CHAPTER XIV. RETAINING WALLS AND ABUTMENTS. Let aa'b'b in Fig. 15 represent the cross section of a wall of masonry which retains a bank of earth having a surface aa. Assume that the portion of the wall and earth under consideration is bounded by two planes parallel to the plane of the paper, and at a unit's distance from each other: then any plane containing the edge of the wall at b, as ba, ba,, etc., cuts this solid in a longitudinal section, which is a rectangle having a width of one unit, and a length ba,, ba,, etc. The resultant of the total pressure distributed over any one of these rectangles of the type ba is applied at onethird of that distance from 6: i.e. the resultant pressure exerted by the earth against the rectangle at ba, is applied at a distance of bk ba, from b. That the resultant is to be applied at this point, is due to the fact that the distributed pressure increases uniformly as we proceed from any point a of the surface toward b: the center of pressure is then at the point stated, as is well known. Again, the direction of the pressures against any vertical plane, as that at ba,, is parallel to the surface aa ̧. This fact is usually overlooked by those who treat this subject, and some arbitrary assumption is made as to the direction of the pressure. That the thrust of the earth against a vertical plane is parallel to the ground surface is proved analytically in Rankine's Applied Mechanics on page 127; which proof may be set forth in an elementary manner by considering the small parallelopiped mn, whose upper and lower surfaces are parallel to the ground surface. Since the pressure on any plane parallel to the surface of the ground is due to the weight of the earth above it, the pressure on such a plane is vertical and uniformly distributed. If mn were a rigid body, it would be held in equilibrium by these vertical pressures, which are, therefore, a system of forces in equilibrium; but as mn is not rigid it must be confined by pressures distributed over each end surface, which last are distributed in the same manner on each end, because each is at the same depth below the surface. Now the vertical pressures and end pressures hold mn in equilibrium they therefore form a system in equilibrium. But the vertical pressures are independently in equilibrium, therefore the end pressures alone form a system which is independently in equilibrium. That this may occur, and no couple be introduced, these must directly oppose each other; i.e. be parallel to the ground line aa,. Draw kp | aa, it then represents the position and direction of the resultant pressure upon the vertical ba.. Draw the horizontal ki, then is the angle ikp called the obliquity of the pressure, it being the angle between the direction of the pressure and the normal to the plane upon which the pressure acts. Let ebc be the angle of friction, i.e. the inclination which the surface of ground would assume if the wall were removed. The obliquity of the pressure exerted by the earth against any assumed plane, such as ba, or ba, must not exceed the angle of friction; for should a greater obliquity occur the prism of earth, a,ba, or a ba, would slide down the plane, ba, or ba,, on which such obliquity is found. For dry earth is usually about 30°; for moist earth and especially moist clay, may be as small as 15°. The inclination of the ground surface aa, cannot be greater than . Now let the points a,, a,, a,, etc., be assumed at any convenient distances along the surface: for convenience we have taken them at equal distances, but this is not essential. With b as a center and any convenient radius, as bc, describe a semi-circumference cutting the lines ba, ba2, etc. at c1, c2, etc. Make eeec; also eeCC1? CC, etc.: then be, has an obliquity with ba, as has also be, with ba, be, with ba,, etc.; for a be =a,be,=a,be,=90° + Þ. 0 Lay off bb, bb, bb, etc., proportional to the weights of the prisms of earth aba, a ba, a ba,, etc.: we have effected this most easily by making a,a,=bb,, aa=bb„‚a ̧à ̧=bb,, etc. Through b, b, b, etc., draw parallels to kp; these will intersect be, be, be„, etc., at b, t1, t, etc. 19 Then is bb,t, the triangle of forces holding the prism aba, in equilibrium, just as it is about to slide down the plane ba,, for bb, represents the weight of the prism, bt, is the known direction of the thrust against ba, and bt, is the direction of the thrust against ba, when it is just on the point of sliding: then is t,b, the greatest pressure which the prism can exert against ba. Similarly tb, is the greatest pressure which the prism a ba, can exert.. Now draw the curve t,t,t,, etc., and a vertical tangent intersecting the parallel to the surface through b at t; then is to the greatest pressure which the earth can exert against ba.. This greatest pressure is exerted approximately by the prism or wedge of earth cut off by the plane ba, for the pressure which it exerts against the vertical plane through b is almost exactly bt=bt. This is Coulomb's "wedge of maximum thrust" correctly obtained: previous determinations of it have been erroneous when the ground surface was not level, for in that case the direction of the pressure has not been ordinarily assumed to be parallel to the ground surface. In case the ground surface is level the wedge of maximum thrust will always be cut off by a plane bisecting the angle cbc,, as may be shown analytically, which fact will simplify the construction of that case, and enable us to dispense with drawing the thrust curve tt. The pressure tb is to be applied at k, and may tend either to overturn the wall or to cause it to slide. In order to discuss the stability of the wall under this pressure, let us find the weight of the wall and of the prism of earth aba. Let us assume that the specific gravity of the masonry composing the wall is twice that of earth. Make a'h bb', then the area abb'a'= abh=abh,; and if ah,=2ah, then ah, represents the weight of the wall reduced to the same scale as the prisms of earth before used. Since aa, is the weight of aba, ah, is the weight of the mass on the right of the vertical ba against which the pressure is exerted. 0 Make bq=ah,, and draw tq, which then represents the direction and amount of the resultant to be applied at o where the resultant pressure applied at k intersects the vertical gw through the center of gravity g of the mass aa bb'a'. The wall will sustain without sliding up some plane such as b'a' or b'a', etc. The difference in the two cases is that in the former case friction hindered the earth from sliding down, while it now hinders it from sliding up the plane on which it rests. Lay off e'eee; then taking any points a'a', etc. on the ground surface, make e '=c'c,', etc. center of gravity g is constructed in the following manner. Lay off a'h bb', and bl=aa'; and join hl. Join also the middle points of ab and a'b': the line so drawn intersects hl at g, the center of gravity of aa'b'b. Find also the center of gravity g,, of aba,, which lies at the intersection of a line parallel to aa, and cutting ba, at a distance of ba, from a and of a line from b bisecting aa.. Through g, and g, draw parallels, and lay off gf, and g,f, on them proportional to the weights applied at g, and g, respectively. We have found it convenient to make g,f,ah,, and gf, and the wall adds to the stability of the aa. Then ff, divides g,g, inversely as wall, and can be made to enter the conthe applied weights; and g, the point of struction if desired, in the same manner intersection, is the required center of as did aba ̧. gravity. 2 2 = 2 Let or be parallel to tq; since it intersects bb' so far within the base, the wall has sufficient stability against overturning. The base of the wall is so much greater than is necessary for the support of the weight resting upon it, that engineers have not found it necessary that the resultant pressure should intersect the base within the middle third of the joint. The practice of English engineers, as stated by Rankine, is to permit this intersection to approach as near b' as bb', while French engineers permit it to approach as near as bb' only. In all cases of buttresses, piers, chimneys, or other structures which call into play some fraction of the ultimate strength of the material, or ultimate resistance of the foundation as great as one tenth, or one fifteenth, the point should not approach b' nearer than bb'. Again, let the angle of friction between the wall and the earth under it be ': then in order that the thrust at k may not cause the wall to slide, the angle wor must be less than '. Lay off b'b'a'a,, etc., and drawing parallels through b, b', etc., we obtain the thrust curve t't, etc. The small prism of earth between b'a' The vertical tangent through s' shows us that the earth in front of the wall can withstand a thrust having a horizontal component b's' measured on a scale such that b'b'a'a,' is the weight of theprism of earth a 'b'a,. This scale is different from that used on the left. To reduce them to the same scale lay off from b', the distances b'd, and b'd' proportional to the perpendiculars from b on aa, and b' on a'a' respectively. In the case before us, as the ground surfaces are parallel, we have made b'd=ba and b'd'=b'a'. 0 Then from any convenient point on b'b', as v, draw vd, and vd': these lines will reduce from one scale to the other. We find then that x'd is the thrust on the scale at the left corresponding to xd=b's' on the right: i.e., the earth under the surface assumed at the right can withstand something over one fourth of the thrust sb at the left. It will be found that a certain small portion of the earth near a' has a thrust curve on the left of b', but as it is not needed in our solution it is omitted. If any pressure is required in pounds, as for example sb, it is founds as follows: the length of ah, is to that of sb as the weight of bb'aa' in lbs. is to the pressure sb in lbs. When, however, the angle D'is less than wor it becomes necessary to gain additional stability by some means, as for example by continuing the wall below the surface of the ground lying in front of it. Let a'a' be the surface of the ground Frequently the ground surface is not a which is to afford a passive resistance to plane, and when this is the case it often the thrust of the wall: then in a manner consists of two planes as ad, da, Fig. 16. precisely analogous to that just employed In that case, draw some convenient line for finding the greatest active pressure as ad,, and lay off ad,, dd,, etc. at will, which earth can exert against a vertical which for convenience we have made plane, we now find the least passive equal. Draw d,a,, d,a,, etc. parallel to pressure which the earth in front of the bd, and join ba,, ba,, etc.: then are the triangles bda, bda,, bda,, bda,, etc. proportional in area to the lines ea, ea,, etc. Hence the weights of the prisms of earth baa,, baa,, etc., are proportional to ad,, ad, etc. In case ab slopes backward the part of the wall at the left of the vertical ba, rests upon the earth below it sufficiently to produce the same pressure which would be produced if baa, were a prism of earth. The weights of the wedges which produce pressures, and which are to be laid off below b, are then proportional to dd=bb1, dd=bb2, etc. The direction of the pressures of the prisms at the right of ba are parallel to ad; but upon taking a larger prism the direction may be assumed to be parallel to a,a,, aa, etc., which is very approximately correct. Now draw bt, aa, bt || aa, etc.; and complete the construction for pressure precisely as in Fig. 15, using for resultant pressure the direction and amount of that due to the wedge of maximum pressure thus obtained, In finding the stability of the wall, it will be necessary to find the weight and center of gravity of the wall itself, minus a prism of earth baa,, instead of plus this prism as in Fig. 15; for it is now sustained by the earth back of the wall. When the back of the wall has any other form than that above treated, the vertical plane against which the pressure is determined should still pass through the lower back edge of the wall. In case the wall is found to be likely to slide upon its foundations when these are level, a sloping foundation is frequently employed, such that it shall be nearly perpendicular to the resultant pressure upon the base of the wall. The construction employed in Fig. 15 applies equally to this case. The investigation of the stability of any abutment, buttress, or pier, against overturning and against sliding, is the same as that of the retaining wall in Fig. 15. As soon as the amount, direction, and point of application, of the pressure exerted against such a structure is determined, it is to be treated precisely as was the resultant pressure kp in Fig. 15. In the case of a reservoir wall or dam, the construction is simplified from the fact that, since the surface of water is level and the angle of friction vanishes, the resultant pressure is perpendicular to the surface upon which the water presses. It is useful to examine this as a case of our previous construction. In Fig. 17, let abb' be the cross-section of the dam; then the wedge of maximum pressure against ba, is cut off by the = plane ba, when cbu, 45°, i.e. ba, bisects load which it sustains. Now consider a cba, as before stated. This produces a horizontal resultant pressure at k equal to the weight of the wedge. Now the total pressure on ab is the resultant of this pressure, and the weight of the wedge aba. The forces to be compounded are then proportional to the lines a,a=by and aa. By similarity of triangles it is seen that ro the resultant is perpendicular to ab. It is seen that by making the inclination of ab small, the direction of ro can be made so nearly vertical that the dam will be retained in place by the pressure of the water alone, even though the dam be a wooden frame, whose weight may be disregarded. 5 vertical plane of one unit in height, say, We can now construct the actual pressures to which the arch of a tunnel surcharged with water or earth is sub- Next, it must be determined what pasjected. Suppose, for example, we wish sive pressure the earth at the left of bb, to find the pressure of such a surcharge can support. The passive resistance of on the voussoir adda, Fig. 14. Find the earth under the surface a against the resultant pressure against a vertical the plane ab as well as that against the plane extending from d, to the upper plane ab, can be found exactly as that surface of the surface and call it p.. was previously found under the surface Draw a horizontal through d, and a'. The difference of these resistances is let its intersection with the vertical the resistance which it is possible for bb ̧ just mentioned he called d". Find to support. Indeed bb, could support the resultant pressure against the verti- this pressure and afford this resistance cal plane extending from d" to the sur- even if the active pressure against ab face, and call it p'. Now let p" were, at the limit of its resistance, which P.-P. and let it be applied at such a point it is not. The limiting resistance which of dd" that p shall be the resultant of p' is thus obtained, is then so far within and p". Then will the resultant press- the limits of stability, that ordinarily, no ure against the voussoir be the resultant further factor of safety is needed, and of p," and the weight of that part of the the stability of the foundation is secured, surcharge directly above it. if the active pressure against bb, does not exceed the passive resistance. This construction should be made on the basis of the smallest angle of friction which the earth assumes when wet; that being smaller than for dry earth, and hence giving a greater active pressure at the right, and a less resistance at the left. FOUNDATIONS IN EARTH. = A method similar to that employed in the determination of the pressure of earth against a retaining wall, or a tunnel arch, enables us to investigate the stability of the foundations of a wall standing in earth. CHAPTER XV. Suppose in Fig. 15 that the wall abb'a' is a foundation wall, and that the pressSPHERICAL DOME OF METAL. ure which it exerts upon the plane bb' The dome which will be treated in the is vertical, being due to its own weight following construction is hemispherical and the weight of the building or other in shape; but the proposed construction |