Calculating Tensile Lap Shear Strength: How Can I Determine the Shear Area?

Click For Summary
SUMMARY

The discussion focuses on calculating tensile lap shear strength in a lap-shear test involving bonded copper foils. The breaking force recorded was 500N, but confusion arose regarding the correct shear area for the calculation. The consensus is that the shear area should be determined by multiplying the bond's length by its width. Additionally, it was established that if the copper fails before the bond, the shear strength of the bond cannot be accurately calculated, as the bond's strength is likely higher than the measured shear stress.

PREREQUISITES
  • Understanding of tensile testing principles
  • Familiarity with shear strength concepts
  • Knowledge of material properties, specifically copper
  • Basic mathematical skills for area calculations
NEXT STEPS
  • Research "Tensile testing standards and methods" for comprehensive testing protocols
  • Learn about "Shear strength testing of adhesive bonds" to understand bond evaluation
  • Explore "Material failure analysis techniques" to assess failure modes in tests
  • Investigate "Copper alloy properties and applications" for insights into material behavior
USEFUL FOR

Materials engineers, mechanical engineers, and researchers involved in adhesive bonding and tensile testing will benefit from this discussion.

Javaid
Messages
2
Reaction score
0
Hi,
I am new to the tensile testing and still learning and would really appreciate the answer to a problem that I have been having.
I performed a lap-shear test on two foils of copper bonded in single lap joint. The copper broke before the bond does giving me a force of say 500N. I want to calculate the tensile lap shear strength. As per my understanding it should be achieved by by dividing the breaking force, in Newtons (500N), by the shear area, in square millimetres. But which is the shear area? Is it the width of the foil/bond multiplied by the thickness of the bonded area or is it the width of the foil/bond multiplied by the length of the bond?
Also, can this test give me the shear strength of the bond? (As the answer to this question is based on the answer to the previous question, I will ask it afterwards but it is here in case someone is kind enough to solve this problem for me)
Thanks in advance :)
 
Engineering news on Phys.org
Javaid said:
Hi,
But which is the shear area?
Hi; welcome aboard.
While I can't help with the actual problem, I have to ask you about terminology. Regardless of what you're trying to test, I would consider the point at which the foil tears to be the "shear area". I'm not educated in the field, though, so I might be missing something.
 
By shear area (cross-sectional area in square millimetres), I meant the value that will be used to divide with the breaking force to calculate tensile lap shear strength.
The foil tears outside the bonded region and I am confused how to calculate the tensile lap shear strength which should be be achieved by dividing the breaking force, in Newtons (500N), by the shear area, in square millimetres. Hence the question...
 
Okay. Good luck with it. I haven't a clue, but someone here will be able to help.
 
Javaid said:
By shear area (cross-sectional area in square millimetres), I meant the value that will be used to divide with the breaking force to calculate tensile lap shear strength.
The foil tears outside the bonded region and I am confused how to calculate the tensile lap shear strength which should be be achieved by dividing the breaking force, in Newtons (500N), by the shear area, in square millimetres. Hence the question...
You can't calculate the shear strength of the bond if the metal itself already failed. The shear area of the bond you want is length times width. The tensile breaking strength of the copper divided by the shear area gives you the shear stress in the bond at the time of failure of the copper itself. The shear strength of the bond will thus be higher, so if you want to test it, you have to use thicker copper or a smaller shear area (reduced bond length) I would think.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
18K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
11K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K