Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 4 (Q.No. 16)
16.
When a body is subjected to bi-axial stress i.e. direct stresses (σx) and (σy) in two mutually perpendicular planes accompanied by a simple shear stress (τxy), then minimum normal stress is
Discussion:
4 comments Page 1 of 1.
Tyagi said:
4 years ago
Minimum Stress will be the shortest distance from origin to the nearest point on Mohr's circle.
Minimum Stress = Distance of Mohr circle center from origin - Radius of Mohr circle.
Distance of Mohr circle center from origin = (σx + σy)/2.
Radius of Mohr circle = (((σx - σy)^2 + 4Txy^2)^(1/2))/2.
Minimum Stress = Distance of Mohr circle center from origin - Radius of Mohr circle.
Distance of Mohr circle center from origin = (σx + σy)/2.
Radius of Mohr circle = (((σx - σy)^2 + 4Txy^2)^(1/2))/2.
Anil Kumar said:
6 years ago
Please explain.
Omkar gund said:
7 years ago
Please Explain the answer.
R k said:
8 years ago
Any one please explain it.
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