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 1 of 2.
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.
Vishal said:
3 years ago
Yes, the answer given is right, we have to use 1+cos2θ = 2cos^2θ.
Yashodanand said:
4 years ago
Normal stress σ/2(1+cos2θ)
Shear stress σ/2sin2θ
Resulant = √(normal stress 2+shear stress 2) then answer become σcosθ.
Shear stress σ/2sin2θ
Resulant = √(normal stress 2+shear stress 2) then answer become σcosθ.
(2)
SURAJ PRASAD KESHARI said:
4 years ago
Thank you all for explaining this.
Abhishek singh said:
5 years ago
Question is for resultant stress not for normal stress. So the given answer is right.
Kunal said:
5 years ago
Right, Thanks @Sanjib.
Sanjib mondal said:
5 years ago
Normal stress ==σcosθ.
Tangential =σ/2 sin2θ.
Resultant =root (σcosθ)2 +(σ/2 sin2θ)2.
By solving properly you will get;
=σcosθ.
Tangential =σ/2 sin2θ.
Resultant =root (σcosθ)2 +(σ/2 sin2θ)2.
By solving properly you will get;
=σcosθ.
Raju said:
6 years ago
@Rajkotha is right.
Jitendra said:
6 years ago
Here, the Tangential stress = 0.
Rajkotha said:
6 years ago
Normal stress = σ(cos θ)^2.
Tangential stress =σ sin 2θ.
Tangential stress =σ sin 2θ.
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