TABLE 2.(1)(2)(3)Dry sand70.700.80 to 0.9626 to 56Wet sand400.851.20 " 1.9073 " 133Very wet sand311.702.00 " 2.40107 " 226Fresh earth901.632.60 " 4.40150 " 242The values ofare given in kilogrammes per square meter. It is seen that suppositions (2) and (3) give discordant results, whereas (1), for each kind of filling, gave identical values ofand offor various heights; hence it may fairly be concluded that, before motion, cohesion and friction both acted at the same time. As to the high values found for,for the coherent fresh earth, Leygue states that Collin found, by an independent method, for clayey earth and clay of little consistency,and,respectively. As a further verification of the values ofandgiven in (1), it is found that, on using them in the formula for computing the height at which the wet sand or earth will stand vertically, the results agree with experiments.The values of,in pounds per square foot, given in Column (1), with the values ofcorresponding to thegiven, are as follows:It is possible that the method used by Leygue may prove of service to experimenters in obtaining more accurately than hitherto the coefficient of internal friction. Increasing pressures could be obtained by adding weights on top of the sand in the box; but, unless the total weight sustained by friction along the sides of the box is determined carefully for each weight used, the results can have but little value. Further, for coherent earths, the method of Leygue is open to objections.Admitting the hypothesis that cohesion and friction act at the same time, a general graphical method[Footnote6]will now be given to find the thrust against the inner face,,of a retaining wall or board,Fig. 11, caused by the earth,,tending to slide down some plane of rupture,,,..., the resistance along this plane being due both to friction and cohesion.Graphical method to find the thrust against the inner face of a retaining wall or boardFig. 11.Supposeto be the plane of rupture, and call the weight of the prism of rupture,,for a thickness of one unit,.The weight of the prism causes the tendency to slide along the planes,and.This tendency is resisted by the reactions of the wall,,and the plane,.The reaction of the wall consists of the normal component,,acting to the right, and the friction resistance,,acting up. The resultant of these two forces,,which is equal and opposed to the earth thrust on,thus makes the angle,,with the normal to the wall. Its direction is given byinFig. 11. The reaction ofis made up of the cohesion,,acting up along,the normal component,,acting up, and the friction,,acting up along.The two forces,andwhen combined, give a resultant,,making an anglewith the normal,,to.Hence, if the angle,,thengives the direction of theresultant,.The prism,,is thus in equilibrium under its own weight,,the cohesive force,,acting up along,the reaction,,of,acting to the right, and the force,,acting up. On drawing, to the scale of force,vertical and equal to;thenparallel toand equal to;thenparallel toto the intersection with,the sides of the closed polygon,,in order, will represent the four forces,,,,and,in equilibrium.A similar investigation pertains to any other supposed prism of rupture. To find the true one, a number of trial planes of rupture,,,..., are assumed, and each is treated in turn as a true one (though there can be only one true one). As seen above, the resultant of the normal reaction and friction on any trial plane of rupture must be inclined below the normal to the plane at the angle,.To lay off the directions of these resultants, from any convenient point,,say, in the vertical through,as a center, describe an arc,,with a convenient radius,.With the same radius andas a center, describe the arc,,cutting the trial planes (produced if necessary) at,,...,.Letbe the point where the line of natural slope fromcuts the arc,.On laying off the chords,,,..., equal to the chords,,,..., respectively, it will follow that,,..., will make angles,,below the normals to the planes,,,..., respectively. To prove this, take any trial plane, as,which makes the angle,,with,and drop a perpendicular,,fromon(produced if necessary); then, because the sides of the angles are perpendicular,,and if,it follows that,as was to be proved. Hence the chord,the chord,,,etc., as stated.The weights, in pounds, of the prisms,,,..., are,,,..., respectively, whereis the length of the perpendicular fromupon(produced if necessary), the footbeing the unit of length andbeing the weight, in pounds, of 1 cu. ft. of earth. The prisms are supposed to be 1 ft. in length perpendicular to the plane of the paper.These weights are now laid off to the scale of force, vertically downward from,to points 1, 2, 3, ..., and from these points, lines are drawn parallel to,,,..., respectively, of lengths equal to,,..., to represent the forces of cohesion, acting upward along,,..., wheretheforce of cohesion, in pounds per square foot. From the extremities of these lines, lines are drawn parallel to the direction of the earth thrust on(inclined at the angle,,below the normal to), to the intersections,,,,..., with,,,..., respectively. With dividers, it is found, for this figure, thatis the longest of these lines; whence,to the scale of force, measures the earth thrust against,in pounds. This follows, because, for any less thrust, sinceis fixed, whenbecomes less,falls to the left of the first position, and the new,representing the thrust on,due to the normal reaction on it and friction only, will make a greater angle thanto the normal to,which is inconsistent with the laws of stability of a granular mass. In fact, ifis the normal component of the thrust on the plane,,is all the friction that can be exerted on it. The resultant ofandthus makes an angle,,with,and this angle cannot be exceeded. The true thrust on the wall,,is thus the greatest of the trial thrusts,.The prism of rupture,,is in equilibrium under the four forces represented by the sides of the closed polygon,;,representing its weight;,the cohesion acting along;,the reaction of(opposed and equal to the earth thrust); andthe reaction of the plane,,due to the normal component and friction on it only. The full reaction of the plane can be found by combining the forces, given in magnitude and direction byand,but it is not needed.It is to be noted thatis the least thrust for which equilibrium is possible. The other trial thrusts should now be lengthened to equal,since this is the true thrust or reaction of the wall,.All the new points,,,,,,will now lie to the right of the old points; hence the new,,,,,will all make less angles thanwith the normals to the planes,,etc.; hence stability everywhere in the earth mass is assured.Special casesFig. 12.The solution represented byFig. 11is general, and applies whetheris inclined to the right or left of the vertical throughor coincides with it, and whether the earth surfaceis horizontal or inclined above or below the horizontal. It can likewise be easily adapted to the case shown inFig. 1.[Footnote7]The construction ofFig. 11has been used in evaluating the thrust and determining (approximately) the plane of rupture in the experiments (recorded below) of Leygue on retaining boards,,that could be rotated about,and thus placed at any inclination to the vertical. In all the experiments, the vertical height ofwas 0.656 ft.; the length of the board was 1.3 ft. The value of the moment of the earth thrust aboutwas found by use of suitable apparatus, corresponding to dry sand with a natural slope of 3 base to 2 rise, or,,andlb. per cu. ft. By use of the partition board mentioned previously, the side friction of the sand on the glass sides of the box containing it was estimated, and the moments corrected, so as to give the true moment when there is no side friction. The notation used to express results is partly given inFig. 12, for the general case where,The resultant,,ofand,evidently makes the angle,,with the normal to,The moment of this resultant about,if we put.From the last formula, it is seen thatis the moment of the thrust about,for;also from,it follows thatis the normal component of the thrust for.When cohesion is included,is not exactly,but it is very slightly less foror 2. It will be assumed at,and, from the values ofgiven byLeygue,will be derived from the formula above,Thus, for the case represented byFig. 11,,therefore,;.Experiment gave,therefore.Neglecting cohesion, theory gives,or twice the amount given by experiment. If, however, the construction ofFig. 11is made for the actual height of the retaining board,ft.,,(cohesion, in pounds per square foot), we find.On substituting this in the formula,,we have,By a comparison of the values, it is evident that, if the cohesion was assumed at a little less than 1 lb. per sq. ft., the theoretical and experimental values could be made to agree exactly. The case just examined exhibits the most pronounced difference between the ordinary theory (corresponding to)and experiment, of any shown inTable 3. Further, it will be observed, that, for an assumed cohesion of about 1 lb. per sq. ft., the theoretical and experimental values for all the cases given by Leygue very nearly agree.The value,,in place of Leygue’s,,was used, which would alter the results somewhat, but not the general conclusions. The construction ofFig. 11will giveand its normal component,,with practical accuracy, but it is not readily adaptable in finding the plane of rupture. In most of the drawings a small scale was used, in order to limit the drawing to a sheet of writing paper, hence, on both accounts,cannot be counted on to nearer than 1° or 2°, except for,whenwas found by computation, or by the construction ofFig. 1.TABLE 3...Cohesion,,in poundspersquarefoot.Angle of rupture withthe horizontal,.Coefficientof thenormal component ofthe thrust.Theory.Experiment.Theory.Experiment.+⅓0050° 12′0.060151°51° 30′0.0420.043252° 30′0.026+⅓⅔033° 41′0.182144°47°0.0840.091249°0.04300056° 36′0.111157°56° 30′0.0930.090258°0.077358° 30′0.0620½047° 30′0.178150°51°0.1480.141253°0.121355°0.0980⅔033° 41′0.345144°49°0.2050.195246° 30′0.150350°0.111–⅓0060° 21′0.185163°61°0.1710.179263°0.155–⅓⅔033° 41′0.660157°57°0.2670.387250°0.236The results inTable 3are remarkable, and explain quite satisfactorily how Leygue, Darwin, and others found, by experiments on small models, results differing so much from the ordinary theory, where cohesion is neglected.It should be remarked that the values ofgiven inTable 3under “Experiment,” are not exactly those given by Leygue in his tables, but are the averages obtained from the two sets of drawings given by him in the plates, and represent the inclinations of the chords of the really curved surfaces of rupture. His experiments with the spring apparatusare not considered, as the results are open to doubt, because the prism of rupture, in descending, could not slide down freely, but as it advanced would rub over the floor, thus lessening the thrust there considerably.FromTable 3, the results given by experiment are seen to differ widely from the ordinary theory in which.The discrepancies are largely, or almost entirely, due to the very small models used, as will be evident from the following considerations: Suppose the height,,of the wall,,to be 10 times the height given inFig. 11, or 6.56 ft.; then, as the areas of triangles such as,etc., vary as the squares of the heights, but the lengths of sides, asetc., vary only as the first power of the heights, the weights of the successive trial prisms of rupture will beor 100 times as great as before, whereas the corresponding cohesive forces, acting along the planes,,etc., will be only 10 times the first values. Hence, if we use a scale of forceof the former scale, the weights of the prisms,,,etc., will be represented, as before, by,,etc., but the lines representing the cohesive forces will be onlyof the former lengths. Thus the new,Fig. 11, will be laid off from 4 onlyof the length shown in the figure.The relative decrease in the lines representing cohesive forces will be still more marked for a wallft. high, the weights of prisms being 400 times as great, but the cohesive forces only 20 times as great as before. It is evident from this reasoning that, for,the cohesive forces are practically negligible for walls, say, 10 ft. high, especially if the earth surface is level. In fact, a little examination of the original drawings showed, for walls about 6 ft. high, that the earth thrust, neglecting cohesion, was only from 1 to 5% in excess over that for.The smaller percentages referring to,or,while the larger percentages referred to,for earth surface horizontal or sloping at the angle of repose.Such results should be of great service to future experimenters as proving two things: (1) that dry sand, with as small a coefficient of cohesion as possible, should be used (perhaps grain would be a more suitable material), and (2) that no experimental wall should be less than from 6 to 10 ft. high.Even if the wall is, say, 6 ft. high, if damp clayey earth is usedas the filling, with a coefficient of adhesion,,then all the diagrams of forces, as inFig. 11, will be the same as before, or similar figures, and the discrepancies noted inTable 3, will be as pronounced as ever. All the experiments on retaining boards, except some of Curie’s, have been with very small models, and the results have brought the common theory under suspicion, if not into disrepute.The writer hopes that the foregoing investigation and results may be instrumental in establishing more confidence in the theory, and in showing when cohesive forces may be practically neglected and when they must be included.As an illustration, the results for a vertical wall 10 ft. high are presented inTable 4, takingand.In the first wall, the surface of the earth was horizontal; in the second wall its slope was 1 rise to 2 base.
TABLE 2.(1)(2)(3)Dry sand70.700.80 to 0.9626 to 56Wet sand400.851.20 " 1.9073 " 133Very wet sand311.702.00 " 2.40107 " 226Fresh earth901.632.60 " 4.40150 " 242
TABLE 2.
The values ofare given in kilogrammes per square meter. It is seen that suppositions (2) and (3) give discordant results, whereas (1), for each kind of filling, gave identical values ofand offor various heights; hence it may fairly be concluded that, before motion, cohesion and friction both acted at the same time. As to the high values found for,for the coherent fresh earth, Leygue states that Collin found, by an independent method, for clayey earth and clay of little consistency,and,respectively. As a further verification of the values ofandgiven in (1), it is found that, on using them in the formula for computing the height at which the wet sand or earth will stand vertically, the results agree with experiments.
The values of,in pounds per square foot, given in Column (1), with the values ofcorresponding to thegiven, are as follows:
It is possible that the method used by Leygue may prove of service to experimenters in obtaining more accurately than hitherto the coefficient of internal friction. Increasing pressures could be obtained by adding weights on top of the sand in the box; but, unless the total weight sustained by friction along the sides of the box is determined carefully for each weight used, the results can have but little value. Further, for coherent earths, the method of Leygue is open to objections.
Admitting the hypothesis that cohesion and friction act at the same time, a general graphical method[Footnote6]will now be given to find the thrust against the inner face,,of a retaining wall or board,Fig. 11, caused by the earth,,tending to slide down some plane of rupture,,,..., the resistance along this plane being due both to friction and cohesion.
Graphical method to find the thrust against the inner face of a retaining wall or boardFig. 11.
Fig. 11.
Supposeto be the plane of rupture, and call the weight of the prism of rupture,,for a thickness of one unit,.The weight of the prism causes the tendency to slide along the planes,and.This tendency is resisted by the reactions of the wall,,and the plane,.The reaction of the wall consists of the normal component,,acting to the right, and the friction resistance,,acting up. The resultant of these two forces,,which is equal and opposed to the earth thrust on,thus makes the angle,,with the normal to the wall. Its direction is given byinFig. 11. The reaction ofis made up of the cohesion,,acting up along,the normal component,,acting up, and the friction,,acting up along.The two forces,andwhen combined, give a resultant,,making an anglewith the normal,,to.Hence, if the angle,,thengives the direction of theresultant,.
The prism,,is thus in equilibrium under its own weight,,the cohesive force,,acting up along,the reaction,,of,acting to the right, and the force,,acting up. On drawing, to the scale of force,vertical and equal to;thenparallel toand equal to;thenparallel toto the intersection with,the sides of the closed polygon,,in order, will represent the four forces,,,,and,in equilibrium.
A similar investigation pertains to any other supposed prism of rupture. To find the true one, a number of trial planes of rupture,,,..., are assumed, and each is treated in turn as a true one (though there can be only one true one). As seen above, the resultant of the normal reaction and friction on any trial plane of rupture must be inclined below the normal to the plane at the angle,.To lay off the directions of these resultants, from any convenient point,,say, in the vertical through,as a center, describe an arc,,with a convenient radius,.With the same radius andas a center, describe the arc,,cutting the trial planes (produced if necessary) at,,...,.Letbe the point where the line of natural slope fromcuts the arc,.On laying off the chords,,,..., equal to the chords,,,..., respectively, it will follow that,,..., will make angles,,below the normals to the planes,,,..., respectively. To prove this, take any trial plane, as,which makes the angle,,with,and drop a perpendicular,,fromon(produced if necessary); then, because the sides of the angles are perpendicular,,and if,it follows that,as was to be proved. Hence the chord,the chord,,,etc., as stated.
The weights, in pounds, of the prisms,,,..., are,,,..., respectively, whereis the length of the perpendicular fromupon(produced if necessary), the footbeing the unit of length andbeing the weight, in pounds, of 1 cu. ft. of earth. The prisms are supposed to be 1 ft. in length perpendicular to the plane of the paper.
These weights are now laid off to the scale of force, vertically downward from,to points 1, 2, 3, ..., and from these points, lines are drawn parallel to,,,..., respectively, of lengths equal to,,..., to represent the forces of cohesion, acting upward along,,..., wheretheforce of cohesion, in pounds per square foot. From the extremities of these lines, lines are drawn parallel to the direction of the earth thrust on(inclined at the angle,,below the normal to), to the intersections,,,,..., with,,,..., respectively. With dividers, it is found, for this figure, thatis the longest of these lines; whence,to the scale of force, measures the earth thrust against,in pounds. This follows, because, for any less thrust, sinceis fixed, whenbecomes less,falls to the left of the first position, and the new,representing the thrust on,due to the normal reaction on it and friction only, will make a greater angle thanto the normal to,which is inconsistent with the laws of stability of a granular mass. In fact, ifis the normal component of the thrust on the plane,,is all the friction that can be exerted on it. The resultant ofandthus makes an angle,,with,and this angle cannot be exceeded. The true thrust on the wall,,is thus the greatest of the trial thrusts,.The prism of rupture,,is in equilibrium under the four forces represented by the sides of the closed polygon,;,representing its weight;,the cohesion acting along;,the reaction of(opposed and equal to the earth thrust); andthe reaction of the plane,,due to the normal component and friction on it only. The full reaction of the plane can be found by combining the forces, given in magnitude and direction byand,but it is not needed.
It is to be noted thatis the least thrust for which equilibrium is possible. The other trial thrusts should now be lengthened to equal,since this is the true thrust or reaction of the wall,.All the new points,,,,,,will now lie to the right of the old points; hence the new,,,,,will all make less angles thanwith the normals to the planes,,etc.; hence stability everywhere in the earth mass is assured.
Special casesFig. 12.
Fig. 12.
The solution represented byFig. 11is general, and applies whetheris inclined to the right or left of the vertical throughor coincides with it, and whether the earth surfaceis horizontal or inclined above or below the horizontal. It can likewise be easily adapted to the case shown inFig. 1.[Footnote7]The construction ofFig. 11has been used in evaluating the thrust and determining (approximately) the plane of rupture in the experiments (recorded below) of Leygue on retaining boards,,that could be rotated about,and thus placed at any inclination to the vertical. In all the experiments, the vertical height ofwas 0.656 ft.; the length of the board was 1.3 ft. The value of the moment of the earth thrust aboutwas found by use of suitable apparatus, corresponding to dry sand with a natural slope of 3 base to 2 rise, or,,andlb. per cu. ft. By use of the partition board mentioned previously, the side friction of the sand on the glass sides of the box containing it was estimated, and the moments corrected, so as to give the true moment when there is no side friction. The notation used to express results is partly given inFig. 12, for the general case where,
The resultant,,ofand,evidently makes the angle,,with the normal to,The moment of this resultant about,if we put.From the last formula, it is seen thatis the moment of the thrust about,for;also from,it follows thatis the normal component of the thrust for.
When cohesion is included,is not exactly,but it is very slightly less foror 2. It will be assumed at,and, from the values ofgiven byLeygue,will be derived from the formula above,
Thus, for the case represented byFig. 11,,therefore,;.Experiment gave,therefore.
Neglecting cohesion, theory gives,or twice the amount given by experiment. If, however, the construction ofFig. 11is made for the actual height of the retaining board,ft.,,(cohesion, in pounds per square foot), we find.On substituting this in the formula,,we have,
By a comparison of the values, it is evident that, if the cohesion was assumed at a little less than 1 lb. per sq. ft., the theoretical and experimental values could be made to agree exactly. The case just examined exhibits the most pronounced difference between the ordinary theory (corresponding to)and experiment, of any shown inTable 3. Further, it will be observed, that, for an assumed cohesion of about 1 lb. per sq. ft., the theoretical and experimental values for all the cases given by Leygue very nearly agree.
The value,,in place of Leygue’s,,was used, which would alter the results somewhat, but not the general conclusions. The construction ofFig. 11will giveand its normal component,,with practical accuracy, but it is not readily adaptable in finding the plane of rupture. In most of the drawings a small scale was used, in order to limit the drawing to a sheet of writing paper, hence, on both accounts,cannot be counted on to nearer than 1° or 2°, except for,whenwas found by computation, or by the construction ofFig. 1.
TABLE 3...Cohesion,,in poundspersquarefoot.Angle of rupture withthe horizontal,.Coefficientof thenormal component ofthe thrust.Theory.Experiment.Theory.Experiment.+⅓0050° 12′0.060151°51° 30′0.0420.043252° 30′0.026+⅓⅔033° 41′0.182144°47°0.0840.091249°0.04300056° 36′0.111157°56° 30′0.0930.090258°0.077358° 30′0.0620½047° 30′0.178150°51°0.1480.141253°0.121355°0.0980⅔033° 41′0.345144°49°0.2050.195246° 30′0.150350°0.111–⅓0060° 21′0.185163°61°0.1710.179263°0.155–⅓⅔033° 41′0.660157°57°0.2670.387250°0.236
TABLE 3.
The results inTable 3are remarkable, and explain quite satisfactorily how Leygue, Darwin, and others found, by experiments on small models, results differing so much from the ordinary theory, where cohesion is neglected.
It should be remarked that the values ofgiven inTable 3under “Experiment,” are not exactly those given by Leygue in his tables, but are the averages obtained from the two sets of drawings given by him in the plates, and represent the inclinations of the chords of the really curved surfaces of rupture. His experiments with the spring apparatusare not considered, as the results are open to doubt, because the prism of rupture, in descending, could not slide down freely, but as it advanced would rub over the floor, thus lessening the thrust there considerably.
FromTable 3, the results given by experiment are seen to differ widely from the ordinary theory in which.
The discrepancies are largely, or almost entirely, due to the very small models used, as will be evident from the following considerations: Suppose the height,,of the wall,,to be 10 times the height given inFig. 11, or 6.56 ft.; then, as the areas of triangles such as,etc., vary as the squares of the heights, but the lengths of sides, asetc., vary only as the first power of the heights, the weights of the successive trial prisms of rupture will beor 100 times as great as before, whereas the corresponding cohesive forces, acting along the planes,,etc., will be only 10 times the first values. Hence, if we use a scale of forceof the former scale, the weights of the prisms,,,etc., will be represented, as before, by,,etc., but the lines representing the cohesive forces will be onlyof the former lengths. Thus the new,Fig. 11, will be laid off from 4 onlyof the length shown in the figure.
The relative decrease in the lines representing cohesive forces will be still more marked for a wallft. high, the weights of prisms being 400 times as great, but the cohesive forces only 20 times as great as before. It is evident from this reasoning that, for,the cohesive forces are practically negligible for walls, say, 10 ft. high, especially if the earth surface is level. In fact, a little examination of the original drawings showed, for walls about 6 ft. high, that the earth thrust, neglecting cohesion, was only from 1 to 5% in excess over that for.The smaller percentages referring to,or,while the larger percentages referred to,for earth surface horizontal or sloping at the angle of repose.
Such results should be of great service to future experimenters as proving two things: (1) that dry sand, with as small a coefficient of cohesion as possible, should be used (perhaps grain would be a more suitable material), and (2) that no experimental wall should be less than from 6 to 10 ft. high.
Even if the wall is, say, 6 ft. high, if damp clayey earth is usedas the filling, with a coefficient of adhesion,,then all the diagrams of forces, as inFig. 11, will be the same as before, or similar figures, and the discrepancies noted inTable 3, will be as pronounced as ever. All the experiments on retaining boards, except some of Curie’s, have been with very small models, and the results have brought the common theory under suspicion, if not into disrepute.
The writer hopes that the foregoing investigation and results may be instrumental in establishing more confidence in the theory, and in showing when cohesive forces may be practically neglected and when they must be included.
As an illustration, the results for a vertical wall 10 ft. high are presented inTable 4, takingand.In the first wall, the surface of the earth was horizontal; in the second wall its slope was 1 rise to 2 base.