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.
Kunal said:
5 years ago
Right, Thanks @Sanjib.
Abhishek singh said:
5 years ago
Question is for resultant stress not for normal stress. So the given answer is right.
SURAJ PRASAD KESHARI said:
4 years ago
Thank you all for explaining this.
Vishal said:
3 years ago
Yes, the answer given is right, we have to use 1+cos2θ = 2cos^2θ.
AKULA ABHILASH said:
3 years ago
Shear stress = σ/2 sin2θ.
Normal stress = σcos^2 θ.
Resultant stress = σcosθ.
Normal stress = σcos^2 θ.
Resultant stress = σcosθ.
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