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Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?
- a)2 σ
- b)0.5 σ
- c)0.25σ
- d)zero
Correct answer is option 'D'. Can you explain this answer?
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Degrees to both the directions of the normal stresses?
The resultant normal stress on a plane inclined at 45 degrees to both the directions of the normal stresses can be found using the principle of superposition.
Let us assume that the two normal stresses acting on the point are σx and σy, where σx is positive and σy is negative.
At the point, the stress state can be represented by a stress element as shown below:
σx τxy
τxy σy
where τxy is the shear stress acting on the plane perpendicular to the normal stresses.
Now, let us consider a plane inclined at 45 degrees to both the directions of the normal stresses, as shown below:
σx τ
τ σy
where σ and τ are the normal and shear stresses on the inclined plane, respectively.
Using the Mohr's circle construction, we can find the magnitudes of σ and τ as follows:
1. Draw the Mohr's circle for the stress element σx, τxy, and σy.
2. Draw a line at 45 degrees from the origin of the Mohr's circle.
3. The intersection of the line with the Mohr's circle gives the coordinates of the stress element on the inclined plane.
4. The magnitude of the normal stress σ on the inclined plane is the distance between the intersection point and the origin of the Mohr's circle.
5. The magnitude of the shear stress τ on the inclined plane is the distance between the two intersection points.
The Mohr's circle construction yields the following results:
- The center of the Mohr's circle is at (σx + σy)/2, which is zero in this case since σx = -σy.
- The radius of the Mohr's circle is |σx - σy|/2, which is equal to the magnitude of each of the normal stresses.
- The intersection of the 45-degree line with the Mohr's circle is at (0, -a), where a is the magnitude of each of the normal stresses.
Therefore, the magnitudes of the normal and shear stresses on the inclined plane are:
σ = a
τ = a
Thus, the resultant normal stress on the inclined plane is equal to the magnitude of each of the normal stresses, and it acts at an angle of 45 degrees to both the directions of the normal stresses.
The resultant normal stress on a plane inclined at 45 degrees to both the directions of the normal stresses can be found using the principle of superposition.
Let us assume that the two normal stresses acting on the point are σx and σy, where σx is positive and σy is negative.
At the point, the stress state can be represented by a stress element as shown below:
σx τxy
τxy σy
where τxy is the shear stress acting on the plane perpendicular to the normal stresses.
Now, let us consider a plane inclined at 45 degrees to both the directions of the normal stresses, as shown below:
σx τ
τ σy
where σ and τ are the normal and shear stresses on the inclined plane, respectively.
Using the Mohr's circle construction, we can find the magnitudes of σ and τ as follows:
1. Draw the Mohr's circle for the stress element σx, τxy, and σy.
2. Draw a line at 45 degrees from the origin of the Mohr's circle.
3. The intersection of the line with the Mohr's circle gives the coordinates of the stress element on the inclined plane.
4. The magnitude of the normal stress σ on the inclined plane is the distance between the intersection point and the origin of the Mohr's circle.
5. The magnitude of the shear stress τ on the inclined plane is the distance between the two intersection points.
The Mohr's circle construction yields the following results:
- The center of the Mohr's circle is at (σx + σy)/2, which is zero in this case since σx = -σy.
- The radius of the Mohr's circle is |σx - σy|/2, which is equal to the magnitude of each of the normal stresses.
- The intersection of the 45-degree line with the Mohr's circle is at (0, -a), where a is the magnitude of each of the normal stresses.
Therefore, the magnitudes of the normal and shear stresses on the inclined plane are:
σ = a
τ = a
Thus, the resultant normal stress on the inclined plane is equal to the magnitude of each of the normal stresses, and it acts at an angle of 45 degrees to both the directions of the normal stresses.
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Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer?
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Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer?.
Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer?.
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Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Normal stresses of equal magnitude a but of opposite signs act at a point of a strained material in perpendicular direction. What would be the resultant normal stress on a plane inclined at 45° to the applied stresses?a)2 σb)0.5 σc)0.25σd)zeroCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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