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 And for horizontal pressure: _P_{h}_ = the horizontal pressure at any abscissa (10), Fig. 2, = _A_{10}_ at depth of water _D_{W1}_ is _A_{10}_ x 90 1 _P_{h}_ = --------------- + _D_{W1}_ x 62--- x 0.4 = tan. [phi] 2 _A_{10}_ x 90 --------------- + _D_{W1}_ x 25. tan. [phi] The only question of serious doubt is at just what depth the sand is incapable of arching itself, but, for purposes of safety, the writer has put this at the point, _F_, as noted above, = _D_{X}_, although he believes that experiments on a large scale would show it to be nearer 0.67._D_{X}_, above which the placing of additional back-fill will lighten the load on the structure. We have, then, for _D_{E}_ < _D_{X}_, the weight of the total prism of the earth plus the water in the voids, plus the added pressure of the water above the earth prism, that is: The pressure per square foot at the base of any vertical ordinate = _V_{P}_ 1 _V_{P}_ = _D_{E}_ x 90 + _D_{E}_ x 62--- x 0.40 + 2 1 ( _D_{W}_ - _D_{E}_ ) x 62---. 2