Mohr's Circle for Stress
Trending Questions
Q. For the plane stress shown in figure, the maximum shear stress and the plane on which it acts are :

- -50 MPa on a plane 45o clockwise w.r.t. to x-axis
- -50 MPa on a plane 45o anti - clockwise w.r.t to x-axis
- 50 MPa at all orientation
- Zero at all orientation.
Q. Assertion (A): In the analysis of two dimensional stresses, the normal stress on a plane will be greater than the average principal stress, if the inclination of the plane with the plane of maximum principal stress is less than 45°.
Reason (R): The horizontal coordinate of the point of the Mohr's circle represents the stresses on the given plane which is greater than the coordinate of the centre of Mohr's circle.
Reason (R): The horizontal coordinate of the point of the Mohr's circle represents the stresses on the given plane which is greater than the coordinate of the centre of Mohr's circle.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true and R is not a correct explanation of A
- A is true but R is false
- A is false but R is true
Q. The Mohr's circle give below corresponds to which one of the following stress conditions.


Q. Mohr's stress circle helps in determining which of the following?
1. Normal stresses on one plane.
2. Normal and tangential stresses on two planes.
3. Principal stresses in all three directions.
4. Inclination of principal planes.
Select the correct answer using the codes given below:
1. Normal stresses on one plane.
2. Normal and tangential stresses on two planes.
3. Principal stresses in all three directions.
4. Inclination of principal planes.
Select the correct answer using the codes given below:
- 1 and 2 only
- 2 and 3 only
- 3 and 4 only
- 2 and 4 only
Q. What is the diameter of Mohr's circle of stress for the state of stress shown below?


- 20
- 10√2
- 10
- Zero
Q. The state of stress at a point, for a body in plane stress, is shown in the figure below. If the minimum principal stress is 10 kPa, then the normal stress σy (in kPa) is


- 9.45
- 18.88
- 37.78
- 75.50
Q. A specimen is subjected to a pure shear stress regime of intensity τ. The resulting tensile and compressive stresses σ, which occur on planes inclined at 45° to the direction of the shear stresses, would be
- τ
- τ2
- √2τ
- τ√2
Q. In a plane stress condition, the components of stress at a point are σx=20 MPa, σy=80 MPa and τxy=40 MPa. The maximum shear stress (in MPa) at the point is
- 20
- 25
- 50
- 100
Q. The state of plane-stress at a point is given by σx=200 MPa, σy=100 MPa and τxy=100 MPa. The maximum shear stress (in MPa) is
- 111.8
- 150.1
- 180.3
- 223.6
Q. If the principal stresses at a point in a stressed body are 150 kN/m2 tensile and 50 kN/m2 compressive, then maximum shear stress at this point will be
- 100 kN/m2
- 150 kN/m2
- 200 kN/m2
- 250 kN/m2