The fact that compressed air can be used at all in tunnel work is evidence that semi-aqueous materials have arching properties, and the fact that “blows” usually occur in light cover is further evidence of it.

When air pressure is used to hold back the water in faces of large area, bracing has to be resorted to.  This again shows that while full hydrostatic pressure is required to hold back the water, the pressure of the earth is in a measure independent of it.

In a peaty or boggy material there is a condition somewhat different, but sufficiently allied to the soft clayey or soupy sands to place it under the same head in ordinary practice.  It is undoubtedly true that piles can be driven to an indefinite depth in this material, and it is also true that the action of the pile is to displace rather than compress, as shown by the fact of driving portions of the tunnels under the North River for long distances without opening the doors of the shield or removing any of the material.  The case of filling in bogs or marshes, causing them to sink at the point of filling and rise elsewhere, is readily explained by the fact that the water is confined in the interstices of the material, admitting of displacement but no compression.

The application of the above to pressures over tunnels in materials of Class A is that the sand or solid matter is virtually assumed to be a series of columns with their bases in such intimate contact with the tunnel roof that water cannot exert pressure on the tunnel or buoyancy on the sand at the point of contact, and that if these columns are sufficiently deep to have their upper portions wholly or partly carried by the arching or wedging action, the pressure of any water on their surfaces is not transferred to the tunnel, and the only aqueous pressure is that which acts on the tunnel between the assumed columns or through the voids.

Let l = exterior width of tunnel,
    d = depth of cover, as: 

D{W}_ = depth, water to roof,
D{E}_ =   "    earth to roof,
D{X}_ =   "    of cover of earth necessary to arching stability,

that is: 

l          / 90 deg. — [phi] \
D{X}_ = ----- ( tan. { ------------- } + [phi] ) =
2           \     2       /

l                [phi]
----- tan.  (45 deg. + ------- ),
2                   2

where [phi] = angle of repose,
and D{W}_ > D{E}_ > D{X}_.

Then the pressure on any square foot of roof, as V{P}_ as at the base of any vertical ordinate, as 9 in Fig. 2, = V{O}_,

   W{E}_ = weight per cubic foot of earth (90 lb.),
   W{W}_ = " " " " " water (621/2 lb.), we have

V{P}_ = V{O}_ x W{E}_ + D{W}_ x W{W}_ x 0.40 =

1
V{O}_ x 90 + D{W}_ x 62—–­ x 0.4 = V{O}_ 90 + D{W}_ x 25.
2