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In the case of bi-axial state of normal stresses, the normal stress on 45° plane is equal to
- a)The sum of the normal stresses
- b)Difference of the normal stresses
- c)Half the sum of the normal stresses
- d)Half the difference of the normal stresses
Correct answer is option 'C'. Can you explain this answer?
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Bi-axial state of normal stresses refers to a situation where a body experiences two normal stresses in perpendicular directions. In this case, the normal stress on the 45-degree plane is equal to half the sum of the normal stresses.
Explanation:
Let us consider a body subjected to two normal stresses, σ1 and σ2 in perpendicular directions x and y, respectively. The stress on the 45-degree plane is represented by σ45. By using the equations of statics, we can determine the value of σ45.
The stress on the 45-degree plane can be resolved into two components, one along the x-direction and the other along the y-direction. The components of σ45 along x and y-direction are given by:
σx = (σ1 + σ2)/2cos45 = (σ1 + σ2)/√2
σy = (σ1 + σ2)/2sin45 = (σ1 + σ2)/√2
Since the stress on the 45-degree plane is perpendicular to both x and y-directions, it does not contribute to the shear stress. Therefore, the shear stress on the 45-degree plane is zero.
The normal stress on the 45-degree plane is given by:
σ45 = σx + σy = (σ1 + σ2)/√2 + (σ1 + σ2)/√2
σ45 = (σ1 + σ2)
Hence, the normal stress on the 45-degree plane is equal to half the sum of the normal stresses, i.e., σ45 = (σ1 + σ2)/2.
Conclusion:
In the bi-axial state of normal stresses, the normal stress on the 45-degree plane is equal to half the sum of the normal stresses. This is because the stress on the 45-degree plane can be resolved into two components along the x and y-directions, and the sum of these components gives the normal stress on the 45-degree plane.
Explanation:
Let us consider a body subjected to two normal stresses, σ1 and σ2 in perpendicular directions x and y, respectively. The stress on the 45-degree plane is represented by σ45. By using the equations of statics, we can determine the value of σ45.
The stress on the 45-degree plane can be resolved into two components, one along the x-direction and the other along the y-direction. The components of σ45 along x and y-direction are given by:
σx = (σ1 + σ2)/2cos45 = (σ1 + σ2)/√2
σy = (σ1 + σ2)/2sin45 = (σ1 + σ2)/√2
Since the stress on the 45-degree plane is perpendicular to both x and y-directions, it does not contribute to the shear stress. Therefore, the shear stress on the 45-degree plane is zero.
The normal stress on the 45-degree plane is given by:
σ45 = σx + σy = (σ1 + σ2)/√2 + (σ1 + σ2)/√2
σ45 = (σ1 + σ2)
Hence, the normal stress on the 45-degree plane is equal to half the sum of the normal stresses, i.e., σ45 = (σ1 + σ2)/2.
Conclusion:
In the bi-axial state of normal stresses, the normal stress on the 45-degree plane is equal to half the sum of the normal stresses. This is because the stress on the 45-degree plane can be resolved into two components along the x and y-directions, and the sum of these components gives the normal stress on the 45-degree plane.
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In the case of bi-axial state of normal stresses, the normal stress on 45° plane is equal toa)The sum of the normal stressesb)Difference of the normal stressesc)Half the sum of the normal stressesd)Half the difference of the normal stressesCorrect answer is option 'C'. Can you explain this answer?
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