Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 3 (Q.No. 50)
50.
In Mohr's circle, the centre of circle from Y-axis is taken as
Discussion:
13 comments Page 1 of 2.
Ajay anand said:
6 years ago
This question is not complete.
If two stress occurs perpendicular to each other and both tensile then option will correct A.
If two stress occurs perpendicular to each other but one tensile and other compressive then option will correct B.
If two stress occurs perpendicular to each other and both tensile then option will correct A.
If two stress occurs perpendicular to each other but one tensile and other compressive then option will correct B.
Ankita said:
6 years ago
The distance between Y-axis (vertical line) and the centre of the circle is measured along X-axis(horizontal line), which is σ avg = (σX+σY)/2.
Divesh Kumar said:
4 years ago
The question is incomplete.
The nature of both stresses is not given and also not mentioned which stress is greater. Anyone, please clarify it.
The nature of both stresses is not given and also not mentioned which stress is greater. Anyone, please clarify it.
Shruti said:
8 years ago
Option B is correct.
Option A is nothing but the radius of mohr's circle or Max shear stress.
Option A is nothing but the radius of mohr's circle or Max shear stress.
DHARMVIR said:
8 years ago
If principal stresses are alike than answer A.
If unlike than answer B.
If unlike than answer B.
Sumit kumar said:
7 years ago
σ X + ( σy -σX)/2= (X+y)/2.
From y-axis.
From y-axis.
Vijay said:
3 years ago
I think it's sx-sy/2 is correct.
Ref; R.S Khurmi.
Ref; R.S Khurmi.
Keshav said:
8 years ago
Answer is right but it must be from X axis.
Sumit kumar said:
7 years ago
Option B is absolutely correct. I agree.
Bharat khot said:
8 years ago
I too think A is the correct answer.
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