Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 3 (Q.No. 6)
6.
The ratio of endurance limit in shear to the endurance limit in flexure is
Discussion:
4 comments Page 1 of 1.
Awais said:
5 years ago
Not getting this, please anyone explain in detail.
Paatel said:
5 years ago
Endurance limit is the opposite force to the applying force, actually endurance limit is 100 in bending and shearing or twisting depending upon the material endurance limit in flexure means moving flexible so endurance limit is 100/2 50, numerically equal to 55.
Manoj suthar said:
7 years ago
While calculating the endurance limit we take some constants so that it can meet the actual working conditions where one is;
Ka for size.
Kb for surface finish.
Kc for the type of load.
Kc=1 for bending.
Kc =.56 for twisting.
Kc = 74 for axial loading.
Ka for size.
Kb for surface finish.
Kc for the type of load.
Kc=1 for bending.
Kc =.56 for twisting.
Kc = 74 for axial loading.
Shivakumar said:
7 years ago
Can anyone explain the solution?
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