Methods of finding stresses on a plane of a soil element

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In this article, I would show two methods of finding stresses on a plane in a soil element under stress. These two methods are:

1. The Mohr circle method, and
2. The pole method

A. The Mohr Circle method

Figure 1 (a & b) show a soil element in which it is required to determine the stresses on the plane EF of the soil.

The stresses can be determined from the equilibrium of forces acting on the wedge on Figure 1(b).

From geometry, EB = EF cos θ and FB = EF sin θ

From Figure 1b, resolving the forces in the x-direction

σ_n (EF sin θ) = σ_x FB + τ EB + τ_n (EF cos θ)

σ_n sin θ = σ_x sin θ + τ cos θ + τ_n cos θ                                    (1)

Resolving the forces in the y-direction

σ_n (EF cos θ) + τ_n (EF sin θ) = τFB + σ_yEB

σ_n cos θ + τ_n sin θ = τ sin θ + σ_y cos θ

σ_n cos θ = τ sin θ + σ_y cos θ – τ_n sin θ                                     (2)

Multiply Equation (1) by sin θ

σ_n sin² θ = σ_x sin² θ + τ sin θ cos θ + τ_n sin θ cos θ                 (3)

Multiply Equation (2) by cos θ

σ_n cos² θ = τ sin θ cos θ + σ_y cos² θ – τ_n sin θ cos θ                (4)

Add Equations (3) and (4)

(σ_n sin² θ + σ_n cos² θ) = σ_x sin² θ + τ sin θ cos θ + τ_n sin θ cos θ + τ sin θ cos θ + σ_ycos² θ – τ_n sin θ cos θ

Collect like terms

σ_n (sin² θ + cos² θ) = σ_xsin² θ + σ_y cos² θ + τ sin θ cos θ+ τ sin θ cos θ + τ_n sin θ cosθ– τ_n sin θ cos θ

Recall from Trigonometry that:

1. sin² θ + cos² θ = 1
2. 2 sin θ cos θ = sin 2θ

Therefore, σ_n = σ_x sin² θ + σ_y cos² θ + 2τ sin θ cos θ              (5)

Recall also that:

1. cos 2θ = cos² θ – 1 = 1 – 2sin² θ
2. cos 2θ + 1 = cos² θ » cos² θ = (1 + cos 2θ)/2
3. 2sin² θ = 1 – cos² θ » sin² θ = (1 – cos 2θ)/2

Replace the terms in Equation (5) above

Thus, σ_n = σ_x ((1 – cos 2θ)/2) + σ_y ((1 + cos 2θ)/2) + τ sin 2θ

Simplifying further,

σ_n = σ_x /2 – (σ_x 2θ)/2 + σ_y /2 + (σ_y cos 2θ)/2 + τ sin 2θ

σ_n = σ_x /2 + σ_y /2 + (σ_y cos 2θ)/2 – (σ_x 2θ)/2 + τ sin 2θ

σ_n = (σ_x + σ_y) /2 + ((σ_{y –} σ_x) cos 2θ)/2 + τ sin 2θ                     (6)

Equation (6) gives the NORMAL STRESS on the plane EF

Substitute Eqn (6) into (1)

[(σ_x + σ_y) /2 + ((σ_{y –} σ_x) cos 2θ)/2 + τ sin 2θ] sin θ = σ_x sin θ + τ cos θ + τ_n cos θ

Simplifying,

τ_n cos θ = (σ_x sin θ) /2 + (σ_y sin θ) /2 + (σ_y sin θ cos 2θ) /2 – (σ_x sin θ cos 2θ)/2 + τ sin θ sin 2θ – σ_x sin θ + τ cos θ

τ_n = (σ_x sin θ) /2 cos θ + (σ_y sin θ) /2 cos θ + (σ_y sin θ cos 2θ) /2 cos θ – (σ_x sin θ cos 2θ)/2 cos θ + (τ sin θ sin 2θ)/ cos θ – (σ_x sin θ)/ cos θ + (τ cos θ)/ cos θ

Simplifying further gives,

τ_n = (σ_y – σ_x) sin 2θ /2 – τ cos 2θ                                             (7)

Equation (7) gives the SHEAR STRESS on the plane EF.

To determine the value of the stresses on the principal planes, choose the value of θ such that τ_n = 0, because on the principal planes, the shear stress is zero.

Solving equation (7) for τ_n = 0 yields the following solutions,

tan 2θ = 2τ / (σ_y – σ_x)

sin 2θ = τ / √ ((σ_y – σ_x)/2)² + τ²

cos 2θ = ((σ_y – σ_x)/2) / √ ((σ_y – σ_x)/2)² + τ²

Substituting the values of sin 2θ and cos 2θ above into Equation (6) yields

σ₁ = (σ_x + σ_y) /2 + [((σ_{y –} σ_x)/2)² + τ²]^1/2

σ₃ = (σ_x + σ_y) /2 – [((σ_{y –} σ_x)/2)² + τ²]^1/2

σ₁ and σ₃ are the major and minor principal stresses respectively.

B. The pole method

This is another useful method of finding stresses on the plane that involves some level of construction following the methods outline below. Using Figures 1 (a & b) as well

1. Plot points of stresses on the soil element on plane AD (σ_x, τ) at point R and plane AB (σ_y, -τ) at point M on Cartesian plane (See Figure 2).
2. Joint these points and draw a Mohr circle with RM as the diameter.
3. Joint M to P, parallel to AB (where the line cuts the Mohr circle is named P which is the Pole and which stands as the unique point for stresses under consideration.
4. Draw a line from the pole, P, that is line PQ, parallel to the plane, EF. Where the line being drawn from P intersects the Mohr circle is named Q and this point represents the stresses being sought.
5. By geometry, angle QOM is equal to 2 times angle QPM.

Thanks for reading!