Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 2 (Q.No. 25)
25.
The resultant stress on an inclined plane which is inclined at an angle θ to the normal cross-section of a body which is subjected to a direct tensile stress (σ) in one plane, is
Discussion:
15 comments Page 2 of 2.
Sauru said:
8 years ago
Yes, I agree @Shiva.
NAITIK Trivedi said:
8 years ago
6cos is the right answer.
S. Ata said:
8 years ago
No answer is correct.
Right answer is= σ(cos θ)^2.
Right answer is= σ(cos θ)^2.
Shiva said:
8 years ago
Answer should be =σ*(cos)^2.
Daka Hitesh said:
9 years ago
Normal stress : 6 * cos2 and Shear strees : 6 * sin * cos.
Resultant : Root(Normal2 + shear2)
Resultant : 6 * cos.
Here, 6 = Sigma.
Resultant : Root(Normal2 + shear2)
Resultant : 6 * cos.
Here, 6 = Sigma.
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