Applied Elasticity for Engineers -Iisc

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Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju

Module 1: Elasticity

1.1.1 INTRODUCTION

If the external forces producing deformation do not exceed a certain limit, the deformation disappearswith the removal of the forces. Thus the elastic behavior implies the absence of any permanentdeformation. Elasticity has been developed following the great achievement of Newton in stating the laws

of motion, although it has earlier roots. The need to understand and control the fracture of solids seems tohave been a first motivation. Leonardo da Vinci sketched in his notebooks a possible test of the tensile

strength of a wire. Galileo had investigated the breaking loads of rods under tension and concluded thatthe load was independent of length and proportional to the cross section area, this being the first steptoward a concept of stress.

Every engineering material possesses a certain extent of elasticity. The common materials of constructionwould remain elastic only for very small strains before exhibiting either plastic straining or brittle failure.

However, natural polymeric materials show elasticity over a wider range (usually with time or rate effectsthus they would more accurately be characterized as viscoelastic), and the widespread use of natural

rubber and similar materials motivated the development of finite elasticity. While many roots of thesubject were laid in the classical theory, especially in the work of Green, Gabrio Piola, and Kirchhoff inthe mid-1800's, the development of a viable theory with forms of stress-strain relations for specificrubbery elastic materials, as well as an understanding of the physical effects of the nonlinearity in simple

problems such as torsion and bending, was mainly the achievement of the British-born engineer and

applied mathematician Ronald S. Rivlin in the1940's and 1950's.

1.1.2 THE GENERAL THEORY OF ELASTICITY Linear elasticity as a general three-dimensional theory has been developed in the early 1820's based onCauchy's work. Simultaneously, Navier had developed an elasticity theory based on a simple particlemodel, in which particles interacted with their neighbours by a central force of attraction between

neighboring particles. Later it was gradually realized, following work by Navier, Cauchy, and Poisson inthe 1820's and 1830's, that the particle model is too simple. Most of the subsequent developments of this

subject were in terms of the continuum theory. George Green highlighted the maximum possible numberof independent elastic moduli in the most general anisotropic solid in 1837. Green pointed out that theexistence of elastic strain energy required that of the 36 elastic constants relating the 6 stress components

to the 6 strains, at most 21 could be independent. In 1855, Lord Kelvin showed that a strain energyfunction must exist for reversible isothermal or adiabatic response and showed that temperature changes

are associated with adiabatic elastic deformation. The middle and late 1800's were a period in which

many basic elastic solutions were derived and applied to technology and to the explanation of natural phenomena. Adhémar-Jean-Claude Barré de Saint-Venant derived in the 1850's solutions for the torsion

of noncircular cylinders, which explained the necessity of warping displacement of the cross section inthe direction parallel to the axis of twisting, and for the flexure of beams due to transverse loading; the

latter allowed understanding of approximations inherent in the simple beam theory of Jakob Bernoulli,Euler, and Coulomb. Heinrich Rudolf Hertz developed solutions for the deformation of elastic solids asthey are brought into contact and applied these to model details of impact collisions. Solutions for stress

and displacement due to concentrated forces acting at an interior point of a full space were derived by

Kelvin and those on the surface of a half space by Boussinesq and Cerruti. In 1863 Kelvin had derived the basic form of the solution of the static elasticity equations for a spherical solid, and this was applied infollowing years for calculating the deformation of the earth due to rotation and tidal force and measuring



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the effects of elastic deformability on the motions of the earth's rotation axis.

1.1.3 ASSUMPTIONS OF LINEAR ELASTICITY

In order to evaluate the stresses, strains and displacements in an elasticity problem, one needs to derive aseries of basic equations and boundary conditions. During the process of deriving such equations, one canconsider all the influential factors, the results obtained will be so complicated and hence practically no

solutions can be found. Therefore, some basic assumptions have to be made about the properties of the body considered to arrive at possible solutions. Under such assumptions, we can neglect some of theinfluential factors of minor importance. The following are the assumptions in classical elasticity.

The Body is Continuous

Here the whole volume of the body is considered to be filled with continuous matter, without any void.Only under this assumption, can the physical quantities in the body, such as stresses, strains anddisplacements, be continuously distributed and thereby expressed by continuous functions of coordinatesin space. However, these assumptions will not lead to significant errors so long as the dimensions of the

body are very large in comparison with those of the particles and with the distances between neighbouring particles.

The Body is Perfectly Elastic

The body is considered to wholly obey Hooke's law of elasticity, which shows the linear relations between the stress components and strain components. Under this assumption, the elastic constants will be independent of the magnitudes of stress and strain components.

The Body is Homogenous

In this case, the elastic properties are the same throughout the body. Thus, the elastic constants will beindependent of the location in the body. Under this assumption, one can analyse an elementary volumeisolated from the body and then apply the results of analysis to the entire body.

The Body is IsotropicHere, the elastic properties in a body are the same in all directions. Hence, the elastic constants will beindependent of the orientation of coordinate axes.

The Displacements and Strains are Small

The displacement components of all points of the body during deformation are very small in comparison

with its original dimensions and the strain components and the rotations of all line elements are much

smaller than unity. Hence, when formulating the equilibrium equations relevant to the deformed state, the

lengths and angles of the body before deformation are used. In addition, when geometrical equations

involving strains and displacements are formulated, the squares and products of the small quantities are

neglected. Therefore, these two measures are necessary to linearize the algebraic and differential

equations in elasticity for their easier solution.

1.1.4 APPLICATIONS OF LINEAR ELASTICITY

The very purpose of application of elasticity is to analyse the stresses and displacements of elements

within the elastic range and thereby to check the sufficiency of their strength, stiffness and stability.

Although, elasticity, mechanics of materials and structural mechanics are the three branches of solid

mechanics, they differ from one another both in objectives and methods of analysis.



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Mechanics of materials deals essentially with the stresses and displacements of a structural or machine

element in the shape of a bar, straight or curved, which is subjected to tension, compression, shear,

bending or torsion. Structural mechanics, on the basis of mechanics of materials, deals with the stresses

and displacements of a structure in the form of a bar system, such as a truss or a rigid frame. The

structural elements that are not in form of a bar, such as blocks, plates, shells, dams and foundations, they

are analysed only using theory of elasticity. Moreover, in order to analyse a bar element thoroughly and

very precisely, it is necessary to apply theory of elasticity.

Although bar shaped elements are studied both in mechanics of materials and in theory of elasticity, the

methods of analysis used here are not entirely the same. When the element is studied in mechanics of

materials, some assumptions are usually made on the strain condition or the stress distribution. These

assumptions simplify the mathematical derivation to a certain extent, but many a times inevitably reduce

the degree of accuracy of the results obtained. However, in elasticity, the study of bar-shaped element

usually does not need those assumptions. Thus the results obtained by the application of elasticity theory

are more accurate and may be used to check the appropriate results obtained in mechanics of materials.

While analysing the problems of bending of straight beam under transverse loads by the mechanics of

materials, it is usual to assume that a plane section before bending of the beam remains plane even after

the bending. This assumption leads to the linear distribution of bending stresses. In the theory of

elasticity, however one can solve the problem without this assumption and prove that the stress

distribution will be far from linear variation as shown in the next sections.

Further, while analysing for the distribution of stresses in a tension member with a hole, it is assumed in

mechanics of materials that the tensile stresses are uniformly distributed across the net section of the

member, whereas the exact analysis in the theory of elasticity shows that the stresses are by no means

uniform, but are concentrated near the hole; the maximum stress at the edge of the hole is far greater than

the average stress across the net section.The theory of elasticity contains equilibrium equations relating to stresses; kinematic equations relating

the strains and displacements; constitutive equations relating the stresses and strains; boundary conditions

relating to the physical domain; and uniqueness constraints relating to the applicability of the solution.



Module 2: Analysis of Stress

2.1.1 INTRODUCTION

body under the action of external forces, undergoes distortion and the effect due to thissystem of forces is transmitted throughout the body developing internal forces in it. To

examine these internal forces at a point O in Figure 2.1 (a), inside the body, consider a planeMN passing through the point O. If the plane is divided into a number of small areas, as inthe Figure 2.1 (b), and the forces acting on each of these are measured, it will be observed

that these forces vary from one small area to the next. On the small area AD at point O, a

force F D will be acting as shown in the Figure 2.1 (b). From this the concept of stress as theinternal force per unit area can be understood. Assuming that the material is continuous, the

term "stress" at any point across a small area AD can be defined by the limiting equation as

below.

(a) (b)

Figure 2.1 Forces acting on a body

Stress =

A

F

AD

D

®D 0

lim (2.0)

where DF is the internal force on the area D A surrounding the given point. Stress is

sometimes referred to as force intensity.

A



2.1.2 NOTATION OF STRESS

Here, a single suffix for notation s , like z y x s s s ,, , is used for the direct stresses and

double suffix for notation is used for shear stresses like ,, xz xy t t etc. xy

t means a stress,

produced by an internal force in the direction of Y, acting on a surface, having a normal in

the direction of X.

2.1.3 CONCEPT OF DIRECT STRESS AND SHEAR STRESS

Figure 2.2 Force components of F acting on small area centered on point O

Figure 2.2 shows the rectangular components of the force vector DF referred to

corresponding axes. Taking the ratios x

z

x

y

x

x

A

F

A

F

A

F

D

D

D

D

D

D,, , we have three quantities that

establish the average intensity of the force on the area D A x. In the limit as D A®0, the aboveratios define the force intensity acting on the X-face at point O. These values of the threeintensities are defined as the "Stress components" associated with the X-face at point O.

The stress components parallel to the surface are called "Shear stress components" denoted

by t . The shear stress component acting on the X-face in the y-direction is identified as t xy.

The stress component perpendicular to the face is called "Normal Stress" or "Direct stress"

component and is denoted by s . This is identified as s x along X-direction.



From the above discussions, the stress components on the X -face at point O are defined asfollows in terms of force intensity ratios

x

x

A x

A

F

x D

D=

®D 0lims

x

y

A xy

A

F

x DD=

®D 0limt (2.1)

x

z

A xz

A

F

x D

D=

®D 0limt

The above stress components are illustrated in the Figure 2.3 below.

Figure 2.3 Stress components at point O



2.1.4 STRESS TENSOR

Let O be the point in a body shown in Figure 2.1 (a). Passing through that point, infinitelymany planes may be drawn. As the resultant forces acting on these planes is the same, thestresses on these planes are different because the areas and the inclinations of these planes

are different. Therefore, for a complete description of stress, we have to specify not only itsmagnitude, direction and sense but also the surface on which it acts. For this reason, the stress is called a "Tensor".

Figure 2.4 Stress components acting on parallelopiped

Figure 2.4 depicts three-orthogonal co-ordinate planes representing a parallelopiped onwhich are nine components of stress. Of these three are direct stresses and six are shear

stresses. In tensor notation, these can be expressed by the tensor t ij, where i = x, y, z and j =

x, y, z. In matrix notation, it is often written as



t ij =

úúú

û

ù

êêê

ë

é

zz zy zx

yz yy yx

xz xy xx

t t t

t t t

t t t

(2.2)

It is also written as

S =

úúú

û

ù

êêê

ë

é

z zy zx

yz y yx

xz xy x

s t t

t s t

t t s

(2.3)

2.1.5 SPHERICAL AND DEVIATORIAL STRESS TENSORS

A general stress-tensor can be conveniently divided into two parts as shown above. Let us

now define a new stress term (s m) as the mean stress, so that

s m =3

z y x s s s ++ (2.4)

Imagine a hydrostatic type of stress having all the normal stresses equal to s m, and all

the shear stresses are zero. We can divide the stress tensor into two parts, one having

only the "hydrostatic stress" and the other, "deviatorial stress". The hydrostatic type of

stress is given by

úúú

û

ù

êêê

ë

é

m

m

m

s

s

s

00

00

00

(2.5)

The deviatorial type of stress is given by

úúú

û

ù

êêê

ë

é

-

-

-

m z yz xz

yzm y xy

xz xym x

s s t t

t s s t

t t s s

(2.6)

Here the hydrostatic type of stress is known as "spherical stress tensor" and the other is

known as the "deviatorial stress tensor".

It will be seen later that the deviatorial part produces changes in shape of the body andfinally causes failure. The spherical part is rather harmless, produces only uniform volume

changes without any change of shape, and does not necessarily cause failure.



2.1.6 INDICIAL NOTATION

An alternate notation called index or indicial notation for stress is more convenient forgeneral discussions in elasticity. In indicial notation, the co-ordinate axes x, y and z are

replaced by numbered axes x1, x2 and x3 respectively. The components of the force DF of

Figure 2.1 (a) is written as DF 1, DF 2 and DF 3, where the numerical subscript indicates the

component with respect to the numbered coordinate axes.

The definitions of the components of stress acting on the face x1can be written in indicial

form as follows:

1

1

011

1

lim A

F

A D

D=

®Ds

1

2

012

1

lim A

F

A D

D=

®Ds (2.7)

1

3

013

1

lim A

F

A D

D=

®Ds

Here, the symbol s is used for both normal and shear stresses. In general, all components ofstress can now be defined by a single equation as below.

i

j

Aij

A

F

i D

D=

®D 0lims (2.8)

Here i and j take on the values 1, 2 or 3.

2.1.7 TYPES OF STRESS

Stresses may be classified in two ways, i.e., according to the type of body on which they act,or the nature of the stress itself. Thus stresses could be one-dimensional, two-dimensional orthree-dimensional as shown in the Figure 2.5.

(a) One-dimensional Stress



(b) Two-dimensional Stress (c) Three-dimensional Stress

Figure 2.5 Types of Stress

2.1.8 TWO-DIMENSIONAL STRESS AT A POINT

A two-dimensional state-of-stress exists when the stresses and body forces are independent

of one of the co-ordinates. Such a state is described by stresses y x

s s , and t xy and the X

and Y body forces (Here z is taken as the independent co-ordinate axis).

We shall now determine the equations for transformation of the stress components

y x s s , and t xy at any point of a body represented by infinitesimal element as shown in the

Figure 2.6.

Figure 2.6 Thin body subjected to stresses in xy plane



Figure 2.7 Stress components acting on faces of a small

wedge cut from body of Figure 2.6

Consider an infinitesimal wedge as shown in Fig.2.7 cut from the loaded body in Figure 2.6.

It is required to determine the stresses x¢s and y x ¢¢t , that refer to axes y x ¢¢, making an

angle q with axes X , Y as shown in the Figure. Let side MN be normal to the x¢axis.

Considering x¢s and y x ¢¢t as positive and area of side MN as unity, the sides MP and PN have

areas cosq and sinq , respectively.

Equilibrium of the forces in the x and y directions requires that

T x = s x cosq + t xy sinq T y = t xy cosq + s y sinq (2.9)



where T x and T y are the components of stress resultant acting on MN in the x and y

directions respectively. The normal and shear stresses on the x' plane (MN plane) are

obtained by projecting T x and T y in the x¢ and y¢ directions.

x¢s = T x cosq + T y sinq (2.10)

y x ¢¢t = T y cosq - T x sinq

Upon substitution of stress resultants from Equation (2.9), the Equations (2.10) become

x¢s = s x cos2q + s y sin2q + 2t xy sinq cosq

y x ¢¢t = xyt (cos2q - sin2q )+(s y -s x) sinq cosq (2.11)

The stress y¢s is obtained by substituting ÷ ø

öçè

æ +

2

p q for q in the expression for x¢s .

By means of trigonometric identities

cos2q =21 (1+cos2q ), sinq cosq =

21 sin2q , (2.12)

sin2q =2

1(1-cos2q )

The transformation equations for stresses are now written in the following form:

( ) ( ) q t q s s s s s 2sin2cos2

1

2

1 xy y x y x x +-++=¢ (2.12a)

( ) ( ) q t q s s s s s 2sin2cos2

1

2

1 xy y x y x y ---+=¢ (2.12b)

( ) q t q s s t 2cos2sin21 xy y x y x +--=¢¢ (2.12c)

2.1.9 PRINCIPAL STRESSES IN TWO DIMENSIONS

To ascertain the orientation of y x ¢¢ corresponding to maximum or minimum x¢s , the

necessary condition 0=¢

q

s

d

d x , is applied to Equation (2.12a), yielding

-(s x -s y) sin2q + 2t xy cos2q = 0 (2.13)

Therefore, tan 2q = y x

xy

s s

t

-

2 (2.14)

As 2q = tan (p +2q ), two directions, mutually perpendicular, are found to satisfy equation

(2.14). These are the principal directions, along which the principal or maximum andminimum normal stresses act.



When Equation (2.12c) is compared with Equation (2.13), it becomes clear that 0=¢¢ y x

t on

a principal plane. A principal plane is thus a plane of zero shear. The principal stresses aredetermined by substituting Equation (2.14) into Equation (2.12a)

s 1,2 = 2

y x s s +

±

2

2

2 xy

y x

t

s s +

÷÷ ø

ö

ççè

æ -

(2.15)

Algebraically, larger stress given above is the maximum principal stress, denoted by s 1.

The minimum principal stress is represented by s 2.

Similarly, by using the above approach and employing Equation (2.12c), an expression forthe maximum shear stress may also be derived.

2.1.10 CAUCHY’S STRESS PRINCIPLE

According to the general theory of stress by Cauchy (1823), the stress principle can be statedas follows:

Consider any closed surface S ¶

within a continuum of region B that separates the region B into subregions B1 and B2. The interaction between these subregions can be represented by a

field of stress vectors ( )nT ˆ defined on S ¶ . By combining this principle with Euler’s

equations that expresses balance of linear momentum and moment of momentum in any kind

of body, Cauchy derived the following relationship.

T ( )n̂ = -T ( )n̂-

T ( )n̂ = s T ( )n̂ (2.16)

where ( )n̂ is the unit normal to S ¶ and s is the stress matrix. Furthermore, in regions

where the field variables have sufficiently smooth variations to allow spatial derivatives upto

any order, we have r A = div s + f (2.17)

where r = material mass density

A = acceleration field

f = Body force per unit volume.

This result expresses a necessary and sufficient condition for the balance of linearmomentum. When expression (2.17) is satisfied,

s = s T (2.18)

which is equivalent to the balance of moment of momentum with respect to an arbitrary

point. In deriving (2.18), it is implied that there are no body couples. If body couples and/orcouple stresses are present, Equation (2.18) is modified but Equation (2.17) remainsunchanged.

Cauchy Stress principle has four essential ingradients



(i) The physical dimensions of stress are (force)/(area).

(ii) Stress is defined on an imaginary surface that separates the region under considerationinto two parts.

(iii) Stress is a vector or vector field equipollent to the action of one part of the material onthe other.

(iv) The direction of the stress vector is not restricted.

2.1.11 DIRECTION COSINES

Consider a plane ABC having an outward normal n. The direction of this normal can be

defined in terms of direction cosines. Let the angle of inclinations of the normal with x, y and

z axes be b a , and g respectively. Let ( ) z y xP ,, be a point on the normal at a radial

distance r from the origin O.

Figure 2.8 Tetrahedron with arbitrary plane





Figure 2.9 Stresses acting on face of the tetrahedron

The area of the perpendicular plane PAB, PAC, PBC may now be expressed in terms of A,the area of ABC, and the direction cosines.

Therefore, Area of PAB = APAB = A x = A.i

= A (li + mj + nk ) i

Hence, APAB = Al

The other two areas are similarly obtained. In doing so, we have altogether

APAB = Al, APAC = Am, APBC = An (2.21)

Here i, j and k are unit vectors in x, y and z directions, respectively.

Now, for equilibrium of the tetrahedron, the sum of forces in x, y and z directions must

be zero.

Therefore, T x A = s x Al + t xy Am + t xz An (2.22)



Dividing throughout by A, we get

T x = s x l + t xy m + t xz n (2.22a)

Similarly, for equilibrium in y and z directions,

T y = t xy l + s y m + t yz n (2.22b)

T z = t xz l + t yz m + s z n (2.22c)

The stress resultant on A is thus determined on the basis of known stresses ,,, z y x

s s s

zx yz xy t t t ,, and a knowledge of the orientation of A.

The Equations (2.22a), (2.22b) and (2.22c) are known as Cauchy’s stress formula. These

equations show that the nine rectangular stress components at P will enable one to determine

the stress components on any arbitrary plane passing through point P.

2.1.13 STRESS TRANSFORMATION

When the state or stress at a point is specified in terms of the six components with reference

to a given co-ordinate system, then for the same point, the stress components with referenceto another co-ordinate system obtained by rotating the original axes can be determined usingthe direction cosines.

Consider a cartesian co-ordinate system X, Y and Z as shown in the Figure 2.10. Let this

given co-ordinate system be rotated to a new co-ordinate system z,y,x ¢¢¢ where

in x¢ lie on an oblique plane. z,y,x ¢¢¢ and X, Y, Z systems are related by the direction

cosines.

l1 = cos ( x¢ , X )

m1 = cos ( x¢ , Y ) (2.23)

n1 = cos ( x¢ , Z )

(The notation corresponding to a complete set of direction cosines is shown in

Table 1.0).

Table 1.0 Direction cosines relating different axes

X Y Z

'x

l1 m1 n1

y¢ l2 m2 n2

z¢ l3 m3 n3



Figure 2.10 Transformation of co-ordinates

The normal stress x¢s is found by projecting y x

T T , and T z in the x¢ direction and adding:

x¢s = T x l1 + T y m1 + T z n1 (2.24)

Equations (2.22a), (2.22b), (2.22c) and (2.24) are combined to yield

x¢s = s x l 2

1 + s y m2

1 + s z n2

1 + 2(t xy l1 m1 +t yz m1 n1 + t xz l1 n1) (2.25)

Similarly by projecting z y xT T T ,, in the y¢ and z¢ directions, we obtain, respectively

y x ¢¢t =s x l1 l2+s y m1 m2+s z n1 n2+t xy (l1 m2+ m1 l2)+t yz (m1 n2 + n1 m2 ) + t xz (n1l2 + l1n2)

(2.25a)

z x ¢¢t =s x l1 l3 +s y m1 m3+s z n1 n3 +t xy (l1 m3 + m1 l3)+t yz (m1 n3 + n1 m3)+t xz (n1 l3+ l1 n3)

(2.25b)

Recalling that the stresses on three mutually perpendicular planes are required to specify thestress at a point (one of these planes being the oblique plane in question), the remainingcomponents are found by considering those planes perpendicular to the oblique plane. For

one such plane n would now coincide with y¢ direction, and expressions for the stresses



z y y y ¢¢¢¢ t t s ,, would be derived. In a similar manner the stresses y z x z z ¢¢¢¢¢ t t s ,, are

determined when n coincides with the z¢ direction. Owing to the symmetry of stress tensor,

only six of the nine stress components thus developed are unique. The remaining stresscomponents are as follows:

y¢s = s x l2

2 + s y m2

2 + s z n2

2 + 2 (t xy l2 m2 + t yz m2 n2 + t xz l2 n2) (2.25c)

z¢s = s x l2

3 + s y m2

3 + s z n2

3 + 2 (t xy l3 m3 + t yz m3 n3 + t xz l3 n3) (2.25d)

z y ¢¢t = s x l2 l3 +s y m2 m3 +s z n2 n3+t xy (m2 l3 + l2 m3)+t yz (n2 m3 + m2 n3)+t xz (l2 n3 + n2 l3)

(2.25e)

The Equations (2.25 to 2.25e) represent expressions transforming the quantities

xz yz xy y x t t t s s ,,,, to completely define the state of stress.

It is to be noted that, because x¢ , y¢ and z¢ are orthogonal, the nine direction cosines must

satisfy trigonometric relations of the following form.

l 2

i + m 2

i + n 2

i = 1 (i = 1,2,3)

and l1 l2 + m1 m2 + n1 n2 = 0

l2 l3 + m2 m3 + n2 n3 = 0 (2.26)

l1 l3 + m1 m3 + n1 n3 = 0





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n

T

m

T

l

T z y x == (2.27c)

These proportionalities indicate that the stress resultant must be parallel to the unit normal

and therefore contains no shear component. Therefore from Equations (2.22a), (2.22b),

(2.22c) we can write as below denoting the principal stress by Ps

T x = s P l T y = s P m T z = s P n (2.27d)

These expressions together with Equations (2.22a), (2.22b), (2.22c) lead to

(s x - s P)l + t xy m + t xz n = 0

t xy l+(s y - s P) m + t yz n = 0 (2.28)

t xz l + t yz m + (s z - s P) n = 0

A non-trivial solution for the direction cosines requires that the characteristic determinantshould vanish.

0

)(

)(

)(

=

-

-

-

P z yz xz

yzP y xy

xz xyP x

s s t t

t s s t

t t s s

(2.29)

Expanding (2.29) leads to 032

2

1

3 =-+- I I I PPP

s s s (2.30)

where I 1 = s x + s y + s z (2.30a)

I 2 = s x s y + s y s z + s zs x - t 2

xy - t 2yz -t 2xz (2.30b)

I 3 =

z yz xz

yz y xy

xz xy x

s t t

t s t

t t s

(2.30c)

The three roots of Equation (2.30) are the principal stresses, corresponding to which arethree sets of direction cosines that establish the relationship of the principal planes to theorigin of the non-principal axes.

2.2.2 STRESS INVARIANTS

Invariants mean those quantities that are unexchangeable and do not vary under differentconditions. In the context of stress tensor, invariants are such quantities that do not change

with rotation of axes or which remain unaffected under transformation, from one set of axes



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to another. Therefore, the combination of stresses at a point that do not change with the

orientation of co-ordinate axes is called stress-invariants. Hence, from Equation (2.30)

s x + s y + s z = I 1 = First invariant of stress

s xs y + s ys z + s zs x - t 2

xy - t 2yz - t 2zx = I 2 = Second invariant of stress

s xs ys z - s xt 2

yz - s yt 2xz - s zt

2

xy + 2t xy t yz t xz = I 3 = Third invariant of stress

2.2.3 EQUILIBRIUM OF A DIFFERENTIAL ELEMENT

Figure 2.11(a) Stress components acting on a plane element

When a body is in equilibrium, any isolated part of the body is acted upon by an equilibrium

set of forces. The small element with unit thickness shown in Figure 2.11(a) represents part



Module2/Lesson2

4


of a body and therefore must be in equilibrium if the entire body is to be in equilibrium. It is

to be noted that the components of stress generally vary from point to point in a stressed

body. These variations are governed by the conditions of equilibrium of statics. Fulfillment

of these conditions establishes certain relationships, known as the differential equations of

equilibrium. These involve the derivatives of the stress components.

Assume that s x, s y, t xy, t yx are functions of X, Y but do not vary throughout the thickness

(are independent of Z ) and that the other stress components are zero.

Also assume that the X and Y components of the body forces per unit volume, F x and F y,

are independent of Z , and that the Z component of the body force F z = 0. As the element is

very small, the stress components may be considered to be distributed uniformly over each

face.

Now, taking moments of force about the lower left corner and equating to zero,

( ) ( )

( ) 02

)(2

2

1

22

22

1

2

=D

DD-D

DD

+D-D

D+D

D÷ ø

öçè

æ D

¶

¶++DD÷÷

ø

öççè

æ D

¶

¶+-

DD÷

÷

ø

ö

ç

ç

è

æ D

¶

¶++

DD÷

÷

ø

ö

ç

ç

è

æ D

¶

¶+-D+

DD-

x y xF

y x yF

x x

x y

y x x

y x x x

y x y y

x x y

y y

y y

y x

yx y

x

x

xy

xy

yx

yx

y

y xy x

t s s

s t

t

t t

s s t s

Neglecting the higher terms involving D x, and D y and simplifying, the above expression is

reduced to

t xy D x D y = t yx D x D y

or t xy = t yx

In a like manner, it may be shown that

t yz = t zy and t xz = t zx

Now, from the equilibrium of forces in x-direction, we obtain

-s x D y + 0=DD+D-D÷÷ ø

öççè

æ D

¶

¶++D÷

ø

öçè

æ D

¶

¶+ y xF x x y

y y x

x x yx

yx

yx x

x t t

t s

s

Simplifying, we get

0=+¶

¶

+¶

¶ x

yx x

F y x

t s

or 0=+¶

¶+

¶

¶ x

xy x F y x

t s



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A similar expression is written to describe the equilibrium of y forces. The x and y equations

yield the following differential equations of equilibrium.

0=+¶

¶+

¶

¶ x

xy x F y x

t s

or 0=+¶

¶+

¶

¶ y

xy yF

x y

t s (2.31)

The differential equations of equilibrium for the case of three-dimensional stress may begeneralized from the above expressions as follows [Figure 2.11(b)].

0=+¶

¶+

¶

¶+

¶

¶ x

xz xy x F z y x

t t s

0=+¶

¶+

¶

¶+

¶

¶ y

yz xy yF

z x y

t t s (2.32)

0=+¶

¶+¶

¶+¶

¶ z

yz xz z F y x z

t t s

Figure 2.11(b) Stress components acting on a three dimensional element



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2.2.4 OCTAHEDRAL STRESSES

A plane which is equally inclined to the three axes of reference, is called the octahedral plane

and its direction cosines are

3

1,

3

1,

3

1±±± . The normal and shearing stresses acting

on this plane are called the octahedral normal stress and octahedral shearing stress

respectively. In the Figure 2.12, X, Y, Z axes are parallel to the principal axes and the

octahedral planes are defined with respect to the principal axes and not with reference to an

arbitrary frame of reference.

(a) (b)

Figure 2.12 Octahedral plane and Octahedral stresses

Now, denoting the direction cosines of the plane ABC by l, m, and n, the equations (2.22a),

(2.22b) and (2.22c) with 0,1 === xz xy x t t s s etc. reduce to

T x = 1s l, T y = s 2 m and T z = s 3 n (2.33)

The resultant stress on the oblique plane is thus2222

3

22

2

22

1

2 t s s s s +=++= nmlT

\ T 2 = s 2 + t 2 (2.34)



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The normal stress on this plane is given by

s = s 1 l2 + s 2 m

2 + s 3 n2 (2.35)

and the corresponding shear stress is

( ) ( ) ( )[ ]2

1222

13

222

32

222

21 lnnmml s s s s s s t -+-+-= (2.36)

The direction cosines of the octahedral plane are:

l = ± 3

1 ,m = ±

3

1 , n = ±

3

1

Substituting in (2.34), (2.35), (2.36), we get

Resultant stress T = )(3

1 2

3

2

2

2

1 s s s ++ (2.37)

Normal stress = s =3

1 (s 1+s 2+s 3) (2.38)

Shear stress = t = 2

13

2

32

2

21 )()()(3

1s s s s s s -+-+- (2.39)

Also, t = )(6)(23

1313221

2

321 s s s s s s s s s ++-++ (2.40)

t = 2

2

1 623

1 I I - (2.41)

2.2.5 MOHR'S STRESS CIRCLE

A graphical means of representing the stress relationships was discovered byCulmann (1866) and developed in detail by Mohr (1882), after whom the graphical methodis now named.

2.2.6 MOHR CIRCLES FOR TWO DIMENSIONAL STRESS SYSTEMS

Biaxial Compression (Figure 2.13a)

The biaxial stresses are represented by a circle that plots in positive (s , t) space, passing

through stress points s 1 , s 2 on the t = 0 axis. The centre of the circle is located on the



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t = 0 axis at stress point ( )212

1s s + . The radius of the circle has the magnitude

( )212

1s s - , which is equal to t max.

Figure 2.13 Simple Biaxial stress systems: (a) compression,

(b) tension/compression, (c) pure shear

(c)

t zy

t zy

.t zy

ss2 s1

- +. t

yz

-

+t

(a)

s1

s1

s2 s2 .+t

-

-

s2 s1

+

s

(b)

s1

s1

s2 s2 .-

+t

s1s2 s

+2q

-



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Biaxial Compression/Tension (Figure 2.13b)

Here the stress circle extends into both positive and negative s space. The centre of the

circle is located on the t = 0 axis at stress point ( )212

1s s + and has radius ( )21

2

1s s - .

This is also the maximum value of shear stress, which occurs in a direction at 45

o

to the s 1 direction. The normal stress is zero in directions ±q to the direction of s 1, where

cos2q = -21

21

s s

s s

-

+

Biaxial Pure Shear (Figure 2.13c)

Here the circle has a radius equal to t zy, which is equal in magnitude to , yzt but opposite in

sign. The centre of circle is at s = 0, t = 0. The principal stresses s 1 , s 2 are equal in

magnitude, but opposite in sign, and are equal in magnitude to t zy. The directions of s 1 , s 2

are at 45o to the directions of yz zy t t ,

2.2.7 CONSTRUCTION OF MOHR’S CIRCLE FOR TWO-

DIMENSIONAL STRESS

Sign Convention

For the purposes of constructing and reading values of stress from Mohr’s circle, the signconvention for shear stress is as follows.

If the shearing stresses on opposite faces of an element would produce shearing forces that result in a clockwise couple, these stresses are regarded as "positive".

Procedure for Obtaining Mohr’s Circle

1) Establish a rectangular co-ordinate system, indicating +t and +s . Both stress scalesmust be identical.

2) Locate the centre C of the circle on the horizontal axis a distance ( )Y X

s s +2

1 from the

origin as shown in the figure above.

3) Locate point A by co-ordinates xy x t s -,

4) Locate the point B by co-ordinates xy y t s ,

5) Draw a circle with centre C and of radius equal to CA.

6) Draw a line AB through C .



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Figure 2.14 Construction of Mohr’s circle

An angle of 2q on the circle corresponds to an angle of q on the element. The state of stress

associated with the original x and y planes corresponds to points A and B on the circle

respectively. Points lying on the diameter other than AB, such as A¢ and B¢ , define state of

stress with respect to any other set of x¢ and y¢ planes rotated relative to the original set

through an angle q .

q

s y

s y

s x

s x

t xy

A

B

C

Ty

Tx

x¢ y¢

s x

t xy

s y

s x¢

t xy

q

. y¢

.

..

... .

s¢= ( )s +s x y2

1

s2

yB( )s t y xy,

s1

E

tmax

s

D

O2q

C

B¢

B1

-tmax

A1

A¢

A( )s -t x xy,

x¢

t

x



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It is clear from the figure that the points A1 and B1 on the circle locate the principal stresses

and provide their magnitudes as defined by Equations (2.14) and (2.15), while D and E

represent the maximum shearing stresses. The maximum value of shear stress (regardless of

algebraic sign) will be denoted by t max and are given by

t max = ± ( )212

1s s - = ± 2

2

2 xy

y xt

s s +÷÷

ø

öççè

æ - (2.42)

Mohr’s circle shows that the planes of maximum shear are always located at 45 o from planesof principal stress.

2.2.8 MOHR’S CIRCLE FOR THREE-DIMENSIONAL STATE OF

STRESS

When the magnitudes and direction cosines of the principal stresses are given, then thestresses on any oblique plane may be ascertained through the application of Equations (2.33)

and (2.34). This may also be accomplished by means of Mohr’s circle method, in which theequations are represented by three circles of stress.

Consider an element as shown in the Figure 2.15, resulting from the cutting of a small cube by an oblique plane.

(a)



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(b)

Figure 2.15 Mohr's circle for Three Dimensional State of Stress

The element is subjected to principal stresses s 1 , s 2 and s 3 represented as coordinate axes

with the origin at P. It is required to determine the normal and shear stresses acting at point

Q on the slant face (plane abcd). This plane is oriented so as to be tangent at Q to a quadrant

of a spherical surface inscribed within a cubic element as shown. It is to be noted that PQ,

running from the origin of the principal axis system to point Q, is the line of intersection of

the shaded planes (Figure 2.15 (a)). The inclination of plane PA2QB3 relative to the s 1 axis

is given by the angle q (measured in the s 1 , s 3 plane), and that of plane PA3QB1, by the

angle F (measured in the s 1 and s 2 plane). Circular arcs A1 B1 A2 and A1 B3 A3 are located on

the cube faces. It is clear that angles q and F unambiguously define the orientation of PQ

with respect to the principal axes.



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Procedure to determine Normal Stress ( ) and Shear Stress ( )

1) Establish a Cartesian co-ordinate system, indicating +s and +t as shown. Lay off the

principal stresses along thes -axis, with s 1 > s 2 > s 3 (algebraically).

2) Draw three Mohr semicircles centered at C 1, C 2 and C 3 with diameters A1 A2, A2 A3 and A1 A3.

3) At point C 1, draw line C 1 B1 at angle 2f ; at C 3, draw C 3 B3 at angle 2q . These lines cut

circles C 1 and C 3 at B1 and B3 respectively.

4) By trial and error, draw arcs through points A3 and B1 and through A2 and B3, with their

centres on the s -axis. The intersection of these arcs locates point Q on the s , t plane.

In connection with the construction of Mohr’s circle the following points are of

particular interest:

a) Point Q will be located within the shaded area or along the circumference of circles C 1,

C 2 or C 3, for all combinations of q and f .

b) For particular case q = f = 0, Q coincides with A1.

c) When q = 450, f = 0, the shearing stress is maximum, located as the highest point

on circle C 3 (2q = 900). The value of the maximum shearing stress is therefore

( )31max2

1s s t -= acting on the planes bisecting the planes of maximum and minimum

principal stresses.

d) When q = f = 450, line PQ makes equal angles with the principal axes. The oblique plane is, in this case, an octahedral plane, and the stresses along on the plane, theoctahedral stresses.

2.2.9 GENERAL EQUATIONS IN CYLINDRICAL CO-ORDINATES

While discussing the problems with circular boundaries, it is more convenient to use the

cylindrical co-ordinates, r, q , z. In the case of plane-stress or plane-strain problems, we have

0== z zr q t t and the other stress components are functions of r and q only. Hence the

cylindrical co-ordinates reduce to polar co-ordinates in this case. In general, polar

co-ordinates are used advantageously where a degree of axial symmetry exists. Examplesinclude a cylinder, a disk, a curved beam, and a large thin plate containing a circular hole.



Module2/Lesson2

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2.2.10 EQUILIBRIUM EQUATIONS IN POLAR CO-ORDINATES:

(TWO-DIMENSIONAL STATE OF STRESS)

Figure 2.16 Stresses acting on an element

The polar coordinate system (r , q ) and the cartesian system ( x, y) are related by the following

expressions:

x = r cosq , r 2 = x

2+y

2

y = r sinq , ÷ ø

öçè

æ = -

x

y1tanq (2.43)

Consider the state of stress on an infinitesimal element abcd of unit thickness described by

the polar coordinates as shown in the Figure 2.16. The body forces denoted by F r and F q aredirected along r and q directions respectively.

Resolving the forces in the r -direction, we have for equilibrium, S F r = 0,





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Module 2: Analysis of Stress

2.3.1 GENERAL STATE OF STRESS IN THREE-DIMENSION IN

CYLINDRICAL CO-ORDINATE SYSTEM

Figure 2. 17 Stresses acting on the element

In the absence of body forces, the equilibrium equations for three-dimensional state aregiven by



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10r r r zr

r r z r (2.47)

210r z r

r r z r (2.48)

01

r zr r

zr z z zr (2.49)

2.25 NUMERICAL EXAMPLES

Example 2.1

When the stress tensor at a point with reference to axes (x, y, z) is given by the array,

802

061

214

MPa

show that the stress invariants remain unchanged by transformation of the axes by 450

about the z-axis,

Solution: The stress invariants are

I 1 = 4 + 6 + 8 = 18 MPa

I 2 = 4 6+6 8+4 8-1 1-2 2-0 = 99 MPa

I 3 = 4 48-1 8+2 (-12) = 160 MPa

The direction cosines for the transformation are given by

x y z

x

2

1

2

1

0

y -

2

1

2

1

0

z 0 0 1

Using Equations (2.21a), (2.21b), (2.21c), (2.21d), (2.21e), (2.21f), we get

MPa

x

6

002

1120

2

16

2

14



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MPa

y

4

002

1120

2

16

2

14

MPa

z

8

0001800

MPa

y x

1

002

1

2

110

2

16

2

14

MPa

z y

2

2

1200000

MPa

z x

2

2

1200000

Hence the new stress tensor becomes

822

241

216

MPa

Now, the new invariants are

188461 I MPa

992218684462 I MPa

1602

521013063 I MPa

which remains unchanged. Hence proved.

Example 2.2

The state-of-stress at a point is given by the following array of terms

423

256

369

MPa

Determine the principal stresses and principal directions.

Solution: The principal stresses are the roots of the cubic equation



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3 – I 1 2 + I 2 - I 3 = 0

Here 184591 I MPa

52326494559222

2 I MPa

27326236495494593 I MPa

The cubic equation becomes

3 - 18 2 + 52 - 27 = 0

The roots of the cubic equation are the principal stresses. Hence the three principal

stresses are

1 = 14.554 MPa; 2 = 2.776 MPa and 3 = 0.669 MPa

Now to find principal directions for major principal stress 1

)554.144(23

2)554.145(6

36)554.149(

=

554.1023

2554.96

36554.5

A =554.102

2554.9=100.83 - 4 = 96.83

B = 554.103

26

= -(-63.324 - 6) = 69.324

C =23

554.96= 12 + 28.662 = 40.662

222C B A

=222

662.40324.6983.96

= 125.83

l1 =222

C B A

A =

83.125

53.96 = 0.769





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300

Y

X

y14 10

6

x28 10

6

Figure 2.20

The planes on which these stresses act are represented by000 15.734515.28s

and015.163s

Figure 2.19 Mohr’s stress circle

Example 2.4

The stress (in N/m2) acting on an element of a loaded body is shown in Figure 2.20.

Apply Mohr’s circle to determine the normal and shear stresses acting on a plane

defined by = 300 .

Solution: The Mohr’scircle drawn belowdescribes the state ofstress for the givenelement. Points A1 and

y

x

66.31

16.52

28.150

x

y

24.9max 41.4

73.150

X

Y

O

E

D

.

.

FA1

B1 C

B(27.6, 20.7)

A(55.2, 20.7)

2 s

2 p

.

.

...



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B1 represent thestress components on the x and y faces, respectively.

The radius of the circle is6

6

10212

102814 . Corresponding to the 300

plane within

the element, it is necessary to rotate through 60

0

counterclockwise on the circle to locate point A . A 2400 counterclockwise rotation locates point B .

(a)

(b)

Figure 2.21 Mohr’s stress circle

O C .60

0Y

XB ( 14 10 ,0)1

6

A (28 10 ,0)1

6

.

.. ..

A

By

x

300

y x

x y18.186 10

6

x17.5 10

6 y

3.5 106



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From the above Mohr’s circle,2660 /105.171060cos217 m N x

26 /105.3 m N y

2606 /1086.1860sin1021 m N y x

Example 2.5

A rectangular bar of metal of cross-section 30 mm 25 mm is subjected to an axial tensile

force of 180KN. Calculate the normal, shear and resultant stresses on a plane whose

normal has the following direction cosines:

(i) 0and 2

1nml

(ii)3

1nml

Solution: Let normal stress acting on the cross-section is given by y .

areasectionalcross

load Axial y

2530

10180 3

2/240 mm N

Now, By Cauchy’s formula, the stress components along x, y and z co-ordinates are

nmlT

nmlT

nmlT

z yz xz z

yz y xy y

xz xy x x

(a)

And the normal stress acting on the plane whose normal has the direction cosines l, m and n

is,

nT mT lT z y x (b)

Case (i) For 02

1nand ml

Here2/240,0,0 mm N y xy x

0,0,0 z yz xz

Substituting the above in (a), we get

0,2

240,0 z y y x T mT T





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Example 2.6

A body is subjected to three-dimensional forces and the state of stress at a point in it is

represented as

MPa

100200200

200100200

200200200

Determine the normal stress, shearing stress and resultant stress on the octahedral

plane.

Solution: For the octahedral plane, the direction cosines are

3

1nml

Here MPa x 200

MPa y

100

MPa y 100

MPa zx yz xy 200

Substituting the above in Cauchy’s formula, we get

MPaT x 41.346

3

1200

3

1200

3

1200

MPaT y 20.173

3

1200

3

1100

3

1200

MPaT z

20.1733

1100

3

1200

3

1200

Normal stress on the plane is given by

nT mT lT z y x ..

3

120.173

3

120.173

3

141.346

MPa400

Resultant Stress =222

z y x T T T T

222

20.17320.17341.346

MPaT 26.424

Also, Tangential Stress =22

40026.424



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MPa41.141

Example 2.7

The state of stress at a point is given as follows:

kPakPakPa

kPakPakPa

zx yz xy

z y x

500,600,400

400,1200,800

Determine (a) the stresses on a plane whose normal has direction cosines2

1,

4

1ml

and (b) the normal and shearing stresses on that plane.

Solution: We have the relation,

4

11

12

1

4

1

1

2

22

222

n

n

nml

(a) Using Cauchy’s formula,

kPaT x 60.4144

11500

2

1400

4

1800

kPaT y 51.2024

11600

2

11200

4

1400

kPaT z 66.5064

11400

2

1600

4

1500

(b) Normal stress,

nT mT lT z y x

=4

1166.506

2

151.202

4

160.414

kPa20.215

Resultant Stress on the Plane =222

66.50651.20260.414T

= 685.28 MPa



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Shear Stress on the plane =22

20.21528.685

= 650.61 kPa

Example 2.8

Given the state of stress at a point as below

4000

06090080100

kPa

Considering another set of coordinate axes, z y x in which z coincides with z and x

is rotated by 300 anticlockwise from x-axis, determine the stress components in the new

co-ordinates system.

Solution: The direction cosines for the transformation are given by

X y z

x 0.866 0.5 0

y -0.5 0.866 0 z 0 0 1

Figure 2.22 Co-ordinate system

Now using equations 2.21(a), 2.21(b), 2.21(c), 2.21(d), 2.21(e) and 2.21(f), we get

300

300

Z z

X

Y

y

x



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005.0866.080205.060866.010022

1 x

kPa x

3.129

00866.05.08020866.0605.010022

y

kPa y 3.89

000214000 2

z

kPa z 40

005.05.0866.0866.0800866.05.0605.0866.0100 y x

kPa y x

3.29

0 z y and 0 x z

Therefore the state of stress in new co-ordinate system is

4000

03.893.29

03.293.129

(kPa)

Example 2.9

The stress tensor at a point is given by the following array

)(

301040

102020

402050

kPa

Determine the stress-vectors on the plane whose unit normal has direction cosines

21,

21,

21

Solution: The stress vectors are given by

nmlT xz xy x x

(a)

nmlT yz y xy y (b)

nmlT z yz xz z (c)

Substituting the stress components in (a), (b) and (c) we get

2

140

2

120

2

150 x

T = kPa35.45

2

110

2

120

2

120

yT = kPa858.0



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2

130

2

110

2

140 zT = kPa28.48

Now, Resultant Stress is given by

kPak jiT ˆ28.48ˆ858.0ˆ35.45

Example 2.10

The Stress tensor at a point is given by the following array

)(

204030

403020

302040

kPa

Calculate the deviator and spherical stress tensors.

Solution: Mean Stress = z y xm3

1

2030403

1

kPa30

Deviator stress tensor =

m z yz xz

yzm y xy

xz xym x

=

30204030

40303020

30203040

= kPa

104030

40020

302010

Spherical Stress tensor =

m

m

m

0000

00



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15


= kPa

3000

0300

0030

Example 2.11

The Stress components at a point in a body are given by

z y y x

xy z xy y xyz

x z xy

z

xz yz y

xy x

22

2

2

2335

0,23

Determine whether these components of stress satisfy the equilibrium equations or not

as the point (1, -1, 2). If not then determine the suitable body force required at this

point so that these stress components are under equilibrium.

Solution: The equations of equilibrium are given by

0 z y x

xz xy x (a)

0 z y x

yz y xy (b)

0 z y x

z yz xz (c)

Differentiating the stress components with respective axes, we get

22 3,0,23 xy z y

z y x

xz xy x

Substituting in (a), 22 3023 xy z y

At point (1, -1, 2), we get 111132213 which is not equal to zero

Similarly,

03,35 2 xy

z

xz

y

yz y

(ii) becomes23350 xy xz

At point (1, -1, 2), we get 161133215 which is not equal to zero



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16


And y z y x

x xyz y

y z

xz yz z 23,26, 22

Therefore (iii) becomes22 2623 y x xyz y z y

At the point (1, -1, 2), we get 2

112211612213 = -5 which

is not equal to zero.

Hence the given stress components does not satisfy the equilibrium equations.

Recalling (a), (b) and (c) with body forces, the equations can be modified as below.

0 x

xz xy x F z y x

(d)

0 y

yz y xyF

z y x (e)

0 z z yz xz F

z y x (f)

Where F x , F y and F z are the body forces.

Substituting the values in (d), (e) and (f), we get body forces so that the stress components become under equilibrium.

Therefore,

11

01132213

x

x

F

F

Also, 01133215 yF

16 yF

and 0)1(122)1(16)1(2213 2

zF

5 zF

The body force vector is given by

k jiF ˆ5ˆ16ˆ11

Example 2.12

The rectangular stress components at a point in a three dimensional stress systemare as follows.



Module2 / Lesso n3

17


222

222

/20/60/40

/80/40/20

mm N mm N mm N

mm N mm N mm N

zx yz xy

z y x

Determine the principal stresses at the given point.

Solution: The principal stresses are the roots of the cubic equation

032

2

1

3 I I I

The three dimensional stresses can be expressed in the matrix form as below.

2/

806020

604040

204020

mm N

z yz xz

yz y xy

xz xy x

Here z y x I 1

= )804020(

= 60

zx yz xy x z z y y x I 222

2

=222 )20()60()40()20(80)80)(40()40(20

= -8000

xz yz xy xy z zx y yz x z y x I 2222

3

= 20(-40)(80)-(20)(-60)2-(-40)(20)

2-80(40)

2+2(40)(-60)(20)

= -344000

Therefore Cubic equation becomes

0344000800060 23 (a)

Now cos3cos43cos 3

Or 03cos4

1cos

4

3cos3

(b)

Put3

cos 1 I r

i.e.,3

60cosr

20cosr

Substituting in (a), we get

034400020cos800020cos6020cos23

r r r



Module2 / Lesso n3

18


034400020cos800020cos6020cos20cos22

r r r r

0344000160000cos8000

cos40400cos6020coscos40400cos 2222

r

r r r r r

0168000cos9200cos33r r

0168000

cos92001

cos.,.32

3

r r ei (c)

Hence equations (b) and (c) are identical if

4

392002

r

3

49200r

755.110

and3

168000

4

3cos

r

3755.110

41680003cos = 0.495

or 495.03cos

0

1 9.3965.1193 or

0

2 1.80 and0

3 9.159

3cos 1

11 I r

3

60)9.39cos(755.110

2/96.104 mm N

3cos 1

222

I r

3

60)1.80cos(755.110

2

2 /04.39 mm N

3cos 1

333

I r





Module2 / Lesso n3

20


628.008.1027

29.645

2221

C B A

Al

628.0

08.1027

646

2221

C B A

Bm

458.008.1027

3.470

2221

C B A

C n

(b) To find principal plane for Stress 2

201010102020

102020

)1010(101010)1010(20

10201010

3001004002010

1020 A

300)100400(2010

1020 B

0)200200(1010

2020C

26.424)0()300(300 222222C B A

707.026.424

300

2222

C B A

Al

707.026.424

300

2222

C B A

Bm

0222

2

C B A

C n



Module2 / Lesso n3

21


(c) To find principal plane for Stress 3

3.71010

103.720

10203.7

)7.210(1010

10)7.210(20

10207.210

71.4610029.533.710

103.7 A

46)100146(3.710

1020 B

127)73200(1010

3.720C

92.142)127()46(71.46 222222C B A

326.092.142

71.46

2223

C B A

Al

322.092.142

46

2223

C B A

Bm

888.092.142

127

2223

C B A

C n



Module3/Lesson1

1


Module 3 : Analysis of Strain

3.1.1 INTRODUCTION

o define normal strain, refer to the following Figure 3.1 where line AB of an axially

loaded member has suffered deformation to become B A ¢¢ .

Figure 3.1 Axially loaded bar

The length of AB is D x. As shown in Figure 3.1(b), points A and B have each been displaced,

i.e., at point A an amount u, and at point B an amount u+ Du. Point B has been displaced by

an amount Du in addition to displacement of point A, and the length D x has been increased

by Du. Now, normal strain may be defined as

dx

du

x

u

x x =

D

D=

®D 0lime (3.0)

In view of the limiting process, the above represents the strain at a point. Therefore "Strain

is a measure of relative change in length, or change in shape".

T



Module3/Lesson1

2


3.1.2 TYPES OF STRAIN

Strain may be classified into direct and shear strain.

Figure 3.2 Types of strains

(a)

(b)

(c) (d)



Module3/Lesson1

3


Figure 3.2(a), 3.2(b), 3.2(c), 3.2(d) represent one-dimensional, two-dimensional,three-dimensional and shear strains respectively.

In case of two-dimensional strain, two normal or longitudinal strains are given by

e x = xu¶¶ , e y =

yv

¶¶ (3.1)

+ ve sign applies to elongation; –ve sign, to contraction.

Now, consider the change experienced by right angle DAB in the Figure 3.2 (d). The total

angular change of angle DAB between lines in the x and y directions, is defined as the

shearing strain and denoted by g xy.

\ g xy =a x + a y = y

u

¶

¶ +

x

v

¶

¶ (3.2)

The shear strain is positive when the right angle between two positive axes decreasesotherwise the shear strain is negative.

In case of a three-dimensional element, a prism with sides dx, dy, dz as shown in Figure

3.2(c) the following are the normal and shearing strains:

z

w

y

v

x

u z y x

¶

¶=

¶

¶=

¶

¶= e e e ,, (3.3)

z

u

x

w

y

w

z

v

x

v

y

u zx yz xy

¶

¶+

¶

¶=

¶

¶+

¶

¶=

¶

¶+

¶

¶= g g g ,,

The remaining components of shearing strain are similarly related:

xz zx zy yz yx xy g g g g g g === ,, (3.4)



Module3/Lesson1

4


3.1.3 DEFORMATION OF AN INFINITESIMAL LINE ELEMENT

Figure 3.3 Line element in undeformed and deformed body

Figure 3.3 Line element in undeformed and deformed body

Consider an infinitesimal line element PQ in the undeformed geometry of a medium as

shown in the Figure 3.3. When the body undergoes deformation, the line element PQ passes

into the line element QP ¢¢ . In general, both the length and the direction of PQ are changed.

Let the co-ordinates of P and Q before deformation be ( ) ( ) z z y y x x z y x D+D+D+ ,,,,,

respectively and the displacement vector at point P have components (u, v, w).

The co-ordinates of P, P¢ and Q are

( ) z y xP ,,:

( )w zv yu xP +++¢ ,,:



Module3/Lesson1

5


( ) z z y y x xQ D+D+D+ ,,:

The displacement components at Q differ slightly from those at point P since Q is away

from P by y x DD , and zD .

\ The displacements at Q are

vvuu D+D+ , and ww D+

Now, if Q is very close to P, then to the first order approximation

z z

u y

y

u x

x

uu D

¶

¶+D

¶

¶+D

¶

¶=D (a)

Similarly, z z

v y

y

v x

x

vv D

¶

¶+D

¶

¶+D

¶

¶=D (b)

And z zw y

yw x

xww D

¶¶+D

¶¶+D

¶¶=D (c)

The co-ordinates of Q¢ are, therefore,

( )ww z zvv y yuu x xQ D++D+D++D+D++D+¢ ,,

Before deformation, the segment PQ had components y x DD , and zD along the three axes.

After deformation, the segment QP ¢¢ has components v yu x +D+D , and w z +D along the

three axes.

Here the terms like y

u

x

u

¶¶

¶¶ , and

z

u

¶¶ etc. are important in the analysis of strain. These are the

gradients of the displacement components in x, y and z directions. These can be represented

in the form of a matrix called the displacement-gradient matrix such as

ú

úúúúú

û

ù

ê

êêêêê

ë

é

¶

¶

¶

¶

¶

¶¶

¶

¶

¶

¶

¶¶

¶

¶

¶

¶

¶

=úúû

ù

êêë

é

¶

¶

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

u

x

u

j

i



Module3/Lesson1

6


3.1.4 CHANGE IN LENGTH OF A LINEAR ELEMENT

When the body undergoes deformation, it causes a point P( x, y, z) in the body under

consideration to be displaced to a new position P¢with co-ordinates ( )w zv yu x +++ ,,

where u, v and w are the displacement components. Also, a neighbouring point Q with co-

ordinates ( ) z z y y x x D+D+D+ ,, gets displaced to Q¢with new co-ordinates

( )ww z zvv y yuu x x D++D+D++D+D++D+ ,, .

Now, let S D be the length of the line element PQ with its components ( ) z y x DDD ,, .

( ) ( ) ( ) ( ) ( )22222 z y xPQS D+D+D==D\

Similarly, S ¢D be the length QP ¢¢ with its components

( )w z zv y yu x x D+D=¢DD+D=¢DD+D=¢D ,,

( ) ( ) ( ) ( ) ( )22222w zv yu xQPS D+D+D+D+D+D=¢¢=¢D\

From equations (a), (b) and (c),

z z

u y

y

u x

x

u x D

¶

¶+D

¶

¶+D÷

ø

öçè

æ

¶

¶+=¢D 1

z z

v y

y

v x

x

v y D

¶

¶+D÷÷

ø

öççè

æ

¶

¶++D

¶

¶=¢D 1

z z

w y y

w x x

w z D÷ ø

öçè

æ

¶

¶

++D¶

¶

+D¶

¶

=¢D 1

Taking the difference between ( )2S ¢D and ( )2

S D , we get

( ) ( ) ( ) ( )2222S S PQQP D-¢D=-¢¢

( ) ( ) ( ) ) ( ) ( ) ( ) ){ }222222 z y x z y x D+D+D-¢D+¢D+¢D

( ) z x z y y x z y x zx yz xy z y x DD+DD+DD+D+D+D= e e e e e e 2222 (3.5)

where

úúû

ù

êêë

é÷ ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶+

¶

¶=

222

2

1

x

w

x

v

x

u

x

u xe (3.5a)



Module3/Lesson1

7


úúû

ù

êêë

é÷÷ ø

öççè

æ

¶

¶+÷÷

ø

öççè

æ

¶

¶+÷÷

ø

öççè

æ

¶

¶+

¶

¶=

222

2

1

y

w

y

v

y

u

y

v ye (3.5b)

úúû

ù

êêë

é÷ ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶+

¶

¶=

222

2

1

z

w

z

v

z

u

z

w

ze (3.5c)

úû

ùêë

é

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶+

¶

¶==

y

w

x

w

y

v

x

v

y

u

x

u

y

u

x

v yx xy e e (3.5d)

úû

ùêë

é

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶+

¶

¶==

z

w

y

w

z

v

y

v

z

u

y

u

z

v

y

w zy yz e e (3.5e)

úû

ùêë

é

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶

¶

¶+

¶

¶+

¶

¶==

x

w

z

w

x

v

z

v

x

u

z

u

x

w

z

u xz zx e e (3.5f)

Now, introducing the notation

S

S S PQ

D

D-¢D=e

which is called the relative extension of point P in the direction of point Q, now,

( ) ( ) ( )( )

( )2

2

222

22S

S

S S

S

S S S S D÷

÷ ø

öççè

æ

D

D-¢D+

D

D-¢D=

D-¢D

( ) ( )22

2

1S PQPQ Dúû

ùêë

é+= e e

( )2

2

11 S PQPQ Dú

û

ùêë

é+= e e

From Equation (3.5), substituting for ( ) ( )22S S D-¢D , we get

( ) ( ) ( ) ( ) z x z y y x z y xS zx yz xy z y xPQPQ DD+DD+DD+D+D+D=D÷ ø

öçè

æ + e e e e e e e e

2222

2

11

If l, m, and n are the direction cosines of PQ, then

S

zn

S

ym

S

xl

D

D=

D

D=

D

D= ,,

Substituting these quantities in the above expression,

nlmnlmnml zx yz xy z y xPQPQ e e e e e e e e +++++=÷ ø

ö

çè

æ

+222

2

1

1

The above equation gives the value of the relative displacement at point P in the direction

PQ with direction cosines l, m and n.





Module3/Lesson1

9


e ij = ÷÷

ø

ö

çç

è

æ

¶

¶+

¶

¶

i

j

j

i

x

u

x

u

2

1 (i , j = x , y , z) (3.7)

The factor 1/2 in the above Equation (3.7) facilitates the representation of the strain

transformation equations in indicial notation. The longitudinal strains are obtained when

i = j; the shearing strains are obtained when i ¹ j and jiij e e = .

It is clear from the Equations (3.2) and (3.3) that

e xy = 2

1g xy , e yz =

2

1g yz , e xz =

2

1g xz (3.8)

Therefore the strain tensor (e ij = e ji ) is given by

úúúúúú

û

ù

êêêêêê

ë

é

=

z zy zx

yz y yx

xz xy x

ij

e g g

g e g

g g e

e

2

1

2

121

21

2

1

2

1

(3.9)

3.1.7 STRAIN TRANSFORMATION

If the displacement components u, v and w at a point are represented in terms of known

functions of x , y and z respectively in cartesian co-ordinates, then the six strain components

can be determined by using the strain-displacement relations given below.

z

w

y

v

x

u z y x

¶¶=

¶¶=

¶¶= e e e ,,

y

w

z

v

x

v

y

u yz xy

¶

¶+

¶

¶=

¶

¶+

¶

¶= g g , and

z

u

x

w zx

¶

¶+

¶

¶=g

If at the same point, the strain components with reference to another set of co-ordinates axes

y x ¢¢, and z¢are desired, then they can be calculated using the concepts of axis

transformation and the corresponding direction cosines. It is to be noted that the

above equations are valid for any system of orthogonal co-ordinate axes irrespective

of their orientations.Hence



Module3/Lesson1

10


z

w

y

v

x

u z y x ¢¶

¶=

¢¶

¶=

¢¶

¶= ¢¢¢ e e e ,,

z

u

x

w

y

w

z

v

x

v

y

u x z z y y x ¢¶

¶+

¢¶

¶=

¢¶

¶+

¢¶

¶=

¢¶

¶+

¢¶

¶= ¢¢¢¢¢¢ g g g ,,

Thus, the transformation of strains from one co-ordinate system to another can be written inmatrix form as below:

úúúúúú

û

ù

êêêêêê

ë

é

¢¢¢¢¢

¢¢¢¢¢

¢¢¢¢¢

z y z x z

z y y x y

z x y x x

e g g

g e g

g g e

2

1

2

12

1

2

12

1

2

1

úúú

û

ù

êêê

ë

é

´

úúúúúú

û

ù

êêêêêê

ë

é

´

úúú

û

ù

êêê

ë

é

=

321

321

321

333

222

111

2

1

2

12

1

2

12

1

2

1

nnn

mmm

lll

nml

nml

nml

z zy zx

yz y yx

xz xy x

e g g

g e g

g g e

In general, [ ] [ ][ ][ ]T aa e e =¢

3.1.8 SPHERICAL AND DEVIATORIAL STRAIN TENSORS

Like the stress tensor, the strain tensor is also divided into two parts, the spherical and thedeviatorial as,

E = E ¢¢ + E ¢

where E ¢¢ =

úúú

û

ù

êêê

ë

é

e

e

e

00

00

00

= spherical strain (3.10)

E ¢ =

úúú

û

ù

êêê

ë

é

-

-

-

)(

)(

)(

e

e

e

z xy zx

yz y yx

xz xy x

e e e

e e e

e e e

= deviatorial strain (3.11)

and e = 3

z y x e e e ++

It is noted that the spherical component E ¢¢ produces only volume changes without any

change of shape while the deviatorial component E ¢ produces distortion or change of shape.

These components are extensively used in theories of failure and are sometimes known as"dilatation" and "distortion" components.



Module3/Lesson1

11


3.1.9 PRINCIPAL STRAINS - STRAIN INVARIANTS

During the discussion of the state of stress at a point, it was stated that at any point in a

continuum there exists three mutually orthogonal planes, known as Principal planes, onwhich there are no shear stresses.

Similar to that, planes exist on which there are no shear strains and only normal strainsoccur. These planes are termed as principal planes and the corresponding strains are known

as Principal strains. The Principal strains can be obtained by first determining the three

mutually perpendicular directions along which the normal strains have stationary values.

Hence, for this purpose, the normal strains given by Equation (3.6b) can be used.

i.e., nlmnlmnml zx yz xy z y xPQ g g g e e e e +++++= 222

As the values of l, m and n change, one can get different values for the strain PQe .

Therefore, to find the maximum or minimum values of strain, we are required to equate

nml

PQPQPQ

¶¶

¶¶

¶¶ e e e ,, to zero, if l, m and n were all independent. But, one of the direction

cosines is not independent, since they are related by the relation.

1222 =++ nml

Now, taking l and m as independent and differentiating with respect to l and m, we get

022

022

=¶

¶+

=¶

¶+

m

nnm

l

nnl

(3.12)

Now differentiating PQe with respect to l and m for an extremum, we get

( ) z zy zx zx xy x

nmll

nnml e g g g g e 220 ++

¶

¶+++=

( ) z zy zx yz xy y

nmlm

nnlm e g g g g e 220 ++

¶

¶+++=

Substituting forl

n

¶

¶ and

m

n

¶

¶ from Equation 3.12, we get

n

nml

l

nml z zy zx zx xy x e g g g g e 22 ++=++





Module3/Lesson1

13


( )( ) 0

0

0)(

=-++

=+-+

=++-

nml

nml

nm

z zy zx

yz y yx

xz xy x

e e e e

e e e e

e e e e

(3.12d)

The above set of equations is homogenous in l, m and n. In order to obtain a nontrivialsolution of the directions l, m and n from Equation (3.12d), the determinant of the

co-efficients should be zero.

i.e.,

( )( )

( )e e e e

e e e e

e e e e

-

-

-

z zy zx

yz y yx

xz xy x

= 0

Expanding the determinant of the co-efficients, we get

032

2

1

3 =-+- J J J e e e (3.12e)

where z y x J e e e ++=1

x xz

zx z

z zy

yz y

y yx

xy x J

e e

e e

e e

e e

e e

e e ++=2

z zy zx

yz y yx

xz xy x

J

e e e

e e e

e e e

=3

We can also write as

( )

( )222

3

222

2

1

4

1

4

1

xy z zx yz x zx yz xy z y x

zx yz xy x z z y y x

z y x

J

J

J

g e g e g e g g g e e e

g g g e e e e e e

e e e

---+=

++-++=

++=

Hence the three roots 21,e e and 3e of the cubic Equation (3.12e) are known as the

principal strains and J 1 , J 2 and J 3 are termed as first invariant, second invariant and third

invariant of strains, respectively.

Invariants of Strain Tensor

These are easily found out by utilizing the perfect correspondence of the components of

strain tensor e ij with those of the stress tensor t ij. The three invariants of the strain are:



Module3/Lesson1

14


J 1 = e x + e y + e z (3.13)

J 2 = e x e y+ e y e z+e ze x – 4

1 222

zx yz xy g g g ++ (3.14)

J 3 = e x e y e z +4

1

222

xy z zx y yz x zx yz xy g e g e g e g g g --- (3.15)

3.1.10 OCTAHEDRAL STRAINS

The strains acting on a plane which is equally inclined to the three co-ordinate axes are

known as octahedral strains. The direction cosines of the normal to the octahedral plane are ,

.3

1,

3

1,

3

1

The normal octahedral strain is:

(e n)oct = e 1 l2 + e 2 m

2 + e 3 n

2

\ (e n)oct =31 (e 1 + e 2 + e 3) (3.16)

Resultant octahedral strain = (e R)oct = ( ) ( ) ( )2

3

2

2

2

1 nml e e e ++

= ( )2

3

2

2

2

13

1e e e ++ (3.17)

Octahedral shear strain = g oct = 2

13

2

32

2

21 )()()(3

2e e e e e e -+-+- (3.18)



Module3/Lesson2

1 Applied Elasticity for Engineers T.G.Sitharam & L.GovindaRaju

Module 3: Analysis of Strain

3.2.1 MOHR’S CIRCLE FOR STRAIN

The Mohr’s circle for strain is drawn and that the construction technique does not differ fromthat of Mohr’s circle for stress. In Mohr’s circle for strain, the normal strains are plotted on

the horizontal axis, positive to right. When the shear strain is positive, the point representing

the x-axis strains is plotted at a distance2

g below the e -line; and the y-axis point a distance

2

g above the e -line; and vice versa when the shear strain is negative.

By analogy with stress, the principal strain directions are found from the equations

tan 2q = y x

xy

e e

g

- (3.19)

Similarly, the magnitudes of the principal strains are

e 1,2 = 2

y x e e + ±

22

22 ÷÷ ø

öççè

æ +÷÷

ø

öççè

æ - xy y x g e e

(3.20)

3.2.2 EQUATIONS OF COMPATABILITY FOR STRAIN

Expressions of compatibility have both mathematical and physical significance. From a

mathematical point of view, they assert that the displacements u, v, w are single valued and

continuous functions. Physically, this means that the body must be pieced together.

The kinematic relations given by Equation (3.3) connect six components of strain to only

three components of displacement. One cannot therefore arbitrarily specify all of the strains

as functions of x, y, z. As the strains are not independent of one another, in what way they

are related? In two dimensional strain, differentiation of e x twice with respect to y, e y twice

with respect to x, and g xy with respect to x and y results in

2

2

y

x

¶

¶ e =

2

3

y x

u

¶¶

¶,

2

2

x

y

¶

¶ e =

y x

v

¶¶

¶2

3

y x

xy

¶¶

¶ g 2

=2

3

y x

u

¶¶

¶ +

y x

v

¶¶

¶2

3

(3.21)



Module3/Lesson2


or2

2

y

x

¶

¶ e +

2

2

x

y

¶

¶ e =

y x

xy

¶¶

¶ g 2

This is the condition of compatibility of the two dimensional problem, expressed in terms of

strain. The three-dimensional equations of compatibility are derived in a similar manner:

Thus, in order to ensure a single-valued, continuous solution for the displacement

components, certain restrictions have to be imposed on the strain components.These resulting equations are termed the compatibility equations.

Suppose if we consider a triangle ABC before straining a body [Figure 3.4(a)] then the sametriangle may take up one of the two possible positions Figure 3.4(b) and Figure 3.4(c)) afterstraining, if an arbitrary strain field is specified. A gap or an overlapping may occur, unlessthe specified strain field obeys the necessary compatibility conditions.

Fig. 3.4 Strain in a body

Now,

x

u x

¶

¶=e (3.23)

y

v y

¶

¶=e (3.23a)

z

w z

¶

¶=e (3.23b)

y

u

x

v xy

¶

¶+

¶

¶=g (3.23c)

z

v

y

w yz

¶¶+

¶¶=g (3.23d)



Module3/Lesson2


x

w

z

u zx

¶

¶+

¶

¶=g (3.23e)

Differentiating Equation (3.23) with respect to y and Equation (3.23a) with respect to x

twice, we get

2

3

2

2

y xu

y x

¶¶¶=¶¶ e (3.23f)

2

3

2

2

x y

v

x

y

¶¶

¶=

¶

¶ e (3.23g)

Adding Equations (3.23f) and (3.23g), we get

2

3

2

3

2

2

2

2

x y

v

y x

u

x y

y x

¶¶

¶+

¶¶

¶=

¶

¶+

¶

¶ e e (3.23h)

Taking the derivative of Equation (3.23c) with respect to x and y together, we get

2

3

2

32

y x

u

x y

v

y x

xy

¶¶

¶

+¶¶

¶

=¶¶

¶ g (3.23i)

From equations (3.23h) and (3.23i), we get

y x x y

xy y x

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

(3.23j)

Similarly, we can get

z y y z

yz z y

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

(3.23k)

z x z x

zx x z

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

(3.23l)

Now, take the mixed derivative of Equation (3.23) with respect to z and y,

z y x

u

z y

x

¶¶¶

¶=

¶¶

¶\

32e (3.23m)

And taking the partial derivative of Equation (3.23c) with respect to z and x, we get

2

332

x z

v

z y x

u

z x

xy

¶¶

¶+

¶¶¶

¶=

¶¶

¶ g (3.23n)

Also taking the partial derivative of Equation (3.23d) with respect to x twice, we get

z x

v

y x

w

x

yz

¶¶

¶+

¶¶

¶=

¶

¶2

3

2

3

2

2g (3.23p)



Module3/Lesson2


And take the derivative of Equation (3.23e) with respect to y and x

Thus, y x

w

z y x

u

y x

zx

¶¶

¶+

¶¶¶

¶=

¶¶

¶2

332g (3.23q)

Now, adding Equations (3.23n) and (3.23q) and subtracting Equation (3.23p), we get

z y x

u

z x y x x

xy xz yz

¶¶¶

¶=

¶¶

¶+

¶¶

¶+

÷÷

ø

ö

çç

è

æ

¶

¶-

322

2

22g g g

(3.23r)

By using Equation (3.23m), we get

úû

ùêë

é

¶

¶+

¶

¶+

¶

¶-

¶

¶=

¶¶

¶

z y x x z y

xy xz yz x g g g e 22

(3.23s)

Similarly, we can get

úû

ùêë

é

¶

¶+

¶

¶+

¶

¶-

¶

¶=

¶¶

¶

x z y y z x

yz yx zx y g g g e 22 (3.23t)

úû

ùêë

é

¶

¶+

¶

¶+

¶

¶-

¶

¶=

¶¶

¶

y x z z y x

zx yz xy z g g g e 22

(3.23u)

Thus the following are the six compatibility equations for a three dimensional system.

y x x y

xy y x

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

z y y z

yz z y

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

x z z x

zx x z

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

(3.24)

÷÷ ø

öççè

æ

¶

¶+

¶

¶+

¶

¶-

¶

¶=

¶¶

¶

z y x x z y

xy xz yz x g g g e 22

÷÷ ø

öççè

æ

¶

¶+

¶

¶-

¶

¶

¶

¶=

¶¶

¶

z y x y x z

xy zx yz y g g g e 22

÷÷ ø

öççè

æ

¶

¶-

¶

¶+

¶

¶

¶

¶=

¶¶

¶

z y x z y x

xy zx yz z g g g e 22





Module3/Lesson2


11 2sin2

2cos221

q g

q e e e e

e q xy y x y x

+÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ += (a)

22 2sin2

2cos222

q g

q e e e e

e q xy y x y x

+÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ += (b)

33 2sin2

2cos223

q g

q e e e e

e q xy y x y x

+÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ += (c)

For a rectangular rosette,0

21 45,0 == q q and0

3 90=q

Substituting the above in equations (a), (b) and (c),

We get

022

0 +÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ +=

y x y x e e e e

e

= ( ) y x y x e e e e -++

2

1

xe e =\ 0

( )2

022

45

xy y x y x g e e e e

e +÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ +=

( ) xy y x g e e ++=2

1

or xy y x g e e e ++=452

) y x xy e e e g +-=\452

Also, ( ) ( )00

90 180sin2

180cos22

xy y x y x g e e e e

e +÷÷ ø

öççè

æ -+÷÷

ø

öççè

æ +=

= ÷÷ ø

öççè

æ --÷÷

ø

öççè

æ +

22

y x y x e e e e

= ( ) y x y x e e e e +-+

2

1

ye e =\ 90

Therefore, the components of strain are given by

and ( )900452 e e e g +-= xy

For an equiangular rosette,

0900 , e e e e == y x



Module3/Lesson2


0

3

0

21 120,60,0 === q q q

Substituting the above values in (a), (b) and (c), we get

( )0120600 223

1, e e e e e e -+== y x

and ( )120603

2 e e g -= xy

Hence, using the values of y x e e , and xy

g , the principal strains maxe and mine can be

computed.

3.2.4 NUMERICAL EXAMPLES

Example 3.1

A sheet of metal is deformed uniformly in its own plane that the strain components

related to a set of axes xy are

x = -200 10-6

y = 1000 10-6

xy = 900 10-6

(a) Find the strain components associated with a set of axes y x ¢¢ inclined at an angle of

30o clockwise to the x y set as shown in the Figure 3.5. Also find the principal

strains and the direction of the axes on which they act.

Figure 3.5

x ¢

300

300

y¢

y



Module3/Lesson2


Solution: ( a)

The transformation equations for strains similar to that for stresses can be written as below:

x¢e =2

y x e e +

+2

y x e e -

cos2q +2

xyg

sin2q

y¢e =2

y x e e + -

2

y x e e -

cos2q -2

xyg

sin2q

2

'' y xg = ÷÷

ø

öççè

æ --

2

y x e e sin2q +

2

xyg cos2q

Using Equation (3.19), we find

2q = tan-1 ÷

ø

öçè

æ

600

450= 36.80

Radius of Mohr’s circle = R =22 )450()600( + = 750

Therefore ,

x¢e = ( ) ( ) ( )0066 8.3660cos1075010400 -´-´ --

=610290 -´-

y¢e = ( ) ( ) ( )0066 8.3660cos1075010400 -´+´ --

6101090 -´=

Because point x¢ lies above the e axis and point y¢ below e axis, the shear strain y x ¢¢g isnegative.

Therefore ,

2

'' y xg

=006 8.3660sin10750 -´- -

610295 -´-=

hence, y x ¢¢g = 610590 -´-

Solution: (b) From the Mohr’s circle of strain, the Principal strains are

6

1 101150 -´=e 6

2 10350 -´-=e



Module3/Lesson2


Figure 3.6 Construction of Mohr’s strain circle



Module3/Lesson2


The directions of the principal axes of strain are shown in figure below.

Figure 3.7

Example 3.2

By means of strain rosette, the following strains were recorded during the test on a

structural member.

mmmmmmmmmmmm /1013,/105.7,/1013 6

90

6

45

6

0

--- ´=´=´-= e e e

Determine (a) magnitude of principal strains

(b) Orientation of principal planes

Solution: (a) We have for a rectangular strain rosette the following:( )90045900 2 e e e g e e e e +-=== xy y x

Substituting the values in the above relations, we get66 10131013 -- ´=´-= y x

e e

( ) 6666 101510131012105.72 ---- ´=\´+´--´´= xy xy g g

The principal strains can be determined from the following relation.

maxe or ( ) 22

min2

1

2 xy y x

y xg e e

e e e +-±÷÷

ø

öççè

æ +=

maxe \ or ( )[ ] ( )26266min 1015101313

2110

21313 --- ´+--±÷

ø öç

è æ +-=e

maxe \ or6

min 1015 -´±=e

6.71

1e

2e



Module3/Lesson2


Hence6

max 1015 -´=e and6

min 1015 -´-=e

(b) The orientation of the principal strains can be obtained from the following relation

( ) y x

xy

e e

g q

-=2tan

( ) 6

6

101313

1015-

-

--´=

577.02tan -=q 01502 =\ q

075=\q

Hence the directions of the principal planes are0

1 75=q and0

2 165=q

Example 3.3

Data taken from a 450 strain rosette reads as follows:

7500 =e micrometres/m

11045 -=e micrometres/m

21090 =e micrometres/m

Find the magnitudes and directions of principal strains.

Solution: Given6

0 10750 -´=e

6

45 10110 -´-=e

6

90 10210 -´=e

Now, for a rectangular rosette,

60 10750 -´== e e x

6

90 10210 -´== e e y

( )900452 e e e g +-= xy

666 1021010750101102 --- ´+´-´-=

6101180 -´-= xyg

\The magnitudes of principal strains are

maxe or ( ) 22

min21

2 xy y x

y xg e e e e e +-±÷÷

ø öçç

è æ +=



Module3/Lesson2


i.e., maxe or ( )[ ] ( )[ ]26266

min 101180102107502

110

2

210750 --- -+-±÷ ø

öçè

æ +=e

( ) 66 107.12972

110480 -- ±´=

66 1085.64810480 -- ´±´=

6

1max 1085.1128 -´==\ e e

6

2min 1085.168 -´-== e e

The directions of the principal strains are given by the relation

( ) y x

xy

e e

g q

-=2tan

( ) 185.210210750

1011802tan 6

6

-=-

´-=\ -

-

q

06.1142 =\ q

0

1 3.57=\q and0

2 3.147=q

Example 3.4

If the displacement field in a body is specified as ( ) 3232 103,103 -- ´=+= z yv xu and

( ) ,103 3-´+= z xw determine the strain components at a point whose coordinates

are (1,2,3)

Solution: From Equation (3.3), we have

,102 3-´=¶

¶= x

x

u xe

,106 3-´=¶

¶= yz

y

v ye

3103 -´=¶

¶=

z

w ze

( ) ( ) úû

ù

êë

é´

¶

¶+´+

¶

¶= -- 3232 103103 z y

x x

y xy

g

0= xyg



Module3/Lesson2


( ) ( ) úû

ùêë

é´+

¶

¶+´

¶

¶= -- 332 103103 z x

y z y

z yzg

32 103 -´= y yzg

and ( ) ( ) úûùê

ëé +

¶¶++

¶¶= -- 323 103103 x

z z x

x zxg

3101 -´= zxg

Therefore at point (1, 2, 3), we get

33

3333

101,1012,0

,103,103610326,102

--

----

´=´==

´=´=´´´=´=

zx yz xy

z y x

g g g

e e e

Example 3.5

The strain components at a point with respect to x y z co-ordinate system are

160.0,30.0,20.0,10.0 ====== xz yz xy z y x g g g e e e

If the coordinate axes are rotated about the z-axis through 450 in the anticlockwise

direction, determine the new strain components.

Solution: Direction cosines

x y z

x¢

2

1

2

1

0

y¢

2

1-

2

1

0

z¢ 0 0 1

Here 0,2

1,

2

1111 =-== nml

0,2

1,

2

1222 === nml

1,0,0333

=== nml

Now, we have, Figure 3.8

450

z(Z )¢

y¢

x¢

y

x



Module3/Lesson2


[ ] [ ][ ][ ]T aa e e =¢

[ ][ ]

úúú

û

ù

êêê

ë

é

úúú

úúú

û

ù

êêê

êêê

ë

é

-=

3.008.008.0

08.02.008.0

08.008.01.0

100

0

2

1

2

1

02

1

2

1

e a

úúú

û

ù

êêê

ë

é

-=

3.008.008.0

0085.0014.0

113.0198.0127.0

[ ]

úúú

úúú

û

ù

êêê

êêê

ë

é-

úú

ú

û

ù

êê

ê

ë

é

-=¢

100

0

2

1

2

1

02

1

2

1

3.008.008.0

0085.0014.0

113.0198.0127.0

e

[ ]úúú

û

ù

êêê

ë

é

=¢

3.03.0113.0

007.005.0

113.005.023.0

e

Therefore, the new strain components are

3.0,07.0,23.0 === z y x e e e

05.02

1 = xyg or 1.0205.0 =´= xy

g

226.02113.0,0 =´== zx yz g g

Example 3.6

The components of strain at a point in a body are as follows:

08.0,1.0,3.0,05.0,05.0,1.0 -====-== xz yz xy z y x g g g e e e

Determine the principal strains and the principal directions.

Solution: The strain tensor is given by



Module3/Lesson2


úúú

û

ù

êêê

ë

é

-

-

-

=

ú

úúúúú

û

ù

ê

êêêêê

ë

é

=

05.005.004.0

05.005.015.0

04.015.01.0

22

22

22

z

yz xz

yz

y

xy

xz xy

x

ij

e g g

g e

g

g g e

e

The invariants of strain tensor are

( ) zx yz xy x z z y y x

z y x

J

J

222

2

1

4

1

1.005.005.01.0

g g g e e e e e e

e e e

++-++=

=+-=++=

( )( ) ( )( ) ( )( ) ( ) ( ) ( )[ ]22208.01.03.0

4

11.005.005.005.005.01.0 -++-+-+-=

0291.02 -=\ J

( )( )( ) ( )( )[ ]2(0.3)0.052(0.08)0.052(0.1)0.10.08)(0.10.34

10.050.050.1

3J -+--+-=

002145.03 -= J

\The cubic equation is

0002145.00291.01.0 23 =+-- e e e (i)

Now q q q cos3cos43cos 3 -=

Or 03cos4

1cos

4

3cos3 =-- q q q (ii)

Let3

cos 1 J r += q e

=3

1.0cos +q r

033.0cos += q e r

\(i) can be written as

( ) ( ) ( ) 0002145.0033.0cos0291.02

033.0cos1.03

033.0cos =++-+-+ q q q r r r

( )( ) ( )0002145.000096.0

cos0291.02

033.0cos1.02

033.0cos033.0cos

=+-

-+-++ q q q q r r r r



Module3/Lesson2


( )

( ) 0002145.000096.0cos0291.000109.0cos067.02cos21.0

00109.0cos067.02cos2033.0cos

=+--++-

+++

q q q

q q q

r r r

r r r

0002145.0

00096.0cos0291.0000109.0cos0067.02cos21.0000036.0

cos0022.02cos2033.0cos00109.02cos2067.03cos3

=

+-----

+++++

q q q

q q q q q

r r r

r r r r r

i.e., 000112.0cos03251.03cos3 =-- q q r r

or 03

00112.0cos

2

03251.03cos =--r r

q q (iii)

Hence Equations (ii) and (iii) are identical if

4

303251.02

=r

i.e., 2082.03

03251.04=

´=r

and3

00112.0

4

3cos

r =

q

or( )

5.0496.02082.0

00112.043cos

3 @=

´=q

0603 =\ q or0

1 203

60==q

0

3

0

2 140100 == q q

3

1.020cos2082.0

3cos

0

1111

+=

+=\ J

r q e

228.01 =e

126.03

1.0140cos2082.03cos

0031.03

1.0100cos2082.0

3cos

01333

01222

-=+=+=

-=+=+=

J r

J r

q e

q e



Module3/Lesson2


To find principal directions

(a) Principal direction for 1e

( )( )

( )

( )( )

( )úúú

û

ù

êêê

ë

é

--

--

--

=

úú

ú

û

ù

êê

ê

ë

é

--

--

--

228.005.005.004.0

05.0228.005.015.0

04.015.0228.01.0

05.005.004.0

05.005.015.0

04.015.01.0

1

1

1

e

e

e

úúú

û

ù

êêê

ë

é

--

-

--

=

178.005.004.0

05.0278.015.0

04.015.0128.0

Now, ( )( ) ( )( )05.005.0178.0278.0178.005.0

05.0278.01 ---=

-

-= A

046984.01 =\ A

( ) ( )( )[ ]04.005.0178.015.0178.004.0

05.015.01 +-´-=

---= B

0247.01 =\ B

04.0278.005.015.005.004.0

278.015.0

1 ´-´=

-

-=

C

00362.01 -=\C

( ) ( ) ( )2222

1

2

1

2

1 00362.00247.0046984.0 -++=++ C B A

= 0.0532

883.00532.0

046984.0

2

1

2

1

2

1

11 ==

++=\

C B A

Al

464.00532.0

0247.0

2

1

2

1

2

1

11

==++= C B A

B

m



Module3/Lesson2


068.00532.0

00362.0

2

1

2

1

2

1

11 -=

-=

++=

C B A

C n

Similarly, the principal directions for 2e can be determined as follows:

( ) ( )( )ú

úú

û

ù

êêê

ë

é

+-

+-

-+

0031.005.005.004.0

05.00031.005.015.004.015.00031.01.0

úúú

û

ù

êêê

ë

é

-

-

-

=

0531.005.004.0

05.00469.015.0

04.015.01031.0

00499.00025.000249.00531.005.0

05.00469.02 -=--=

-= A

009965.0)002.0007965.0(0531.004.005.015.0

2 -=+-=-

= B

00562.000188.00075.005.004.0

0469.015.02 =-=

-

-=C

Now, ( ) ( ) ( ) 0125.000562.0009965.000499.02222

2

2

2

2

2 =+-+-=++ C B A

399.00125.0

00499.0

2

2

2

2

2

2

22 -=

-=

++=\

C B A

Al

797.00125.0

009965.0

22

22

22

2

2

-=-

=++

=C B A

Bm

450.00125.0

00562.0

2

2

2

2

2

2

22 ==

++=

C B A

C n

And for 126.03 -=e

( )( )

( )126.005.005.004.0

05.0126.005.015.0

04.015.0126.01.0

+-

+-

-+

176.005.004.0

05.0076.015.0

04.015.0226.0

-

-

=



Module3/Lesson2


Now, 0109.00025.00134.0176.005.0

05.0076.03 =-== A

0284.0)002.00264.0(176.004.0

05.015.03 -=+-=

--= B

01054.000304.00075.005.004.0

076.015.03 =+=

--=C

Now, ( ) ( ) ( ) 0322.001054.00284.00109.02222

3

2

3

2

3 =+-+=++ C B A

338.00322.0

0109.0

2

3

2

3

2

3

33 ==

++=\

C B A

Al

882.00322.0

0284.0

2

3

2

3

2

3

33 -=

-=

++=

C B A

Bm

327.00322.0

01054.02

3

2

3

2

3

33 ==

++=

C B A

C n

Example 3.7

The displacement components in a strained body are as follows:22322 05.001.0,01.002.0,02.001.0 z xyw y z xv y xyu +=+=+=

Determine the strain matrix at the point P (3,2, -5)

Solution: y x

u x

01.0=¶

¶=e

301.0 z yv

y =¶¶=e

z z

w z

1.0=¶

¶=e

y x x y

u

x

v xy

04.001.004.0 ++=¶

¶+

¶

¶=g

y z xy z

v

y

w yz

203.002.0 +=¶

¶+

¶

¶=g

201.00 y

x

w

z

u zx +=

¶

¶+

¶

¶=g

At point P (3, 2, -5), the strain components are

5.0,25.1,02.0 -=-== z y x e e e

04.0,62.1,23.0 === zx yz xy g g g



Module3/Lesson2


Now, the strain tensor is given by

úúúúúú

û

ù

êêêêêê

ë

é

=

z zy zx

yz y yx

xz xy x

ij

e g g

g e g

g g e

e

21

21

2

1

2

12

1

2

1

\Strain matrix becomes

úúú

û

ù

êêê

ë

é

-

-=

50.081.002.0

81.025.1115.0

02.0115.002.0

ije

Example 3.8

The strain tensor at a point in a body is given by

úúú

û

ù

êêê

ë

é

=

0005.00004.00005.0

0004.00003.00002.0

0005.00002.00001.0

ije

Determine (a) octahedral normal and shearing strains. (b) Deviator and Spherical

strain tensors.

Solution: For the octahedral plane, the direction cosines are3

1=== nml

(a) octahedral normal strain is given by

( ) ( )nlmnlmnml zx yz xy z y xoct n e e e e e e e +++++= 2222

Here yz yz xy xy g e g e 2

1

,2

1

== and zx zx g e 2

1

=

( )

úû

ùêë

é÷ ø

öçè

æ +÷

ø

öçè

æ +÷

ø

öçè

æ

+÷÷ ø

öççè

æ +÷÷

ø

öççè

æ +÷÷

ø

öççè

æ =\

3

10005.0

3

10004.0

3

10002.02

3

10005.0

3

10003.0

3

10001.0

222

oct ne

( ) 001.0=\oct n

e

Octahedral Shearing Strain is given by

( ) ( )222 oct noct Roct e e g -=

where ( )oct R

e = Resultant strain on octahedral plane





Module3/Lesson2


Solution: For the compatibility condition of the strain field, the system of strains mustsatisfy the compatibility equations

i.e., y x x y

xy y x

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

Now, using the given strain field,

zc y

yzc y

x x12

2

1 2,2 =¶

¶=

¶

¶ e e

z x

xz x

y y2,2

2

2

=¶

¶=

¶

¶ e e

zc y x

yzc x

xy xy

2

2

2 2,2 =¶¶

¶=

¶

¶ g g

( )112

2

2

2

1222 c z z zc x y

y x +=+=¶

¶+

¶

¶\

e e and zc

y x

xy

2

2

2=¶¶

¶ g

Since y x x y

xy y x

¶¶¶¹

¶¶+

¶¶ g e e

2

2

2

2

2, the strain field is not compatible.

Example 3.10

Under what conditions are the following expressions for the components of strain at a

point compatible?

cxybyaxy x

22 22 ++=e

bxax y += 2e

yax xy y x xy hb a g +++= 22

Solution: For compatibility, the strain components must satisfy the compatibility equation.

i.e., y x x y

xy y x

¶¶

¶=

¶

¶+

¶

¶ g e e 2

2

2

2

2

(i)

or 0

2

2

2

2

2

=¶¶

¶-

¶

¶+

¶

¶

y x x y

xy y x g e e

(ii)

Now, cxybyaxy x 22 22 ++=e

cxbyaxy

y

x 224 ++=

¶

¶\

e

bax y

x 242

2

+=¶

¶ e



Module3/Lesson2


bxax y += 2e

bax x

y+=

¶

¶2

e

a x

y2

2

2

=¶

¶ e

yax xy y x xy hb a g +++= 22

ax y xy x

xy22 ++=

¶

¶b a

g

b a g

+=¶¶

¶ x

y x

xy2

2

\(i) becomes

b a +=++ xabax 2224

( ) b a +=++ xbaax 224

xax a 24 =\

or a2=a

and ( )ba += 2b

Example 3.11

For the given displacement field

( ) x xcu 2

2

+= ( ) z y xcv ++= 224

24czw =

where c is a very small constant, determine the strain at (2,1,3), in the direction

0,2

1,

2

1-

Solution: cc y

u

x

vcx

x

u xy x

404,2 =+=¶

¶+

¶

¶==

¶

¶= g e

cc

z

v

y

wcy

y

v yz y =+=

¶

¶+

¶

¶==

¶

¶= 0,4 g e

cc x

w

z

ucz

z

w zx z

202,8 =+=¶

¶+

¶

¶==

¶

¶= g e

\At point (2,1,3),



Module3/Lesson2


ccc xy x

4,422 ==´= g e

ccc yz y ==´= g e ,414

ccc zx z 2,2438 ==´= g e

\The Resultant strain in the direction2

1,

2

1,0 =-== nml is given by

nlmnlmnml zx yz xy z y xr g g g e e e e +++++= 222

( ) ( )022

1

2

104

2

124

2

140

22

ccccc +÷ ø

öçè

æ ÷ ø

öçè

æ -++÷

ø

öçè

æ +÷

ø

öçè

æ -+=

cr

5.13=\e

Example 3.12

The strain components at a point are given by

01.0,02.0,015.0,03.0,02.0,01.0 -====-== xz yz xy z y x g g g e e e

Determine the normal and shearing strains on the octahedral plane.Solution: An octahedral plane is one which is inclined equally to the three principal

co-ordinates. Its direction cosines are1 1 1

, ,3 3 3

Now, the normal strain on the octahedral plane is

( ) nlmnlmnml zx yz xy z y xoct n g g g e e e e +++++= 222

[ ]01.002.0015.003.002.001.03

1-+++-=

( ) 015.0=\oct ne

The strain tensor can be written as

÷÷÷

ø

ö

ççç

è

æ

-

-

-

=

÷÷÷÷÷÷

ø

ö

çççççç

è

æ

-

-

-

=÷÷÷

ø

ö

ççç

è

æ

03.001.0005.0

01.002.00075.0

005.00075.001.0

03.02

02.0

2

01.02

02.002.0

2

015.02

01.0

2

015.001.0

z yz xz

yz y xy

xz xy x

e e e

e e e

e e e

Now, the resultant strain on the octahedral plane is given by

( ) ( ) ( ) ( ){ }222

3

1 z yz xz yz y xy xz xy xoct R

e e e e e e e e e e ++++++++=

( ) ( ) ( ){ }22203.001.0005.001.002.00075.0005.00075.001.0

3

1 ++-++-+-+=

0004625.0=



Module3/Lesson2


( ) 0215.0=\ oct Re

and octahedral shearing strain is given by

( ) ( ) ( )222 n Roct S e e e -= ( ) ( )22

015.00215.02 -=

( ) 031.0=\oct S

e

Example 3.13

The displacement field is given by

( ) ( ) 222 4,24,2 Kzw z y xK v z xK u =++=+=

where K is a very small constant. What are the strains at (2,2,3) in directions

1 1( ) 0, , , ( ) 1, 0, ( ) 0.6, 0, 0.8

2 2a l m n b l m n c l m n= = = = = = = = =

Solution: Kz z

wKy

y

vKx

x

u z y x

8,4,2 =¶

¶==

¶

¶==

¶

¶= e e e

K K yu

xv

xy 404 =+=¶¶+

¶¶=g

K K z

v

y

w yz =+=

¶

¶+

¶

¶= 0g

K K x

w

z

u zx

202 =+=¶

¶+

¶

¶=g

\At point (2,2,3),

K K K z y x 24,8,4 === e e e

K K K zx yz xy

2,,4 === g g g

Now, the strain in any direction is given by

nlmnlmnml zx yz xy z y xr g g g e e e e +++++= 222

(i)

Case ( a) Substituting the values in expression (i), we get

( ) ( ) ( )022

1

2

104

2

124

2

1804

22

K K K K K K r +÷ ø

öçè

æ ÷ ø

öçè

æ ++÷

ø

öçè

æ +÷

ø

öçè

æ +=e

K K K r 2

1124 ++=\e

K r 5.16=\e

Case ( b)

( ) ( ) ( ) ( ) ( ) ( )0200402408142

K K K K K r +++++=e

K r 4=\e



Module3/Lesson2


Case ( c)

( ) ( ) ( ) ( ) ( ) ( )( )6.08.020048.024086.0422

K K K K K r +++++=e

K r 76.17=\e



1


Module4/Lesson1

Module : 4 Stress-Strain Relations

4.1.1 INTRODUCTION

n the previous chapters, the state of stress at a point was defined in terms of sixcomponents of stress, and in addition three equilibrium equations were developed to relate

the internal stresses and the applied forces. These relationships were independent of thedeformations (strains) and the material behaviour. Hence, these equations are applicableto all types of materials.

Also, the state of strain at a point was defined in terms of six components of strain. Thesesix strain-displacement relations and compatibility equations were derived in order to relateuniquely the strains and the displacements at a point. These equations were also independentof the stresses and the material behavior and hence are applicable to all materials.

Irrespective of the independent nature of the equilibrium equations and strain-displacementrelations, in practice, it is essential to study the general behaviour of materials under appliedloads including these relations. This becomes necessary due to the application of a load,

stresses, deformations and hence strains will develop in a body. Therefore in a generalthree-dimensional system, there will be 15 unknowns namely 3 displacements, 6 strains and

6 stresses. In order to determine these 15 unknowns, we have only 9 equations such as 3equilibrium equations and 6 strain-displacement equations. It is important to note that thecompatibility conditions as such cannot be used to determine either the displacements or

strains. Hence the additional six equations have to be based on the relationships between sixstresses and six strains. These equations are known as "Constitutive equations" because they

describe the macroscopic behavior of a material based on its internal constitution.

4.1.2 LINEAR ELASTICITY-GENERALISED HOOKE’S LAW

There is a unique relationship between stress and strain defined by Hooke’s Law, which isindependent of time and loading history. The law assumes that all the strain changesresulting from stress changes are instantaneous and the system is completely reversible andall the input energy is recovered in unloading.

In case of uniaxial loading, stress is related to strain as

x x E e s = (4.0)

where E is known as "Modulus of Elasticity".

The above expression is applicable within the linear elastic range and is called

Hooke’s Law.

In general, each strain is dependent on each stress. For example, the strain xe written as a

function of each stress is

I



2


Module4/Lesson1

xe = C 11 x

s + C 12 ys + C 13 z

s + C 14 xyt + C 15 yzt + C 16 zxt + C 17 xzt + C 18 zy

t + C 19 yxt

(4.1)

Similarly, stresses can be expressed in terms of strains stating that at each point in a material,

each stress component is linearly related to all the strain components. This is known as

"Generalised Hook’s Law".

For the most general case of three-dimensional state of stress, equation (4.0) can be written

as

) { }klijklij D e s = (4.2)

where ijkl D = Elasticity matrix

ijs = Stress components

{ }kle = Strain components

Since both stress ijs and strain ij

e are second-order tensors, it follows that ijkl D is a fourth

order tensor, which consists of 34 = 81 material constants if symmetry is not assumed.Therefore in matrix notation, the stress-strain relations would be

Now, from jiij s s = and jiij e e = the number of 81 material constants is reduced to 36

under symmetric conditions of jilk ijlk jiklijkl D D D D ===

Therefore in matrix notation, the stress – strain relations can be

)3.4(

999897969594939291

898887868584838281

797877767574737271

696867666564636261

595857565554535251

494847464544434241

393837363534333231

292827262524232221

191817161514131211

ïïï

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ïïïïïï

ý

ü

ïïï

ïïï

î

ïïïïïï

í

ì

úúú

úúúúúúúúú

û

ù

êêê

êêêêêêêêê

ë

é

=

ïïï

ïïï

ïïïïïï

ý

ü

ïïï

ïïï

î

ïïïïïï

í

ì

yx

zy

xz

zx

yz

xy

z

y

x

yx

zy

xz

zx

yz

xy

z

y

x

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

D D D D D D D D D

g

g

g

g

g

g

e

e

e

t

t

t

t

t

t

s

s

s



3


Module4/Lesson1

Equation (4.4) indicates that 36 elastic constants are necessary for the most general form ofanisotropy (different elastic properties in all directions). It is generally accepted, however,

that the stiffness matrix ij D is symmetric, in which case the number of independent elastic

constants will be reduced to 21. This can be shown by assuming the existence of a strainenergy function U .

It is often desired in classical elasticity to have a potential function

)ijU U e = (4.5)

with the property that

ij

ij

U s

e =

¶

¶ (4.6)

Such a function is called a "strain energy" or "strain energy density function".

By equation (4.6), we can write

jiji

i

DU

e s

e

==

¶

¶ (4.7)

Differentiating equation (4.7) with respect to je , then

ij

ji

DU

=¶¶

¶

e e

2

(4.8)

The free index in equation (4.7) can be changed so that

i ji j

j

DU

e s e

==¶

¶ (4.9)

Differentiating equation (4.9) with respect to ie , then,

ji

i j

DU

=¶

¶

e e

2

(4.10)

)4.4(

666564636261

565554535251

464544434241

363534333231

2625242322

161514131211

21

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ý

ü

ïïïï

î

ïïïï

í

ì

úúúúúúúú

û

ù

êêêêêêêê

ë

é

=

ïïïï

ïïïï

ý

ü

ïïïï

î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D D D D D D

D D D D D D

D D D D D D

D D D D D D

D D D D D D

D D D D D D

g

g

g

e

e

e

t

t

t

s

s

s



4


Module4/Lesson1

Hence, equations (4.8) and (4.10) are equal, or jiij D D =

which implies that ij D is symmetric. Then most general form of the stiffness matrix or

array becomes

Or

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î

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í

ì

úúúúúúúúú

û

ù

êêêêêêêêê

ë

é

=

ïïïï

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ý

ü

ïïïï

î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D

D D

D D D

D D D D

D D D D D D D D D D D

g

g

g

e

e

e

t

t

t

s

s

s

66

5655

464544

36353433

2625242322

161514131211

(4.12)

Further, a material that exhibits symmetry with respect to three mutually orthogonal planes is

called an "orthotropic" material. If the xy, yz and zx planes are considered planes of

symmetry, then equation (4.11) reduces to 12 elastic constants as below.

Also, due to orthotropic symmetry, the number of material constants for a linear elasticorthotropic material reduces to 9 as shown below.

)11.4(

665646362616

565545352515

464544342414

363534332313

262524232212

161514131211

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í

ì

úúúúúúúú

û

ù

êêêêêêêê

ë

é

=

ïïïï

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ý

ü

ïïïï

î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D D D D D D

D D D D D D

D D D D D D

D D D D D D

D D D D D D

D D D D D D

g

g

g

e

e

e

t

t

t

s

s

s

)13.4(

00000

00000

00000

000

000

000

66

55

44

333231

232221

131211

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ý

ü

ïïïï

î

ïïïï

í

ì

úúúúúúúú

û

ù

êêêêêêêê

ë

é

=

ïïïï

ïïïï

ý

ü

ïïïï

î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D

D

D

D D D

D D D

D D D

g

g

g

e

e

e

t

t

t

s

s

s



5


Module4/Lesson1

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ý

ü

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î

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í

ì

úúúúúúú

û

ù

êêêêêêê

ë

é

=

ïïïï

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ý

ü

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î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D D

D

D

D D

D D D

g g

g

e

e

e

t t

t

s

s

s

66

55

44

33

2322

131211

0

00

000

000

000

(4.14)

Now, in the case of a transversely isotropic material, the material exhibits a rationally elastic

symmetry about one of the coordinate axes, x , y , and z. In such case, the material constants

reduce to 8 as shown below.

(4.15)Further, for a linearly elastic material with cubic symmetry for which the properties along

the x , y and z directions are identical, there are only 3 independent material constants.

Therefore, the matrix form of the stress – strain relation can be expressed as:

(4.16)

4.1.3 ISOTROPY

For a material whose elastic properties are not a function of direction at all, only two

independent elastic material constants are sufficient to describe it’s behavior completely.

This material is called "Isotropic linear elastic". The stress- strain relationship for this

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û

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êêêêêêêê

ë

é

-=

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ü

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î

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í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D

DSymmetry

D D

D

D D

D D D

g

g

g

e

e

e

t

t

t

s

s

s

66

55

1211

33

2322

131211

0

00)(21

000

000

000

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í

ì

úúúúúúúú

û

ù

êêêêêêêê

ë

é

=

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í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D

DSymmetry

D

D

D D

D D D

g

g

g

e

e

e

t

t

t

s

s

s

44

44

44

11

1211

121211

0

00

000

000

000



6


Module4/Lesson1

material is thus written as an extension of that of a transversely isotropic material

as shown below.

( )

( )

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ý

ü

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í

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úúúúúúúúú

û

ù

êêêêêêêêê

ë

é

-

-

-=

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ü

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î

ïï

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í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

D D

D D

D D

D

D

D

D

D

Symmetry

D

g

g

g e

e

e

t

t

t s

s

s

1211

1211

1211

11

12

12

11

1211

2

1

0

0

0

0

0

2

1

0

0

0

0

2

1

0

0

0

(4.17)

Thus, we get only 2 independent elastic constants.

Replacing 12 D and ( )12112

1 D D - respectively by l and G which are called "Lame’s

constants", where G is also called shear modulus of elasticity, equation (4.17) can be written

as:

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û

ù

êêêêêêêê

ë

é

+

+

+

=

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ü

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î

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í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

G

GSymmetry

G

G

G

G

g

g

g

e

e

e

l

l l

l l l

t

t

t

s

s

s

0

00

0002

0002

0002

(4.18)

Therefore, the stress-strain relationships may be expressed as

ïïïï

ïïïï

ý

ü

ïïïï

î

ïïïï

í

ì

úúúúúúúú

û

ù

êêêêêêêê

ë

é

+

+

+

=

ïïïï

ïïïï

ý

ü

ïïïï

î

ïïïï

í

ì

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

G

G

G

G

G

G

g

g

g

e

e

e

l l l

l l l

l l l

t

t

t

s

s

s

00000

00000

00000

0002

0002

0002

(4.19)

Therefore,

xs = ( )l +G2 xe + ) z y e e l +

( ) ( ) x z y y

G e e l e l s +++= 2

( ) ) y x z z

G e e l e l s +++= 2 (4.20)



7


Module4/Lesson1

Also, xy xy Gg t =

yz yzGg t =

zx zx Gg t =

Now, expressing strains in terms of stresses, we get

( ) ( )( )

z y x xGGGG

Gs s

l

l s

l

l e +

+-

+

+=

23223

( ) ( )( )

x z y yGGGG

Gs s

l

l s

l

l e +

+-

+

+=

23223

( ) ( )( )

y x z zGGGG

Gs s

l

l s

l

l e +

+-

+

+=

23223 (4.21)

G

xy

xy

t g =

G

yz

yz

t

g =

G

zx

zx

t g =

Now consider a simple tensile test

Therefore,

xe = E

xs =

)23( GG

G

+

+

l

l s x

or

E

1 =

)23( GG

G

+

+

l

l

or E =)(

)23(

G

GG

+

+

l

l (4.22)

where E = Modulus of Elasticity

Also,

e y = - ne x = - n E

xs

e z = - ne x = - n

E

xs

where n = Poisson’s ratio

For s y = s z = 0 , we get from equation (4.21)



8


Module4/Lesson1

)23(2 GG +-

l

l s x =

E

n - s x

Therefore, E

n =

)23(2 GG +l

l (4.23)

Substituting the value of E from equation (4.22), we get

)23(

)(

GG

G

+

+

l

l n =

)23(2 GG +l

l

Therefore, 2n (l +G) = l

or n =)(2 G+l

l (4.24)

Solving for l from equations (4.22) and (4.23), we get

l =)3()2(

G E

E GG

-- =

( )n n G E

G6

42

-

or G = )1(2 n +

E (4.25)

For a hydrostatic state of stress, i.e., all round compression p,

s x = s y = s z = - p

Therefore, e x+e y+e z = E

p)21(3 n --

or - p = )21(3

)(n

e e e -

++ z y x E

= (l +3

2G )(e x+e y+e z)

or - p = K (e x+e y+e z)

Hence, K = ÷ ø

öçè

æ +

3

2Gl (4.26)

where K = Bulk modulus of elasticity.

Also,

- p = K (e x+e y+e z)

úû

ùêë

é --=-

E

pK p

)21(3 n



9


Module4/Lesson1

or ( )[ ]vK E 213 -=

Therefore, K =)21(3 n -

E (4.27)



Module4/Lesson2

1


Module 4: Stress-Strain Relations

4.2.1 ELASTIC STRAIN ENERGY FOR UNIAXIAL STRESS

Figure 4.1 Element subjected to a Normal stress

In mechanics, energy is defined as the capacity to do work, and work is the product of forceand the distance, in the direction, the force moves. In solid deformable bodies,stresses multiplied by their respective areas, results in forces, and deformations are distances.

The product of these two quantities is the internal work done in a body by externallyapplied forces. This internal work is stored in a body as the internal elastic energy ofdeformation or the elastic strain energy.

Consider an infinitesimal element as shown in Figure 4.1a, subjected to a normal stress s x.

The force acting on the right or the left face of this element is s x dydz. This force causes an

elongation in the element by an amount e x dx , where e x is the strain in the direction x.

The average force acting on the element while deformation is taking place is2

dzdy x

s .

This average force multiplied by the distance through which it acts is the work done on theelement. For a perfectly elastic body no energy is dissipated, and the work done on theelement is stored as recoverable internal strain energy. Therefore, the internal elastic strain

energy U for an infinitesimal element subjected to uniaxial stress is

dU =2

1s x dy dz ´ e x dx

(a) (b)



Module4/Lesson2

2


=2

1 s x e x dx dy dz

Therefore, dU =2

1s x e x dV

where dV = volume of the element.Thus, the above expression gives the strain energy stored in an elastic body per unit volume

of the material, which is called strain-energy density 0U .

Hence,dV

dU = 0U =

2

1s x e x

The above expression may be graphically interpreted as an area under the inclined lineon the stress-strain diagram as shown in Figure (4.1b). The area enclosed by the inclinedline and the vertical axis is called the complementary energy. For linearly elastic materials,the two areas are equal.

4.2.2 STRAIN ENERGY IN AN ELASTIC BODY

(a) (b)



Module4/Lesson2

3


Figure 4.2. Infinitesimal element subjected to: uniaxial tension (a), with resultingdeformation (b); pure shear (c), with resulting deformation (d)

When work is done by an external force on certain systems, their internal geometric statesare altered in such a way that they have the potential to give back equal amounts of workwhenever they are returned to their original configurations. Such systems are calledconservative, and the work done on them is said to be stored in the form of potential energy.For example, the work done in lifting a weight is said to be stored as a gravitational potentialenergy. The work done in deforming an elastic spring is said to be stored as elastic potentialenergy. By contrast, the work done in sliding a block against friction is not recoverable; i.e.,friction is a non-conservative mechanism.

Now we can extend the concept of elastic strain energy to arbitrary linearly elastic bodiessubjected to small deformations.

Figure 4.2(a) shows a uniaxial stress component s x acting on a rectangular element,and Figure 4.2(b) shows the corresponding deformation including the elongation due

to the strain component e x. The elastic energy stored in such an element is commonly calledstrain energy.

In this case, the force s x dydz acting on the positive x face does work as the element

undergoes the elongation dx xe . In a linearly elastic material, strain grows in proportion to

stress. Thus the strain energy dU stored in the element, when the final values of stress and

strain are xs and xe is

dU =2

1 (s x dydz) (e xdx)

(c)(d)



Module4/Lesson2

4


=2

1s x e x dV (4.28)

where dV = dx dy dz = volume of the element.

If an elastic body of total volume V is made up of such elements, the total strain energy U isobtained by integration

U =2

1òv

s x ve dV (4.29)

Taking s x = A

P and e x =

L

d

where P = uniaxial load on the member

d = displacement due to load P

L = length of the member,

A = cross section area of the member

We can write equation (4.28) as

ò÷ ø öçè æ ÷ ø öçè æ =v

dV L A

PU d 21

Therefore, U =2

1P.d since V = L ´ A (4.30)

Next consider the shear stress component t xy acting on an infinitesimal element in

Figure 4.2(c). The corresponding deformation due to the shear strain component g xy is

indicated in Figure 4.2(d). In this case the force t xy dxdz acting on the positive y face does

work as that face translates through the distance g xy dy. Because of linearity, g xy and t xy grow

in proportion as the element is deformed.

The strain energy stored in the element, when the final values of strain and stress are g xy andt xy is

( )( )dydxdzdU xy xy g t 2

1=

dxdydz xy xy

g t 2

1=

Therefore, dU =2

1 t xy g xy dV (4.31)

The results are analogous to equation (4.28) and equation (4.31) can be written for any other

pair of stress and strain components (for example, s y and e y or t yz and g yz) whenever thestress component involved is the only stress acting on the element.

Finally, we consider a general state of stress in which all six stress components are present.The corresponding deformation will in general involve all six strain components. The total



Module4/Lesson2

5


strain energy stored in the element when the final stresses are s x , s y , s z , t xy , t yz , t zx and the

final strains are e x , e y , e z , g xy , g yz , g zx is thus

dU =2

1 (s xe x + s ye y + s ze z + t xy g xy + t yz g yz + tzx gzx) dV (4.32)

In general, the final stresses and strains vary from point to point in the body. The strainenergy stored in the entire body is obtained by integrating equation (4.32) over the volumeof the body.

U =2

1 ò

v

(s xe x + s ye y + s ze z + t xy g xy + t yz g yz + t zx g zx) dV (4.33)

The above formula for strain energy applies to small deformations of any linearlyelastic body.

4.2.3 BOUNDARY CONDITIONS

The boundary conditions are specified in terms of surface forces on certain boundaries of a body to solve problems in continuum mechanics. When the stress components vary over the

volume of the body, they must be in equilibrium with the externally applied forces on the boundary of the body. Thus the external forces may be regarded as a continuation of internal

stress distribution.

Consider a two dimensional body as shown in the Figure 4.3

Figure 4.3 An element at the boundary of a body



Module4/Lesson2

6


Take a small triangular prism ABC , so that the side BC coincides with the boundary of the

plate. At a point P on the boundary, the outward normal is n. Let X and Y be the

components of the surface forces per unit area at this point of boundary. X and Y must be a

continuation of the stresses y x s s , and xy

t at the boundary. Now, using Cauchy’s equation,

we have

ml X T xy x x t s +== (a)

mlY T y xy y s t +==

in which l and m are the direction cosines of the normal n to the boundary.

For a particular case of a rectangular plate, the co-ordinate axes are usually taken parallel tothe sides of the plate and the boundary conditions (equation a) can be simplified. For

example, if the boundary of the plate is parallel to x-axis, at point 1P , then the boundary

conditions become

xy X t = and yY s = (b)

Further, if the boundary of the plate is parallel to y-axis, at point 2P , then the boundary

conditions become

x X s = and xyY t = (c)

It is seen from the above that at the boundary, the stress components become equal to thecomponents of surface forces per unit area of the boundary.

4.2.4 ST. VENANT’S PRINCIPLE

For the purpose of analysing the statics or dynamics of a body, one force system may be

replaced by an equivalent force system whose force and moment resultants are identical.Such force resultants, while equivalent need not cause an identical distribution of strain,

owing to difference in the arrangement of forces. St. Venant’s principle permits the use of an

equivalent loading for the calculation of stress and strain.

St. Venant’s principle states that if a certain system of forces acting on a portion of the

surface of a body is replaced by a different system of forces acting on the same portion of the

body, then the effects of the two different systems at locations sufficiently far distant from

the region of application of forces, are essentially the same, provided that the two systems of

forces are statically equivalent (i.e., the same resultant force and the same resultant moment).

St. Venant principle is very convenient and useful in obtaining solutions to many

engineering problems in elasticity. The principle helps to the great extent in prescribing the

boundary conditions very precisely when it is very difficult to do so. The following figures

4.4, 4.5 and 4.6 illustrate St. Venant principle.



Module4/Lesson2

7


Figure 4.4 Surface of a body subjected to (a) Concentrated load and

(b) Strip load of width b/2

y

P

+b

-b

x

0,14+ q

-1,31q

-5,12q

-1,31q

0,14+ q

0,16+ q

0,16+ q

-0,72q

-1,72q

-2,64q

-0,72q

-1,72q

0,11- q

0,11- q

0,50- q

0,95- q

1,53- q

1,85- q

0,44- q

0,44- q

0,69- q

1,00- q

1,33- q

1,48- q

0,85- q

0,85- q

0,91- q

1,00- q

1,10- q

1,14- q

0,99- q

0,99- q

0,98- q

1,00- q

1,02- q

1,03- q

0,17+ q

-0,37q

-1,95q

-3,39q

0,17+ q

0,14+ q

0,14+ q

-0,83q

-1,73q

-2,29q

0,16- q

0,16- q

-0,99q

-1,49q

-1,74q

-0,54q

0,48- q

0,48- q

-1,02q

-1,30q

-0,72q

0,87- q

0,87- q

-1,00q

-1,10q

-0,92q

-1,13q

0,99- q

0,99- q

-1,00q

-1,02q

-0,99q

y

P

+b

-b

x

b2

1

4 x b=

b2

x = 3

4 x b=

3

2 x b= x b= x b= 2

1

2bq =



Module4/Lesson2

8


Figure 4.5 Surface of a body subjected to (a) Strip load of width b and

(b) Strip load of width 1.5 b

0,08+ q

-0,72q

-1,25q

-1,33q

-1,34q

0,08+ q

0,32- q

-0,76q

-1,12q

-1,27q

-1,30q

0,32- q

0,80- q

-0,89q

-1,02q

-1,12q

-1,15q

0,80- q

0,95- q

-0,98q

-1,00q

-1,03q

-1,05q

0,95- q

1,00- q

-1,00q

-1,00q

-1,00q

-1,00q

1,00- q

P

+b

-b

x

y

3

2b

b

4 x =

b

2 x =

3

4 x b=

3

2 x b= x b= x b= 2

P

2bq =

0,61- q

-0,83q

-1,05q

-1,19q

-1,23q

0,61- q

0,18+ q

-0,20q

-1,03q

-1,84q

-1,95q

0,18+ q

0,03+ q

-0,48q

-1,06q

-1,56q

-1,73q

0,03+ q

0,32- q

-0,66q

-1,05q

-1,37q

-1,49q

0,32- q

0,60- q

-0,79q

-1,03q

-1,23q

-1,30q

0,60- q

0,90- q

-0,95q

-1,00q

-1,07q

-1,10q

0,90- q

0,99- q

-0,99q

-1,00q

-1,01q

-1,02q

0,99- q

P

+b

-b

x

y

b



Module4/Lesson2

9


Figure 4.6 Surface of a body subjected to (a) Two strip load and

(b) Inverted parabolic two strip loads

Figures 4.4, 4.5 and 4.6 demonstrate the distribution of stresses (q) in the body when

subjected to various types of loading. In all the cases, the distribution of stress throughout the body is altered only near the regions of load application. However, the stress distribution is

not altered at a distance b x 2= irrespective of loading conditions.

4.2.5 EXISTENCE AND UNIQUENESS OF SOLUTION (UNIQUENESS

THEOREM)

This is an important theorem in the theory of elasticity and distinguishes elastic deformationsfrom plastic deformations. The theorem states that, for every problem of elasticity defined by

a set of governing equations and boundary conditions, there exists one and only one solution.This means that “elastic problems have a unique solution” and two different solutions cannotsatisfy the same set of governing equations and boundary conditions.

0,13+ q

-2,06q

-0,28q

0,13+ q

-2,06q

0,45- q

-0,78q

0,45- q

-1,35q

-1,35q

0,80- q

-0,96q

0,80- q

-1,11q

-1,11q

0,93- q

-1,01q

0,93- q

-1,03q

-1,03q

0,99- q

-1,00q

0,99- q

-1,01q

-1,01q

1,00- q

-1,00q

1,00- q

-1,00q

-1,00q

+b

-b

x

y

P

2

P

2

0,16+ q

-2,20q

-0,25q

0,16+ q

-2,20q

0,41- q

-1,39q

-0,74q

0,41- q

-1,39q

0,80- q

-0,74q

0,80- q

-1,12q

-1,12q

0,93- q

-1,01q

0,93- q

-1,03q

-1,03q

0,99- q

-1,02q

0,99- q

-1,01q

-1,01q

1,00- q

P

2

P

2

+b

-b

x

y

-1,00q

1,00- q

-1,00q

-1,00q

14

x = b1

2 x = b

3

4 x b=

3

2 x b= x b= x b= 2

P

2q = b



Module4/Lesson2

10


Proof

In proving the above theorem, one must remember that only elastic problems are dealt with

infinitesimal strains and displacements. If the strains and displacements are not infinitesimal,

the solution may not be unique.

Let a set of stresses zx y x t s s ¢¢¢ ,........, represents a solution for the equilibrium of a body undersurface forces X , Y , Z and body forces F x , F y , F z. Then the equations of equilibrium and

boundary conditions must be satisfied by these stresses, giving

0 xy x xz

xF

x y z

t s t ¢¶¢ ¢¶ ¶+ + + =

¶ ¶ ¶; ( x , y , z)

and ),,(; z y xF nml x xz xy x =¢+¢+¢ t t s

where ( x , y , z) means that there are two more equations obtained by changing the suffixes y for x and z for y, in a cyclic order.

Similarly, if there is another set of stresses zx y x

t s s ¢¢¢¢¢¢ ,...., which also satisfies the boundary

conditions and governing equations we have,

),,(;0 z y x x z y x

xz xy x =+¶

¢¢¶+

¶

¢¢¶+

¶

¢¢¶ t t s

and ),,(; z y xF nml x xz xy x

=¢¢+¢¢+¢¢ t t s

By subtracting the equations of the above set from the corresponding equations of the previous set, we get the following set,

( ) ( ) ( ) ),,(;0 z y x

z y x

xz xz xy xy x x =¢¢-¢

¶

¶+¢¢-¢

¶

¶+¢¢-¢

¶

¶t t t t s s

and ( ) ) ( ) ),,(;0 z y xnml xz xz xy xy x x

=¢¢-¢+¢¢-¢+¢¢-¢ t t t t s s

In the same way it is shown that the new strain components ( e ' x -e '' x) ,

(e ' y -e '' y)…. etc. also satisfy the equations of compatibility. A new solution (s ' x - s '' x) ,

(s ' y - s '' y) ,….. (t ' xz -t '' xz) represents a situation where body forces and surface forces both are

zero. The work done by these forces during loading is zero and it follows that the total

strain energy vanishes, i.e.,

ò ò ò V o dxdydz = 0

where V o = (s xe x + s ye y + s ze z + t xy g xy + t yz g yz + t zx g zx)



Module4/Lesson2

11


The strain energy per unit volume V o is always positive for any combination of strains and

stresses. Hence for the integral to be zero, V o must vanish at all the points, giving all the

stress components (or strain components) zero, for this case of zero body and surface forces.

Therefore (s ' x-s '' x)=(s ' y-s '' y )=(s ' z-s '' z)= 0

and (t ' xy-t '' xy) = (t ' yz-t '' yz) = (t ' zx-t '' zx) = 0

This shows that the set s ' x , s ' y , s ' z ,…. t ' zx is identical to the set s '' x , s '' y , s '' z , …. t '' zx and

there is one and only one solution for the elastic problem.


Example 4.1

The following are the principal stress at a point in a stressed material. Taking2/210 mmkN E = and 3.0=n , calculate the volumetric strain and the Lame’s

constants.222

/120,/150,/200 mm N mm N mm N z y x === s s s

Solution: We have

( )[ ] z y x x

E s s n s e +-=

1

( )[ ]1201503.020010210

13

+-´

=

41067.5 -´=\ x

e

( )[ ] x z y y E s s n s e +-=

1

( )[ ]2001203.015010210

13

+-´

=

41057.2 -´=\ ye

( )[ ] y x z z

E s s n s e +-=

1

( )[ ]1502003.0120

10210

13

+-

´

=

51014.7 -´=\ z

e

Volumetric strain = ) z y xv

e e e e ++=



Module4/Lesson2

12


544 1014.71057.21067.5 --- ´+´+´=

310954.8 -´=\v

e

To find Lame’s constants

We have, ( )n += 12

E

G

( )3.012

10210 3

+´

=G

23 /1077.80 mm N G ´=\

( )( )G E

E GG

3

2

--

=l

( )( )33

333

1077.80310210

102101077.8021077.80

´´-´

´-´´´=

23

/1014.121 mm N ´=\l

Example 4.2

The state of strain at a point is given by

001.0,004.0,0,003.0,001.0 =-===-== yz xz xy z y x g g g e e e

Determine the stress tensor at this point. Take26 /10210 mkN E ´= ,

Poisson’s ratio = 0.28. Also find Lame’s constant.

Solution: We have

( )n +=

12

E G

( )28.012

10210 6

+´

=

26 /1003.82 mkN G ´=\

But( )

( )G E

E GG

3

2

--

=l

( )( )66

666

1003.82310210

102101603.8221003.82

´´-´

´-´´´=

26 /1042.104 mkN ´=\l

Now,

( ) ) z y x x

G e e l e l s +++= 2



Module4/Lesson2

13


( ) ( )0003.01042.104001.01042.10403.822 66 +-´+´+´= 2/44780 mkN

x -=\s

or MPa x

78.44-=s

( ) ( ) x z y y

G e e l e n s +++= 2

( ) ( ) ( )001.001042.104003.01042.10403.822 66 +´+-´´+´=

2/701020 mkN y -=\s

or MPa y 02.701-=s

( ) ) y x z z G e e l e l s +++= 2

= ( ) ( ) ( )003.0001.01042.10401042.10403.822 66 -´++´

2/208840 mkN z -=\s

or MPa z

84.208-=s

xy xy Gg t =

= 01003.82 6 ´´

0=\ xy

t

26 /82030001.01003.82 mkN G yz yz =´´== g t

or MPa yz

03.82=t

( ) 26 /328120004.01003.82 mkN G xz xz -=-´´== g t

MPaor xz 12.328-=t

\ The Stress tensor is given by

÷÷÷

ø

ö

ççç

è

æ

--

-

--

=÷÷÷

ø

ö

ççç

è

æ

=

84.20803.8212.328

03.8202.7010

12.328078.44

z yz xz

yz y xy

xz xy x

ij

s t t

t s t

t t s

s

Example 4.3

The stress tensor at a point is given as

2

/160100120100240160

120160200

mkN ÷÷÷

ø

ö

ççç

è

æ

--

-

Determine the strain tensor at this point. Take 3.0/10210 26 =´= n and mkN E



Module4/Lesson2

14


Solution: ( )[ ] z y x x

E s s n s e +-=

1

= ( )[ ]1602403.020010210

16

+--´

6

10067.1

-

´=\ xe

( )[ ] x z y y

E s s n s e +-=

1

= ( )[ ]2001603.024010210

16

+--´

610657.1 -´-=\ ye

( )[ ] y x z z

E s s n s e +-=

1

= ( )[ ]2402003.0160

10210

16

--

´

61082.0 -´=\ ze

Now,( ) ( )

266

/1077.803.012

10210

12mkN

E G ´=

+´

=+

=n

xy xy xy G g g t ´´== 61077.80

6

610981.1

1077.80

160 -´=´

==\G

xy

xy

t g

6

61024.1

1077.80

100 -´=´

==G

yz

yz

t g

66 10486.1

1077.80120 -´-=´-== G

zx zx t g

Therefore, the strain tensor at that point is given by

÷÷÷÷÷÷÷

ø

ö

ççççççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

=

z

zy zx

yz

y

xy

xz xy

x

z zy zx

yz y xy

xz xy x

ij

e g g

g e

g

g g e

e e e

e e e

e e e

e

22

22

22

610

82.062.0743.0

62.0657.19905.0

743.09905.0067.1

-´÷÷÷

ø

ö

ççç

è

æ

-

-

-

=\ ije



Module4/Lesson2

15


Example 4.4

A rectangular strain rosette gives the data as below.

msmicrometre /6700 =e



Find the principal stresses 21 s s and if 3.0,102 5 =´= n MPa E

Solution: We have6

0 10670 -´== e e x

6

90 10150 -´== e e y

( )900452 e e e g +-= xy = ( )666 1015010670103302 --- ´+´-´´

610160 -´-=\ xyg

Now, the principal strains are given by

( ) 22

minmax21

2 xy y x

y xor g e e e e e e +-±÷÷

ø öçç

è æ +=

i.e., ( )[ ] ( )26266

minmax 10160101506702

110

2

150670 --- ´-+-±÷ ø

öçè

æ +=e e or

66

minmax 1003.27210410 -- ´±´=\ e e or

6

1max 103.682 -´==\ e e

6

2min 1097.137 -´== e e

The principal stresses are determined by the following relations

( ) ( )( )

5

2

6

2

211 102

3.011097.1373.003.682.

1´´

-´+=

-+=

-

E n ne e s

MPa1591 =\s

Similarly,( ) ( )

( )5

2

6

2

122 102

3.01

1003.6823.097.137.

1´´

-

´+=

-

+=

-

E n

ne e s

MPa3.752 =\s



Module5/Lesson1

1

Applied Elasticity for Engineer T.G.Sitharam & L.GovindaRaju

Module 5: Two Dimensional Problems in

Cartesian Coordinate System

5.1.1 INTRODUCTION

Plane Stress Problems

In many instances the stress situation is simpler than that illustrated in Figure 2.7. An

example of practical interest is that of a thin plate which is being pulled by forces in the

plane of the plate. Figure 5.1 shows a plate of constant thickness, t subjected to axial and

shear stresses in the x and y directions only. The thickness is small compared to the other

two dimensions of plate. These stresses are assumed to be uniformly distributed over the

thickness t . The surface normal to the z-axis is stress free.

Figure 5.1 General case of plane stress

The state of stress at a given point will only depend upon the four stress componentssuch as

úû

ù

êë

é

y yx

xy x

s t

t s (5.0)



Module5/Lesson1

2


in which the stress components are functions of only x and y. This combination of stress

components is called "plane stress" in the xy plane. The stress-strain relations for plane stressis given by

( ) y x x

E ns s e -=

1

( ) x y y

E ns s e -=

1 (5.1)

G

xy

xy

t g =

and ( ) y x z yz xz

E

vs s e g g +-=== ,0

Compatibility Equation in terms of Stress Components (Plane stress case)

For two dimensional state of strain, the condition of compatibility (Eq. 3.21) is given by

2

2

y

x

¶

¶ e +

2

2

x

y

¶

¶ e =

y x

xy

¶¶

¶ g 2

(5.1a)

Substituting Eq. 5.1 in Eq. 5.1a

y x x y

xy

x y y x¶¶

¶+=-

¶

¶+-

¶

¶ t n ns s ns s

2

2

2

2

2

)1(2)()( (5.1b)

Further equations of equilibrium are given by

0=+¶

¶+

¶

¶ x

xy x F y x

t s (5.1c)

0=+¶

¶+

¶

¶ y

xy yF

x y

t s (5.1d)

Differentiate (5.1c) with respect to x and (5.1d) with respect to y and adding the two, we get

úû

ùêë

é

¶

¶+

¶

¶-=

¶¶

¶+

¶

¶+

¶

¶

y

F

x

F

y x y x

y x xy y x t s s

2

2

2

2

2

2 (5.1e)

Substituting Eq. (5.1e) in Eq. (5.1b), we get



Module5/Lesson1

3


( ) ( ) ÷÷ ø

öççè

æ

¶

¶+

¶

¶+-=+÷÷

ø

öççè

æ

¶

¶+

¶

¶

y

F

x

F v

y x

y x

y x 12

2

2

2

s s (5.2)

If the body forces are constant or zero, then

( ) 02

2

2

2

=+÷÷ ø

öççè

æ

¶

¶+

¶

¶ y x

y xs s (5.2 a)

This equation of compatibility, combined with the equations of equilibrium, represents a

useful form of the governing equations for problems of plane stress. The constitutive

relation for such problems is given by

( ) ï

ïý

ü

ïî

ïí

ì

úúúúú

û

ù

êêêêê

ë

é

÷ ø

öçè

æ --=

ï

ïý

ü

ïî

ïí

ì

xy

y

x

xy

y

x E

g

e e

n n

n

n t

s s

2

100

0101

1 2 (5.3)

Plane Strain Problems

Problems involving long bodies whose geometry and loading do not vary significantly in thelongitudinal direction are referred to as plane-strain problems. Some examples of practical

importance, shown in Figure 5.2, are a loaded semi-infinite half space such as a strip footingon a soil mass, a long cylinder; a tunnel; culvert; a laterally loaded retaining wall; and a longearth dam. In these problems, the dependent variables can be assumed to be functions of onlythe x and y co-ordinates, provided a cross-section is considered some distance away from theends.



Module5/Lesson1

4


(a) Strip Footing (b) Long cylinder

(c) Retaining wall (d) Earth Dam

Figure 5.2 Examples of practical plane strain problems

Hence the strain components will be

xe = x

u

¶

¶,

ye =

y

v

¶

¶,

xyg =

x

v

y

u

¶

¶+

¶

¶ (5.4)



Module5/Lesson1

5


ze = z

w

¶

¶ = 0 , xzg =

z

u

x

w

¶

¶+

¶

¶ =0 , yzg =

z

v

y

w

¶

¶+

¶

¶ = 0 (5.5)

Moreover, from the vanishing of ze , the stress z

s can be expressed in terms of xs and y

s

as

) y x z s s n s += (5.6)

Compatibility Equation in terms of Stress Components (Plane strain case)

Stress-strain relations for plane strain problems are

( )[ ] y x x

E s n n s n e )1(1

1 2 +--=

( )[ ] x y y E

s n n s n e )1(11 2+--= (5.6 a)

G

xy

xy

t g =

The equilibrium equations, strain-displacement elations and compatibility conditions are thesame as for plane stress case also. Therefore substituting Eq. (5.6 a) in Eq. (5.1 a), we get

( ) y x x y x y

xy x y y x

¶¶

¶=

ú

ú

û

ù

ê

ê

ë

é

¶

¶+

¶

¶-

ú

ú

û

ù

ê

ê

ë

é

¶

¶+

¶

¶-

t s s n

s s n

2

2

2

2

2

2

2

2

2

21 (5.6 b)

Now, differentiating the equilibrium equations (5.1 c) and (5.1 d) and adding the results as before and then substituting them in Eq. (5.6 b), we get

( ) ÷÷ ø

öççè

æ

¶

¶+

¶

¶

--=+÷÷

ø

öççè

æ

¶

¶+

¶

¶

y

F

x

F

y x

y x y x

n s s

1

12

2

2

2

(5.6 c)

If the body forces are constant or zero, then

( ) 02

2

2

2

=+÷÷ ø

öççè

æ

¶

¶+

¶

¶ y x

y xs s (5.6 d)

It can be noted that equations (5.6 d) and (5.2 a) are identical. Hence, if the body

forces are zero or constant, the differential equations for plane strain will be same as

that for plane stress. Further, it should be noted that neither the compatibility



Module5/Lesson1

6


equations nor the equilibrium equations contain the elastic constants. Hence, the

stress distribution is same for all isotropic materials in two dimensional state of

stress. Also, the constitutive relation for plane strain problems is given by

( )( )

( )( )

ï

ïý

ü

ïî

ïí

ì

úúúúú

û

ù

êêêêê

ë

é

÷ ø

öçè

æ --

-

-+=

ï

ïý

ü

ïî

ïí

ì

xy

y

x

z

y

x E

g

e

e

n

n n

n n

n n s

s

s

2

2100

01

01

211

Relationship between plane stress and plane strain

(a) For plane-stress case

From the stress-strain relationship (equation 4.20), we have

( ) ) z y x x G e e l e l s +++= 2

or ) z y x x x G e e l e l e s +++= 2

or ) x z y x x Ge e e e l s 2+++=

Similarly, y z y x y Ge e e e l s 2+++=

02 =+++= z z y x z Ge e e e l s

00 ===== zx zx yz yz xy xy r Gr Gr G t t t

Denoting ) ==++ 1 J z y x e e e First invariant of strain, then

02,2,2 111 =+=+=+= z z y y x x G J G J G J e l s e l s e l s (a)

From, 0= zs , we get

( )( )

y x zG

e e l

l e +

+-=

2

Using the above value of ze , the strain invariant 1 J becomes

( ) y xG

G J e e

l ++= 2

21 (b)

Substituting the value of 1 J in equation (a), we get



Module5/Lesson1

7


( ) x y x x G

G

Ge e e

l

l s 2

2

2++

+=

( ) y y x y G

G

Ge e e

l

l s 2

2

2++

+=

(b) For plane-strain case

Here 0= ze

) x y x x x GG J e e e l e l s 221 ++=+=\

) y y x y y GG J e e e l e l s 221 ++=+=

1 J z l s = ) y x

e e l +=

If the equations for stress xs for plane strain and plane stress are compared, it can be

observed that they are identical except for the comparison of co-efficients of the term

) y x e e + .

i.e.,

( )

( )ïî

ïí

ì

+++

++

=stress plane2

2

2

strain plane2

x y x

x y x

xG

G

G

G

e e e l

l

e e e l

s

Since all the equations for stresses in plane-stress and plane-strain solutions are identical, the

results from plane strain can be transformed into plane stress by replacing l in plane-strain

case byG

G

22+l

l in plane-stress case. This is equivalent to replacingn

n -1

in plane strain

case by n in plane stress case. Similarly, a plane-stress solution can be transformed into a

plane-strain solution by replacingG

G

2

2

+l

l in plane-stress case by l in plane-strain case.

This is equivalent to replacing n in plane-stress case byn

n

-1in plane-strain case.



Module5/Lesson2

1


Module 5: Two Dimensional Problems in

Cartesian Coordinate System

5.2.1 THE STRESS FUNCTION

For two-dimensional problems without considering the body forces, the equilibriumequations are given by

0=¶

¶+

¶

¶

y x

xy xt s

0=¶

¶+

¶

¶

x y

xy y t s

and the equation of compatibility is

ø

öççè

æ

¶

¶+

¶

¶2

2

2

2

y x) 0=+

y x s s

The equations of equilibrium are identically satisfied by the stress function, ( ) y x,f ,

introduced by G. B. Airy for the two dimensional case. The relationships between the stress

function f and the stresses are as follows:

y x x y xy y x

¶¶

¶-=

¶

¶=

¶

¶=

f t

f s

f s

2

2

2

2

2

,, (5.11)

Substituting the above expressions into the compatibility equation, we get

4

4

x¶

¶ f + 2

22

4

y x ¶¶

¶ f +

4

4

y¶

¶ f = 0 (5.12)

Further, equilibrium equations are automatically satisfied by substituting the aboveexpressions for stress components.

In general, 04 =Ñ f

where4Ñ =

4

4

x¶

¶ +

22

42

y x ¶¶

¶ +

4

4

y¶

¶

The above Equation (5.12) is known as "Biharmonic equation" for plane stress and plane

strain problems. Since the Biharmonic equation satisfies all the equilibrium and

compatibility equations, a solution to this equation is also the solution for a two-dimensional



Module5/Lesson2

2


problem. However, the solution in addition to satisfying the Biharmonic equation also has to

satisfy the boundary conditions.

To solve the derived equations of elasticity, it is suggested to use polynomial functions,

inverse functions or semi-inverse functions. The use of polynomial functions for solving

two-dimensional problems is discussed in the next article. The inverse method requires

examination of the assumed solutions with a view towards finding one which will satisfy the

governing equations and the boundary conditions.

The semi-inverse method requires the assumption of a partial solution, formed by expressing

stress, strain, displacement, or stress function in terms of known or undetermined

coefficients. The governing equations are thus rendered more manageable.



1

Module 5: Two Dimensional

Problems in Cartesian Coordinate System

5.3.1 SOLUTIONS OF TWO-DIMENSIONAL PROBLEMS BY THE USE

OF POLYNOMIALS

The equation given by

÷÷ ø

öççè

æ

¶

¶+

¶

¶2

2

2

2

y x ÷÷ ø

öççè

æ

¶

¶+

¶

¶2

2

2

2

y x

f f =

4

4

x¶

¶ f + 2

22

4

y x ¶¶

¶ f +

4

4

y¶

¶ f = 0 (5.13)

will be satisfied by expressing Airy’s function ( ) y x,f in the form of homogeneous

polynomials.

(a) Polynomial of the First Degree

Let yb xa 111 +=f

Now, the corresponding stresses are

xs =2

1

2

y¶

¶ f = 0

ys =

2

1

2

x¶

¶ f = 0

xyt =

y x¶¶

¶- 1

2f = 0

Therefore, this stress function gives a stress free body.

(b) Polynomial of the Second Degree

Let 2f =22

2

22

22 y

c xyb x

a++

The corresponding stresses are

xs =

2

2

2

y¶

¶ f = 2c

ys =

2

2

2

x¶

¶ f = 2a

xyt = y x¶¶

¶- f

2

= 2b-

This shows that the above stress components do not depend upon the co-ordinates x and y,

i.e., they are constant throughout the body representing a constant stress field. Thus, the





3

and At y = +h, hd x 3+=s

The variation of xs with y is linear as shown in the Figure 5.4.

Figure 5.4 Variation of Stresses

Similarly, if all the coefficients except 3b are zero, then we get

0= xs

yb y 3=s

xb xy 3-=t

The stresses represented by the above stress field will vary as shown in the Figure 5.5.



4

Figure 5.5 Variation of Stresses

In the Figure 5.5, the stress ys is constant with x (i.e. constant along the span L of

the beam), but varies with y at a particular section. At y = +h, hb y 3=s (i.e., tensile),

while at y = -h, hb y 3-=s (i.e. compressive). x

s is zero throughout. Shear stress xyt is

zero at 0= x and is equal to Lb3- at x = L. At any other section, the shear stress is

proportional to x.

(d) Polynomial of the Fourth Degree

Let 4f = 44342243444

1262612 ye xyd y xc y xb xa ++++

The corresponding stresses are given by

2

44

2

4 ye xyd xc x

++=s

2

44

2

4 yc xyb xa y ++=s

244

24

22

2 y

d xyc x

b xy ÷

ø

öçè

æ --÷

ø

öçè

æ -=t

Now, taking all coefficients except4

d equal to zero, we find

,4 xyd x =s ,0=

ys

24

2 y

d xy

-=t



5

Assuming 4d positive, the forces acting on the beam are shown in the Figure 5.6.

Figure 5.6 Stresses acting on the beam

On the longitudinal sides, y = ±h are uniformly distributed shearing forces. At the ends, the

shearing forces are distributed according to a parabolic distribution. The shearing forcesacting on the boundary of the beam reduce to the couple.

Therefore, M = hLhd

h Lhd

223

12

2

2

4

2

4 -

Or M = Lhd 3

43

2

This couple balances the couple produced by the normal forces along the side x = L ofthe beam.

(e) Polynomial of the Fifth Degree

5 4 3 2 2 3 4 55 5 5 5 5 55

20 12 6 6 12 20

a b c d e f Let x x y x y x y xy yj = + + + + +

The corresponding stress components are given by

3

55

2

55

2

5

35

2

5

2

)2(3

1)32(

3 yd b xyac y xd x

c

y x +-+-+=

¶

¶=

f s

352

5

2

5

3

52

5

2

3 yd xyc y xb xa

x y +++=

¶¶= f s

3

55

2

5

2

5

3

5

5

2

)32(3

1

3

1 yac xyd y xc xb

y x xy ++---=

¶¶

¶-= f

t



6

Here the coefficients 5555 ,,, d cba are arbitrary, and in adjusting them we obtain solutions

for various loading conditions of the beam.

Now, if all coefficients, except 5d , equal to zero, we find

÷ ø

öçè

æ -= 32

5

3

2 y y xd xs

2

5

3

53

1

xyd

yd

xy

y

-=

=

t

s

Case (i)The normal forces are uniformly distributed along the longitudinal sides of the beam.

Case (ii)

Along the side x = L, the normal forces consist of two parts, one following a linear law and

the other following the law of a cubic parabola. The shearing forces are proportional to x on

the longitudinal sides of the beam and follow a parabolic law along the side x = L.

The distribution of the stresses for the Case (i) and Case (ii) are shown in the Figure 5.7.

Case (i)



7

Case (ii)

Figure 5.7 Distribution of forces on the beam

5.3.2 BENDING OF A NARROW CANTILEVER BEAM SUBJECTED TO

END LOAD

Consider a cantilever beam of narrow rectangular cross-section carrying a load P at the endas shown in Figure 5.8.

Figure 5.8 Cantilever subjected to an end load

The above problems may be considered as a case of plane stress provided that the thickness

of the beam t is small relative to the depth 2h.



8

Boundary Conditions

0=±= h y At xy

t

0=±= h y At ys (5.14)

These conditions express the fact that the top and bottom edges of the beam are not loaded.Further, the applied load P must be equal to the resultant of the shearing forces distributedacross the free end.

Therefore, P = - ò+

-

h

h xydyb2t (5.14a)

By Inverse Method

As the bending moment varies linearly with x, and xs at any section depends upon y, it is

reasonable to assume a general expression of the form

xs = xyc y

12

2

=¶

¶ f (5.14b)

where c1 = constant. Integrating the above twice with respect to y, we get

f = )()(6

121

3

1 x f x yf xyc ++ (5.14c)

where f 1( x) and f 2( x) are functions of x to be determined. Introducing the f thus obtained

into Equation (5.12), we have

y 04

2

4

4

1

4

=+dx

f d

dx

f d (5.14d)

Since the second term is independent of y, there exists a solution for all x and y provided that

04

1

4

=

dx

f d and 0

4

2

4

=

dx

f d

Integrating the above, we get

f 1( x) = c2 x3+c3 x

2+c4 x+c5

f 2( x) = c6 x3+c7 x

2+c8 x+c9

where c2 , c3……., c9 are constants of integration.

Therefore, (5.14c) becomes

f = 98

2

7

3

654

2

3

3

2

3

1 )(6

1c xc xc xc yc xc xc xc xyc ++++++++

Now, by definition,

( ) ( )73622

2

26 c yc xc yc x

y +++=

¶

¶=

f s



9

t xy = 43

2

2

2

1

2

232

1c xc xc yc

y x----=÷÷

ø

öççè

æ

¶¶

¶-

f (5.14e)

Now, applying boundary conditions to (5.14e), we get

c2 = c3 = c6 = c7 = 0 and c4 =21- c1h2

Also, ò ò+

- - =-=-

h

h

h

h xyPdyh ybcdyb )(2

2

12 22

1t

Solving, c1 = - ÷ ø

öçè

æ -=÷÷

ø

öççè

æ

I

P

hb

P34

3

where I = 3

4 bh3 is the moment of inertia of the cross-section about the neutral axis.

From Equations (5.14b) and (5.14e), together with the values of constants, the stresses arefound to be

xs = - )(

2,0, 22

yh I

P

I

Pxy xy y -

-==÷

ø

öçè

æ t s

The distribution of these stresses at sections away from the ends is shown in Figure 5.8 b

By Semi-Inverse Method

Beginning with bending moment M z = Px, we may assume a stress field similar to the caseof pure bending:

xs = y I

Px÷

ø

öçè

æ -

xyt = ( ) y x xy ,t (5.14f)

0==== yz xz z y t t s s

The equations of compatibility are satisfied by these equations. On the basis ofequation (5.14f), the equations of equilibrium lead to

0=¶

¶+

¶

¶

y x

xy x t s

, 0=¶

¶

x

xyt

(5.14g)

From the second expression above, t xy depends only upon y. The first equation of (5.14g)

together with equation (5.14f) gives

I

Py

dy

d xy=

t



10

or t xy = c

I

Py+

2

2

Here c is determined on the basis of (t xy) y=±h = 0

Therefore , c = - I

Ph

2

2

Hence , t xy = I

Ph

I

Py

22

22

-

Or t xy = - )(2

22 yh I

P-

The above expression satisfies equation (5.14a) and is identical with the result previouslyobtained.

5.3.3 PURE BENDING OF A BEAM

Consider a rectangular beam, length L, width 2b, depth 2h, subjected to a pure couple M

along its length as shown in the Figure 5.9

Figure 5.9 Beam under pure bending

Consider a second order polynomial such that its any term gives only a constant state of

stress. Therefore

f = 2a22

2

22

2 yc xyb

x++

By definition,



11

xs =

2

2

y¶

¶ f , ys =

2

2

x¶

¶ f , t xy = ÷÷

ø

öççè

æ

¶¶

¶-

y x

f 2

\ Differentiating the function, we get

s x = 2

2

y¶

¶ f = c2 , s y = 2

2

x¶

¶ f = a2 and t xy = ÷÷ ø

ö

ççè

æ

¶¶

¶

- y x

f 2

= -b2

Considering the plane stress case,

s z = t xz = t yz = 0

Boundary Conditions

(a) At y = ± h, = y

s 0

(b) At y = ± h, t xy = 0

(c) At x = any value,

2b ò+

-

a

a

x ydys = bending moment = constant

\2bx ò+

-

+

-

=úû

ùêë

é=

h

h

h

h

y xbc ydyc 0

22

2

22

Therefore, this clearly does not fit the problem of pure bending.

Now, consider a third-order equation

f =6226

3

3

2

323

3

3 yd xyc y x

b xa+++

Now, xs = yd xc y

332

2

+=¶

¶ f (a)

ys = a3 x + b3 y (b)

t xy = -b3 x - c3 y (c)

From (b) and boundary condition (a) above,

0 = a3 x ± b3a for any value of x

\ a3 = b3 = 0

From (c) and the above boundary condition (b),

0 = -b3 x ± c3a for any value of x

therefore c3 = 0

hence, xs = d 3 y ys = 0

t xy = 0

Obviously, Biharmonic equation is also satisfied.



12

i.e., 024

4

22

4

4

4

=¶

¶+

¶¶

¶+

¶

¶

y y x x

f f f

Now, bending moment = M = 2b ò+

-

h

h x ydys

i.e. M = 2b ò

+

-

h

h dy yd 2

3

= 2bd 3 ò+

-

h

hdy y 2

= 2bd 3

h

h

y +

-

úû

ùêë

é

3

3

M = 4bd 33

3h

Or d3 =34

3

bh

M

I

M d =3 where

3

4 3bh I =

Therefore, xs = y

I

M



13

5.3.4 BENDING OF A SIMPLY SUPPORTED BEAM BY A DISTRIBUTED

LOADING (UDL)

Figure 5.10 Beam subjected to Uniform load

Consider a beam of rectangular cross-section having unit width, supported at the ends and

subjected to a uniformly distributed load of intensity q as shown in the Figure 5.10.

It is to be noted that the bending moment is maximum at position x = 0 and decreases with

change in x in either positive or negative directions. This is possible only if the stress

function contains even functions of x. Also, it should be noted that ys various from zero at

y = -c to a maximum value of -q at y = +c. Hence the stress function must contain odd

functions of y.

Now, consider a polynomial of second degree with

022 == cb

222

2 x

a=\f

a polynomial of third degree with 033 == ca

33233

62 y

d y x

b+=\f

and a polynomial of fifth degree with 05555 ==== ecba

úû

ùêë

é-=-=\ 55

553255

3

2

306d f y

d y x

d f

532 f f f f ++=\

or55325332322

306622 y

d y x

d y

d y x

b x

a-+++=f (1)

Now, by definition,



14

÷ ø

öçè

æ -+=

¶

¶= 32

532

2

3

2 y y xd yd

y x

f s (2)

35322

2

3 y

d yba

x y

++=¶

¶=

f s (3)

2

53 xyd xb xy --=t (4)

The following boundary conditions must be satisfied.

(i) 0=±= c y xyt

(ii) 0=+= c y ys

(iii) qc y y -=

-=s

(iv) ( )ò+

-

±= =c

c

L x x dy 0s

(v) ( )ò+

-±= ±=

c

c

L x xy qLdyt

(vi) ( )ò+

-

±= =c

c

L x x ydy 0s

The first three conditions when substituted in equations (3) and (4) give

02

53 =-- cd b

03

3532 =++ c

d cba

qcd cba -=--35

323

which gives on solving

35324

3,

4

3,

2 c

qd

c

qb

qa -==-=

Now, from condition (vi), we have

ò+

-

=úû

ùêë

é÷

ø

öçè

æ -+

c

c

ydy y y xd yd 03

2 32

53

Simplifying,

÷ ø

öçè

æ --= 22

535

2h Ld d



15

÷ ø

öçè

æ -=

5

2

4

32

2

h

L

h

q

÷ ø

öçè

æ --÷

ø

öç

è

æ -=\ 32

32

2

3

2

4

3

5

2

4

3 y y x

h

q y

h

L

h

q x

s

3

344

3

2 y

h

q y

h

qq y

-+÷ ø

öçè

æ -=s

2

34

3

4

3 xy

h

q x

h

q xy

+÷ ø

öçè

æ -=t

Now, ( ) 3

33

3

2

12

8

12

21h

hh I ==

´=

where I = Moment of inertia of the unit width beam.

( ) ÷ ø

öçè

æ -+-=\

532

23

22 yh y

I

q y x L

I

q x

s

÷ ø

öçè

æ +-÷

ø

öçè

æ -= 32

3

3

2

32h yh

y

I

q y

s

( )22

2 yh x

I

q xy

-÷ ø

öçè

æ -=t


Example 5.1

Show that for a simply supported beam, length 2 L, depth 2 a and unit width, loaded by

a concentrated load W at the centre, the stress function satisfying the loading condition

is cxy xyb

+= 2

6f the positive direction of y being upwards, and x = 0 at midspan.



16

Figure 5.11 Simply supported beam

Treat the concentrated load as a shear stress suitably distributed to suit this function, and so

that ò+

-

÷ ø

öçè

æ -=

a

a

x

W dy

2s on each half-length of the beam. Show that the stresses are

÷ ø

öçè

æ -= xy

a

W x 34

3s

0= y

s

úû

ùêë

é÷÷

ø

öççè

æ --=

2

2

18

3

a

y

a

W xyt

Solution: The stress components obtained from the stress function are

bxy y

x =¶

¶=

2

2f s

02

2

=¶

¶=

x y

f s

cby

y x xy +÷÷

ø

öççè

æ -=

¶¶

¶-=

2

22f t



17

Boundary conditions are

(i) a y for y ±== 0s

(ii) a y for xy

±== 0t

(iii) ò

+

-±==-

a

a

xy L x for

W

dy 2t

(iv) ò+

-

±==a

a

x L x for dy 0s

(v) ò+

-

±==a

a

x L x for ydy 0s

Now,

Condition (i)

This condition is satisfied since 0= y

s

Condition (ii)

cba

+÷÷ ø

öççè

æ -=

20

2

2

2ba

c =\

Condition (iii)

( )ò+

-

--=a

a

dy yabW 22

22

÷÷ ø

öççè

æ --=

3

22

2

33 a

ab

÷÷ ø

öççè

æ -=\

3

2

2

3baW

or ÷ ø

öçè

æ -=

34

3

a

W b

and ÷ ø

ö

çè

æ

-= a

W

c 8

3

Condition (iv)



18

ò+

-

=÷ ø

öçè

æ -

a

a

xydya

W 0

4

33

Condition (v)

ò

+

-=

a

a

x ydy M s

ò+

-

÷ ø

öçè

æ -=

a

a

dy xya

W 2

34

3

2

Wx M =\

Hence stress components are

xya

W x ÷

ø

öçè

æ -=

34

3s

0= ys

÷ ø

öçè

æ -÷÷

ø

öççè

æ =

a

W y

a

W xy

8

3

24

3 2

3t

úû

ùêë

é÷÷

ø

öççè

æ --=\

2

2

18

3

a

y

a

W xyt

Example 5.2

Given the stress function ÷ ø

öçè

æ ÷

ø

öçè

æ = -

z

x z

H 1tanp

f . Determine whether stress function f is

admissible. If so determine the stresses.

Solution: For the stress function f to be admissible, it has to satisfy bihormonic equation.

Bihormonic equation is given by

024

4

22

4

4

4

=¶

¶+

¶¶

¶+

¶

¶

z z x x

f f f (i)

Now, úû

ùêë

é÷

ø

öçè

æ +÷

ø

öçè

æ

+-=

¶

¶ -

z

x

z x

xz H

z

1

22tan

p

f

( ) [ ]32322

2222

2

21

x xz x xz xz z x

H

z----

+÷

ø

öçè

æ =

¶

¶

p

f

( ) úúû

ù

êêë

é

+÷

ø

öçè

æ -=

¶

¶\

222

3

2

2 2

z x

x H

z p

f





20

Example 5.3

Given the stress function: ( ) zd xzd

F 232

3 -÷

ø

öçè

æ -=f .

Determine the stress components and sketch their variations in a region included in z =

0, z = d , x = 0, on the side x positive.

Solution: The given stress function may be written as

3

3

2

2

23 xz

d

F xz

d

F ÷

ø

öçè

æ +÷

ø

öçè

æ -=f

xzd

F

d

Fx

z÷

ø

öçè

æ +÷

ø

öçè

æ -=

¶

¶\

322

2 126f

and 02

2

=¶

¶

x

f

also2

32

2 66 z

d

F

d

Fz

z x÷

ø

öçè

æ +÷

ø

öçè

æ -=

¶¶

¶ f

Hence xzd

F

d

Fx x ÷

ø

öçè

æ +÷

ø

öçè

æ -=

32

126s (i)

0= zs (ii)

2

32

2 66 z

d

F

d

Fz

z x xz ÷

ø

öçè

æ +÷

ø

öçè

æ -=

¶¶

¶-= j

t (iii)

VARIATION OF STRESSES AT CERTAIN BOUNDARY POINTS

(a) Variation of x

From (i), it is clear that xs varies linearly with x, and at a given section it varies linearly

with z.

\ At x = 0 and z = ± d , xs = 0

At x = L and z = 0, ÷ ø

öçè

æ -=

2

6

d

FL xs

At x = L and z = +d ,232

6126

d

FL Ld

d

F

d

FL x =÷

ø

öçè

æ +÷

ø

öçè

æ -=s

At x = L and z = -d , ÷ ø

öçè

æ -=÷

ø

öçè

æ -÷

ø

öçè

æ -=

232

18126

d

FL Ld

d

F

d

FL xs

The variation of xs is shown in the figure below



21

Figure 5.12 Variation of x

(b) Variation of z

zs is zero for all values of x.

(c) Variation of xz

We have xzt =

2

32.

66 z

d

F

d

Fz÷

ø

öçè

æ -÷

ø

öçè

æ

From the above expression, it is clear that the variation of xzt is parabolic with z. However,

xzt is independent of x and is thus constant along the length, corresponding to a given valueof z.

\At z = 0, xzt = 0

At z = +d , 066 2

32 =÷

ø

öçè

æ -÷

ø

öçè

æ = d

d

F

d

Fd xzt

At z = -d , ÷ ø

öçè

æ -=-÷

ø

öçè

æ -÷

ø

öçè

æ -=

d

F d

d

F d

d

F xz

12)(

66 2

32t

The variation of xzt is shown in figure below.



22

Figure 5.13 Variation of xz

Example 5.4

Investigate what problem of plane stress is satisfied by the stress function3

2

2

3

4 3 2

F xy p xy y

d d j

é ù= - +ê ú

ë û

applied to the region included in y = 0, y = d, x = 0 on the side x positive.

Solution: The given stress function may be written as3

2

3

3 1

4 4 2

F Fxy p xy y

d d j

æ öæ ö æ ö= - +ç ÷ç ÷ ç ÷

è ø è øè ø

02

2

=¶

¶\

x

f

2

2 3 3

3 2 2. 1.5

4 2

Fxy p F p xy

y d d

j ¶ ´æ ö æ ö= - + = -ç ÷ ç ÷¶ è ø è ø

and3

22

4

3

4

3

d

Fy

d

F

y x-=

¶¶

¶ f

Hence the stress components are2

2 31.5 x

F p xy

y d

j s

¶= = -¶

02

2

=¶

¶=

x y

f s



23

d

F

d

Fy

y x xy

4

3

4

33

22

-=¶¶

¶-= f

t

(a) Variation of x

31.5 x

F p xyd s æ ö

= - ç ÷è ø

When x = 0 and y = 0 or , xd ps ± = (i.e., constant across the section)

When x = L and y = 0, x ps =

When x = L and y = +d ,2

1.5 x

FL p

d s

æ ö= - ç ÷

è ø

When x = L and y = -d ,2

5.1d

FLP

x +=s

Thus, at x = L, the variation of xs is linear with y.

The variation of xs is shown in the figure below.

Figure 5.14 Variation of stress x

(b) Variation of z

02

2

=¶

¶=

x y

f s

\ ys is zero for all value of x and y



24

(c) Variation of xy

÷ ø

öçè

æ -÷÷

ø

öççè

æ =

d

F

d

Fy xy

4

3

4

33

2

t

Thus, xyt varies parabolically with z. However, it is independent of x, i.e., it's value is the

same for all values of x.

\At ÷ ø

öçè

æ -==

d

F y xy

4

3,0 t

At 04

3)(

4

3, 2

3 =úû

ùêë

é-úû

ùêë

é=±=

d

F d

d

F d y xyt

Figure 5.15 Variation of shear stress xy

The stress function therefore solves the problem of a cantilever beam subjected to point loadF at its free end along with an axial stress of p.

Example 5.5

Show that the following stress function satisfies the boundary condition in a beam of

rectangular cross-section of width 2 h and depth d under a total shear force W .

úû

ùêë

é--= )23(

2

2

3 yd xy

hd

W f

Solution: 2

2

y x

¶

¶=

f s

Y

d

d

L

Xo

t xy

- 3F

4d

- 3F

4d



25

Now, [ ]2

366

2 xy xyd

hd

W

y--=

¶

¶f

[ ] xy xd hd

W

y126

2 32

2

--=¶

¶ f

[ ] xy xd hd

W x

633

--=\s

02

2

=¶

¶=

x y

f s

and y x

xy¶¶

¶-= f

t 2

= [ ]2

366

2 y yd

hd

W -

= [ ]2

3 33 y yd hd

W

-

Also, 02

22

4

4

4

4

44 =ú

û

ùêë

é

¶¶

¶+

¶

¶+

¶

¶=Ñ f f

y x y x

Boundary conditions are

(a) d and y for y 00 ==s

(b) d and y for xy

00 ==t

(c) Land x for W dyh

d

xy 0.2.0

==òt

(d) WL M L xand x for dyh M

d

x ===== ò ,00.2.0

s

(e) L xand x for dy yh

d

x ===ò 00..2.0

s

Now, Condition (a)

This condition is satisfied since 0= y

s

Condition (b)

[ ] 033 22

3 =- d d

hd

W



26

Hence satisfied.

Condition (c)

[ ] hdy y yd

hd

W d

233

0

2

3ò -

[ ]dy y yd d

W d

ò -=0

2

333

2

d

yd y

d

W

0

32

3 2

32úû

ùêë

é-=

úû

ùêë

é-= 3

3

3 2

32d

d

d

W

2.

2 3

3

d

d

W

=

= W

Hence satisfied.

Condition (d)

[ ] hdy xy xd hd

W d

2630

3ò --

[ ]d xy xyd

d

W 0

2

333

2--=

= 0

Hence satisfied.

Condition (e)

[ ] ydyh xy xd hd

W d

.2630 3ò --

d

xy xdy

d

W

0

32

32

2

32úû

ùêë

é--=

úûùê

ëé --= 3

3

32

2

32 xd

xd

d

W



27

úû

ùêë

é--= 3

3 2

12 xd

d

W

Wx=

Hence satisfied



1


Module6/Lesson1

Module 6: Two Dimensional Problems in Polar

Coordinate System

6.1.1 INTRODUCTION

n any elasticity problem the proper choice of the co-ordinate system is extremelyimportant since this choice establishes the complexity of the mathematical expressions

employed to satisfy the field equations and the boundary conditions.

In order to solve two dimensional elasticity problems by employing a polar co-ordinate

reference frame, the equations of equilibrium, the definition of Airy’s Stress function,and one of the stress equations of compatibility must be established in terms of PolarCo-ordinates.

6.1.2 STRAIN-DISPLACEMENT RELATIONS

Case 1: For Two Dimensional State of Stress

Figure 6.1 Deformed element in two dimensions

Consider the deformation of the infinitesimal element ABCD, denoting r and q displacements

by u and v respectively. The general deformation experienced by an element may be

I



2


Module6/Lesson1

regarded as composed of (1) a change in the length of the sides, and (2) rotation of the sidesas shown in the figure 6.1.

Referring to the figure, it is observed that a displacement " u" of side AB results in both radial

and tangential strain.

Therefore, Radial strain = e r =r u

¶¶ (6.1)

and tangential strain due to displacement u per unit length of AB is

(e q )u=r

u

rd

rd d ur =

-+

q

q q )( (6.2)

Tangential strain due to displacement v is given by

(e q )v = q q

q q

¶

¶

=

÷ ø

öçè

æ

¶

¶

v

r rd

d v

1 (6.3)

Hence, the resultant strain is

e q = (e q )u + (e q )v

e q = ÷ ø

öçè

æ

¶

¶+

q

v

r r

u 1 (6.4)

Similarly, the shearing strains can be calculated due to displacements u and v as below.

Component of shearing strain due to u is

)ur q g ÷

ø

öçè

æ

¶

¶=

÷ ø

öçè

æ

¶

¶

=q q

q q u

r rd

d u

1 (6.5)

Component of shearing strain due to v is

(g r q )v = ÷ ø

öçè

æ -

¶

¶

r

v

r

v (6.6)

Therefore, the total shear strain is given by

) )vr ur r q q q g g g +=



3


Module6/Lesson1

g r q = ÷ ø

öçè

æ -

¶

¶+÷

ø

öçè

æ

¶

¶

r

v

r

vu

r q

1 (6.7)

Case 2: For Three -Dimensional State of Stress

Figure 6.2 Deformed element in three dimensions

The strain-displacement relations for the most general state of stress are given by

e r = z

w

r

uv

r r

u

z¶

¶=÷

ø

öçè

æ +÷

ø

öçè

æ

¶

¶=

¶

¶e

q e

q

,1

,

g r q = ÷ ø

öçè

æ -÷

ø

öçè

æ

¶

¶+

¶

¶

r

vu

r r

v

q

1 (6.8)



4


Module6/Lesson1

g q z = ÷ ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶

z

vw

r q

1

g zr = ÷ ø

öçè

æ

¶

¶+

¶

¶

r

w

z

u

6.1.3 COMPATIBILITY EQUATION

We have from the strain displacement relations:

Radial strain,r

ur

¶

¶=e (6.9a)

Tangential strain, ÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ =

r

uv

r q e q

1 (6.9b)

and total shearing strain,q

g q ¶

¶÷ ø

öçè

æ +÷

ø

öçè

æ -

¶

¶=

u

r r

v

r

vr

1 (6.9c)

Differentiating Equation (6.9a) with respect to q and Equation (6.9b) with respect to r , weget

q q

e

¶¶

¶=

¶

¶

r

ur 2

(6.9d)

q q

e q

¶

¶÷ ø

öçè

æ -

¶¶

¶+÷

ø

öçè

æ -

¶

¶÷ ø

öçè

æ =

¶

¶ v

r r

v

r u

r r

u

r r .

1.

1112

2

2

úû

ùêë

é

¶

¶÷ ø

öçè

æ +-

¶¶

¶÷ ø

öçè

æ +=

q q

e v

r r

u

r r

v

r r

r 111 2

q

q

e q

e e

÷ ø

ö

çè

æ -

¶¶

¶

÷ ø

ö

çè

æ +=

¶

¶\

r r

v

r r r

r 1

.

1 2

(6.9e)

Now, Differentiating Equation (6.9c) with respect to r and using Equation (6.9d), we get

q q

g q

¶

¶÷ ø

öçè

æ -

¶¶

¶÷ ø

öçè

æ ++

¶

¶÷ ø

öçè

æ -

¶

¶=

¶

¶ u

r r

u

r r

v

r

v

r r

v

r

r

2

2

22

2 111

q q ¶¶

¶+÷

ø

öçè

æ

¶

¶+-

¶

¶-

¶

¶=

r

u

r

u

r r

v

r

v

r r

v 2

2

2 111

q

e g

g q

q

¶

¶÷ ø

öçè

æ +÷

ø

öçè

æ -

¶

¶=

¶

¶\ r

r

r

r r r

v

r

112

2

(6.9f)

Differentiating Equation (6.9e) with respect to r and Equation (6.9f) with respect to q , we

get,

q q q e

e

q q e

e e 2

2

22

3

22

2111111

r r r r

v

r r

v

r r r r r r

r +¶

¶÷ ø

öçè

æ --

¶¶

¶÷ ø

öçè

æ -

¶¶

¶÷ ø

öçè

æ +÷

ø

öçè

æ -

¶

¶÷ ø

öçè

æ =

¶

¶ (6.9g)



5


Module6/Lesson1

and2

2

2

32 11

q

e

q

g

q q

g q q

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ -

¶¶

¶=

¶¶

¶ r r r

r r r

v

r

or2

2

222

321111

q

e

q

g

q q

g q q

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ -

¶¶

¶÷ ø

öçè

æ =

¶¶

¶÷ ø

öçè

æ r r r

r r r

v

r r r (6.9h)

Subtracting Equation (6.9h) from Equation (6.9g) and using Equation (6.9e), we get,

22

2

22

2

22

2

2

2111111

r r r r r r

v

r r r r r r r

r r r r r q q q q q e

q

e

q

g e

q

e e

q

g e +

¶

¶÷

ø

öçè

æ -

¶

¶÷

ø

öçè

æ +

¶

¶÷

ø

öçè

æ -

¶¶

¶÷

ø

öçè

æ -÷

ø

öçè

æ -

¶

¶÷

ø

öçè

æ =

¶¶

¶÷

ø

öçè

æ -

¶

¶

÷÷ ø

öççè

æ

¶

¶+

¶

¶-

¶

¶-÷÷

ø

öççè

æ -

¶¶

¶+-÷

ø

öçè

æ

¶

¶=

2

22 1.

11111

q

e

q

g e e

q

e e q q q r r r r

r r r r r r

v

r r r r r

2

2

22

11111

q

e

q

g e e e q q q

¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ = r r r

r r r r r r r r

2

2

22

1121

q

e

q

g e e q q

¶

¶

÷ ø

ö

çè

æ

-¶

¶

÷ ø

ö

çè

æ

+¶

¶

÷ ø

ö

çè

æ

-¶

¶

÷ ø

ö

çè

æ

=r r r

r r r r r r

2

2

22

22

2

11211

q

e e e e

q

g

q

g q q q q

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ +

¶

¶=

¶¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ \ r r r r

r r r r r r r r r

6.1.4 STRESS-STRAIN RELATIONS

In terms of cylindrical coordinates, the stress-strain relations for 3-dimensional state of stress

and strain are given by

e r = )]([1

zr E

s s n s q +-

e q = )]([1 zr

E s s n s q +- (6.10)

e z = )]([1

q s s n s +- r z E

For two-dimensional state of stresses and strains, the above equations reduce to,

For Plane Stress Case

e r = )(1

q ns s -r E

e q = )(1

r

E

ns s q - (6.11)

g r q = q t r G

1



6


Module6/Lesson1

For Plane Strain Case

e r = ])1[()1(

q gs s n n

--+

r E

e q = ])1[()1(

r

E

gs s n n

q --+

(6.12)

g r q = q t r G

1

6.1.5 AIRY’S STRESS FUNCTION

With reference to the two-dimensional equations or stress transformation [Equations (2.12a)

to (2.12c)], the relationship between the polar stress components q s s ,r and q t r and the

Cartesian stress components y x s s , and xy

t can be obtained as below.

q t q s q s s 2sinsincos 22

xy y xr ++=

q t q s q s s q 2sinsincos22

xy x y -+= (6.13)

q t q q s s t q 2coscossin xy x yr +-=

Now we have,

y x x y xy y x

¶¶

¶-=

¶

¶=

¶

¶=

f t

f s

f s

2

2

2

2

2

(6.14)

Substituting (6.14) in (6.13), we get

q f

q f

q f

s 2sinsincos2

2

2

22

2

2

y x x yr

¶¶

¶-

¶

¶+

¶

¶=

q f

q f

q f

s q 2sinsincos2

2

2

22

2

2

y x y x ¶¶

¶+

¶

¶+

¶

¶= (6.15)

q f

q q f f

t q 2coscossin2

2

2

2

2

y x y xr

¶¶

¶-÷÷

ø

öççè

æ

¶

¶-

¶

¶=

The polar components of stress in terms of Airy’s stress functions are as follows.

2

2

2

11

q

f f s

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ =

r r r r (6.16)

2

2

r ¶

¶=

f s q and

q

f

q

f t q

¶¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ =

r r r r

2

2

11 (6.17)

The above relations can be employed to determine the stress field as a function of r and q .



7


Module6/Lesson1

6.1.6 BIHARMONIC EQUATION

As discussed earlier, the Airy’s Stress function f has to satisfy the Biharmonic equation

,04 =Ñ f provided the body forces are zero or constants. In Polar coordinates the stress

function must satisfy this same equation; however, the definition of 4Ñ operator must be

modified to suit the polar co-ordinate system. This modification may be accomplished by

transforming the4Ñ operator from the Cartesian system to the polar system.

Now, we have, q q sin,cos r yr x ==

÷ ø

öçè

æ =+= -

x

yand y xr 1222 tanq (6.18)

where r and q are defined in Figure 6.3

Differentiating Equation (6.18) gives

q q

coscos

===¶¶

r

r

r

x

x

r

q q

sinsin

===¶

¶

r

r

r

y

y

r

÷ ø

öçè

æ -==÷

ø

öçè

æ -=

¶

¶

r r

r

r

y

x

q q q sinsin22

r r

r

r

x

y

q q q coscos22

===¶

¶



8


Module6/Lesson1

Figure.6.3

dy y xdxdr r 222 +=

dyr

ydx

r

xdr ÷

ø

öçè

æ +÷

ø

öçè

æ =\

Also, ÷ ø

ö

çè

æ +

÷ ø

ö

çè

æ -=

x

dy

xy x

y

d 2

2

sec q q

x x

r

r x ¶

¶

¶

¶+

¶

¶

¶

¶=

¶

¶ q

q

f f f

q

f

q

f

¶

¶÷÷ ø

öççè

æ ÷ ø

öçè

æ -+

¶

¶

+=

2222 sec

1

x

y

r y x

x

÷ ø

öçè

æ

¶

¶-÷

ø

öçè

æ

¶

¶=

¶

¶\

q

f q f q

f

r r x

sincos

Similarly,

y y

r

r y ¶

¶

¶

¶+

¶

¶

¶

¶=

¶

¶ q

q

f f f

÷ ø

öçè

æ

¶

¶+÷

ø

öçè

æ

¶

¶=

¶

¶\

q

f q f q

f

r r y

cossin



9


Module6/Lesson1

Now,

2

2

2 sincos ÷

ø

öçè

æ ÷ ø

öçè

æ

¶

¶-÷

ø

öçè

æ

¶

¶=

¶

¶

q

f q f q

f

r r x

÷

ø

öç

è

æ

¶

¶+÷

ø

öç

è

æ

¶

¶+÷÷

ø

öçç

è

æ

¶

¶+

¶¶

¶÷

ø

öç

è

æ -

¶

¶=

r r r r r r r

f q

q

f q q

q

f q

q

f q q f q

2

22

2

2

22

2

22 sincossin2sincossin2

cos

(i)Similarly,

2

2

2

22

2

2

2

22

2

2 coscoscossin2cossin2sin

q

f q f q

q

f q q

q

f q q f q

f

¶

¶+÷

ø

öçè

æ

¶

¶+

¶

¶÷ ø

öçè

æ -

¶¶

¶+

¶

¶=

¶

¶

r r r r r r r y

(ii)

And,

2

2

22

2

2

22cossin2cos2cos

cossincossin

q

f q q

q

f q

q

f q f q q

f q q f

¶

¶÷

ø

öçè

æ -

¶

¶÷

ø

öçè

æ -

¶¶

¶+

¶

¶+

¶

¶÷

ø

öçè

æ -=

¶¶

¶

r r r r r r r y x

(iii)Adding (i) and (ii), we get

2

2

22

2

2

2

2

2 11

q

f f f f f

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ +

¶

¶=

¶

¶+

¶

¶

r r r r y x

2

2

22

2

2

2

2

22 11

,.q

f f f f f f

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ +

¶

¶=

¶

¶+

¶

¶=Ñ

r r r r y xei

or2 2 2 2

4 2 2

2 2 2 2 2 2

1 1 1 1( ) 0

r r r r r r r r

j j j j j

q q

æ öæ ö¶ ¶ ¶ ¶ ¶ ¶Ñ = Ñ Ñ = + + + + =ç ÷ç ÷

¶ ¶ ¶ ¶ ¶ ¶è øè ø

The above Biharmonic equation is the stress equation of compatibility in terms of Airy’sstress function referred in polar co-ordinate system.



Module6/Lesson2

1



Coordinate System

6.2.1 AXISYMMETRIC PROBLEMS

Many engineering problems involve solids of revolution subjected to axially symmetric

loading. The examples are a circular cylinder loaded by uniform internal or external

pressure or other axially symmetric loading (Figure 6.4a), and a semi-infinite half space

loaded by a circular area, for example a circular footing on a soil mass (Figure 6.4b). It is

convenient to express these problems in terms of the cylindrical co-ordinates. Because of

symmetry, the stress components are independent of the angular (q ) co-ordinate; hence, all

derivatives with respect to q vanish and the components v, g r q , g q z , t r q and t q z are zero.

The non-zero stress components are s r ,s q ,, s z and t rz.

The strain-displacement relations for the non-zero strains become

e r = z

w

r

u

r

u z

¶

¶==

¶

¶e e q ,,

g rz =r

w

z

u

¶

¶+

¶

¶ (6.19)

and the constitutive relation is given by

( )( )

( )

( )

ïï

þ

ïï

ý

ü

ïï

î

ïï

í

ì

úúúúúú

û

ù

êêêêêê

ë

é

-

-

--

-+=

ïï

þ

ïï

ý

ü

ïï

î

ïï

í

ì

rz

z

r

rz

z

r

Symmetry

vvv

E

g

e

e

e

n

n

n n

n n

t

s

s

s

q q

2

)21(

01

0)1(01

211





Module6/Lesson2

3


6.2.2 THICK-WALLED CYLINDER SUBJECTED TO INTERNAL AND

EXTERNAL PRESSURES

Consider a cylinder of inner radius ‘a’ and outer radius ‘b’ as shown in the figure 6.5.

Let the cylinder be subjected to internal pressure i p and an external pressure 0 p .

This problem can be treated either as a plane stress case (s z = 0) or as a plane straincase (e z = 0).

Case (a): Plane Stress

Figure 6.5 (a) Thick-walled cylinder (b) Plane stress case (c) Plane strain case

Consider the ends of the cylinder which are free to expand. Let s z = 0. Owing to uniform

radial deformation, t rz = 0. Neglecting the body forces, equation of equilibrium reduces to

0=÷ ø

öçè

æ -+

¶

¶

r r

r r q s s s (6.20)

Hereq

s andr

s denote the tangential and radial stresses acting normal to the sides of the

element.

Since r is the only independent variable, the above equation can be written as

0)( =- q s s r

r dr

d (6.21)

From Hooke’s Law,

e r = )(1

),(1

r r E E

ns s e ns s q q q

-=-





Module6/Lesson2

5


Hence, ( ) úû

ùêë

é÷ ø

öçè

æ --+

- 2212

11

)1( aC C

E n n

n = i

p-

and ( ) úû

ùêë

é÷ ø

öçè

æ --+

- 2212

11

)1( bC C

E n n

n = 0 p-

where the negative sign in the boundary conditions denotes compressive stress.

The constants are evaluated by substitution of equation (6.23a) into (6.23)

C 1 = ÷÷ ø

öççè

æ

-

-÷ ø

öçè

æ -

)(

122

0

22

ab

pb pa

E

in

C 2 = ÷÷ ø

öççè

æ

-

-÷ ø

öçè

æ +

)(

)(122

0

22

ab

p pba

E

in

Substituting these in Equations (6.22) and (6.23), we get

s r = ÷÷

ø

öçç

è

æ

-

--÷

÷

ø

öçç

è

æ

-

-222

22

0

22

0

22

)(

)(

r ab

ba p p

ab

pb pa ii (6.24)

s q = ÷÷ ø

öççè

æ

-

-+÷÷

ø

öççè

æ

-

-222

22

0

22

0

22

)(

)(

r ab

ba p p

ab

pb pa ii (6.25)

u =r ab

ba p p

E ab

r pb pa

E

ii

)(

)(1

)(

)(122

22

0

22

0

22

-

-÷ ø

öçè

æ ++

-

-÷ ø

öçè

æ - n n (6.26)

These expressions were first derived by G. Lambe.

It is interesting to observe that the sum (s r + s q ) is constant through the thickness of the wall

of the cylinder, regardless of radial position. Hence according to Hooke’s law, the

stresses s r and s q produce a uniform extension or contraction in z-direction.The cross-sections perpendicular to the axis of the cylinder remain plane. If two adjacentcross-sections are considered, then the deformation undergone by the element does not

interfere with the deformation of the neighbouring element. Hence, the elements areconsidered to be in the plane stress state.

Special Cases

(i) A cylinder subjected to internal pressure only: In this case, 0 p = 0 and i p = p.

Then Equations (6.24) and (6.25) become

s r = ÷÷

ø

öçç

è

æ -

-

2

2

22

2

1

)( r

b

ab

pa (6.27)

s q = ÷÷ ø

öççè

æ +

- 2

2

22

2

1)( r

b

ab

pa (6.28)



Module6/Lesson2

6


Figure 6.6 shows the variation of radial and circumferential stresses across the thickness of

the cylinder under internal pressure.

Figure 6.6 Cylinder subjected to internal pressure

The circumferential stress is greatest at the inner surface of the cylinder and is given by

(s q )max = 22

22 )(

ab

ba p

-

+ (6.29)

(ii) A cylinder subjected to external pressure only: In this case, i p = 0 and 0 p = p.

Equation (6.25) becomes

s r = - ÷÷ ø

öççè

æ -÷÷

ø

öççè

æ

- 2

2

22

2

1r

a

ab

pb (6.30)

s q = - ÷÷ ø

öççè

æ +÷÷

ø

öççè

æ

- 2

2

22

2

1r

a

ab

pb (6.31)

Figure 6.7 represents the variation of s r and s q across the thickness.

However, if there is no inner hole, i.e., if a = 0, the stresses are uniformly distributed in the

cylinder as

s r = s q = - p



Module6/Lesson2

7


Figure 6.7 Cylinder subjected to external pressure

Case (b): Plane Strain

If a long cylinder is considered, sections that are far from the ends are in a state of planestrain and hence s z does not vary along the z-axis.

Now, from Hooke’s Law,

e r = ( )[ ] zr

E s s n s q +-

1

e q = ( )[ ] zr E

s s n s q +-1

e z = ( )[ ]q s s n s +- r z E

1

Since e z = 0, then

0 = ( )[ ]q s s n s +- r z E

1



Module6/Lesson2

8


s z = n (s r +s q )

Hence,

e r = ( )[ ]q ns s n n

--+

r E

1)1(

e q = ( )[ ]r

E ns s n

n q --

+1

)1(

Solving for s q and s r ,

s q = ( )[ ]q e n ne n n

-++-

1)1)(21(

r

E

s r = ( )[ ]q ne e n n n

+-+-

r

E 1

)1)(21(

Substituting the values of e r

and e q , the above expressions for s q and s r can be written as

s q = ( ) úû

ùêë

é-+

+- r

u

dr

du E n n

n n 1

)1)(21(

s r = ( ) úû

ùêë

é+-

+- r

vu

dr

du E n

n n 1

)1)(21(

Substituting these in the equation of equilibrium (6.21), we get

( ) 0)1(1 =---úû

ùêë

é+-

r

u

dr

duu

dr

dur

dr

d n n n n

or 02

2

=-+r

u

dr

ud r

dr

du

01

22

2

=-+r

u

dr

du

r dr

ud

The solution of this equation is the same as in Equation (6.22)

u = C 1r + C 2 / r

where C 1 and C 2 are constants of integration. Therefore,s q and s r are given by

s q = ( ) úûù

êëé -+

+- 2

21 21

)1)(21( r

C C E n n n



Module6/Lesson2

9


s r = ( ) úû

ùêë

é--

+- 2

21 21

)1)(21( r

C C

E n

n n

Applying the boundary conditions,

s r =i

p- when r = a

s r = 0 p- when r = b

Therefore, ( ) i p

a

C C

E -=ú

û

ùêë

é--

+- 2

21 21

)1)(21(n

n n

( ) o pb

C C

E -=ú

û

ùêë

é--

+- 2

21 21

)1)(21(n

n n

Solving, we get

C 1 = ÷÷ ø ö

ççè æ

--+-

22

220)1)(21(

ba

a pb p

E

in n

and C 2 = ÷÷ ø

öççè

æ

-

-+22

22

0 )()1(

ba

ba p p

E

in

Substituting these, the stress components become

s r = 2

22

22

0

22

2

0

2

r

ba

ab

p p

ab

b pa pii ÷

ø

öçè

æ

-

--÷÷

ø

öççè

æ

-

- (6.32)

s q = 2

22

22

0

22

2

0

2

r

ba

ab

p p

ab

b pa p ii ÷ ø

öçè

æ

-

-+÷÷

ø

öççè

æ

-

- (6.33)

s z = 2n ÷÷ ø

öççè

æ

-

-22

22

0

ab

b pa pi (6.34)

It is observed that the values of s r and s q are identical to those in plane stress case. But in

plane stress case, s z = 0, whereas in the plane strain case, s z has a constant value given by

equation (6.34).



Module6/Lesson2

10


6.2.3 ROTATING DISKS OF UNIFORM THICKNESS

The equation of equilibrium given by

0=+÷ ø

öçè

æ -+ r

r r F r dr

d q s s s (a)

is used to treat the case of a rotating disk, provided that the centrifugal "inertia force" isincluded as a body force. It is assumed that the stresses induced by rotation are distributed

symmetrically about the axis of rotation and also independent of disk thickness.

Thus, application of equation (a), with the body force per unit volume F r equated to the

centrifugal force r w2 r , yields

r wr dr

d r r 2 r s s s q +÷

ø

öçè

æ -+ = 0 (6.35)

where r is the mass density and w is the constant angular speed of the disk in rad/sec. The

above equation (6.35) can be written as

0)( 22 =+- r wr dr

d r r s s q (6.36)

But the strain components are given by

e r = dr

du and e q =

r

u (6.37)

From Hooke’s Law, with s z = 0

e r = )(1

q ns s -r E

(6.38)

e q = )(1

r E ns s

q - (6.39)

From equation (6.37),

u = r e q

dr

du = e r = )( q e r

dr

d

Using Hooke’s Law, we can write equation (6.38) as

úû

ùêë

é-=- )(

1)(

1r r r r

dr

d

E E s n s ns s q q (6.40)

Let r s r = y (6.41)

Then from equation (6.36)

s q = 22

r wdr

dy r + (6.42)



Module6/Lesson2

11


Substituting these in equation (6.40), we obtain

r 2 0)3( 32

2

2

=+++-+ r w ydr

dyr

dr

yd r n

The solution of the above differential equation is

y = Cr + C 1 32

8

31r w

r r

n ÷ ø öç

è æ +-÷

ø öç

è æ (6.43)

From Equations (6.41) and (6.42), we obtain

s r = C + C 1 22

2 8

31r w

r r

n ÷ ø

öçè

æ +-÷

ø

öçè

æ (6.44)

s q = C - C 1 22

2 8

311r w

r r

n ÷ ø

öçè

æ +-÷

ø

öçè

æ (6.45)

The constants of integration are determined from the boundary conditions.

6.2.4 SOLID DISK

For a solid disk, it is required to take C 1 = 0, otherwise, the stresses s r and s q becomes

infinite at the centre. The constant C is determined from the condition at the periphery

(r = b) of the disk. If there are no forces applied, then

(s r )r=b = C - 08

3 22 =÷ ø

öçè

æ +bw r

n

Therefore, C = 22

8

3bw r

n

÷ ø

öçè

æ + (6.46)

Hence, Equations (6.44) and (6.45) become,

s r = )(8

3 222 r bw -÷ ø

öçè

æ + r

n (6.47)

s q = 2222

8

31

8

3r wbw r

n r

n ÷ ø

öçè

æ +-÷

ø

öçè

æ + (6.48)

The stresses attain their maximum values at the centre of the disk, i.e., at r = 0.

Therefore,s r = s q = 22

8

3bw r

n ÷ ø

öçè

æ + (6.49)



Module6/Lesson2

12


6.2.5 CIRCULAR DISK WITH A HOLE

Let a = Radius of the hole.

If there are no forces applied at the boundaries a and b, then

(s r )r=a = 0, (s r )r=b = 0

from which we find that

C = )(8

3 222abw +÷

ø

öçè

æ + r

n

and C 1 = - 222

8

3baw r

n ÷ ø

öçè

æ +

Substituting the above in Equations (6.44) and (6.45), we obtain

s r = ÷÷ ø

öççè

æ -÷÷

ø

öççè

æ -+÷

ø

öçè

æ + 2

2

22222

8

3r

r

baabw r

n (6.50)

s q = ÷÷ ø

öççè

æ ÷ ø

öçè

æ

+

+-÷÷

ø

öççè

æ ++÷

ø

öçè

æ + 2

2

22222

3

31

8

3r

r

baabw

n

n r

n (6.51)

The radial stress s r reaches its maximum at r = ab , where

(s r )max = 22 )(

8

3abw -÷

ø

öçè

æ + r

n (6.52)

The maximum circumferential stress is at the inner boundary, where

(s q )max = ÷

ø

öç

è

æ ÷

ø

öç

è

æ

+

-+÷

ø

öç

è

æ + 222

3

1

4

3abw

n

n r

n (6.53)

The displacement ur for all the cases considered can be calculated as below:

ur = r e q = )(r

E

r ns s q - (6.54)

6.2.6 STRESS CONCENTRATION

While discussing the case of simple tension and compression, it has been assumed that the bar has a prismatical form. Then for centrally applied forces, the stress at some distance fromthe ends is uniformly distributed over the cross-section. Abrupt changes in cross-section giverise to great irregularities in stress distribution. These irregularities are of particular

importance in the design of machine parts subjected to variable external forces and toreversal of stresses. If there exists in the structural or machine element a discontinuity that

interrupts the stress path, the stress at that discontinuity may be considerably greater than thenominal stress on the section; thus there is a “Stress Concentration” at the discontinuity.The ratio of the maximum stress to the nominal stress on the section is known as the





Module6/Lesson2

14


Figure 6.8 Irregularities in Stress distribution

A A

(a)

B B

(b)

CC

(c)

A A

(d)

B B

(e)

CC

(f)



Module6/Lesson2

15


6.2.7 THE EFFECT OF CIRCULAR HOLES ON STRESS

DISTRIBUTIONS IN PLATES

Figure 6.9 Plate with a circular hole

Consider a plate subjected to a uniform tensile stress P as shown in the Figure 6.9. The plate

thickness is small in comparison to its width and length so that we can treat this problem as a

plane stress case. Let a hole of radius 'a' be drilled in the middle of the plate as shown in the

figure. This hole will disturb the stress field in the neighbourhood of the hole. But from

St.Venant's principle, it can be assumed that any disturbance in the uniform stress field will

be localized to an area within a circle of radius 'b'. Beyond this circle, it is expected that the

stresses to be effectively the same as in the plate without the hole.

Now consider the equilibrium of an element ABC at r = b and angle q with respect to

x-axis.

÷ ø öç

è æ =\

AC BC Pr

q s cos.

q 2cos.P=

( )q s 2cos12

+=\P

r (6.56)

and ÷ ø

öçè

æ -=

AC BC Pr

q t q

sin.

q q cossin.P-=

q t q 2sin

2

Pr -=\ (6.57)

A B

C

q

P.BC P.BC

tr q

sr

q

P X

Y

X P

Y

m

n

b

a



Module6/Lesson2

16


These stresses, acting around the outside of the ring having the inner and outer radii r = a

and r = b, give a stress distribution within the ring which may be regarded as consisting of

two parts.

(a) A constant radial stress2

P at radius b. This condition corresponds to the ordinary thick

cylinder theory and stresses q s s ¢¢ and r at radius r is given by

÷ ø

öçè

æ +=¢

2r

B Ar s and ÷

ø

öçè

æ -=¢

2r

B Aq s

Constants A and B are given by boundary conditions,

(i) At 0, == r ar s

(ii) At2

,P

br r == s

On substitution and evaluation, we get

( ) ÷÷

ø

öçç

è

æ -

-

=¢2

2

22

2

1

2 r

a

ab

Pbr

s

( ) ÷÷ ø

öççè

æ +

-=¢

2

2

22

2

12 r

a

ab

Pbq s

(b) The second part of the stress q s s ¢¢¢¢ and r are functions of q . The boundary conditions

for this are:

q s 2cos2

Pr =¢¢ for br =

q t q 2sin2 ÷ ø

öçè

æ -=¢

Pr for br =

These stress components may be derived from a stress function of the form,( ) q f 2cosr f =

because with

r r r r

¶

¶÷ ø

öçè

æ +

¶

¶÷ ø

öçè

æ =¢¢

f

q

f s

112

2

2

andq

f

q

f s q

¶¶

¶÷ ø

öçè

æ -

¶

¶÷ ø

öçè

æ =¢¢

r r r

2

2

11

Now, the compatibility equation is given by,

( ) 02cos11

2

2

22

2

=÷÷

ø

öçç

è

æ

¶

¶+

¶

¶+

¶

¶q

q

r f

r r r r

But ( ) ( ) ( ) q q

q q 2cos1

2cos1

2cos2

2

22

2

r f r

r f r r

r f r ¶

¶+

¶

¶+

¶

¶



Module6/Lesson2

17


( ) ( ) ( )þýü

îíì

-¶

¶+

¶

¶= r f

r r f

r r r f

r 22

2 412cos q

Therefore, the compatibility condition reduces to

( )0

412cos

2

22 =

þý

ü

îí

ì-

¶

¶+

¶

¶r f

r r r r q

As q 2cos is not in general zero, we have

( ) 04

.1

2

22 =

þýü

îíì

-¶

¶+

¶

¶r f

r r r r

We find the following ordinary differential equation to determine ( )r f

i.e., 04141

22

2

22

2

=þýü

îíì

-+þýü

îíì

-+r

f

dr

df

r dr

f d

r dr

d

r dr

d

i.e.,

0164484

1112416422.

1

432

2

243

32

2

23

3

432

2

22

2

233

3

4

4

=+--+-

-++-+--++

r

f

dr

df

r dr

f d

r r

f

dr

df

r dr

df

r dr

f d

r dr

f d

r r

f

dr

df

r dr

f d

r dr

f d

r dr

df

r dr

f d

r dr

f d

or 0992

32

2

23

3

4

4

=+-+dr

df

r dr

f d

r dr

f d

r dr

f d

This is an ordinary differential equation, which can be reduced to a linear differential

equation with constant co-efficients by introducing a new variable t such thatt er = .

Also,

dt

df

r dr

dt

dt

df

dr

df 1==

÷÷ ø

öççè

æ -=

dt

df

dt

f d

r dr

f d 2

2

22

2 1

÷÷ ø

öççè

æ +-=

dt

df

dt

f d

dt

f d

r dr

f d 23

12

2

3

3

33

3

÷÷ ø

öççè

æ -+-=

dt

df

dt

f d

dt

f d

dt

f d

r dr

f d 6116

12

2

3

3

4

4

44

4

on substitution, we get

099

232

61161

42

2

42

2

3

3

42

2

3

3

4

4

4 =÷ ø

öçè

æ +÷÷ ø

ö

ççè

æ

--÷÷ ø

ö

ççè

æ

+-+÷÷ ø

ö

ççè

æ

-+- dt

df

r dt

df

dt

f d

r dt

df

dt

f d

dt

f d

r dt

df

dt

f d

dt

f d

dt

f d

r

or 016442

2

3

3

4

4

=+--dt

df

dt

f d

dt

f d

dt

f d





Module6/Lesson2

19


046

224

=++a

D

a

C A

2

462

24

P

b

D

b

C A -=++

026

6224

2 =--+a

D

a

C Ba A

2

2662

24

2 P

b

D

b

C Bb A -=--+

Solving the above, we get

( ) úúû

ù

êêë

é

--=

322

22

2 ba

bPa B

If ‘a’ is very small in comparison to b, we may write 0@ B

Now, taking approximately,

2

2 Pa D =

÷÷ ø

öççè

æ -=

4

4 PaC

÷ ø

öçè

æ -=

4

P A

Therefore the total stress can be obtained by adding part (a) and part (b). Hence, we have

q s s s 2cos4

3

1212 2

2

4

4

2

2

÷÷ ø

ö

ççè

æ

-++÷÷ ø

ö

ççè

æ

-=¢¢+¢= r

a

r

aP

r

aPr r r (6.58)

q s s s q q q 2cos3

12

12 4

4

2

2

÷÷ ø

öççè

æ +-÷÷

ø

öççè

æ +=¢¢+¢=

r

aP

r

aP (6.59)

and q t t q q 2sin23

12 2

2

4

4

÷÷ ø

öççè

æ +--=¢¢=

r

a

r

aPr r (6.60)

Now, At 0, == r ar s

q s 2cos2PPr -=\

When2

3

2

p p q or =

P3=q s



Module6/Lesson2

20


When p q q == or 0

P-=q s

Therefore, we find that at points m and n, the stress q s is three times the intensity of applied

stress. The peak stress 3P rapidly dies down as we move from r = a to r = b since at 2

p

q =

÷÷ ø

öççè

æ ++=

4

4

2

2 32

2 r

a

r

aPq s

which rapidly approaches P as r increases.

From the above, one can conclude that the effect of drilling a hole in highly stressed elementcan lead to serious weakening.

Now, having the solution for tension or compression in one direction, the solution for tension

or compression in two perpendicular directions can be obtained by superposition. However, by taking, for instance, tensile stresses in two perpendicular directions equal to p, we find at

the boundary of the hole a tensile stress .2 p=q s Also, by taking a tensile stress p in the x-direction and compressive stress –p in the y-direction as shown in figure, we obtain the

case of pure shear.

Figure 6.10 Plate subjected to stresses in two directions



Module6/Lesson2

21


Therefore, the tangential stresses at the boundary of the hole are obtained from

equations (a), (b) and (c).

i.e., ( )[ ]p q q s q ----= 2cos22cos2 p p p p

For2

3

2

p q

p q == or that is, at the points n and m,

p4=q

s

For ,0 p q q == or that is, at the points pmn 4,11

and -=q

s

Hence, for a large plate under pure shear, the maximum tangential stress at the boundary ofthe hole is four times the applied pure shear stress.



Module6/Lesson3

1



Coordinate System

6.3.1 BARS WITH LARGE INITIAL CURVATURE

There are practical cases of bars, such as hooks, links and rings, etc. which have large initial

curvature. In such a case, the dimensions of the cross-section are not very small in

comparison with either the radius of curvature or with the length of the bar. The treatment

that follows is based on the theory due to Winkler and Bach.

6.3.2 WINKLER’S – BACH THEORY

Assumptions

1. Transverse sections which are plane before bending remain plane even after bending.

2. Longitudinal fibres of the bar, parallel to the central axis exert no pressure on each other.

3. All cross-sections possess a vertical axis of symmetry lying in the plane of the centroidalaxis passing through C (Figure 6.11)

4. The beam is subjected to end couples M . The bending moment vector is normal

throughout the plane of symmetry of the beam.

Winkler-Bach Formula to Determine Bending Stress or Normal Stress (Also known as

Circumferential Stress)

Figure 6.11 Beam with large initial curvature

(a)

(b)



Module6/Lesson3

2


Consider a curved beam of constant cross-section, subjected to pure bending produced bycouples M applied at the ends. On the basis of plane sections remaining plane, we can statethat the total deformation of a beam fiber obeys a linear law, as the beam element rotatesthrough small angle Dd q . But the tangential strain e q does not follow a linear relationship.

The deformation of an arbitrary fiber, gh = e c Rd q + yDd q

where e c denotes the strain of the centroidal fiber

But the original length of the fiber gh = ( R + y) d q

Therefore, the tangential strain in the fiber gh = e q =q

q q e

d y R

d y Rd c

)(

][

+

D+

Using Hooke’s Law, the tangential stress acting on area dA is given by

s q =( )

E y R

d d y Rc

+D+ )/( q q e (6.61)

Let angular strain l q

q =

D

d

d

Hence, Equation (6.61) becomes

s q =( )

E y R

y Rc

+

+ l e (6.62)

Adding and subtracting e c y in the numerator of Equation (6.62), we get,

s q =( )

E y R

y y y R ccc

+-++ e e l e

Simplifying, we get

s q =( )

E y R

ycc ú

û

ùêë

é

+-+ )( e l e (6.63a)

The beam section must satisfy the conditions of static equilibrium,

F z = 0 and M x = 0, respectively:

\

ò ò == M ydAdA

q q

s s and 0 (6.63b)

Substituting the above boundary conditions (6.63b) in (6.63a), we get







Module6/Lesson3

5


Table 6.1. Value m for various shapes of cross-section

Formula for ‘ ’m

m = -1 + [ (R C ) ( ).1

(R C ) (R C)]

t.1n b t n

b.1n

+ + -

- - -

1

2

R A

m = -1 +

+ +

[ (R C ) ( ).

(R C ) ). (R C ) R C)]

b .1n t b 1n

b t 1n b.1n

1 1 1

3 2

+ + -

( - - - ( -

R

A

Cross-section

A

C

R

B

C

R

C1

D

bR

C2

C1C3

C

b1

t

t

t

E

C

R b

h

b1

C1

C

C

R

t

hC1

C

C2

R

t

2

b

t

2

1 2 2m = - + - - 1R C

R2

CR

2

C ö ø

æ è

ö ø

æ è

ö ø

æ è

For Rectangular Section: ;= b bC C1 1=For Triangular Section: 0b1 =

m = -1 + R/A {[ (R C )( )]

1 ( ) }

h b h b b

n b b h

1 1 1

1

+ + -

- -

1m = - +

1

2R

C C2 2- R

2 2 2 2- - -C R C

1

éë

ù û

R C

R C

+

-

1

ø æ è



Module6/Lesson3

6


Sign Convention

The following sign convention will be followed:

1. A bending moment M will be taken as positive when it is directed towards the concave

side of the beam (or it decreases the radius of curvature), and negative if it increases the

radius of curvature.

2. 'y' is positive when measured towards the convex side of the beam, and negative when

measured towards the concave side (or towards the centre of curvature).

3. With the above sign convention, if s q is positive, it denotes tensile stress while negative

sign means compressive stress.

The distance between the centroidal axis ( y = 0) and the neutral axis is found by setting the

tangential stress to zero in Equation (5.15)

\ 0 = úû

ùêë

é

++

)(1

y Rm

y

AR

M

or 1 = -)(

n

n

y Rm

y

-

where yn denotes the distance between axes as indicated in Figure 5.2. From the above,

yn =( )1+

-m

mR

This expression is valid for pure bending only.However, when the beam is acted upon by a normal load P acting through the centriod of

cross-sectional area A, the tangential stress given by Equation (5.15) is added to the stress

produced by this normal load P. Therefore, for this simple case of superposition, we have

s q = úûùê

ëé +++

)(1

y Rm y

AR M

AP (6.65)

As before, a negative sign is associated with a compressive load P.

6.3.3 STRESSES IN CLOSED RINGS

Crane hook, split rings are the curved beams that are unstrained at one end or both ends. For

such beams, the bending moment at any section can be calculated by applying the equations

of statics directly. But for the beams having restrained or fixed ends such as a close ring,

equations of equilibrium are not sufficient to obtain the solution, as these beams are statically

indeterminate. In such beams, elastic behaviour of the beam is considered and an additionalcondition by considering the deformation of the member under given load is developed as in

the case of statically indeterminate straight beam.



Module6/Lesson3

7


Now, consider a closed ring shown in figure 6.12 (a), which is subjected to a concentrated

load P along a vertical diametrical plane.

Figure 6.12 Closed ring subjected to loads

The distribution of stress in upper half of the ring will be same as that in the lower half dueto the symmetry of the ring. Also, the stress distribution in any one quadrant will be same asin another. However, for the purposes of analysis, let us consider a quadrant of the circularring as shown in the Figure 6.12 (c), which may be considered to be fixed at the section BB

(a) (b)

(c)(d)



Module6/Lesson3

8


and at section AA subjected to an axial load2

P and bending moment A M . Here the

magnitude and the sign of the moment A M are unknown.

Now, taking the moments of the forces that lie to the one side of the section, then we get,

( ) x RP M M Amn -+-=

2

But from Figure, q cos R x =

( )q cos2

R RP

M M Amn -+-=\

( )q cos12

-+-=\PR

M M Amn (a)

The moment mn M at the section MN cannot be determined unless the magnitude of A M is

known. Resolving

2

P into normal and tangential components, we get

Normal Component, producing uniform tensile stress = N = q cos2

1P

Tangential component, producing shearing stress = q sin2

1PT =



Module6/Lesson3

9


Determination of A

M

Figure 6.13 Section PQMN

Consider the elastic behavior of the two normal sections MN and PQ, a differential distance

apart. Let the initial angle q d between the planes of these two sections change by an

amount q d D when loads are applied.

Therefore, the angular strain =q

q w

d

d D=

i.e., .q q d d =D

Therefore, if we are interested in finding the total change in angle between the sections, that

makes an angle 1q and 2q with the section AA, the expression ò1

2

q

q q w d will give that angle.

But in the case of a ring, sections AA and BB remain at right angles to each other before andafter loading. Thus, the change in the angle between these planes is equal to zero. Hence

ò =2 0p

q w o

d (b)



Module6/Lesson3

10


In straight beams the rate of change of slope of the elastic curve is given by EI

M

dx

yd =

2

2

.

Whereas in initially curved beam, the rate of change of slope of the elastic curve isq

q

Rd

d D,

which is the angle change per unit of arc length.

Now, EI

M

EI

M

R Rd

d mn===

D w

q

q for curved beams

Or EI

M Rw mn=

Substituting the above in equation (b), we get

ò =2 0.

p

q o

mn d EI

M R

since R, E and I are constants,

ò =\ 2 0p

q o

mnd M

From Equation (a), substituting the value of ,mn

M we obtain

ò òò =-+- 2

0

2

0

2

00cos

2

1

2

1p p p

q q q q d PRd PRd M A

Integrating, we get

[ ] [ ] [ ] 0sin2

1

2

120

20

20 =-+-

p p p

q q q PRPR M A

02sin2

1

22

1

2 =÷ ø

öçè

æ -÷ ø

öçè

æ +÷ ø

öçè

æ -

p p p PRPR M A

Thus ÷ ø

öçè

æ -=

p

21

2

PR M A

Therefore, knowing A M , the moment at any section such as MN can be computed and then

the normal stress can be calculated by curved beam formula at any desired section.



Module6/Lesson3

11



Example 6.1

Given the following stress function

q q p

f cosr P=

Determine the stress components q q t s s r r

and ,

Solution: The stress components, by definition of f , are given as follows

2

2

2

11

q

f f s

¶

¶÷

ø

öçè

æ +

¶

¶÷

ø

öçè

æ =

r r r r (i)

2

2

r ¶

¶=

f s q (ii)

q

f

q

f

t q ¶¶

¶

÷ ø

ö

çè

æ

-¶

¶

÷ ø

ö

çè

æ

= r r r r

2

2

11

(iii)

The various derivatives are as follows:

q q p

f cos

P

r =

¶

¶

02

2

=¶

¶

r

f

( )q q q p q

f cossin +-=

¶

¶r

P

( )q q q p q

f

sin2cos2

2

+-=¶

¶r

P

( )q q q p q

f cossin

2

+-=¶¶

¶ P

r

Substituting the above values in equations (i), (ii) and (iii), we get

( )q q q p

q q p

s sin2cos1

cos1

2 +÷

ø

öçè

æ -÷

ø

öçè

æ = r

P

r

P

r r

q p

q q p

q q p

sin21

cos1

cos1 P

r

P

r

P

r ÷

ø

öçè

æ -÷

ø

öçè

æ -÷

ø

öçè

æ =

q p

s sin2 Pr

r -=\

02

2

=¶

¶=

r

f s q



Module6/Lesson3

12


( ) ( )q q q p

q q q p

t q cossin1

cossin1

2 +-÷

ø

öçè

æ -+-÷

ø

öçè

æ =

P

r r

P

r r

0=\ q t r

Therefore, the stress components are

q p

s sin2 P

r r ÷

ø öç

è æ -=

0=q s

0=q t r

Example 6.2

A thick cylinder of inner radius 10 cm and outer radius 15 cm is subjected to an internal

pressure of 12 MPa. Determine the radial and hoop stresses in the cylinder at the inner

and outer surfaces.

Solution: The radial stress in the cylinder is given by

s r = 2

22

2222

22

r

ba

ab

p p

ab

b pa poioi ÷ ø

öçè

æ

-

--÷÷

ø

öççè

æ

-

-

The hoop stress in the cylinder is given by

s q = 2

22

2222

22

r

ba

ab

p p

ab

b pa poioi ÷ ø

öçè

æ

-

-+÷÷

ø

öççè

æ

-

-

As the cylinder is subjected to internal pressure only, the above expressions

reduce to

s r = 2

22

2222

2

r

ba

ab

p

ab

a pii ÷

ø

öçè

æ

--÷÷

ø

öççè

æ

-

and s q = 2

22

2222

2

r

ba

ab

p

ab

a pii ÷

ø

öçè

æ

-+÷÷

ø

öççè

æ

-

Stresses at inner face of the cylinder (i.e., at r = 10 cm):

Radial stress = s r = úû

ùêë

é

-úû

ùêë

é-ú

û

ùêë

é

-

´222

22

22

2

)1.0()15.0(

12

)1.0(

)1.0()15.0(

)1.0()15.0(

)1.0(12

= 9.6 – 21.6

or s r = -12 MPa



Module6/Lesson3

13


Hoop stress = s q = úû

ùêë

éúû

ùêë

é

-+ú

û

ùêë

é

-

´2

22

2222

2

)1.0(

)1.0()15.0(

)1.0()15.0(

12

)1.0()15.0(

)1.0(12

= 9.6 + 21.6

or s q = 31.2 MPa

Stresses at outerface of the cylinder (i.e., at r = 15 cm):

Radial stress = s r = úû

ùêë

éúû

ùêë

é

--ú

û

ùêë

é

-

´2

22

2222

2

)15.0(

)15.0()1.0(

)1.0()15.0(

12

)1.0()15.0(

)1.0(12

s r = 0

Hoop stress = s q = úû

ùêë

é

-úû

ùêë

é+ú

û

ùêë

é

-

´222

22

22

2

)1.0()15.0(

12

)15.0(

)15.0()1.0(

)1.0()15.0(

)1.0(12

= 9.6 + 9.6

or s q = 19.2 MPa

Example 6.3

A steel tube, which has an outside diameter of 10cm and inside diameter of 5cm, is

subjected to an internal pressure of 14 MPa and an external pressure of 5.5 MPa.

Calculate the maximum hoop stress in the tube.

Solution: The maximum hoop stress occurs at r = a.

Therefore, Maximum hoop stress = (s q )max = úû

ùêë

éúû

ùêë

é

-

-+ú

û

ùêë

é

-

-2

22

22

0

22

2

0

2

a

ba

ab

p p

ab

b pa pii

= 2220

22

2

0

2

bab p p

abb pa p ii

úûùê

ëé --+ú

û

ùêë

é

--

=22

2

0

22

0

2

ab

b pb pb pa p ii

-

-+-

(s q )max =22

2

0

22 2)(

ab

b pba pi

-

-+

Therefore, (s q )max = 2)05.0(

2)1.0(

2)1.0(5.52]

2)1.0(

2)05.0[(14

-

´´-+

Or (s q )max = 8.67 MPa





Module6/Lesson3

15


=( ) ú

û

ùêë

é

-

-++22

2222

2

1

ab

a pb pb pa piiii

35 = )( 22

2

ab

b pi

-

i.e., 35 = )(

822

2

ab

b

-´

35b2-35a2

= 8b2

35b2-8b2

= 35a2

35b2-8b2

=35(0.5)2

Therefore, b = 0.5693

If t is the thickness of the cylinder, then

b = 0.5+ t = 0.5693

\t = 0.0693 m or 69.3 mm.

Example 6.5

The circular link shown in Figure 6.14 has a circular cross-section 3cm in diameter.

The inside diameter of the ring is 4cm. The load P is 1000 kg. Calculate the stress at A

and B. Compare the values with those found by the straight beam formula. Assume

that the material is not stressed above its elastic strength.

Solution:

Cross-sectional area = A=4

p (3)

2 = 7.06 cm

2.

For circular cross-section m is given by

m = -1+2

2

÷ ø

öçè

æ

c

R –2 ÷

ø

öçè

æ

c

R1

2

-÷ ø

öçè

æ

c

R

Here R = 2+1.5 = 3.5 cm

c = 1.5 cm. (Refer Table 6.1)

Therefore,Figure 6.14 Loaded circular link

m = 15.1

5.3

5.1

5.32

5.1

5.321

22

-÷ ø

öçè

æ ÷

ø

öçè

æ -÷

ø

öçè

æ +-

m = 0.050

.. AB

P

R



Module6/Lesson3

16


At section AB, the load is resolved into a load P and a bending couple whose moment is

positive. The stress at A and B is considered to be the sum of the stress due to axial load P,

and the stress due to the bending moment M .

Therefore, Stress at point A is

s q A= s A = ú

û

ùêë

é+

++)(

1 A

A

y Rm

y

AR

M

A

P

= - úû

ùêë

é

-

-+

´

´+

)5.15.3(050.0

)5.1(1

5.306.7

)10005.3(

06.7

1000

or s A = -2124.65 kg/cm2 (compressive).

The stress at point B is given by

Bq s = s B = + úû

ùêë

é

+++

B

B

y Rm

y

AR

M

A

P

(1

= úû

ùêë

é

++

´+

-

)5.15.3(050.0

5.11

5.306.7

3500

06.7

1000

\s B = 849.85 kg/cm2 (Tensile)

Comparison by Straight Beam Formula

The moment of inertia of the ring cross-section about the centroidal axis is

I = 444

976.364

)3(

64cm

d ==

p p

If the link is considered to be a straight beam, the corresponding values are

s A = I

My

A

P+

= -976.3

)5.1)(3500(

06.7

1000 -++

\s A = -1462.06 kg/cm2 (compressive)

& s B =976.3

5.13500

06.7

1000 ´+

-

s B = 1178.8 kg/cm2 (tensile)



Module6/Lesson3

17


Figure 6.15 Stresses along the cross-section

Example 6.6

An open ring having T -Section as shown in the Figure 6.16 is subjected to a

compressive load of 10,000 kg. Compute the stresses at A and B by curved beam

formula.

Figure 6.16 Loaded open ring

1178.8

849.85

Straight beam

Curved beam

1462.06

2124.65

+

-

.

.

Centroid axis

Neutral axis

A

B

d

1 0 c

m .. AB

1 8 c m

P=10,000Kg

10.34cm

R

14cm

2cm

2cm

.



Module6/Lesson3

18


Solution:

Area of the Section = A = 2 ´ 10 + 2 ´ 14 = 48 cm2

The value of m can be calculated from Table 6.1 by substituting b1 = 0 for the unsymmetric I -section.

From Figure,

R = 18+5.66 = 23.66 cm

c1 = c3 = 10.34 cm

c2 = 3.66 cm, c = 5.66 cm

t = 2 cm

b1 = 0 , b = 10 cm

m is given by

( ) ( ) ( ) ( ) ( ) ( )[ ]c Rbc Rt bc Rbt c Rb

A

Rm ----++-+++-= ln.ln.ln.ln.1 23111

( ) ( ) ( ) ( ) ( )[ ]66.566.23ln1066.366.23ln21034.1066.23ln02048

66.231 ----++-++-=

Therefore, m = 0.042

Now, stress at A,

s A = úû

ùêë

é

+++

)(1

A

A

y Rm

y

AR

M

A

P

= - úû

ùêë

é

-

-++

)66.566.23(042.0

)66.5(1

66.23x48

)66.23x10000(

48

10000

\s A = -1559.74 kg/cm2 (compressive)

Similarly, Stress at B is given by

s B = úû

ùêë

é

+++

)(1

B

B

y Rm

y

AR

M

A

P

= úû

ùêë

é

++

´

´+-

)34.1066.23(042.0

34.101

66.2348

66.2310000

48

10000

\ s B = 1508.52 kg/cm2

(tensile)



Module6/Lesson3

19


Example 6.7

A ring shown in the Figure 6.17, with a rectangular section is 4cm wide and 2cm thick.

It is subjected to a load of 2,000 kg. Compute the stresses at A and B and at C and D by

curved beam formula.

Figure 6.17 Loaded ring with rectangular cross-section

Solution: Area of the section2824 cm A =´=

The Radius of curvature of the centroidal axis = .624 cm R =+=

From Table 6.1, the m value for trapezoidal section is given by,

( )( )[ ] ( )þýü

îíì

--÷ ø

öçè

æ

-

+-+++-= hbb

c R

c Rbbc Rhb

Ah

Rm 1

1111 ln1

But for rectangular section, ,, 11 bbcc ==

Therefore ( )( )[ ] ( )þýü

îí

ì

-÷ ø

öçè

æ

-

+

+++-= 0ln01 c R

c Rc Rbh Ah

Rm

( )( )[ ]þýü

îíì

÷ ø

öçè

æ

-

+++´

´+-=

26

26ln02642

48

61m

Therefore 0397.0=m

Now, stress at( )ú

û

ùêë

é

+++==

A

A

A y Rm

y

AR

M

A

P A 1s

( )( )ú

û

ùêë

é

-

-+

´

´+-=

260397.0

21

68

62000

8

2000

2/6.3148 cmkg A -=\s (Compression)

P=2000kg

.30

0 A

C

B

D

4 c m

4cm

2cm

Cross section



Module6/Lesson3

20


Stress at( )ú

û

ùêë

é

+++==

B

B

B y Rm

y

AR

M

A

P B 1s

( )úû

ùêë

é

++

´

´+

-=

260397.0

21

68

62000

8

2000

Therefore,2/31.1574 cmkg B +=s (Tension)

To compute the stresses at C and D

Figure 6.18

At section CD, the bending moment,030cosPR M =

i.e.,030cos62000 ´´= M

cmkg -= 10392

Component of P normal to CD is given by,

.173230cos200030cos 00 kgP N ===

Therefore, stress at( )ú

û

ùêë

é

+++==

A

A

c y Rm

y

AR

M

A

N C 1s

( )( )ú

û

ùêë

é

-

-+

´+

-=

260397.0

21

68

10392

8

1732

2/7.2726 cmkgc

-=\s (Compression)

Stress at( )ú

û

ùêë

é

+++==

B

B

D y Rm

y

AR

M

A

N D 1s

..

O.

6 c m

300

300

C

D

P=2000kg



Module6/Lesson3

21


( )úû

ùêë

é

++

´+

-=

260397.0

21

68

10392

8

1732

Therefore, 2/4.1363 cmkg D =s (Tension)

Example 6.8The dimensions of a 10 tonne crane hook are shown in the Figure 6.19. Find the

circumferential stresses B A and s s on the inside and outside fibers respectively at the

section AB.

Figure 6.19 Loaded crane hook

Solution: Area of the section27212

2

39cm A =´

+==

Now, .539

3293

12 cm y A =úû

ùêë

é

+

´+=

Therefore ( ) .7512 cm y B =-=

Radius of curvature of the centroidal axis .1257 cm R =+== For Trapezoidal cross section, m is given by the Table 6.1 as,

( ) ( )( )[ ] ( )þýü

îíì

--÷ ø

öçè

æ

-

+-++´

´+-= 1239

512

712ln.39712123

1272

121m

080.0=\ m

Moment cmkgPR M -=´=== 12000012000,10

Now,

Stress at( )ú

û

ùêë

é

+++==

A

A

A y Rm

y

AR

M

A

P A 1s

7 c m 10tA

R

B

.

12cm

9cm3cm

YB

Section AB

YA



Module6/Lesson3

22


( )( )ú

û

ùêë

é

-

-+

´+

-=

51208.0

51

1272

120000

72

10000

2/1240 cmkg A -=\s (Compression)

Stress at( )ú

ûùê

ëé

+++==

B

B B

y Rm y

AR M

AP B 1s

( )úû

ùêë

é

++

´+

-=

71208.0

71

1272

120000

72

000,10

2/62.639 cmkg B =\s (Tension)

Example 6.9

A circular open steel ring is subjected to a compressive force of 80 kN as shown

in the Figure 6.20. The cross-section of the ring is made up of an unsymmetrical

I-section with an inner radius of 150 mm. Estimate the circumferential stressesdeveloped at points A and B.

Figure 6.20 Loaded circular ring with unsymmetrical I-section

.A B O

1 5 0

W

160

.80 100

160

YBYA

20

20

20R



Module6/Lesson3

23


Solution:

From the Table 6.1, the value of m for the above section is given by

( ) ( ) ( ) ( ) ( ) ( )[ ]c Rbc Rt bc Rbt c Rb A

Rm ----++-+++-= lnlnlnln1 23111

Hence = R Radius of curvature of the centroidal axis.

Now,2600020802012010020 mm A =´+´+´=

( ) ( ) ( ).33.75

6000

150208080201201020100mm y B =

´´+´´+´´=

( ) .67.8433.75160 mm y A =-=\

Also, ( ) .33.22533.75150 mm R =+=

( ) ( ) ( )( ) ( ) ( )úû

ùêëé

----++-+++-=\

33.7533.225ln10033.5533.225ln2010067.6433.225ln802067.8433.225ln80

600033.2251m

.072.0=\ m

Moment = .10803.133.225100080 7mm N PR M -´=´´==

Now, Stress at point( )ú

û

ùêë

é

+++==

B

B

B y Rm

y

AR

M

A

P B 1s

( )

( )úû

ù

êë

é

-

-+

´

´+-=\

33.7533.225072.0

33.75

133.2256000

10803.1

6000

80000 7

Bs

2/02.93 mm N B

-=\s (Compression)

Stress at point( )ú

û

ùêë

é

+++==

A

A

A y Rm

y

AR

M

A

P A 1s

( )úû

ùêë

é

++

´

´+-=

67.8433.225072.0

67.841

33.2256000

10803.1

6000

80000 7

2/6.50 mm N A

=\s (Tension)

Hence, the resultant stresses at A and B are,

2/6.50 mm N A

=s (Tension),2/02.93 mm N

B -=s (Compression)



Module6/Lesson3

24


Example 6.10

Calculate the circumferential stress on inside and outside fibre of the ring at A and B,

shown in Figure 6.21. The mean diameter of the ring is 5 cm and cross-section is circular

with 2 cm diameter. Loading is within elastic limit.

Figure 6.21 Loaded closed ring

Solution: For circular section, from Table 6.1

1221

22

-÷ ø

öçè

æ ÷

ø

öçè

æ -÷

ø

öçè

æ +-=

c

R

c

R

c

Rm

1

1

5.2

1

5.22

1

5.221

22

-÷

ø

öç

è

æ ÷

ø

öç

è

æ -÷

ø

öç

è

æ +-=

0435.0=\ m

We have, ÷ ø

öçè

æ -=

p

21PR M A

PR364.0= P5.2364.0 ´=

P M A 91.0=\

Now,( )ú

û

ùêë

é

+++÷

ø

öçè

æ -=

A

i A A

y Rm

y

AR

M

A

Pi

1s

( )

( )úû

ù

êë

é

-

-+

´+

÷ ø

ö

çè

æ -=

15.20435.0

1

15.2

91.0

A

P

A

P

÷ ø

öçè

æ -÷

ø

öçè

æ -=

A

P

A

P21.5

. R = 2. 5 c

mBi

AO

BO

Ai

2P 1000kg=

MA

m

n

q

P

Pcosq

Psinq

Ai AoP



Module6/Lesson3

25


÷ ø

öçè

æ -=\

A

Pi A 21.6s (Compressive)

( )úû

ùêë

é

+++÷

ø

öçè

æ -=

B

o A

y Rm

y

AR

M

A

P1

0s

( )úû

ùêë

é+

+´

+÷ ø

öçè

æ -=15.20435.0

11

5.2

91.0

A

P

A

P

÷ ø

öçè

æ =\

A

Po A 755.1s (Tension)

Similarly, ( )PR M M A B -=

( )PRPR -= 364.0 PR636.0-=

P5.2636.0 ´-=

P M B

59.1-=\

Now,( )ú

û

ùêë

é

++=

i

i B Bi

y Rm

y

AR

M 1s

( )( )ú

û

ùêë

é

-

-+

´-=

15.20435.0

11

5.2

59.1

A

P

÷ ø

öçè

æ =\

A

P Bi 11.9s (Tension)

and( )

( )úû

ùêë

é

+

++÷

ø

öçè

æ

´-=

15.20435.0

11

5.2

59.1

A

P Bo

s

÷ ø

öçè

æ -=

A

P81.4 (Compression)

Now, substituting the values of ,500kgP =

( ) ,14159.31 22cm A == p above stresses can be calculated as below.

2/988500

21.6 cmkgi A

-=´-=p

s

2/32.279500

755.10

cmkg A =´=p

s

2

/1450

500

11.9 cmkgi B =´=

p s

2/54.765500

81.40

cmkg B -=´-=p

s



Module6/Lesson3

26


Example 6.11

A ring of 200mm mean diameter has a rectangular cross-section with 50mm in the

radial direction and 30mm perpendicular to the radial direction as shown in Figure

6.22. If the maximum tensile stress is limited to ,/120 2mm N determine the tensile

load that can be applied on the ring.

Figure 6.22 Closed ring with rectangular cross-section

Solution: ,100mm R = Area of cross-section =215005030 mm A =´=

From Table 6.1, the value of m for the rectangular section is given by

[ ]þýü

îíì -÷

ø öç

è æ

-++´

´+-= 0

2510025100ln05030

5015001001m

0217.0=\ m

.

W

C

A

1 0 0 m

m

B

W

D

50mm

.

50mm

YA YB

30mm

R

Section AB



Module6/Lesson3

27


To find AB M

Figure 6.23

The Bending moment at any section MN can be determined by

( )q cos12

-+-= WR M M AB MN

ABmn M M At -==\ ,0q

But ÷ ø

öçè

æ -=

p

21

2

WR M AB

W W

M AB 17.182

12

100=÷

ø

öçè

æ -

´=\

p

Now,( )ú

û

ùêë

é

+++=

A

A A

A y Rm

y

AR

M

A

P1s

( )úû

ùêë

é

+++=

A

A A

y Rm

y

AR

M

A

W 1

2

( ) ( )( )÷÷

ø

öççè

æ

-

-+

´

-+

´=

251000217.0

251

1001500

17.18

15002

W W

W A002073.0=\s (Tensile)

and( )ú

û

ùêë

é

+++=

B

B A

B y Rm

y

AR

M

A

P1s

( )úû

ù

êë

é

++

´-

´=

251000217.0

25

11001500

17.18

15002

W w

W B00090423.0-=\s (Compression)

W

A B

W2

W2

RM

N

q

MABMAB



Module6/Lesson3

28


To find stresses at C and D

We have, ( )q cos12

-+-=WR

M M ABmn

2,900 WR

M M M At ABCDmn

+-===\ q

W W W M CD 83.312

10017.18 =´+-=\

Now, stress at( )ú

û

ùêë

é

+++==

C

C CD

C y Rm

y

AR

M

A

PC 1s

( )( )ú

û

ùêë

é

-

-+

´+=

251000217.0

251

1001500

83.310

W

W 00305.0-= (Compression)

and stress at( )ú

û

ùêë

é

+++==

D

DCD

D y Rm

y

AR

M

A

P D 1s

( )úû

ùêë

é

++

´+=

251000217.0

251

1001500

83.310

W

W D 00217.0=\s (Tensile)

By comparison, the tensile stress is maximum at Point D.

12000217.0 =\ W kN or N W 3.5554.55299=\

Example 6.12A ring of mean diameter 100mm is made of mild steel with 25mm diameter. The ring is

subjected to four pulls in two directions at right angles to each other passing throughthe center of the ring. Determine the maximum value of the pulls if the tensile stress

should not exceed2/80 mm N



Module6/Lesson3

29


Figure 6.24 Closed ring with circular cross-section

Solution: Here mm R 50=

From Table 6.1, the value of m for circular section is given by,

1221

22

-÷ ø

öçè

æ ÷

ø

öçè

æ -÷

ø

öçè

æ +-=

C

R

C

R

C

Rm

15.12

505.12

5025.12

5021

22

-÷ ø öç

è æ ÷

ø öç

è æ -÷

ø öç

è æ +-=

016.0=\ m

Area of cross-section = ( ) 2287.4905.12 mm A == p

We have, ÷ ø

öçè

æ -=

p

21

2

WR M A

÷ ø

öçè

æ -´=

p

2150

2

W

W M A 085.9=

Now,( )ú

û

ùêë

é

+++=

A

A A A

y Rm

y

AR

M

A

P1s

.

W

C

A

1 0 0 m

m B WW

W

D

MAB

M

N

q

W2

W

2

W2



Module6/Lesson3

30


( )( )ú

û

ùêë

é

-

-+

´-

´=

5.1250016.0

5.121

5087.490

085.9

87.4902

W W

W 0084.0= (Tensile)

( )úû

ù

êë

é

++÷ ø

ö

çè

æ

-+=\ B

B A

B y Rm

y

AR

M

A

P

1s

( )úû

ùêë

é

++

´-

´=

5.1250016.0

5.121

5087.490

085.9

87.4902

W W

W B00398.0-=\s (Compression)

Also, ( ) ÷ ø

öçè

æ ´+-=-= 50

2085.9

W PR M M ACD

W M CD

915.15+=\

Now,( )ú

û

ùêë

é-

+=C

C CD

C y Rm

y

AR

M 1s

( )( )ú

û

ùêë

é

-

-+

´+=

5.1250016.0

5.121

5087.490

918.15 W

W C 013.0-=\s (Compression)

and( )ú

û

ùêë

é

++

´+=

5.1250016.0

5.121

5087.490

918.15 W Ds

W 0088.0= (Tension)

Stresses at Section CD due to horizontal Loads

We have, moment at any section MN is given by

( )q cos12

-+-=PR

M M A MN

At section CD, ,0=q

( )0cos12

-+-=\ RW

M M ACD

W M M ACD 085.9-=-=



Module6/Lesson3

31


( )úû

ùêë

é

+++=\

C

C CD

C y Rm

y

AR

M

A

P1s

( ) ( )

( )

ú

û

ùê

ë

é

-

-+

´

-+

´=

5.1250016.0

5.121

5087.490

085.9

87.4902

W W

W C

00836.0=\s (Tensile)

and( )ú

û

ùêë

é

+++=

D

DCD

D y Rm

y

AR

M

A

P1s

( )( )ú

û

ùêë

é

++

´

-+

´=

5.1250016.0

5.121

5087.490

085.9

87.4902

W W

W D00398.0-=\s (Compression)

Resultant stresses are( ) W W W C 00464.000836.0013.0 -=+-=s (Compression)

( ) W W W D00482.000398.00088.0 =-=s (Tension)

In order to limit the tensile stress to2/80 mm N in the ring, the maximum value of the force

in the pulls is given by

0.00482W = 80

kN or N W 598.1651.16597=\

6.3.5 EXERCISES

1. Is the following function a stress function?

q q p

f sinr P

÷ ø

öçè

æ -=

If so, find the corresponding stress. What is the problem solved by this function?

2. Investigate what problem of plane stress is solved by the following stressfunctions

(a) q q f sinr K

P=

(b) q q p f sinr P

-=

3. Derive the equilibrium equation for a polar co-ordinate system.

4. Derive the expressions for strain components in polar co-ordinates.







Module6/Lesson3

34


Figure 6.26

26. A semicircular curved bar is loaded as shown in figure and has a trapezoidal cross-

section. Calculate the tensile stress at point A if kN P 5=

Figure 6.27

27. A curved beam with a circular centerline has a T-section shown in figure below. It is

subjected to pure bending in its plane of symmetry. The radius of curvature of theconcave face is 60mm. All dimensions of the cross-section are fixed as shown except the

thickness t of the stem. Find the proper value of the stem thickness so that the extreme

fiber stresses are bending will be numerically equal.

20mm

40mm A B

P

P

20mm

AB

20mm

20mm

20mm

W

a

b2 b1

c

m n



Module6/Lesson3

35


Figure 6.28

28. A closed ring of mean diameter 200mm has a rectangular section 50mm wide by a 30mm

thick, is loaded as shown in the figure. Determine the circumferential stress on the inside

and outside fiber of the ring at A and B. Assume2/210 mmkN E =

Figure 6.29

29. A hook has a triangular cross-section with the dimensions shown in figure below.

The base of the triangle is on the inside of the hook. The load of 20kN applied along aline 50mm from the inner edge of the shank. Compute the stress at the inner and

outer fibers.

60mm

80mm

20mm

t

. A

1 0 0 m m

B

5 0 m m .

50mm

30mmA1

B1

50KN

50KN



Module6/Lesson3

36


Figure 6.30

30. A circular ring of mean radius 40mm has a circular cross-section with a diameter of25mm. The ring is subjected to diametrical compressive forces of 30kN along thevertical diameter. Calculate the stresses developed in the vertical section under theload and the horizontal section at right angles to the plane of loading.

65mm

50mm



Module7/Lesson1

1


Module: 7 Torsion of Prismatic Bars

7.1.1 INTRODUCTION

rom the study of elementary strength of materials, two important expressions related tothe torsion of circular bars were developed. They are

t = J

r M t (7.1)

and q =GJ

dz M

L

t

Lò1

(7.2)

Here t represents the shear stress, M t the applied torque, r the radius at which the stress is

required, G the shear modulus, q the angle of twist per unit longitudinal length, L the length,

and z the axial co-ordinate.

Also, J = Polar moment of inertia which is defined by A A

ò d r 2

The following are the assumptions associated with the elementary approach in deriving (7.1)

and (7.2).

1. The material is homogeneous and obeys Hooke’s Law.

2. All plane sections perpendicular to the longitudinal axis remain plane following the

application of a torque, i.e., points in a given cross-sectional plane remain in that plane after

twisting.

3. Subsequent to twisting, cross-sections are undistorted in their individual planes, i.e., theshearing strain varies linearly with the distance from the central axis.

4. Angle of twist per unit length is constant.

In most cases, the members that transmit torque, such as propeller shaft and torque tubes of

power equipment, are circular or turbular in cross-section.

But in some cases, slender members with other than circular cross-sections are used. These

are shown in the Figure 7.0.

F



Module7/Lesson1

2


Figure 7.0 Non-Circular Sections Subjected to Torque

While treating non-circular prismatic bars, initially plane cross-sections [Figure 7.0 (a)]

experience out-of-plane deformation or "Warping" [Figure 7.0(b)] and therefore assumptions

2. and 3. are no longer appropriate. Consequently, a different analytical approach isemployed, using theory of elasticity.

7.1.2 GENERAL SOLUTION OF THE TORSION PROBLEM

The correct solution of the problem of torsion of bars by couples applied at the ends was

given by Saint-Venant. He used the semi-inverse method. In the beginning, he made certain

assumptions for the deformation of the twisted bar and showed that these assumptions could

satisfy the equations of equilibrium given by

0=+¶

¶+

¶

¶+

¶

¶ x

xz xy x F z y x

t t s

0=+¶

¶+

¶

¶+

¶

¶ y

yz xy yF

z x y

t t s

0=+¶

¶+

¶

¶+

¶

¶ z

yz xz z F y x z

t t s

and the boundary conditions such as

X = s x l + t xym + t xzn

Y = s ym+ t yzm + t xy l

Z = s zn+ t xz l + t yzm

in which F x , F y , F z are the body forces, X , Y , Z are the components of the surface forces per

unit area and l, m, n are the direction cosines.



Module7/Lesson1

3


Also from the uniqueness of solutions of the elasticity equations, it follows that the torques

on the ends are applied as shear stress in exactly the manner required by the

solution itself.

Now, consider a prismatic bar of constant arbitrary cross-section subjected to equal and

opposite twisting moments applied at the ends, as shown in the Figure 7.1(a).

Figure 7.1 Bars subjected to torsion

Saint-Venant assumes that the deformation of the twisted shaft consists of

1. Rotations of cross-sections of the shaft as in the case of a circular shaft and

2. Warping of the cross-sections that is the same for all cross-sections.

The origin of x , y , z in the figure is located at the center of the twist of the cross-section,

about which the cross-section rotates during twisting. Figure 7.1(b) shows the partial endview of the bar (and could represent any section). An arbitrary point on the cross-section,

point P( x , y), located a distance r from center of twist A, has moved to P¢ ( x-u , y+v) as a

result of torsion. Assuming that no rotation occurs at end z = 0 and that q is small, the x and

y displacements of P are respectively:

u = - (r q z) sina

But sina = r y /

Therefore, u = -(r q z) y / r = - yq z (a)

Similarly, v = (r q z) cosa = (r q z) x / r = xq z (b)

where q z is the angle of rotation of the cross-section at a distance z from the origin.



Module7/Lesson1

4


The warping of cross-sections is defined by a function as

w = q y ( x , y) (c)

Here, the equations (a) and (b) specify the rigid body rotation of any cross-section through a

small angle zq . However, with the assumed displacements (a), (b) and (c), we calculate the

components of strain from the equations given below.

e x = x

u

¶¶

, e y = y

v

¶¶

, e z = z

w

¶¶

g xy = x

v

y

u

¶¶

+¶¶

, g yz = y

w

z

v

¶¶

+¶¶

and g zx = z

u

x

w

¶¶

+¶¶

,

Substituting (a), (b) and (c) in the above equations, we obtain

e x = e y = e z = g xy = 0

g xz = y x

w-

¶¶

q = ÷ ø ö

çè æ -

¶¶

q y

q y x

or gxz = ÷ ø

öçè

æ -¶¶

y x

y q

and gyz = x y

w+

¶¶

q = ÷÷ ø

öççè

æ +

¶¶

q y

q x y

or gyz = ÷÷ ø

ö

ççè

æ

+¶

¶ x y

y

q

Also, by Hooke’s Law, the stress-strain relationships are given by

s x = 2Ge x + l e , t xy = Gg xy

s y = 2Ge y + l e , t yz = Gg yz

s z = 2Ge z + l e , t xz = g xz

where e = e x + e y + e z

and l = )21)(1( n n

n -+

E

Substituting (a), (b) and (c) in the above equations, we obtain





Module7/Lesson1

6


Therefore the stress in a bar of arbitrary section may be determined by solving Equations(7.3) and (7.4) along with the given boundary conditions.

7.1.3 BOUNDARY CONDITIONS

Now, consider the boundary conditions given by

X = s x l + t xy m + t xzn

Y = s ym+ t yzn + t xy l

Z = s zn+ t xz l + t yzm

For the lateral surface of the bar, which is free from external forces acting on the boundary

and the normal n to the surface is perpendicular to the z-axis, we have

X = Y = Z = 0 and n = 0. The first two equations are identically satisfied and the third

gives,

t xz l + yzt m = 0 (7.5)

which means that the resultant shearing stress at the boundary is directed along the tangent to

the boundary, as shown in the Figure 7.2.

Figure 7.2 Cross-section of the bar & Boundary conditions



Module7/Lesson1

7


Considering an infinitesimal element abc at the boundary and assuming that S is increasing

in the direction from c to a,

= cos ( N, x) =dS

dy

m = cos( N , y) = -dS

dx

\Equation (7.5) becomes

÷ ø

öçè

æ -÷ ø

öçè

æ dS

dx

dS

dy yz xz t t = 0

or 0=÷ ø

öçè

æ ÷÷

ø

öççè

æ +

¶¶

-÷ ø

öçè

æ ÷

ø

öçè

æ -¶¶

dS

dx x

ydS

dy y

x

y y (7.6)

Thus each problem of torsion is reduced to the problem of finding a function satisfying

equation (7.3a) and the boundary condition (7.6).

7.1.4 STRESS FUNCTION METHOD

As in the case of beams, the torsion problem formulated above is commonly solved byintroducing a single stress function. This procedure has the advantage of leading to simpler boundary conditions as compared to Equation (7.6). The method is proposed by Prandtl.In this method, the principal unknowns are the stress components rather than thedisplacement components as in the previous approach.

Based on the result of the torsion of the circular shaft, let the non-vanishing components be

t zx and t yz. The remaining stress components s x , s y and s z and t xy are assumed to be zero.

In order to satisfy the equations of equilibrium, we should have

,0=¶

¶

z

xzt

,0=¶

¶

z

yzt

0=¶

¶+

¶

¶

y x

yz xz t t

The first two are already satisfied since t xz and t yz, as given by Equations (d) and (e) are

independent of z.

In order to satisfy the third condition, we assume a function f ( x , y) called Prandtl stress

function such that

t xz = y¶

¶f , t yz =

x¶¶

- f

(7.7)

With this stress function, (called Prandtl torsion stress function), the third condition is alsosatisfied. The assumed stress components, if they are to be proper elasticity solutions, haveto satisfy the compatibility conditions. We can substitute these directly into the stress





Module7/Lesson1

9


hence, ò ò =dxdy yzt 0 (7.11)

Thus the resultant of the forces distributed over the ends of the bar is zero, and these forcesrepresent a couple the magnitude of which is

M t =

ò ò - dxdy y x

xz yz

)( t t (7.12)

= - ò ò ¶¶

+¶¶

dxdy y

y x

x )( f f

Therefore,

M t = - ò ò ò ò ¶¶

-¶¶

dxdy y

ydxdy x

x f f

Integrating by parts, and observing that f = 0 at the boundary, we get

M t = ò ò ò ò+ dxdydxdy f f (7.13)

\ M t = 2 ò ò dxdyf (7.14)

Hence, we observe that each of the integrals in Equation (7.13) contributing one half of the

torque due to t xz and the other half due to t yz.

Thus all the differential equations and boundary conditions are satisfied if the stress function

f obeys Equations (7.8) and (7.14) and the solution obtained in this manner is the exactsolution of the torsion problem.

7.1.5 TORSION OF CIRCULAR CROSS SECTION

The Laplace equation is given by

02

2

2

2

=¶¶+

¶¶

y x

y y

where = warping function.

The simplest solution to the above equation is

= constant = C

But the boundary condition is given by the Equation (7.6) is

0=÷ ø

öçè

æ ÷÷

ø

öççè

æ +

¶¶

-÷ ø

öçè

æ ÷

ø

öçè

æ -¶¶

dS

dx x

ydS

dy y

x

y y

Therefore, with = C , the above boundary condition becomes

(0- y) (dy / dS ) – (0+ x) (dx/dS ) = 0



Module7/Lesson1

10


- y 0=-dS

dx x

dS

dy

or 02

22

=+ y x

dS

d

i.e., x

2

+ y

2

= constantwhere ( x , y) are the co-ordinates of any point on the boundary. Hence the boundary

is a circle.

From Equation (c), we can write

w = qy ( x , y)

i.e., w = q C

The polar moment of inertia for the section is

J = ò ò =+P

I dxdy y x )( 22

But M t = GI Pq

or q =P

t

GI

M

Therefore, w =

P

t

GI

C M

which is a constant. Since the fixed end has zero w at least at one point, w is zero at every

cross-section (other than the rigid body displacement). Thus the cross-section does not warp.

Further, the shear stresses are given by the Equations (d) and (e) as

t xz = G ÷ ø

öçè

æ -¶¶

=÷ ø

öçè

æ -¶¶

y x

G y x

w y q q

t yz = G ÷÷ ø ö

ççè æ +

¶¶=÷÷

ø ö

ççè æ +

¶¶ x

yG x

y

w y q q

\ t xz = -Gq y

and t yz = Gq x

or t xz = -G yGI

M

P

t

t xz = -P

t

I

x M

and t yx = Gq x

= G xGI

M

P

t



Module7/Lesson1

11


hence, t yz =P

t

I

x M

Therefore, the direction of the resultant shear stress t is such that, from Figure 7.3

tan a = y x

I y M

I x M

Pt

Pt

xz

yz/

/

/-=

-

=t

t

Figure 7.3 Circular bar under torsion

Hence, the resultant shear stress is perpendicular to the radius.

Further,

t 2 = t 2 yz + t 2

xz

t 2 = M 2

t ( x2+ y2

) / 2

p I

or t =22 y x

I

M

P

t +

Therefore, t =P

t

I

r M .

or t = J

r M t . (since J = I P)

where r is the radial distance of the point ( x , y). Hence all the results of the elementary

analysis are justified.



Module: 7 Torsion of Prismatic Bars

7.2.1 TORSION OF ELLIPTICAL CROSS-SECTION

Let the warping function is given by Axy= (7.15)

where A is a constant. This also satisfies the Laplace equation. The boundary

condition gives

( Ay - y) 0)( =+-dS

dx x Ax

dS

dy

or y ( A-1) 0)1( =+-dS

dx A x

dS

dy

i.e., ( A+1)2 x 02)1( =--dS

dy y A

dS

dx

or 0])1()1[( 22 =--+ y A x AdS

d

Integrating, we get

(1+ A) x2+(1- A) y2 = constant.

This is of the form

12

2

2

2

=+b

y

a

x

These two are identical if

A

A

b

a

+

-=

1

12

2

or A =22

22

ab

ab

+

-

Therefore, the function given by

= xyab

ab22

22

+

- (7.16)

represents the warping function for an elliptic cylinder with semi-axes a and b under torsion.The value of polar moment of inertia J is

J = ò ò -++ dxdy Ay Ax y x )( 2222 (7.17)



= ( A+1) ò ò ò ò-+ dxdy y Adxdy x 22 )1(

J = ( A+1) I y+(1- A) I x (7.18)

where I x =

4

3abp

and I y =

4

3bap

Substituting the above values in (7.18), we obtain

J = 22

33

ba

ba

+

p

But q =GJ

M

GI

M t

P

t =

Therefore, M t = GJ q

= Gq 22

33

ba

ba

+

p

or q = 33

22

ba

ba

G

M t

p

+

The shearing stresses are given by

t yz = Gq ÷÷ ø

öççè

æ +

¶

¶ x

y

y

= M t 33

22

ba

ba

p

+ x

ab

ab÷÷

ø

öççè

æ +

+

-1

22

22

or t yz =ba x M t 32

p

Similarly, t xz = 3

2

ab

y M t

p

Therefore, the resultant shearing stress at any point ( x , y) is

t = 22

xz yz t t + =33

2

ba

M t

p [ ]2

12424 ya xb + (7.19)

Determination of Maximum Shear Stress

To determine where the maximum shear stress occurs, substitute for x2 from

12

2

2

2

=+b

y

a

x,

or x2 = a2

(1- y2 / b2

)



and t = [ ] 2

1

222242

33)(

2 ybaaba

ba

M t -+p

Since all terms under the radical (power 1/2) are positive, the maximum shear stress occurs

when y is maximum, i.e., when y = b. Thus, maximum shear stress t max occurs at the ends of

the minor axis and its value is

t max =2/124

33)(

2ba

ba

M t

p

Therefore, t max = 2

2

ab

M t

p (7.20)

For a = b , this formula coincides with the well-known formula for circular cross-section.

Knowing the warping function, the displacement w can be easily determined.

Therefore, w = q = xyGba

ab M t 33

22 )(

p

- (7.21)

The contour lines giving w = constant are the hyperbolas shown in the Figure 7.4 having the principal axes of the ellipse as asymptotes.

Figure 7.4 Cross-section of elliptic bar and contour lines of w

7.2.2 PRANDTL’S MEMBRANE ANALOGY

It becomes evident that for bars with more complicated cross-sectional shapes, moreanalytical solutions are involved and hence become difficult. In such situations, it is





Similarly, the components of the forces Fdx acting on face AB and CD are

-Fdx y

z

¶

¶ and Fdx ú

û

ùêë

é

¶

¶

¶

¶+

¶

¶dy

y

z

y y

z)(

Therefore, the resultant force in z-direction due to tension F

= úûùê

ëé ¶¶+¶¶+¶¶-ú

ûùê

ëé ¶¶+¶¶+¶¶- dy

y z

y zFdx

y zFdxdx

x z

x zFdy

x zFdy

2

2

2

2

= F dxdy y

z

x

z÷÷

ø

öççè

æ

¶

¶+

¶

¶2

2

2

2

But the force p acting upward on the membrane element ABCD is p dxdy, assuming that themembrane deflection is small.

Hence, for equilibrium,

F ÷÷ ø

öççè

æ

¶

¶+

¶

¶2

2

2

2

y

z

x

z = - p

or2

2

2

2

y

z

x

z

¶

¶+

¶

¶ = - p / F (7.22)

Now, if the membrane tension F or the air pressure p is adjusted in such a way that p / F becomes numerically equal to 2Gq , then Equation (7.22) of the membrane becomes identical

to Equation (7.8) of the torsion stress function f . Further if the membrane height z remainszero at the boundary contour of the section, then the height z of the membrane becomes

numerically equal to the torsion stress function f = 0. The slopes of the membrane are thenequal to the shear stresses and these are in a direction perpendicular to that of the slope.

Further, the twisting moment is numerically equivalent to twice the volume under themembrane [Equation (7.14)].

Table 7.1 Analogy between Torsion and Membrane Problems

Membrane problem Torsion Problem

Z f

S

1

G

P 2q

y

z

x

z

¶

¶

¶

¶- , zx zy

t t ,

2 (volume

beneath membrane)t M



The membrane analogy provides a useful experimental technique. It also serves as the basisfor obtaining approximate analytical solutions for bars of narrow cross-section as well as formember of open thin walled section.

7.2.3 TORSION OF THIN-WALLED SECTIONS

Consider a thin-walled tube subjected to torsion. The thickness of the tube may not beuniform as shown in the Figure 7.6.

Figure 7.6 Torsion of thin walled sections

Since the thickness is small and the boundaries are free, the shear stresses will be essentially

parallel to the boundary. Let t be the magnitude of shear stress and t is the thickness.

Now, consider the equilibrium of an element of length D l as shown in Figure 7.6. The areas

of cut faces AB and CD are t 1 D l and t 2 D l respectively. The shear stresses (complementary

shears) are t 1 and t 2.

For equilibrium in z-direction, we have

-t 1 t 1 D l + t 2 t 2 D l = 0

Therefore, t 1 t 1 = t 2 t 2 = q = constant

Hence the quantity t t is constant. This is called the shear flow q, since the equation is

similar to the flow of an incompressible liquid in a tube of varying area.



Determination of Torque Due to Shear and Rotation

Figure 7.7 Cross section of a thin-walled tube and torque due to shear

Consider the torque of the shear about point O (Figure 7.7).

The force acting on the elementary length dS of the tube = DF = t t dS = q dS

The moment arm about O is h and hence the torque = D M t = (qdS ) h

Therefore, D M t = 2qdA

where dA is the area of the triangle enclosed at O by the base dS .

Hence the total torque is

M t = S 2qdA+

Therefore, M t = 2qA (7.23)

where A is the area enclosed by the centre line of the tube. Equation (7.23) is generally

known as the "Bredt-Batho" formula.

To Determine the Twist of the Tube

In order to determine the twist of the tube, Castigliano's theorem is used. Referring to Figure

7.7(b), the shear force on the element is t t dS = qdS . Due to shear strain g , the force does

work equal to DU

i.e., d t )(21 tdS U =D

= ltdS D.)(2

1g t



=G

ltdS t

t .).(2

1D (since g t G= )

=Gt

ldS t

2

22 Dt

=Gt

ldS q2

2

D

=t

dS

G

lq.

2

2 D

t

dS

G A

l M U t .

8 2

2D=D

Therefore, the total elastic strain energy is

U = òD

t

dS

G A

l M t 2

2

8

Hence, the twist or the rotation per unit length ( lD = 1) is

q =t M

U

¶

¶= ò t

dS

G A

M t 24

or q = ò t

dS

G A

qA24

2

or q = ò t

dS

AG

q

2 (7.24)



7.2.4 TORSION OF THIN-WALLED MULTIPLE-CELL CLOSED

SECTIONS

Figure 7.8 Torsion of thin-walled multiple cell closed section

Consider the two-cell section shown in the Figure 7.8. Let A1 and A2 be the areas of the cells

1 and 2 respectively. Consider the equilibrium of an element at the junction as shown in the

Figure 7.8(b). In the direction of the axis of the tube, we can write

-t 1 t 1 lD + t 2 t 2 lD + t 3 t 3 lD = 0

or t 1 t 1 = t 2 t 2 + t 3 t 3

i.e., q1 = q2 + q3

This is again equivalent to a fluid flow dividing itself into two streams. Now, choose

moment axis, such as point O as shown in the Figure 7.9.



Figure. 7.9 Section of a thin walled multiple cell beam and moment axis

The shear flow in the web is considered to be made of q1 and – q2, since q3 = q1 - q2.

Moment about O due to q1 flowing in cell 1 (including web) is

1t M = 2q1 A1

Similarly, the moment about O due to q2 flowing in cell 2 (including web) is

M t 2 = 2q2 ( A2+ A1) - 2q2 A1

The second term with the negative sign on the right hand side is the moment due to shear

flow q2 in the middle web.

Therefore, The total torque is

M t = M t 1 + M t 2

M t = 2q1 A1 + 2q2 A2 (a)

To Find the Twist ( )

For continuity, the twist of each cell should be the same.

We have

q = ò t

dS

AG

q

2

or 2Gq = ò t

qdS

A

1





Solution: The above figure shows the membrane surface ABCD

Now, the Applied torque =M t = 2qA

56,500 = 2q(0.5x0.25)

56,500 = 0.25q

hence , q = 226000 N/m.

Now, the shearing stresses are

t 1 =26

1

/10833.18012.0

226000m N

t

q´==

t 2 =26

2

/10667.37006.0

226000m N

t

q´==

t 3 =26 /106.22

01.0

226000m N ´=

Now, the angle of twist per unit length is

q = ò t

ds

GA

q

2

Therefore,

q =úû

ùêë

é++

01.0

25.0)2(

006.0

5.0

012.0

25.0

125.0x10x6.27x2

2260009

or q = 0.00696014 rad/m

Example 7.2

The figure below shows a two-cell tubular section as formed by a conventional airfoilshape, and having one interior web. An external torque of 10,000 Nm is acting in a

clockwise direction. Determine the internal shear flow distribution. The cell areas

are as follows:

A1 = 680 cm2 A2 = 2000 cm2

The peripheral lengths are indicated in Figure

Solution:

For Cell 1, a1 = ò (t

dS including the web)

=09.0

33

06.0

67

+

therefore, a1 = 148.3



For Cell 2,

a2 =08.0

67

09.0

48

09.0

63

09.0

33+++

Therefore, a2 = 2409For web,

a12 = 36609.0

33=

Now, for Cell 1,

2Gq = )(1

21211

1

qaqa A

-

= )3661483(680

1

21 qq -

Therefore, 2Gq = 2.189q1 – 0.54q2 (i)

For Cell 2,

2Gq = )(1

11222

2

qaqa A

-

= )3662409(2000

112 qq -

Therefore, 2Gq = 1.20q2 – 0.18q1 (ii)

Equating (i) and (ii), we get

2.18 q1 – 0.54q2 = 1.20q2 – 0.18q1

or 2.36q1 – 1.74q2 = 0

or q2 = 1.36q1

The torque due to shear flows should be equal to the applied torque

Hence, from Equation (a),

M t = 2q1 A1 + 2q2 A2

10,000´ 100 = 2q1 x 680 + 2q

2 x 2000

= 1360q1 + 4000q2

Substituting for q2, we get

10000´ 100 = 1360q1 + 4000 ´ 1.36q1



Therefore,

q1 = 147 N and q2 = 200 N

Figure 7.11

Example 7.3

A thin walled steel section shown in figure is subjected to a twisting moment T .

Calculate the shear stresses in the walls and the angle of twist per unit length of the

box.

Figure 7.12

Solution: Let A1 and 2 A be the areas of the cells (1) and (2) respectively.

2

2

1

a A

p =\

( ) 2

2 422 aaa A =´=

For Cell (1),

t

dsa ò=1 (Including the web)

÷ ø

öçè

æ +=

t

aaa

21

p

For Cell (2),

t

dsa ò=2

q 2

q 1 0.09cm

S = 6 3 c m

0.09cm S = 6 7 c m

0.08cm

S=67cm

S = 4 8 c m

S = 3 3 c m

Cell-1

Cell-2

0.09cm

0.06cm

2a

2a

a

A1

q2

q1

A2

t

t

t

t



t

a

t

a

t

a

t

a 2222+++=

÷ ø

öçè

æ =\

t

aa

82

For web,

÷ ø

öçè

æ =

t

aa

212

Now,For Cell (1),

( )21211

1

12 qaqa

AG -=q

( )úû

ùêë

é÷

ø

öçè

æ -

+= 212

222q

t

aq

t

aa

a

p

p

( )[ ]212 222 qqtaa -+= p

p

( )[ ]21 222

2 qqat

G -+=\ p p

q )1(

For Cell (2),

( )11222

2

12 qaqa

AG -=q

úû

ùêë

é-= 122

28

4

1q

t

aq

t

a

a

[ ]1224

4

2qq

t a

a-=

[ ]1242

12 qq

at G -=\ q )2(

Equating (1) and (2), we get,

( )[ ] [ ]1221 42

122

2qq

at qq

at -=-+p

p

or ( )[ ] [ ]1221 421222 qqqq -=-+p

p



( )[ ] [ ]1221 4224

qqqq -=-+p p

( )04

8241221 =+--

+\ qqqq

p p

p

( )04

81

2421 =úû

ùêë

é+-úû

ùêë

é+

+qq

p p

p

( )0

482421 =úû

ùêë

é +-úû

ùêë

é ++qq

p

p

p

p p

or ( ) ( ) 21 4884 qq p p p +=++

1284

85qq ÷

ø

öçè

æ

+

+=\

p

p

But the torque due to shear flows should be equal to the applied torque.

i.e., 2211 22 Aq AqT += )3(

Substituting the values of 12 , Aq and 2 A in (3), we get,

2

1

2

1 4.84

852

22 aq

aqT ÷

ø

öçè

æ

+

++÷÷

ø

öççè

æ =

p

p p

1

2

1

2

84

858 qaqa ÷

ø

öçè

æ

+

++=

p

p p

( )( ) 1

22

2

1612q

aT ú

û

ùêë

é

+

++=\

p

p p

( )( )1612

2221

++

+=\

p p

p

a

T q

Now, from equation (1), we have,

( ) ( )( )

( )( )úû

ùêë

é

++

+÷

ø

öçè

æ

+

+-

++

++=

1612

2

84

852

1612

22

22

2222 p p

p

p

p

p p

p p

p q

a

T

a

T

at G

Simplifying, we get the twist as( )( )úû

ùêë

é

++

+=

16122

3223 p p

p q

t Ga

T



Example 7.4

A thin walled box section having dimensions t aa ´´2 is to be compared with a solid

circular section of diameter as shown in the figure. Determine the thickness t so that the

two sections have:

(a) Same maximum shear stress for the same torque.

(b) The same stiffness.

Figure 7.13

Solution: (a) For the box section, we have

aat T

At

qAT

´==

=

2...2

...2

2

t

t

t a

T 24

=\t )(a

Now, For solid circular section, we have

r I

T

p

t =

Where I p = Polar moment of inertia

÷ ø öç

è æ

=

÷÷ ø öçç

è æ

\

232

4

aa

T t

p

aa

T or

t

p

2324

=

÷ ø

öçè

æ =\

3

16

a

T

p t )(b

Equating (a) and (b), we get

32

16

4 a

T

t a

T

p = T atT a 3264 p =\

64at p =\

(b) The stiffness of the box section is given by

a.

2a

ta



t

ds

GA

qò=

2q

Here T = 2qA A

T q

2=\

úûùêë

é +++=\t a

t a

t a

t a

GAT 22

4 2q

( ) t aG

aT

t GA

aT

22

2

24

6

4

6

=

=

Gt a

aT 416

6=\q )(c

The stiffness of the Solid Circular Section is

44

32

32

aG

T

aG

T

GI

T

p p p q =

÷÷ ø

öççè

æ == )(d

Equating (c) and (d), we get

44

32

16

6

aG

T

Gt a

aT

p =

p

32

16

6=

t

a

32166

´=\ at p

÷ ø

öçè

æ =\

644

3 at

p







i.e., [ ]02.1656083.25013015000

12 ´´´=q G

lengthmmradians /10824.1100083

7.22706

15000

1 5-´=´

´=\q

04.1=q or degrees/m length

Example 7.6

A tubular section having three cells as shown in the figure is subjected to a torque of

113 kN-m. Determine the shear stresses developed in the walls of the section.

All dimensions in mm

Figure 7.15

Solution: Let 654321 ,,,,, qqqqqq be the shear flows in the various walls of the tube as

shown in the figure. 321 ,, Aand A A be the areas of the three cells.

( ) 22

1 25322127

2

mm A ==\ p

2

2 64516254254 mm A =´= 2

3 64516mm A =

Now, From the figure,

q1 = q2 + q4

q2 = q3 + q5

q3 = q6

or 4422111 t t t q t t t +==

66333

5533222

t t q

t t t q

t t

t t t

==

+== (1)

Where 654321 ,,,, t t t t t t and are the Shear Stresses in the various walls of the tube.

Now, The applied torque is

254 254

2 5 4

q 1

q 6

q 3

q 3

q 4

q 2

q 2

(1) (2) (3)

0.8

0.8

1.3 1.0

1 2 7

0.6

q 5



( )333222111

332211

2

222

t At At A

q Aq Aq A M t

t t t ++=

++=

i.e., ( ) ( ) ( )[ ]8.0645168.0645168.025322210113 21

6 ´+´+´=´ t t

( ) 3718397.3 321 =++\ t t t (2)

Now, considering the rotations of the cells and 654321 ,,,, S and S S S S S as the length of cell

walls,

We have,

3663355

2552244

14411

22

22

2

AGS S S

AGS S S

AGS S

q t t t

q t t t

q t t

=++-

=++-

=+

(3)

Here ( ) mmS 3981271 =´= p

mmS S S S S 25465432 =====

\(3) can be written as

q t t t

q t t t

q t

G

G

GS

645162542542254

645162542542254

25322254398

632

522

41

=+´´+-

=+´´+-

=+

(4)

Now, Solving (1), (2) and (4) we get

2

1 /4.40 mm N =t

22 /2.55 mm N =t

2

3 /9.48 mm N =t

2

4 /7.12 mm N -=t

2

6 /6.36 mm N =t



Module 8/Lesson 1

1


Module 8: Elastic Solutions and Applications

in Geomechanics

8.1.1 INTRODUCTION

ost of the elasticity problems in geomechanics were solved in the later part ofnineteenth century and they were usually solved not for application to geotechnical

pursuits, but simply to answer basic questions about elasticity and behavior of elastic bodies.With one exception, they all involve a point load. This is a finite force applied at a point: asurface of zero area. Because of stress singularities, understanding point-load problems willinvolve limiting procedures, which are a bit dubious in regard to soils. Of all the point-load

problems, the most useful in geomechanics is the problem of a point load acting normal tothe surface of an elastic half-space.

The classical problem of Boussinesq dealing with a normal force applied at the plane boundary of a semi-infinite solid has found practical application in the study ofthe distribution of foundation pressures, contact stresses, and in other problemsof soil mechanics. Solutions of the problems of Kelvin, Flamant, Boussinesq, Cerrutti andMindlin related to point load are discussed in the following sections.

8.1.2 KELVIN’S PROBLEM

It is the problem of a point load acting in the interior of an infinite elastic body as shown inthe Figure 8.1.

Figure 8.1 Kelvin’s Problem

M



Module 8/Lesson 1

2


Consider a point load of magnitude 2P acting at a point in the interior of an infinite

elastic body.

In the cylindrical coordinate system, the following displacements can be obtained by

Kelvin’s solution.

Displacement in radial direction = ur = 3)1(8 RG

Prz

n p -

Tangential displacement = uq = 0

Vertical displacement = u z = úû

ùêë

é++

-- 3

21)21(2

)1(8 R

z

R RG

P n

n p (8.1)

Similarly, the stresses are given by

s r = -

úû

ù

êë

é-

-

- 5

2

3

3)21(

)1(4 R

zr

R

zP n

n p

s q = 3)1(4

)21(

R

zP

n p

n

-

-

s z = úû

ùêë

é+

-- 5

3

3

3)21(

)1(4 R

z

R

zP n

n p (8.2)

t rz = úû

ùêë

é+

-- 5

2

3

3)21(

)1(4 R

rz

R

r P n

n p

t r q = t q r = t q z = t zq = 0

Here R =22

r z +

It is clear from the above expressions that both displacements and stresses die out for larger

values of R. But on the plane z = 0, all the stress components except for t rz vanish, at all

points except the origin.

Vertical Tractions Equilibrating the applied Point Load

Consider the planar surface defined by z = h± , as shown in the Figure 8.2.



Module 8/Lesson 1

3


Figure 8.2 Vertical stress distributions on horizontal planes above and below point load

The vertical component of traction on this surface is s z. If we integrate s z, over this entire

surface, we will get the resultant force. To find this resultant force, consider a horizontal

circle centered on the z-axis over which s z is constant (Figure 8.3).

Therefore, the force acting on the annulus shown in Figure 8.3 will be s z ´ 2p rdr .

Now, the total resultant force on the surface z = h is given by,

Resultant upward force = ò¥

o z rdr )2( p s

= )2(3)21(

)1(45

3

3rdr

R

h

R

hPo

p n

n p úû

ùêë

é+

--ò

¥

=úû

ùêë

é+

--ò

¥

5

3

3

3)21(

)1(2 R

h

R

hPo

n

n r dr

To simplify the integration, introduce the angle y as shown in the Figure 8.2.



Module 8/Lesson 1

4


Here

r = h tany and dr = h sec2y d y

Therefore, Resultant upward force = ò +--

2/2 ]sincos3sin)21[(

)1(2

p

y y y y n n o

d P

Solving, we get resultant upward force on the lower plane = P which is exactly one-half theapplied load. Further, if we consider a similar surface z = - h, shown in Figure 8.2, we will

find tensile stresses of the same magnitude as the compressive stresses on the lower plane.

Hence, Resultant force on the upper plane = -P (tensile force). Combining the two resultant

forces, we get 2P which exactly equilibrate the applied load.

Figure 8.3 Geometry for integrating vertical stress







Module 8/Lesson 1

7


Now, ]sincos[1

q q q p q

f +=

¶¶ q

r

01

=÷ ø

öçè

æ ¶¶

¶¶

q

f

r r

Hence, t r q = 0

The stress function assumed in Equation (8.3) will satisfy the compatibility equation

÷÷ ø

öççè

æ

¶

¶+

¶¶

+¶

¶÷÷

ø

öççè

æ

¶

¶+

¶¶

+¶

¶2

2

22

2

2

2

22

2 1111

q

f f f

q r r r r r r r r = 0

Here s r and s q are the major and minor principal stresses at point P. Now, using the

above expressions for s r , s q and t r q , the stresses in rectangular co-ordinate system(Figure 8.5) can be derived.

Therefore,

s z = s r cos2q +s q sin2q - 2t r q sinq cosq

Here, s q = 0 and t r q = 0

Hence, s z = s r cos2q

)(coscos2 2q q p r

q

s z = q p

3cos2

r

q

But from the Figure 8.5,

r = 22 z x +

cosq = )( 22 z x

z

+ , sinq =

22 z x

x

+

Therefore,

s z = )(

2

22 z x

q

+p

( )322

3

z x

z

+

s z = 222

3

)(

2

z x

qz

+p

Similarly,s x = s r sin2q + s q cos2q + 2t r q sinq

= 00sincos2 2 ++q q p r

q



Module 8/Lesson 1

8


=22

2

z x

q

+p

22 z x

z

+

)( 22

2

z x

x

+

or s x =

( )222

22

z x

zqx

+p

and t xz = -s q sinq cosq + s r sinq cosq + t r q (cos2q sin2q )

= 0 + q q q p

cossincos2

r

q+ 0

= q q p

2cossin2

r

q

=)(

222

2

22 z x

z

z x

x

r

q

++p

or t xz =222

2

)(

2

z x

qxz

+p

But for the plane strain case,

s y = n (s x + s z)

where, n = Poisson’s ratio

Substituting the values of s x and s z in s y, we get

s y = n p p ú

û

ùêë

é

++

+ 222

3

222

2

)(

2

)(

2

z x

qz

z x

zqx

= ][

)(

2 22

22 z x

z x

zq+

+p

n

or s y = )(

222

z x

zq

+p

n

Therefore according to Flamant’s solution, the following are the stresses due to a verticalline load on the surface of an half-space.

s x = 222

2

)(

2

z x

zqx

+p

s y = )(

222

z x

zq

+p

n

s z = 222

3

)(

2

z x

qz

+p (8.7)



Module 8/Lesson 1

9


t xz = 222

2

)(

2

z x

qxz

+p

and t xy = t yx = t zy = t yz = 0

8.1.4 ANALYSIS TO FIND THE TRACTIONS THAT ACT ON THECYLINDRICAL SURFACE ALIGNED WITH LINE LOAD

Figure 8.6 Cylindrical surface aligned with line load

One can carry out an analysis to find the tractions that act on the cylindrical surface by usingthe stress components in Equation (8.7).

Here, the traction vector is given by T = nb

qzˆ

22p

(8.8)

where n̂ is the unit normal to the cylindrical surface. This means to say that the cylindricalsurface itself is a principal surface. The major principal stress acts on it.

Hence, s 1 = 2

2

b

qz

p (8.9)

The intermediate principal surface is defined by n̂ = {0, 1, 0} and the intermediate principal

stress is s 2 = ns 1.

The minor principal surface is perpendicular to the cylindrical surface and to the

intermediate principal surface and the minor principal stress is exactly zero.The other interesting characteristic of Flamant’s problem is the distribution of the principalstress in space.



Module 8/Lesson 1

10


Now, consider the locus of points on which the major principal stress s 1 is a constant. FromEquation (8.9), this will be a surface for which

hqb

z

2

1

21

2 == ps

where C is a constant.

But b2 = x2 + z2

Therefore,h z x

z

2

1

)( 22 =

+

or b2 = ( x2+ z2) = 2h z

which is the equation of a circle with radius C centered on the z-axis at a depth C beneath theorigin, as shown in Figure 8.7.

Figure 8.7 Pressure bulb on which the principal stresses are constant

At every point on the circle, the major principal stress is the same. It points directly at the

origin. If a larger circle is considered, the value of s 1 would be smaller. This result gives usthe idea of a "pressure bulb" in the soil beneath a foundation.



Module 8/Lesson 1

11


8.1.5 BOUSSINESQ’S PROBLEM

The problem of a point load acting normal to the surface of an elastic half-space was solved by the French mathematician Joseph Boussinesq in 1878. The problem geometry isillustrated in Figure 8.8. The half-space is assumed to be homogeneous, isotropic and elastic.The point load is applied at the origin of co-ordinates on the half-space surface. Let P be the

magnitude of the point load.

Figure 8.8 Boussinesq’s problem

Consider the stress function

( )2

122 zr B +=f (8.10)

where B is a constant.

The stress components are given by

s r = ÷÷ ø

öççè

æ

¶

¶-Ñ

¶¶

2

22

r z

f f n

s q = ÷ ø

öçè

æ ¶¶

-Ñ¶¶

r r z

f f n

12 (8.11)

s z = úû

ùêë

é

¶

¶-Ñ-

¶

¶2

22)2(

z z

f f n



Module 8/Lesson 1

12


t rz = úû

ùêë

é

¶

¶-Ñ-

¶¶

2

22)1(

zr

f f n

Therefore, by substitution, we get

( ) ( ) ( ) úû

ù

êë

é+-+-=

--2

52222

322 321 zr zr zr zv Br

s

[ ] ( ) 2

32221

-+-= zr zv Bq s (8.12)

( ) ( ) ( ) úû

ùêë

é+++--

--2

52232

322 321 zr z zr zv B z

s

( ) ( ) ( ) úû

ùêë

é+++--=

--2

52222

322 321 zr rz zr r v Brz

t

Now, the shearing forces on the boundary plane z = 0 is given by

t rz = 2

)21(

r

B n -- (a)

In polar co-ordinates, the distribution of stress is given by

2,

3

R

dR

d

R

A R R R

s s s s q +==

or32

1

R

A-=q s

where A is a constant and22

zr R +=

In cylindrical co-ordinates, we have the following expressions for the stress components:

s r = s R sin2 y + s q cos2 y

s z = s R cos2 y + s q sin2 y (8.13)

t rz =2

1(s R -s q ) sin2y

32

1

R

A-=q s

But from Figure 8.8

( )2

122sin

-+= zr r y

( ) 2

122cos

-+= zr zy

Substituting the above, into s r , s z , t rz and s q we get



Module 8/Lesson 1

13


( ) 2

52222

2

1 -+÷

ø

öçè

æ -= zr zr Ar s

( ) 2

52222

2

1 -+÷

ø

öçè

æ -= zr r z A zs

t rz = ( ) 2

522

2

3 -+ zr ( A r z) (8.14)

s q = ( ) 2

322

2

1 -+- zr A

Suppose now that centres of pressure are uniformly distributed along the z-axis from z = 0 to

z = -¥ . Then by superposition, the stress components produced are given by

( )ò¥ -

+÷ ø

öçè

æ -= z

r dz zr zr A 2

52222

12

1s

= úû

ùêë

é+-+-

--2

3

222

1

22

22

1 )()(1

2 zr z zr

r

z

r

A

( )ò¥

-+÷

ø

öçè

æ -= z

z dz zr r z A 2

52222

12

1s (8.15)

= 2

3

221 )(2

-+ zr z

A

( ) dz zr rz A z

rz

ò

¥-

+=2

522

12

3

t =2

3

221

)(2

-

+ zr r

A

( )ò¥

-+-=

z

dz zr A 2

322

12

1q s

= úû

ùêë

é+--

-2

1

22

22

1 )(1

2 zr

r

z

r

A

On the plane z = 0, we find that the normal stress is zero and the shearing stress is

t rz = 2

1

2

1

r

A (b)

From (a) and (b), it is seen that the shearing forces on the boundary plane are eliminated if,

( ) 02

21 1 =+-- A

v B

Therefore, ( )v B A 2121 -=



Module 8/Lesson 1

14


Substituting the value of A1 in Equation (8.15) and adding together the stresses (8.12) and

(8.15), we get

( ) ( ) ( ) ýü

îíì

+-úû

ùêë

é+--= --

2

52222

122

223

121 zr zr zr

r

z

r Br n s

( ) 2

5

2233 -

+-= zr Bz zs

( ) ( ) ( ) úû

ùêë

é++++--=

--2

322

2

122

22

121 zr z zr

r

z

r B n s q

( ) 2

52223

-+-= zr Brz

rzt (8.16)

The above stress distribution satisfies the boundary conditions, since s z = t rz = 0 for z = 0.

To Determine the Constant B

Consider the hemispherical surface of radius ‘a’ as illustrated in the Figure 8.9. For any

point on this surface let R = a = constant. Also, y be the angle between a radius of the

hemisphere and the z-axis.

Figure 8.9 Vertical tractions acting on the hemispherical surface

The unit normal vector to the surface at any point can be written as

n̂ =úúúû

ù

êêêë

é

y

y

cos

0

sin

while r and z components of the point are



Module 8/Lesson 1

15


z = a cos y , r = a sin y

The traction vector that acts on the hemispherical surface is,

úú

ú

û

ù

êê

ê

ë

é

+

+

=

úú

ú

û

ù

êê

ê

ë

é

úú

ú

û

ù

êê

ê

ë

é

=

úú

ú

û

ù

êê

ê

ë

é

=

y s y t

y t y s

y

y

s t

s

t s

q q

cossin

0

cossin

cos

0

sin

0

00

0

zrz

rzr

zrz

rzr

z

r

T

T

T

T

Considering the component of stress in the z-direction on the hemispherical surface,

we have

T z = -(t rz siny + s z cosy )

Substituting the values of t rz , s z , siny and cosy , in the above expression, we get

T z = 3B z2 (r

2 + z

2)-2

Integrating the above, we get the applied load P.

Therefore,

( )ò +=2

0

2

1222

p

y p d zr r T P z

= ( )[ ] ( )ò úû

ùêë

é++

-2

0

2

1222222 23

p

y p d zr r zr Bz

= ò2/

0

2 sincos)6(p

y y y p d B

= 6p B ò2/

2 sincosp

y y y o

d

Now, solving for ò2/

2 sincosp

y y y o

d , we proceed as below

Put cosy = t

i.e., -siny , d y = dt

If cos0 = 1, then t = 1

If cos2

p = 0, then t = 0

Hence, ò =úûùê

ëé=ú

ûùê

ëé-=-1

0

1

0

30

1

3

2

31

33t t dt t Therefore, P = 2p B

Or B =p 2

P



Module 8/Lesson 1

16


Substituting the value of B in Equation (8.16), we get

s z = - ( ) 2

5223

2

3 -+ zr z

π

P

s r = ( ) ( ) ( ) ý

ü

îí

ì

+-úû

ù

êë

é

+--

--2

5222

2

122

22 3

1

212 zr zr zr r

z

r

P

n p

s q = ( ) ( ) ýü

îíì

++++-- --

2

322

2

122

22

1)21(

2 zr z zr

r

z

r

Pn

p

t rz = ( ) 2

5222

2

3 -+- zr rz

P

p

Putting R = 22

zr + and simplifying, we can write

s z = 5

3

2

3

R

Pz

p

-

s r = úû

ùêë

é-

+-

5

23

)(

)21(

2 R

zr

z R R

P n

p

s q = úû

ùêë

é

+-

-)(

1

2

)21(3 z R R R

zP

p

n

t rz = 5

2

2

3

R

rzP

p - (8.16a)

Also, Boussinesq found the following displacements for this case of loading.

ur = úûù

êëé

+--

z R

r

R

rz

GR

P )21(

4 2

n

p

uq = 0 (8.16b)

u z = úû

ùêë

é+-

2

2

)1(24 R

z

GR

Pn

p

8.1.6 COMPARISON BETWEEN KELVIN’S AND BOUSSINESQ’S

SOLUTIONS

On the plane z = 0, all the stresses given by Kelvin vanish except t rz. For the special casewhere Poisson’s ratio n = 1/2 (an incompressible material), then t rz will also be zero on this

surface, and that part of the body below the z = 0 plane becomes equivalent to the half space



Module 8/Lesson 1

17


of Boussinesq’s problem. Comparing Kelvin’s solution (with n = 1 / 2) with Boussinesq’s

solution (with n = 1/2), it is clear that for all z ³ 0, the solutions are identical. For z £ 0, we

also have Boussinesq’s solution, but with a negative load – P. The two half-spaces,

which together comprise the infinite body of Kelvin’s problem, act as if they are

uncoupled on the plane z = 0, where they meet.

Further, a spherical surface is centered on the origin, we find a principal surface on which

the major principal stress is acting. The magnitude of the principal stress is given by

s 1 = 32

3

R

Pz

p (8.17)

where R is the sphere radius. It can be observed that the value of s 1 changes for negative

values of z, giving tensile stresses above the median plane z = 0.

8.1.7 CERRUTTI’S PROBLEM

Figure 8.10 shows a horizontal point load P acting on the surface of a semi-infinite

soil mass.

Figure 8.10 Cerrutti’s Problem



Module 8/Lesson 1

18


The point load represented by P acts at the origin of co-ordinates, pointing in the x-direction.

This is a more complicated problem than either Boussinesq’s or Kelvin problem due to the

absence of radial symmetry. Due to this a rectangular co-ordinate system is used in

the solution.

According to Cerrutti’s solution, the displacements are given by

ýü

îíì

úû

ùêë

é

+-

+-++=

2

2

2

2

)()21(1

4 z R

x

z R

R

R

x

GR

Pu x n

p (8.18)

u y = ýü

îíì

+--

22 )()21(

4 z R

xy

R

xy

GR

Pn

p (8.18a)

u z = ýü

îíì

+-+

z R

x

R

xz

GR

P)21(

4 2 n

p (8.18b)

and the stresses are

s x = - ( ) ( )

ýü

îíì

úû

ùêë

é

+--

+

-+-

z R

Ry y R

z R R

x

R

Px 222

22

2

3

2)21(3

2

n

p (8.19)

s y = - ( ) ( )

ýü

îíì

úû

ùêë

é

+--

+

-+-

z R

Rx x R

z R R

y

R

Px2

22

22

2

3

23

)21(3

2

n

p (8.19a)

s z = 5

2

2

3

R

Pxz

p (8.19b)

t xy = - ý

ü

îí

ì

úû

ù

êë

é

+++-+

-

+- )(

2

)(

)21(3

2

222

22

2

3 z R

Rx

x R z R R

x

R

Py n

p (8.19c)

t yz = 52

3

R

Pxyz

p (8.19d)

t zx = 5

2

2

3

R

zPx

p (8.19e)

Here, R2 = x2+y2+z2

It is observed from the above stress components that the stresses approach to zero for large

value of R. Inspecting at the x-component of the displacement field, it is observed thatthe particles are displaced in the direction of the point load. The y-component of

displacement moves particles away from the x-axis for positive values of x and towards

the x-axis for negative x. The plot of horizontal displacement vectors at the surface z = 0 is



Module 8/Lesson 1

19


shown in Figure 8.11 for the special case of an incompressible material. Vertical

displacements take the sign of x and hence particles move downward in front of the load

and upward behind the load.

Figure 8.11 Distribution of horizontal displacements surrounding the point load



Module 8/Lesson 1

20


8.1.8 MINDLIN’S PROBLEM

Figure 8.12 Mindlin’s Problem

The two variations of the point-load problem were solved by Mindlin in 1936. These are the

problems of a point load (either vertical or horizontal) acting in the interior of an elastic

half space. Mindlin’s problem is illustrated in Figure 8.12. The load P acts at a point

located a distance z beneath the half-space surface. Such problems are more complex thanBoussinesq’s or Kelvin or Cerrutti’s. They have found applications in the computations of

the stress and displacement fields surrounding an axially loaded pile and also in the study of

interaction between foundations and ground anchors.

It is appropriate to write Mindlin’s solution by placing the origin of co-ordinates a distance

C above the free surface as shown in the Figure 8.12. Then the applied load acts at the

point z = 2h.

From Figure 8.12,

R2 = r 2 + z2

R2

1 = r 2

+ z2

1 where z1 = z – 2h

Here z1 and R1 are the vertical distance and the radial distance from the point load.



Module 8/Lesson 1

21


For the vertical point load, Mindlin’s solution is most conveniently stated in terms ofBoussinesq’s solution. For example, consider the displacement and stress fields in

Boussinesq’s problem in the region of the half-space below the surface z = C . These

displacements and stresses are also found in Mindlin’s solution, but with additional terms.The following equations will give these additional terms.

To obtain the complete solution, add them to Equations (8.7a) and (8.7b)

Therefore,

s r =( )

( ) ( ) ( ) ( )

îíì +--

----

+-

-- 5

222

33

1

1

5

1

1

224276311221213

18 R

zccz zr

R

c z

R

z

R

zr P n n n n

n p

( )

ýü-

-7

230

R

c zcz (8.20)

s q =

( )

( ) ( )( ) ( )5

22

33

1

1 621662121

18 R

zccz

R

c z

R

zP ---

+-+

ýü

î

íì --

-n n n

n p (8.20a)

s z = ( )

( ) ( )( ) ( )

îíì --+

---

--

+- 5

223

33

1

1

5

1

3

1 182123221213

18 R

zccz z

R

c z

R

z

R

zP n n n

n p

( )

ýü-

+7

230

R

c zcz (8.20b)

t rz=( )

( ) ( ) ( ) ( )

ýü

îíì -

+--+

--

--

+7

2

5

22

33

1

5

1

2

1 306236321213

R

c zcz

R

ccz z

R R R

z

-18

Pr n n n

n p

(8.20c)

and t r q = t qr = t qz = t zq = 0 (8.20d)

Mindlin’s solution for a horizontal point load also employs the definitions for z1 and

R1. Now introduce rectangular coordinate system because of the absence of

cylindrical symmetry.

Replace r 2 by x2+y2, and assume (without any loss of generality) that the load acts in the

x-direction at the point z = c. Here the solution is conveniently stated in terms of Cerrutti’s

solution, just as the vertical point load was given in terms of Boussinesq’s solution.Therefore, the displacements and stresses to be superposed on Cerrutti’s solution are

u x =

( )

( ) ( ) ( )( ) ( )( )

ýü

îíì -

--+-

+-

--

+- 5

2

3

2

131

2 624343

116 R

c zcx

R

c zc x

R R R

x

G

P n n

n p

(8.21)

u y =( )

( )

ýü

îíì -

--- 533

1

6

116 R

c zcxy

R

xy

R

xy

G

P

n p (8.21a)



Module 8/Lesson 1

22


u z =( )

( ) ( )

ýü

îíì -

--+

-- 533

1

1 6432

116 R

c zcxz

R

cz xz

R

xz

G

P n

n p (8.21b)

s x =

( )

( ) ( ) ( ) ( )

ýü

î

íì -

-+--

--

--

+

-

7

2

5

22

33

1

5

1

2 3018236321213

18 R

c zcx

R

ccz x

R R R

xPx n n n

n p

(8.21c)

s y = ( )

ýü

îíì -

-+--

--

+-

-- 7

2

5

22

33

1

5

1

2 )(306)21(63)21()21(3

18 R

c zcy

R

ccz y

R R R

yPx n n n

n p

(8.21d)

s z = ( )

( ) ( ) ( ) ( )

ýü

îíì -

-+-+

--

+-

-- 7

2

5

22

33

1

5

1

2

1 306216321213

18 R

c zcz

R

ccz z

R R R

zPx n n n

n p

(8.21e)

t xy = ( )

( ) ( ) ýü

îíì --------+

- 7

2

5

2

33

1

5

1

2

30632121318 R

c zcx R

c zc x R R R

xPy n n n p

(8.21f)

t yz = ( )

( ) ( )

ýü

îíì -

--+

-- 755

1

1 3021633

18 R

c zcz

R

c z

R

zPxy n

n p (8.21g)

t zx = ( )

( ) ( )( ) ( ) ( )

îíì ---+

---

--

+- 5

22

33

1

1

5

1

12

62163221213

18 R

c zczcx z x

R

c z

R

z

R

z xP n n n

n p

( )

ýü-

-7

230

R

c z zcx (8.21h)

8.1.9 APPLICATIONS

The mechanical response of naturally occurring soils are influenced by a variety of factors.These include (i) the shape, size and mechanical properties of the individual soil particles,(ii) the configuration of the soil structure, (iii) the intergranular stresses and stress history,and (iv) the presence of soil moisture, the degree of saturation and the soil permeability.These factors generally contribute to stress-strain phenomena, which display markedlynon-linear, irreversible and time dependent characteristics, and to soil masses, which exhibitanisotropic and non-homogeneous material properties. Thus, any attempt to solve asoil-foundation interaction problem, taking into account all such material characteristics,is clearly a difficult task. In order to obtain meaningful and reliable information for practical

problems of soil-foundation interaction, it becomes necessary to idealise the behaviour ofsoil by taking into account specific aspects of its behavior. The simplest type of idealised soilresponse assumes linear elastic behaviour of the supporting soil medium.



Module 8/Lesson 1

23


In general, one can divide the foundation problems into two classes, (1) interactive problems

and (2) noninteractive problems. In the case of interactive problems, the elasticity of the

foundation plays an important role. For example, a flexible raft foundation supporting a

multistorey structure, like that illustrated in Figure 8.13 interacts with the soil. In terms of

elasticity and structural mechanics, the deformation of the raft and the deformation of the

soil must both obey requirements of equilibrium and must also be geometrically compatible.If a point on the raft is displaced relative to another point, then it can be realised that the

bending stresses will develop within the raft and there will be different reactive pressures in

the soil beneath those points. The response of the raft and the response of the soil are

coupled and must be considered together.

Figure 8.13 Flexible raft foundation supporting a multistorey structure

But non-interactive problems are those where one can reasonably assume the elasticity of the

foundation itself is unimportant to the overall response of the soil. Examples of non-

interactive problems are illustrated in Figure 8.14.

The non-interactive problems are the situations where the structural foundation is either very

flexible or very rigid when compared with the soil elasticity. In non-interactive problems, itis not necessary to consider the stress-strain response of the foundation. The soil

deformations are controlled by the contact exerted by the foundation, but the response of the

soil and the structure are effectively uncoupled.



Module 8/Lesson 1

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