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.
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)
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.
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.
Abhishek singh said:
5 years ago
Question is for resultant stress not for normal stress. So the given answer is right.
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.
S. Ata said:
8 years ago
No answer is correct.
Right answer is= σ(cos θ)^2.
Right answer is= σ(cos θ)^2.
Rajkotha said:
6 years ago
Normal stress = σ(cos θ)^2.
Tangential stress =σ sin 2θ.
Tangential stress =σ sin 2θ.
Shiva said:
8 years ago
Answer should be =σ*(cos)^2.
SURAJ PRASAD KESHARI said:
4 years ago
Thank you all for explaining this.
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