46 THEORY OF EARTH PRESSURE. Denoting the angle ABC by 0, and by W the weight in lb of the prism ABC of length 1, then W = {wlh2 tan 0, w denoting the heaviness of the substance; and resolving perpendicular to R, P=W cot(0+€) = wlh2 tan 0 cot(0+€) = {wth? sin(20+e)—sin e = }wth2 {1 2 sin € sin(20+e)+sin e For different directions of the plane BC, or different values of 0, P will be greatest when sin(20+€)=1, or 0=1π—že; and then P=wlh2 1-sin e 1 + sin e = wlh2 tan2(12€). This is the greatest thrust the wall can on this theory be called upon to support, supposing the loose substance to crack and slide along a plane BC through the foot of the wall; and it is the same as the hydrostatic thrust of a liquid of heaviness w tan2(1𗧀). If the friction of the vertical wall AB is taken into account, the theory is more complicated. For a substance like ice, in which we may suppose the planes of cleavage perfectly smooth, e=0; so that the complete hydrostatic thrust will be restored if ice, frozen up to the level AC, cracks along planes of cleavage; as, for instance, in a glacier. *31. Surcharged Retaining Walls. Suppose the substance is retained by a parallel vertical wall DE, of less height than the level AC, and that the RETAINING WALLS AND REVETMENTS. 47 surface is sloped down to D in the plane FD, called the talus; the wall DE is then said to be surcharged to the height of ACF above D; and the slope a of FD to the horizon cannot exceed e, the angle of repose. To determine the horizontal thrust Q on DE, suppose the wall DE to yield horizontally a slight distance, and in consequence the substance to crack along a plane of cleavage EM or EN, making an angle with the vertical wall DE. H N F M Fig. 21. We suppose that the prism of material DEM or DENF begins to slide down the plane EM or EN; and then as before, if W denotes the weight in lb of the material in the prism, Q= W cot(0+€). If the plane EM meets the talus DF in M, and we put DE=a, then W = wla2 sin cos a cos(0+a) sin e cos a cos(+€), { sin(0+e) {wla2 ( 1 + sin(0—e)}, sin(0+) if the slope of the talus is the angle of repose, or α π ε. 48 THRUST OF A GRANULAR SUBSTANCE. As increases from zero, the thrust Q also increases from zero; and when 0+c=7, or the plane EM is parallel to the talus DF, Q=wla2cos2e, the same as for liquid of density w cos2. But this implies that the wall DE is surcharged to an infinite height; but if surcharged to a finite height b, then when the plane of cleavage EN meets the horizontal level surface in N, W=wl{(a+b)2tan 0— b2cot a}, and the corresponding value of Q will become a maximum for a value of 0 depending not only one, but also on the ratio of b to a. The determination of this maximum value must be deferred; but now it is important to notice that P and Q are not equal, the difference between them being taken up by the frictional resistance of the ground BE. *32. The Thrust due to an Aggregation of Cylindrical Particles or of Spherules. An exact Theory of Earth Pressure can be constructed if we suppose the substance which is held up by a retaining wall to be composed of individual particles or atoms of cylindrical form, such as canisters, pipes, barrels, or cylindrical projectiles, regularly stacked; or else to be composed of spherules, such as lead shot, billiard balls, or spherical shot and shell, piled in regular order, as common formerly in forts and arsenals. It will be necessary to begin by supposing that the lowest layer of cylinders or spheres is imbedded in the ground; as otherwise a wedging action takes place, due to the slightest variation of level, and the problem is to a certain extent indeterminate, as in the preceding article on Earth Pressure. AGGREGATION OF CYLINDERS. 49 Now in the case of cylindrical bodies, regularly packed as close as possible (fig. 22), the slope of the talus DF is 60°; and if EN is drawn through E the foot of the retaining wall DE parallel to the talus DF, the thrust between the cylinders across the plane EN will also make an angle of 60° with the horizon; so that considering the equilibrium of DENF, of weight W, the thrust Q on the retaining wall DE is given by Q = W cot 60°. A N F E Fig. 22. Also, if w denotes the apparent heaviness of the substance, measured in lb/ft3, W = {wl (h2 — b2)cot 60° = wl({a2+ab)cot 60°, so that Q=wl(a2+ab)cot260° = {wl(±a2+ab), the same as the hydrostatic thrust of liquid, of heaviness w, on the portion DE of a vertical wall, of which the top edge D is submerged to a depth b in the liquid. If the cylindrical particles are of diameter d, and composed of solid metal of density p, then since the triangular prism formed by the axes of three adjacent cylinders is of cross section 1/3d2, of which only the area d2 is occupied by solid metal, therefore so that }πpd2= {√3wd2, or w/p=}π√3; the same as for liquid of density π/3p. 50 THRUST OF AGGREGATION Now, suppose the cylinders in the lowest layer are imbedded in the ground, and regularly separated so as to be at a distance x from axis to axis; the slope a of the talus DF is given by and cos a= x/d, Q= W cot a=wl({a2+ab)cot2a; but now w, the apparent heaviness of the substance, is When x>d/3, vertical planes of contact come into existence; and alternate vertical columns descend, so that w/p=πd/x. The thrust Q will become very large when the cylinders are nearly in square order. The thrust Q is theoretically infinite when the cylinders are in square order; but this arrangement being unstable, a seismic rearrangement takes place, and the original triangular order is regained, except that the talus now appears stepped. Suppose the wall AB or DE to yield horizontally a slight distance; the cylinders in ABC or DENF will roll and wedge down along planes of cleavage BC or EN. The lowest layer of cylinders being imbedded, no further motion is possible; but if they were free to roll sideways, a molecular rearrangement would take place, and the cylinders would appear wedged against the walls AB and DE in close order, except along two planes of cleavage. |