applied elasticity_t.g. sitharam n l.govinda raju

8/19/2019 Applied Elasticity_T.G. Sitharam n L.govinda Raju

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Module1/Lesson1

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

Module 1: Elasticity1.1.1 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 lawsof 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 tensilestrength 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 naturalrubber 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 andapplied mathematician Ronald S. Rivlin in the1940's and 1950's.

1.1.2 T HE G ENERAL T HEORY OF E LASTICITY 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 betweenneighboring 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 thissubject 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 componentsto 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 changesare associated with adiabatic elastic deformation. The middle and late 1800's were a period in whichmany 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 torsionof 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; thelatter 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 stressand displacement due to concentrated forces acting at an interior point of a full space were derived byKelvin 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 L INEAR E LASTICITY

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 nosolutions 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 comparisonwith its original dimensions and the strain components and the rotations of all line elements are muchsmaller than unity. Hence, when formulating the equilibrium equations relevant to the deformed state, thelengths and angles of the body before deformation are used. In addition, when geometrical equationsinvolving strains and displacements are formulated, the squares and products of the small quantities areneglected. Therefore, these two measures are necessary to linearize the algebraic and differential

equations in elasticity for their easier solution.

1.1.4 A PPLICATIONS OF L INEAR E LASTICITY

The very purpose of application of elasticity is to analyse the stresses and displacements of elementswithin 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 solidmechanics, 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 machineelement 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 stressesand displacements of a structure in the form of a bar system, such as a truss or a rigid frame. Thestructural elements that are not in form of a bar, such as blocks, plates, shells, dams and foundations, theyare analysed only using theory of elasticity. Moreover, in order to analyse a bar element thoroughly andvery 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, themethods of analysis used here are not entirely the same. When the element is studied in mechanics ofmaterials, some assumptions are usually made on the strain condition or the stress distribution. Theseassumptions simplify the mathematical derivation to a certain extent, but many a times inevitably reducethe degree of accuracy of the results obtained. However, in elasticity, the study of bar-shaped elementusually 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 ofmaterials, it is usual to assume that a plane section before bending of the beam remains plane even afterthe bending. This assumption leads to the linear distribution of bending stresses. In the theory ofelasticity, however one can solve the problem without this assumption and prove that the stressdistribution 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 inmechanics of materials that the tensile stresses are uniformly distributed across the net section of themember, whereas the exact analysis in the theory of elasticity shows that the stresses are by no meansuniform, 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 relatingthe strains and displacements; constitutive equations relating the stresses and strains; boundary conditionsrelating to the physical domain; and uniqueness constraints relating to the applicability of the solution.



Module 2: Analysis of Stress

2.1.1 I NTRODUCTION

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 observedthat these forces vary from one small area to the next. On the small area AD at point O, aforce 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, theterm "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

A D

D

®D 0lim (2.0)

where DF is the internal force on the area D A surrounding the given point. Stress issometimes referred to as force intensity.

A



2.1.2 N OTATION 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. xyt means a stress,

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

2.1.3 C ONCEPT 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 DD

DD

DD

,, , 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 AF

x DD

=®D 0

lims

x

y

A xy AF

x DD=

®D 0limt (2.1)

x

z

A xz AF

x DD

=®D 0

limt

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

Figure 2.3 Stress components at point O



2.1.4 S TRESS T ENSOR

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 shearstresses. 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 S PHERICAL AND DEVIATORIAL STRESS T ENSORS

A general stress-tensor can be conveniently divided into two parts as shown above. Let usnow 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 allthe shear stresses are zero. We can divide the stress tensor into two parts, one havingonly the "hydrostatic stress" and the other, "deviatorial stress". The hydrostatic type ofstress 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 isknown 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 volumechanges without any change of shape, and does not necessarily cause failure.



2.1.6 I NDICIAL 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 arereplaced by numbered axes x1, x2 and x3 respectively. The components of the force DF ofFigure 2.1 (a) is written as DF 1, DF 2 and DF 3, where the numerical subscript indicates thecomponent with respect to the numbered coordinate axes.The definitions of the components of stress acting on the face x1can be written in indicialform as follows:

1

1

0111

lim AF

A DD

=®D

s

1

2

0121

lim AF

A DD

=®D

s (2.7)

1

3

0131

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 DD

=®D 0

lims (2.8)

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

2.1.7 T YPES 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 T WO -DIMENSIONAL STRESS AT A P OINT

A two-dimensional state-of-stress exists when the stresses and body forces are independentof one of the co-ordinates. Such a state is described by stresses y x s s , and t xy and the Xand 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 smallwedge 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) areobtained 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 cos 2q + s y sin 2q + 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 öç

è æ +

2p

q for q in the expression for x¢s .

By means of trigonometric identities

cos 2q =21

(1+ cos2 q ), sinq cosq =21 sin2q , (2.12)

sin 2q =21

(1-cos2 q )

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

( ) ( ) q t q s s s s s 2sin2cos21

21

xy y x y x x +-++=¢ (2.12a)

( ) ( ) q t q s s s s s 2sin2cos21

21

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 P RINCIPAL STRESSES IN T WO 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 cos2 q = 0 (2.13)

Therefore, tan 2 q = 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 xt 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 C AUCHY’S STRESS P RINCIPLE

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 afield of stress vectors ( )nT ˆ defined on S ¶ . By combining this principle with Euler’sequations that expresses balance of linear momentum and moment of momentum in any kindof 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 regionswhere 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 consideration

into two parts.(iii) Stress is a vector or vector field equipollent to the action of one part of the material on

the other.

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

2.1.11 D IRECTION C OSINES

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



From figure,

r y

r x == b a cos,cos and

r z=g cos

or b a cos,cos r yr x == and g cosr z =

Let ml == b a cos,cos and n=g cos

Therefore, mr y

lr

x == , and nr z =

Here, l, m and n are known as direction cosines of the line OP . Also, it can be written as2222 r z y x =++ (since r is the polar co-ordinate of P )

or 12

2

2

2

2

2

=++r z

r y

r x

1222 =++ nml 2.1.12 S TRESS COMPONENTS ON AN ARBITRARY P LANE

Consider a small tetrahedron isolated from a continuous medium (Figure 2.9) subjected to ageneral state of stress. The body forces are taken to be negligible. Let the arbitrary plane

ABC be identified by its outward normal n whose direction cosines are l, m and n.

In the Figure 2.9, z y x T T T ,, are the Cartesian components of stress resultant T , acting on

oblique plane ABC. It is required to relate the stresses on the perpendicular planesintersecting at the origin to the normal and shear stresses acting on ABC.

The orientation of the plane ABC may be defined in terms of the angle between a unit

normal n to the plane and the x, y, z directions. The direction cosines associated with theseangles are

cos (n, x ) = lcos (n, y ) = m and (2.19)cos (n, z ) = n

The three direction cosines for the n direction are related by

l2 + m 2 + n 2 = 1 (2.20)



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, A PAC = Am, A PBC = 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 getT 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. Theseequations show that the nine rectangular stress components at P will enable one to determinethe stress components on any arbitrary plane passing through point P .

2.1.13 S TRESS T RANSFORMATION

When the state or stress at a point is specified in terms of the six components with referenceto 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 thisgiven 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 directioncosines.

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 inTable 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 21 + s y m 2

1 + s z n 21 + 2(t xy l1 m1 + t yz m1 n1 + t xz l1 n1) (2.25)

Similarly by projecting z y x T 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+ m 1 l2)+ t yz (m1 n2 + n 1 m2 ) + t xz (n1l2 + l 1n2)(2.25a)

z x ¢¢t = s x l1 l3 + s y m1 m3+ s z n 1 n3 + t xy (l1 m3 + m 1 l3)+ t yz (m1 n3 + n 1 m3)+ t xz (n1 l3+ l 1 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. Forone 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 l22 + s y m

22 + s z n

22 + 2 (t xy l2 m2 + t yz m2 n2 + t xz l2 n2) (2.25c)

z¢s = s x l23 + s y m 2

3 + s z n 23 + 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 + m 2 n3)+t xz (l2 n3 + n 2 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 2i + n 2

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

and l1 l2 + m 1 m2 + n 1 n2 = 0

l2 l3 + m 2 m3 + n 2 n3 = 0 (2.26)

l1 l3 + m 1 m3 + n 1 n3 = 0



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2.2.1 P RINCIPAL STRESS IN T HREE DIMENSIONS

For the three-dimensional case, for principal stresses it is required that three planes of zeroshear stress exist, that these planes are mutually perpendicular, and that on these planes thenormal stresses have maximum or minimum values. As discussed earlier, these normal

stresses are referred to as principal stresses, usually denoted by s 1 , s 2 and s 3. The largest

stress is represented by s 1 and the smallest by s 3.

Again considering an oblique plane x¢, the normal stress acting on this plane is given by theEquation (2.25).

x¢s = s x l2

+ s y m2

+ s z n2

+ 2 (t xy lm + t yz mn + t xz ln ) (2.27)The problem here is to determine the extreme or stationary values of x¢s . To accomplish

this, we examine the variation of x¢s relative to the direction cosines. As l, m and n are not

independent, but connected by l2 + m 2 + n 2 = 1 , only l and m may be regarded asindependent variables.

Thus,

l x

¶¶

's = 0 ,

m x

¶¶

's = 0 (2.27a)

Differentiating Equation (2.27), in terms of the quantities in Equations (2.22a), (2.22b),(2.22c), we obtain

T x+ T z ln

¶¶

= 0,

T y + T z mn

¶¶

= 0, (2.27b)

From n2 = 1 - l 2 - m 2, we have

nl

ln -=

¶¶

andnm

mn -=

¶¶

Introducing the above into Equation (2.27b), the following relationship between thecomponents of T and n is determined



Module2/Lesson2


n

T

m

T

l

T z y x == (2.27c)

These proportionalities indicate that the stress resultant must be parallel to the unit normaland 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 0322

13 =-+- 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 2xy - t 2

yz -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 S TRESS 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 changewith rotation of axes or which remain unaffected under transformation, from one set of axes



Module2/Lesson2


to another. Therefore, the combination of stresses at a point that do not change with theorientation 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 2xy - t 2

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

s xs ys z - s xt 2yz - s yt 2

xz - s zt 2

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

2.2.3 E QUILIBRIUM 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 equilibriumset of forces. The small element with unit thickness shown in Figure 2.11(a) represents part



Module2/Lesson2


of a body and therefore must be in equilibrium if the entire body is to be in equilibrium. It isto 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. Fulfillmentof these conditions establishes certain relationships, known as the differential equations ofequilibrium. 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 isvery small, the stress components may be considered to be distributed uniformly over eachface.

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

( ) ( )

( ) 02

)(2

21

22

22

1

2

=DDD-DDD

+D-DD+DD÷ ø öç

è æ D

¶¶++DD÷÷ ø

öççè

æ D

¶¶

+-

DD÷÷ ø

ö

ççè

æ D

¶

¶++DD

÷÷ ø

ö

ççè

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 isreduced 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 öç

è 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 equationsyield the following differential equations of equilibrium.

0=+¶

¶+

¶¶

x xy x F

y x

t s

or 0=+¶

¶+

¶¶

y xy y F

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 xt t s

0=+¶

¶+

¶¶

+¶

¶ y

yz xy y F 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



Module2/Lesson2


2.2.4 O CTAHEDRAL STRESSES

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

and its direction cosines are3

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

322

222

12 t s s s s +=++= nmlT

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



Module2/Lesson2


The normal stress on this plane is given by

s = s 1 l2 + s 2 m2 + s 3 n2 (2.35)

and the corresponding shear stress is

( ) ( ) ( )[ ]21

22213

22232

22221 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 = )(31 2

322

21 s s s ++ (2.37)

Normal stress = s =31

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

Shear stress = t = 213

232

221 )()()(

31

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

Also, t = )(6)(231

3132212

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

t = 22

1 6231

I I - (2.41)

2.2.5 M OHR'S STRESS C IRCLE

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 M OHR C IRCLES FOR T WO DIMENSIONAL STRESS SYSTEMS

Biaxial Compression (Figure 2.13a)

The biaxial stresses are represented by a circle that plots in positive ( s , t ) space, passingthrough stress points s 1 , s 2 on the t = 0 axis. The centre of the circle is located on the



Module2/Lesson2


t = 0 axis at stress point ( )2121

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

( )2121

s 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 s

1

+s

(b)

s1

s 1

s2

s2 .-

+t

s1

s2 s

+2q

-



Module2/Lesson2


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 ( )2121

s s + and has radius ( )2121

s s - .

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

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

cos2 q = -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 inmagnitude, but opposite in sign, and are equal in magnitude to t zy. The directions of s 1 , s 2 are at 45 o to the directions of yz zy t t ,

2.2.7 C ONSTRUCTION OF M OHR’S C IRCLE FOR T WO -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 +21

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 .



Module2/Lesson2


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 stressassociated with the original x and y planes corresponds to points A and B on the circlerespectively. Points lying on the diameter other than AB, such as A¢ and B¢, define state ofstress 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

T x

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

O2qC

B¢

B1-t

max

A1

A¢

A( )s - t x xy,

x ¢

t

x



Module2/Lesson2


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 = ± ( )2121

s s - = ± 2

2

2 xy y x t

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 M OHR’S C IRCLE FOR T HREE -DIMENSIONAL STATE OFSTRESS

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)



Module2/Lesson2


(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 PA 2QB 3 relative to the s 1 axis

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



Module2/Lesson2


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

( )31max 21

s 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 G ENERAL E QUATIONS IN C YLINDRICAL C O-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


2.2.10 E QUILIBRIUM E QUATIONS IN POLAR C O-ORDINATES :(T WO -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 followingexpressions: x = r cosq , r 2 = x 2+y 2

y = r sinq , ÷ ø öç

è 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,



Module2/Lesson2


( )

02

cos2

cos2

sin

2sin

=÷ ø öç

è æ

¶¶

++-

÷ ø öç

è æ

¶¶

+-+-+÷ ø öç

è æ

¶¶

++´-

q q

q t

t q

t q

q q

s s

q s q

s s q s

q q q

q q q

d dr d

d dr

d

dr d F d

dr d dr r dr r

rd

r r r

r r

r r

Since d q is very small,

22sin

q q d d = and 12

cos =q d

Neglecting higher order terms and simplifying, we get

0=¶

¶+-+¶

¶q

q t

q s q s q s q

q d dr d dr d dr d dr r

r r r

r

on dividing throughout by rd q dr , we have

01 =+

-+

¶¶

+¶

¶r

r r r F r r r

q q s s q

t s (2.44)

Similarly resolving all the forces in the q - direction at right angles to r - direction, we have

( ) 02

sin

2sin

2cos

2cos

=+÷ ø ö

çè æ

¶¶+++-

÷ ø öç

è æ

¶¶+++÷

ø öç

è æ

¶¶++-

q q

q q

q q q

q q q

t t q q t

q

q q

t t

q t

q q

q s

s q

s

F dr r

d dr r rd d

dr d d

dr d

dr d d

dr

r r r

r r r

On simplification, we get

0=÷ ø öç

è æ

¶¶+++

¶¶

dr d r

r r r r q

t t t

q s q

q q q

Dividing throughout by rd q dr , we get

02

.1 =++¶+

¶¶

q q q q t t

q s

F r dr r

r r (2.45)

In the absence of body forces, the equilibrium equations can be represented as:

01 =

-+

¶¶

+¶

¶r r r

r r r q q s s q

t s (2.46)

021 =+

¶¶

+¶

¶r r r

r r q q q t t q

s



Module2 / Le sso n 3



2.3.1 GENERAL STATE OF STRESS IN T HREE -DIMENSION INC YLINDRICAL C O-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





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 N UMERICAL E XAMPLES

Example 2.1When 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 45 0 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

0021

12021

621

4





MPa

y

4

0021

12021

621

4

MPa

z

8

0001800

MPa

y x

1

0021

21

1021

621

4

MPa

z y

2

2

1200000

MPa

z x

2

2

1200000

Hence the new stress tensor becomes

822241

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





3 – I 1 2 + I 2 - I 3 = 0

Here 184591 I MPa

52326494559 2222 I MPa

27326236495494593 I MPa

The cubic equation becomes3 - 18 2 + 52 - 27 = 0

The roots of the cubic equation are the principal stresses. Hence the three principalstresses 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(636)554.149(

=

554.10232554.96

36554.5

A =554.102

2554.9= 100.83 - 4 = 96.83

B = 554.10326

= -(-63.324 - 6) = 69.324

C =23554.96

= 12 + 28.662 = 40.662

222 C B A

= 222 662.40324.6983.96

= 125.83

l1 =222 C B A

A =

83.12553.96

= 0.769





m1 =222 C B A

B =

83.125324.69

= 0.550

n1 =222 C B A

C =

84.125662.40

= 0.325

Similarly, the principal stress directions for 2 stress and 3 stress are calculated.

Therefore, l2 = 0.596 l3 = - 0.226 m2 = - 0.800 m3 = - 0.177 n2 = 0.057 n3 = 0.944

Example 2.3At a point in the structural member, the stresses (in MPa) are represented as in Figure2.18. Employ Mohr’s circle to determine:(a) the magnitude and orientation of the principal stresses(b) the magnitude and orientation of the maximum shearing stresses and associated

normal stresses.In each case show the results on a properly orientedelement.

Solution: Centre of the Mohr’s circle = OC

=2

2.556.27 = 41.4 MPa

(a) Principal stresses are represented by points A1 and B1.Hence the maximum and minimum principal stresses,referring to the circle are Figure 2.18

1,2= 41.422

7.206.272.5541

1 = 66.3 MPa and 2 = 16.5 MPa

The planes on which the principal stresses act are given by

p2 = tan -1 030.568.137.20

and 030.23618030.562 p

Hence, 015.28 p and 015.118 p

Mohr’s circle clearly indicates that p locates the 1 plane.

(b) The maximum shearing stresses are given by points D and E . Thus

max = 22 7.206.272.5541

= 24.9 MPa

X

Y 27.6 y

20.7 xy

55.2 x





300

Y

X

y 14 106

x 28 106

Figure 2.20

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

and 015.163s

Figure 2.19 Mohr’s stress circle

Example 2.4 The stress (in N/m 2) 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 planedefined by = 30 0 .

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

..

FA1B1 C

B(27.6, 20.7)

A(55.2, 20.7)

2 s

2 p

.

.

...





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

The radius of the circle is 66

10212

102814 . Corresponding to the 300 plane within

the element, it is necessary to rotate through 600

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)16

A (28 10 ,0)16

... ..

A

By

x

300

y x

x y 18.186 106

x 17.5 106

y 3.5 106





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 2.5A rectangular bar of metal of cross-section 30 mm 25 mm is subjected to an axial tensileforce of 180KN. Calculate the normal, shear and resultant stresses on a plane whosenormal 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 .

areasectionalcrossload Axial

y

253010180 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 nis,

nT mT lT z y x (b)

Case (i) For 02

1nand ml

Here 2/240,0,0 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





Substituting in (b), we get

2/12002

1

2

2400 mm N

Resultant Stress on the plane is

222 z y x T T T T

= 02

2400

2

2/706.169 mm N T

But shear stress can be determined from the relation

222T

or 22T

22 120706.169

2/120 mm N

Case (ii) For3

1nml

Again from (a),

0,3

240,0 z y y x T mT T

Normal Stress = 2/00.8003

1

3

2400 mm N

Resultant Stress on the plane is222

z y x T T T T

03

2400

2

T

2/13.113 mm N

Shear Stress = 22 8056.138 2/13.113 mm N





Example 2.6A body is subjected to three-dimensional forces and the state of stress at a point in it isrepresented as

MPa100200200

200100200

200200200

Determine the normal stress, shearing stress and resultant stress on the octahedralplane.

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 41.3463

1200

31

2003

1200

MPaT y 20.1733

1200

31

1003

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





MPa41.141

Example 2.7The 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 cosines21

,41

ml

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

Solution: We have the relation,

411

121

41

1

222

222

n

n

nml

(a) Using Cauchy’s formula,

kPaT x 60.414411

50021

40041

800

kPaT y 51.202411

60021

120041

400

kPaT z 66.506411

40021

60041

500

(b) Normal stress,

nT mT lT z y x

=411

66.50621

51.20241

60.414

kPa20.215

Resultant Stress on the Plane = 222 66.50651.20260.414T

= 685.28 MPa





Shear Stress on the plane = 22 20.21528.685 = 650.61 kPa

Example 2.8Given the state of stress at a point as below

400006090080100

kPa

Considering another set of coordinate axes, z y x in which z coincides with z and x is rotated by 30 0 anticlockwise from x-axis, determine the stress components in the newco-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





005.0866.080205.060866.0100 221 x

kPa x 3.129

00866.05.08020866.0605.0100 22 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

400003.893.29

03.293.129

(kPa )

Example 2.9The stress tensor at a point is given by the following array

)(301040102020

402050kPa

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

21,

21,

21

Solution: The stress vectors are given bynmlT 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

21

4021

202

150 xT = kPa35.45

21

1021

202

120 yT = kPa858.0





21

3021

102

140 zT = kPa28.48

Now, Resultant Stress is given bykPak jiT ˆ28.48ˆ858.0ˆ35.45

Example 2.10The Stress tensor at a point is given by the following array

)(204030403020

302040kPa

Calculate the deviator and spherical stress tensors.

Solution: Mean Stress = z y xm 31

20304031

kPa30

Deviator stress tensor =

m z yz xz

yzm y xy

xz xym x

=

302040304030302030203040

= kPa

10403040020

302010

Spherical Stress tensor =

m

m

m

0000

00





= kPa

3000

0300

0030

Example 2.11The 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 notas 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) becomes 23350 xy xz

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





And y z y x

x xyz y

y z

xz yz z 23,26, 22

Therefore (iii) becomes 22 2623 y x xyz y z y

At the point (1, -1, 2), we get 2112211612213 = -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 xy F

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.





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

0322

13 I I I

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

2/806020604040

204020mm 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 22223

= 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 03cos41

cos43

cos 3 (b)

Put3

cos 1 I r

i.e.,3

60cosr

20cosr

Substituting in (a), we get

034400020cos800020cos6020cos 23 r r r





034400020cos800020cos6020cos20cos 22 r r r r

0344000160000cos8000

cos40400cos6020coscos40400cos 2222

r

r r r r r

0168000cos9200cos 33 r r

0168000

cos92001

cos.,. 323

r r ei (c)

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

439200

2r

349200

r

755.110

and3

1680004

3cosr

3755.1104168000

3cos = 0.495

or 495.03cos

01 9.3965.1193 or

02 1.80 and 0

3 9.159

3cos 1

11 I r

360

)9.39cos(755.110

2/96.104 mm N

3cos 1

222

I r

360

)1.80cos(755.110

22 /04.39 mm N

3cos 1

333

I r





360

)9.159cos(755.110

23 /84 mm N

Example 2.13At a point in a given material, the three dimensional state of stress is given by

22 /20,/10 mm N mm N xy z y x and 2/10 mm N zx yz

Compute the principal planes if the corresponding principal stresses are2

32

22

1 /7.2,/10,/3.37 mm N mm N mm N

Solution : The principal planes can be obtained by their direction Cosines l, m and n associated with each of the three principal stresses, ., 321 and

(a) To find Principal plane for Stress 1

3.271010103.272010203.27

)3.3710(101010)3.3710(2010203.3710

Now,3.2710

103.27 A = 745.29-100

A = 645.29

646

)100546(

3.2710

1020

B

B

10103.2720

C

= 200 + 270.3

C = 470.3

08.1027

)3.470()646(29.645 222222 C B A





628.008.102729.645

2221C B A

Al

628.0

08.1027

6462221

C B A

Bm

458.008.10273.470

2221C B A

C n

(b) To find principal plane for Stress 2

201010102020

102020

)1010(101010)1010(20

10201010

30010040020101020

A

300)100400(2010

1020 B

0)200200(10102020

C

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

707.026.424

3002222

C B A

Al

707.026.424

3002222

C B A

Bm

02222

C B A

C n





(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 222222 C B A

326.092.14271.46

2223C B A

Al

322.092.142

462223

C B A

Bm

888.092.142

1272223

C B A

C n



Module3/Lesson1


Module 3 : Analysis of Strain

3.1.1 I NTRODUCTION

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

dxdu

xu

x x =

DD=

®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


3.1.2 T YPES OF STRAIN

Strain may be classified into direct and shear strain.

Figure 3.2 Types of strains

(a)

(b)

(c) (d)



Module3/Lesson1


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 = yu

¶¶

+ xv

¶¶

(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 Figure3.2(c) the following are the normal and shearing strains:

zw

yv

xu

z y x ¶¶=

¶¶=

¶¶= e e e ,, (3.3)

zu

xw

yw

zv

xv

yu

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


3.1.3 D EFORMATION OF AN INFINITESIMAL L INE E LEMENT

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


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

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

y yu

x xu

u D¶¶+D

¶¶+D

¶¶=D (a)

Similarly, z zv

y yv

x xv

v 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 yu

xu

¶¶

¶¶ , and

zu

¶¶ 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 representedin the form of a matrix called the displacement-gradient matrix such as

úúúúúú

û

ù

êêêêêê

ë

é

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

¶¶

=úúû

ù

êêë

é¶¶

zw

yw

xw

zv

yv

xv

zu

yu

xu

xu

j

i



Module3/Lesson1


3.1.4 C HANGE IN L ENGTH OF A L INEAR E LEMENT

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 ,,

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

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

z zu

y yu

x xu

x D¶¶+D

¶¶+D÷

ø öç

è æ

¶¶+=¢D 1

z zv

y yv

x xv

y D¶¶+D÷÷ ø

öççè

æ ¶¶++D

¶¶=¢D 1

z zw

y yw

x xw

z D÷ ø ö

çè æ

¶¶

++D¶¶

+D¶¶

=¢D 1

Taking the difference between ( )2S ¢D and ( )2S D , we get

( ) ( ) ( ) ( )2222 S 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

ùêêë

é÷ ø öç

è æ

¶¶+÷

ø öç

è æ

¶¶+÷

ø öç

è æ

¶¶+

¶¶=

222

21

xw

xv

xu

xu

xe (3.5a)



Module3/Lesson1


úúû

ù

êêë

é÷÷ ø

öççè

æ ¶¶+÷÷

ø

öççè

æ ¶¶+÷÷

ø

öççè

æ ¶¶+

¶¶=

222

21

yw

yv

yu

yv

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¶¶

¶¶+

¶¶

¶¶+

¶¶

¶¶+

¶¶+

¶¶==

yw

xw

yv

xv

yu

xu

yu

xv

yx xy e e (3.5d)

úû

ùêë

é¶¶

¶¶+

¶¶

¶¶+

¶¶

¶¶+

¶¶+

¶¶==

zw

yw

zv

yv

zu

yu

zv

yw

zy yz e e (3.5e)

úûù

êëé

¶¶

¶¶+

¶¶

¶¶+

¶¶

¶¶+

¶¶+

¶¶==

xw

zw

xv

zv

xu

zu

xw

zu

xz zx e e (3.5f)

Now, introducing the notation

S S S

PQ DD-¢D=e

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

( ) ( ) ( )( )

( )22

222

22S

S

S S S

S S S S D÷÷ ø

öççè

æ

DD-¢D+

DD-¢D=D-¢D

( ) ( )22

21

S PQPQ Dúû

ùêëé += e e

( )22

11 S PQPQ

Dúû

ùêë

é += e e

From Equation (3.5), substituting for ( ) ( )22 S 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 öç

è æ + e e e e e e e e 2222

21

1

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

S z

nS y

mS x

lDD=

DD=

DD= ,,

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

21

1

The above equation gives the value of the relative displacement at point P in the directionPQ with direction cosines l, m and n.



Module3/Lesson1


3.1.5 C HANGE IN L ENGTH OF A L INEAR E LEMENT -L INEARC OMPONENTS

It can be observed from the Equation (3.5a), (3.5b) and (3.5c) that they contain linear

terms like .,,,, etc zw

yv

xu ----

¶¶

¶¶

¶¶

as well as non-linear terms like

.,.,2

etc yu

xu

xu ----÷÷ ø

öççè

æ ¶¶

¶¶

÷ ø öç

è æ

¶¶

If the deformation imposed on the body is small, the terms

like etc,, yv

xu

¶¶

¶¶

are extremely small so that their squares and products can be neglected.

Hence retaining only linear terms, the linear strain at point P in the direction PQ can beobtained as below.

zw

yv

xu z y x ¶

¶=¶

¶=¶

¶= e e e ,, (3.6)

zu

xw

yw

zv

xv

yu

zx yz xy ¶¶+

¶¶=

¶¶+

¶¶=

¶¶+

¶¶= g g g ,, (3.6a)

and nlmnlmnml zx yz xy z y xPQPQ g g g e e e e e +++++=@ 222 (3.6b)

If however, the line element is parallel to x axis, then l = 1, m = 0, n = 0 and the linearstrain is

xu

xPQ ¶¶== e e

Similarly, for element parallel to y axis, then l = 0, m = 1, n = 0 and the linear strain is

yv

yPQ ¶¶== e e

and for element parallel to z axis, then l = 0, m = 0, n = 1 and the linear strain is

zw

zPQ ¶¶== e e

The relations expressed by equations (3.6) and (3.6a) are known as the strain displacementrelations of Cauchy.

3.1.6 S TRAIN T ENSOR

Just as the state of stress at a point is described by a nine-term array, the strain can berepresented tensorially as below:



Module3/Lesson1


e ij = ÷÷ ø

öççè

æ ¶¶

+¶¶

i

j

j

i

x

u

xu

21

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

The factor 1/2 in the above Equation (3.7) facilitates the representation of the straintransformation 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 = 21

g xy , e yz = 21

g yz , e xz = 21

g 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

21

21

21

21

21

21

(3.9)

3.1.7 S TRAIN T RANSFORMATION

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.

zw

yv

xu

z y x ¶¶=

¶¶=

¶¶= e e e ,,

yw

zv

xv

yu

yz xy ¶¶+

¶¶=

¶¶+

¶¶= g g , and

zu

xw

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

of their orientations.Hence



Module3/Lesson1


zw

yv

xu

z y x ¢¶¶=

¢¶¶=

¢¶¶= ¢¢¢ e e e ,,

zu

xw

yw

zv

xv

yu

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

21

21

21

21

21

21

úúú

û

ù

êêê

ë

é´

úúúúúú

û

ù

êêêêêê

ë

é

´úúú

û

ù

êêê

ë

é=

321

321

321

333

222

111

21

21

21

21

21

21

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 S PHERICAL AND DEVIATORIAL STRAIN T ENSORS

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 anychange 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


3.1.9 P RINCIPAL STRAINS - STRAIN INVARIANTS

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

nmlPQPQPQ

¶¶

¶¶

¶¶ 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

ln

nl (3.12)

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

( ) z zy zx zx xy x nmlln

nml e g g g g e 220 ++¶¶+++=

( ) z zy zx yz xy y nmlmn

nlm e g g g g e 220 ++¶¶+++=

Substituting forln

¶¶

andmn

¶¶

from Equation 3.12, we get

nnml

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



Module3/Lesson1


n

nml

m

nlm z zy zx yz xy y e g g g g e 22 ++=

++

Denoting the right hand expression in the above two equations by e 2 ,

022

022

022

and =-++

=-++

=-++

nnml

mnml

lnml

z zy zx

yz y xy

xz xy x

e e g g

e g e g

e g g e

(3.12a)

Using equation (3.12a), we can obtain the values of l, m and n which determine the direction

along which the relative extension is an extremum. Now, multiplying the first Equation by l,the second by m and the third by n, and adding them,

We get

( ) ( )222222 22 nmlnlmnlmnml zx yz xy z y x ++=+++++ e g g g e e e (3.12b)

Here nlmnlmnml zx yz xy z y xPQ g g g e e e e +++++= 222

1222 =++ nml

Hence Equation (3.12b) can be written as

e e =PQ

which means that in Equation (3.12a), the values of l, m and n determine the direction alongwhich the relative extension is an extremum and also, the value of e is equal to thisextremum. Hence Equation (3.12a) can be written as

( )

( )

( ) 021

21

021

21

021

21

=-++

=+-+

=++-

nml

nml

nml

z zy zx

yz y yx

xz xy x

e e g g

g e e g

g g e e

(3.12c)

Denoting,

,21

xy xy e g = ,21

yz yz e g = zx zx e g =

21

then

Equation (3.12c) can be written as



Module3/Lesson1


( )( ) 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 theco-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 get032

21

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

( )

( )2223

2222

1

41

41

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



Module3/Lesson2


Module 3: Analysis of Strain

3.2.1 M OHR’S C IRCLE 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 onthe horizontal axis, positive to right. When the shear strain is positive, the point representing

the x-axis strains is plotted at a distance2g

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

2g

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 2 q = 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 E QUATIONS OF C OMPATABILITY FOR STRAIN

Expressions of compatibility have both mathematical and physical significance. From amathematical point of view, they assert that the displacements u, v, w are single valued andcontinuous 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 onlythree components of displacement. One cannot therefore arbitrarily specify all of the strainsas 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 xu

¶¶¶

, 2

2

x y

¶¶ e

= y xv¶¶

¶2

3

y x xy

¶¶¶ g 2

= 2

3

y xu

¶¶¶

+ y xv¶¶

¶2

3

(3.21)



Module3/Lesson2


or 2

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

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

xu

x ¶¶=e (3.23)

yv

y ¶¶=e (3.23a)

zw

z ¶¶=e (3.23b)

yu

xv

xy ¶¶+

¶¶=g (3.23c)

zv

yw

yz ¶¶+

¶¶=g (3.23d)



Module3/Lesson2


xw

zu

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 yv

x y

¶¶¶=

¶¶ e

(3.23g)

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

2

3

2

3

2

2

2

2

x yv

y xu

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 xu

x yv

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

(3.23 l)

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

z y xu

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 xv

y xw

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 xu

z x y x x xy xz yz

¶¶¶¶=

¶¶¶

+¶¶

¶+÷

÷ ø

öççè

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


3.2.3 M EASUREMENT OF SURFACE STRAINS - STRAIN R OSETTES

Strain Rosettes

Whenever a material is subjected to plane stress, it is desirable to obtain the stresses by directmeasurement. As the stresses cannot be measured directly, it is essential to measure thestrains or deformation that takes place in the material during loading. These strains ordeformations are measured with sensitive strain gauges attached to the surface of the body

before it is loaded so that these gauges can record the amount of strain that takes placeduring loading. It is more accurate and easier to measure in the neighbourhood of a chosen

point on the surface of the body, the linear strains in different directions and then computefrom these measurements the magnitudes and directions of the principal strains 1e and 2e .Such a group of strain gauges is called a strain rosette.

Strain Transformation Laws

If the components of strain at a point in a body are represented as y x e e , and xyg with

reference to the rectangular co-ordinate axes OX and OY, then the strain components withreference to a set of axes inclined at an angle q with axis OX can be expressed as

q g

q e e e e

e q 2sin2

2cos22

xy y x y x +÷÷ ø ö

ççè æ -+÷÷ ø

öççè æ += (3.25)

) q g q e e g q 2cos2sin xy x y +-= (3.26)

and the principal strains are given by

maxe or ( ) 22min 2

12 xy y x

y x g e e e e

e +-±÷÷ ø

öççè

æ += (3.27)

The direction of the principal strains are defined by the angle q as

÷÷ ø

ö

ççè

æ

-=

y x

xy

e e

g q 2tan (3.28)

Also, the maximum shear strain at the point is given by following relation.

( ) 22max xy y x g e e g +-= (3.29)

Measurement of Strains using Rosettes

In a rectangular rosette, the strains are measured at angles denoted by 021 45,0 == q q and

03 90=q . In an equiangular rosette (also called Delta Rosette)

03

02

01 120,60,0 === q q q

Let 21 , q q e e and 3q e be the strains measured at three different angles 2,1 q q and 3q respectively. Now, using the section transformation laws, we can write the threesimultaneous equations as follows:



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


öççè

æ -+÷÷ ø

öççè

æ += (b)

33 2sin2

2cos223

q g

q e e e e


öççè

æ -+÷÷ ø

öççè

æ += (c)

For a rectangular rosette,0

21 45,0 == q q and 03 90=q

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

We get

0220

+÷÷ ø

öççè

æ -+÷÷ ø

öççè

æ += y x y x e e e e

e

= ( ) y x y x e e e e -++21

xe e =\0

( )2

02245

xy y x y x g e e e e e +÷÷ ø

öççè

æ -+÷÷ ø

öççè

æ +=

( ) xy y x g e e ++=21

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

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

Also , ( ) ( )0090 180sin

2180cos

22 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 +-+21

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


03

021 120,60,0 === q q q

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

( )0120600 2231

, 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 xyg , the principal strains maxe and mine can be

computed.

3.2.4 N UMERICAL E XAMPLES

Example 3.1A sheet of metal is deformed uniformly in its own plane that the strain componentsrelated 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

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

cos2 q +2 xyg

sin2 q

y¢e =2

y x e e + -

2 y x e e -

cos2 q -2 xyg

sin2 q

2'' y xg

= ÷÷ ø

öççè

æ --

2 y x e e

sin2 q + 2 xyg

cos2 q

Using Equation (3.19), we find

2q = tan -1 ÷ ø öç

è æ

600450

= 36.8 0

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

61 101150 -´=e

62 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 3.2By means of strain rosette, the following strains were recorded during the test on astructural member.

mmmmmmmmmmmm /1013,/105.7,/1013 690

645

60

--- ´=´=´-= 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 ( ) 22min 2

12 xy y x

y x g e e e e

e +-±÷÷ ø

öççè

æ +=

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

2110

21313 --- ´+--±÷

ø öç

è æ +-=e

maxe \ or 6min 1015 -´±=e

6.71

1e

2e



Module3/Lesson2


Hence 6max 1015 -´=e and 6

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

1013131015

--

--´=

577.02tan -=q 01502 =\ q

075=\ q Hence the directions of the principal planes are 0

1 75=q and 02 165=q

Example 3.3Data taken from a 45 0 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 : Given 60 10750 -´=e

645 10110 -´-=e

690 10210 -´=e

Now, for a rectangular rosette,

60 10750 -´== e e x

690 10210 -´== e e y

( )900452 e e e g +-= xy

666 1021010750101102 --- ´+´-´-=

6101180 -´-= xyg

\ The magnitudes of principal strains are

maxe or ( ) 22min 2

12 xy y x

y x g e e e e e +-±÷÷ ø öççè

æ +=



Module3/Lesson2


i.e., maxe or ( )[ ] ( )[ ]26266min 10118010210750

21

102

210750 --- -+-±÷ ø öç

è æ +=e

( ) 66 107.129721

10480 -- ±´=

66 1085.64810480 -- ´±´= 6

1max 1085.1128 -´==\ e e

62min 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

01 3.57=\ q and 0

2 3.147=q Example 3.4If 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 xu

xe

,106 3-´=¶¶= yz yv

ye

3103 -´=¶¶= zw

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 ( ) ( ) úûùêë

é +¶¶++

¶¶= -- 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 45 0 in the anticlockwisedirection, 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

02

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

[ ]

úúúúúú

û

ù

êêêêêê

ë

é -

úú

ú

û

ù

êê

ê

ë

é-=¢

100

02

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

05.021 =

xyg or 1.0205.0 =´= xyg

226.02113.0,0 =´== zx yz g g

Example 3.6The 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

--

-=

úúúúúú

û

ù

êêêêêê

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

2222

1

41

1.005.005.01.0

g g g e e e e e e

e e e

++-++=

=+-=++=

( )( ) ( )( ) ( )( ) ( ) ( ) ( )[ ]222 08.01.03.041

1.005.005.005.005.01.0 -++-+-+-=

0291.02 -=\ J

( )( )( ) ( )( )[ ]2(0.3)0.052(0.08)0.052(0.1)0.10.08)(0.10.341

0.050.050.13J -+--+-=

002145.03 -= J

\ The cubic equation is0002145.00291.01.0 23 =+-- e e e (i)

Now q q q cos3cos43cos 3 -=

Or 03cos41

cos43

cos 3 =-- q q q (ii)

Let3

cos 1 J r += q e

=31.0

cos +q r

033.0cos += q e r

\ (i) can be written as

( ) ( ) ( ) 0002145.0033.0cos0291.02033.0cos1.03033.0cos =++-+-+ q q q r r r

( )( ) ( )0002145.000096.0

cos0291.02033.0cos1.02033.0cos033.0cos=+-

-+-++ 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 0300112.0

cos203251.03cos =--

r r q q (iii)

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

4303251.0

2 =

r

i.e., 2082.0303251.04 =´=r

and 3

00112.043cos

r =q

or( )

5.0496.02082.0

00112.043cos 3

@=´=q

0603 =\ q or 01 20

360 ==q

03

02 140100 == q q

31.0

20cos2082.0

3cos

0

1111

+=

+=\ J r q e

228.01 =e

126.031.0

140cos2082.03cos

0031.031.0

100cos2082.03

cos

01333

01222

-=+=+=

-=+=+=

J r

J r

q e

q e



Module3/Lesson2


To find principal directions

(a) Principal direction for 1e

( )( )

( )( )

( )( )ú

úú

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

278.015.01

´-´=-

-=

C

00362.01 -=\ C

( ) ( ) ( )22221

21

21 00362.00247.0046984.0 -++=++ C B A

= 0.0532

883.00532.0

046984.02

12

12

1

11

==++

=\C B A

Al

464.00532.00247.0

21

21

21

11 ==++= C B A

Bm



Module3/Lesson2


068.00532.000362.0

21

21

21

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(0531.004.0

05.015.02

-=+-=-

= B

00562.000188.00075.005.004.0

0469.015.02

=-=-

-=C

Now, ( ) ( ) ( ) 0125.000562.0009965.000499.0 22222

22

22

=+-+-=++ C B A

399.00125.000499.0

22

22

22

22

-=-=++

=\C B A

Al

797.00125.0

009965.022

22

22

2

2 -=-=

++=

C B A

Bm

450.00125.0

00562.022

22

22

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, ( ) ( ) ( ) 0322.001054.00284.00109.0 22223

23

23

=+-+=++ C B A

338.00322.00109.0

23

23

23

33

==++

=\C B A

Al

882.00322.00284.0

23

23

23

33

-=-=++

=C B A

Bm

327.00322.0

01054.023

23

23

33

==++

=C B A

C n

Example 3.7The 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 xu

x 01.0=¶¶=e

301.0 z yv y =¶¶=e

z zw

z 1.0=¶¶=e

y x x yu

xv

xy 04.001.004.0 ++=¶¶+

¶¶=g

y z xy zv

yw

yz203.002.0 +=

¶¶+

¶¶=g

201.00 y

x

w

z

u zx

+=

¶

¶+

¶

¶=g

At point P (3, 2, -5), the strain components are5.0,25.1,02.0 -=-==

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

21

21

21

21

\ Strain matrix becomes

úúú

û

ù

êêê

ë

é

--=

50.081.002.0

81.025.1115.0

02.0115.002.0

ije

Example 3.8The 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 Sphericalstrain 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 21

,21

== and zx zx g e 21

=

( )

úûù

êëé

÷ ø öç

è æ +÷

ø öç

è æ +÷

ø öç

è æ

+÷÷ ø

öççè

æ +÷÷ ø

öççè

æ +÷÷ ø

öççè

æ =\

31

0005.031

0004.031

0002.02

3

10005.0

3

10003.0

3

10001.0

222

oct ne

( ) 001.0=\oct ne

Octahedral Shearing Strain is given by

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

where ( )oct Re = Resultant strain on octahedral plane



Module3/Lesson2


( ) ( ) ( ) ( )[ ]222

31

y yz xz yz y xy xz xy xoct R e e e e e e e e e e ++++++++=\

( ) ( ) ( )[ ]222 0005.00004.00005.00004.00003.00002.00005.00002.00001.031 +++++++=

( ) 001066.0=\ oct Re

( ) ( )22 001.000106.02 -=\ oct g

000739.0=\ oct g

(b) Deviator and Spherical strain tensors.

Here Mean Strain =3

z y xm

e e e e

++=

3

0005.00003.00001.0 ++=

0003.0=\ me

\ Deviator Strain tensor =

( )( )

( )úúú

û

ù

êêê

ë

é

--

-

0003.00005.00004.00005.0

0004.00003.00003.00002.0

0005.00002.00003.00001.0

i.e.,úúú

û

ù

êêê

ë

é-=¢

0002.00004.00005.0

0004.000002.0

0005.00002.00002.0

E

and spherical strain tensor =úúú

û

ù

êêê

ë

é=¢¢

m

m

m

E

e

e e

00

0000

i.e.,úúú

û

ù

êêê

ë

é=¢¢

0003.000

00003.00

000003.0

E

Example 3.9The components of strain at a point in a body are as follows:

221 y x zc x +=e

z x y2=e

xyzc xy 22=g where c1 and c2 are constants. Check whether the strain field is compatible one?



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 y 2,2 2

2

=¶

¶=

¶¶ e e

zc y x

yzc x

xy xy2

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 3.10Under what conditions are the following expressions for the components of strain at apoint 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 02

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

y 22

2

=¶

¶ e

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

ax y xy x

xy 22 ++=¶

¶b a

g

b a g +=¶¶

¶ x

y x xy 2

2

\(i) becomes

b a +=++ xabax 2224

( ) b a +=++ xbaax 224

xax a 24 =\ or a2=a and ( )ba += 2b

Example 3.11For the given displacement field

( ) x xcu 22

+= ( ) 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 yu

xv

cx xu

xy x 404,2 =+=¶¶+

¶¶==

¶¶= g e

cc z

v

y

wcy

y

v yz y

=+=¶

¶+¶

¶==¶

¶= 0,4 g e

cc xw

zu

cz zw

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 öç

è æ ÷

ø öç

è æ -++÷

ø öç

è æ +÷

ø öç

è æ -+=

cr 5.13=\ e

Example 3.12The 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

[ ]01.002.0015.003.002.001.031 -+++-=

( ) 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

03.0202.0

201.0

202.0

02.02015.0

201.0

2015.0

01.0

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

31

z yz xz yz y xy xz xy xoct R e e e e e e e e e e ++++++++=

( ) ( ) ( ){ }222 03.001.0005.001.002.00075.0005.00075.001.031 ++-++-+-+=

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 3.13The 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 zw

Ky yv

Kx xu

z y x 8,4,2 =¶¶==

¶¶==

¶¶= e e e

K K yu

xv xy 404 =+=¶¶+¶¶=g

K K zv

yw

yz =+=

¶¶+

¶¶= 0g

K K xw

zu

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

12

104

21

242

1804

22

K K K K K K r +÷

ø öç

è æ ÷

ø öç

è æ ++÷

ø öç

è æ +÷

ø öç

è æ +=e

K K K r 21

124 ++=\ e

K r 5.16=\ e

Case ( b)( ) ( ) ( ) ( ) ( ) ( )020040240814 2 K K K K K r

+++++=e K r 4=\ e



Module3/Lesson2


Case ( c)( ) ( ) ( ) ( ) ( ) ( )( )6.08.020048.024086.04 22 K K K K K r

+++++=e K r 76.17=\ e




Module4/Lesson1

Module : 4 Stress-Strain Relations

4.1.1 I NTRODUCTION

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 and6 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 orstrains. 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 theydescribe the macroscopic behavior of a material based on its internal constitution.

4.1.2 L INEAR E LASTICITY -G ENERALISED H OOKE’S L AW

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 calledHooke’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




Module4/Lesson1

xe = C 11 xs + C 12 ys + C 13 zs + C 14 xyt + C 15 yzt + C 16 zxt + C 17 xzt + C 18 zyt + 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 writtenas

) { }klijklij D e s = (4.2)

where ijkl D = Elasticity matrix

ijs = Stress components

{ }kle = Strain components

Since both stress ijs and strain ije are second-order tensors, it follows that ijkl D is a fourth

order tensor, which consists of 3 4 = 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

ïïïïïï

ïïïïïï

ý

ü

ïïïïïï

î

ïïïïïï

í

ì

úúúúúúúúúúúú

û

ù

êêêêêêêêêêêê

ë

é

=

ïïïïïï

ïïïïïï

ý

ü

ïïïïïï

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




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

ijij

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

jijii

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,

jii j

DU =

¶¶

e e

2

(4.10)

)4.4(

666564636261

565554535251

464544434241

363534333231

2625242322

161514131211

21

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

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




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

ïïïï

ïïïï

ý

ü

ïïïï

î

ïïïï

í

ì

úúúúúúúúú

û

ù

êêêêêêêêê

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

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 iscalled an "orthotropic" material. If the xy, yz and zx planes are considered planes ofsymmetry, 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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úúúúúúúú

û

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ë

é

=

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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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î

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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




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

=

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ý

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 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 elasticsymmetry about one of the coordinate axes, x , y , and z. In such case, the material constantsreduce to 8 as shown below.

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

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

000000

000




Module4/Lesson1

material is thus written as an extension of that of a transversely isotropic materialas shown below.

( )( )

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û

ù

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ë

é

--

-=

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ý

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 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

21

0

0

0

0

0

21

0

0

0

0

21

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 writtenas:

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++

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=

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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

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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)




Module4/Lesson1

Also, xy xy Gg t =

yz yz Gg t =

zx zx Gg t =

Now, expressing strains in terms of stresses, we get

( ) ( )( ) z y x x GGGG

Gs s

l l

s l

l e +

+-

++=

23223

( ) ( )( ) x z y y GGGG

Gs s

l l

s l

l e +

+-

++=

23223

( ) ( )( ) y x z z GGGG

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( GGG

++

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 ElasticityAlso,

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)




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 öç

è æ +

32G

l (4.26)

where K = Bulk modulus of elasticity.Also,

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

úûù

êëé --=-

E p

K p)21(3 n




Module4/Lesson1

or ( )[ ]vK E 213 -=

Therefore, K =)21(3 n -

E (4.27)



Module4/Lesson2


Module 4: Stress-Strain Relations

4.2.1 E LASTIC STRAIN E NERGY 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 anelongation 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 is2dzdy

xs .

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 strainenergy U for an infinitesimal element subjected to uniaxial stress is

dU =21

s x dy dz ´ e x dx

(a) (b)



Module4/Lesson2


=21

s x e x dx dy dz

Therefore, dU =21

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

Hence,dV dU

= 0U =21

s 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 E NERGY IN AN E LASTIC BODY

(a) (b)



Module4/Lesson2


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 dueto 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 elementundergoes the elongation dx xe . In a linearly elastic material, strain grows in proportion tostress. Thus the strain energy dU stored in the element, when the final values of stress andstrain are xs and xe is

dU =21 (s x dydz ) (e xdx)

(c)(d)



Module4/Lesson2


=21

s 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 =21 ò

v

s x ve dV (4.29)

Taking s x = AP

and e x = Ld

where P = uniaxial load on the memberd = 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 APU d 21

Therefore, U =21

P .d since V = L ´ A (4.30)

Next consider the shear stress component t xy acting on an infinitesimal element inFigure 4.2(c). The corresponding deformation due to the shear strain component g xy isindicated in Figure 4.2(d). In this case the force t xy dxdz acting on the positive y face doeswork as that face translates through the distance g xy dy. Because of linearity, g xy and t xy growin 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 21=

dxdydz xy xyg t 21=

Therefore, dU =21

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


strain energy stored in the element when the final stresses are s x , s y , s z , t xy , t yz , t zx and thefinal strains are e x , e y , e z , g xy , g yz , g zx is thus

dU =21

(s xe x + s ye y + s ze z + t xy g xy + t yz g yz + t zx 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 =21

òv

(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 B OUNDARY C ONDITIONS

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 thevolume 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 internalstress 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


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 acontinuation of the stresses y x s s , and xyt at the boundary. Now, using Cauchy’s equation,

we haveml 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. Forexample, if the boundary of the plate is parallel to x-axis, at point 1P , then the boundaryconditions 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 boundaryconditions 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 S T . 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 anequivalent 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 thesurface 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 fromthe region of application of forces, are essentially the same, provided that the two systems offorces 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 manyengineering 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 figures4.4, 4.5 and 4.6 illustrate St. Venant principle.



Module4/Lesson2


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

- 1,31 q

- 5,12 q

- 1,31 q

0,14+ q

0,16+ q

0,16+ q

- 0,72 q

- 1,72 q

- 2,64 q

- 0,72 q

- 1,72 q

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

- 0,37 q

- 1,95 q

- 3,39 q

0,17+ q

0,14+ q

0,14+ q

- 0,83 q

- 1,73 q

- 2,29 q

0,16- q

0,16- q

- 0,99 q

- 1,49 q

- 1,74 q

- 0,54 q

0,48- q

0,48- q

- 1,02 q

- 1,30 q

- 0,72 q

0,87- q

0,87- q

- 1,00 q

- 1,10 q

- 0,92 q

- 1,13q

0,99- q

0,99- q

- 1,00 q

- 1,02 q

- 0,99 q

y

P

+ b

- b

x

b2

14

x b= b2

x = 34

x b= 32

x b= x b= x b= 21

2 bq =



Module4/Lesson2


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

- 0,72 q

- 1,25 q

- 1,33 q

- 1,34 q

0,08+ q

0,32- q

- 0,76 q

- 1,12 q

- 1,27 q

- 1,30 q

0,32- q

0,80- q

- 0,89 q

- 1,02 q

- 1,12 q

- 1,15 q

0,80- q

0,95- q

- 0,98 q

- 1,00 q

- 1,03 q

- 1,05 q

0,95- q

1,00- q

- 1,00 q

- 1,00 q

- 1,00 q

- 1,00 q

1,00- q

P

+ b

- b

x

y

3

2b

b 4

x = b2

x = 3

4 x b=

3

2 x b= x b= x b= 2

P

2 bq =

0,61- q

- 0,83 q

- 1,05 q

- 1,19 q

- 1,23 q

0,61- q

0,18+ q

- 0,20 q

- 1,03 q

- 1,84 q

- 1,95 q

0,18+ q

0,03+ q

- 0,48 q

- 1,06 q

- 1,56 q

- 1,73 q

0,03+ q

0,32- q

- 0,66 q

- 1,05 q

- 1,37 q

- 1,49 q

0,32- q

0,60- q

- 0,79 q

- 1,03 q

- 1,23 q

- 1,30 q

0,60- q

0,90- q

- 0,95 q

- 1,00 q

- 1,07 q

- 1,10 q

0,90- q

0,99- q

- 0,99 q

- 1,00 q

- 1,01 q

- 1,02 q

0,99- q

P

+ b

- b

x

y

b



Module4/Lesson2


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 isnot altered at a distance b x 2= irrespective of loading conditions.

4.2.5 E XISTENCE AND UNIQUENESS OF SOLUTION (UNIQUENESS

T HEOREM )

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 bya 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

- 2,06 q

- 0,28 q

0,13+ q

- 2,06 q

0,45- q

- 0,78 q

0,45- q

- 1,35 q

- 1,35 q

0,80- q

- 0,96 q

0,80- q

- 1,11 q

- 1,11 q

0,93- q

- 1,01 q

0,93- q

- 1,03 q

- 1,03 q

0,99- q

- 1,00 q

0,99- q

- 1,01 q

- 1,01 q

1,00- q

- 1,00 q

1,00- q

- 1,00 q

- 1,00 q

+ b

- b

x

y

P2

P2

0,16+ q

- 2,20 q

- 0,25 q

0,16+ q

- 2,20 q

0,41- q

- 1,39 q

- 0,74 q

0,41- q

- 1,39 q

0,80- q

- 0,74 q

0,80- q

- 1,12 q

- 1,12 q

0,93- q

- 1,01 q

0,93- q

- 1,03 q

- 1,03 q

0,99- q

- 1,02 q

0,99- q

- 1,01 q

- 1,01 q

1,00- q

P2

P2

+ b

- b

x

y

- 1,00 q

1,00- q

- 1,00 q

- 1,00 q

14 x = b

12 x = b

34 x b=

32 x b= x b= x b= 2

P2q = b



Module4/Lesson2


Proof

In proving the above theorem, one must remember that only elastic problems are dealt withinfinitesimal 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 arezero. The work done by these forces during loading is zero and it follows that the totalstrain energy vanishes, i.e.,

ò ò òV 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


The strain energy per unit volume V o is always positive for any combination of strains andstresses. Hence for the integral to be zero, V o must vanish at all the points, giving all thestress 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 andthere is one and only one solution for the elastic problem .


Example 4.1The following are the principal stress at a point in a stressed material. Taking

2/210 mmkN E = and 3.0=n , calculate the volumetric strain and the Lame’sconstants.

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 -´=\ xe

( )[ ] 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 -´=\ ze

Volumetric strain = ) z y xv e e e e ++=



Module4/Lesson2


544 1014.71057.21067.5 --- ´+´+´=

310954.8 -´=\ ve

To find Lame’s constants

We have, ( )n += 12 E

G

( )3.01210210 3

+´=G

23 /1077.80 mm N G ´=\ ( )

( )G E E GG

32-

-=l

( )( )33

333

1077.80310210102101077.8021077.80

´´-´´-´´´=

23

/1014.121 mm N ´=\ l

Example 4.2The 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. Take 26 /10210 mkN E ´= ,Poisson’s ratio = 0.28. Also find Lame’s constant.

Solution : We have

( )n +=

12 E

G

( )28.01210210 6

+´=

26 /1003.82 mkN G ´=\

But( )

( )G E E GG

32-

-=l

( )( )66

666

1003.82310210102101603.8221003.82

´´-´´-´´´=

26 /1042.104 mkN ´=\ l Now,

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



Module4/Lesson2


( ) ( )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=\ xyt 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 4.3The 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


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

1021012

mkN E

G ´=+´=

+=

n

xy xy xy G g g t ´´== 61077.80

66 10981.1

1077.80160 -´=

´==\

G xy

xy

t g

66 1024.1

1077.80100 -´=

´==

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

çççè

æ

--

-=\ ije



Module5/Lesson1

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

Module 5: Two Dimensional Problems inCartesian Coordinate System

5.1.1 I NTRODUCTION Plane Stress Problems

In many instances the stress situation is simpler than that illustrated in Figure 2.7. Anexample 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 andshear stresses in the x and y directions only. The thickness is small compared to the othertwo dimensions of plate. These stresses are assumed to be uniformly distributed over thethickness 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


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

s 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 y F

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


( ) ( ) ÷÷ 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 auseful form of the governing equations for problems of plane stress. The constitutiverelation for such problems is given by

( ) ï

ïý

ü

ïî

ïí

ì

úúúúú

û

ù

êêêêê

ë

é

÷ ø öç

è æ --

=ï

ïý

ü

ïî

ïí

ì

xy

y

x

xy

y

x E

g

e e

n n

n n

t

s s

21

00

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 practicalimportance, 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


(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 = xu

¶¶

, ye = yv

¶¶

, xyg = xv

yu

¶¶+

¶¶

(5.4)



Module5/Lesson1


ze = zw

¶¶

= 0 , xzg = zu

xw

¶¶+

¶¶

=0 , yzg = zv

yw

¶¶+

¶¶

= 0 (5.5)

Moreover, from the vanishing of ze , the stress zs can be expressed in terms of xs and ys

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

¶¶

¶=

úúû

ù

êêë

é

¶

¶+

¶

¶-

úúû

ù

êêë

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

xF

y x y x

y x n s s

11

2

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 bodyforces are zero or constant, the differential equations for plane strain will be same asthat for plane stress. Further, it should be noted that neither the compatibility



Module5/Lesson1


( ) x y x x GG

Ge e e

l l

s 22

2 +++

=

( ) y y x y GG

Ge e e

l l

s 22

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

) y x e e + .

i.e.,

( )( )ï

î

ïíì

+++

++=

stress plane22

2

strain plane2

x y x

x y x

x GG

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

case byG

G2

2+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

G2

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 -1

in plane-strain case.



Module5/Lesson2


Module 5: Two Dimensional Problems inCartesian Coordinate System

5.2.1 T HE STRESS F UNCTION

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 stressfunction 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

where 4Ñ = 4

4

x¶¶

+ 22

42 y x ¶¶

¶ + 4

4

y¶¶

The above Equation (5.12) is known as "Biharmonic equation" for plane stress and planestrain problems. Since the Biharmonic equation satisfies all the equilibrium andcompatibility equations, a solution to this equation is also the solution for a two-dimensional



2

stress function 2f represents a state of uniform tensions (or compressions) in two perpendicular directions accompanied with uniform shear, as shown in theFigure 5.3 below.

Figure 5.3 State of stresses

Figure 5.3 Constant Stress field

(c) Polynomial of the Third Degree

Let 3f = 33232333

6226 y

d xy

c y x

b x

a +++

The corresponding stresses are

xs = 23

2

y¶¶ f

= yd xc 33 +

ys = 23

2

x¶¶ f = yb xa 33 +

xyt = y x¶¶

¶- 32f

= yc xb 33 --

This stress function gives a linearly varying stress field. It should be noted that themagnitudes of the coefficients 333 ,, cba and 3d are chosen freely since the expression for

3f is satisfied irrespective of values of these coefficients.

Now, if 0333 === cba except 3d , we get from the stress components

yd x 3=s

0= ys and 0= xyt

This corresponds to pure bending on the face perpendicular to the x-axis.\ At y = - h, hd x 3

-=s



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). xs 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 by2

442

4 ye xyd xc x ++=s

244

24 yc xyb xa y

++=s

244

24

22

2 y

d xyc x

b xy ÷

ø öç

è æ --÷

ø öç

è 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, theshearing 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

24

24 -

Or M = Lhd 3

432

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 5

5 20 12 6 6 12 20a b c d e f

Let x x y x y x y xy yj = + + + + +

The corresponding stress components are given by

355

255

25

3525

2

)2(31

)32(3

yd b xyac y xd xc

y x +-+-+=

¶¶= f

s

3525

25

352

52

3 yd xyc y xb xa

x y +++=

¶¶= f s

355

25

25

35

52

)32(31

31

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 solutionsfor various loading conditions of the beam.

Now, if all coefficients, except 5d , equal to zero, we find

÷ ø

öçè

æ -= 325

3

2 y y xd xs

25

353

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 B ENDING OF A NARROW C ANTILEVER BEAM SUBJECTED TO

E ND L OAD 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 thicknessof the beam t is small relative to the depth 2h.



8

Boundary Conditions

0=±= h y At xyt

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 xy dyb2t (5.14a)

By Inverse Method

As the bending moment varies linearly with x, and xs at any section depends upon y, it isreasonable 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 = )()(61

213

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 obtainedinto Equation (5.12), we have

y 042

4

41

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

041

4

=

dx

f d and 04

24

=

dx

f d

Integrating the above, we get

f 1( x) = c 2 x3+c 3 x2+c 4 x+c 5

f 2( x) = c 6 x3+c 7 x2+c 8 x+c 9

where c2 , c3……., c9 are constants of integration.

Therefore, (5.14c) becomes

f = 982

73

6542

33

23

1 )(61

c 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 = 432

22

1

2

2321

c 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 xy Pdyh ybcdyb )(221

2 221t

Solving, c1 = - ÷ ø öç

è æ -=÷÷ ø

öççè æ

I P

hbP

343

where I = 34 bh 3 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 öç

è 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 öç

è 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

Py22

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 P URE BENDING OF A BEAM

Consider a rectangular beam, length L, width 2b, depth 2h, subjected to a pure couple Malong 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

22

2

2 yc xyb

x ++

By definition,



11

xs = 2

2

y¶¶ f

, ys = 2

2

x¶¶ f

, t xy = ÷÷ ø

öççè

æ ¶¶

¶- y xf 2

\ Differentiating the function, we get

s x = 2

2

y¶¶ f

= c2 , s y = 2

2

x¶¶ f

= a 2 and t xy = ÷÷ ø

öççè

æ ¶¶

¶- y x

f 2

= - b2

Considering the plane stress case,s z = t xz = t yz = 0

Boundary Conditions

(a) At y = ± h , = ys 0

(b) At y = ± h, t xy = 0 (c) At x = any value,

2b ò+

-

a

a x ydys = bending moment = constant

\ 2bx ò+

-

+

-

=úûù

êëé=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

33

2323

33 yd xyc

y xb xa +++

Now, xs = yd xc y 332

2

+=¶¶ f

(a)

ys = a 3 x + b3 y (b)

t xy = - b3 x - c 3 y (c)

From (b) and boundary condition (a) above,

0 = a 3 x ± b3a for any value of x \ a 3 = 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., 02 4

4

22

4

4

4

=¶¶+

¶¶¶+

¶¶

y y x xf 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 +

-úûù

êëé

3

3

M = 4bd 33

3h

Or d3 =34

3bh M

I M

d =3 where

34 3bh

I =

Therefore, xs = y I

M



13

5.3.4 B ENDING OF A SIMPLY SUPPORTED BEAM BY A DISTRIBUTED

L OADING (UDL )

Figure 5.10 Beam subjected to Uniform load

Consider a beam of rectangular cross-section having unit width, supported at the ends andsubjected 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 withchange in x in either positive or negative directions. This is possible only if the stressfunction 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 oddfunctions of y.

Now, consider a polynomial of second degree with022

== cb

222 2

xa=\ f

a polynomial of third degree with 033 == ca

33233 62

yd

y xb +=\ f

and a polynomial of fifth degree with 05555 ==== ecba

úûù

êëé -=-=\ 55

553255 3

2306

d f yd

y xd

f

532 f f f f ++=\

or 55325332322

306622 y

d y x

d y

d y x

b x

a -+++=f (1)

Now, by definition,



14

÷ ø öç

è æ -+=

¶¶= 32

532

2

32

y y xd yd y x

f s (2)

35322

2

3 y

d yba

x y ++=

¶¶= f

s (3)

253 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) give02

53 =-- cd b

03

3532

=++ cd

cba

qcd cba -=-- 3532 3

which gives on solving

3532 43

,43

,2 c

qd

cq

bq

a -==-=

Now, from condition (vi), we have

ò+

-

=úûù

êëé

÷ ø öç

è æ -+

c

c

ydy y y xd yd 032 32

53

Simplifying,

÷ ø öç

è æ --= 22

53 52

h Ld d



15

÷ ø öç

è æ -=

52

43

2

2

h L

hq

÷ ø öç

è æ --÷

ø

öç

è

æ -=\ 32

32

2

32

43

52

43

y y xhq

yh L

hq

xs

3

3443

2 y

hq

yhqq

y -+÷

ø öç

è æ -=s

2

343

43

xyhq

xhq

xy +÷

ø öç

è æ -=t

Now, ( ) 333

32

128

1221

hhh

I ==´=

where I = Moment of inertia of the unit width beam.

( ) ÷ ø öç

è æ -+-=\

532

2322 yh y

I q

y x L I q

xs

÷ ø öç

è æ +-÷

ø öç

è æ -= 32

3

32

32h yh

y I q

ys

( )22

2 yh x

I q

xy -÷ ø öç

è æ -=t

5.3.5 N UMERICAL E XAMPLES Example 5.1Show that for a simply supported beam, length 2 L , depth 2 a and unit width, loaded bya 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 öç

è æ -=

a

a x

W dy

2s on each half-length of the beam. Show that the stresses are

÷ ø öç

è æ -= xy

aW

x 343

s

0= ys

úû

ùêë

é÷÷ ø

öççè

æ --=2

2

183

a y

aW

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=

ys

Condition (ii)

cba +÷÷ ø

öççè

æ -=2

02

2

2bac =\

Condition (iii)

( )ò+

-

--=a

a

dy yabW 22

22

÷÷ ø

öççè

æ --=3

22

2

33 a

ab

÷÷ ø

öççè

æ -=\3

22

3baW

or ÷ ø öç

è æ -=

343

aW

b

and ÷ ø ö

çè æ

-= aW

c 83

Condition (iv)



18

ò+

-

=÷ ø öç

è æ -

a

a

xydyaW

043

3

Condition (v)

ò+

-=

a

a x ydy M s

ò+

-

÷ ø öç

è æ -=

a

a

dy xyaW 2

343

2Wx

M =\

Hence stress components are

xyaW

x ÷ ø öç

è æ -=

343

s

0= ys

÷ ø öç

è æ -÷÷ ø

öççè

æ =a

W yaW

xy 83

243 2

3t

úû

ùêë

é÷÷ ø

öççè

æ --=\2

2

183

a y

aW

xyt

Example 5.2

Given the stress function ÷ ø öç

è æ ÷

ø öç

è 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

02 4

4

22

4

4

4

=¶¶+

¶¶¶+

¶¶

z z x xf f f

(i)

Now, úûù

êëé

÷ ø öç

è æ +÷

ø öç

è æ

+-=

¶¶ -

z x

z x xz H

z1

22 tanp

f

( ) [ ]32322

2222

2

21

x xz x xz xz z x

H

z----

+÷ ø

öçè

æ =¶

¶

p

f

( ) úúû

ù

êêë

é

+÷ ø öç

è æ -=

¶¶\

222

3

2

2 2

z x

x H z p

f



19

Also,( ) ú

úû

ù

êêë

é

+=

¶¶

322

3

3

3 8

z x

z x H z p

f

( ) ú

ú

û

ù

ê

ê

ë

é

+

-=¶¶

422

235

4

4 408

z x

z x x H

z p

f

( ) úúû

ù

êêë

é

+-

÷ ø öç

è æ -=

¶¶¶

322

423

2

3 32

z x

x z x H x z p

f

( ) úúû

ù

êêë

é

+--=

¶¶¶

422

5423

22

4 82464

z x

x xz z x H x z p f

Similarly,

( )úûù

êë

é+

=¶¶

22

2

z x z H

x p f

( ) úúûù

êêëé

+÷ ø öç

è æ -=

¶¶

222

2

2

2 2

z x

xz H x p

f

( )( ) ú

úû

ù

êêë

é

+-=

¶¶

322

222

3

3 32

z x

z x z

H x p

f

( ) úúû

ù

êêë

é

--=

¶¶

422

234

4

4 2424

z x

z x xz H x p

f

Substituting the above values in (i), we get

( ) [5423234

422824642424

14 x xz z x z x xz

z x--+-

+p ]0408 235 =-+ z x x

Hence, the given stress function is admissible.Therefore, the stresses are

( ) úúû

ù

êêë

é

+÷ ø öç

è æ -=

¶¶=

222

3

2

2 24

z x

x z x p

f s

( ) úúû

ù

êêë

é

+÷ ø öç

è æ -=

¶¶=

222

2

2

2 24

z x

x x y p

f s

and ( ) úúûù

êêë

é

+÷ ø ö

çè æ

-=¶¶¶

= 222

22 24

z x

z x z x xy p f

t



20

Example 5.3

Given the stress function: ( ) zd xzd F

2323

-÷ ø öç

è 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

33

22

23 xz

d F

xzd F ÷

ø öç

è æ +÷

ø öç

è æ -=f

xzd

F d Fx

z÷ ø öç

è æ +÷

ø öç

è æ -=

¶¶\

322

2 126f

and 02

2

=¶¶ x

f

also 232

2 66 z

d F

d Fz

z x÷ ø öç

è æ +÷

ø öç

è æ -=

¶¶¶ f

Hence xzd

F d Fx

x ÷ ø öç

è æ +÷

ø öç

è æ -=

32

126s (i)

0= zs (ii)

232

2 66 z

d F

d Fz

z x xz ÷ ø öç

è æ +÷

ø öç

è 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 linearlywith z.\ At x = 0 and z = ± d , xs = 0

At x = L and z = 0, ÷ ø öç

è æ -=

2

6d FL

xs

At x = L and z = +d , 232

6126d FL

Ld d

F d FL

x =÷ ø öç

è æ +÷

ø öç

è æ -=s

At x = L and z = -d , ÷ ø öç

è æ -=÷

ø öç

è æ -÷

ø öç

è æ -=

232

18126d 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 = 232 .

66 z

d F

d Fz ÷

ø öç

è æ -÷

ø öç

è 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 öç

è æ -÷

ø öç

è æ = d

d F

d Fd

xzt

At z = -d , ÷ ø öç

è æ -=-÷

ø öç

è æ -÷

ø öç

è æ -=

d F

d d F

d d F

xz12

)(66 2

32t

The variation of xzt is shown in figure below.



22

Figure 5.13 Variation of xz

Example 5.4Investigate what problem of plane stress is satisfied by the stress function

32

2

34 3 2F xy p

xy yd d

j é ù

= - +ê úë 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

23

3 14 4 2F Fxy p

xy yd 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 2Fxy p F

p xy y d d

j ¶ ´æ ö æ ö= - + = -ç ÷ ç ÷¶ è ø è ø

and 3

22

43

43

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 43

43

3

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 , 21.5 x

FL p

d s

æ ö= - ç ÷è ø

When x = L and y = -d , 25.1d FL

P 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 öç

è æ -÷÷ ø

öççè

æ =

d F

d Fy

xy 43

43

3

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 öç

è æ -==

d F

y xy 43

,0 t

At 043

)(43

, 23

=úûù

êëé-úû

ùêëé=±=

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 ofrectangular cross-section of width 2 h and depth d under a total shear force W .

úûù

êëé --= )23(

22

3 yd xyhd W

f

Solution : 2

2

y x ¶¶= f

s

Y

d

d

L

Xo

t xy

- 3F

4d

- 3F

4d



25

Now, [ ]23 66

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

= [ ]23 66

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 yhd

x ===ò 00..2.

0

s

Now, Condition (a)

This condition is satisfied since 0= ys

Condition (b)

[ ] 033 223

=- d d hd W



26

Hence satisfied.

Condition (c)

[ ]hdy y yd

hd

W d

2330

23ò -

[ ]dy y yd d W d

ò -=0

23 33

2

d

yd y

d W

0

32

3 232

úû

ùêë

é-=

úû

ùêë

é-= 3

3

3 232

d 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

02

3 332 --=

= 0

Hence satisfied.

Condition (e)

[ ] ydyh xy xd hd W d

.2630 3ò --

d

xy xdy

d W

0

32

3 22

32úû

ùêë

é--=

úûùê

ëé --= 33

3 22

32 xd xd d W



27

úûù

êëé--= 3

3 212

xd d W

Wx=

Hence satisfied




Module6/Lesson1

Module 6: Two Dimensional Problems in PolarCoordinate System

6.1.1 I NTRODUCTION

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-ordinatereference 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 S TRAIN -DISPLACEMENT R ELATIONS

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




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 radialand 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 öç

è æ

¶¶

vr rd

d v

1 (6.3)

Hence, the resultant strain is

e q = (e q )u + (e q )v

e q = ÷ ø öç

è 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 öç

è æ

¶¶=

÷ ø öç

è 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 öç

è æ -

¶¶

r v

r v

(6.6)

Therefore, the total shear strain is given by

) )vr ur r q q q g g g +=




Module6/Lesson1

g r q = ÷ ø öç

è æ -

¶¶+÷

ø öç

è 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 öç

è æ -÷

ø öç

è æ

¶¶+

¶¶

r vu

r r v

q 1

(6.8)




Module6/Lesson1

g q z = ÷ ø öç

è æ

¶¶+÷

ø öç

è æ

¶¶

zvw

r q 1

g zr = ÷ ø öç

è æ

¶¶+

¶¶

r w

zu

6.1.3 C OMPATIBILITY E QUATION

We have from the strain displacement relations:

Radial strain,r u

r ¶¶=e (6.9a)

Tangential strain, ÷ ø öç

è æ +

¶¶

÷ ø öç

è æ =

r uv

r q e q

1 (6.9b)

and total shearing strain,q

g q ¶¶

÷ ø öç

è æ +÷

ø öç

è æ -

¶¶= u

r r v

r v

r 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 öç

è æ -

¶¶¶+÷

ø öç

è æ -

¶¶

÷ ø öç

è æ =

¶¶ v

r r v

r u

r r u

r r .

1.

1112

2

2

úûù

êëé

¶¶

÷ ø öç

è æ +-

¶¶¶

÷ ø öç

è 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 öç

è æ -

¶¶¶

÷ ø öç

è æ ++

¶¶

÷ ø öç

è æ -

¶¶=

¶¶ u

r r u

r r v

r v

r r v

r r

2

2

22

2 111

q q ¶¶¶+÷

ø öç

è æ

¶¶+-

¶¶-

¶¶=

r u

r u

r r v

r v

r r v 2

2

2 111

q e

g g

q q

¶¶

÷ ø öç

è æ +÷

ø öç

è æ -

¶¶=

¶¶

\ r r

r

r r r v

r 11

2

2

(6.9f)

Differentiating Equation (6.9e) with respect to r and Equation (6.9f) with respect to q , weget,

q q q e

e q q

e e e

2

2

22

3

22

2 111111r r r r

vr r

vr r r r r r

r +¶

¶÷ ø öç

è æ --

¶¶¶

÷ ø öç

è æ -

¶¶¶

÷ ø öç

è æ +÷

ø öç

è æ -

¶¶

÷ ø öç

è æ =

¶¶

(6.9g)




Module6/Lesson1

and 2

2

2

32 11q e

q g

q q g q q

¶¶

÷ ø ö

çè æ +

¶¶

÷ ø ö

çè æ -

¶¶¶=

¶¶¶

r r r

r r r v

r

or 2

2

222

32 1111q e

q g

q q g q q

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ -

¶¶¶

÷ ø öç

è æ =

¶¶¶

÷ ø öç

è 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

2 111111r r r r r r

vr 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 öç

è æ -

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ -

¶¶¶

÷ ø öç

è æ -÷

ø öç

è æ -

¶¶

÷ ø öç

è æ =

¶¶¶

÷ ø öç

è æ -

¶¶

÷÷ ø

öççè

æ ¶¶

+¶

¶-

¶¶

-÷÷ ø

öççè

æ -

¶¶¶+-÷

ø öç

è æ

¶¶

=2

22 1.

11111q 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

11111q e

q g e e e q q q

¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ = r r r

r r r r r r r r

2

2

22

1121q 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

11211q e e e e

q g

q g q q q q

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ +

¶¶

=¶¶

¶÷ ø öç

è æ +

¶¶

÷ ø öç

è æ \ r r r r

r r r r r r r r r

6.1.4 S TRESS -STRAIN R ELATIONS

In terms of cylindrical coordinates, the stress-strain relations for 3-dimensional state of stressand 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 G1




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 G1

6.1.5 A IRY’S STRESS F UNCTION

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 xyt 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

22

22

2

2

y x x yr ¶¶¶-

¶¶+

¶¶=

q f

q f

q f

s q 2sinsincos2

22

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

11q

f f s ¶

¶÷ ø öç

è æ +

¶¶

÷ ø öç

è æ =

r r r r (6.16)

2

2

r ¶¶= f

s q andq

f q f

t q ¶¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è 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 .




Module6/Lesson1

6.1.6 B IHARMONIC E QUATION

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 bemodified to suit the polar co-ordinate system. This modification may be accomplished by

transforming the 4Ñ operator from the Cartesian system to the polar system.

Now, we have, q q sin,cos r yr x ==

÷ ø öç

è æ =+= -

x y

and 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

xr

q q

sinsin ===

¶¶

r r

r y

yr

÷ ø ö

çè æ -==÷

ø ö

çè æ -=

¶¶

r r r

r y

xq q q sinsin

22

r r r

r x

yq q q coscos

22 ===

¶¶




Module6/Lesson1

Figure.6.3

dy y xdxdr r 222 +=

dyr

ydx

r x

dr ÷ ø öç

è æ +÷

ø öç

è æ =\

Also, ÷ ø ö

çè æ +

÷ ø ö

çè æ -=

xdy

xy x y

d 2

2

sec q q

x xr

r x ¶¶

¶¶+

¶¶

¶¶=

¶¶ q

q f f f

q f

q f

¶¶

÷÷ ø

öççè

æ ÷ ø öç

è æ -+

¶¶

+=

2222 sec1

x y

r y x

x

÷ ø öç

è æ

¶¶-÷

ø öç

è æ

¶¶=

¶¶\

q f q f

q f

r r xsin

cos

Similarly, y y

r

r y ¶

¶

¶

¶+¶

¶

¶

¶=¶

¶ q

q

f f f

÷ ø öç

è æ

¶¶+÷

ø öç

è æ

¶¶=

¶¶\

q f q f

q f

r r ycos

sin




Module6/Lesson1

Now,2

2

2 sincos ÷

ø öç

è æ

÷ ø öç

è æ

¶¶-÷

ø öç

è 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 öç

è æ

¶¶+

¶¶

÷ ø öç

è æ -

¶¶¶+

¶¶=

¶¶

r r r r r r r y

(ii)And,

2

2

22

2

2

22 cossin2cos2coscossin

cossinq

f q q q f q

q f q f

q q f q q f

¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ -

¶¶¶+

¶¶+

¶¶

÷ ø öç

è 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 11q f f f f f

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ +

¶¶=

¶¶+

¶¶

r r r r y x

2

2

22

2

2

2

2

22 11

,.q

f f f f f f

¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è æ +

¶¶=

¶¶+

¶¶=Ñ

r r r r y xei

or2 2 2 2

4 2 22 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



6.2.1 AXISYMMETRIC PROBLEMS

Many engineering problems involve solids of revolution subjected to axially symmetricloading. 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 spaceloaded by a circular area, for example a circular footing on a soil mass (Figure 6.4b). It isconvenient to express these problems in terms of the cylindrical co-ordinates. Because ofsymmetry, the stress components are independent of the angular ( q ) co-ordinate; hence, allderivatives 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 = zw

r u

r u

z ¶¶==

¶¶

e e q ,,

g rz =r w

zu

¶¶+

¶¶

(6.19)

and the constitutive relation is given by

( )( )

( )

( )ïïþ

ïïý

ü

ïïî

ïïí

ì

úúúúúú

û

ù

êêêêêê

ë

é

--

--

-+=

ïïþ

ïïý

ü

ïïî

ïïí

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


(a) Cylinder under axisymmetric loading

(b) Circular Footing on Soil mass

Figure 6.4 Axisymmetric Problems



Module6/Lesson2


6.2.2 T HICK -W ALLED C YLINDER SUBJECTED TO INTERNAL AND E XTERNAL 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 öç

è æ -

+¶

¶r r

r r q s s s (6.20)

Here q s and r s denote the tangential and radial stresses acting normal to the sides of theelement.

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


Therefore , e r =dr du

and e q =r u

or the stresses in terms of strains are

s r = )()1( 2 q ne e

n +

- r

E

s q = )()1( 2 r

E ne e n q +-

Substituting the values of e r and e q in the above expressions, we get

s r = ( ) ÷ ø öç

è æ +

- r u

dr du E

n n 21

s q = ( ) ÷ ø öç

è æ +

- dr du

r u E

n n 21

Substituting these in the equilibrium Equation (6.21)

0=÷ ø ö

çè æ

+-÷ ø ö

çè æ

+ dr du

r u

udr du

r dr d

n n

02

2

=--++dr du

r u

dr du

dr ud

r dr du

n n

or 01

22

2

=-+r u

dr du

r dr ud

The above equation is called equidimensional equation in radial displacement. The solutionof the above equation is

u = C 1r + C 2 / r (6.22)

where C 1 and C 2 are constants.

The radial and tangential stresses are written in terms of constants of integration C 1 and C 2.

Therefore,

s r = ( ) úûù

êëé

÷ ø öç

è æ --+

- 2212

11

)1( r C C

E n n

n

s q = ( ) úûù

êëé

÷ ø öç

è æ -++

- 2212

11

)1( r C C

E n n

n (6.23)

The constants are determined from the boundary conditions.

when r = a , s r = i p-

r = b , s r = 0 p- (6.23a)



Module6/Lesson2


Hence, ( ) úûù

êëé

÷ ø ö

çè æ --+

- 2212

11

)1( aC C

E n n

n = i p-

and ( ) úûù

êëé

÷ ø öç

è 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 öç

è æ -

)(1

220

22

ab

pb pa

E in

C 2 = ÷÷ ø

öççè

æ -

-÷ ø öç

è æ +

)(

)(122

022

ab

p pba

E in

Substituting these in Equations (6.22) and (6.23), we get

s r = ÷÷ ø

öççè

æ

-

--÷÷

ø

öççè

æ

-

-222

220

220

22

)(

)(

r ab

ba p p

ab

pb pa ii (6.24)

s q = ÷÷ ø

öççè

æ -

-+÷÷ ø

öççè

æ --

222

220

220

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

220

220

22

--

÷ ø öç

è æ ++

--

÷ ø öç

è 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 wallof 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 notinterfere 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

bab

pa (6.28)



Module6/Lesson2


Figure 6.6 shows the variation of radial and circumferential stresses across the thickness ofthe 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 )(abba 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 thecylinder as

s r = s q = - p



Module6/Lesson2


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


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 = ( ) úûù

êëé -+

+- r u

dr du E

n n n n

1)1)(21(

s r = ( ) úûù

êëé +-

+- 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 =---úûù

êëé +-

r u

dr du

udr du

r 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 = ( ) úûùêë

é -++- 2

21 21

)1)(21( r C C E n

n n



Module6/Lesson2


s r = ( ) úûù

êëé --

+- 22

1 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 pa

C C

E -=úûù

êëé --

+- 22

1 21)1)(21(

n n n

( ) o pb

C C

E -=úûù

êëé --

+- 22

1 21)1)(21(

n n n

Solving, we get

C 1 = ÷÷ ø öççè

æ --+-

22

220)1)(21(ba

a pb p E

in n

and C 2 = ÷÷ ø

öççè

æ -

-+22

220 )()1(

ba

ba p p

E in

Substituting these, the stress components become

s r = 2

22

220

22

20

2

r ba

ab

p p

ab

b pa p ii ÷ ø öç

è æ

--

-÷÷ ø

öççè

æ --

(6.32)

s q = 2

22

220

22

20

2

r ba

ab

p p

ab

b pa p ii ÷ ø öç

è æ

--

+÷÷ ø

öççè

æ --

(6.33)

s z = 2n ÷÷ ø

öççè

æ --

22

220

ab

b pa p i (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 byequation (6.34).



Module6/Lesson2


6.2.3 R OTATING DISKS OF UNIFORM T HICKNESS

The equation of equilibrium given by

0=+÷ ø öç

è 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 distributedsymmetrically 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 thecentrifugal force r w 2 r , yields

r wr dr

d r r 2 r s s s q +÷

ø öç

è æ -+ = 0 (6.35)

where r is the mass density and w is the constant angular speed of the disk in rad/sec. Theabove 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

úûù

êëé -=- )(

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


Substituting these in equation (6.40), we obtain

r 2 0)3( 322

2

=+++-+ r w ydr dy

r dr

yd r n

The solution of the above differential equation is

y = Cr + C 1 32

831 r 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 1 222 8

31r w

r r

n ÷ ø ö

çè æ +-÷

ø ö

çè æ

(6.44)

s q = C - C 1 222 8

311r w

r r

n ÷ ø öç

è æ +-÷

ø öç

è æ

(6.45)

The constants of integration are determined from the boundary conditions.

6.2.4 S OLID 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 öç

è æ +

bw r n

Therefore, C = 228

3bw r

n ÷ ø ö

çè æ +

(6.46)

Hence, Equations (6.44) and (6.45) become,

s r = )(8

3 222 r bw -÷ ø öç

è æ +

r n

(6.47)

s q = 2222

831

83

r wbw r n

r n ÷

ø öç

è æ +-÷

ø öç

è 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

83

bw r n ÷

ø öç

è æ +

(6.49)



Module6/Lesson2


6.2.5 C IRCULAR DISK WITH A H OLE

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

3 222 abw +÷ ø öç

è æ +

r n

and C 1 = - 222

83

baw r n ÷

ø öç

è æ +

Substituting the above in Equations (6.44) and (6.45), we obtain

s r = ÷÷ ø

öççè

æ -÷÷

ø

öççè

æ -+÷

ø öç

è æ + 2

2

22222

83

r r ba

abw r n (6.50)

s q = ÷÷ ø

öççè

æ ÷ ø ö

çè æ

++-÷÷ ø

öççè

æ ++÷

ø ö

çè æ + 2

2

22222

331

83

r r ba

abwn n

r n (6.51)

The radial stress s r reaches its maximum at r = ab , where

(s r )max = 22 )(8

3abw -÷

ø öç

è 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 u r for all the cases considered can be calculated as below:

u r = r e q = )( r E r

ns s q - (6.54)

6.2.6 S TRESS C ONCENTRATION

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 thatinterrupts 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


'Stress Concentration Factor'. Thus, the expression for the maximum normal stress in acentrically loaded member becomes

÷ ø öç

è æ = AP

K s (6.55)

where A is either gross or net area (area at the reduced section), K = stress concentrationfactor and P is the applied load on the member. In Figures 6.8 (a), 6.8(b) and 6.8(c), the typeof discontinuity is shown and in Figures 6.8(d), 6.8(e) and 6.8(f) the approximate distributionof normal stress on a transverse plane is shown.

Stress concentration is a matter, which is frequently overlooked by designers. The high stressconcentration found at the edge of a hole is of great practical importance. As an example,holes in ships decks may be mentioned. When the hull of a ship is bent, tension orcompression is produced in the decks and there is a high stress concentration at the holes.Under the cycles of stress produced by waves, fatigue of the metal at the overstressed

portions may result finally in fatigue cracks.



Module6/Lesson2


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


6.2.7 T HE E FFECT OF C IRCULAR H OLES ON STRESS

DISTRIBUTIONS IN P LATES

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 platethickness 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 thefigure. This hole will disturb the stress field in the neighbourhood of the hole. But fromSt.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 thestresses 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

+=\ Pr (6.56)

and ÷ ø öç

è 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

qP X

Y

X P

Y

m

n

b

a



Module6/Lesson2


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 oftwo parts.

(a) A constant radial stress2P

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 öç

è æ +=¢

2r B

Ar s and ÷ ø öç

è 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

öççè

æ -

-=¢

2

2

22

2

12 r

a

ab

Pbr s

( ) ÷÷ ø

öççè

æ +

-=¢

2

2

22

2

12 r

aab

Pbq s

(b) The second part of the stress q s s ¢¢¢¢and r are functions of q . The boundary conditionsfor this are:

q s 2cos2P

r =¢¢ for br =

q t q 2sin2

÷ ø öç

è æ -=¢ P

r 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 öç

è æ +

¶¶

÷ ø öç

è æ =¢¢ f

q f

s 11

2

2

2

andq

f q f

s q ¶¶¶

÷ ø öç

è æ -

¶¶

÷ ø öç

è æ =¢¢

r r r

2

2

11

Now, the compatibility equation is given by,

( ) 02cos11

2

2

22

2

=÷÷ ø

öççè

æ

¶

¶+¶

¶+¶

¶q

q r f

r r r r

But ( ) ( ) ( ) q q

q q 2cos1

2cos1

2cos 2

2

22

2

r f r

r f r r

r f r ¶

¶+¶¶+

¶¶



Module6/Lesson2


( ) ( ) ( )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 differentialequation with constant co-efficients by introducing a new variable t such that t 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

231

2

2

3

3

33

3

÷÷ ø

öççè

æ -+-=

dt df

dt f d

dt f d

dt f d

r dr f d

61161

2

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 ö

çè æ

+÷÷ ø

öççè

æ --÷÷ ø

öççè

æ +-+÷÷ ø

öççè

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 01644 2

2

3

3

4

4

=+--dt df

dt f d

dt f d

dt f d



Module6/Lesson2


046

2 24 =++a D

aC

A

246

2 24

Pb D

bC

A -=++

026

62 242 =--+

a D

aC

Ba A

226

62 242 P

b D

bC

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 öç

è æ -=

4P

A

Therefore the total stress can be obtained by adding part (a) and part (b). Hence, we have

q s s s 2cos43

1212 2

2

4

4

2

2

÷÷ ø

öççè

æ -++÷÷ ø

öççè

æ -=¢¢+¢= r

ar aP

r aP

r 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 aP

r r (6.60)

Now, At 0, ==r ar s

q s 2cos2PPr -=\

When2

32

p p q or =

P3=q s



Module6/Lesson2


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 3 P rapidly dies down as we move from r = a to r = b since at 2p

q =

÷÷ ø

öççè

æ ++=

4

4

2

2 32

2 r a

r aP

q 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 tensionor 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 thecase of pure shear.

Figure 6.10 Plate subjected to stresses in two directions



Module6/Lesson3



6.3.1 B ARS WITH L ARGE INITIAL C URVATURE

There are practical cases of bars, such as hooks, links and rings, etc. which have large initialcurvature. In such a case, the dimensions of the cross-section are not very small incomparison with either the radius of curvature or with the length of the bar. The treatmentthat follows is based on the theory due to Winkler and Bach.

6.3.2 W INKLER’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 normalthroughout the plane of symmetry of the beam.

Winkler-Bach Formula to Determine Bending Stress or Normal Stress (Also known asCircumferential Stress)

Figure 6.11 Beam with large initial curvature

(a)

(b)



Module6/Lesson3


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 y

cc úûù

êëé

+-+ )( 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


0 =( )ò úû

ùêëé

+-+ dA

y R y

cc )( e l e

or( )ò ò +

--= dA y R

ydA cc )( e l e

or( )ò ò +

--= dA y R

ydA cc )( e l e (6.63c)

Also,

M =( )

E dA y R

y ydA cc ú

û

ùêë

é+

-+ò ò2

)( e l e (6.63d)

Here ò = AdA , and since y is measured from the centroidal axis, ò = .0 ydA

Let ( )ò -=+ mAdA y R y

Or m = -( )ò +

dA y R

y A1

Therefore,( ) ( )ò ò ÷÷ ø

öççè æ

+-=

+dA

y R Ry

ydA y R

y 2

=( )ò ò +

- dA y R

Ry ydA

= 0 – R [-mA]

\ ( )ò =

+ mRAdA y R

y 2

Substituting the above values in (6.63c) and (6.63d), we get,

e c = ( l - e c)m

and M = E ( l -e c) mAR

From the above, we get

e c = AER M

and l ÷ ø ö

çè æ

+= mR M

R M

AE 1

(6.63e)

Substitution of the values of Equation (6.63e) into Equation (6.63a) gives an expression forthe tangential stress in a curved beam subject to pure bending.



Module6/Lesson3


Therefore, s q = úû

ùêë

é+

+)(

1 y Rm

y AR M

(6.64)

The above expression for s q is generally known as the "Winkler-Bach formula". Thedistribution of stress s q is given by the hyperbolic (and not linear as in the case of straight

beams) as shown in the Figure 6.11 (b).

In the above expression, the quantity m is a pure number, and is the property of each particular shape of the cross-section. Table 6.1 gives the formula for m for various shapes ofthe cross-section.



Module6/Lesson3


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

2RC C

2 2- R

2 2 2 2- - -C R C1

éë

ù û

R CR C

+-

1

ø æ è



Module6/Lesson3


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 concaveside 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 whenmeasured 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 negativesign means compressive stress.

The distance between the centroidal axis ( y = 0) and the neutral axis is found by setting thetangential stress to zero in Equation (5.15)

\ 0 = úûù

êëé

++

)(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+-

mmR

This expression is valid for pure bending only.However, when the beam is acted upon by a normal load P acting through the centriod ofcross-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 S TRESSES IN C LOSED R INGS

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


Now, consider a closed ring shown in figure 6.12 (a), which is subjected to a concentratedload 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


and at section AA subjected to an axial load2P

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 cos21

P

Tangential component, producing shearing stress = q sin21

PT =



Module6/Lesson3


Determination of A M

Figure 6.13 Section PQMN

Consider the elastic behavior of the two normal sections MN and PQ, a differential distanceapart. Let the initial angle q d between the planes of these two sections change by anamount 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


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 mn d M

From Equation (a), substituting the value of ,mn M we obtain

ò òò =-+- 20

20

20

0cos21

21p p p

q q q q d PRd PRd M A

Integrating, we get

[ ] [ ] [ ] 0sin21

21 2

020

20

=-+-p p p

q q q PRPR M A

02sin21

221

2 =÷ ø ö

çè æ

-÷ ø ö

çè æ

+÷ ø ö

çè æ

- p p p

PRPR M A

Thus ÷ ø öç

è æ -=

p 2

12

PR M A

Therefore, knowing A M , the moment at any section such as MN can be computed and thenthe normal stress can be calculated by curved beam formula at any desired section.



Module6/Lesson3



Example 6.1Given 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

11q f f

s ¶¶

÷ ø öç

è æ +

¶¶

÷ ø öç

è 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

Pr

=¶¶

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

+-=¶¶

¶ Pr

Substituting the above values in equations (i), (ii) and (iii), we get

( )q q q p

q q p

s sin2cos1

cos1

2 +÷

ø öç

è æ -÷

ø öç

è æ = r

Pr

Pr r

q p

q q p

q q p

sin21

cos1

cos1 P

r P

r P

r ÷

ø öç

è æ -÷

ø öç

è æ -÷

ø öç

è æ =

q p

s sin2 Pr r -=\

02

2

=¶¶=

r f

s q



Module6/Lesson3


( ) ( )q q q p

q q q p

t q cossin1

cossin1

2 +-÷

ø öç

è æ -+-÷

ø öç

è æ = P

r r

Pr r

0=\ q t r

Therefore, the stress components are

q p

s sin2 Pr 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 internalpressure of 12 MPa . Determine the radial and hoop stresses in the cylinder at the innerand outer surfaces.

Solution : The radial stress in the cylinder is given by

s r = 2

22

2222

22

r ba

ab p p

abb pa p oioi ÷

ø öç

è æ

---÷÷ ø

öççè

æ --

The hoop stress in the cylinder is given by

s q = 2

22

2222

22

r ba

ab p p

abb pa p oioi ÷

ø öç

è æ

--+÷÷ ø

öççè

æ --

As the cylinder is subjected to internal pressure only, the above expressionsreduce to

s r = 2

22

2222

2

r ba

ab p

aba p ii ÷

ø öç

è æ

--÷÷ ø

öççè

æ -

and s q = 2

22

2222

2

r ba

ab p

aba p ii ÷

ø öç

è æ

-+÷÷ ø

öççè

æ -

Stresses at inner face of the cylinder (i.e., at r = 10 cm):

Radial stress = s r = úûù

êëé

-úû

ùêë

é-úû

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


Hoop stress = s q = úû

ùêë

éúû

ùêë

é-

+úû

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-

-úû

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-ú

û

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 6.3A steel tube, which has an outside diameter of 10cm and inside diameter of 5cm, issubjected 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

éúûù

êëé

--+ú

û

ùêë

é--

2

22

220

22

20

2

a

ba

ab

p p

ab

b pa p ii

= 2220

22

2

0

2

bab p p

abb pa p ii úûùêëé --+ú

û

ùêë

é--

= 22

20

220

2

ab

b pb pb pa p ii

--+-

(s q )max = 22

20

22 2)(ab

b pba p i

--+

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


Example 6.4A steel cylinder which has an inside diameter of 1m is subjected to an internal pressureof 8 MPa . Calculate the wall thickness if the maximum shearing stress is not to exceed35 MPa .

Solution : The critical point lies on the inner surface of the cylinder, i.e., at r = a .

We have, Radial stress = s r = 2

22

220

22

20

2

r

ba

ab

p p

ab

b pa p iiúûù

êëé

---ú

û

ùêë

é--

At r = a and 0 p = 0,

s r = 2

22

2222

2 00a

baab

pab

a p iiúûù

êëé

---ú

û

ùêë

é-

-

= 22

22

ab

b pa p ii

--

=)(

)(22

22

abab p i

---

Therefore, s r = i p-

Similarly,

Hoop stress = s q = 2

22

220

22

20

2

r

ba

ab

p p

ab

b pa p ii úûù

êëé

--+ú

û

ùêë

é--

At r = a and 0 p = 0,

s q = 2

22

2222

2 00a

baab

pab

a p iiúûùêë

é--+ú

ûùê

ëé

--

s q = )(

)(22

22

ab

ba p i

-+

Here the maximum and minimum stresses are

s 3 = i p- and s 1 = s q

But the maximum shear stress = t max = ( )31

2

1s s -

i.e. t max =( )

( ) úû

ùêë

é+

-+

ii p

ab

ba p22

22

21



Module6/Lesson3


= ( ) úû

ùêë

é-

-++22

2222

21

ab

a pb pb pa p iiii

35 = )( 22

2

abb p i

-

i.e ., 35 = )(

822

2

abb

-´

35b2-35a 2 = 8b2

35b2-8b2 = 35a 2

35b2-8b2 = 35 (0.5 )2 Therefore, b = 0.5693

If t is the thickness of the cylinder, thenb = 0.5 + t = 0.5693

\ t = 0.0693 m or 69.3 mm.

Example 6.5The 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. Assumethat the material is not stressed above its elastic strength.

Solution :

Cross-sectional area = A=4p

(3)2 = 7.06 cm2.

For circular cross-section m is given by

m = -1+2

2

÷ ø öç

è æ

c R

–2 ÷ ø öç

è æ

c R

12

-÷ ø öç

è 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


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

ûùê

ëé

+++

)(1

A

A

y Rm y

AR M

AP

= - úû

ùêë

é-

-+´

´+)5.15.3(050.0

)5.1(1

5.306.7)10005.3(

06.71000

or s A = -2124.65 kg/cm 2 (compressive).

The stress at point B is given by

Bq s = s B = + úû

ùêë

é+

++ B

B

y Rm y

AR M

AP

(1

= úû

ùêë

é+

+´

+-)5.15.3(050.0

5.11

5.306.73500

06.71000

\ s B = 849.85 kg/cm 2 (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(64

cmd == p p

If the link is considered to be a straight beam, the corresponding values are

s A = I

My AP +

= -976.3

)5.1)(3500(06.7

1000 -++

\ s A = -1462.06 kg/cm 2 (compressive)

& s B =976.3

5.1350006.7

1000 ´+-

s B = 1178.8 kg/cm 2 (tensile)



Module6/Lesson3


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 acompressive load of 10,000 kg. Compute the stresses at A and B by curved beamformula.

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


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

AP

= - úû

ùêë

é

-

-++)66.566.23(042.0

)66.5(1

66.23x48

)66.23x10000(

48

10000

\ s A = -1559.74 kg/cm 2 (compressive)

Similarly, Stress at B is given by

s B = úû

ùêë

é+

++)(

1 B

B

y Rm y

AR M

AP

= úûù

êëé

++

´´+-

)34.1066.23(042.034.10

166.2348

66.231000048

10000

\ s B = 1508.52 kg/cm2

(tensile)



Module6/Lesson3


Stress at( )úû

ùêë

é+

++== B

B B y Rm

y

AR M

AP

B 1s

( )úûù

êëé

++

´´+-=

260397.02

168

620008

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( )úû

ùêë

é+

++== A

Ac y Rm

y

AR M

A N

C 1s

( )( )úû

ùêëé

--+

´+-=

260397.02

168

103928

1732

2/7.2726 cmkgc -=\ s (Compression)

Stress at( )úû

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


( )úûù

êëé

++

´+-=

260397.02

168

103928

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 thecircumferential 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 section 272122

39cm A =´+==

Now, .539 329312 cm y A =úûù

êëé

+´+

=

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 öç

è æ

-+-++´

´+-= 1239

512712

ln.397121231272

121m

080.0=\ m

Moment cmkgPR M -=´=== 12000012000,10

Now,

Stress at( )úû

ùêë

é+

++== A

A A y Rm

y

AR M

AP

A 1s

7 c m 10tA

R

B.

12cm

9cm3cm

YB

Section AB

YA



Module6/Lesson3


( )( )úû

ùêëé

--+

´+-=

51208.05

11272

12000072

10000

2/1240 cmkg A -=\ s (Compression)

Stress at( )úû

ùêëé +++==

B

B B y Rm

y AR M

AP B 1s

( )úûù

êëé

++

´+-=

71208.07

11272

12000072

000,10

2/62.639 cmkg B =\ s (Tension)

Example 6.9A circular open steel ring is subjected to a compressive force of 80 kN as shownin the Figure 6.20. The cross-section of the ring is made up of an unsymmetricalI-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

160YBY A

20

20

20R



Module6/Lesson3


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 R

m ----++-+++-= lnlnlnln1 23111

Hence = R Radius of curvature of the centroidal axis.

Now, 2600020802012010020 mm A =´+´+´=

( ) ( ) ( ).33.75

6000150208080201201020100

mm 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 7 mm N PR M -´=´´==

Now, Stress at point( )úû

ùêë

é+

++== B

B B y Rm

y

AR M

AP

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( )úû

ùêë

é+

++== A

A A y Rm

y

AR M

AP

A 1s

( )úûù

êëé

++

´´+-=

67.8433.225072.067.84

133.2256000

10803.16000

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


Example 6.10Calculate 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 circularwith 2 cm diameter. Loading is within elastic limit.

Figure 6.21 Loaded closed ring

Solution : For circular section, from Table 6.1

122122

-÷ ø öç

è æ ÷

ø öç

è æ -÷

ø öç

è æ +-=

c R

c R

c R

m

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 öç

è æ -=

p 2

1PR M A

PR364.0= P5.2364.0 ´=

P M A 91.0=\

Now,( )úû

ùêë

é+

++÷ ø öç

è æ -=

A

i A A y Rm

y AR M

AP

i1s

( )( )úû

ù

êë

é

-

-+´

+÷ ø ö

çè æ -=

15.20435.01

15.291.0

A

P

A

P

÷ ø öç

è æ -÷

ø öç

è æ -=

AP

AP

21.5

. R = 2. 5 c

mBi

AO

BO

A i

2P 1000kg=

MA

m

n

q

P

Pcos q

Psinq

A i AoP



Module6/Lesson3


÷ ø öç

è æ -=\ AP

i A 21.6s (Compressive)

( )úûù

êë

é+

++÷ ø öç

è æ -=

B

o A y Rm

y AR M

AP

10

s

( )úûù

êëé

++

´+÷

ø öç

è æ -=

15.20435.01

15.2

91.0 A

P AP

÷ ø öç

è æ =\ AP

o A 755.1s (Tension)

Similarly, ( )PR M M A B -=

( )PRPR -= 364.0 PR636.0-=

P5.2636.0 ´-=

P M B

59.1-=\

Now,( )úû

ùêë

é+

+=i

i B Bi y Rm

y AR M

1s

( )( )úû

ùêëé

--+

´-=

15.20435.01

15.2

59.1 A

P

÷ ø öç

è æ =\ AP

Bi 11.9s (Tension)

and( )

( )úûù

êëé

+++÷

ø öç

è æ

´-=

15.20435.01

15.2

59.1 A

P Bos

÷ ø öç

è æ -= AP

81.4 (Compression)

Now, substituting the values of ,500 kgP =

( ) ,14159.31 22 cm 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

/1450500

11.9 cmkgi B =´=

p s 2/54.765

50081.4

0cmkg B

-=´-=p

s



Module6/Lesson3


Example 6.11A ring of 200mm mean diameter has a rectangular cross-section with 50mm in theradial direction and 30mm perpendicular to the radial direction as shown in Figure6.22. If the maximum tensile stress is limited to ,/120 2mm N determine the tensileload that can be applied on the ring.

Figure 6.22 Closed ring with rectangular cross-section

Solution : ,100 mm 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


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 öç

è æ -=

p 2

12

WR M AB

W W

M AB 17.182

12100 =÷

ø öç

è æ -´=\

p

Now,( )úû

ùêë

é+

++= A

A A A y Rm

y

AR

M

AP

1s

( )úûù

êëé

+++=

A

A A

y Rm y

AR M

AW

12

( ) ( )( )÷

÷ ø

öççè

æ -

-+´

-+´

=251000217.0

251

100150017.18

15002W W

W A 002073.0=\ s (Tensile)

and( )úû

ùêë

é+

++= B

B A B y Rm

y

AR

M

AP

1s

( )úûù

êë

é

++

´-

´=

251000217.0

25

11001500

17.18

15002

W w

W B 00090423.0-=\ s (Compression)

W

A B

W2

W2

RM

N

qMAB

MAB



Module6/Lesson3


To find stresses at C and D

We have, ( )q cos12

-+-= WR M M ABmn

2,90 0 WR

M M M At ABCDmn +-===\ q

W W W M CD 83.312

10017.18 =´+-=\

Now, stress at( )úû

ùêë

é+

++==C

C CDC y Rm

y

AR

M

AP

C 1s

( )( )úû

ùêëé

--+

´+=

251000217.025

11001500

83.310

W

W 00305.0-= (Compression)

and stress at( )úû

ùêë

é+

++== D

DCD D y Rm

y

AR

M

AP

D 1s

( )úûù

êëé

++

´+=

251000217.025

11001500

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 stressshould not exceed 2/80 mm N



Module6/Lesson3


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,

122122

-÷ ø öç

è æ ÷

ø öç

è æ -÷

ø öç

è æ +-=

C R

C R

C R

m

15.12

505.12

5025.12

502122

-÷ ø öç

è æ ÷

ø öç

è æ -÷

ø öç

è æ +-=

016.0=\ m

Area of cross-section = ( ) 22 87.4905.12 mm A == p

We have, ÷ ø öç

è æ -=

p 2

12

WR M A

÷ ø öç

è æ -´=

p 2

1502

W

W M A 085.9=

Now,( )úû

ùêë

é+

++= A

A A A y Rm

y AR M

AP

1s

.

W

C

A

1 0 0 m

m B WW

W

D

MAB

M

N

q

W2

W

2

W2



Module6/Lesson3


( )( )úû

ùêëé

--+

´-

´=

5.1250016.05.12

15087.490

085.987.4902

W W

W 0084.0= (Tensile)

( )úûù

êë

é++÷ ø

öçè æ

-+=\ B

B A B y Rm

y AR

M AP

1s

( )úûù

êëé

++

´-

´=

5.1250016.05.12

15087.490

085.987.4902

W W

W B 00398.0-=\ s (Compression)

Also, ( ) ÷ ø öç

è æ ´+-=-= 50

2085.9

W PR M M ACD

W M CD 915.15+=\

Now,( )úû

ùêëé

-+=

C

C CDC y Rm

y AR

M 1s

( )( )úû

ùêëé

--+

´+=

5.1250016.05.12

15087.490

918.15 W

W C 013.0-=\ s (Compression)

and( )úû

ùêëé

++

´+=

5.1250016.05.12

15087.490

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


( )úûù

êë

é+

++=\C

C CDC y Rm

y

AR

M

AP

1s

( ) ( )

( )úû

ùêë

é-

-+´

-+´

=

5.1250016.0

5.121

5087.490

085.9

87.4902

W W

W C 00836.0=\ s (Tensile)

and( )úû

ùêë

é+

++= D

DCD D y Rm

y

AR

M

AP

1s

( )( )úû

ùêëé

++

´-+

´=

5.1250016.05.12

15087.490

085.987.4902

W W

W D 00398.0-=\ s (Compression)

Resultant stresses are( ) W W W C 00464.000836.0013.0 -=+-=s (Compression)

( ) W W W D 00482.000398.00088.0 =-=s (Tension)

In order to limit the tensile stress to 2/80 mm N in the ring, the maximum value of the forcein the pulls is given by

0.00482 W = 80

kN or N W 598.1651.16597=\

6.3.5 E XERCISES

1. Is the following function a stress function?

q q p

f sinr P ÷

ø öç

è 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


5. Starting from the stress function DCr r Br r A 22 loglog ++=f , obtain the stress

components r s and q s in a pipe subjected to internal pressure i p and external pressure

o p . Obtain the maximum value of q s when 0=o p and indicate where it occurs.

6. Check whether the following is a stress function

( ) a q q a f tancos2222

r r r c -+-= where a is a constant.

7. Starting from the stress function 321

8)3(

r C r

C C r r

++-+=f , derive expressions for

r s and q s in case of a rotating disk of inner radius ' a ' and outer radius ' b '. Obtain the

maximum values of r s and q s .

8. Show that the stress function DCr r Br r A +++= 22 loglogf solves the problem of

axisymmetric stress distribution, obtain expressions for r s and q s in case of a pipe

subjected to internal pressure i r and external pressure 0 r .9. Show that the following stress function solves the problem of axisymmetric stress

distribution in polar coordinates

DCr r Br r A +++= 22 loglogf 10. Explain axisymmetric problems with examples.11. Derive the general expression for the stress function in the case of axisymmetric stress

distribution.12. Derive the expression for radial and tangential stress in a thick cylinder subjected to

internal and external fluid pressure.13. A curved bar bent into a arc of a circle having internal radius ‘a’ and external radius

‘b’ is subjected to a bending couple M at its end. Determine the stresses

q s s ,r and q t r .

14. For the stress function, r Ar log2=f , where A is a constant, compute the stresscomponents q s s ,r and q t r .

15. A thick cylinder of inner radius 150 mm and outer radius 200 mm is subjected to aninternal pressure of 15 MN/m 2. Determine the radial and hoop stresses in the cylinder atinner and outer surfaces.

16. The internal and external diameters of a thick hollow cylinder are 80 mm and 120 mm respectively. It is subjected to an external pressure of 40 MN/m 2, when the internal

pressure is 120 MN/m 2. Calculate the circumferential stresses at the external and internalsurfaces and determine the radial and circumferential stresses at the mean radius.

17. A thick-wall cylinder is made of steel ( E = 200 GPa and 29.0=n ), has an insidediameter of 20 mm, and an outside diameter of 100 mm. The cylinder is subjected to an



Module6/Lesson3


internal pressure of 300 MPa . Determine the stress components r s and q s at

r = a = 10 mm, r = 25 mm and r = b = 50 mm. 18. A long closed cylinder has an internal radius of 100 mm and an external radius of 250 mm.

It is subjected to an internal pressure of 80 MPa . Determine the maximum radial,

circumferential and axial stresses in the cylinder.19. A solid disc of radius 200 mm is rotating at a speed of 3000 rpm . Determine the radial and

hoop stresses in the disc if 3.0=n and r = 8000 kg/m 3. Also determine the stresses in

the disc if a hole of 30 mm is bored at the centre of the disc.20. A disc of 250 mm diameter has a central hole of 50 mm diameter and runs at 4000 rpm .

Calculate the hoop stresses. Take 25.0=n and r = 7800 kg/m3.

21. A turbine rotor 400 mm external diameter and 200 mm internal diameter revolves at1000 rpm . Find the maximum hoop and radial stresses assuming the rotor to be thin disc.Take the weight of the rotor as 7700 kg/m3 and poisson’s ratio 0.3.

22. Investigate what problem of plane stress is solved by the following stress function2

2

3

2343

yP

C

xy xy

C F +

þýü

îíì

-=f . Check whether the following is a stress function

q f 2cos222 ÷

ø öç

è æ +++= D

r C

Br Ar

23. Show that x DyeCye Be Ae y y y y a a a a a sin-- +++ represents stress function.24. The curved beam shown in figure has a circular cross-section 50mm in diameter. The

inside diameter of the curved beam is 40mm. Determine the stress at B whenkN P 20= .

Figure 6.25

25. A crane hook carries a load kN W 20= as shown in figure. The cross-section mn of thehook is trapezoidal as shown in the figure. Find the total stresses at points m and n. Use

the data as given mmammbmmb 30,10,40 21 === and mmc 120=

B

P

C



Module6/Lesson3


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 6.2727. 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 thethickness t of the stem. Find the proper value of the stem thickness so that the extremefiber 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


Figure 6.28

28. A closed ring of mean diameter 200mm has a rectangular section 50mm wide by a 30mmthick, is loaded as shown in the figure. Determine the circumferential stress on the inside

and outside fiber of the ring at A and B. Assume 2/210 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 20 kN applied along aline 50mm from the inner edge of the shank. Compute the stress at the inner andouter fibers.

60mm

80mm

20mm

t

. A

1 0 0 m m B

5 0 m m .

50mm

30mmA1

B1

50KN

50KN



Module6/Lesson3


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 30 kN 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


Module: 7 Torsion of Prismatic Bars

7.1.1 I NTRODUCTION

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 isrequired, 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 Aò 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 theapplication of a torque, i.e., points in a given cross-sectional plane remain in that plane aftertwisting.

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. Theseare shown in the Figure 7.0.

F



Module7/Lesson1


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 G ENERAL SOLUTION OF THE T ORSION P ROBLEM

The correct solution of the problem of torsion of bars by couples applied at the ends wasgiven by Saint-Venant. He used the semi-inverse method. In the beginning, he made certainassumptions for the deformation of the twisted bar and showed that these assumptions couldsatisfy the equations of equilibrium given by

0=+¶

¶+¶

¶+

¶¶

x xz xy x F

z y xt t s

0=+¶

¶+

¶¶

+¶

¶ y

yz xy y F 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 perunit area and l, m, n are the direction cosines.



Module7/Lesson1


Also from the uniqueness of solutions of the elasticity equations, it follows that the torqueson the ends are applied as shear stress in exactly the manner required by thesolution 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 aresult 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


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 asmall angle zq . However, with the assumed displacements (a), (b) and (c), we calculate the

components of strain from the equations given below.

e x = xu

¶¶

, e y = yv

¶¶

, e z = zw

¶¶

g xy = xv

yu

¶¶+

¶¶

, g yz = yw

zv

¶¶+

¶¶

and g zx = zu

xw

¶¶+

¶¶

,

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

e x = e y = e z = g xy = 0

g xz = y xw -

¶¶

q = ÷ ø öç

è æ -

¶¶

q y

q y x

or gxz = ÷ ø öç

è æ -

¶¶

y xy

q

and gyz = x yw +

¶¶

q = ÷÷ ø ö

ççè æ +

¶¶

q y

q x y

or gyz = ÷÷ ø ö

ççè æ

+¶¶

x yy

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


s x = s y = s z = t xy = 0

t xz = G ÷ ø öç

è æ -

¶¶=÷

ø öç

è æ -

¶¶

y x

G y xw y

q q (d)

t yz = G ÷÷ ø ö

ççè æ

+¶¶

=÷÷ ø ö

ççè æ

+¶¶

x yG x yw y

q q (e)

It can be observed that with the assumptions (a), (b) and (c) regarding deformation, there will be no normal stresses acting between the longitudinal fibers of the shaft or in the longitudinaldirection of those fibers. Also, there will be no distortion in the planes ofcross-sections, since e x , e y and g xy vanish. We have at each point, pure shear defined by thecomponents t xz and t yz.

However, the stress components should satisfy the equations of equilibrium given by:

0=+¶

¶+¶

¶+

¶¶

x xz xy x F

z y xt t s

0=+¶

¶+¶

¶+¶

¶ y

yz xy y F z x y

t t s 0=+

¶¶+

¶¶+

¶¶

z yz xz z F

y x z

t t s

Assuming negligible body forces, and substituting the stress components into equilibriumequations, we obtain

0=¶

¶ z

xzt , 0=

¶¶

z zyt

, 0=¶

¶+

¶¶

y x zy xz

t t (7.3)

Also, the function ( x , y) , defining warping of cross-section must be determined by theequations of equilibrium.

Therefore, we find that the function must satisfy the equation

02

2

2

2

=¶¶+

¶¶

y xy y

(7.3a)

Now, differentiating equation (d) with respect to y and the equation (e) with respect to x, andsubtracting we get an equation of compatibility

Hence, q t

G y

xz -=¶

¶

q t

G x

yz =¶

¶

x y yz

xz¶

¶-¶

¶ t t = -Gq - G q = -2 Gq = H

Therefore, x y

yz xz

¶¶

-¶

¶ t t = H (7.4)



Module7/Lesson1


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 B OUNDARY C ONDITIONS

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 boundaryand 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 thirdgives,

t xz l + yzt m = 0 (7.5)

which means that the resultant shearing stress at the boundary is directed along the tangent tothe boundary, as shown in the Figure 7.2.

Figure 7.2 Cross-section of the bar & Boundary conditions



Module7/Lesson1


Considering an infinitesimal element abc at the boundary and assuming that S is increasingin the direction from c to a ,

= cos ( N, x ) =dS dy

m = cos ( N , y) = -dS dx

\ Equation (7.5) becomes

÷ ø öç

è æ -÷

ø öç

è æ

dS dx

dS dy

yz xz t t = 0

or 0=÷ ø öç

è æ

÷÷ ø ö

ççè æ +

¶¶-÷

ø öç

è æ ÷

ø öç

è æ -

¶¶

dS dx

x ydS

dy y

xy 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 S TRESS FUNCTION M ETHOD

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 bet 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) areindependent of z.

In order to satisfy the third condition, we assume a function f ( x , y) called Prandtl stressfunction 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


equations of compatibility. Alternately, we can determine the strains corresponding to theassumed stresses and then apply the strain compatibility conditions.

Therefore from Equations (7.7), (d) and (e), we have

)( y

x

G

y

-¶

¶=¶

¶ y q

f - )( x

y

G

x

+¶

¶=¶

¶ y q

f

Eliminating y by differentiating the first with respect to y, the second with respect to x, andsubtracting from the first, we find that the stress function must satisfy the differentialequation

q f f

G y x

22

2

2

2

-=¶¶+

¶¶

or H y x

=¶¶+

¶¶

2

2

2

2 f f (7.8)

where H = - 2Gq

The boundary condition (7.5) becomes, introducing Equation. (7.7)

0==¶¶+

¶¶

dS d

dS dx

xdS dy

yf f f

(7.9)

This shows that the stress function f must be constant along the boundary of the cross-section. In the case of singly connected sections, example, for solid bars, this constant can

be arbitrarily chosen. Since the stress components depend only on the differentials of f , fora simply connected region, no loss of generality is involved in assuming f = 0 on S .However, for a multi-connected region, example shaft having holes, certain additionalconditions of compatibility are imposed. Thus the determination of stress distribution over across-section of a twisted bar is used in finding the function f that satisfies

Equation (7.8) and is zero at the boundary.Conditions at the Ends of the Twisted bar

On the two end faces, the resultants in x and y directions should vanish, and the momentabout A should be equal to the applied torque M t . The resultant in the x-direction is

òò òò ¶¶= dxdy y

dxdy z xf

t = ò ò ¶¶

dy y

dx f

Therefore, òò =dxdy xzt 0 (7.10)

Since f is constant around the boundary. Similarly, the resultant in the y-direction is

òò òò ¶¶-= dxdy x

dxdy yz f t

= ò ò ¶¶- dx x

dy f



Module7/Lesson1


hence, òò =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 =

òò - dxdy y x

xz yz)( t t (7.12)

= - òò ¶¶+

¶¶

dxdy y

y x

x )( f f

Therefore,

M t = - òò òò ¶¶-

¶¶

dxdy y

ydxdy x

x f f

Integrating by parts, and observing that f = 0 at the boundary, we get

M t = òò òò+ dxdydxdy f f (7.13)

\ M t = 2 òòdxdyf (7.14)

Hence, we observe that each of the integrals in Equation (7.13) contributing one half of thetorque 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 functionf obeys Equations (7.8) and (7.14) and the solution obtained in this manner is the exactsolution of the torsion problem.

7.1.5 T ORSION OF C IRCULAR C ROSS SECTION

The Laplace equation is given by

02

2

2

2 =¶¶+

¶¶

y xy 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 öç

è æ

÷÷ ø ö

ççè æ +

¶¶-÷

ø öç

è æ ÷

ø öç

è æ -

¶¶

dS dx

x ydS

dy y

xy y

Therefore, with = C , the above boundary condition becomes(0- y) (dy / dS ) – (0+ x) (dx/dS ) = 0



Module7/Lesson1


- y 0=-dS dx

xdS dy

or 02

22

=+ y xdS d

i.e., x2

+ y2

= constantwhere ( x , y) are the co-ordinates of any point on the boundary. Hence the boundaryis a circle.

From Equation (c), we can writew = qy ( x , y) i.e., w = q C

The polar moment of inertia for the section is J = òò =+ 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 everycross-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 öç

è æ -

¶¶=÷

ø öç

è æ -

¶¶

y x

G y xw y

q q

t yz = G ÷÷ ø öççè

æ +¶¶=÷÷ ø

öççè æ +

¶¶ x

yG x

yw 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


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 2t ( 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 elementaryanalysis are justified.



Module: 7 Torsion of Prismatic Bars

7.2.1 T ORSION OF E LLIPTICAL C ROSS -SECTION

Let the warping function is given by Axy= (7.15)

where A is a constant. This also satisfies the Laplace equation. The boundarycondition gives

( Ay - y) 0)( =+-dS dx

x AxdS dy

or y ( A-1) 0)1( =+-dS dx

A xdS 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

ba

+-=

11

2

2

or A = 22

22

abab

+-

Therefore, the function given by

= xyabab

22

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 = òò -++ dxdy Ay Ax y x )( 2222 (7.17)



= ( A+ 1) òò òò-+ 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

baba

+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

baba

G M t

p

+

The shearing stresses are given by

t yz = Gq ÷÷ ø

öççè

æ +¶¶

x yy

= M t 33

22

baba

p

+ x

abab

÷÷ ø

öççè

æ +

+-

122

22

or t yz =ba x M t 32

p

Similarly , t xz = 3

2ab

y M t p

Therefore, the resultant shearing stress at any point ( x , y) is

t = 22 xz yz t t + = 33

2ba

M t p

[ ]21

2424 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 = a 2 (1- y2 / b2)



and t = [ ]21

22224233 )(

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 = 2/12433 )(

2ba

ba M t

p

Therefore, t max = 2

2ab 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 P RANDTL’S M EMBRANE 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



desirable to use other techniques – experimental or otherwise. The membrane analogyintroduced by Prandtl has proved very valuable in this regard.

Let a thin homogeneous membrane, like a thin rubber sheet be stretched with uniformtension fixed at it’s edge which is a given curve (the cross-section of the shaft) in the xy-plane as shown in the figure 7.5.

Figure 7.5 Stretching of a membrane

When the membrane is subjected to a uniform lateral pressure p , it undergoes a smalldisplacement z where z is a function of x and y.

Consider the equilibrium of an infinitesimal element ABCD of the membrane afterdeformation. Let F be the uniform tension per unit length of the membrane. The value of the

initial tension F is large enough to ignore its change when the membrane is blown up by thesmall pressure p . On the face AD, the force acting is F.dy . This is inclined at an angle b to

the x-axis. Also, tan b is the slope of the face AB and is equal to x z¶¶

. Hence the component

of Fdy in z-direction is ÷ ø öç

è æ

¶¶- x z

Fdy . The force on face BC is also Fdy but is inclined at an

angle ( b + Db ) to the x-axis. Its slope is, therefore,

dx x z

x x z ÷

ø öç

è æ

¶¶

¶¶+

¶¶

and the component of the force in the z-direction is

úûù

êëé

÷ ø öç

è æ

¶¶

¶¶+

¶¶

dx x z

x x z

Fdy



Similarly, the components of the forces Fdx acting on face AB and CD are

-Fdx y z¶¶

and Fdx úû

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

or 2

2

2

2

yz

xz

¶¶+

¶¶

= - 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 identicalto 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 becomesnumerically 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 ProblemZ 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 T ORSION OF T HIN -W ALLED 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 Dl as shown in Figure 7.6. The areasof cut faces AB and CD are t 1 Dl and t 2 Dl respectively. The shear stresses (complementaryshears) 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 issimilar 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 dSThe 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 generallyknown 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 Figure7.7(b), the shear force on the element is t t dS = qdS . Due to shear strain g , the force doeswork equal to DU

i.e., d t )(21 tdS U =D

= ltdS D.)(21

g t



=G

ltdS t

t .).(21 D (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 = òDt

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 AqA

242

or q = ò t dS

AGq

2 (7.24)



7.2.4 T ORSION OF T HIN -W ALLED M ULTIPLE -C ELL C LOSED 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 cells1 and 2 respectively. Consider the equilibrium of an element at the junction as shown in theFigure 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, choosemoment 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 shearflow 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

AGq

2

or 2Gq = ò t qdS

A1



Let a 1 = ò t dS

for Cell 1 including the web

a 2 = ò t dS

for Cell 2 including the web

a 12 = ò t dS

for the web only

Then for Cell 1

2Gq = )(1

212111

qaqa A

- (b)

For Cell 2

2Gq = )(1

112222

qaqa A

- (c)

Equations (a), (b) and (c) are sufficient to solve for q1 , q2 and q .


Example 7.1A hollow aluminum tube of rectangular cross-section shown in Figure below, issubjected to a torque of 56,500 m-N along its longitudinal axis. Determine the shearingstresses and the angle of twist. Assume G = 27.6x10 9 N/m 2.

Figure 7.10

0.25

0.5

t 1 t 3

0.006t 2=

0.006t 4=0.012

All Dimensions in metre

Membrane Surface

A

B C p

D

Shear Flowq=

0.01



Solution: The above figure shows the membrane surface ABCD

Now, the Applied torque =M t = 2qA

56,500 = 2 q(0.5x0.25)

56,500 = 0.25 q

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 /106.2201.0

226000m N ´=

Now, the angle of twist per unit length isq = ò t

dsGAq

2

Therefore,

q = úûù

êëé ++

01.025.0

)2(006.0

5.0012.025.0

125.0x10x6.27x2226000

9

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 aclockwise direction. Determine the internal shear flow distribution. The cell areasare as follows:

A 1 = 680 cm 2 A 2 = 2000 cm 2

The peripheral lengths are indicated in Figure

Solution :

For Cell 1, a 1 = ò (t

dS including the web)

= 09.033

06.067

+

therefore, a 1 = 148.3



For Cell 2,

a 2 =08.0

6709.0

4809.0

6309.0

33 +++

Therefore, a 2 = 2409For web,

a 12 = 36609.0

33 =

Now, for Cell 1,

2Gq = )(1

212111

qaqa A

-

= )3661483(680

121 qq

-

Therefore, 2Gq = 2.189 q1 – 0.54 q2 (i)

For Cell 2,

2Gq = )(1

112222

qaqa A

-

= )3662409(2000

112 qq -

Therefore, 2Gq = 1.20 q2 – 0.18 q1 (ii)Equating (i) and (ii), we get

2.18 q1 – 0.54 q2 = 1.20 q2 – 0.18 q1

or 2.36 q1 – 1.74 q2 = 0

or q2 = 1.36 q1

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 = 2 q

1 x 680 + 2 q

2 x 2000

= 1360 q1 + 4000 q2

Substituting for q2, we get10000 ´ 100 = 1360 q1 + 4000 ´ 1.36 q1



Therefore,q1 = 147 N and q2 = 200 N

Figure 7.11

Example 7.3A 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 =\

( ) 22 422 aaa A =´=

For Cell (1),

t ds

a ò=1 (Including the web)

÷ ø öç

è æ +=

t aa

a2

1p

For Cell (2),

t ds

a 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

q 2

q 1

A2

t

t

t

t



t a

t a

t a

t a 2222 +++=

÷ ø öç

è æ =\

t a

a8

2

For web,÷ ø öç

è æ =

t a

a2

12

Now,For Cell (1),

( )212111

12 qaqa

AG -=q

( )úûù

êëé

÷ ø öç

è æ -+=

212

222q

t a

qt

aaa

p p

( )[ ]212 222 qqtaa -+= p

p

( )[ ]21 222

2 qqat

G -+=\ p p

q )1(

For Cell (2),

( )112222

12 qaqa

AG -=q

úû

ùêë

é -= 122

28

4

1q

t

aq

t

a

a

[ ]122 442

qqt a

a -=

[ ]12421

2 qqat

G -=\ q )2(

Equating (1) and (2), we get,

( )[ ] [ ]1221 421

222

qqat

qqat

-=-+p p

or ( )[ ] [ ]1221 421222 qqqq -=-+p

p



( )[ ] [ ]1221 4224

qqqq -=-+p p

( )04

8241221

=+--+\ qqqqp p

p

( ) 048

124

21 =úû

ùêëé +-úû

ùêëé ++

qqp p

p

( )0

482421 =úû

ùêëé +-úû

ùêëé ++

qqp

p p

p p

or ( ) ( )21 4884 qq p p p +=++

12 8485

qq ÷ ø öç

è 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,

21

2

1 4.8485

22

2 aqa

qT ÷ ø öç

è æ

+++÷÷ ø

öççè

æ =

p p p

12

12

8485

8 qaqa ÷ ø öç

è æ

+++=

p p

p

( )( ) 1

22

21612

qa

T úû

ùêë

é+

++=\p

p p

( )( )1612

2221 ++

+=\p p

p a

T q

Now, from equation (1), we have,

( ) ( )( )

( )( )úû

ùêë

é++

+÷ ø öç

è æ

++-

++++=

16122

8485

21612

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 GaT



Example 7.4 A thin walled box section having dimensions t aa ´´2 is to be compared with a solidcircular section of diameter as shown in the figure. Determine the thickness t so that thetwo sections have:

(a) Same maximum shear stress for the same torque.

(b) The same stiffness.

Figure 7.13Solution : (a) For the box section, we have

aat T At

qAT

´==

=

2...2...2

2

t t

t aT

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 öççè

æ \

232

4

aa

T t

p

aaT

or t

p 232

4 =

÷ ø öç

è æ =\

3

16aT

p t )(b

Equating (a) and (b), we get

32

164 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

GAq ò=

2q

Here T = 2qA AT

q2

=\

úûùêë

é +++=\t a

t a

t a

t a

GAT 22

4 2q

( )t aG

aT t GA

aT

22

2

24

646

=

=

Gt aaT

4166=\ q )(c

The stiffness of the Solid Circular Section is

44

32

32aGT

aG

T GI

T

p p p q =

÷÷ ø

öççè

æ == )(d

Equating (c) and (d), we get

44

3216

6aGT

Gt aaT

p =

p 32

166 =

t a

32166´

=\ at p

÷ ø öç

è æ =\

6443 a

t p



Example 7.5 A two-cell tube as shown in the figure is subjected to a torque of 10 kN-m . Determinethe Shear Stress in each part and angle of twist per metre length. Take modulus ofrigidity of the material as 83 kN/mm 2 .

All dimensions in mm

Figure 7.14

Solution : For Cell 1Area of the Cell = A1=

215000100150 mm=´

t ds

a ò=1 (including web)

1305

1005.2

1505

1005

150

=

+++=

For Cell 2

Area of the cell = ( ) ( )222 75125150

21 -´´= A

= 7500 mm2

t ds

a ò=\2 (including web)

5.2125

5.2125

5.2150 ++=

1602 =\ a

For the web,

605.2

15012

==a

1 5 0

100

1 2 5

1 2 5 5

2.5

2.5

2.5q 1

q 2

M t



For Cell (1)

( )212111

12 qaqa

AG -=q

( )21 6013015000

12 qqG -=\ q )(a

For Cell (2)

( )112222

12 qaqa

AG -=q

( )12 601607500

1qq -= )(b

Equating (a) and (b), we get

( )1221 601607500

1)60130(

150001

qqqq -=-

Solving, 21 52.1 qq = )(c

Now, the torque due to shear flows should be equal to the applied torque.

i.e., 2211 22 Aq Aq M t +=

)7500(2)15000(21010 216 qq +=´ )(d

Substituting (c) in (d), we get

)7500(2)52.1(1500021010 226 qq +´=´

N q 02.1652 =\

N q 83.25002.16552.11 =´=\

Shear flow in the web = ( ) ( )02.16583.250213 -=-= qqq

N q 81.853 =\

2

1

11 /17.50

583.250

mm N t q ===\ t

2

2

22 /01.66

5.202.165

mm N t q ===t

2

3

33 /32.345.2 81.85 mm N t q ===t

Now, the twist q is computed by substituting the values of q1 and q2 in equation (a)



i.e., [ ]02.1656083.25013015000

12 ´´´=q G

lengthmmradians /10824.1100083

7.2270615000

1 5-´=´

´=\ q

04.1=q or degrees/m length

Example 7.6A tubular section having three cells as shown in the figure is subjected to a torque of113 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.

( ) 221 25322127

2

mm A ==\ p

22 64516254254 mm A =´=

23 64516 mm 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 216 ´+´+´=´ 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 cellwalls,

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 get2

1 /4.40 mm N =t

22 /2.55 mm N =t

23 /9.48 mm N =t

24 /7.12 mm N -=t

26 /6.36 mm N =t



Module 8/Lesson 1

1

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

Module 8: Elastic Solutions and Applicationsin Geomechanics

8.1.1 I NTRODUCTION

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 K ELVIN’S P ROBLEM

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 2 P acting at a point in the interior of an infiniteelastic body.

In the cylindrical coordinate system, the following displacements can be obtained byKelvin’s solution.

Displacement in radial direction = u r = 3)1(8 RGPrz

n p -

Tangential displacement = u q = 0

Vertical displacement = u z = úûù

êëé ++-

- 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 = úûù

êëé +-

- 5

3

3

3)21()1(4 R

z R

zP n n p

(8.2)

t rz = úûù

êëé +-

- 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

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 od

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 willfind 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 resultantforces, we get 2 P which exactly equilibrate the applied load.

Figure 8.3 Geometry for integrating vertical stress



Module 8/Lesson 1

5


8.1.3 F LAMANT’S PROBLEM

Figure 8.4 shows the case of a line load of intensity ‘q’ per unit length acting on the surfaceof a homogeneous, elastic and isotropic half-space.

Figure 8.4 Vertical line load on Surface of an half-space

Figure 8.5 Stresses due to a vertical line load in rectangular coordinates



Module 8/Lesson 1

6


The stresses at a point P (r, q ) can be determined by using the stress function

f = q q p

sinr q

(8.3)

In the polar co-ordinate system, the expressions for the stresses are as follows:

s r = 2

2

211

q f f

¶¶+

¶¶

r r r (8.4)

and s q = 2

2

r ¶¶ f

(8.5)

t r q = ÷ ø öç

è æ

¶¶

¶¶-

q f

r r 1

(8.6)

Now, differentiating Equation (8.3) with respect to r , we get

q q p

f sin

qr

=¶¶

Similarly 02

2

=¶¶

r f

and s q = 0

Also, differentiating equation (8.3) with respect to q , we get

]sincos[ q q q p q

f +=¶¶

r q

p q

p q q

q f sincos qr qr +=

¶¶

Similarly q p

q q q p q

f cos]cos)sin([

2

2 qr qr ++-=¶¶

q p

q q

p q

p q f

sincoscos2

2 qr qr qr -+=¶¶

Therefore, equation (8.4) becomes

s r = ÷ ø öç

è æ -++÷

ø öç

è æ

q q p

q p

q p

q q p

sincoscos1

sin1

2 r q

r q

r q

r

q

r

= q q p

q p

q q p

sincos2

sinr

qr q

r q -+

Or s r =r

qp

q cos2



Module 8/Lesson 1

7


Now, ]sincos[1

q q q p q

f +=¶¶ q

r

01 =÷

ø öç

è æ

¶¶

¶¶

q f

r r

Hence, t r q = 0

The stress function assumed in Equation (8.3) will satisfy the compatibility equation

÷÷ ø ö

ççè æ

¶¶+

¶¶+

¶¶

÷÷ ø ö

ççè æ

¶¶+

¶¶+

¶¶

2

2

22

2

2

2

22

2 1111q

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 theabove 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 cos 2q + s q sin 2q - 2t r q sinq cosq

Here, s q = 0 and t r q = 0

Hence, s z = s r cos 2q

)(coscos2 2 q 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 = )(

222 z x

q

+p ( )322

3

z x

z

+

s z = 222

3

)(2

z xqz+p

Similarly,s x = s r sin 2q + s q cos 2q + 2t r q sinq

= 00sincos2 2 ++q q p r

q



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

21

21

2 == ps

where C is a constant.

But b2 = x 2 + z 2 Therefore,

h z x z

21

)( 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 B OUSSINESQ’S P ROBLEM

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

( )21

22 zr B +=f (8.10)where B is a constant.

The stress components are given by

s r = ÷÷ ø ö

ççè æ

¶¶-Ñ

¶¶

2

22

r zf

f n

s q = ÷ ø öç

è æ

¶¶-Ñ

¶¶

r r zf

f n 12 (8.11)

s z = úûù

êëé

¶¶-Ñ-

¶¶

2

22)2(

z zf

f n



Module 8/Lesson 1

12


t rz = úûù

êëé

¶¶-Ñ-

¶¶

2

22)1(

zr f

f n

Therefore, by substitution, we get

( ) ( ) ( ) úû

ù

êë

é +-+-= --25

22223

22 321 zr zr zr zv Br s

[ ] ( )23

2221 -+-= zr zv Bq s (8.12)

( ) ( ) ( ) úûù

êëé +++-- --

25

22323

22 321 zr z zr zv B zs

( ) ( ) ( ) úûù

êëé +++--= --

25

22223

22 321 zr rz zr r v Brzt

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

RdR

d

R A R

R R

s s s s q +==

or 321

R A-=q s

where A is a constant and 22 zr R +=

In cylindrical co-ordinates, we have the following expressions for the stress components:

s r = s R sin2 y + s q cos 2 y

s z = s R cos2 y + s q sin 2 y (8.13)

t rz =21

(s R -s q ) sin2 y

321

R A-=q s

But from Figure 8.8

( )21

22sin -+= zr r y

( )21

22cos -+= zr zy

Substituting the above, into s r , s z , t rz and s q we get



Module 8/Lesson 1

13


( )25

2222

21 -+÷

ø öç

è æ -= zr zr Ar s

( )25

2222

21 -+÷

ø öç

è æ -= zr r z A zs

t rz = ( )25

22

23 -+ zr ( A r z ) (8.14)

s q = ( )23

22

21 -+- 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 öç

è æ -=

zr dz zr zr A 25

22221 2

1s

= úû

ùêë

é+-+-

--23

2221

2222

1 )()(1

2 zr z zr

r z

r A

( )ò¥

-+÷ ø öç

è æ -=

z z dz zr r z A 2

52222

1 21

s (8.15)

= 23

221 )(2

-+ zr z

A

( ) dz zr rz A zrz ò

¥-

+=25

22

123

t =23

221

)(2

-

+ zr r A

( )ò¥

-+-= z

dz zr A 23

2212

1q s

= úû

ùêë

é+--

-21

2222

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 = 21

21

r A

(b)

From (a) and (b), it is seen that the shearing forces on the boundary plane are eliminated if,

( ) 02

21 1 =+-- Av 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

ùêëé +--= --

25

22221

2222 3

121 zr zr zr

r z

r Br n s

( )2

5

2233 -

+-= zr Bz zs

( ) ( ) ( ) úûù

êëé ++++--= --

23

2221

2222

121 zr z zr

r z

r B n s q

( )25

2223 -+-= zr Brzrzt (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

cos0

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

cossin0

cossin

cos0

sin

000

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

021

222p

y p d zr r T P z

= ( )[ ] ( )ò úûù

êëé ++ -2

021

222222 23p

y p d zr r zr Bz

= ò 2/

0

2 sincos)6(p

y y y p d B

= 6p B ò2/ 2 sincos

p y y y

od

Now, solving for ò2/ 2 sincos

p y y y

od , we proceed as below

Put cosy = t

i.e., -siny , d y = dt

If cos0 = 1, then t = 1

If cos2p

= 0, then t = 0

Hence, ò =úûùê

ëé=ú

ûùê

ëé-=-1

0

1

0

30

1

32

31

33t t dt t Therefore, P = 2p B

Or B =p 2P



Module 8/Lesson 1

16


Substituting the value of B in Equation (8.16), we get

s z = - ( )25

223

23 -+ zr z

π

P

s r = ( ) ( ) ( ) ýü

îíì

+-úûù

êëé

+-- --

25

22221

22

22 31

212 zr zr zr r z

r P

n p

s q = ( ) ( ) ýü

îíì ++++-- --

23

2221

2222

1)21(

2 zr z zr

r z

r P

n p

t rz = ( )25

222

23 -+- zr rz

Pp

Putting R = 22 zr + and simplifying, we can write

s z = 5

3

2

3

R

Pz

p

-

s r = úûù

êëé -

+-

5

23)(

)21(2 R

zr z R R

P n p

s q = úûù

êëé

+--

)(1

2)21(

3 z R R R zP

p n

t rz = 5

2

23

RrzP

p - (8.16a)

Also, Boussinesq found the following displacements for this case of loading.

u r = úûù

êëé

+--

z Rr

Rrz

GRP )21(

4 2

n p

uq = 0 (8.16b)

u z = úûù

êëé +-

2

2

)1(24 R

zGRP

n p

8.1.6 C OMPARISON BETWEEN K ELVIN’S AND BOUSSINESQ’SSOLUTIONS

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 whichthe major principal stress is acting. The magnitude of the principal stress is given by

s 1 = 323

RPz

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 C ERRUTTI’S P ROBLEM

Figure 8.10 shows a horizontal point load P acting on the surface of a semi-infinitesoil 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 theabsence of radial symmetry. Due to this a rectangular co-ordinate system is used inthe solution.

According to Cerrutti’s solution, the displacements are given by

ýü

îíì

úûù

êëé

+-

+-++=

2

2

2

2

)()21(1

4 z R x

z R R

R x

GRP

u x n p

(8.18)

u y = ýü

îíì

+--

22 )()21(

4 z R xy

R xy

GRP

n p

(8.18a)

u z = ýüîíì

+-+

z R x

R xz

GRP

)21(4 2 n p

(8.18b)

and the stresses are

s x = - ( ) ( )

ýü

îíì

úûù

êëé

+--

+-+-

z R Ry

y R z R R

x RPx 2

2222

2

3

2)21(32

n p

(8.19)

s y = - ( ) ( )

ýü

îíì

úûù

êëé

+--

+-+-

z R Rx

x R z R R

y

R

Px 222

22

2

3

23

)21(32

n p

(8.19a)

s z = 5

2

23

RPxzp

(8.19b)

t xy = - ýü

îíì

úûù

êëé

+++-+-

+- )(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 = 523

RPxyz

p (8.19d)

t zx = 5

2

23

R zPx

p (8.19e)

Here, R2 = x 2+y 2+z 2

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 ofdisplacement moves particles away from the x-axis for positive values of x and towardsthe 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. Verticaldisplacements take the sign of x and hence particles move downward in front of the loadand upward behind the load.

Figure 8.11 Distribution of horizontal displacements surrounding the point load



Module 8/Lesson 1

20


8.1.8 M INDLIN’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 elastichalf 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 ofthe stress and displacement fields surrounding an axially loaded pile and also in the study ofinteraction between foundations and ground anchors.

It is appropriate to write Mindlin’s solution by placing the origin of co-ordinates a distanceC 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 + z 2

R21 = r

2

+ z21

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 inBoussinesq’s problem in the region of the half-space below the surface z = C . Thesedisplacements 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

331

151

12 24276311221213

18 R zccz zr

Rc 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 62166212118 R

zccz R

c z R

zP ---+-+

ýü

îíì --

-n n n

n p (8.20a)

s z = ( )

( ) ( )( ) ( )îíì --+-----+

- 5

223

331

151

31 182123221213

18 R zccz z

Rc 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

331

51

21 306236321213

Rc zcz

Rccz 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 ofcylindrical symmetry.

Replace r 2 by x2+y 2, 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’ssolution, 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 ---

- 5331

6116 R

c zcxy R xy

R xy

GP

n p (8.21a)



Module 8/Lesson 1

22


u z =( )

( ) ( ) ýü

îíì ---+-

- 5331

1 6432116 R

c zcxz

R

cz xz

R

xz

GP n

n p (8.21b)

s x =

( )

( ) ( ) ( ) ( )

ýü

îíì --+------+

-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

331

51

2 )(306)21(63)21()21(318 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

331

51

21 306216321213

18 Rc zcz

Rccz z

R R R

zPx n n n n p

(8.21e)

t xy = ( ) ( ) ( )

ýü

îíì --------+- 7

2

5

2

331

51

2

30632121318 R

c zcx R

c zc x R R R

xPy n n n p

(8.21f)

t yz = ( )

( ) ( ) ýü

îíì ---+-

- 7551

1 302163318 R

c zcz R

c z R

zPxy n n p

(8.21g)

t zx = ( )

( ) ( )( ) ( ) ( )îíì ---+-----+

- 5

22

331

151

12 62163221213

18 Rc zczcx z x

Rc z

R

z

R

z xP n n n n p

( )

ýü--

7

230 R

c z zcx (8.21h)

8.1.9 A PPLICATIONS

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 problemsand (2) noninteractive problems. In the case of interactive problems, the elasticity of thefoundation plays an important role. For example, a flexible raft foundation supporting amultistorey structure, like that illustrated in Figure 8.13 interacts with the soil. In terms ofelasticity 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 inthe soil beneath those points. The response of the raft and the response of the soil arecoupled 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 thefoundation 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 soildeformations are controlled by the contact exerted by the foundation, but the response of thesoil and the structure are effectively uncoupled.



Module 8/Lesson 1

applied elasticity_t.g. sitharam n l.govinda raju

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