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.
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.
Shiva said:
8 years ago
Answer should be =σ*(cos)^2.
S. Ata said:
8 years ago
No answer is correct.
Right answer is= σ(cos θ)^2.
Right answer is= σ(cos θ)^2.
NAITIK Trivedi said:
8 years ago
6cos is the right answer.
Sauru said:
8 years ago
Yes, I agree @Shiva.
Rajkotha said:
6 years ago
Normal stress = σ(cos θ)^2.
Tangential stress =σ sin 2θ.
Tangential stress =σ sin 2θ.
Jitendra said:
6 years ago
Here, the Tangential stress = 0.
Raju said:
6 years ago
@Rajkotha is right.
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.
Kunal said:
5 years ago
Right, Thanks @Sanjib.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers