A piece of material is subjected to two perpendicular tensile stresses (σ_x = 100 MPa. and σ_y = 60 MPa). What will be inclination (θ) of the plane on which the resultant stress (R) has maximum obliquity (φ) with the normal?

Explanation:

Determining the Inclination of the Plane for Maximum Obliquity:

When a piece of material is subjected to two perpendicular tensile stresses, σ_x and σ_y, the inclination of the plane on which the resultant stress (R) has maximum obliquity (φ) with the normal can be determined using the principles of stress transformation in mechanics of materials.

In this problem, we are given:

• σ_x = 100 MPa
• σ_y = 60 MPa

The maximum obliquity of the resultant stress occurs when the shear stress (τ) on the plane is maximized. This typically happens at an angle θ where the plane is oriented such that the normal stress is balanced by the shear stress. Using Mohr's circle of stress, the inclination θ of the plane can be found.

The formula for the inclination θ of the plane on which the resultant stress R has maximum obliquity φ with the normal is given by:

θ = tan^-1 (τ / σ)

Here, τ is the shear stress and σ is the normal stress on the plane. For the given problem, we need to derive the correct expression for θ by considering the relationship between the normal and shear stresses.

From the theory of Mohr's circle, the inclination θ for the plane on which the resultant stress has maximum obliquity can be derived as:

θ = tan^-1 (σ_x / σ_y)

Plugging in the given values:

θ = tan^-1 (100 / 60)

Simplifying this expression, we get:

\( \theta=\tan ^{-1} \sqrt{\frac{5}{3}}\)

Therefore, the correct inclination θ of the plane on which the resultant stress has maximum obliquity is:

\( \theta=\tan ^{-1} \sqrt{\frac{5}{3}}\)