In a disk test performed on a specimen 40-mm in diameter and 5 m thick, the specimen fractures at a...

Question

In a disk test performed on a specimen 40-mm in diameter and 5 m thick, the specimen fractures at a stress of 500 MPa. What was the load on the disk at fracture?

Show that, according to the distortion-energy criterion, the yield stress in plane strain is 1.15Y where Y is the uniaxial yield stress of the material.

Dr Shweta · Accepted Answer

Ans.1 Given:  When a disk test was executed on a specimen which was 40-mm in diameter and has a thickness of 5 m then the specimen gets fracture at the stress of 500 MPa. 
To find: The load on the disk at the time of fracture 
Calculation: In this case, the compression force is applied and the stress is calculated as: 
Stress = 2P/πDt ------[1]
Here, P is the load applied =?

In a disk test performed on a specimen 40-mm in diameter and 5 m thick, the specimen fractures at a stress of 500 MPa. What was the load on the disk at fracture? Show that, according to the...

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