Let l = exterior width of tunnel,
d = depth of cover, as:
DW = depth, water to roof,
DE = " earth to roof,
DX = " of cover of earth necessary to arching stability,
that is:
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where φ = angle of repose,
and DW > DE > DX.
Then the pressure on any square foot of roof, as VP as at the base of any vertical ordinate, as 9 in [Fig. 2], = VO,
WE = weight per cubic foot of earth (90 lb.),
WW = " " " " " water (62½ lb.), we have
VP = VO × WE + DW × WW × 0.40 = VO × 90 + DW × 62½ × 0.4 = VO 90 + DW × 25.
And for horizontal pressure:
Ph = the horizontal pressure at any abscissa (10), [Fig. 2], = A10 at depth of water DW1 is
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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, = DX, although he believes that experiments on a large scale would show it to be nearer 0.67·DX, above which the placing of additional back-fill will lighten the load on the structure.
We have, then, for DE < DX, 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 = VP
VP = DE × 90 + DE × 62½ × 0.40 + ( DW - DE ) × 62½.
To those who may contend that water acting through so shallow a prism of earth would exert full pressure over the full area of the tunnel, it may be stated that the water cannot maintain pressure over the whole area without likewise giving buoyancy to the sand previously assumed to be in columns, in which case there is the total weight of the water plus the weight of the prism of earth, less its buoyancy in water, that is
VP = DW × 62½ + DE × ( 90 - 62½ ),