Arching in Soils and Braced Cuts

Arching in soil is the transfer of pressure from a yielding part of a soil mass to the adjacent non-yielding (stationary) parts through the shear resistance of the soil. When one part of the support moves away, the soil above tries to move with it; the shear stresses along the boundaries of the moving mass hold it up, so the pressure on the yielding part falls and the pressure on the neighbouring stationary part rises.

   ground
   ~~~~~~~~~~~~~~~~~~~~~~~~
   | stationary |  moving  | stationary |
   |  (stress   | (stress  |  (stress   |
   |  increases)| reduces) |  increases)|
   |____________|__________|____________|
                 yielding support

Essential requirements

1. Relative movement: one part of the soil support must yield (move or deflect) relative to the neighbouring part; arching does not occur in a rigid, unyielding support.
2. Shear strength in the soil: the soil must be able to mobilise shear resistance (friction and/or cohesion) along the boundaries of the moving mass; therefore it works best in dense sand, and in clay in the short term.
3. Sufficient depth of soil cover: a sufficient height of soil above the yielding part, so that the arch can form (arching fails if the cover is too thin).
4. Limited movement: the yielding must be small. Large movement breaks the arch, and the soil then loses its supporting effect.
5. Stable adjacent parts: the adjacent soil and supports must be stiff enough to resist the extra load transferred to them.

Examples: pressure on tunnel lining, buried conduits, braced cuts, flexible retaining walls and piles in soil.

Terzaghi's trap door experiment

A long rectangular box is filled with dry sand. At its base there is a narrow strip (trap door) of width $2b$ that can be lowered while the rest of the base is fixed. The vertical pressure on the door is measured as it is lowered.

        sand surface
   ~~~~~~~~~~~~~~~~~~~~~~~~~~~
   |  \                 /  |
   |   \  sliding wedge /   |   shear planes
   |    \             /    |   (nearly vertical,
   |_____\___________/_____|   curving at the top)
   fixed  |  trap door |  fixed
          |<---- 2b -->|
             v moves down

Observations

1. Before any movement the pressure on the door is the full overburden $\gamma z$.
2. As the door moves down by a small amount, the pressure on it falls rapidly to a minimum (about 0.4 to 0.5 of the overburden pressure or less).
3. The pressure on the adjacent fixed base rises correspondingly, so the total load is unchanged.
4. If the door continues to move, the pressure on it stays nearly constant at this reduced value (the load is carried by the arch).

Terzaghi's arching theory

When the strip yields, the soil above the door tries to settle. The adjacent soil resists it by shear along the vertical planes through the door edges. These shear forces carry part of the weight, so the load on the door is less than the weight of soil above it. For a door of half-width $b$ at depth $z$ with a surcharge $q$:

$$\sigma_v = \frac{b\,(\gamma - c/b)}{K\tan\phi}\left(1 - e^{-K\tan\phi\, z/b}\right) + q\,e^{-K\tan\phi\, z/b}$$

where $K$ is the lateral pressure coefficient (about 1.0) and $\phi$ the angle of friction. For large $z$ the pressure no longer increases with depth: it reaches a limit.

Importance in geotechnical engineering

• Design of tunnels and underground conduits (loads on linings and pipes).
• Pressure on braced excavations: apparent pressure diagrams differ from Rankine's triangle.
• Flexible retaining structures such as sheet piles and anchored walls.
• Loads on piles and pile caps, bin and silo pressures, and loads on culverts or buried pipes.

Peck's (1969) apparent pressure diagrams are envelopes of the measured strut loads in braced cuts, so they are used only to find strut, wale and sheeting forces. $H$ is the depth of the cut.

1. Cuts in sand

Uniform (rectangular) pressure

$$p_a = 0.65\,K_a\gamma H,\qquad K_a = \tan^2(45^\circ-\phi/2)$$

2. Cuts in firm (stiff) clay ($\gamma H/c \le 4$)

Trapezoid: pressure rises from zero at the top over $0.25H$, remains constant for $0.5H$, and then reduces to zero over the last $0.25H$.

$$p_a = 0.2\gamma H\ \text{to}\ 0.4\gamma H\quad(\text{average }0.3\gamma H)$$

3. Cuts in soft to medium clay ($\gamma H/c > 4$)

Trapezoid: pressure rises over the top $0.25H$ and remains constant below.

$$p_a = \gamma H\left[1 - m\frac{4c}{\gamma H}\right]\ \ge 0.3\gamma H$$

with $m = 1$ in general ($m = 0.4$ for very soft clay below the cut).

   SAND            STIFF CLAY        SOFT-MEDIUM CLAY
 |----|            |    /----\       |    /-----|
 |    |  H         |   /      \      |   /      |
 |    |            |  /        \     |  /       |
 |----|            | /          \    | /        |
  0.65 Ka g H     0.25H|0.5H|0.25H    0.25H | 0.75H
                   p=0.2-0.4 g H      p = g H(1 - 4mc/gH)

Braced cut: an excavation with vertical or nearly vertical sides whose walls (sheet piles, or soldier piles with lagging) are held in place by horizontal bracing members (wales and struts) installed as the excavation proceeds, so the sides do not collapse.

Strut load: the compressive force carried by a strut, equal to the reaction per metre of the sheeting at the strut level multiplied by the horizontal (centre to centre) spacing of the struts: $P = R \times s$.

A braced excavation consists of the following parts.

        ground surface
   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     |                       |
   --|=====[ strut ]=========|--  level 1
     |   wale          wale  |
     |                       |
   --|=====[ strut ]=========|--  level 2
     |                       |
     | sheeting     sheeting |
     |_____ excavation ______|
     |<--- width of cut --->|
        embedded toe below base

1. Sheeting (sheet piles, or soldier piles with timber lagging): the vertical wall in contact with the soil; it carries soil pressure to the wales.
2. Wales (walers): horizontal beams (steel or timber) fixed along the sheeting at the strut levels; they collect the load from the sheeting and pass it to the struts.
3. Struts: horizontal compression members across the cut at one or more levels; they transfer the load from one wall to the other. Diagonal struts (rakers) are used for wide cuts.
4. Anchors or rakers (alternative): tie-backs or inclined struts supported from the ground or base.
5. Embedment (toe) of sheeting below the base level, to resist bottom heave and provide support.
6. Dewatering system and surcharge control, where needed.

When the retained soil is layered with sand and clay, Peck's diagrams (single homogeneous soil) cannot be used directly. The usual criteria (Peck, 1969) are:

• Sand diagram if the cut is in sand, or sand is the dominant stratum.
• Clay diagram if clay is the dominant soil.
• For layered soils, use the weighted average values over the full depth $H$ of the cut:

$$\gamma_{avg} = \frac{1}{H}\sum \gamma_i H_i,\qquad (q_u)_{avg} = \frac{1}{H}\sum (q_u)_i H_i,\qquad c_{avg} = \tfrac12(q_u)_{avg}$$

and

$$(\tan\phi)_{avg}\ \text{or}\ \phi_{avg} = \tan^{-1}\left[\frac{1}{H}\sum H_i\tan\phi_i\right]$$

• Then choose the diagram from $\gamma H/c_{avg}$: if the clay is soft to medium ($>4$), use the soft clay diagram; if stiff, the stiff clay diagram.
• Where a layer carries thick sand over clay, use the sand diagram with the equivalent $\phi$ and sand weight in the upper part and clay in the lower part, and design for the more severe case.

Arching in soils

Arching is the transfer of pressure from a yielding part of a soil mass to the neighbouring stationary parts by shear resistance of the soil. It reduces the load on the yielding part and increases it on the adjacent parts (shown by Terzaghi's trap door experiment). It explains the non-triangular pressure on braced cuts and tunnel linings.

Heave of the bottom of a cut in soft clay

In a deep cut in soft or medium clay, the weight of the soil beside the cut ($\gamma H$) acts as a surcharge on the soil at the base level. If this exceeds the bearing capacity of the clay below the base, the clay is pushed into the excavation, so the bottom heaves up and the ground behind the wall settles.

   ~~~~~~~ ground ~~~~~~~
   |  gamma*H  |   |  gamma*H  |
   |  surcharge| cut | surcharge |
   |           |_____|           |
   |      ,-""  ^ heave "-.       |
   |     (   failure arc    )     |

• Cause: $\gamma H$ at the cut base level compared with bearing capacity $\approx 5.7c$ (Terzaghi) when $\gamma H/c$ is large (above about 5.7).
• Factor of safety (Terzaghi):

$$FS = \frac{5.7c}{\gamma H - \sqrt{2}\,cH/B}$$

($B$ = width of the cut).

• Effects: heaving of the cut bottom, increased pressure on sheeting and struts, settlement of ground outside, possible failure of struts.
• Remedies: deeper embedment of sheet piles into a stiffer layer, a stiff base (concrete slab or grouted block), lowering water level, cutting in stages or with berms.

Data: cut depth $H = 10$ m, $\gamma = 18$ kN/m$^3$, $c = 30$ kN/m$^2$ (clay), struts at A, B, C, D at 1.5, 4.0, 6.5 and 9.0 m, horizontal spacing 2.0 m, hinges at B and C.

1. Apparent pressure (Peck, clay)

$$\frac{\gamma H}{c} = \frac{18\times10}{30} = 6 > 4\ \Rightarrow\ \text{soft to medium clay}$$ $$p_a = \gamma H\left(1 - \frac{4c}{\gamma H}\right) = 180\left(1 - \frac{120}{180}\right) = 60.0\ \text{kN/m}^2\ (> 0.3\gamma H = 54)$$

The diagram rises from 0 at the top to 60 kN/m$^2$ at $0.25H = 2.5$ m, then stays 60 kN/m$^2$ to the base (total load $= 525$ kN/m).

 0   |
     |\
 2.5 | 60
 A 1.5
 B 4.0  ---- 60 constant
 C 6.5
 D 9.0
10   |_____ 60

2. Reactions per metre run (hinge method)

• Top part (0 to 4.0 m): load $= 165.0$ kN/m, moment about A gives the hinge reaction at B: $R_{B1} = 68.0$ kN/m, $R_A = 165.0 - 68.0 = 97.0$ kN/m.
• Middle part (B to C, 2.5 m) simply supported, uniform 60: $R_{B2} = R_{C1} = 60\times2.5/2 = 75.0$ kN/m.
• Bottom part (6.5 to 10 m): load $= 210$ kN/m; moments about D give $R_{C2} = 63.0$ kN/m and $R_D = 210 - 63.0 = 147.0$ kN/m.

Check: $\sum R = 525$ kN/m $= 525$ kN/m.

Answer: strut loads A = 194 kN, B = 286 kN, C = 276 kN, D = 294 kN.

Arching effect

Arching is the transfer of pressure from a yielding part of a soil mass to the adjacent stationary parts through the shear resistance of the soil. The pressure on the yielding part decreases and that on the stationary part increases (Terzaghi's trap door experiment).

Factor of safety against bottom heave (Terzaghi)

Rapid excavation means undrained conditions, so $\phi_u = 0$, $c = c_u = 35$ kN/m$^2$. The heave is checked as a bearing failure of the base under the weight of soil beside the cut, resisted by shear on vertical planes. The bearing capacity of the base is taken as $5.7c$.

(a) Deep clay (no hard layer; $T \ge 0.7B$): the width of the failing strip is $B' = B/\sqrt2 = 0.7B$:

$$FS = \frac{5.7c}{\gamma H - \sqrt{2}\,cH/B} = \frac{5.7\times35}{18\times10 - \sqrt2\times35\times10/5} = \frac{199.5}{81.01} = 2.46$$

(b) Hard rock 2.5 m below the base: here $T = 2.5\ \text{m} < 0.7B = 3.5$ m, so the failure zone is limited by the rock and $B' = T = 2.5$ m:

$$FS = \frac{5.7c}{\gamma H - cH/T} = \frac{199.5}{180 - 35\times10/2.5} = \frac{199.5}{40.0} = 4.99$$

Answer: $FS = 2.46$ for deep clay (marginal, below the usual 1.5 to 2 for temporary works, so the sides must be braced/embedded); $FS = 4.99$ with rock 2.5 m below the trench bottom, which is adequate.

Strut loads

Data: $H = 8$ m, $\phi = 30^\circ$, $\gamma = 16$ kN/m$^3$, three struts at 2, 4 and 6 m below the ground, horizontal spacing $s = 4$ m. The width of the cut (5 m) does not affect the Peck envelope.

Apparent pressure (Peck, sand):

$$K_a = \tan^2(45^\circ-15^\circ) = 0.3333,\qquad p = 0.65K_a\gamma H = 0.65\times0.3333\times16\times8 = 27.73\ \text{kN/m}^2$$

The pressure is uniform over the full depth of 8 m (total 221.9 kN/m).

   0  |          |
   2m |--A-------|  p = 27.73
   4m |--B-------|
   6m |--C-------|
   8m |__________|

Reactions (tributary-area method, each strut takes the load halfway to the next support; the bottom strut takes the load to the base):

Check: $\sum R = 221.9$ kN/m.

Answer: design strut loads A = 333 kN, B = 222 kN, C = 333 kN (the same maximum, 333 kN, can be used for the top and bottom struts).

Terzaghi's trap door experiment

A box of dry sand has a base strip (trap door) that can be lowered.

        sand surface
   ~~~~~~~~~~~~~~~~~~~~~~
   |  \              /  |
   |   \  yielding  /   |
   |____\__ wedge __/____|
   fixed |  trap  | fixed
         | door  |
           v moves down

When the door moves down slightly the pressure on it falls sharply to about half of the overburden pressure, and the pressure on the neighbouring fixed base rises by the same total amount. Shear stresses along the nearly vertical planes carry part of the weight: this is arching. It explains why the earth pressure on braced cuts is not triangular but nearly uniform (Peck's envelope).