Numerical Analysis of Mechanical Test Methods for Evaluating Shear Strength of Joint by Using Interface Element

Abstract:

As examples of the most typical methods to determine the shear strength of SiC/SiC composite joints, the tensile test of lap joined composite and the asymmetrical four point bending test of butt joined composite were analyzed by using finite element method with the interface element. From the calculation results, it was revealed that the strength in the tensile test was strongly influenced by the residual stress as the increase of the joint layer thickness. In the case of asymmetrical bending test, it was found that the crack initiation point would move due to the residual stress and the strength was also affected by the joint layer thickness.

• Periodical: Key Engineering Materials (Volumes 345-346)
• Pages: 1489-1492
• DOI: https://doi.org/10.4028/www.scientific.net/KEM.345-346.1489
• Online since: August 2007
• Authors: Hisashi Serizawa, Kazuaki Katayama, Charles Lewinsohn, Mrityunjay Singh, Hidekazu Murakawa
• Keywords: Ceramic, Finite Element Analysis Method, Interface, Joining, Shear Strength

References

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DOI: 10.1016/s0022-3115(00)00247-6

[10] -4.

[10] -3.

[10] -2.

[10] -1.

[10] [0] [10] [1] [10] [2] Scale Parameter r 0 (µ m).

[10] [1] [10] [2] [10] [3] [10] [4] [10] [5] Fracture Load (N) Without Inherent Strain With 5. 0 x 10 -4 Inherent Strain With 1. 0 x 10 -3 Inherent Strain Thickness of Joint = 1 µ m -1 Fig. 2 Effect of r0 and inherent strain on fracture load in lap joint (1 µm thickness).

DOI: 10.1080/09507116.2012.708501

[10] -4.

[10] -3.

[10] -2.

[10] -1.

[10] [0] [10] [1] [10] [2] Scale Parameter r 0 (µ m).

[10] [1] [10] [2] [10] [3] [10] [4] [10] [5] Fracture Load (N) Without Inherent Strain With 5. 0 x 10 -4 Inherent Strain With 1. 0 x 10 -3 Inherent Strain Thickness of Joint = 10 µ m -1 Fig. 3 Effect of r0 and inherent strain on fracture load in lap joint (10 µm thickness).

DOI: 10.1080/09507116.2012.708501

[10] -4.

[10] -3.

[10] -2.

[10] -1.

[10] [0] [10] [1] [10] [2] Scale Parameter r 0 (µ m).

[10] [1] [10] [2] [10] [3] [10] [4] [10] [5] Fracture Load (N) Without Inherent Strain With 5. 0 x 10 -4 Inherent Strain With 1. 0 x 10 -3 Inherent Strain Thickness of Joint = 100 µ m -1 Fig. 4 Effect of r0 and inherent strain on fracture load in lap joint (100 µm thickness).

DOI: 10.1080/09507116.2012.708501

[10] -3.

[10] -2.

[10] -1.

[10] [0] [10] [1] [10] [2] [10] [3] Scale Parameter r 0 (µ m).

[10] -1.

[10] [0] [10] [1] [10] [2] [10] [3] [10] [4] [10] [5] [10] [6] Fracture Load (N) Thickness of Joint = 50 µ m Thickness of Joint =100 µ m Thickness of Joint =200 µ m With 1. 0 x 10 -3 Inherent Strain -1 Fig. 5 Effect of r0 and joint thickness on fracture load in asymmetrical bending test.

DOI: 10.31390/gradschool_dissertations.2025