at a point in a strained material the principal stresses are 60MPa and 40MPa.find the position of the plane across which the resultant stress is most inclined to the normal and determine the value of this stress.
To find the position of the plane across which the resultant stress is most inclined to the normal, we need to determine the principal directions of stress. The principal stresses are given as 60 MPa and 40 MPa.
The formula to calculate the principal stress is:
σ₁,₂ = (σ₁ + σ₂) / 2 ± √((σ₁ - σ₂) / 2)² + τ²
Where:
σ₁,₂ = Principal Stress
σ₁ = Larger stress magnitude
σ₂ = Smaller stress magnitude
τ = Shear stress (in this case, it is assumed to be 0)
Let's calculate the principal stresses:
σ₁,₂ = (60 MPa + 40 MPa) / 2 ± √((60 MPa - 40 MPa) / 2)² + 0²
= 50 MPa ± √((20 MPa) / 2)²
= 50 MPa ± √(200 MPa²)
= 50 MPa ± 14.14 MPa
Thus, the principal stresses are approximately 35.86 MPa and 64.14 MPa.
Now, to find the position of the plane across which the resultant stress is most inclined to the normal, we need to determine the angle θ between the normal and the principal stress direction.
θ = 0.5 * arctan(2 * τ / (σ₁ - σ₂))
Since we assume there is no shear stress in this case, τ = 0, the angle θ can be calculated as:
θ = 0.5 * arctan(0 / (64.14 MPa - 35.86 MPa))
= 0
The angle θ is 0, which means the resultant stress is most inclined to the normal, indicating that the principal stress direction is aligned with the normal direction of the plane.
Therefore, the position of the plane across which the resultant stress is most inclined to the normal is the same as the normal direction of the plane. The value of the stress in this direction is equal to the principal stress, so the value of this stress is 64.14 MPa.