Displacementandstressdistributionsunderauniforminclined...

INTERNATIONAL JOURNAL FOR NUMERICAL AND ANALYTICAL METHODS IN GEOMECHANICSInt. J. Numer. Anal. Meth. Geomech. 2009; 33:709–748Published online 26 August 2008 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/nag.738

Displacement and stress distributions under a uniform inclinedrectangular load on a cross-anisotropic geomaterial

Cheng-Der Wang∗,†,‡, Zhi-Qi Ye and Zheng-Wei Ruan

Department of Civil and Disaster Prevention Engineering, National United University, Miao-Li,Taiwan 36003, Taiwan

SUMMARY

In practical engineering, an applied rectangular area load is not often horizontally or vertically distributedbut is frequently inclined at a certain angle with respect to the horizontal and vertical axes. Thus, thesolutions of displacements and stresses due to such a load are essential to the design of foundations. Thisarticle yields the analytical solutions of displacements and stresses subjected to a uniform rectangularload that inclines with respect to the horizontal and vertical axes, resting on the surface of a cross-anisotropic geomaterial. The planes of cross-anisotropy are assumed to be parallel to the horizontal groundsurface. The procedures to derive the solutions can be integrated the modified point load solutions, whichare represented by several displacement and stresses elementary functions. Then, upon integrations, thedisplacement and stress integral functions resulting from a uniform inclined rectangular load for (1) thedisplacements at any depth, (2) the surface displacements, (3) the average displacements in a given layer,(4) the stresses at any depth, and (5) the average stresses in a given layer are yielded. The proposedsolutions are clear and concise, and they can be employed to construct a series of calculation charts. Inaddition, the present solutions clarify the load inclinations, the dimensions of a loaded rectangle, and theanalyzed depths, and the type and degree of geomaterial anisotropy profoundly affect the displacementsand stresses in a cross-anisotropic medium. Parametric results show that the load inclination factor shouldbe considered when an inclined rectangular load uniformly distributed on the cross-anisotropic material.Copyright q 2008 John Wiley & Sons, Ltd.

Received 23 February 2008; Accepted 18 June 2008

KEY WORDS: displacements; stresses; inclined rectangular load; cross-anisotropic geomaterial; loadinclinations

∗Correspondence to: Cheng-Der Wang, Department of Civil and Disaster Prevention Engineering, National UnitedUniversity, No. 1, Lien Da, Kung-Ching Li, Miao-Li, Taiwan 36003, Taiwan.

†E-mail: [email protected], [email protected]‡Professor.

Copyright q 2008 John Wiley & Sons, Ltd.

710 C.-D. WANG, Z.-Q. YE AND Z.-W. RUAN

INTRODUCTION

Generally, the estimations of the magnitudes of displacements and stresses in materials are predictedby using solutions that model materials such as homogeneous, linearly elastic, and isotropic media.However, there are natural soils that have deposited through a geological process of sedimentationover a long period of time, such as flocculated clays, varved silts or sands, or rock massescut by discontinuities, such as cleavages, foliations, stratifications, schistosities, and joints; theircharacteristics should account for anisotropy. From the standpoint of practical considerations ingeotechnical engineering, anisotropic bodies are often modeled as orthotropic or cross-anisotropicgeomaterials. Hence, in this work, an inclined loading problem for a cross-anisotropic medium isconcerned.

The computation of displacements and stresses resulting from a footing load is an importantfactor in the design of foundations. Footings and foundation structures very often tend to be of arectangular shape in plane [1], and this kind of load uniformly distributed on the horizontal groundsurface plays a predominant role for practicing engineers [2]. The systemized closed-form solu-tions for displacements and stresses due to three-dimensional buried uniform and linear varyingrectangular loads in a cross-anisotropic half-space can be found in [3]. In that reference, theycomprehensively summarized the existing loading solutions relating to the cross-anisotropic media.Sequentially, they extended the uniform rectangular loading solutions to derive the displacements[4] and stresses [5] subjected to three-dimensional embedded parabolic rectangular foundationloads. In addition, the solutions under linearly varying and nonlinearly varying loads with distinctintensities at each corner of a rectangle were obtained [6]. The presentations and usages of afore-mentioned solutions for a rectangular loaded area are concise and easy. However, it is frequentlyencountered that the loading is not horizontally or vertically distributed, but is inclined at a certainangle with respect to the horizontal or vertical axis. In addition, the displacement and stressdistributions are commonly utilized owing to inclined loads, viz., retaining walls, foundations onsloping ground, or batter piles, etc. [7]. Therefore, to derive the solutions of displacements andstresses beneath the corner of a uniform inclined rectangular load on a cross-anisotropic solidcould be virtually applied to the design of foundations. Yet, only Ramiah and Chikkanagappa [7]have presented the vertical normal stress distribution in an isotropic half-space under an inclinedpoint load by combining Boussinesq’s (a vertical point load) [8] and Cerruti’s solutions (a hori-zontal point load) [9]. They also extended the solutions to include a uniform inclined infinite-and finite-length line loads. Furthermore, they prepared a table and showed the positions of thestress bulbs for different inclinations of the concentrated and line loads inclined to the vertical.Eventually, they integrated Cerruti’s solution [9] for a horizontal point load over a rectangular areaand presented the influence chart, as well as combining [10] influence chart for easy computa-tion of vertical normal stress at any point in an isotropic medium when the applied loads wereinclined. However, it is noteworthy to point out that the frequent treatment of material as anisotropic medium is no more suitable for that with anisotropic characteristic. To the best of ourknowledge, no solutions for the displacements and stresses cased by an inclined rectangular-shapedload have been addressed. Hence, we intend to yield them in this paper for the following fivecases:

(1) the displacements at any depth;(2) the surface displacements;(3) the average displacements in a given layer;

Copyright q 2008 John Wiley & Sons, Ltd. Int. J. Numer. Anal. Meth. Geomech. 2009; 33:709–748DOI: 10.1002/nag

DISPLACEMENT AND STRESS DISTRIBUTIONS 711

(4) the stresses at any depth;(5) the average stresses in a given layer.

The assumptions of an inclined rectangular load on the cross-anisotropic material are as follows:

(1) The load is perfectly flexible.(2) The material behaves as a homogeneous, linearly elastic, and cross-anisotropic continuum. Its

mechanical properties are defined by five independent elastic constants, E,E ′,�,�′, and G ′.(3) The planes of cross-anisotropy are parallel to the horizontal ground surface.(4) There are no body forces.

Thus, the explicit solutions of displacements and stresses under the action of an inclined rectan-gular load can be obtained by integrating the modified point load solutions, which are presented interms of several displacement/stress elementary functions in this article. This means that the desiredaforementioned solutions with the consideration of load inclinations are regrouped as the forms ofthe modified point load solutions except for the load intensity (force) and the displacement/stresselementary functions are replaced, respectively, by the applied pressure (force per unit area) andfive cases of displacement/stress integral functions. The presentations of the derived solutions arevery clear and concise, and they can construct a series of calculation charts, which are convenientwhen computers or calculators are unavailable. Additionally, the proposed solutions demonstratethat the load inclinations (�,�), the dimensions of a loaded rectangle (B, L), the analyzed depths(z, or H ), and the type and degree of material anisotropy (the combinations of E,E ′, �,�′, G ′)significantly influence the displacements and stresses in a cross-anisotropic medium. By a para-metric study, three examples are illustrated to investigate the effect of load inclination(s) on thesurface displacements and stresses as follows:

(1) effect of load inclination, �, on the normalized surface displacements with fixed �;(2) effect of load inclination, �, on the nondimensional stresses with fixed z and �;(3) effect of load inclinations, � and �, on the normalized vertical surface displacement.

The obtained cross-anisotropic solutions of vertical normal stress in a limiting process areidentical with the combinations of Steinbrenner’s [10], and Ramiah and Chikkanagappa’s [7]isotropic solutions. Moreover, the yielded solutions for the displacements at any depth, the surfacedisplacements, and the stresses at any depth are all in excellent agreement with Wang and Liao’scross-anisotropic solutions [3] for the following simplified circumstances:

(1) �=0 and �=0 (only a horizontal rectangular load distributed toward the x direction);(2) �=0 and �=90◦ (only a horizontal rectangular load distributed toward the y direction);(3) �=90◦ (only a vertical rectangular load distributed toward the z direction).

Parametric results reveal that the load inclination factor should be taken into account whenestimating the induced displacements and stresses by an inclined rectangular area load, resting ona cross-anisotropic geomaterial.

INCLINED POINT LOADING SOLUTIONS FOR DISPLACEMENTS AND STRESSES

In this article, the solutions for displacements and stresses in a cross-anisotropic half-spacedue to a uniform inclined surface rectangular load are directly integrated from the surface



Pz

Px

Py

x

y

z

Figure 1. A three-dimensional point load acting on the surface of a cross-anisotropic half-space.

point load solutions in a Cartesian co-ordinate system, which are modified from Wang andLiao [3]. Figure 1 depicts a three-dimensional point load, (Px , Py, Pz), acting at the origin ofthe co-ordinates. The modified closed-form solutions for displacements (ux ,uy,uz) and stresses(�xx ,�yy,�zz,�xy,�xz,�yz) resulting from (Px , Py, Pz) resting on the surface of a cross-anisotropicmaterial are presented as

ux = Px4�

(g1∗ pd11−g2∗ pd12+ 2

u3A44∗ pd23

)

+ Py4�

(−g1∗ pd31+g2∗ pd32+ 2

u3A44∗ pd33

)

− Pz4�

(g3∗ pd41−g4∗ pd42) (1)

uy = Px4�

(−g1∗ pd31+g2∗ pd32+ 2

u3A44∗ pd33

)

+ Py4�

(g1∗ pd21−g2∗ pd22+ 2

u3A44∗ pd13

)

− Pz4�

(g3∗ pd51−g4∗ pd52) (2)

uz = Px4�

(− f1∗g1∗ pd41+ f2∗g2∗ pd42)

+ Py4�

(− f1∗g1∗ pd51+ f2∗g2∗ pd52)

− Pz4�

( f1∗g3∗ pd61− f2∗g4∗ pd62) (3)



�xx = Px4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13)∗g1∗ ps11

−(A11−u2∗ f2∗A13)∗g2∗ ps12

−2∗A66∗(g1∗ ps71−g2∗ ps72)+4∗u3∗ ps73

⎤⎥⎥⎦

+ Py4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13−2∗A66)∗g1∗ ps21

−(A11−u2∗ f2∗A13−2∗A66)∗g2∗ ps22

+2∗A66∗(g1∗ ps81−g2∗ ps82)−4∗u3∗ ps83

⎤⎥⎥⎦

+ Pz4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13−2∗A66)∗g3∗ ps31

−(A11−u2∗ f2∗A13−2∗A66)∗g4∗ ps32

+2∗A66∗(g3∗ ps51−g4∗ ps52)

⎤⎥⎥⎦ (4)

�yy = Px4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13−2∗A66)∗g1∗ ps11

−(A11−u2∗ f2∗A13−2∗A66)∗g2∗ ps12

+2∗A66∗(g1∗ ps71−g2∗ ps72)−4∗u3∗ ps73

⎤⎥⎥⎦

+ Py4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13)∗g1∗ ps21

−(A11−u2∗ f2∗A13)∗g2∗ ps22

−2∗A66∗(g1∗ ps81−g2∗ ps82)+4∗u3∗ ps83

⎤⎥⎥⎦

+ Pz4�

⎡⎢⎢⎣

(A11−u1∗ f1∗A13−2∗A66)∗g3∗ ps31

−(A11−u2∗ f2∗A13−2∗A66)∗g4∗ ps32

+2∗A66∗(g3∗ ps61−g4∗ ps62)

⎤⎥⎥⎦ (5)

�zz = Px4�

[(A13−u1∗ f1∗A33)∗g1∗ ps11−(A13−u2∗ f2∗A33)∗g2∗ ps12]

+ Py4�

[(A13−u1∗ f1∗A33)∗g1∗ ps21−(A13−u2∗ f2∗A33)∗g2∗ ps22]

+ Pz4�

[(A13−u1∗ f1∗A33)∗g3∗ ps31−(A13−u2∗ f2∗A33)∗g4∗ ps32] (6)

�xy = Px4�

[2∗A66∗(g1∗ ps81−g2∗ ps82)−2∗u3∗(2∗ ps83− ps23)]

+ Py4�

[2∗A66∗(g1∗ ps71−g2∗ ps72)−2∗u3∗(2∗ ps73− ps13)]

− Pz2�

A66∗(g3∗ ps41−g4∗ ps42) (7)



�xz = Px4�

[(u1+ f1)∗A44∗g1∗ ps51−(u2+ f2)∗A44∗g2∗ ps52+2∗ ps63]

− Py4�

[(u1+ f1)∗A44∗g1∗ ps41−(u2+ f2)∗A44∗g2∗ ps42−2∗ ps43]

− Pz4�

A44∗[(u1+ f1)∗g3∗ ps11−(u2+ f2)∗g4∗ ps12] (8)

�yz = − Px4�

[(u1+ f1)∗A44∗g1∗ ps41−(u2+ f2)∗A44∗g2∗ ps42−2∗ ps43]

+ Py4�

[(u1+ f1)∗A44∗g1∗ ps61−(u2+ f2)∗A44∗g2∗ ps62+2∗ ps53]

− Pz4�

A44∗[(u1+ f1)∗g3∗ ps21−(u2+ f2)∗g4∗ ps22] (9)

where Ai j (i, j =1–6) are the elastic moduli or elasticity constants of the medium and can beexpressed in terms of five engineering elastic constants of a cross-anisotropic material as

A11 =E

(1− E

E ′ �′2)

(1+�)

(1−�− 2E

E ′ �′2) , A13= E�′

1−�− 2E

E ′ �′2, A33= E ′(1−�)

1−�− 2E

E ′ �′2

A44 = G ′, A66= E

2(1+�)

(10)

where E and E ′ are Young’s moduli in the plane of cross-anisotropy and in a direction normal toit, respectively; � and �′ are Poisson’s ratios characterizing the lateral strain response in the planeof cross-anisotropy to a stress acting parallel or normal to it, respectively; G ′ is the shear modulusin planes normal to the plane of cross-anisotropy. u3=√

A66/A44, u1 and u2 are the roots of thefollowing characteristic equation:

u4−su2+c=0 (11)

where s=(A11A33−A13(A13+2A44))/A33A44, c= A11/A33. Since the strain energy is assumedto be positive definite in the medium, the values of elastic constants are restricted. Hence, thereare three categories of the characteristic roots, u1 and u2, as follows:

Case 1: u1,2=±√

12 [s±

√s2−4c] are two real distinct roots when s2−4c>0.

Case 2: u1,2=±√s/2,±√

s/2 are double equal real roots when s2−4c=0.Case 3: u1= 1

2

√s+2

√c− i 12

√−s+2√c=− i ,u2=+ i are two complex conjugate roots

(where cannot be equal to zero) when s2−4c<0.

f j = (A13+A44)∗u j

A33∗u2j −A44= A11−A44∗u2j

(A13+A44)∗u j( j=1,2), k= A13+A44

A33∗A44(u21−u22)



g0 = 2∗k∗( f2∗u1− f1∗u2)u1−u2

, g1= g0∗u1f1∗ f2∗( f1+u1)

, g2= g0∗u2f1∗ f2∗( f2+u2)

g3 = g0f1+u1

, g4= g0f2+u2

, pd1i = 1

Ri +zi− x2

Ri (Ri +zi )2

pd2i = 1

Ri +zi− y2

Ri (Ri +zi )2, pd3i = xy

Ri (Ri +zi )2, pd4i = x

Ri (Ri +zi )

pd5i = y

Ri (Ri +zi ), pd6i = 1

Ri, ps1i = x

R3i

, ps2i = y

R3i

, ps3i = ziR3i

ps4i = xy(2Ri +zi )

R3i (Ri +zi )2

, ps5i = 1

Ri (Ri +zi )− x2(2Ri +zi )

R3i (Ri +zi )2

ps6i = 1

Ri (Ri +zi )− y2(2Ri +zi )

R3i (Ri +zi )2

, ps7i = x

R3i

− 3x

Ri (Ri +zi )2+ x3(3Ri +zi )

R3i (Ri +zi )3

ps8i = y

R3i

− 3y

Ri (Ri +zi )2+ y3(3Ri +zi )

R3i (Ri +zi )3

, Ri=√x2+ y2+z2i , zi =ui ∗z (i=1,2,3)

We define pd1i–pd6i and ps1i–ps8i in Equations (1)–(3) and (4)–(9) as, respectively, the displace-ment and stress elementary functions for derivations in this work.

As mentioned above, the desired inclined rectangular loading solutions in this paper could beobtained by integrating the inclined point loading solutions with two inclination angles, � and �,as shown in Figure 2. In Figure 2, an inclined point load with intensity P can be decomposed intoP ∗cos�∗cos�, P ∗cos�∗sin�, and P ∗sin� with respect to the co-ordinates x, y, and z. Hence,Px , Py , Pz in Equations (1)–(9) should be, respectively, replaced by P ∗cos�∗cos�, P ∗cos�∗sin�,and P ∗sin�. The procedures to obtain the inclined rectangular loading solutions for displacementsand stresses over a cross-anisotropic geomaterial will be presented below.

P

Px=P*cos *cos

Pz=P*siny

Py=P*cos *sin

P

x

z

Figure 2. An inclined point load with intensity P can be decomposed into three-dimensional components(Px , Py, Pz) acting on the surface of a cross-anisotropic half-space.



INCLINED RECTANGULAR LOADING SOLUTIONS FORDISPLACEMENTS AND STRESSES

An inclined flexible area load, q (force per unit area), distributed on a rectangle of width B andlength L on the surface of a cross-anisotropic half-space, as depicted in Figure 3, is investigated.The induced displacement and stress solutions at point A below the corner of a rectangular loadedarea are derived in this paper. Observing from Figure 3, q can also be decomposed into (qx ,qy,qz)with respect to the co-ordinates x , y, and z. In addition, considering an elementary area dA=dx dyon a flexible loaded region, the total load on the elementary area is as follows:

dPj =q j dx dy ( j = x, y, z) (12)

This means that the point load intensities (Px , Py, Pz) in Equations (1)–(9) should be substi-tuted by qx =q ∗cos�∗cos�,qy =q ∗cos�∗sin�, and qz =q ∗sin� for an inclined rectangular load.However, it should be mentioned that there are three special loading cases for the load inclinations,� and �, as follows:

(1) when �=0 and �=0, q=qx signifies only a horizontal rectangular load distributed towardthe x direction;

(2) when �=0 and �=90◦,q=qy signifies only a horizontal rectangular load distributed towardthe y direction;

(3) when �=90◦,q=qz signifies only a vertical rectangular load distributed toward the zdirection.

In other words, the explicit solutions of displacements and stresses under the action of an inclinedrectangular load can be regrouped as the forms of Equations (1)–(9) except for the displacementelementary functions pd1i–pd6i , and stress elementary functions ps1i–ps8i are replaced by thefollowing five cases of displacement integral functions and stress integral functions, which include(1) the displacements at any depth, (2) the surface displacements, (3) the average displacements ina given layer, (4) the stresses at any depth, and (5) the average stresses in a given layer. Therefore,only the deriving procedures and their integral functions are expressed in the subsections.

B

L

qz

qx

qy

Oq

x

y

z

A

Figure 3. A uniform inclined rectangular load with intensity q can be decomposed into three-dimensionalcomponents (qx ,qy,qz) acting on the surface of a cross-anisotropic half-space.



DISPLACEMENT INTEGRAL FUNCTIONS FOR DISPLACEMENTS AT ANY DEPTH

As mentioned above, the displacements and stresses at point A (Figure 3) beneath the groundsurface due to an inclined rectangular load can be yielded by integrating the displacement elemen-tary functions pd1i–pd6i and stress elementary functions ps1i–ps8i in Equations (1)–(3) and(4)–(9), with respect to the co-ordinates x and y, from 0 to B and 0 to L . It implies that theexact solutions of displacements/stresses owing to an inclined rectangular-shaped load are thesame as Equations (1)–(3)/Equations (4)–(9) except for the displacement elementary functionspd1i–pd6i/stress elementary functions ps1i–ps8i are replaced by distinct displacement integralfunctions/stress integral functions. Hence, only the displacement and stress integral functionsaccording to five desired cases are proposed. Firstly, for the induced displacements at pointA, the displacement integral functions at any depth, denoted by rd1i–rd6i (i=1,2,3), can beobtained by

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rd1i

rd2i

rd3i

rd4i

rd5i

rd6i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=∫ L

0

∫ B

0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

pd1i

pd2i

pd3i

pd4i

pd5i

pd6i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦dx dy (13)

Integration yields rd1i–rd6i as

rd1i = B

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

mi

⎛⎝tan−1 mi

ni− tan−1 mini√

m2i +n2i +1

− tan−1 mi

ni√m2

i +n2i +1

⎞⎠

+ ln

∣∣∣∣∣∣√m2

i +n2i +1+ni√m2

i +1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(14)

rd2i = B

⎡⎣ 1

mi

⎛⎝tan−1 mi

ni√m2

i +n2i +1− tan−1 mi

ni

⎞⎠+ ni

miln

∣∣∣∣∣∣√m2

i +n2i +1+mi√n2i +1

∣∣∣∣∣∣⎤⎦ (15)

rd3i = B1

mi

⎡⎢⎢⎢⎢⎢⎣

−1+√m2

i +1+√n2i +1−

√m2

i +n2i +1

− ln

∣∣∣∣∣∣√m2

i +1+1

2

∣∣∣∣∣∣+ ln

∣∣∣∣∣∣√m2

i +n2i +1+1√n2i +1+1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎦ (16)



rd4i = B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

tan−1 nimi

− tan−1 ni

mi

√m2

i +n2i +1

− 1

mi

⎛⎝ln |

√m2

i +1|− ln

∣∣∣∣∣∣√m2

i +n2i +1+ni√n2i +1+ni

∣∣∣∣∣∣−ni ln

∣∣∣∣∣∣√m2

i +n2i +1+1√n2i +1+1

∣∣∣∣∣∣⎞⎠

⎤⎥⎥⎥⎥⎥⎥⎥⎦

(17)

rd5i = B

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

nimi

⎛⎝tan−1 mi

ni− tan−1 mi

ni√m2

i +n2i +1

⎞⎠+ ln

∣∣∣∣∣∣√m2

i +n2i +1+1√m2

i +1+1

∣∣∣∣∣∣

− 1

mi

⎛⎝ln |

√n2i +1|− ln

∣∣∣∣∣∣√m2

i +n2i +1+mi√m2

i +1+mi

∣∣∣∣∣∣⎞⎠

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(18)

rd6i = B

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− 1

mitan−1 mini√

m2i +n2i +1

+ ln

∣∣∣∣∣∣√m2

i +n2i +1+ni√m2

i +1

∣∣∣∣∣∣

+ nimi

ln

∣∣∣∣∣∣√m2

i +n2i +1+mi√n2i +1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(19)

where mi = B/ui z, ni = L/ui z (i=1,2,3).It is found from Equations (14)–(19) that rd1i–rd6i (i=1,2,3) are the functions of the following

factors:

rd1i–rd6i = f (B,mi ,ni )= f (B, L , z,ui ) (20)

In other words, the solutions of displacements at the corner (point A) due to a uniform inclinedrectangular load on a cross-anisotropic medium are governed by (a) the dimensions of a loadedrectangle, B and L; (b) the depth, z; and (c) the five elastic engineering constants, ui (i=1,2,3), asdefined in Equation (11). In addition, based on Equations (13)–(18), a set of calculation charts fordisplacement integral functions at any depth, rd1i–rd6i , can be prepared. Figures 4–9, respectively,show the calculation charts for rd1i/B–rd6i/B (i=1,2,3), with two nondimensional variables,mi = B/ui z and ni = L/ui z, in which mi and ni are, respectively, in the ranges of 0.1–10 and0.05–10 (or 0.1–10). It can be observed from Figures 4–9 that rd1i/B–rd6i/B increase with theincrease in ni . Figures 4–9 can be utilized when computers or calculators are unavailable; however,they are only appropriate for the root type of the characteristic equation (Equation (11)) belongingto Case 1 (i.e. where u1 and u2 are two real distinct roots). Therefore, if the root type belongs toCase 2 or Case 3, the addressed calculation charts (Figures 4–9) cannot be applied to estimate thedisplacements at any depth resulting from an inclined rectangular load acting on a cross-anisotropichalf-space.



1010.1 5320.50.2 0.3

mi=Buiz

0

0.5

1

1.5

2

2.5

rd1i

B

0.1

0.5

1

0.3

0.05

0.20.4

0.7

2

5

10

ni=L

uiz

Figure 4. Calculation chart for rd1i with the variations of mi and ni .

1010.1 5320.2 0.3 0.5

mi=Buiz

0

0.2

0.4

0.6

0.8

1

rd2i

B

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

5

10ni=

Luiz




1010.1 5320.2 0.3 0.5

mi=Buiz

0

0.1

0.2

0.3

0.4

rd3i

B

0.10.5

10.7

2

5

10

ni=L

uiz


1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.2

0.4

0.6

0.8

1

rd4i

B

0.1

0.5

1

0.3

0.05

0.20.4

0.7

2

5

10

ni= L

uiz




1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.4

0.8

1.2

1.6

2

rd5i

B

0.10.5

1

0.7

2

5

10

ni= L

uiz


1010.1 5320.2 0.3 0.5

mi= Buiz

0

1

2

3

rd6i

B

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

5

10 ni= L

uiz




DISPLACEMENT INTEGRAL FUNCTIONS FOR SURFACE DISPLACEMENTS

Since the distributions of surface displacements are valuable for geotechnical engineering, thesolutions can be acquired by setting z=0 in the above displacement integral functions at any depth(rd1i–rd6i , Equations (14)–(19)) as

rd1i(z=0) = B ln

∣∣∣∣∣√1+ 1

t2+ 1

t

∣∣∣∣∣ (21)

rd2i(z=0) = B1

tln |√1+ t2+ t | (22)

rd3i(z=0) = B

[1+ 1

t− 1

t

√1+ t2

](23)

rd4i(z=0) = B

[tan−1 1

t+ 1

tln |√1+ t2|

](24)

rd5i(z=0) = B

[1

ttan−1 t+ ln

∣∣∣∣∣√1+ 1

t2

∣∣∣∣∣]

(25)

rd6i(z=0) = B

[ln

∣∣∣∣∣√1+ 1

t2+ 1

t

∣∣∣∣∣+ 1

tln |√1+ t2+ t |

](26)

1010.1 5320.50.30.2

t=BL

0

1

2

3

4

r d

1i(z

=0)

B,

r d

2i(z

=0)

B,

r d

3i(z

=0

)

B,

r d

4i(z

=0)

B,

r d

5i(z

=0)

B,

r d

6i(z

=0)

B

rd1i(z=0)

rd2i(z=0)

rd3i(z=0)

rd4i(z=0)

rd5i(z=0)

rd6i(z=0)

Figure 10. Calculation chart for rd1i(z=0)–rd6i(z=0) with the variation of t .



where rd1i(z=0)–rd6i(z=0) represent the displacement integral functions for surface displacements,and t= B/L .

It is also discovered from Equations (21)–(26) that rd1i(z=0)–rd6i(z=0) are the functions of thefollowing factors:

rd1i(z=0)–rd6i(z=0) = f (B, t)= f (B, L) (27)

Similarly, we can plot a calculation chart, as displayed in Figure 10, for the normalizedrd1i(z=0)/B–rd6i(z=0)/B vs the dimensionless ratio, t (from 0.1 to 10). Carefully observingEquations (21)–(26) that, since z=0, this calculation chart (Figure 10) for the surface displace-ments is unconcerned with ui (i=1,2,3); hence, it would not be restricted to the root type of thecharacteristic equation belonging to Case 1, but is suitable for all cross-anisotropic materials.

DISPLACEMENT INTEGRAL FUNCTIONS FOR AVERAGEDISPLACEMENTS IN A GIVEN LAYER

In the cases of practical engineering, it is rather convenient to find the average displacement/stresscomponents in a given layer, below the corner of an inclined rectangular loaded area, with limitsof z=0 to z=H (the thickness of a given cross-anisotropic layer below the surface), as depictedin Figure 11. Therefore, the solutions for average displacements can be achieved by integratingrd1i–rd6i (Equations (14)–(19)) as follows:⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rd1i(avg)

rd2i(avg)

rd3i(avg)

rd4i(avg)

rd5i(avg)

rd6i(avg)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= 1

H

∫ H

0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rd1i

rd2i

rd3i

rd4i

rd5i

rd6i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦dz (28)

displacements and stresses

B

L

qz

qx

qy

Oq

x

y

z

A

H

z

average values

dz

Figure 11. Average displacements and stresses in a given layer with thickness H due to a uniform inclinedrectangular load acting on the surface of a cross-anisotropic half-space.



where rd1i(avg)–rd6i(avg) represent the displacement integral functions for average displacementsin a given layer, and these displacement integral functions can be yielded as

rd1i(avg) = B2

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1

a2i

⎛⎝tan−1 ai

bi− tan−1 ai

bi√a2i +b2i +1

− tan−1 aibi√a2i +b2i +1

⎞⎠

− tan−1 bi

ai√a2i +b2i +1

+ 2

ailn

∣∣∣∣∣∣√a2i +b2i +1+bi√

a2i +1

∣∣∣∣∣∣+biai

ln

∣∣∣∣∣∣√a2i +b2i +1+1√

a2i +b2i

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(29)

rd2i(avg) = B2

2

⎡⎢⎢⎢⎢⎢⎢⎣

− 1

a2itan−1 ai

bi+ 1−b2i

a2itan−1 ai

bi√a2i +b2i +1

+biai

ln

∣∣∣∣∣∣√a2i +b2i +1+1√

a2i +b2i

∣∣∣∣∣∣+2bia2i

ln

∣∣∣∣∣∣√a2i +b2i +1+ai√

b2i +1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎦

(30)

rd3i(avg) = B2

4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− 3

a2i(1−

√a2i +1−

√b2i +1+

√a2i +b2i +1)

+ ln

∣∣∣∣∣∣√a2i +b2iai

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√a2i +1+1

∣∣∣∣∣∣+ 2

a2i

⎛⎝ln

∣∣∣∣∣∣√a2i +b2ibi

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +1+1

2

∣∣∣∣∣∣⎞⎠

+2−b2ia2i

⎛⎝ln

∣∣∣∣∣∣√a2i +b2i +1+1√b2i +1+1

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2ibi

∣∣∣∣∣∣⎞⎠

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(31)

rd4i(avg) = B2

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− bia2i

(bi −√a2i +b2i −

√b2i +1+

√a2i +b2i +1)

+ 2

ai

⎛⎝tan−1 bi

ai− tan−1 bi

ai√a2i +b2i +1

⎞⎠

+ ln

∣∣∣∣∣∣√a2i +b2i +bi

ai

∣∣∣∣∣∣−1

a2iln |√b2i +1+bi |

−(1− 1

a2i

)ln

∣∣∣∣∣∣√a2i +b2i +1+bi√

a2i +1

∣∣∣∣∣∣+2bia2i

ln

∣∣∣∣∣∣√a2i +b2i +1+1√b2i +1+1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(32)



rd5i(avg) = B2

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− 1

ai(ai −

√a2i +1−

√a2i +b2i +

√a2i +b2i +1)

+2bia2i

⎛⎝tan−1 ai

bi− tan−1 ai

bi√a2i +b2i +1

⎞⎠

+ 2

ailn

∣∣∣∣∣∣√a2i +b2i +1+1√a2i +1+1

∣∣∣∣∣∣−1

a2iln |√a2i +1+ai |

+b2ia2i

ln

∣∣∣∣∣∣√a2i +b2i +ai

bi

∣∣∣∣∣∣+1−b2ia2i

ln

∣∣∣∣∣∣√a2i +b2i +1+ai√

b2i +1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(33)

rd6i(avg) = B2

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

− tan−1 bi

ai√a2i +b2i +1

− 1

a2itan−1 aibi√

a2i +b2i +1

−b2ia2i

tan−1 ai

bi√a2i +b2i +1

+ 2

ailn

∣∣∣∣∣∣√a2i +b2i +1+bi√

a2i +1

∣∣∣∣∣∣+2biai

ln

∣∣∣∣∣∣√a2i +b2i +1+1√

a2i +b2i

∣∣∣∣∣∣

+2bia2i

ln

∣∣∣∣∣∣√a2i +b2i +1+ai√

b2i +1

∣∣∣∣∣∣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(34)

where ai = B/ui H , bi = L/ui H (i=1,2,3). Hence, rd1i(avg)–rd6i(avg) are the functions of

rd1i(avg)–rd6i(avg) = f (B,ai ,bi )= f (B, L ,H,ui ) (35)

Furthermore, Equations (29)–(34) can be utilized to prepare another set of calculation chartsfor the average displacements caused by a uniform inclined loaded rectangle. They are plottedrd1i(avg)/B2–rd6i(avg)/B2 (i=1,2,3) vs the nondimensional variable ai = B/ui H (from 0.1 to 10)with the variations of bi = L/ui H (from 0.05 to 10), as, respectively, shown in Figures 12–17.Again, since rd1i(avg)–rd6i(avg) are associated with ui ; therefore, Figures 12–17 are only proper tothe characteristic root (Equation (11)) belonging to Case 1.

Nevertheless, in reality, it might be encountered to calculate the average displacements in aspecific layer in the interior of cross-anisotropic layered media. The present average displacementsolutions can also be applied to solve such problems by using the principle of superposition.



1010.1 5320.2 0.3 0.5

ai= B

uiH

0

10

20

30

rd1i(avg)

B2

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

5

10bi=

LuiH

Figure 12. Calculation chart for rd1i(avg) with the variations of ai and bi .

1010.1 5320.2 0.3 0.5

ai= B

uiH

0

2

4

6

8

10

rd2i(avg)

B2

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

5

10

bi= L

uiH




1010.1 5320.2 0.3 0.5

ai= B

uiH

0

0.2

0.4

0.6

0.8

1

rd3i(avg)

B2

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

5

10

bi= L

uiH


1010.1 5320.2 0.3 0.5

ai= B

uiH

0

0.4

0.8

1.2

1.6

2

rd4i(avg)

B2

0.1

0.5

1

0.3

0.05

0.2

0.4

0.7

2

510

bi= L

uiH




1010.1 5320.2 0.3 0.5

ai= B

uiH

0

10

20

30

rd5i(avg)

B2

0.1

0.5

1

0.3

0.050.20.4

0.7

2

5

10

bi= L

uiH


1010.1 5320.2 0.3 0.5

ai= B

uiH

0

10

20

30

40

rd6i(avg)

B2

0.1

0.5

1

0.3

0.05

0.2

0.40.7

2

5

10

bi= L

uiH




STRESS INTEGRAL FUNCTIONS FOR STRESSES AT ANY DEPTH

Identically, the stress fields increased by an inclined rectangular-shaped load at point A, as exhibitedin Figure 3, can be derived by the integrations of stress elementary functions, ps1i–ps8i (Equations(4)–(9)), with respect to the co-ordinates x and y, from 0 to B and 0 to L , as⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rs1irs2irs3irs4irs5irs6irs7irs8i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=∫ L

0

∫ B

0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

ps1ips2ips3ips4ips5ips6ips7ips8i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦dx dy (36)

Consequently, the stress integral functions for stresses at any depth, rs1i–rs8i (i=1,2,3), areexpressed as

rs1i = ln |√m2

i +1|− ln

∣∣∣∣∣∣√m2

i +n2i +1+ni√n2i +1+ni

∣∣∣∣∣∣ (37)

rs2i = ln |√n2i +1|− ln

∣∣∣∣∣∣√m2

i +n2i +1+mi√m2

i +1+mi

∣∣∣∣∣∣ (38)

rs3i = tan−1 mini√m2

i +n2i +1(39)

rs4i = ln

∣∣∣∣∣∣√m2

i +1+1

2

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√m2

i +n2i +1+1√n2i +1+1

∣∣∣∣∣∣ (40)

rs5i = − tan−1 mi

ni+ tan−1 mini√

m2i +n2i +1

+ tan−1 mi

ni√m2

i +n2i +1(41)

rs6i = tan−1 mi

ni− tan−1 mi

ni√m2

i +n2i +1(42)

rs7i = ni

⎛⎝ 1√

n2i +1+1− 1√

m2i +n2i +1+1

⎞⎠ (43)

rs8i =mi

⎛⎝ 1√

m2i +1+1

− 1√m2

i +n2i +1+1

⎞⎠ (44)

where mi = B/ui z and ni = L/ui z (i=1,2,3) are the same as defined in Equations (14)–(19).



According to Equations (37)–(44), the calculation charts for the stress integral functions, rs1i–rs8i (i=1,2,3), are sketched, respectively, in Figures 18–25. The principles of constructing Figures18–25 are similar to those of Figures 4–9, and they are unsuitable for the characteristic root to beCase 2 or Case 3.

STRESS INTEGRAL FUNCTIONS FOR AVERAGE STRESSES IN A GIVEN LAYER

Although the presentation of the stress integral functions for stresses at any depth (Equations(37)–(44)) is concise and easy to be computed. However, as mentioned above, it is considerablypractical to use the average stress concept [11, 12] irrespective of the depth, z. That is to say, thestress integral functions for this case can be obtained by integrating rs1i–rs8i (i=1,2,3), withrespect to z, from 0 to H , as

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rs1i(avg)

rs2i(avg)

rs3i(avg)

rs4i(avg)

rs5i(avg)

rs6i(avg)

rs7i(avg)

rs8i(avg)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= 1

H

∫ H

0

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rs1i

rs2i

rs3i

rs4i

rs5i

rs6i

rs7i

rs8i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

dz (45)

Upon integrations, the stress integral functions for average stresses in a given layer are demonstratedbelow:

rs1i(avg) = ai tan−1 bi

ai√a2i +b2i +1

+ ln |√a2i +1|− ln

∣∣∣∣∣∣√a2i +b2i +1+bi√b2i +1+bi

∣∣∣∣∣∣

+bi

⎛⎝ln

∣∣∣∣∣∣√a2i +b2ibi

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√b2i +1+1

∣∣∣∣∣∣⎞⎠ (46)

rs2i(avg) = bi tan−1 ai

bi√a2i +b2i +1

+ ln |√b2i +1|− ln

∣∣∣∣∣∣√a2i +b2i +1+ai√a2i +1+ai

∣∣∣∣∣∣

+ai

⎛⎝ln

∣∣∣∣∣∣√a2i +b2iai

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√a2i +1+1

∣∣∣∣∣∣⎞⎠ (47)



rs3i(avg) = tan−1 aibi√a2i +b2i +1

+ai

⎛⎝ln

∣∣∣∣∣∣√a2i +1

ai

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+bi√a2i +b2i +bi

∣∣∣∣∣∣⎞⎠

+bi

⎛⎝ln

∣∣∣∣∣∣√b2i +1

bi

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+ai√a2i +b2i +ai

∣∣∣∣∣∣⎞⎠ (48)

rs4i(avg) = 1+ai +bi −√a2i +1−

√b2i +1−

√a2i +b2i +

√a2i +b2i +1

− ln

∣∣∣∣∣∣√a2i +1+1

2

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√b2i +1+1

∣∣∣∣∣∣ (49)

rs5i(avg) = − tan−1 aibi

+ tan−1 ai

bi√a2i +b2i +1

+ tan−1 aibi√a2i +b2i +1

+ai

⎛⎝ln

∣∣∣∣∣∣√a2i +1

ai

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+bi√a2i +b2i +bi

∣∣∣∣∣∣⎞⎠ (50)

rs6i(avg) = tan−1 aibi

− tan−1 ai

bi√a2i +b2i +1

+bi

⎛⎝ln

∣∣∣∣∣∣√b2i +1

bi

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+ai√a2i +b2i +ai

∣∣∣∣∣∣⎞⎠ (51)

rs7i(avg) = bi2

⎛⎝ 1√

b2i +1+1− 1√

a2i +b2i +1+1+ ln

∣∣∣∣∣∣√a2i +b2ibi

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√b2i +1+1

∣∣∣∣∣∣⎞⎠ (52)

rs8i(avg) = ai2

⎛⎝ 1√

a2i +1+1− 1√

a2i +b2i +1+1+ ln

∣∣∣∣∣∣√a2i +b2iai

∣∣∣∣∣∣− ln

∣∣∣∣∣∣√a2i +b2i +1+1√a2i +1+1

∣∣∣∣∣∣⎞⎠ (53)

where ai = B/ui H ,bi = L/ui H (i=1,2,3) are also the same as defined in Equations (29)–(34).Regarding for the solutions of average stresses in a specific layer of cross-anisotropic layeredgeomaterials, the principle of superposition can also be employed.

The displacements and stresses at any point inside/outside of a uniform rectangular loaded areaare found by dividing the region into 2–4 rectangles, in which the desired point is at the corner ofeach rectangle. As for the displacements and stresses due to an inclined arbitrary-shaped loading,the loading area can be divided into many rectangular sub-areas; influences from these sub-areasare then superimposed.



1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.5

1

1.5

2

2.5

rs1i

0.1

0.5

1

0.30.2

0.4

0.7

2

5

10

ni= Luiz

Figure 18. Calculation chart for rs1i with the variations of mi and ni .

1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.5

1

1.5

2

2.5

rs2i

0.1

0.5

1

0.3

0.7

2

5

10

ni= Luiz




1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.4

0.8

1.2

1.6

rs3i

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10ni= Luiz


1010.1 5320.50.30.2

mi= Buiz

0

0.4

0.8

1.2

1.6

rs4i

0.1

0.5

1

0.7

2

5

10

ni= Luiz

0.3




1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.2

0.4

0.6

0.8

1

rs5i

0.1

0.5

1

0.30.2

0.4

0.7

2

5

10

ni= Luiz


1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.2

0.4

0.6

0.8

1

rs6i

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10

ni= L

uiz




1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.1

0.2

0.3

0.4

0.5

rs7i

0.1

0.5

1

0.3

0.2

0.4

0.7

25

10

ni= Luiz


1010.1 5320.2 0.3 0.5

mi= Buiz

0

0.1

0.2

0.3

0.4

0.5

rs8i

0.10.5

1

0.3

0.7

2

5

10ni= Luiz




The present cross-anisotropic solutions for vertical normal stress due to an inclined rectangularload by a limiting procedure are in accordance with the conjunction of Steinbrenner’s [10], andRamiah and Chikkanagappa’s [7] isotropic equations. In addition, the solutions of displacementsat any depth, the surface displacements, and the stresses at any depth are all in perfect agreementwith Wang and Liao’s solutions [3] when (1) �=0 and �=0 (only qx applied), (2) �=0 and�=90◦ (only qy applied), and (3) �=90◦ (only qz applied). Eventually, the calculation chartsbased on Equations (46)–(53) are drawn in Figures 26–33. A flowchart that illustrates the proposedsolutions for computing the displacements and stresses due to an inclined rectangular load uniformlydistributed on the surface of a cross-anisotropic medium is presented in Figure 34.

ILLUSTRATIVE EXAMPLE

This section presents practical examples to confirm the addressed solutions and elucidate the effectof the load inclinations, � and �, and the type and degree of geomaterial anisotropy, on the surfacedisplacements and stresses. Seven types of isotropic and cross-anisotropic soils are considered toconstitute the foundation geomaterials. Their elastic properties and root type specified by the ratiosE/E ′, G ′/E ′, and �/�′ are listed in Table I. The selected domains of anisotropic variation in TableI basically follow the suggestion of Gazetas [13] with E/E ′ ranging from 0.6 to 4 for clays and aslow as 0.2 for sands. Therefore, the range of ratio E/E ′ is 1.35–2.37,G ′/E ′ is 0.23–0.44 [14–17],and �/�′ is hypothetically assumed varying 0.75–1.5. The values of E and � adopted in Table Iare 50MPa [18] and 0.3 [19]. Therefore, a parametric study by three examples is illustrated asfollows:

(1) Effect of load inclination, �, on the normalized surface displacements when �=30◦: FromFigure 34, it is shown that the displacements at any depth, the surface displacements, and theaverage displacements in a given layer can utilize Equations (1)–(3) with the combinations ofEquations (14)–(19), (21)–(26), and (29)–(34). In this example, the surface displacements, ux ,uy ,and uz , are computed by using the properties of isotropic/cross-anisotropic soils (Soil 1/Soils 2–7,Table I), and assuming �=30◦, the width (B) and the length (L) of a rectangle are, respectively,4 and 6m. Figure 35 shows the effect of load inclination, �, from 0 to 360◦, on the normalizedsurface displacements, ux/q (Figure 35(a)), uy/q (Figure 35(b)), and uz/q (Figure 35(c)), forSoils 1–7. In Figure 35(a) and (b), a dashed circle represents zero-displacement to distinguish thepositive- and negative-value displacements. From Figure 35(a), it is found that, when � is withinthe range of 97.5–277.5◦, the induced normalized ux/q are all negative. In addition, when � is atabout 0–97.5◦, the increase in ratio E/E ′ from 1 (Soil 1) →1.35 (Soil 2) →2.37 (Soil 3), ux/qwould be increased. Evidently, the ratios E/E ′ (Soil 2 and 3) and G ′/E ′ (Soil 4) strongly affect theinduced surface displacement, ux , but the effect of ratio �/�′ (Soils 6 and 7) on it is slight. Figure35(b) depicts the normalized surface displacement, uy/q , of all soils. The trends of the calculatedresults in this figure are quite different from those in Figure 35(a). Also, the magnitudes estimatedfrom Soils 2, 3, and 4 have more obviously distinct results from the isotropic soil (Soil 1). As forthe very interesting quantity for civil engineers is to determine the vertical surface displacement,uz ; hence, the computed results are presented in Figure 35(c). In Figure 35(c), it is apparentthat there are no negative-value displacements occurred in all soil media, and the influences ofall ratios of soil anisotropy (E/E ′,G ′/E ′, and �/�′) considerably affect the normalized surfacedisplacement, uz/q .



1010.1 5320.2 0.3 0.5

ai= B

uiH

0

1

2

3

4

rs1i(avg)

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10

bi= L

uiH

Figure 26. Calculation chart for rs1i(avg) with the variations of ai and bi .

0110.1 5320.2 0.3 0.5

ai= B

uiH

0

1

2

3

4

rs2i(avg)

0.1

0.5

1

0.30.2

0.4

0.7

2

5

10

bi= L

uiH




1010.1 5320.2 0.3 0.5

ai= B

uiH

0

0.4

0.8

1.2

1.6

rs3i(avg)

0.1

0.5

1

0.3

0.2

0.4

0.7

2

510

bi= L

uiH


1010.1 5320.50.30.2

ai= B

uiH

-4

-3

-2

-1

0

1

rs4i(avg)

0.1

0.5

1

0.3

0.7

2

5

10

bi= L

uiH




1010.1 5320.2 0.3 0.5

ai= B

uiH

0

0.4

0.8

1.2

rs5i(avg)

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10

bi= L

uiH


1010.1 5320.2 0.3 0.5

ai=B

uiH

0

0.4

0.8

1.2

rs6i(avg)

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10

bi=L

uiH




1010.1 5320.50.30.2

ai= B

uiH

0

0.2

0.4

0.6

0.8

rs7i(avg)

0.1

0.5

1

0.3

0.2

0.4

0.7

2

5

10

bi= L

uiH


1010.1 5320.50.30.2

ai= B

uiH

0

0.2

0.4

0.6

0.8

rs8i(avg)

0.1

0.5

1

0.30.2

0.4

0.7

2

5

10

bi= L

uiH




Figure 34. Flowchart to compute the induced displacements and stresses for a cross-anisotropic geomaterialby a uniform inclined rectangular load.



Table I. Elastic properties for the isotropic/cross-anisotropic soils (E=50MPa,�=0.3).

Soil type E/E ′ G′/E ′ �/�′

Soil 1: isotropy 1 0.385 1Soil 2: cross-anisotropy 1.35 0.385 1Soil 3: cross-anisotropy 2.37 0.385 1Soil 4: cross-anisotropy 1 0.230 1Soil 5: cross-anisotropy 1 0.440 1Soil 6: cross-anisotropy 1 0.385 0.75Soil 7: cross-anisotropy 1 0.385 1.5

(2) Effect of load inclination, �, on the nondimensional stresses when z=3m and �=30◦: Thesecond example is illustrated to explore the effect of � and soil anisotropy on the stress fieldscaused by an inclined rectangular area load with fixed sides B=4m, L=6m, and z=3m,�=30◦.Again, from Figure 34, the stress components can be calculated by employing Equations (4)–(9),(37)–(44) for stresses at any depth, and Equations (4)–(9), (46)–(53) for average stresses in agiven layer. In this example, the normal stresses (�xx ,�yy,�zz) and shear stresses (�xy,�xz,�yz)are estimated for Soils 1–7 (Table I). Figure 36 displays the effect of � (from 0 to 360◦) on thenondimensional stresses, �xx/q (Figure 36(a)), �yy/q (Figure 36(b)), �zz/q (Figure 36(c)), �xy/q(Figure 36(d)), �xz/q (Figure 36(e)), and �yz/q (Figure 36(f)), for Soils 1–7. From Figure 36(a)–(c),the calculated normal stresses are also influenced by the type and degree of material anisotropy;especially, the increase in ratio E/E ′ from 1 (Soil 1) to 2.37 (Soil 3) do have a conspicuousdissimilarity than other cross-anisotropic soils (Soils 2, 4–7). In addition, the magnitude of Figure36(a)–(c) shows that the order induced normal stresses as follows: �zz>�yy>�xx . Figure 36(d)–(f)reveals the distributions of shear stresses owing to an inclined rectangular-shaped load for Soils1–7 vs the load inclination, �. It can be noted that the effect of soil anisotropy on the three shearstresses for Soils 2, 3, 4 is more obvious than others (Soils 5–7). Moreover, the magnitude ofFigure 36(d)–(f) demonstrates that it obeys the order: �xz>�yz>�xy .

(3) Effect of load inclinations, � and �, on the normalized vertical surface displacement: Asknown in the first example, the estimation of vertical surface displacement (uz) is very important forgeotechnical engineering. Additionally, the results of Figure 35 exhibit that the magnitude estimatedfrom Soil 3 (Figure 35(c)) is quite distinct from other cross-anisotropic soils. Consequently, in thethird example, we represent the distribution of normalized vertical surface displacement (uz/q)with the variations of both inclined angles, � (from 0 to 360◦) and � (from 0 to 90◦), as depictedin Figure 37(a) and (b), respectively, for the isotropic Soil 1 and cross-anisotropic Soil 3. Thesetwo figures reveal that the magnitude and distribution of the computed uz/q are highly differedfrom each other. That interprets the importance of the effect of load inclination, �, on the verticalsurface displacement.

Three examples are demonstrated above to show the effect of the type and degree of geomaterialanisotropy and the load inclinations, on the normalized surface displacements and nondimensionalnormal and shear stresses. Results present that the displacements and stresses in the isotropic/cross-anisotropic soils (Soil 1/Soils 2–7) due to a uniform inclined rectangular load are rather differentfrom those solutions by treating the loaded area horizontally or vertically distributed on the groundsurface. Hence, it is imperative to consider the load inclination factor when calculating the induceddisplacements and stresses in a cross-anisotropic medium by an inclined rectangular load.



015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

-0.08 -0.04 0 0.04 0.08

uq (

mMPa)

Effect of load inclination, α on u when z=0 and β=30

Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

α( ) α( )

-0.12 -0.08 -0.04 0 0.04 0.08

uq (

mMPa)


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(a) (b)

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

α( )

0.01 0.02 0.03 0.04 0.05 0.06

uq (

mMPa)


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(c)

Figure 35. Effect of load inclination � on the normalized surface displacementswhen �=30◦ for (a) ux ; (b) uy ; and (c) uz .



015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

-0.04 0 0.04 0.08 0.12

σq

Effect of load inclination, α on σ when z=3 m and β=30

Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(a)

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

α( )

0

α( )α( )α( )

α( )α( )α( )α( )

-0.08 -0.04 0 0.04 0.08 0.12 0.16

σq


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(b)

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

-0.1 0 0.1 0.2 0.3

σq


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(c) (d)

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

-0.08 -0.04 0 0.04 0.08

τq

Effect of load inclination, α on τ when z=3 m and β=30

Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

Figure 36. Effect of load inclination � on the nondimensional stresses when z=3m and �=30◦ for(a) �xx ; (b) �yy ; (c) �zz ; (d) �xy ; (e) �xz ; and (f) �yz .



015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

α( ) α( )

-0.04 0 0.04 0.08 0.12 0.16

τq


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(e)

015

30

45

60

75

90

105

120

135

150

165180

195

210

225

240

255

270

285

300

315

330

345

-0.12 -0.08 -0.04 0 0.04 0.08 0.12

τq


Soil 1, E/E'=1, G'/E'=0.385, υ/υ'=1

Soil 2, E/E'=1.35, G'/E'=0.385, υ/υ'=1

Soil 3, E/E'=2.37, G'/E'=0.385, υ/υ'=1

Soil 4, E/E'=1, G'/E'=0.23, υ/υ'=1

Soil 5, E/E'=1, G'/E'=0.44, υ/υ'=1

Soil 6, E/E'=1, G'/E'=0.385, υ/υ'=0.75

Soil 7, E/E'=1, G'/E'=0.385, υ/υ'=1.5

(f)

Figure 36. Continued.

CONCLUSIONS

The analytical solutions of displacements and stresses due to a uniform inclined rectangular loadresting on the surface of a cross-anisotropic geomaterial, in which the load is arbitrarily inclinedwith respect to the horizontal and vertical axes, are derived in this article. The assumed loadis perfectly flexible on a linearly elastic, homogeneous cross-anisotropic continuum with cross-anisotropic planes parallel to the horizontal ground surface. The rectangular-shaped loading ischosen since footing or foundation structure is frequently to be such a plane, and this kind ofload uniformly distributed on the surface playing a predominant role for practicing engineers.The present solutions can be obtained by integrating the point load solutions in a Cartesian co-ordinate system, which are modified from Wang and Liao [3] in this paper. The modified point loadsolutions are expressed in terms of several displacement and stresses elementary functions and thenintegrations yield the displacements and stresses integral functions owing to a uniform inclinedrectangular area load for (1) the displacements at any depth, (2) the surface displacements, (3) theaverage displacements in a given layer, (4) the stresses at any depth, and (5) the average stressesin a given layer. Restated, the desired solutions with the consideration of load inclinations can beregrouped as the forms of the modified point load solutions except for the load intensity (force) andthe displacement/stress elementary functions are replaced, respectively, by the applied pressure(force per unit area) and five aforementioned cases of displacement/stress integral functions. Inaddition, since the presentations of the derived solutions are very clear and concise; they can beutilized to construct a series of calculation charts, which are suitable when computers or calculatorsare unavailable. Furthermore, the proposed solutions demonstrate that the load inclinations (�,�),



Figure 37. Effect of load inclinations � and � on the normalized vertical surfacedisplacement for (a) Soil 1 and (b) Soil 3.

the dimensions of a loaded rectangle (B, L), the analyzed depths (z, or H ), and the type and degreeof geomaterial anisotropy (specified by the ratios E/E ′, G ′/E ′, �/�′), remarkably influence theinduced displacements and stresses in a cross-anisotropic medium. A parametric study by threeexamples is shown to illustrate the effect of the load inclinations, and the type and degree of soilanisotropy on the surface displacements and stresses. Some following interesting conclusions canbe drawn from this work.

1. The first example investigates the effect of load inclination, � (from 0–360◦), and the typeand degree of soil anisotropy, on the induced normalized surface displacements (z=0) bya uniform inclined rectangular-shaped load, with fixed B, L , and �, for isotropic/cross-anisotropic soils (Soil 1/Soils 2–7). It is found that the ratios E/E ′ (Soil 2 and 3) andG ′/E ′ (Soil 4) strongly affect the induced surface displacements, ux and uy , but the effectsof ratio �/�′ (Soils 6 and 7) on them are little. However, the influence of all ratios of



soil anisotropy (E/E ′,G ′/E ′, and �/�′) on the vertical surface displacement, uz , is veryevident.

2. The second example is illustrated to explore the effect of load inclination, � (from 0 to 360◦),and the type and degree of soil anisotropy, on the nondimensional stresses caused by aninclined rectangle area load, with fixed B, L , z, and �, for Soils 1–7. It is noted from thecalculated results of normal stresses, �xx ,�yy,�zz , the increase in ratio E/E ′ from 1 (Soil 1)to 2.37 (Soil 3) do have a conspicuous dissimilarity than other cross-anisotropic soils (Soils2, 4–7). In addition, the effect of soil anisotropy on shear stresses, �xy,�xz , and �yz , for Soils2, 3, 4, is more obvious than others (Soils 5–7).

3. The last example shows the variations of load inclinations, � (from 0 to 360◦) and � (from0 to 90◦) in the normalized vertical surface displacement for the isotropic Soil 1 and cross-anisotropic Soil 3. The magnitudes and distributions of the computed results are highlydiffered from each other, and those clarify the importance of the effect of load inclination,�, on the vertical surface displacement.

The calculations of displacements and stresses due to a uniform inclined rectangular load in across-anisotropic half-space are easy and fast by the present solutions. Moreover, the parametricresults emphasize that the induced displacements and stresses by the assumed load should beconsidered as the importance of load inclination factor. The proposed solutions could offer valuablereferences for practical loading circumstances in the fields of geotechnical engineering.

APPENDIX A: NOMENCLATURE

Ai j (i, j =1–6) elastic moduli or elasticity constantsai ,bi dimensionless ratios (functions of B, L ,H,ui )B the width of a rectangular loading areaE,E ′,�,�′,G ′ five engineering elastic constants of a cross-anisotropic geomaterialf1, f2,g0–g4,k coefficientsH the thickness of a given cross-anisotropic layer below the surfacei complex number (=√−1)L the length of a rectangular loading areami ,ni dimensionless ratios (functions of B, L , z,ui )Px , Py, Pz three-dimensional point loads in a Cartesian co-ordinate system,

as depicted in Figures 1 and 2 (force)pd1i–pd6i displacement elementary functions for the action of a point loadps1i–ps8i stress elementary functions for the action of a point loadqx ,qy,qz three-dimensional components of a uniform inclined rectangular load in a

Cartesian co-ordinate system, as depicted in Figure 3 (force per unit area)rd1i–rd6i displacement integral functions for displacements at any depthrd1i(avg)–rd6i(avg) displacement integral functions for average displacements in a given layerrd1i(z=0)–rd6i(z=0) displacement integral functions for surface displacementsrs1i–rs8i stress integral functions for stresses at any depthrs1i(avg)–rs8i(avg) stress integral functions for average stresses in a given layers,c coefficientst dimensionless ratio (functions of B, L)



u1,u2,u3 roots of the characteristic equationux ,uy,uz displacement componentsx, y, z Cartesian co-ordinate system

Greek letters

�,� load inclinations�xx ,�yy,�zz normal stress components�xy,�xz,�yz shear stress components

REFERENCES

1. Stark RF, Booker JR. Surface displacement of non-homogeneous elastic half-space subjected to uniform surfacetractions. Part II: loading on rectangular shaped areas. International Journal for Numerical and Analytical Methodsin Geomechanics 1997; 21:379–395.

2. Giroud JP. Settlement of a linearly loaded rectangular area. Journal of the Soil Mechanics and FoundationsDivision (ASCE) 1968; 94:813–831.

3. Wang CD, Liao JJ. Elastic solutions for a transversely isotropic half-space subjected to buried asymmetric-loads.International Journal for Numerical and Analytical Methods in Geomechanics 1999; 23:115–139.

4. Wang CD, Liao JJ. Elastic solutions of displacements for a transversely isotropic half-space subjected to three-dimensional buried parabolic rectangular loads. International Journal of Solids and Structures 2002; 39:4805–4824.

5. Wang CD, Liao JJ. Elastic solutions for stresses in a transversely isotropic half-space subjected to three-dimensionalparabolic rectangular loads. International Journal for Numerical and Analytical Methods in Geomechanics 2002;26:1449–1476.

6. Wang CD. Three-dimensional non-linearly varying rectangular loads on a transversely isotropic half-space.International Journal of Geomechanics (ASCE) 2004; 4:240–253.

7. Ramiah BK, Chikkanagappa S. Stress distribution under inclined loads. Soils and Foundations 1973; 13:35–47.8. Boussinesq J. Application des Potentiels a L’Etude de L’Equilibre et due Mouvement des Solides Elastiques.

Gauthier-Villars: Paris, 1885.9. Cerruti V. Sulla deformazione di un corpo elastico isotropo per alcune speciali condizioni ai limiti. Estratto da’

Rendiconti della r. Acc. de’Lincei 1888; 4:785–792.10. Steinbrenner W. Tafeln zur setzungsberechnung. Die Straße 1934; 1:121–124.11. Griffiths DV. A chart for estimating the average vertical stress increase in an elastic foundation below a uniformly

loaded rectangular area. Canadian Geotechnical Journal 1984; 21:710–713.12. Wang CD, Ruan ZW, Ye ZQ. Average vertical stress increment due to a uniform embedded rectangular foundation

on a cross-anisotropic geomaterial. Journal of Geotechnical and Geoenvironment Engineering (ASCE) 2008;under review.

13. Gazetas G. Stresses and displacements in cross-anisotropic soils. Journal of the Geotechnical Engineering Division(ASCE) 1982; 108:532–553.

14. Ward WH, Marsland A, Samuels SG. Properties of the London clay at the ashford common shaft. Geotechnique1965; 15:321–344.

15. Gibson RE. The analytical method in soil mechanics. Geotechnique 1974; 24:115–140.16. Lee KM, Rowe RK. Deformation caused by surface loading and tunneling: the role of elastic anisotropy.

Geotechnique 1989; 39:125–140.17. Tarn JQ, Lu CC. Analysis of subsidence due to a point sink in an anisotropic porous elastic half space.

International Journal for Numerical and Analytical Methods in Geomechanics 1991; 15:573–592.18. Chou WI, Bobet A. Predictions of ground deformations in shallow tunnels in clay. Tunnelling and Underground

Space Technology 2002; 17:3–19.19. Karakus M, Fowell RJ. Effects of different tunnel face advance excavation on the settlement by FEM. Tunnelling

and Underground Space Technology 2003; 18:513–523.


Displacementandstressdistributionsunderauniforminclined...

Documents

Transcript of Displacementandstressdistributionsunderauniforminclined...