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Normal stresses of equal magnitude p, but of opposite signs, act at a point of a strained material in perpendicular direction. What is the magnitude of the resultant normal stress on a plane inclined at 45° to the applied stresses?
  • a)
    2 p
  • b)
    p/2
  • c)
    p/4
  • d)
    Zero
Correct answer is option 'D'. Can you explain this answer?
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Resultant Normal Stress on an Inclined Plane

Given:
Magnitude of normal stresses = p
Direction of normal stresses = perpendicular to each other
Angle of inclined plane = 45 degrees

To Find:
Magnitude of resultant normal stress on the inclined plane

Solution:
We can use the formula for resolving stresses on an inclined plane to find the magnitude of resultant normal stress on the inclined plane.

Resolving Stresses on an Inclined Plane:
σn = (σx + σy)/2 + (σx - σy)/2 * cos(2θ) + τxy * sin(2θ)

Where:
σn = normal stress on the inclined plane
σx = normal stress in the x-direction
σy = normal stress in the y-direction
θ = angle between the x-axis and the inclined plane
τxy = shear stress in the xy-plane

In this case, we have two normal stresses of equal magnitude p, but of opposite signs. Therefore, σx = p and σy = -p.

Substituting the given values:
σn = (p - p)/2 + (p + p)/2 * cos(2*45) + 0 * sin(2*45)
σn = 0 + p/2 * cos(90) + 0
σn = 0

Therefore, the magnitude of resultant normal stress on the inclined plane is zero.

Conclusion:
When normal stresses of equal magnitude p, but of opposite signs, act at a point of a strained material in perpendicular direction, the magnitude of resultant normal stress on an inclined plane at 45 degrees to the applied stresses is zero.
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