In this article, I would show two methods of finding stresses on a plane in a soil element under stress. These two methods are:
- The Mohr circle method, and
- 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
σn (EF sin θ) = σx FB + τ EB + τn (EF cos θ)
σn sin θ = σx sin θ + τ cos θ + τn cos θ (1)
Resolving the forces in the y-direction
σn (EF cos θ) + τn (EF sin θ) = τFB + σyEB
σn cos θ + τn sin θ = τ sin θ + σy cos θ
σn cos θ = τ sin θ + σy cos θ – τn sin θ (2)
Multiply Equation (1) by sin θ
σn sin2 θ = σx sin2 θ + τ sin θ cos θ + τn sin θ cos θ (3)
Multiply Equation (2) by cos θ
σn cos2 θ = τ sin θ cos θ + σy cos2 θ – τn sin θ cos θ (4)
Add Equations (3) and (4)
(σn sin2 θ + σn cos2 θ) = σx sin2 θ + τ sin θ cos θ + τn sin θ cos θ + τ sin θ cos θ + σycos2 θ – τn sin θ cos θ
Collect like terms
σn (sin2 θ + cos2 θ) = σxsin2 θ + σy cos2 θ + τ sin θ cos θ+ τ sin θ cos θ + τn sin θ cosθ– τn sin θ cos θ
Recall from Trigonometry that:
- sin2 θ + cos2 θ = 1
- 2 sin θ cos θ = sin 2θ
Therefore, σn = σx sin2 θ + σy cos2 θ + 2τ sin θ cos θ (5)
Recall also that:
- cos 2θ = cos2 θ – 1 = 1 – 2sin2 θ
- cos 2θ + 1 = cos2 θ » cos2 θ = (1 + cos 2θ)/2
- 2sin2 θ = 1 – cos2 θ » sin2 θ = (1 – cos 2θ)/2
Replace the terms in Equation (5) above
Thus, σn = σx ((1 – cos 2θ)/2) + σy ((1 + cos 2θ)/2) + τ sin 2θ
Simplifying further,
σn = σx /2 – (σx 2θ)/2 + σy /2 + (σy cos 2θ)/2 + τ sin 2θ
σn = σx /2 + σy /2 + (σy cos 2θ)/2 – (σx 2θ)/2 + τ sin 2θ
σn = (σx + σy) /2 + ((σy – σx) cos 2θ)/2 + τ sin 2θ (6)
Equation (6) gives the NORMAL STRESS on the plane EF
Substitute Eqn (6) into (1)
[(σx + σy) /2 + ((σy – σx) cos 2θ)/2 + τ sin 2θ] sin θ = σx sin θ + τ cos θ + τn cos θ
Simplifying,
τn cos θ = (σx sin θ) /2 + (σy sin θ) /2 + (σy sin θ cos 2θ) /2 – (σx sin θ cos 2θ)/2 + τ sin θ sin 2θ – σx sin θ + τ cos θ
τn = (σx sin θ) /2 cos θ + (σy sin θ) /2 cos θ + (σy sin θ cos 2θ) /2 cos θ – (σx sin θ cos 2θ)/2 cos θ + (τ sin θ sin 2θ)/ cos θ – (σx sin θ)/ cos θ + (τ cos θ)/ cos θ
Simplifying further gives,
τn = (σy – σx) 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 τn = 0, because on the principal planes, the shear stress is zero.
Solving equation (7) for τn = 0 yields the following solutions,
tan 2θ = 2τ / (σy – σx)
sin 2θ = τ / √ ((σy – σx)/2)2 + τ2
cos 2θ = ((σy – σx)/2) / √ ((σy – σx)/2)2 + τ2
Substituting the values of sin 2θ and cos 2θ above into Equation (6) yields
σ1 = (σx + σy) /2 + [((σy – σx)/2)2 + τ2]1/2
σ3 = (σx + σy) /2 – [((σy – σx)/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
- Plot points of stresses on the soil element on plane AD (σx, τ) at point R and plane AB (σy, -τ) at point M on Cartesian plane (See Figure 2).
- Joint these points and draw a Mohr circle with RM as the diameter.
- 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.
- 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.
- By geometry, angle QOM is equal to 2 times angle QPM.
Thanks for reading!
