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.
Vijay said:
3 years ago
I think it's sx-sy/2 is correct.
Ref; R.S Khurmi.
Ref; R.S Khurmi.
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.
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.
Rajib said:
7 years ago
Option A is correct. I too agree.
Sumit kumar said:
7 years ago
σ X + ( σy -σX)/2= (X+y)/2.
From y-axis.
From y-axis.
Sumit kumar said:
7 years ago
Option B is absolutely correct. I agree.
Keshav said:
8 years ago
Answer is right but it must be from X axis.
DHARMVIR said:
8 years ago
If principal stresses are alike than answer A.
If unlike than answer B.
If unlike than answer B.
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.
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