[Illustration:  PLATE XXIV, FIG. 1.—­INITIAL SETTLEMENT IN 3-FT.  SAND ARCH, DUE TO COMPRESSION OF MATERIAL ON REMOVING SUPPORTS FROM BOTTOM.]

[Illustration:  PLATE XXIV, FIG. 2.—­FINAL SETTLEMENT OF SAND ARCH, DUE TO COMPRESSION IN EXCESS LOADING.]

In this experiment the sand was good and sharp, containing some gravel, and was taken directly from the adjoining excavation.  When thrown loosely in a heap, it assumed an angle of repose of about 45 degrees.  It should be noted that this material when tested was not compacted as much, nor did it possess the same cohesion, as sand in its normal undisturbed condition in a bank, and for this reason it is believed that the depth of key given here is absolutely safe for all except extraordinary conditions, such as non-homogeneous material and others which may require special consideration.

Referring again to the area, A N{1} V J N A_, Fig. 2, it is probable that, while self-sustaining, some at least of the lower portion must derive its initial support from the “centering” below, and the writer has made the arbitrary assumption that the lower half of it is carried by the structure while the upper half is entirely independent of it, and, in making this assumption, he believes he is adding a factor of safety thereto.  The area, then, which is assumed to be carried by an underground structure the depth of which is sufficient to allow the lines, V A and J A, to intersect below the surface, is the lower half of A N{1} V E J N A_, or its equivalent, A V E J A, plus the area, V E J, or A V J A, the angle, A V J, being

1                                   [phi]
[alpha] = --- ( 90 deg. - [phi] ) + [phi] = 45 deg. + -------.
2                                     2

It is not probable that these lines of thrust or pressure transmission, A N, D K, etc., will be straight, but, for purposes of calculation, they will be assumed to be so; also, that they will act along and parallel to the lines of repose of their natural slope, and that the thrust of the earth will therefore be measured by the relation between the radius and the tangent of this angle multiplied by the weight of material affected.  The dead weight on a plane, V J, due to the material above, is, therefore, where

l = span or extreme width of opening = V J,
W = weight per cubic foot of material, and
W{1}_ = weight per linear foot.

2 x (l / 2) tan. [alpha] x W
W{1}_ = ---------------------------------- =
2
1            / 1                       \
—–­ l tan. { —–­ (90 deg. — [phi]) + [phi] } W =
2            \ 2                       /

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

The application of the above to flat-arched or circular tunnels is very simple, except that the question of