Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 7)
7.
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum normal stress will be
Discussion:
59 comments Page 2 of 6.
Vijay said:
10 years ago
Thank you @Yajat Sharma.
Patil Pramod said:
10 years ago
Thanks for all the above answers.
Abhishek said:
10 years ago
Thank you all for the explanation.
Atosh said:
10 years ago
The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses.
Hareshkumar said:
9 years ago
The resultant tensile stress is (1200+600)/2 =900 MPA shear stress =400 Mpa using maximum normal stress theory 1/2(900+ 900^2+4 * 400^2) = 1400Mpa.
Alham said:
10 years ago
If shear stress is applied, maximum normal stress will always be greater than applied maximum stress(1200). So the answer is 1400. No calculation only observation.
Rahul said:
10 years ago
Normal stress = (sum of stress/2) = 900 answer.
Seenu said:
10 years ago
Thanks for all your explanation.
Avneesh said:
10 years ago
You can also use this formula,
Stress = [(x + y)/2] + 0.5 * (x - y)^2 + 4(shear stress)^2.
Stress = [(x + y)/2] + 0.5 * (x - y)^2 + 4(shear stress)^2.
L3FA said:
9 years ago
Thanks for the given explanations.
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