transferred out of the ceramic. In between the two locations where the shear stress in the adhesive decays to zero, the strain in the ceramic will be uniform.

I suggest that you try reading Hart-Smith, L.J., Design and Analysis of Adhesive Bonded Joints, Air Force Conference on Fibrous Composites in Flight Vehicle Design, Dayton OH, Sep 26-28 1972.

Failing that I can send some lecture material I have which explains this better than I can by text alone.

http://www.adhesionassociates.com

Regards blakmax