Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 42)
42.
If a member is subjected to a tensile force P, having its normal cross-section A, the resulting shear stress in an oblique plane inclined at an angle θ to its transverse plane, is
Discussion:
5 comments Page 1 of 1.
Trent said:
1 year ago
For normal stress its P/A cos^2θ.
For shear or tangential stress its P/2A *sin2θ.
For resultant stress its P/A cosθ.
For shear or tangential stress its P/2A *sin2θ.
For resultant stress its P/A cosθ.
(5)
Sanjib said:
6 years ago
Given option is correct.
Normal stress is P/Acos2 θ.
Normal stress is P/Acos2 θ.
Prasenjit said:
7 years ago
Yes, it's true.
And normal stress P/A*cos2.
And normal stress P/A*cos2.
Ankit said:
7 years ago
Shear stress is nothing else it is tangential stress over inclined plane for uniaxial direct strees.
Shakeel said:
8 years ago
Anyone can explain it?
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