Adaptation of graphical method to find pressures of coherent earth against retaining wallsFig. 25.Recurring again to earth pressures, the discussion pertaining toFig. 24suggests the following modification of the graphical method ofFig. 11to adapt it to finding the pressures of coherent earth against retaining walls, for the case supposed above. InFig. 25, letbe theinner face of a retaining board or wall, which is backed by earth with a horizontal surface.The vertical height of the wall is 10 ft. and the physical constants are assumed as follows:,,and.FromEquation (8),Now lay off, vertically downward from the free surface, a distance,ft. (to scale) and throughdraw a line,,parallel to the top surface. ByEquation (7), the thrust on any vertical plane as,,..., is null, and any tension and cohesion that may actually exist along any of these planes will be ignored. This is equivalent to supposing vertical cracks along these planes, which leads to an increase of the thrust and is thus on the side of safety.The weights of the successive trial prisms of rupture,,,..., are now laid off to scale, along some vertical, as.The successive weights are represented by the vertical lines,,,...,.The points,,..., were taken 1 ft. apart; hence the area,the area,etc.; so that, after computing the area,the areas,etc., can be found by successive additions. These areas, multiplied by 100, give the weights, in pounds, of the successive prisms of rupture for 1 ft. length of wall. As before, two arcs are drawn with the same radius (), with centersandand having laid off the angle,,the chords,,..., are laid off equal to chords,,....The lines,,,..., now make angles with the normals to the planes,,,..., respectively, each equal to.Also, lay off the chord,;thengives the direction of the reaction of the wall,(directly opposed to the earth thrust), inclined at the angle,,to its normal.Next, measure, to the scale of distance, the length, in feet, of any line of rupture, as.Multiply this number by,to get the force of cohesion, in pounds, acting up along,and lay it off, to the scale of force, from,on a line parallel to,to.Similarly, lay off the cohesive forces, acting along,,..., at,,..., and from points, such as,lines are drawn parallel to the direction of the thrust,,to the intersections with the corresponding rays,,,.... Supposeto prove the greatest of these segments, then,to the scale of force, gives the earth thrust on,in pounds. In the present instance, the plane of rupture lies midway betweenand,and the corresponding thrust is 1 440 lb. For purposes of illustration, regardas the plane of rupture; then the forces acting on the prism,,are represented by the sides of the closed polygon,representing its weight,,the cohesion acting up along,the reaction of the wall, andthe resultant of the normal reaction,,of the plane,,and the friction,,acting along it. As stated, the actual plane of rupture is found (by drawing one or more additional trial planes) to lie midway betweenand.If it should be deemed desirable to include the cohesion acting upward along the vertical planes,,etc., it is very readily done. At each point, such as,draw vertically upward a line of length (to scale of force)lb., to represent the force of cohesion acting along the corresponding plane. From the extremities of such lines, parallels toare drawn to the intersections with the corresponding rays, of the type,.As before, the greatest of these lines represents the thrust on.Its amount, in this instance, is 1 240 lb. It is seen, especially whenis one-half ofor more, that the change in thrust is quite appreciable.Lastly, if each plane, as,is extended to the surface and considered to offer full cohesive and frictional resistances throughout its whole extent, the construction ofFig. 11applies, and gives a thrust of 1 220 lb., practically the previous amount.As mentioned before, in connection withFig. 24, this supposes sufficient tension in the mass to drag down the triangle,.Suppose, now,to be the face of a trench which will just stand unsupported for the height,,when the cohesion along,,..., is included. Then, if, from the drying out of the earth, causing contraction near the surface and possibly changes inand,cracks occur along some of these planes, the resistance to motion of the corresponding prism of rupture is decreased, and the mass will move down, unless, in the meantime, sufficient bracing has been put in. This well illustrates what constructors tell us of the importance of getting in well-keyed-up bracing in time to prevent any crack from developing, which, as we have seen, largely increases the thrust. Even when cracks do not appear, a heavy rain, shortly after a trench is opened, is a frequent cause of falls; evidently becauseand possiblyhave been very materially decreased.To compare the results of this very general graphical method with those given by analysis, as expressed byEquations (7)and(8), letbe taken vertical or coinciding with,regard the thrust onas horizontal, and neglect any cohesion on the vertical planes,,etc. Then it is found, for the given constants, that the graphical method gives exactly the thrust (560 lb.) obtained fromEquation (7). This result was foreseen.It is likewise interesting to know, if the thrust onis taken as parallel to,or making the angle,,with the normal to,that the thrust on,as given by the construction, is again found to be the same (510 lb.) as that given by a well-known formula forthe thrust on the plane,,for earth devoid of cohesion and having a free surface,.The importance of these conclusions lies in this: that for the wall vertical, the earth surface horizontal, the earth thrust being horizontal or otherwise, the shorter method is available. If preferred, the thrust on the wall,,for the earth devoid of cohesion, with the free surface,,for either direction of the thrust, can be evaluated by the graphical methods (page 404) hitherto given.Since this paper was written, Résal’s“Poussée des Terres,” Deuxième Partie, on Coherent Earths, has appeared.[Footnote27]In it the author gives an exhaustive discussion of lines of rupture for a great number of cases. The equivalent ofEquation (7)is found for the case of the horizontal pressure on a vertical plane when the free surface of earth is horizontal; but it was found to be impracticable to derive a formula for the earth thrust for the general case of the earth surface sloping at an angle to the horizontal, the wall being either inclined or vertical. In fact, for such cases, the intensity of the earth thrust at any depth is not a linear function of the depth, as obtains in the case shown byFig. 24. Hence Résal resorts to the following approximation: Conceive a line drawn parallel to the surface, at a depth,(as given byEquation (8)), below it, and regard this line as the free surface of non-coherent earth of the same specific weight and angle of repose as the given earth; compute the thrust against the wall for such earth, devoid of cohesion, by methods pertaining to such earth; the thrust thus found is assumed to be approximately the true thrust on the wall for the original coherent earth. It is proper to state that Résal rejects the sliding-wedge theory for non-coherent earth, and uses a method of his own, which involves elaborate tables given in his book. The wedge theory is admittedly imperfect, mainly because the surface of rupture is a curve, but we have seen that it agrees with experiments on model walls or retaining boards, when properly interpreted, and it will be used, as before, in computing the earth thrusts,,below, for earths devoid of cohesion. The graphical method has already been indicated.[Footnote28]InTable 6comparative results are given for various cases, including those already examined. Each retaining board was supposed to be 10 ft. high, the earth to have a natural slope of 1 on 1½, and to weigh 100 lb. per cu. ft.TABLE 6.Case...,in poundsper squarefoot.,in pounds.,in pounds.1.–⅓01001 4408802.0010005605603.001005105104.0½1006607505.0⅔1008801 6306.+⅓05018°26′240490It is seen, by comparing the values ofand,inTable 6, that, except for Cases 2 and 3, where the coincidence in the results has been already noted, the thrusts by the two methods differ very widely, hence the second method must be rejected, as in some cases undervaluing the thrust, and in other cases overvaluing it. In Case 5, where the surcharge slopes at the angle of repose, the large excess is due principally to the ordinary theory for computing,involving an infinite plane of rupture, as hitherto noted.In Case 6, where the top of the wall leaned toward the earth,was first assumed equal to 100, but it was found to give no thrust against the wall; which means that the earth would stand unsupported at the slopeor with the face making an angle of 18°26' with the vertical, when this face was 10 ft. high. Hence, a second trial was made, with,with the results shown. It will be noticed thatwas assumed as 18°26', so that the thrust was taken horizontal. In fact, Résal asserts that the thrust against such a leaning wall, makes a less angle thanwith the normal to the wall. According to his tables,,for this wall cannot exceed 20°40'. As the wall approaches the vertical,,approaches,the exact value given for a vertical wall. By comparing the thrusts,,for Cases 1 and 6, the economy of using the latter type of wall is so apparent that it is astonishing that constructors do not adopt it oftener.Surcharged wall with trial prisms of ruptureFig. 26.The general conclusion to be drawn fromTable 6is that, except for Cases 2 and 3, the general graphical method ofFig. 25must be used for accuracy. If applicable to the case in hand, the cohesion along the vertical planes,,etc., can be included with very little additional labor. The graphical treatment given is so general and the theory involved is so apparent to the eye, that it seems to commend itself as a practical treatment of a very complicated problem.The general method illustrated inFig. 25can also be applied to the surcharged wall ofFig. 26. Here, the lines,and,are drawn parallel, respectively, toand,and vertically,ft. below them. The trial prisms of rupture are of the type,.Their weights are laid off as before, on the line,,ofFig. 25, and the further construction is exactly as there indicated. It is assumed here that the slope ofdoes not exceed the natural slope, for, if it does, the construction is somewhat altered.It may be asked whether the construction ofFig. 25, where the cohesion on the vertical planes,,etc., is omitted, if applied to the experiments on rotating boards, will appreciably alter the results given inTable 3. The answer is no; for, when,,thereforeft., or is very small. In fact, as in the experiments, no cracks were formed, the cohesion on the vertical planes should be included, which would lead to the results given.After the thrust, for earth endowed with cohesion, has been computed, the next question is, at what point on the retaining wall does it act? This can be answered at once, when the inner face of the wall is vertical and the earth surface horizontal, for then the earth thrust acts atof the height,,Figs. 24and25, as hitherto proved. The case is not so simple when the wall is inclined, either toward or from the earth. InFig. 25, the thrust,,on the wall,,must be decomposed into horizontal and vertical components. The horizontal component is the resultant of the horizontal forces acting fromtowhich may be assumed to follow the linear law; hence this component will act onat a point, distantfrom,going fromto.The vertical component, similarly, will be regarded as acting on,at a distancefrom.The same approximate rule is suggested whenlies to the right of the vertical,.When the earth surface slopes at the angle of repose, the distribution of stress is not linear, but more like that shown inFig. 10. As theequation of the corresponding pressure curve cannot be found, an approximation only can be suggested. Whenis (say) less than one or two tenths, the factor,,above, can be used; but asincreases toward its limit, 1, the factor increases toward some unknown limit. Probably it does not exceed 0.4 even forFig. 26, and, for want of accurate knowledge on the subject, it may be taken at 0.4 as a rude approximation.Finally, when the surface slope is less than the natural slope of the earth, then as it decreases fromto zero, the factor should. decrease from the extreme value (say 0.4) to.Résal takes this factor atfor all cases, which is certainly not on the safe side.The case of the retaining wall which receives the active thrust of the earth has been hitherto examined, and next the case of the braced trench will be discussed. As the trench (having vertical sides) is dug, the usual sheeting, rangers, and bracing are put in and the bracing is kept well keyed-up, so as to exert an active pressure on the earth. To illustrate the theory, inFig. 25, letrepresent the vertical side of the trench, the earth extending only to the right of.Then from points such as,draw horizontal lines to the intersection with the correspondings;the longest of these lines, to the scale of force, will represent the total force that must be exerted by all the braces, per foot of length of trench, to prevent any motion of the mass. As has been seen, this force is given byEquation (7). If a still greater force is exerted by the braces, less than a certain value which would just cause motion of the earth up some plane of rupture, stability is completely assured.As a numerical illustration, assume a trench 40 ft. deep and 15 ft. wide, the constants for the earth being,,,.Then, byEquation (8),ft. andft. Hence, byEquation (7), the least force the braces must exert per linear foot of trench is,Suppose the braces to be 10 ft. apart horizontally, and that there are six braces in the same vertical plane. The least force that the horizontal braces must exert on 10 lin. ft. of trench is 188 440 lb., and if each carries the same stress, the force to be exerted by one, islb. Assuming a unit stress of 800 lb. per sq. in. for an 8 by 8-in. wooden brace, 15 ft. long, it is seen that onebrace can safely exert a force of 51200 lb., which is greater than the least amount required, as should be the case to allow for changes inanddue to heavy rains. To meet such contingencies, 10 by10-in.braces are suggested. Of course, very little is now known as to the coefficient,,but, from the observed heights of trenches which have stood without sheathing, it is probable that values offrom 100 to 300 lb. per sq. ft. can be counted on for most trenching. For the present, the only safe way is to be guided by experience, such as has been elicited by Mr. Meem’s paper, “The Bracing of Trenches and Tunnels, With Practical Formulas for Earth Pressures.”As to the exact distribution of the stresses, theory cannot speak definitely, for the conditions are different from those in the case of the retaining wall, the passive resistance of which opposes the active earth thrust. For the braced trench, the earth, at first, simply resists the active pressure exerted by the braces, when first put in and keyed-up tight, particularly on that upper portion where the active earth thrust is nothing or very small. As the construction proceeds, the braces will receive more and more of the active earth thrust, which necessarily increases with the depth of trench. In fact, the distribution of stress, indicated by the arrows inFig. 24, although true in the case of a retaining wall, where the earth has been deposited behind it, is not necessarily or generally true in a sheeted trench, because of the manner of its construction. Thus, in digging a trench, the bracing is put in at intervals, but when a brace is inserted near the bottom of the trench, the digging is continued for several feet without bracing, until a depth is attained at which it is thought best to insert another brace. Before the latter is put in, the unprotected face of the trench, say 6 ft. in depth, can exert no pressure, as it would in the case of the retaining wall. The thrust that would be exerted, for this area, on the supposed wall, does not exist for the unsupported face of the trench, for the full horizontal thrust of the earth for the whole depth has been taken up by the braces above the unsupported area. This state of affairs is characteristic of the work as it proceeds, the lowest brace that is put in at first carries only the stress due to the keying-up, but takes more and more stress as the excavation proceeds. It can thus very well happen that the upper or the middle braces may receive more stress in the end, than the lower braces. In fact, this was asserted to be generally true, for well-drained material, by many engineers in the valuable discussion on Mr. Meem’s paper referred to. Other engineers advised caution in accepting this view, and asserted that in wet or saturated ground the lower braces were most severely stressed. If it were possible to force a board of a size equal to the length and depth of a trench, vertically downward, excavate and brace, all in a millionth of a second, then one can conceive that the distribution of stress shown inFig. 24might be realized; but, for trenches as actually constructed, the distribution of stress in the earth massis very much altered from this, though it would appear that the total earth pressure would be given, at least approximately, by the construction ofFig. 25, or by the use ofEquation (7).The writer hopes that Mr. Meem may agree to the foregoing explanation, for the subject of trenches was entered into in great detail, in order to explain, if possible, all the facts, as presented by many engineers who held very diverse views about the explanation of them. Mr. Meem’sFig. 22can illustrate the writer’s view: if bracing only extends, say, fromto,then the braces must exert sufficient horizontal thrust to prevent the descent of the wedge, say,if this gives the maximum thrust for the coherent earth.Turning now to pressures on tunnels, the writer is pleased to note that Mr. Meem has recorded some more of his experiences concerning them. From lack of proper knowledge of the so-called constants,,and,such experience is an aid in leading to more probable values.As the writer proceeds in this investigation, he hopes to make clear the conception of arch action alluded to, and will derive the limiting values ofandin the simple manner outlined briefly in the Appendix. Before considering these matters, however, it may be well to call attention to the fact that a vertical prism of earth can be held up by the cohesion of its sides alone whenis large enough. Take the vertical prism of depthand cross-section area,Fig. 20, its weight being.In a long tunnel, two sides only can be counted on to furnish cohesive forces. Callthe perimeter available in any case. Then the cohesive force is,and this alone (without reference to any friction that may act) will support the prism when,.In the long or completed tunnel,(see notation of Appendix); but if we take a short section at the heading, say,then support can be derived from three sides, and.Thus if we assume the horizontal cross-section of the prism to be 15 by 7.5 ft.,lb. per sq. ft., wherelb. per cu. ft. Thus, such a vertical prism of any height, at the heading, can be sustained by a cohesion of 338 lb. per sq. ft. acting on three of its sides. This refers to a tunnel 15 ft. wide. For the completed tunnel 15 ft. wide, the part vertically over the tunnel can be supported if the two sides can furnish double this unit cohesion. The termwill appear in an equation mentioned later. It is seen now that when this term is zero, cohesion alone can sustainthe prism, and the pressure reduces to zero. When this term is negative, it indicates that that too large a value ofhas been assumed, for stability is assured for.In reference toFig. 16, it was stated that a series of superposed arches or domes were assumed, but that since the reactions of the horizontal laminas ofFig. 20and corresponding arches were the same, the former were substituted for convenience. If each arch has the vertical thickness,,it will have the same volume and weight,,as the horizontal lamina. The reactions marked onFig. 20can be assumed to be those of the corresponding arch, and the resultant of the horizontal and vertical reactions represents the thrust of the arch at the sides; its direction will be that of the tangent line of the arch. Whenis small, the horizontal reaction is much greater than the vertical one, and the arches are all very flat. Now, considering the whole series of arches, it is plain that the greatest or limiting value of(call it simply)would be realized if the weight of any arch is entirely held up by the friction on the sides (induced by the lateral thrust) and the cohesion acting there; for then this same condition of affairs would exist for all lower arches,[Footnote29]and thusis transmitted vertically downward, unchanged, to the tunnel. Stating this condition in algebraic form,This equation strictly holds for large values of,but it is practically true for much smaller values. For such values,is quite small compared withand can be assumed as equal to zero, whencewill also be taken equal to zero and the usual bin formula,,written.[Footnote30]On substituting this value in the equation above, reducing and placing,weobtain,where,.In the Appendix, the writer assumed,,basing this value on the results pertaining to a smooth steel bin, 1 ft. indiameter, filled with sand. From the experiments of Jamieson on large wooden bins, about 12 by 12 ft., filled with wheat,was found to be 0.60. Janssen determinedexperimentally to be 0.67. For wheat, Jamieson gave;whence.His experimental value wasof this; hence, for lack of definite data, there will be assumed for earth,From the last three equations, the values ofandhave been computed for various values of,,,and,and are given inTable 7.
Adaptation of graphical method to find pressures of coherent earth against retaining wallsFig. 25.
Fig. 25.
Recurring again to earth pressures, the discussion pertaining toFig. 24suggests the following modification of the graphical method ofFig. 11to adapt it to finding the pressures of coherent earth against retaining walls, for the case supposed above. InFig. 25, letbe theinner face of a retaining board or wall, which is backed by earth with a horizontal surface.The vertical height of the wall is 10 ft. and the physical constants are assumed as follows:,,and.FromEquation (8),
Now lay off, vertically downward from the free surface, a distance,ft. (to scale) and throughdraw a line,,parallel to the top surface. ByEquation (7), the thrust on any vertical plane as,,..., is null, and any tension and cohesion that may actually exist along any of these planes will be ignored. This is equivalent to supposing vertical cracks along these planes, which leads to an increase of the thrust and is thus on the side of safety.
The weights of the successive trial prisms of rupture,,,..., are now laid off to scale, along some vertical, as.The successive weights are represented by the vertical lines,,,...,.The points,,..., were taken 1 ft. apart; hence the area,the area,etc.; so that, after computing the area,the areas,etc., can be found by successive additions. These areas, multiplied by 100, give the weights, in pounds, of the successive prisms of rupture for 1 ft. length of wall. As before, two arcs are drawn with the same radius (), with centersandand having laid off the angle,,the chords,,..., are laid off equal to chords,,....The lines,,,..., now make angles with the normals to the planes,,,..., respectively, each equal to.Also, lay off the chord,;thengives the direction of the reaction of the wall,(directly opposed to the earth thrust), inclined at the angle,,to its normal.
Next, measure, to the scale of distance, the length, in feet, of any line of rupture, as.Multiply this number by,to get the force of cohesion, in pounds, acting up along,and lay it off, to the scale of force, from,on a line parallel to,to.Similarly, lay off the cohesive forces, acting along,,..., at,,..., and from points, such as,lines are drawn parallel to the direction of the thrust,,to the intersections with the corresponding rays,,,.... Supposeto prove the greatest of these segments, then,to the scale of force, gives the earth thrust on,in pounds. In the present instance, the plane of rupture lies midway betweenand,and the corresponding thrust is 1 440 lb. For purposes of illustration, regardas the plane of rupture; then the forces acting on the prism,,are represented by the sides of the closed polygon,
representing its weight,,the cohesion acting up along,the reaction of the wall, andthe resultant of the normal reaction,,of the plane,,and the friction,,acting along it. As stated, the actual plane of rupture is found (by drawing one or more additional trial planes) to lie midway betweenand.
If it should be deemed desirable to include the cohesion acting upward along the vertical planes,,etc., it is very readily done. At each point, such as,draw vertically upward a line of length (to scale of force)lb., to represent the force of cohesion acting along the corresponding plane. From the extremities of such lines, parallels toare drawn to the intersections with the corresponding rays, of the type,.As before, the greatest of these lines represents the thrust on.Its amount, in this instance, is 1 240 lb. It is seen, especially whenis one-half ofor more, that the change in thrust is quite appreciable.
Lastly, if each plane, as,is extended to the surface and considered to offer full cohesive and frictional resistances throughout its whole extent, the construction ofFig. 11applies, and gives a thrust of 1 220 lb., practically the previous amount.
As mentioned before, in connection withFig. 24, this supposes sufficient tension in the mass to drag down the triangle,.Suppose, now,to be the face of a trench which will just stand unsupported for the height,,when the cohesion along,,..., is included. Then, if, from the drying out of the earth, causing contraction near the surface and possibly changes inand,cracks occur along some of these planes, the resistance to motion of the corresponding prism of rupture is decreased, and the mass will move down, unless, in the meantime, sufficient bracing has been put in. This well illustrates what constructors tell us of the importance of getting in well-keyed-up bracing in time to prevent any crack from developing, which, as we have seen, largely increases the thrust. Even when cracks do not appear, a heavy rain, shortly after a trench is opened, is a frequent cause of falls; evidently becauseand possiblyhave been very materially decreased.
To compare the results of this very general graphical method with those given by analysis, as expressed byEquations (7)and(8), letbe taken vertical or coinciding with,regard the thrust onas horizontal, and neglect any cohesion on the vertical planes,,etc. Then it is found, for the given constants, that the graphical method gives exactly the thrust (560 lb.) obtained fromEquation (7). This result was foreseen.
It is likewise interesting to know, if the thrust onis taken as parallel to,or making the angle,,with the normal to,that the thrust on,as given by the construction, is again found to be the same (510 lb.) as that given by a well-known formula forthe thrust on the plane,,for earth devoid of cohesion and having a free surface,.The importance of these conclusions lies in this: that for the wall vertical, the earth surface horizontal, the earth thrust being horizontal or otherwise, the shorter method is available. If preferred, the thrust on the wall,,for the earth devoid of cohesion, with the free surface,,for either direction of the thrust, can be evaluated by the graphical methods (page 404) hitherto given.
Since this paper was written, Résal’s“Poussée des Terres,” Deuxième Partie, on Coherent Earths, has appeared.[Footnote27]In it the author gives an exhaustive discussion of lines of rupture for a great number of cases. The equivalent ofEquation (7)is found for the case of the horizontal pressure on a vertical plane when the free surface of earth is horizontal; but it was found to be impracticable to derive a formula for the earth thrust for the general case of the earth surface sloping at an angle to the horizontal, the wall being either inclined or vertical. In fact, for such cases, the intensity of the earth thrust at any depth is not a linear function of the depth, as obtains in the case shown byFig. 24. Hence Résal resorts to the following approximation: Conceive a line drawn parallel to the surface, at a depth,(as given byEquation (8)), below it, and regard this line as the free surface of non-coherent earth of the same specific weight and angle of repose as the given earth; compute the thrust against the wall for such earth, devoid of cohesion, by methods pertaining to such earth; the thrust thus found is assumed to be approximately the true thrust on the wall for the original coherent earth. It is proper to state that Résal rejects the sliding-wedge theory for non-coherent earth, and uses a method of his own, which involves elaborate tables given in his book. The wedge theory is admittedly imperfect, mainly because the surface of rupture is a curve, but we have seen that it agrees with experiments on model walls or retaining boards, when properly interpreted, and it will be used, as before, in computing the earth thrusts,,below, for earths devoid of cohesion. The graphical method has already been indicated.[Footnote28]
InTable 6comparative results are given for various cases, including those already examined. Each retaining board was supposed to be 10 ft. high, the earth to have a natural slope of 1 on 1½, and to weigh 100 lb. per cu. ft.
TABLE 6.Case...,in poundsper squarefoot.,in pounds.,in pounds.1.–⅓01001 4408802.0010005605603.001005105104.0½1006607505.0⅔1008801 6306.+⅓05018°26′240490
TABLE 6.
It is seen, by comparing the values ofand,inTable 6, that, except for Cases 2 and 3, where the coincidence in the results has been already noted, the thrusts by the two methods differ very widely, hence the second method must be rejected, as in some cases undervaluing the thrust, and in other cases overvaluing it. In Case 5, where the surcharge slopes at the angle of repose, the large excess is due principally to the ordinary theory for computing,involving an infinite plane of rupture, as hitherto noted.
In Case 6, where the top of the wall leaned toward the earth,was first assumed equal to 100, but it was found to give no thrust against the wall; which means that the earth would stand unsupported at the slopeor with the face making an angle of 18°26' with the vertical, when this face was 10 ft. high. Hence, a second trial was made, with,with the results shown. It will be noticed thatwas assumed as 18°26', so that the thrust was taken horizontal. In fact, Résal asserts that the thrust against such a leaning wall, makes a less angle thanwith the normal to the wall. According to his tables,,for this wall cannot exceed 20°40'. As the wall approaches the vertical,,approaches,the exact value given for a vertical wall. By comparing the thrusts,,for Cases 1 and 6, the economy of using the latter type of wall is so apparent that it is astonishing that constructors do not adopt it oftener.
Surcharged wall with trial prisms of ruptureFig. 26.
Fig. 26.
The general conclusion to be drawn fromTable 6is that, except for Cases 2 and 3, the general graphical method ofFig. 25must be used for accuracy. If applicable to the case in hand, the cohesion along the vertical planes,,etc., can be included with very little additional labor. The graphical treatment given is so general and the theory involved is so apparent to the eye, that it seems to commend itself as a practical treatment of a very complicated problem.
The general method illustrated inFig. 25can also be applied to the surcharged wall ofFig. 26. Here, the lines,and,are drawn parallel, respectively, toand,and vertically,ft. below them. The trial prisms of rupture are of the type,.Their weights are laid off as before, on the line,,ofFig. 25, and the further construction is exactly as there indicated. It is assumed here that the slope ofdoes not exceed the natural slope, for, if it does, the construction is somewhat altered.
It may be asked whether the construction ofFig. 25, where the cohesion on the vertical planes,,etc., is omitted, if applied to the experiments on rotating boards, will appreciably alter the results given inTable 3. The answer is no; for, when,,thereforeft., or is very small. In fact, as in the experiments, no cracks were formed, the cohesion on the vertical planes should be included, which would lead to the results given.
After the thrust, for earth endowed with cohesion, has been computed, the next question is, at what point on the retaining wall does it act? This can be answered at once, when the inner face of the wall is vertical and the earth surface horizontal, for then the earth thrust acts atof the height,,Figs. 24and25, as hitherto proved. The case is not so simple when the wall is inclined, either toward or from the earth. InFig. 25, the thrust,,on the wall,,must be decomposed into horizontal and vertical components. The horizontal component is the resultant of the horizontal forces acting fromtowhich may be assumed to follow the linear law; hence this component will act onat a point, distantfrom,going fromto.The vertical component, similarly, will be regarded as acting on,at a distancefrom.The same approximate rule is suggested whenlies to the right of the vertical,.
When the earth surface slopes at the angle of repose, the distribution of stress is not linear, but more like that shown inFig. 10. As theequation of the corresponding pressure curve cannot be found, an approximation only can be suggested. Whenis (say) less than one or two tenths, the factor,,above, can be used; but asincreases toward its limit, 1, the factor increases toward some unknown limit. Probably it does not exceed 0.4 even forFig. 26, and, for want of accurate knowledge on the subject, it may be taken at 0.4 as a rude approximation.
Finally, when the surface slope is less than the natural slope of the earth, then as it decreases fromto zero, the factor should. decrease from the extreme value (say 0.4) to.Résal takes this factor atfor all cases, which is certainly not on the safe side.
The case of the retaining wall which receives the active thrust of the earth has been hitherto examined, and next the case of the braced trench will be discussed. As the trench (having vertical sides) is dug, the usual sheeting, rangers, and bracing are put in and the bracing is kept well keyed-up, so as to exert an active pressure on the earth. To illustrate the theory, inFig. 25, letrepresent the vertical side of the trench, the earth extending only to the right of.Then from points such as,draw horizontal lines to the intersection with the correspondings;the longest of these lines, to the scale of force, will represent the total force that must be exerted by all the braces, per foot of length of trench, to prevent any motion of the mass. As has been seen, this force is given byEquation (7). If a still greater force is exerted by the braces, less than a certain value which would just cause motion of the earth up some plane of rupture, stability is completely assured.
As a numerical illustration, assume a trench 40 ft. deep and 15 ft. wide, the constants for the earth being,,,.Then, byEquation (8),ft. andft. Hence, byEquation (7), the least force the braces must exert per linear foot of trench is,
Suppose the braces to be 10 ft. apart horizontally, and that there are six braces in the same vertical plane. The least force that the horizontal braces must exert on 10 lin. ft. of trench is 188 440 lb., and if each carries the same stress, the force to be exerted by one, islb. Assuming a unit stress of 800 lb. per sq. in. for an 8 by 8-in. wooden brace, 15 ft. long, it is seen that onebrace can safely exert a force of 51200 lb., which is greater than the least amount required, as should be the case to allow for changes inanddue to heavy rains. To meet such contingencies, 10 by10-in.braces are suggested. Of course, very little is now known as to the coefficient,,but, from the observed heights of trenches which have stood without sheathing, it is probable that values offrom 100 to 300 lb. per sq. ft. can be counted on for most trenching. For the present, the only safe way is to be guided by experience, such as has been elicited by Mr. Meem’s paper, “The Bracing of Trenches and Tunnels, With Practical Formulas for Earth Pressures.”
As to the exact distribution of the stresses, theory cannot speak definitely, for the conditions are different from those in the case of the retaining wall, the passive resistance of which opposes the active earth thrust. For the braced trench, the earth, at first, simply resists the active pressure exerted by the braces, when first put in and keyed-up tight, particularly on that upper portion where the active earth thrust is nothing or very small. As the construction proceeds, the braces will receive more and more of the active earth thrust, which necessarily increases with the depth of trench. In fact, the distribution of stress, indicated by the arrows inFig. 24, although true in the case of a retaining wall, where the earth has been deposited behind it, is not necessarily or generally true in a sheeted trench, because of the manner of its construction. Thus, in digging a trench, the bracing is put in at intervals, but when a brace is inserted near the bottom of the trench, the digging is continued for several feet without bracing, until a depth is attained at which it is thought best to insert another brace. Before the latter is put in, the unprotected face of the trench, say 6 ft. in depth, can exert no pressure, as it would in the case of the retaining wall. The thrust that would be exerted, for this area, on the supposed wall, does not exist for the unsupported face of the trench, for the full horizontal thrust of the earth for the whole depth has been taken up by the braces above the unsupported area. This state of affairs is characteristic of the work as it proceeds, the lowest brace that is put in at first carries only the stress due to the keying-up, but takes more and more stress as the excavation proceeds. It can thus very well happen that the upper or the middle braces may receive more stress in the end, than the lower braces. In fact, this was asserted to be generally true, for well-drained material, by many engineers in the valuable discussion on Mr. Meem’s paper referred to. Other engineers advised caution in accepting this view, and asserted that in wet or saturated ground the lower braces were most severely stressed. If it were possible to force a board of a size equal to the length and depth of a trench, vertically downward, excavate and brace, all in a millionth of a second, then one can conceive that the distribution of stress shown inFig. 24might be realized; but, for trenches as actually constructed, the distribution of stress in the earth massis very much altered from this, though it would appear that the total earth pressure would be given, at least approximately, by the construction ofFig. 25, or by the use ofEquation (7).
The writer hopes that Mr. Meem may agree to the foregoing explanation, for the subject of trenches was entered into in great detail, in order to explain, if possible, all the facts, as presented by many engineers who held very diverse views about the explanation of them. Mr. Meem’sFig. 22can illustrate the writer’s view: if bracing only extends, say, fromto,then the braces must exert sufficient horizontal thrust to prevent the descent of the wedge, say,if this gives the maximum thrust for the coherent earth.
Turning now to pressures on tunnels, the writer is pleased to note that Mr. Meem has recorded some more of his experiences concerning them. From lack of proper knowledge of the so-called constants,,and,such experience is an aid in leading to more probable values.
As the writer proceeds in this investigation, he hopes to make clear the conception of arch action alluded to, and will derive the limiting values ofandin the simple manner outlined briefly in the Appendix. Before considering these matters, however, it may be well to call attention to the fact that a vertical prism of earth can be held up by the cohesion of its sides alone whenis large enough. Take the vertical prism of depthand cross-section area,Fig. 20, its weight being.In a long tunnel, two sides only can be counted on to furnish cohesive forces. Callthe perimeter available in any case. Then the cohesive force is,and this alone (without reference to any friction that may act) will support the prism when,.In the long or completed tunnel,(see notation of Appendix); but if we take a short section at the heading, say,then support can be derived from three sides, and.Thus if we assume the horizontal cross-section of the prism to be 15 by 7.5 ft.,lb. per sq. ft., wherelb. per cu. ft. Thus, such a vertical prism of any height, at the heading, can be sustained by a cohesion of 338 lb. per sq. ft. acting on three of its sides. This refers to a tunnel 15 ft. wide. For the completed tunnel 15 ft. wide, the part vertically over the tunnel can be supported if the two sides can furnish double this unit cohesion. The termwill appear in an equation mentioned later. It is seen now that when this term is zero, cohesion alone can sustainthe prism, and the pressure reduces to zero. When this term is negative, it indicates that that too large a value ofhas been assumed, for stability is assured for.
In reference toFig. 16, it was stated that a series of superposed arches or domes were assumed, but that since the reactions of the horizontal laminas ofFig. 20and corresponding arches were the same, the former were substituted for convenience. If each arch has the vertical thickness,,it will have the same volume and weight,,as the horizontal lamina. The reactions marked onFig. 20can be assumed to be those of the corresponding arch, and the resultant of the horizontal and vertical reactions represents the thrust of the arch at the sides; its direction will be that of the tangent line of the arch. Whenis small, the horizontal reaction is much greater than the vertical one, and the arches are all very flat. Now, considering the whole series of arches, it is plain that the greatest or limiting value of(call it simply)would be realized if the weight of any arch is entirely held up by the friction on the sides (induced by the lateral thrust) and the cohesion acting there; for then this same condition of affairs would exist for all lower arches,[Footnote29]and thusis transmitted vertically downward, unchanged, to the tunnel. Stating this condition in algebraic form,
This equation strictly holds for large values of,but it is practically true for much smaller values. For such values,is quite small compared withand can be assumed as equal to zero, whencewill also be taken equal to zero and the usual bin formula,,written.[Footnote30]
On substituting this value in the equation above, reducing and placing,weobtain,
where,.
In the Appendix, the writer assumed,,basing this value on the results pertaining to a smooth steel bin, 1 ft. indiameter, filled with sand. From the experiments of Jamieson on large wooden bins, about 12 by 12 ft., filled with wheat,was found to be 0.60. Janssen determinedexperimentally to be 0.67. For wheat, Jamieson gave;whence.His experimental value wasof this; hence, for lack of definite data, there will be assumed for earth,
From the last three equations, the values ofandhave been computed for various values of,,,and,and are given inTable 7.