Principle Stresses & Strains

7/28/2019 Principle Stresses & Strains

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Principal Stresses

and StrainsUNIT 3 PRINCIPAL STRESSES AND STRAINS

Structure

3.1 Introduction

Objectives

3.2 State of Stress3.3 Normal and Shear Stresses

3.4 Stress Components on an Arbitrary Plane

3.5 Principal Stresses and Principal Planes

3.5.1 Expressions of Principal Planes and Principal Stresses

3.5.2 Maximum Shear Stress

3.6 Circular Representation of State of Stress

3.7 Mohrs Circle for the Analysis of State of Stress

3.8 State of Stress in Combined Bending and Shear

3.9 State of Stress in Combined Bending and Torsion

3.10 Strain Energy due to Normal Stress

3.11 Strain Energy due to Shear Stress

3.12 Strain Energy in Terms of Principal Stresses

3.12.1 Principal Strains

3.12.2Net Strain Energy Density

3.12.3 Components of Strain Energy Density

3.13 Concept of Failure and Equivalent Stresses

3.13.1 Theories of Failure3.13.2 A Comparison of Different Theories of Failure

3.13.3 Equivalent Stress

3.13.4 Factors of Safety and Design

3.14 Summary

3.15 Answers to SAQs

3.1 INTRODUCTION

In Unit 1, you have already been introduced to simple states of stress. Stress Analysis is

an essential requirement in the evaluation of strength, stiffness, deformations and safetyof solids so that one may produce functionally efficient and economic designs. There is a

large number of ways in which stresses are induced in solids (a few samples ones you

have already learnt), which will engage your attention in the subsequent units. In this

unit, we shall be concerned with the analysis of a given state of stress (expressed in terms

of stress components on selected planes) which will have a bearing on the analysis of

strength and safety of solid components.

Objectives

After studying this unit, you should be able to

define six stress components on mutually perpendicular planes at the

requisite location, describe the principal plane and principal stress,

identify the plane of maximum shear stress,


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Stresses in Solids analyse the state of stress in combined bending and shear and combinedbending and torsion, and

describe various theories of failure.

3.2 STATE OF STRESS

From the point of functional utilization of a solid component we may determine thepossible loads (forces) to which it may be subjected to, so that its equilibrium,

compatibility and stability are satisfied on the whole. But a more critical analysis will

imply the satisfaction of equilibrium at each and every point of the solid. The distribution

of stresses over the volume of the solid is analyzed taking into these requirements. Once

such a distribution has been arrived at it will give the state of stress at each and every

point in the solid in terms of the stress components. Often one is not interested in the

state of stress at each and every point in the solid, but is satisfied with the analysis of the

state of stress at the critical locations of the solid. Description of the general state of

stress involves the definition of six stress components namely, x, y, z, xy, yz and zx onthe three mutually perpendicular planes of a small element at the requisite location.

However, in the initial stages of the course, it is sufficient to master the concepts with

reference to the state of stress in two dimensions. The general state of stress at any pointin a two-dimensional element is given by the stress components x, yand xyas shown inFigure 3.1. Of course, any element could only be three-dimensional, but the state of

stress is two-dimensional due to the absence of any stress components in the pair ofz

planes. Hence, in considering equilibrium of forces, the dimension of the element inz

direction is taken as unity; in whatever units the other two dimensions are expressed.

3.3 NORMAL AND SHEAR STRESSES

You have been already introduced to the concept, definition and description of normal

stress and shear stress. In expressing shear stress components, we use two subscripts,

such as xy, yz, zxetc. Here, the first subscript denotes the direction of normal to theplane and the second subscript denotes the direction in which stress (its resulting force) is

acting. Thus xy is the stress iny direction onx plane, i.e. plane normal tox direction.Logically, all the stress components should have double subscripts. However, as

direction of the stress and direction of the normal to the plane are identically same in the

case of normal stress component, only a single subscript is used, i.e. x really represents

xxand so on. In the case of a shear stress component, two subscripts are necessary to

define it correctly. The second subscript also indicates the plane on which its

complementary component is acting.

We have already stated that among normal stresses, tension is considered positive while

compression is considered negative. In the case of shear stresses, one of the components

tends to rotate the element in the positive, i.e. anticlockwise direction and is consideredpositive, while its complementary component which tends to rotate the element in the

clockwise direction is considered negative. Accordingly, in the state of stress, described

in Figure 1.28, xyis positive, while yxis negative. This definition helps us to determinethe sign of the shear stress on inclined planes also.

3.4 STRESS COMPONENTS ON AN ARBITRARY

PLANE

You have already studied in Section 1.7 in Unit 1, as how to determine the normal and

shear stress components on any arbitrary plane whose inclination is defined by the aspectangle and Eqs. (1.30) and (1.31) furnish expressions for these components. Let us havean example of practical application.

Example 3.1


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Principal Stresses

and StrainsFigure 3.1 shows the projection of a rectangular prismABCD, formed by adhesive

bonding of two triangular prismsABCandACD. The state of stress in the prism is

given by the components x = 40 N/mm2, y = 0 and = 0.

If the tensile and shear strengths of the adhesive are 10 N/mm2

and 12 N/mm2,

verify the safety of the joints and find out the value ofx at which the joint willfail.

SolutionThe aspect angle of the bonding planeAC

o 1 o o5090 tan 90 33.69 123.6975

= + = + =

o

n

C

B

X

D

A75 mm

50 mm

Figure 3.1

Known stress components are as follows :

x = 40 N/mm2, y = 0 and xy = 0

Stress components on planeAC,

Normal stress, 2cosn x =

2 o40 cos 123.69=

= 12.3076 N/mm2

> 10 N/mm2

Shear stress sin 22

xnt

=

o40 sin (2 123.69 )2

=

= 18.4615 N/mm2

> 12 N/mm2

The tensile stress on planeACis not within the tensile strength of the bond. Also,the shear stress on the plane exceeds the shear strength of the bond and hence, the

bond will fail both in tensile and shear.

Let us find the normal stress x that may be safely applied.

Shear strength of the bond = 12 N/mm2

Shear stress on bonding plane sin 22

x =

o12 sin 247.382

x =

2o

12 226 N/mm

sin 247.38x

= =


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So the maximum stress we may apply on the plane CB is 26 N/mm2. Here, you

may note that in strength analysis the sign of the shear stress has no significance,

while the sign of the normal stress is important, since the tensile and compressive

strengths may differ considerably.

Stresses in Solids

SAQ 1

If the prism shown in Figure 3.1 is bonded along the diagonalDB, instead ofAC,

verify the safety of the joint and calculate the magnitude ofx at which the jointwill fail.

3.5 PRINCIPAL STRESSES AND PRINCIPAL

PLANES

In Section 3.4, we have seen that for a given state of stress at a point, the magnitude of

normal stress and shear stress may vary with respect to the inclination of planes. If we are

concerned with the safety of solids under stress, we are required to find on which planes

extreme values of normal and shear stress components are present. Hence, it is essential

to know :

(a) Maximum tensile stress,

(b) Maximum compressive stress, and

(c) Maximum shear stress.

In addition, we may also require to know the planes on which these values occur.

The extreme values of normal stresses are called the Principal Stresses and the planes on

which the principal stresses act are called the principal planes. In two-dimensional

problems, there are two principal stresses, namely the major principal stress and the

minor principal stress which are defined as the maximum and minimum values of the

normal stresses respectively. Here, the maximum or minimum is to be considered

algebraically. For example, if the principal stresses happen to be 20 N/mm2

tensile and 75

N/mm2 compressive, the tensile stress of 20 N/mm2

is to be taken as the major principal

stress denoted by the symbol 1 and the compressive stress of 75 N/mm2

is to be taken as

the minor principal stress (algebraically 75 N/mm2

) and denoted by the symbol 2. Thecorresponding planes are defined as major and minor principal planes.

3.5.1 Expressions of Principal Planes and Principal Stresses

In calculus, you have learnt that when a function reaches maximum or minimum its

derivative with respect to the independent variable becomes zero. Since the normal stress

on an arbitrary plane is a function of the aspect angle as given by the expression,

n=2

yx ++

2

yx cos 2 + xy sin 2, the maxima and minima ofnoccur on the

planes for which

d

d n becomes zero, (similarly, ntwill be maximum on planes

where

d

d nt = 0).

Let us now derive the expression,


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Principal Stresses

and Strains

d

d n=

2

yx ( 2 sin 2) + xy2 cos 2

+= 2sin

22cos2

xyxy = 2 nt

i.e.

d

d n = 2 nt . . . (3.1)

Eq. (3.1) gives an important characteristic of the principal plane, namely, the absence of

shear stress components on the plane. We can, therefore, alternatively define a principal

plane as a plane on which only a normal stress component is acting. When dealing with a

three-dimensional state of stress you will find that the third principal plane is neither

maximum nor minimum. Hence, we will define principal planes as planes on which shear

stresses are zero.

Equating

d

d n to zero, we get

xycos 2 +2

xy sin 2 = 0

oryx

xy

xy

xy

=

=

22

2cos

2sin

Denoting the specific angles defining principal planes by 1 and 2,

or tan 2 =yx

xy

2 . . . (3.2)

Eq. (3.2) gives a condition for the determination of principal planes. Eq. (3.2) will have

two solutions within the range /2 < < /2 and they will give the orientation ofprincipal planes.

Further, the second derivatives,2

2

d

d n , will be negative for the solution 1 (aspect angle

of the major principal plane) and positive for the solution 2 (aspect angle of the minorprincipal plane). Let us obtain these expressions too.

2

22 2cos2 2 sin

2

y xn nxy

d dd

d dd

= =

2

2

24 cos 2 sin

2

y xnxy

d

d

=

2

. . . (3.3)

After obtaining the solutions 1 and 2 of Eq. (3.2), their values may be substituted in the

expression for2

2

d

d n given in Eq. (3.3) and the major and minor principal planes may be

identified. But in practical solutions this step is rarely required. Instead, substitute the

two solutions 1 and 2 in the expression for normal stress and obtain the values ofprincipal stresses 1 and 2 and corresponding principal planes may be identified.

Now, let us derive the general expressions for the principal stresses. Since, we know that

tan 2 =yx

xy

2on a principal plane, we may write as

sin 2 =2

2

2

xy

x yxy

+

. . . (3.4)


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and cos 2 =2

2

2

2

x y

x yxy

+

. . . (3.5)

Stresses in Solids

Substituting Eqs. (3.4) and (3.5) in the expression forn, we obtain

n=2 2

2 2

2

2 2

2 2

x y

x y x y xy xy

x y x yxy xy

+ + + + +

=

2

2

2 2

2 2

2

2

2 2

x y

x y xy

x y x yxy xy

+ + +

+ +

=

2

2

2 2

x y x yxy

+ + +

Since

2

2

2

x yxy

+

will have two roots namely

2

2

2

x yxy

+

, we may

write the final expression for major and minor principal stresses as follows :

1, 2 =2

2

2 2

x y x yxy

+ +

. . . (3.6)

Eqs. (3.2) and (3.6) may be used to readily determine the principal planes and principal

stresses.

Let us now have an example for determination of principal stress and principal planes,

given the state of stress.

Example 3.2

Evaluate the principal stresses and principal planes for the state of stress shown in

Figure 3.2.

60 N/mm2

26 N/mm2

20 N/mm2

Figure 3.2


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Principal Stresses

and StrainsSolution

Given x= 60 N/mm

y= 20 N/mm

xy= 26 N/mm

On substituting in Eq. (3.6), we get

1, 2 =

2260 20 60 20 ( 26)

2 2

+ + +

= 40 32.8

1 = 72.8 N/mm and 2 = 7.2 N/mm

Again substituting the values x, y and xyin Eq. (3.2).

tan 2 =yx

xy

2=

2 ( 26)

60 20

= 1.3

Since is general angle, the specific angles representing the principal planes are

designated as 1 and 2.

2 = 52.43, 127.57

using 2 = 52.43

n=60 20

2

++

60 20

2

cos ( 52.43) 26 sin ( 52.43)

= 72.8 N/mm

Hence, we recognize that 1 =2

43.52 defines the major principal plane and

therefore, 2 =257.127 should define the minor principal plane.

SAQ 2

(a) Derive an expression for the maximum shear stress in a general two

dimensional state of stress and also an expression for the aspect angle of thecorresponding plane.

(b) Evaluate the principal stresses and principal planes for the state of stressshown in Figure 3.3.

40 N/mm2

27 N/mm2

32 N/mm2

Figure 3.3

(c) Also find the normal and shear stress components on the planes whoseaspect angles are given as 30, 45 and 75.


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3.5.2 Maximum Shear StressStresses in Solids

We have the general expression for shear stress as

nt= xy cos 2

2

yxsin 2

Differentiating w.r.t. , and equating the derivative to zero,

d

d nt = 2 xy sin 2 (xy) cos 2 = 0

tan 2 =xy

yx

2

)(. . . (3.7)

Since the planes on which maximum shear stresses occur are specific set of planes we

may denote them distinctly by (instead of general aspect angle ).

Comparing Eqs. (3.2) and (3.7), we conclude that 2 = 2 90 as tan 2. tan 2 = 1.

= 45 . . . (3.8)

Eq. (3.8) indicates that the planes of maximum shear stress bisect the right anglesbetween the major and minor principal planes.

The normals to the major and minor principal planes may now be defined as the major

and minor principal axes. Once the principal stresses and principal planes are known,further analysis may be simplified by expressing the state of stress w.r.t. a new

coordinate system with major and minor principal axes as coordinate axes themselves.These axes are usually called axes 1 and 2 respectively.

The general expressions for stress components on arbitrary planes whose aspect angle

may now be measured with axis-1 as reference axis.

Hence, 1 2 1 2 cos 22 2

n

+ = + . . . (3.9)

1 2 sin 22

nt

= . . . (3.10)

Eq. (3.8) already defines that should be 45 forntto be maximum.

Thus,o1 2

max,min sin ( 90 )2

=

1 2max2

= . . . (3.11)

and 1 2min2

=

Since the sign of maximum shear stress is not significant, expression formin is notgenerally used. Let us have a few examples.

Example 3.3

The state of stress at a critical point of a strained solid is given by

x= 70 kN/mm, y= 50 kN/mm and xy= 45 kN/mm. If the strength of thesolid in tension, compression, and shear are given as 120 kN/mm,

90 kN/mm and 75 kN/mm respectively, verify the safety of the component.


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Principal Stresses

and StrainsSolution

Given x= 70 kN/mm

y= 50 kN/mm

xy= 45 kN/mm

22

1,2

70 ( 50) 70 ( 50)

(45) 10 752 2

+

= + =

= 85 65 N/mm

Maximum shear stress, 1 2max85 ( 65)

2 2

= =

= 75 N/mm

All the stresses are within the strength limits of the solid and hence, the solid is

safe.

Factor of safety in tension120

1.412

85

= =

Factor of safety in compression90

1.384665

= =

Factor of safety in compression75

175

= =

Here, maximum tensile and compressive stresses are well within strength limits,maximum shear stress has reached the strength limit and therefore if the state ofstress is proportionally raised the solid will fail in shear.

Example 3.4

A machine component is made of a material whose ultimate strength in tension;compression and shear are 40 N/mm, 110 N/mm and 55 N/mm respectively. Atthe critical point in the component the state of stress is represented by

x= 25 kN/mm and y= 75 kN/mm.

Find the maximum value of the shear stress xy which will cause failure of thecomponent and also specify the mode of failure.

Solution

Given state of stress : x= 25 kN/mm

y= 75 kN/mm.

We have to find what xyis safe, if1 40 kN/mm, 2110 kN/mm andmax >/ 55 kN/mm.

The above three conditions are to be independently satisfied.

Now,

2

21 40

2 2

x y x yxy

+ = + +

In the limiting case,

2225 ( 75) 25 ( 75)40

2 2 xy

+ = +

+

2 240 25 50 xy= + +


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Stresses in Solids or xy= [ 40 (25) ] 50 = 1725

xy = 41.533 N/mm

Note that the limiting case of1 = 40 N/mm will occur for both the xyvalues of41.533 N/mm and 41.533 N/mm. But the planes of failure will be different.

2x y 2

2110

2 2

x y

xy

+ = +

In the limiting case,

2225 ( 75) 25 ( 75)110

2 2xy

+ = +

2 2110 25 50 xy = +

2 285 50 xy = +

2 285 50 68.7386 N/mmxy = = 2

max = xyyx 2

2

2+

55

i.e. 50 + xy = (55)

xy = 22 5055 = 22.91 N/mm

The permissible value ofxyis different for different limiting criteria, namely

| xy | 41.53 if

1 40 68.74 if2 110

22.91 ifmax >/ 55

Hence, we find that the maximum safe value ofxyis only 22.91 N/mm and thematerial will fail in shearing mode.

Example 3.5

The state of stress at a point in a loaded solid is prescribed on two faces of anelement whose shape is a triangular prism as shown in Figure 3.4. Evaluate theprincipal stresses and principal planes.

C

60o

B

x

30o

A yx

y =48

=36

Figure 3.4

Solution

Here, we have to first obtain the value ofx and subsequent calculations will be astandard set.

Given y = 48 MPa


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Principal Stresses

and Strains yx = 36 MPa xy = 36 MPa

30o = 60 MPa (aspect angle of planeBCis + 30o)

oo o

30cos (2 30 ) sin (2 30 )

2 2

x y x yxy

+ = + +

o o48 4860 cos 60 ( 36) sin 602 2

x x + = + +

o o60 24 cos 60 24 cos 60 36 sin 602 2

x x = + + o

i.e.o o60 (1 cos 60 ) 24 24 cos 60 36 sin 60

2

x= + + o

2

[60 24 ( 24 0.5) ( 36) 0.866]1 0.5

x = +

2[79.177] 105.57 MPa

1.5= =

22

1,2

105.57 48 105.57 48( 36) 76.835 46.093

2 2

+ = + =

1 76.835 46.093 122.928 MPa = + =

2 76.835 46.093 30.742 MPa = =

Let the aspect angles of the principal planes be

o o2 2 ( 36)

tan 2 51.35 or 128.65105.57 48

xy

x y

= = =

o o25.675 or 64.325 =

Substituting in expression foro25.675 (or 2 51.35 ) = = o n

o o105.57 48 105.57 48 cos ( 51.35 ) 36 sin ( 51.35 )

2 2n

+ = +

= 122.92 MPa

o o1 225.675 and 64.325 = =

SAQ 3

(a) In the elementABCshown in Figure 3.4, find the normal and shear stresscomponents on the planeAC.

(b) If the state of stress at a point is defined by the stress component

x= 9 MPa, y= 7 MPa, xy= 5 MPa, find the principal stresses andprincipal planes. Also find the plane on which normal and shear stress

components are equal in magnitude.


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Stresses in Solids3.6 CIRCULAR REPRESENTATION OF STATE OF

STRESS

From a state of stress defined by the components x, y and xy, we express the stresscomponents on an arbitrary plane as

cos 2 sin 22 2

x y x y xy

+ = + + . . . (3.12)

sin 2 cos 22

x yxy

= +

. . . (3.13)

which we may rewrite in general as

. . .

(i)

cos 2 sin 2a b c = + +

. . .

(ii)

sin 2 cos 2b c = +

Let us now try to establish direct relationship between and by eliminating betweenEqs. (i) and (ii).

To simplify the effort, let us take the origin of coordinate at (a, 0), so that the newvariable ( a = ) is considered for developing the relationship.

cos 2 sin 2b c = + . . .(iii)

. . . (iv)sin 2 cos 2b c = +

Squaring and adding, we get

2 2 2 2 2 2cos 2 sin 2 2 cos 2 . sin 2b c bc + = + +

2 2 2 2sin 2 cos 2 2 cos 2 sin 2b c bc+ +

2 2 2 2 2 2(cos 2 sin 2 ) (sin 2 cos 2 )b c= + + +

i.e. 2 2 2b c + = + 2

2

.

Since b and c are constants let 2 2b c r+ =

2 2

r + =

2

. . . (3.14)

Eq. (3.14) shows that if and , the normal and shear stress components on any arbitraryplane are plotted as coordinates, the locus of the point will be a circle whose centre will

be , 02

x y +

and radius will be2

2 2(i.e. )2

x yxy b c

+ +

2 .

In other words, the state of stress at any point may be represented by a circle and a point

on the circle represents the normal and shear stress components, on some plane ashorizontal and vertical coordinates.

Let us now consider the circular representation more closely so as to obtain clear

interpretation of the state of stress.

In Figure 3.5 a coordinate system (, ) has been formed with on horizontal axis and on the vertical downward axis. On this coordinate plane of plane a pointXis


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Principal Stresses

and Strainschoosen whose coordinates are x and xy respectively. Also another point Ychoosen withcoordinates y and yx (ory and xy).

F

E

22 2D

212

Z

A

C (c, c)

10 200

10

20

30

40

50

10

20

30

40

X (x, xy)

(2, 0)B

(y, yx)Y

D(D, D)

(1, 0)

120 130

Figure 3.5 : Circular Representation of State of Stress

These two pointsXand Yrepresent the stress components on two planes and hence,according to Eq. (3.14), should be points on a circle. PontXrepresentsx plane and pointYrepresents they plane. When we join the pointsXand Yby a straight line, we find that

the line passes through the point O on the axis which should be the origin of the circledefined by Eq. (3.14). As O is the origin of circleXand Yare points on the periphery of

the circle the lineX O Y(or simplyXY) should be the diameter of the circle which we are

trying to establish. Hence, draw a circle withXYas diameter.

Now any point of this circle should represent the state of stress on an oblique plane. Let

us consider a few specific points. PointsA andB lie on the horizontal () axis, i.e. = 0and hence they represent the two principal planes. Since is maximum atA, pointArepresents the major principal plane (and hence, coordinator ofA is 1) and pointBrepresents the minor principal plane. XOA and XOB are equal to 12 and 22 respectively where and are the inclinations of the major and minor principal

planes with thex plane.1 2

Consider any arbitrary point say Csuch that the angle XOC is equal to 2 c . Then,the coordinates of the point Cgive the normal and shear stress components on plane

inclined at to thex plane. Similarly, each pointZon this circle may be interpreted to

give the normal and shear stress components on a plane whose inclination withx axis isc

1

2XOZ .

3.7 MOHRS CIRCLE FOR THE ANALYSIS OF

STATE OF STRESS

Mohrs circle is a graphical method of finding normal, tangential and resultant stresses onan oblique plane. Mohrs circle will be drawn for the following cases :

(a) A body subjected to two mutually perpendicular principal tensile stresses ofunequal intensities.

(b) A body subjected to two mutually perpendicular principal stresses which areunequal and unlike (i.e. one is tensile and other is compressive).

(c) A body subjected to two mutually perpendicular principal tensile stressesaccompanied by a simple shear stress.

The circle in Figure 3.5 is called the Mohrs Circle of stress. Mohrs circle is very useful

in graphical analysis of state of stress at a point.


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Stresses in Solids Given the state of stress (defined by x, y and xy), the procedure for construction of theMohrs circle was discussed in Section 3.6. The determination of principal stresses and

principal planes from the Mohrs circle was also indicated. In this section let us learn afew applications of stress analysis with the help of the Mohrs circle.

Suppose we need to find the normal and shear stress components on a plane whose

inclination tox plane is D. We need only to draw a radial line OD making an angle 2Dwith the radial line OX. The coordinates (D, D) of the pointD will give the normal andshear stress components on the plane. Thus, once the Mohrs circle is constructed, thestress components on any plane may be readily obtained.

Example 3.6

The state of stress at a point is given by the stress components x = 70 MPa,y = 10 MPa, and xy = 40 MPa. Using Mohrs circle find (i) Principal stresses,and (ii) Principal planes. Also, determine the normal and shear stress components

on planes making 25o, 40o and 60o respectively with thex plane.

20 20

0

20

40

20

40

40

120o

80o50o

R(28, 48.5)

S(57,

36.8

)

X (70,

40o

40)

60

Y(10,40)

(22,10)

(95, 65)

100

1=90

80

Figure 3.6

Solution

(a) Choose (, ) coordinate system to a suitable scale.

(b) Mark the pointsX(70, 40) and Y(10, + 40).

(c) Draw a circle withXYas diameter. This circle cuts axis atA andB.

(d) Measure the coordinates of A and B to obtain principal stresses

1 2 = 90 MPa, = 10 MPa

(The radius of the Mohrs circle gives max = 50 MPa)

(e) Measure XOA ; Here, .o52.8XOA =

Aspect angle 1 of major principal plane is 26.4o.

(f) Measure XOB ; Here, o127.2XOB = +

Aspect angle 2 of major principal plane is + 63.6o.

(g) Draw radial lines OR, OSand OTmaking angles 50o, 80o and 120o with OX.

(h) Coordinates ofR give the normal and shear stress components on the planewhich makes 25

owith thex plane. Here,


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Principal Stresses

and Strains= 28 MPa, = 48.5 MPan nt

(i) Coordinates ofSgiven the normal and shear stress components on the planewhich aspect angle 40o. Here,

= 5.7 MPa, = 36.8 MPan nt

(j) Coordinates ofTgive the normal and shear stress components on the planewith aspect angle 60

o. Here,

= 9.5 MPa, = 6.5 MPan nt

SAQ 4

Analyse the states of stress shown in Figure 3.7 and find the principal stresses andprincipal planes.

6

6

3

6

6

3

9

15

(a) (b) (c)

Figure 3.7

(a) Verify the theoretical results by drawing Mohrs circles for each case.

(b) Verify the results of Example 3.6 analytically.

(c) Solve the Examples 3.2, 3.3 and 3.4 using Mohrs circle and verify theresults.

(d) Draw a Mohrs circle to represent the state of stress x = y = 0 and xy = 10and then, find 1 and 2.

(e) Draw a Mohrs circle to represent the state of stress x = y and xy = 0.

3.8 STATE OF STRESS IN COMBINED BENDING

AND SHEAR

So far we have been analysing the state of stress defined by stress components x, y andxy. But we have not directed our attention as to how different stress components areinduced and how they vary within the solid body and also how to identify the criticallocations. In this and the next section, we shall consider two specific types of loading and

identify the pattern of stresses induced in these cases.

You would have seen a large number of girders supporting bridges and still a largenumber of beams supporting roof or floor slabs in buildings. You will learn, in later Units4, 5 and 6, how these members are loaded and supported and how the stresses induced in

them are calculated. Here we shall consider a few simple cases.


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74

Consider a beam of rectangular cross-section as shown in Figure 3.8. On any section ofthe beam there may be acting two kinds of reactive resistance, namely, a Bending

MomentMand a Shearing Force Q. In terms of the cross-sectional dimensions (breadthb and height h) the stress components on a layer located at a height y from theneutral axis (centroidal axis here) are given as follows :

Stresses in Solids

3

12=x

My

bd . . .

(i)

=y 0 . . .

(ii)

2

2

1.5 4= 1xy

Q y

bh h

. . .

(iii)

The variations of stresses across the depth as per Eqs. (i) and (iii) are shown inFigures 3.8(b) and (c). (How Eqs. (i) to (iii) are obtained and how to obtaincorresponding expressions for breams or girders of other cross-sections will be dealt with

in detail in Units 3 and 4).

y

hy

2

N A

YY

E

A B

F

b

hy 2

2

h+y 2

2

h

2

h

2

(a) (b) (c)

Figure 3.8

Example 3.7For the cross-section shown in Figure 3.8, let us take b = 100 mm, d= 300 mm and

M= 9 107 N mm and Q = 18 104 N and analyse the state of stress at a layer 50mm below the top layer.

Solution

Using Eqs. (i) and (iii)

7

2

9 12 10=

100 300x

y

Since extreme layers are at 150 mm from neutral axis,

= 150 50 100 mmy =


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75

Principal Stresses

and Strains7

2

3

9 12 10 100= 4

100 300x

=

0 N/mm

4 22

2

1.5 18 10 4 100= 1 5

100 300 300xy

=

N/mm

Hence,

2

21,2

40 40= 52 2

+

i.e. 22 = 40.616 N/mm

21 = 0.616 N/mm +

2max = 20.616 N/mm

Example 3.8

A beam ofI-section as shown in Figure 3.9(a) is subjected to a bending moment of

84.928 kN m and a shear force of 106.1 kN. Examine the state of stress at thejunction of flange and web.

9.7522.8

26.0

28.4

30.0

30.9

30.0

F =106.16 kNTAV =106106/4000

=26.54 N

Top Flange

120 mm

20 mm

400 mm

Y

y/2

10 mm

20 mm

A =8400

I =2.1232 108

AW =4000

WEB

Bottom Flange

y200 ( 20)

2 +

(a) (b)

Figure 3.9

Solution

(In this case also the solution will be provided by stating the expressions for stresscomponents, without their derivation which will be dealt with in Units 5 and 6.)

The neutral axis of the section will be the horizontal axis through the centroid ofthe section.

The bending stress at a layer y above the neutral axis is given as,

=xMy

I

. . . (iv)

where,Iis the moment of inertia of the section andMis the bending moment onthe section.

The shear stress at any layer is given as


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76

Stresses in Solids .=xy

Q Ay

I b . . . (v)

where, Ay is the moment of the area of that portion of the section above or below

the layer about the neutral axis, b is the breadth of the layer and Q, the shear forceat the section.

For the cross-section given, the moment of inertia is given as 2.1232 108 mm4.

Total depth of beam = 400 mm.

Neutral axis is at a depth of 200 mm from top.

The distance of the layer at junction of web and flange from the neutral axis is

180 mm.

Determination of Bending Stress

62

8

84.928 10 180= 7

2.1232 10x

My

I

= =

2 N/mm

Determination of Shear Stress

In this case Ay will be given by the moment of the flange area about the

neutral axis.

5 320= 120 20 180 4.56 10 mm2

Ay + =

3 5

2

8

106.1 10 4.56 10= 2

2.1232 10 10xy

=

2.8 N/mm

21,2

72 72= (

2 2

+

22.8)

Thus, we get,

21 = 6.613 N/mm

22 = 78.613 N/mm

2max = 42.613 N/mm

You may obverse here that the junction between flange and web will become thecritical zone, when the shear force at the section is considerable. However,experience will tell you, where to look for critical zones in different cases.

3.9 STATE OF STRESS IN COMBINED BENDINGAND TORSION

In Section 3.8, we have considered the effect of Bending Moment, a moment which tendsto bend the axis of the beam, in terms of stresses produced by it. Another type of momentwhich tends to rotate the member about its axis is called twisting moment or torque it

produces shear stresses in the member. This phenomenon is called Torsion. Study ofstresses and strains due to torsion is very difficult except for members with circular (solid

or hollow) cross-section. If a torque of magnitude Tis applied on a circular bar, the shear

stress produced at a point located at a radial distance r from the axis of the shaft isgiven by

=T r

J . . . (vi)

where,Jis the polar moment of inertia of the cross-section of the bar.


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77

Principal Stresses

and StrainsWhen a member of circular section is subjected to both bending moment and torque, themaximum stresses due to both load cases are produced only in extreme fibres and hence,

locating the critical section is not a problem. Let us illustrate with a numerical example.

Example 3.9

A prismatic bar of circular section with 80 mm diameter is subjected to bendingmoment of 5 kN-m and a torque of 7 kN-m. Analyse the state of stress at the

critical section.

Solution

Moment of the inertia,4 4

6 480= 2.016 10 mm

64 64

DI

= =

Polar moment of inertia,4

4 6= 80 4.032 1032 32

DJ

= = 4mm

Maximum bending stress,6

2max

6

5 10 40= 99.206 N/mm

2.016 10x

My

I

= =

Maximum shear stress6

26

7 10 40= 69.4444.032 10

xy =

N/mm

Determination of Principal Stress

22

1,2

99.206 99.206= (69.444)

2 2

+

Thus, we get

2 21 2= 134.944 N/mm and 35.737 N/mm =

21 2max = 85.34 N/2 = mm

Note

When the cross-section of the bar is hollow circular with outer andinner diametersD and drespectively the analysis of stresses is to be

carried by the same procedure except for using the expressions,

4 4 4 4= ( ) and (64 32

I D d J D d

= )

SAQ 5

A prismatic bar of hollow circular cross-section has the outer and inner diametersas 100 mm and 60 mm respectively. Find the maximum stresses induced due to theaction of a bending moment of 5 kN-m along with a torque of 7 kN-m.

Example 3.10

A prismatic bar of hollow circular cross-section with outer and inner diameters 100

mm and 80 mm respectively, carries a bending moment of 5 kN-m. If the tensile,compressive and shear strengths of the material are given as 140 N/mm2, 125

N/mm2

and 95 N/mm2

respectively, what is the magnitude of torque that maysafely be applied in addition to the bending moment.

Solution


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78

Here, the criteria to be considered are as follows :Stresses in Solids

21 140 N/mm >/

22 125 N/mm < /

2max 95 N/mm >/

As the magnitudes of1 and 2 for such bars will be equal, we need to satisfyonly.

22 125 N/mm < /

and 2max 95 N/mm >/

Ifor the section4 4 6(100 80 ) 2.898 10 mm

64

4= =

Jfor the section 4 4 6(100 80 ) 5.796 10 mm32

= = 4

Maximum compression due to bending moment,

62

max 6

5 10 5086.266 N/mm

2.898 10

My

I

= = =

IfTbe the torque applied (in kN-m units), then,

62

max 6

10 508.62664 N/mm

5.796 10

TT

= =

(under torsion alone)

22

2

86.266 86.266(8.62664 ) 125

2 2T

= + =

2

2 286.266 86.2668.62664 125 81.8672 2

T + = + =

2 2 243.133 8.62664 ( 81.867)T+ = 2

i.e.

12 2 2

2

81.867 43.1338.066 kN-m

8.62664T

= =

If the applied torque is within 8.066 kN-m, we are certain that the bar will be safein compression as well as tension.

We should now analyse what torque will have to be applied if .2max 95 N/mm >/

We know that,

2

2max as 0

2

x yxy y

= +

=

22

max

86.266(8.62664 ) 95

2T

= +

=

(under combined bending and torsion)


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79

Principal Stresses

and Strains

12 2

2

2

86.26695

29.8119 kN-m

8.62664T

= =

However, we cannot apply this much torque, since it will cause compression

failure. Thus, the safe value of additional torque should be restricted to8.066 kN-m.

SAQ 6

A shaft is to be made of a material with safe strength values in tension,

compression and shear of 125 N/mm2, 105 N/mm

2and 84 N/mm

2respectively.

What should be the diameter of the shaft to carry a bending moment of 6 kN-malong with a torque of 8.5 kN-m.

3.10 STRAIN ENERGY DUE TO NORMAL STRESS

In Unit 1, you have learnt a few of the characteristics of elastic solids. Here, we shalllearn one more important characteristic. Whenever forces are applied on elastic

(deformable) solids, the points of application of the forces move due to deformations inthe solid, and hence they do work, loosing their potential energy in the process. In elastic

solids this energy is fully stored and released when the strains are removed. This storedenergy is called Strain Energy.

Let us now develop an expression for the strain energy stored in the solid, when it is

subjected to a normal stress, say x.

Consider a small element of dimensions dx, dy and dz as shown in Figure 3.10, and

subjected to a normal stress x which produces an elongation d in the direction of thestress.

IfEis the Youngs Modulus of the material,

strain produced, xxE

=

elongation . . xxd dx dxE

= =

Total force applied on the element dF, is given by the stress multiplied by area on whichit is applied.

i.e. . .xdF dy dz=

Work done by dF= Force Average Displacement

i.e. .2

ddU dF

=

1. . .

2

xxdU dy dz dx

E

=


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80

Stresses in Solids21 . . .

2xdU dx dy dz

E=

Since dx . dy . dz represents the volume of the element dv.

2

2

xdU dvE

= . . . (3.15)

(By similar reasoning, due to other normal stress components the energy stored in the

element may be shown to be

2

2

ydv

E

and

2

2

z dvE

.)

dy

x

dx

dz

x

d

Figure 3.10

Total energy stored in the solid,

2

2

x

v

U dE

= v . . . (3.16)

Dividing Eq. (3.15) by dv, we can also obtain an expression for the strain energy densityat any point as

2

2

xuE

= . . . (3.17)

(Also the strain energy density due to other stress components may be shown to be

2

2

y

E

and2

2

z

E

respectively).

The application of Eqs. (3.15), (3.16) and (3.17) will be dealt elaborately in a later unit.However, these expressions are presented here, as they are useful in the study of a given

state of stress and will be explained in Section 3.14.

3.11 STRAIN ENERGY DUE TO SHEAR STRESS

A small element of dimensions dx, dy and dz is subjected to a shear stress component xyand undergoing shear strain xy is shown in Figure 3.11.

G G

Cdy

xy

(dF=yx,dx,dy)

xy

xy.dy

EE

H

D

FF


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81

Principal Stresses

and Strains

Figure 3.11

As shown in Section 3.10, we may calculate the work done as product of force dFand

average displacement2

d.

Displacement, xyd d = y

Force, xydF dx dy=

Shear strain,xy

xyG

=

xyd dG

= y

Work done or energy stored in the element,

1. .

2

xyxydU dx dz dy

G

= . . . (3.18)

21

.2

xydU dv

G

=

Total Strain Energy

2

2

xy

v

U dG

= v . . . (3.19)

Strain energy density at any point,

2

2

xyu

G

= . . . (3.20)

Expressions similar to Eqs. (3.18), (3.19) and (3.20) can be developed for other shear

stress components yz and zx also.

3.12 STRAIN ENERGY IN TERMS OF PRINCIPAL

STRESSES

In Sections 3.10 and 3.11, we have derived expressions for strain energy assuming that at

a time only one stress component is acting. For example, we have taken x as equal to

x

E

which is true only ifx alone is acting. Otherwise, we know that,

y yxx

v vE E E

= ,

and the use of such expressions will result in more complex expressions for strain energyand strain energy density. Hence, in the case of a general stress field, we may be able to


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82

get simpler expressions if we reduce the number of terms to be considered. A general

stress field with six components of stress namely x, y, z, xy, yz and zx may beexpressed in terms of equivalent principal stress components 1, 2 and 3. In terms ofthese three components, let us now derive expressions for total strain energy.

Stresses in Solids

3.12.1 Principal Strains

If the state of stress is given in terms of the principal stress components 1, 2 and 3, the

corresponding strains components may be calculates as,

31 21 v v

E E E

= . . . (3.21)

32 12 v v

E E E

= . . . (3.22)

3 1 23 v v

E E E

= . . . (3.23)

Alternately, we may rewrite as,

1 1

2

3 3

11 1

1

v v

v vE

v v

2

=

. . . (3.24)

Using abbreviations, [] for1

2

3

, [] for1

2

3

and [C] for

11

1

1

v v

v vE

v v

=

[ . . . (3.25)] [ ] [C = ]

Eq. (3.25) will be useful in writing compact expressions and simplified derivations.

3.12.2 Net Strain Energy DensityNet strain energy density in an element subjected to a general stress field can be obtainedby adding (algebraically) the energy density due to all the stress components. That is

1 1 2 2 3 3

1 1[ ]

2 2

Tu = + + =

but = C

1

2

Tu C= . . . (3.26)

Eq. (3.26) may, if desired, be expanded as,

1

1 2 3 2

3

11

[ ] 12

1

v vu v

Ev v

v =

1 2

1 2 3 2 1 3

3 1

( )1

[ ] (2

( )

v

u vE

v

+ 3

2

)

= + +

2 2 21 2 3 1 2 2 3 3 1

1( 2 (

2u v

E))= + + + + . . . (3.27)

3.12.3 Components of Strain Energy DensityYou may recall that the total stress field on an element may be considered as composed

of two component sets, one of which is the dilatation component and the other is the


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83

Principal Stresses

and Strainsdistortion component. Hence, the net strain energy is also divided into two components,namely strain energy of dilatation, us and strain energy of distortion, ud.

If the principal components of stress field are given by 1, 2 and 3, we may write as,

1 1

2 2

3 3

s s

s s

s s

= +

. . . (3.28)

where the dilatation or spherical components of the stress field s is given by

1 2 3 or3 3

x y z + + + +

Now let us obtain an expression for strain energy of distortion, using the form ofEq. (3.27).

2 21 2 3

1( ) ( ) ( )

2d s su

E = + +

2s

[ ]1 2 2 3 3 12 ( ) ( ) ( ) ( ) ( ) ( )s s s s sv + + s

( )2 21 1 2 3 2 1 2 3 3 1 2 31

{3 ( )} {3 ( )} {3 ( )}18E

= + + + + + + + + 2

( )1 1 2 3 2 1 2 32 {3 ( )}{3 ( )} . . .v + + + + +

( 2 21 2 3 2 1 3 3 1 21

(2 ) (2 ) (2 )18E

= + + 2

1 2 3 2 3 1 2 3 1 3 1 22 {(2 ) (2 ) (2 ) (2 )v +

)3 2 1 1 2 3(2 ) (2 )}+

2 2 21 2 3 1 2 2 3 3 1

1 6 ( ) 6 ( )18E

= + + + +

2 2 21 2 3 1 2 2 3 3 16 ( )v + + +

2 2 21 2 3 1 2 2 3 3 1

(1 )

3

v

E

+ = + +

2 21 2 2 3 3 1

2(1 ) 1(( ) ( ) ( ) )

6 2

v

E

+ = + + 2

2 2 2

)1 2 2 3 3 11

( ) ( ) (12E = + + as shear modulus, 2(1 )E

G v

= + . . . (3.29)

Thus, 2 21 2 2 3 3 11

( ) ( ) ( )12

duG

= + + 2

2 222 3 3 11 21

3 2 2 2du

G

= + +

. . . (3.30)

Eq. (3.30) is more significant since the terms 2 3 31 2 , and2 2 2

1 represent

the three extreme values of shear stress components.

The dilatation component of strain energy, us is not of much practical significance.

However, if desired it may easily be obtained using the form of Eq. (3.27), as


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84

Stresses in Solids

( )2 2 21 2 ( )2

s s s s s s s s s su vE

= + + + +

i.e.2 23 ( 2

2)s s su v

E=

or23 (1 2 )

2

ss

vu

E

= . . . (3.31)

or2

2

ssu

K

= . . . (3.32)

where, Kis the Bulk Modulus given by3(1 2 )

E

v.

The expression for strain energy density may be used to obtain, wherever required, to

obtain the total strain energy by the expressions

., andd d s sv v v

U u dv U u dv U u dv= = =

Such expressions have wide applications in structural analysis and theory of elasticity.

However, in analysing the state of stress, energy density expressions have been found tobe adequate.

3.13 CONCEPT OF FAILURE AND EQUIVALENT

STRESSES

When we are testing materials for strength, generally, we apply one uniaxial component

of stress usually in tension or compression and obtain a value for the limiting state ofyield of failure and consider a solid to have failed if that state is reached.

What is meant by reaching the limiting state? Let us go into the question, a little deeper.When we conduct a tension test on a mild steel rod and find that it yields when the stress

reaches a value of 250 MPa, this should be the limiting state as it is obtained byexperimental. But one may put a question whether the material has yielded because the

normal stress reached a limit of 260 MPa or on the contrary because the normal strain hasreached a value of 0.0013. The question need not stop here. If you calculate the

maximum shear stress, you may also ask whether reaching a shear stress of 130 MPa isthe limit for yielding of the material. Further question may be whether there is any othercriteria for yield?

Every independent answer to this question had led to a failure theory and we will now

learn a few of these theories.

3.13.1 Theories of Failure

Different people have prescribed different criteria for failure of a solid and hence, anumber of failure theories (also called strength theories) have been formulated.

In what is known as Principal Stress Theory, if the maximum principal stress (orminimum principal stress in the case of compression) reaches the same value of

maximum principal stress at failure in uniaxial strength test, then the yield of failure limitis considered as reached.

According to Principal Strain Theory, a material is considered to have reached the yieldor failure limit when the maximum principal strain in the material has reached the value

of the maximum principal strain at failure as observed in the uniaxial strength test.

According to Shear Stress Theory, a material is considered to have reached the yield orfailure limit when the maximum shear stress in the material has reached the value of the

maximum shear stress at failure as observed in the uniaxial strength test.


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Principal Stresses

and StrainsApart from the stress or strain limits as governing criteria for failure, the capacity of thematerial to store energy is also considered as criteria for failure and theories have been

formulated based on these criteria.

According to Total Strain Energy Theory, a material is considered to have reached the

yield or failure limit when the total strain energy density (anywhere within the material)has reached the total strain energy density observed at failure in the case of uniaxial

strength test.

Yet another failure theory has been formulated, with the assumption that the distortionenergy density, rather than the total strain energy density, is significant as failure criteria.According to this theory, known asDistortion Energy Theory, a material is considered to

have reached the yield or failure limit when the distortion energy density (anywherewithin the solid) reaches the value of distortion energy density at failure as observed in

the case of uniaxial strength test.

A few other theories are not so popular and hence, are not dealt with here. Now, you mayget a doubt, whether the different theories really set different criteria for failure or arethese theories only different ways of expressing the same criteria and hence, are

essentially the same? Let us examine a few simple cases and ascertain the answer.

3.13.2 A Comparison of Different Theories of Failure

We know that mild steel yields at a stress value of 260 MPa. The normal strain at this

limit is 0.0013 and maximum shear stress at yield is 130 MPa.

Total strain energy density [Eq. (3.27)] at yield

2 2260

2 2

yu

E E

= = .

Distortion energy density at yield [Eq. (3.29)]2 21 ( ( )

12d y )y

G= + u

2 2260

6 6

ydu = =

G G

Example 3.11

Let us consider a few cases of solids of the same material (mild steel) under

different states of stress as shown in Figure 3.12. The given states of stress havealready been reduced in terms of principal stresses. Poissons ratio v is given

as 0.3.

Solution

Case I

First, let us consider the solid shown in Figure 3.12(a).

(a) (b) (c)

Figure 3.12

(a) Since the major principal stress, 1 (300 MPa) is more than

y (260 MPa), the solid will fail according to principal stress theory.

(b) The maximum principal strain, 1300 ( 200)

0.3E E

+ =

200 150 100

300 250200


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86

Stresses in Solids 240 260

E E= <

According to principal strain theory, the solid is safe.

(c) Maximum shear stress, max300 200

50 1302

= = < <

The solid is safe according to shear stress theory too.(d) Total strain energy density in the solid, u is as follows :

22 21 47000 260

[300 200 2 0.3 300 200]2 2

uE E E

= + = >

The solid will fail according to total strain energy theory.

(e) Distortion energy density in the solid, udas follows :

22 2 21 70000 260[(300 200) 200 300 ]

12 6 6du

G G= + + = >

G

Distortion energy theory also predicts that the solid will fail.Case II

Let us now consider the solid shown in Figure 3.12(b).

(a) As the principal stresses are within 260 MPa, the solid is safeaccording to the principal stress theory.

(b) The maximum principal strain 1250 150 295 260

0.3E E E

= = > E

The solid will fail according to principal strain theory.

(c) Maximum shear stress, max250 ( 150)

200 1302

= = >

The solid is not safe as per shear stress theory.

(d) Total strain energy density

2 21 [250 ( 150) 2 0.3 250 ( 150)]2

uE

= +

253750 260

2E E= >

The solid will fail.

(e) Distortion energy density 2 21

({250 ( 150)} 250 150 )12G

= + + 2

2122500 260

6 6G G= >

Distortion energy theory also predicts failure.

Case III

Let us now consider the solid shown in Figure 3.12(c).

(a) By inspection we may see that the solid is safe according to principal

stress theory (1 < 260).

(b) Maximum principal strain 1200 100 230 260

0.3E E E

= =


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87

Principal Stresses

and Strains The solid is safe according to principal strain theory too.

(c) Maximum shear stress max200 ( 100)

150 1302

= = >

Shear stress theory predicts failure.

(d) Total strain energy density is as follows :

2 21 [200 ( 100) 2 0.3 200 ( 100)]2

uE

= +

231000 260

E E= >

The solid is safe according to strain energy theory.

(e) Distortion energy density is as follows :

( )2 21

[200 ( 100)] 200 10012

duG

= + + 2

2

70000 2606 6G G

= >

Distortion energy theory predicts failure.

If what is safe according to one theory is unsafe according to another theory while whatis safe according to the second theory is unsafe according to yet another theory, you maywonder whether to consider the solid as safe or not. Or should the solid be safe according

to all the theories? That will put a very severe condition for safety. Shall we choose atheory which we like best? Certainly not, we have to find out which theory correctly

represents the failure criteria and choose it. It has been found from experience that theshear stress theory (known as Trescas Theory) suits brittle materials, while, thedistortion energy theory (known as Von Mises Theory) is suitable for ductile materials.

So depending on the nature of the material, the designer can choose the appropriatetheory.

You may also note, that it is not enough to be merely safe (irrespective of the theory), but

that there should be a sufficient margin of safety, defined by the Factor of Safety chosendepending on the nature of the problem.

3.13.3 Equivalent Stress

Equivalent stress is a concept useful in the design of components undergoing a stressfield with multiple components of stress. An equivalent stress corresponding to a given

state of stress is the value of uniaxial stress that will produce the same effect (dependingon the theory used) as that produced by the given set of stress components.

Example 3.12

Consider the state of stress given in Figure 3.12(a).

(a) According to principal stress theory the equivalent stress for this case is

simply 1, i.e. 300 MPa.

(b) Let us consider principal strain theory.

The principal strain introduced in this case is as follows :

1

300 200 2400.3

E E E

= =

This much of strain can be produced in uniaxial tension by a stress of

240 MPa and hence, the equivalent stress according to principal straintheory is 240 MPa.


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88

Stresses in Solids(c) Maximum shear stress max

300 20050 MPa

2

= = .

This could be produced by a uniaxial stress of 100 MPa and henceequivalent stress according to shear stress theory is only 100 MPa.

(d) Strain energy density, 2 21

[300 200 2 0.3 300 200]2

uE

= +

47000

E= . . .

(i)

If the equivalent uniaxial stress is e, then2

2

euE

= . . .

(ii)

i.e.2

47000

2

e

E E

=

2 47000 306.6 MPae = =

(e) Distortion energy density

2 2 21 70000[(300 200) 300 200 ]12 6

duG G

= + + =

If equivalent uniaxial stress is e, then

22 21 ( )

12 6

ed e eu

G G

= + =

i.e.

2

700006 6

eG G

=

70000 264.6 MPae = =

Even though we have earlier analysed whether the solid is safe or not according to

different theories, only by evaluating the equivalent stress, we are able to get anidea of the margin of safety according to each of the theories.

SAQ 7

Evaluate the equivalent stress values given by different theories of failure for thestates of stress given in Figures 3.12(b) and (c), taking Poissons Ratio as 0.3.

3.13.4 Factors of Safety and Design

Factor of safety with respect to a state of stress, according to any chosen theory may bedefined as the ratio of the yield/failure stress of the material in uniaxial strength test to

the equivalent stress according to the theory.

Let us illustrate how the concept is applied in design.

Example 3.13


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89

Principal Stresses

and StrainsA mild steel bolt is to be designed to simultaneously carry an axial tensile force of

17 kN along with a shear force of 12 kN. Taking y = 260 MPa and Poissons ratiov = 0.32, find the required diameter of the bolt according to various theories of

failure, if the required factor of safety is 2.0.

Solution

Safe stress260

130 MPaFactor of Safety 2

y= = = .

Accordingly the equivalent stress should be restricted to 130 MPa.

LetA be the area of cross-section of the bolt.

The state of stress will be defined by the components,

17000 12000, 0 andx y xy

A A = = =

Principal stresses2 2

1,2

17000 17000 12000

2 2A A A

= +

1(8500 14705.44)

A=

i.e. 1 223205.44 6205.44

andA A

= =

(a) Design according to Principal Stress Theory

Here, 1 = 130 MPa, as we have,23205.44

130A

=

Area required 223205.44

178.5034 mm130

= =

Required diameter of the bolt178.5034 4

15.075 mm

= =

.

(b) Design according to Principal Strain Theory

According to this theory, we have 1 2130v

E E E

=

i.e.23205.44 0.32 ( 6205.44) 130

AE AE

=

E

25191.181 130AE E

=

225191.181

193.78 mm130

A = =

Required diameter of the bolt4 193.78

15.71 mm

= =

(c) Design according to Shear Strain Theory

According to this theory, we have, 1 2max130

2 2

= =

23205.44 ( 6205.44)130

2 2

A A

=


32/35

90

Stresses in Solidsor

29410.88 130

2 2A=

or229410.88 226.24 mm

130A = =


16.9722 mm

= =

.

(d) Design according to Strain Energy Theory

According to this theory, we have,

22 21 2 1 2 2 3 3 1

1 130[ 2 ( )]

2 2v

E E + + + =

2 21 2 1 22 . 130v + =

2

2 2223205.44 6205.44 23205.44 6205.442 0.32 130

A A A A

+

=

i.e. 22

669159909.2130

A= =

1

2 2

2

669159909.2198.986 mm

130A

= =

Required diameter of the bolt

1

24 193.98615.917 mm

= = .

(e) Design according to Distortion Energy Theory

According to this theory, we have,

22 2 2

1 2 2 3 3 1

1 1[( ) ( ) ( )

12 6G G + + =

30

or2 2 2

21 2 2 1( ) 1302 2 2

+ + =

or 2 2 21 2 2 1( ) 2 13 + + = 2

0

2 2 2223205.44 ( 6205.44) 23205.44 6205.44 2 130

A A A A

+ + =

i.e. 22

14419997942 130

A=

1

22

2

1441999794206.8 mm

2 130A

= =


16.217 mm

= =

.

NoteWhatever be the given state of stress (or even loading) once it is reduced to

the state of principal stress components 1, 2 and 3 the member can bedesigned according to any theory.


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91

Principal Stresses

and StrainsSAQ 8

(a) Design the bolt of Example 3.14 by taking y = 250 MPa,Poissons ratio = 0.333 and Factor of safety = 1.75.

(b) Design a bolt with the material described in Example 3.14 for a tensile force

of 22 kN and a shear force of 14 kN, i.e. with y = 260 MPa, v = 0.32 andFactor of safety = 2.

(c) A bolt of 16 mm diameter made of a material with y = 260 MPa andPoissions ratio = 0.32 is subjected to an axial tensile force of 18 kN and ashear force of 11.6 kN. Evaluate the factor of safety for the bolt according to

various theories of failure.

3.14 SUMMARY

This unit is a vital link in the analysis of solids so as to ensure safe design of differentcomponents of structures or machines or other systems. Here, you were exposed to adeeper insight into the implications of a given state of stress. You have learnt how to

evaluate the stress components on different planes and also to find the extreme values ofstress components. In addition, a cursory treatment of the methods of establishing the

state of stress in simple cases of combined loading is also provided. This study should behelpful in understanding a given loading situation and its bearing on the strength of solidinvolved.

Finally, an introductory treatment of different failure theories has been provided with

suitable illustrative examples of analysis and design. Rightly, we have come to the closeof Block 1 which undertakes a treatment of simple cases of stresses and strains on simple

members.Now, you are armed with the knowledge and skill adequate enough for undertaking a

study of more complex systems of structures and developing the capability for designingsuch systems, an activity which may be considered as backbone of engineering field.

3.15 ANSWERS TO SAQs

SAQ 1

On plane DB, n = 12.31, nt= 18.462 both unsafe; when x = 32.494, the jointwill fail in tension.

SAQ 2

(a) tan 22

x y

xy

=

2

2max

2

x yxy

= +

(b) 1, 2 = 49, 41

1, 2 = 71.565, 18.435

(c) For = 30, n = 9.3827 and nt= 44.677

For = 45, n = 31 and nt= 36


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92

Stresses in Solids For = 75, n = 48.677 and nt= 5.383

SAQ 3

n = 93.57 and nt 42.93

SAQ 4

(a)

o o

1,2 1,29, 3, 45 , 45 = = +

o o1,2 1,245, 45, 13.28 , 76.72 = =

o o1,2 1,219.5, 10.5, 36.65 , 53.35 = =

(d) It will be circle of radius 10 with centre at origin.

(e) It will be just the point (x, 0).

SAQ 5

(max) 58.513x = ,

(max) 40.953xy = , and

1 100.68 = .

SAQ 6

Required diameterD = 92.67 mm.

SAQ 7

Equivalent Stress (MPa) According to

Case as inFigure

PrincipalStress

Theory

PrincipalStrain

Theory

ShearStress

Theory

StrainEnergy

Theory

DistortionEnergy

Theory

3.12(b) 250 295 200 327.872 350

3.12(c) 200 230 150 249 264.6

SAQ 8

(a) and (b)

Required Diameter of Bolt in mm According to

ProblemPrincipalStress

Theory

PrincipalStrain

Theory

ShearStress

Theory

StrainEnergy

Theory

DistortionEnergy

Theory

SAQ 8(a) 14.4 15.01 11.45 14.93 15.47

SAQ 8(b) 16.8 17.42 13.21 17.61 17.91

(c)

Theory

Principal

Stress

Theory

Principal

Strain

Theory

Shear

Stress

Theory

Strain

Energy

Theory

Distortion

Energy

Theory

Equivalent

Stress

177.78 126.82 73.02 129.62 134.16

Factor of

Safety

2.2075 2.05 3.56 2.005 1.938


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Principal Stresses

and Strains

FURTHER READING

Prasad, Jainti, Strength of Materials, CBS Publishers and Distributors.

Timoshenko, Stepen, Strength of Materials Part I and II, CBS Publisher and

Distributors.

Schauns Outline Series, (1989), Strength of Materials,Second Edition, McGraw-HillBook Company.

E. P., Popov, (1993),Mechanics of Materials, Second Edition, Prentice Hall of India Pvt.Ltd.

G. H., Ryder, (1993), Strength of Materials, Educational Low Priced Books Scheme.

Case, John, L., Ross, T. F., Carl, (1993), Strength of Materials and Structures, ThirdEdition, Educational Low Priced Books Scheme.

Principle Stresses & Strains

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Transcript of Principle Stresses & Strains