Determine the stress components acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec. 9.1.

The image of the question is missing, so i have attached it.

σ_x ≈ -387.47 psi

σ_y ≈ 454.65 psi

Explanation:

From the method of equilibrium and looking at the image of the inclined plane, we can sum moments about the X-Plane and equate to zero and also do the same to the y-plane.

Thus;

At Σf_x = 0 we have;

Δf_x - 400(ΔA × cos 60°)cos 60° + 650(ΔA sin 60°)cos 30° = 0

Δf_x - (400(ΔA × 0.5)0.5) + (650(ΔA × 0.866)0.866) = 0

Δf_x - 100ΔA + 487.47ΔA = 0

Where;

Δf_x is change in force along the x - axis and ΔA is change in area along the X-axis.

Δf_x = 100ΔA - 487.47ΔA

Δf_x = -387.47ΔA

Thus, Δf_x/ΔA = -387.47 psi

As ΔA approaches zero, Δf_x/ΔA which will be the stress along the x-component will be;

σ_x = Δf_x/ΔA ≈ -387.47 psi

Similarly for the y-component;

At Σf_y = 0 we have;

Δf_y - 650(ΔA × sin 60°)sin 30° - 400(ΔA × cos 60°)sin 60° = 0

Δf_y - (650(ΔA × 0.866)0.5) - (400(ΔA × 0.5)0.866) = 0

Δf_y - 281.45ΔA - 173.2ΔA = 0

Δf_y = 281.45ΔA + 173.2ΔA

Δf_y = 454.65ΔA

Δf_y/ΔA = 454.65 psi

Now,

As ΔA approaches zero, Δf_x/ΔA which will be the stress along the y -component will be;

σ_y = Δf_y/ΔA ≈ 454.65 psi

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