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PUBLIC ROADS

il 4) er ll, ‘ly. cal UI, call, 2)

A JOURNAL OF HIGHWAY RESEARCH FEDERAL WORKS AGENCY

Pai bea ROADS ADMINISTRATION

Cae Lae 5 IN OS VW @) @ sig) Es ee

ON US 395 IN NEVADA

For sale by the Superintendent of Documents, Washington, D. C. - - - - - - - - - - - - - See page 2 of cover for prices

PUBLIC ROADS

>> > 4 Journal of Highway Research

Issued by the

FEDERAL WORKS AGENCY

PUBLIC ROADS ADMINISTRATION D. M. BEACH, Editor

Volume 20, No. 8

October 1939

The reports of research published in this magazine are necessarily qualified by the conditions of the tests from which the data are obtained. Whenever it is deemed possible to do so, generalizations are drawn from the results of the tests; and, unless this is done, the conclusions formulated must be considered as specifically pertinent only to described conditions.

In This Issue

Design of a Fill Supported by Clay Underlaid by Rock

Significant Trends in Motor-Vehicle Registrations and Receipts

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DESIGN OF A FILL SUPPORTED BY CLAY UNDERLAID BY ROCK

AN APPLICATION OF SOIL MECHANICS IN SOLVING A HIGHWAY FILL PROBLEM BY THE DIVISION OF TESTS, PUBLIC ROADS ADMINISTRATION

Reported by L."-A. PALMER, Associate Chemist

HIS REPORT is a continuation of the theoretical

considerations contained in two previous publica- tions.'* Its purpose is to present in usable form the analytical methods based on the assumption of condi- tions of plane strain * and to extend these analyses to include the problem of determining the supporting power of a clay stratum supporting a symmetr ical earth fill when the clay stratum is underlaid by rock.

As shown in one of the previous publications? a problem involving plane strain conditions is one involy- ing two dimensions. The load is distributed over an area that is quite long as compared to its width and the analytical procedure is applied to a vertical cross section of unit thickness in the direction of the longitudinal axis of the load. This is taken as the Y direction. It is considered that there is no displacement of material in this direction and that whatever soil movements occur are in the Z direction, which is toward the center of the earth, and in the X or horizontal direction, that is, perpendicular to both the Y and 7 directions.

The analytical procedures used in the theoretical solution of the present problem involve two theories, that of elasticity and that of plastic equilibrium, and four principal assumptions are involved. The first three are common to both theories. The fourth is made only when the theory of plastic equilibrium is applied. These are:

1. The strength of the clay stratum depends essen- tially on its cohesion. The str ength due to the element of friction is comparatively small one may be neglected. Hence, whenever and wherever the unit shearing stress becomes equal to the unit cohesion, c, the soil becomes plastic and undergoes plastic flow; that is, the soil fails.

2. The adhesion of the clay to the rock surface is ‘Derfect.’’ No slippage occurs at this surface although there may be lateral movement in the clay at points very near the rock surface.

3. The soil deformations considered in this paper are those that occur at an assumed constant volume. It seems reasonable to assume that the deformations caused by lateral yield in the X direction occur during a period of time that is brief in comparison with the time required for an appreciable degree of consolidation of the stressed clay stratum. When deformations occur

(the

No}

at constant volume, Poisson’s ratio is taken as

approximate value). 4. In applying the method of plastic equilibrium it is considered that the fill acts like an absolutely rigid body in its production of stresses in the clay stratum when the soil is in the plastic state. Thus the fill above and 1 Principles of Soil Mechanics Involved in Fill Construction, L. A. Palmer and E. S. Barber, Proceedings Highway Research Board, Annual "Meeting 1937.

2 Principles of Soil Mechanies Involved in the Design of Retaining Walls and Bridge Abutments, L. A. Palmer, PUBLIC ROADS, vol. 19, No. 10, December 1938.

178157—39

peas ~~ UNIFORM LOAD

| =a =X | +X

} CLAY STRATUM 2h

— ROCK SURFACE

Figure 1.—Unirorm Loap on a Lone Srrie SUPPORTED BY

Ciuay UNDERLAID By Souip Rock.

the solid rock boundary below the clay constitute a “nuteracker.”’

Probably the fourth assumption is the least valid of the four.

Since it is assumed that there is no displacement either in the fill or in the supporting soil in the direction of the longitudinal (Y) axis of the fill, the problem is one of planestrain. One vertical cross section perpendi- cular to the Y axis is the same as any other insofar as stresses and deformations are concerned, assuming, of course, that both the fill material and the supporting clay are, in themselves, homogeneous. Since the rock is supposedly rigid, it follows that there is no vertical displacement of ‘soil at this bound: ary.

STRESSES IN THE CLAY COMPUTED FROM THEORY OF ELASTICITY

Carothers * has shown that for a uniform load p per unit area on a long strip of width 26 (see fig. 1) at the surface, the shearing stress, s,,, at the rock surface is

Dp ax—b sah] sech 5 woe —sech 5 ~57- 2 (1)

“a

where 2h is the thickness of the intervening clay layer.

This expression for s,, for uniform strip loading and other expressions for stresses for other types of surface loading (see for example equation 12) are developed from the theory of elasticity. When these expressions are used it is considered that the clay mass has not been stressed to its ultimate supporting power and is there- fore not reduced to a plastic condition throughout.

In the following discussion equations 2, 3, 4, and 8 are those frequently seen in texts on the theory of elasticity.*

"3 Test Loads on Foundations as Affected by Scale of Tested Area, S. D. Carothers, Proceedings International Mathematical Congress, Toronto, 1924, Sen 527-549.

4 See, for example, pp. 8-20, inclusive, of Theory of Elasticity, by S. Timoshenko. McGraw-Hill Book Co., Ist. ed., 1934.

157

158

The fundamental strain relations are

= Prt.) | ----- = 2) =F Peet) |p —-- a= (3) =F Pe Het Ps) | ----------- (4)

where e;, €,, and e, are the strains and p,, p,, and p, are the normal stresses in the X, Y, and Z_ directions, respectively; LH’ is Young’s modulus; and u is Poisson’s ratio.

P 3 il Since e,=e,=0 at the rock surface and since M5)

equation 3 becomes

Ue a EO

and equation 4 becomes

Pe=Pz------------------ (7)

which is true at the boundary of rock and clay. The maximum shearing stress, Smax., at any point of the undersoil is

which (since p,= pz at the rock surface) becomes

at all points along the rock surface. Hence at the

rock boundary equation 1 becomes

Sine =2| sech 5

which is the expression for the shearing stress at any point T of the rock surface (see fig. 1). For a tri-

araetb Dp ohs

z a sech

angular loading, dp’ pe ,aB (see fig. 2), where B is any

variable horizontal nate from the OZ axis to the slope. By differentiating s with respect to » in equa-

tion 10 and substituting F dB for dp’, there is then obtained

AS max-= ob f,| sech 5

This is the shearing stress at T due to the shaded horizontal element of figure 2. Integration between the limits, 0 and 6, yields for all such elements

rz—B ret+B Oh —sech 5 2h

fs aie)

ih ER 2 2h

4h Sua FE 2 arc tan eé ba

ie a at ded arc tan e 2 2 —arc tan e2 2h {| es er

PUBLYUG ROADS

Vol. 20 No. 8

CLAY STRATUM

ROCK SURFACE

Figure 2.—TRIANGULAR LoAap oN A LONG SrRip SUPPORTED By CLAY UNDERLAID BY Rock.

This is Jiirgenson’s > formula for the shearing stress at a point T of the rock surface when the loading is tri- angular. (See fig. 2.) The use of equations 10 and 12 is not dependent on the relative magnitudes of A and 6.

From equation 12, the greatest value of Smax., de- noted by Soy depends on the ratio of b to h. For ex-

ample, if the depth to the rock surface, 2h, is 5 ae then

Smax-=S,=0.318p at the point z=0.6256. Ifthe clay has no friction, the plastic condition for these relative di- mensions begins to be developed at the point s=0.625b at the rock surface when

Smax-==8,=c=0.318p or when 7 (see fig. 2)=3.14c¢ where ¢ is the unit cohesion. Similarly, for oh= 2), Smax:==8, 206 ¢==0.670 andmona plastic zone begins when

Smax-=8,=¢€=0.22p

or when p (see fig. 2)=4.55c.

For any fixed ratio, b: h, ordinate values of Smax- May be plotted against x as abscissa, using equation 12. The value of x, where Smax. —s,—the greatest shearing stress, is the maximum ordinate of the curve thus obtained.

HENCKY’S METHOD OF PLASTIC EQUILIBRIUM IS FUNDAMENTAL

The application of the method of plastic equilibrium to this problem involving the boundary conditions illus- trated in figures 1, 2, 3, es and 5 is limited to the con- dition that the distance, Dh, must not exceed the dis- tance, 6/2 where 2h is the thickness of the clay layer and b is half the base width of the loaded surface area.

A thin layer of soil between two rigid plates whose surfaces in contact with the soil are rough and which are of great length and of width 26 (see fig. 3) is con- sidered. ‘The soil is supposed to have cohesion and a zero or very small value for its effective angle of internal friction. ‘The method of Hencky ® will now be shown

5 The application of Theories of Elasticity and Plasticity to Foundation yee py rea: Jiirgenson, Journal of the Boston Society of Civil Engineers, vol. 21, No. 3, a 6 ‘Sher Statisch bestimmte Falle des Gleichgewichtes in plastischen Kérpen, H.

Hencky, Zeitschrift fiir ang Mathematik und Mechanik, 1924, vol. 3, p. 291, p. 401. See also Plasticity, Chapter 33, A. Nadai, 1931. McGraw- Hill Book Co.

ORLA GC

October 1939

ROADS 159

as originally devised and applied by Prandtl’ to the problem illustrated by figure 3, the plastic flow of soil from between two rigid plates. Certain equations for stresses will be derived in this application. Then these expressions for the stresses will be used in the solution of the problem of the fill, ABCD, figure 4, supported by a clay stratum underlaid by rock. First of all it is assumed (figs. 3 and 4) that / is either equal to or less than 6/4. In no case in the following development may h be considered as greater than 6/4. The solution follows.

FLOW OF PLASTIC MATERIAL

Sens

SiElP EINES

DISTRIBUTION OF STRESS

Figure 3.—ConpDiTIONS AT FAILURE IN A Puastic MatTpEriau PRESSED BETWEEN Two RovuGH PARALLEL PLATES.

When the material pressed between the plates by a load P (see fig. 3) becomes a plastic mass, flow occurs with a constant maximum shear expressed by the equation,

ae 2 | | Pees | +s,,=the unit cohesion ¢ or

Damm Pe +2Je—s,/ for according to theory, 8mar,=constant=c under these conditions. There are two other equations of equilib- rium, namely,

Os “Or and

oP Joes es

The stresses p,, pz, and s,, may be determined from equations 13, 14, and 15. Differentiating 15 with respect to x and 14 with respect to z and subtracting, there is obtained

Of O*Sre O'Sss ‘ 22a) Dx) = Oe Or eae eee a (10) substituting equation 13 in equation 16, _0°Szz__ O°Sze £259 hee a Ne ea Oa ers ore

7L. Prandtl, Zeitschrift fiir ang., Mathematik und Mechanik, vol. 6, 1923.

CLAY

AREA AED = AREA ABCD g = SUPPORTING POWER = ob FACTOR OF SAFETY = :

FicurEe 4.—SupportTinac Power or Cuay LAayeR UNDERLAID By Rock, Megruop or HENcKY.

CLAY WITH A UNIT COHESION OF 500 POUNDS PER SQUARE FOOT

ROCK SURFACE

Figure 5.—PRoBLEM OF THE SupporTING Power oF A CLAY Stratum SANDWICHED BETWEEN A Fintt, ABCD, anv SoLip Rock.

Equation 17 is now solved by assuming that s,, de- pends on z alone and not on z. When this is true equation 17 reduces to

Ose

Dg U----------------- (18) which is readily integrable, and there is obtained

=kKk,+ kz ee Sa (19)

The shearing stress s,, cannot anywhere exceed ¢, the unit cohesion. If AK, be taken as zero, there are two straight lines (the upper and lower boundaries, fig. 3), the equations of which are z=--A and ae along which the shearing stress s;, becomes Smax,=C since > by equation 9, Smax.=Szz at the rock (rigid) surface. In the present case there are two rigid surfaces, at z=-th, which form natural limits for the plastic mass. The sign of K, in equation 19 depends on whether Sr== te or 8,,==—c tor g=h. If for z=-+h, s,=—-+c,

then for K,=0, equation 19 becomes Sx2—=S max. = +e=Koh or Zia se LG= ry and therefore for any value of z between --/A and —h, cz Sr2= ay ---------------- (20)

by substitution in equation 19.

160 PUBLIC ROADS

Now,

from equation 14,

Op, See

and from equation 15,

Op: Ooms O Cz 9 mm + ; 0) 22 Oz Ou al | ; | ia)

By integration, equations 21 and 22 yield

P2=- + fi (2) (23) and

De= a(x) , (24)

respectively, where /, (2) is a funetion of 2 alone and (x) is a funetion of 2 alone. Both f; (2) and fy (v)

must be so determined that equation 13,

De—Pe= 2 Ve Saat (13) will be satisfied. Equation 18 is called the “condition of plasticity.’ Substituting the values for p, and py as given In equations 23 and 24 and for s,, from equation

20 in equation 13 there results,

A@)+y- fi(2) = + 2Qey1— 2/3 (25) Putting x=0 in equation 25. Then Ai(2)= KF 2c./1—22/h?, where K=f,(0). Putting g=0 in equation 25. Then Ja(e) =A 7 where A=, (0) + 2¢. It may be easily shown that /,(0)cb2e=/o(0). Hence

the symbol Av may denote either value. By substitution in equations 28 and 24 there results,

p= K- S209) 1— 8 an aueu~-~¥ (26) and ,=K- r cea wares (AT)

where JY is a constant.

Equations 26 and 27, together with equation 20, completely determine the stresses at any point in the plastic mass when Av is known. With reference to ficure 38, when g=-+A and xv=b, p.=0 so that by substitution in equation 27

6 0 K 7 hh '

or Ky: md

Vol, 20 No, 8

Therefore

Dy e(— a) r2eyV1 ih". (28) h ‘(h—a py (29) and So, br (20)

At the boundaries, g=--A and 2 h, (bw) Po=Pr h

ANC Soe=8may. = EC,

HENCKYS METHOD APPLICABLE IN FILL DESIGN

rom equation 29 16 is seen that p, is a maximum when ves0, and diminishes as 2 increases (b is positive on the right and negative on the left of OZ), The loading on the surtaces of plastic clay is therefore tri angular as shown in figure 8, although tho load applied bo ‘the reid frames ts uniform,

The problem illustrated in figure 4, a fill, ABCD, supported by a clay stratum undorlaid by rock, is con. sidered next. The computation of the supporting power, g, of the soil layer, figure 4, is based on the assump lon that the structure, ABCD, is: absolutely rigid, ‘This assumption is equivalent to saying that tho soil layer is between two rigid frames, the fill above and the rock below. But in order to use equations 28, 29, and 20, derived for soil between two plates, there must be made another simplifying assumption for the problem illustrated in figure 4, which is that the resist- ance to flow offered by the soil in the clay layer to the left of A and to the right of D (figure 4) is small enough (relatively) to be noglected,

With all these simplifying assumptions, equations 28, 29, and 20 apply in computing the supporting power, qd, of the soil layer, figure 4. Since the structure, ABCD, is rigid, then according to equation 29 the dis- tribution of vertical Pressure, Pa at the upper boundary (figure 4) is triangular, ‘The same vertical stress distri- bution at this boundary would be realized in fact if the load diagram, ABCD, becomes triangular, As’) the area of ABCD and that of AD bene identical since the total load of the fill cross section (1 foot thick in the direction perpendicular to the plane of fig. 4) is the same,

The total vertical force, ?, on a strip of unit width (yee 1, fig. 4) on the plane boundary, gh, is

*D *b 1 <= c(b— way ‘ if pita af, h

or cb? Pe ee aye . (80 h (30) But Pe=pb from figure 4, where p is the maximum surface load per unit area and hence eb? Pm pb

October 0) ae or ch bro (41)

The factor of safety against overloading of the elny

atratum is q/P, / bewmp the supporting power, At the 7} instant of failure, gop if

A comparison of values obtained by the elastic theory on the ofe hand and the theory of plastieity on the other is now considered, It has alrondy been shown

that for Zhe

4b, the plastic zone starts to appear when

the magnitude of p is such that peoBide, Krom equation 41 plastic flow of the entire soil mass below the fill begins when

ch ch ad dapat hana Ae 4

when Yh L, or le Up 7, 4

Hence for a comparison: 1, By the elastic theory, a plastic zone is started when pet ¢, 2, by the theory of plaste equilibrium the wtimnate bearing capacity ¢ of the supporting soil ie gq 4 ¢. h 2 region in the supporting soil mass begins when p te

i AA 7 7 a 4SNO0 or 74.5 percent of the ultimate supporting

Thus for 2h the development of a plastic zone or

ee I ; power, Similarly when 2h 4) the plastic zone js Ab be

ry ZSNOO or 57 percent

of the ultimate beaing capacity or supporting power,

parted when the value of p is

AVITACATION OF THVOMY ILLUWTMATED

Suppose that it js required to know the factor of wifety with respect to the supporting power of the soil below the fill, ABCD, figure 6, when the following conditions obtain:

l, bes he 60 feet,

2, The fill, ABC), ia symmetrical with #2 5S wlope,

4. The height of the fill ie 20 feet and the top width BC ia 40 feet,

4, The unit weight w of fill material ie 100 pounds per cubic foot,

5, The supporting soil ie ewmentially clay, Ite co hesion is 500 pounds per square foot and ite angle of internal friction Ww too small to consider, St ie then assured that all of the supporting power ie due to

cohesion, ‘ , BOL AD The aren of the trapewid, ABCD, ie ~ ‘ , 4A0+-120 : YX height i) YL 1,600 wgunre fleet, The aren of

triangle ALD ia aloo 1,600 square feet and itm height . 1600 IT ia wit

L 2667 «2,667 pounds pet 5 np pe foot, ¥ 4 cm 4 YOO or 2,000 pounds per square foot,

26,67 feet, Then p is equal to wile 100 ie equal

She

PUBLIC

ROADS 161

factor of safely against failure of the undersoil is then

2000, Mm iD Gay Si4,

Therefore the supporting sou will fail under the fill of the proposed dimensions, lor the undersoil to be safe the height //7 of the triangle AM) must be reduced ance p WH inuet be reduced, If the width of the rondwiay (BO, fig, 6) remains 40 feet and the height of the fill, ABCD, is reduced to 12 feet, the aren of ABCD

AQ} 120 pt K 12

is then YOO square feet and the height

. , 960 ; , of the equivalent triangle ie a0 16 feet, The value } pie then 1,600 pounds per square foot and ZOO0 ‘Y / ) Pm dP T Gog 1h

It hine been shown! that for a cohesive soil (with no angle of internal fmetion) extending downward to a yront depth the bearing capneity, g, for the soul support. ing a symmetrical fil, as computed by two Seana methods, is ne follows:

Method Vale of 7 in bering Of will cohesion ‘Verznghi gq 46 (omeuioing fll ie nonrigid), Vrandt qe b Ne Oweuming fi is rigid),

In the forevoing example, if the rock boundary were removed and the olay extended far below it, the value of g necording to Vrandt! would be computed as bemy HAA SHO0--2,570 pounds per square foot whieh is larger than the value, 2,000 pounds per square foot, as foundin the example, On the other fens with different relative yalues of band # and the same fill ae that cone sidered in the example, the supporting power q of the ay stratum could be much greater than 2,000 pounds

aa h ch Keb per square loot, Thus for hb equal to tay Na

404,000 pounds per square foot, a value that ie much yreater than that obtamimg when the rock layer is nonexietent, Tf thie condition had exited in the pre coding example, the fnetor of safely (all other conditions being the wame) would have been

a] AQUGY

y Hm B607

This je in necord with common sense and experience, [It ie obviously more difficult to “squeeze out’ a thin layer of soil from between two egh steel blocks than it is to cause # much thicker layer of the same soil to flow out laterally, There is alwnys the practical con- sideration that as the day Inayer becomes increasingly thin, it is Jews a major item of cost to excavate and place the fill directly on the #olid rock,

BSUMMAMY

Subsequent to construction a new fl tends to cone whidate the supporting day, Vrior to the realization of any appreciable degree of consolidation, the fill load is carried for the most part by water in the supporting ay mass, Thus initially the superimposed fill load theoretically causes no contact pressure between solid mrticles and therefore no frictional force ia developed te the neutral hydrostatic pressure in the supporting clay, It ie during thie early period following con-

iVrindyhes A WA Meannion InvAve4 in FIN Construction, I, A, Valmer wna M4, Satter, Vrowesings VNighway Mesearch Mond, Annual Aeedting 1947

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struction (or possibly during construction) that failure of the supporting soil is most likely to occur. Hence it is entirely on the side of safety to consider only the cohesion in computing the supporting power.

For the case of a supporting layer of cohesive soil underlaid by rock, the author has found no expressions for shearing stresses other than those published by Carothers. Biot * has derived quite complicated ex- pressions for the vertical stress p, for the case of axially symmetric stress distribution and for the case of a line load. For 2h=infinity his derived expressions reduce to those of Boussinesq and Mitchell. The formulas derived by Carothers do not similarly reduce, but this fact in itself indicates nothing insofar as validity is concerned.

There is no flaw in the analytical derivations of the formulas for supporting power as developed by Hencky and Prandtl and extended by Jiirgenson. The limita- tions are inherent in the assumptions. Obviously the less rigid the fill the more untenable is the assumption of rigidity.

A solution called the ‘““Method of Haines” has been indicated by Hough ° for the case of a nonrigid structure.

The cases of partially rigid structures are beyond the borderline of present theoretical knowledge existing in published form and there is therefore opportunity for progress beyond this frontier.

Jiirgenson '° has recently suggested that if the fill is nonrigid, the bearing capacity, g, should be taken as (5 which is half its value when the fill is rigid. This suggested value is only for the case when 2h is less than 6/2.

The method of Haines referred to by Hough requires a more complete presentation and description than has been published to enable the student of theoretical soil mechanics to evaluate properly its utility. The fact that this method follows Jiirgenson’s boundary case up to 2h=0.36 is interesting and adds a degree of confidence in the use of Jiirgenson’s formula,

_ eb h for relatively thin supporting soil strata.

It is the opinion of the author that it is useless to assume a surface of failure in the supporting soil stratum in this problem. The conditions are too variable to warrant this procedure. A surface of failure is not assumed in the method of Hencky as extended and applied by Prandtl and Jiirgenson. The slip lines shown in figure 3 are determinable from equations 20, 26, and 27 and are families of cycloids.

In the absence of rock, q, the supporting power, is taken with reference to the weight of a column of fill

material of height equal to that of the fill and of 1 square foot cross-sectional area. For this case there are obtained by three different analytical methods the following values for q in terms of the unit cohesion c (@ being small enough to be neglected):

By the method of Terzaghi, ix Ac. By the method of Prandtl, =(r+2)e. By the method of Krey, mea

These values are all for a factor of safety of one.

§ Effect of Certain Discontinuities on the Pressure Distribution in a Loaded Soil, M. A. Biot. Publications from the Graduate School of Engineering, Harvard University, No. 172, 1935-36.

9 Stability of Embankment haere B. K. Hough, Jr. Transactions, American Society of Civil Engineers, 1938, p.

10 On the Stability of Monn datioget and Embankments, Leo Jiirgenson, Paper No. G-8, vol. 2, Proceedings, International Conference on Soil Mechanics and Foundation Engineering, 1936.

For the case of a rigid rock boundary below the sup- porting clay, the formula of Jiirgenson is

for a factor of safety of one, where p is the weight of a column of fill material of height equal to that of the equivalent triangle. (See fig. 3.) For 2h equal to or less than 6/2, ¢ is equal to or greater than 4c, according to this formula. For values of 2A greater than 6/2, Jiirgenson’s formula gives such increasingly small values for y as to be obviously in error.

The question arises as to the best procedure to follow when 2A is greater than 6/2. Pending the time that a more general and satisfactory solution of this problem is obtained, the following procedures are believed to be warranted and their use is suggested.

1. For depths to rock less than one-fourth of the base width of the fill, the supporting power, g, is computed directly from Jiirgenson’s formula if the fill has a rigidity and strength such that it resists the shearing stress, Se7==C, ab 1ts Dade.

2. For depths to rock greater than one-fourth and less than three-fourths of the base width of the fill, the value of ¢ is considered as constant and equal to 4c regardless of the rigidity of the fill. In this case also gis considered as equal to ~, the weight of a column of fill material of height equal to that of the