The Three Limit Lateral Earth Pressures: 07-Str-A3, May 2014, Q2

Question 2: The Three Limit Lateral Earth Pressures (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

The single idea behind all three states is that the vertical effective stress on an element of backfill is fixed by the weight of soil above it and does not care what the wall does, whereas the horizontal effective stress depends entirely on how much the wall has moved. Writing $\sigma_h' = K\sigma_v'$, the whole of lateral earth pressure theory is a statement about the value of $K$, and $K$ is a function of wall movement.

Figure 2.1 — The three states as wall-movement cases. In each sketch the vertical effective stress on the element is the same; only the horizontal stress, and hence $K$, changes.

The at-rest state

If the wall does not move at all — a basement wall braced by the floor slabs, a bridge abutment tied into the deck, a massive gravity structure founded on rock — the backfill retains the horizontal stress it acquired as it was deposited or placed. There has been no lateral strain, so the soil is nowhere near failure and the state is not a limit state at all in the plastic sense; it is simply the elastic starting point. For a normally consolidated soil Jaky’s empirical expression gives

$$K_0 = 1 - \sin \phi'$$

which for a typical $\phi' = 30^\circ$ granular fill returns $K_0 = 0.50$. An over-consolidated soil remembers the higher horizontal stress locked in by the removed load, and $K_0 = (1-\sin\phi')\,\mathrm{OCR}^{\sin\phi'}$ can exceed unity in a heavily over-consolidated clay. Because no failure has occurred, the at-rest thrust is the largest of the three that a rigid, unyielding wall must actually be designed to carry.

The active state

Let the wall yield away from the fill — the top of a cantilever wall rotating outward, a sheet pile deflecting under its own retained height — and the soil behind it is free to expand laterally. The horizontal stress falls while the vertical stress stays put, until the soil can shed no more and a wedge of backfill fails, pushing down and outward. That is the active limit, the smallest horizontal stress the soil can sustain:

$$K_a = \frac{1-\sin\phi'}{1+\sin\phi'} = \tan^2\!\left(45^\circ - \frac{\phi'}{2}\right)$$

with $\sigma_h' = K_a\sigma_v' - 2c'\sqrt{K_a}$ when the soil has cohesion. At $\phi' = 30^\circ$, $K_a = 1/3$. Very little movement is required to reach it: roughly $0.001H$ for a dense granular fill and $0.004H$ for a loose one, which is why almost every free-standing retaining wall is designed for active pressure — the movement needed to mobilise it is smaller than the movement the wall will make anyway.

The passive state

Now drive the wall into the fill — the embedded toe of a sheet pile, the face of an anchor block, the buried portion of a bridge abutment under braking load. The soil is compressed laterally, the horizontal stress rises above the vertical stress, and failure occurs when a wedge is pushed up and outward. That is the passive limit, the largest horizontal stress the soil can sustain:

$$K_p = \frac{1+\sin\phi'}{1-\sin\phi'} = \tan^2\!\left(45^\circ + \frac{\phi'}{2}\right) = \frac{1}{K_a}$$

with $\sigma_h' = K_p\sigma_v' + 2c'\sqrt{K_p}$. At $\phi' = 30^\circ$, $K_p = 3$, nine times the active value. Passive resistance is expensive to mobilise: something of the order of $0.02H$ to $0.06H$ of movement is needed, an order of magnitude more than the active case, which is why passive resistance is normally taken with a substantial factor of safety, or with a reduced mobilised $\phi'$, in Canadian practice.

The three states on one Mohr diagram

The Mohr construction makes the ranking self-evident. Draw the effective-stress failure envelope $\tau_f = c' + \sigma'\tan\phi'$, and mark the fixed vertical effective stress $\sigma_v'$ on the normal-stress axis. Every one of the three circles must pass through that point, because $\sigma_v'$ never changes; the three circles differ only in where their second intersection with the axis, $\sigma_h'$, falls.

Figure 2.2 — Mohr circles for the three states, all sharing the same $\sigma_v'$. Active: $\sigma_v'$ is the major principal stress and the circle grows leftward until it touches the envelope. Passive: $\sigma_v'$ becomes the minor principal stress and the circle grows rightward until it touches. At rest: the circle is small and sits inside the envelope, so no failure plane exists.

At rest — the circle runs from $K_0\sigma_v'$ to $\sigma_v'$, is comparatively small, and lies wholly below the envelope. No plane in the element is at failure; the soil is elastic.
Active — $\sigma_h'$ has fallen to $K_a\sigma_v'$, so the circle has grown to the left until it is tangent to the envelope. Here $\sigma_1' = \sigma_v'$ and $\sigma_3' = \sigma_h'$, and the failure planes are inclined at $45^\circ + \phi'/2$ to the horizontal.
Passive — $\sigma_h'$ has risen to $K_p\sigma_v'$, so the principal stresses have swapped: $\sigma_1' = \sigma_h'$ and $\sigma_3' = \sigma_v'$. The circle has grown to the right until it is tangent, and the failure planes now lie at $45^\circ + \phi'/2$ to the vertical, that is at $45^\circ - \phi'/2$ to the horizontal.

Because both limiting circles are tangent to the same envelope while sharing one point on the axis, the ratio between them is fixed by $\phi'$ alone, giving $K_p/K_a = \tan^4(45^\circ + \phi'/2)$ — a factor of 9 at $\phi' = 30^\circ$ and 21 at $\phi' = 40^\circ$. That single ratio is the reason a wall can be pushed over far more easily than it can be pushed into the ground.

Topic: Rankine limit states, the coefficient of lateral earth pressure, and the Mohr-circle representation of active and passive failure.

Key relations: $\sigma_h' = K\sigma_v'$; $K_0 = 1-\sin\phi'$ (normally consolidated) and $K_0 = (1-\sin\phi')\mathrm{OCR}^{\sin\phi'}$ (over-consolidated); $K_a = \tan^2(45^\circ - \phi'/2)$; $K_p = \tan^2(45^\circ + \phi'/2) = 1/K_a$; and the printed formula-sheet relation $\sigma_1' = \sigma_3'\tan^2(45^\circ + \phi'/2) + 2c'\tan(45^\circ + \phi'/2)$, which is the same statement written between principal stresses.

Why this works: The vertical effective stress at a point in the backfill is set by the overburden and is unaffected by the wall. Any lateral strain therefore changes only $\sigma_h'$, which slides the Mohr circle along the axis about a fixed point until it either touches the envelope on the left (active) or on the right (passive). The at-rest condition is not a failure state but the no-strain starting point from which both limits are approached, and it is the only one of the three that depends on stress history rather than on wall movement.

Common pitfall: Assuming that the design pressure is always the active pressure. Active pressure requires the wall to move, and a wall that is prevented from moving — a propped basement wall, an integral abutment, a wall founded on rock — must be designed for $K_0$, which is roughly 50 % higher. The second pitfall is drawing the passive Mohr circle with $\sigma_v'$ still as the major principal stress; in the passive state the principal directions have rotated through $90^\circ$ and $\sigma_h'$ is the major principal stress.

Source: Knappett & Craig, Craig’s Soil Mechanics, 8th ed., Ch. 11 (lateral earth pressure: at-rest, Rankine active and passive states, movement required to mobilise each); Das, Principles of Geotechnical Engineering, 9th ed., Ch. 13; Canadian Geotechnical Society, CFEM, 4th ed., Ch. 24 — the Canadian guidance on when $K_0$ rather than $K_a$ governs and on the factor of safety to apply to passive resistance.