Chapter 6 Combined Stresses...(SOM-201)

8/8/2019 Chapter 6 Combined Stresses...(SOM-201)

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An Introduction to Mechanics of

Materials

Vijay Gupta

Lovely Professional University, Punjab


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shear force which is statically equivalent to the sectional shear

distribution must be applied.

The strain energy in slender beams is dominated by the energydue to bending stresses and is given by . / .

Problems

6 Combined Stresses and Strains

6.1 Introduction

We have in the previous chapters discussed cases of structures where the

loading was simple. We started in Chapter 2 with simple axially-loaded

members which resisted the loads through axial stresses. We dealt inChapter 3 with shafts subjected to torsional loads such that there were

only shear stresses on the members. In Chapter 5 beams were considered.

Of course in these there were tensile as well as shear stresses but, as was

shown, the shear stresses were much smaller than the bending stresses forslender beams. In the cases where longitudinal stresses were dominant we

designed structures such that the stresses within the members did not

exceed the safe level of such stresses. This safe level was determined bysubjecting a specimen of the material to tensile test, and was termed as

the tensile strength of the material. Similarly, a shaft was designed so thatthe maximum shear stress did not exceed the safe level for the material of

the shaft.

Real structures are not that simple. Considerthe structure shown in Fig. 6.1. The structure

near the root is subjected to both torsion andbending (in addition to shear) loading. At a

point near the root, there will be all types of

stresses: tensile and shear, and along manydirections. How does one analyse this case to

Fig. 6.1


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determine if the structure is safe?

We had earlier in Sec. 1.6 introduced a double-index notation for the

stress components recognizing three mutually-perpendicular planes and

three components of

stresses on each plane.This means that there

are a total of nine

components of stresses

(Fig. 6.2).It will be shown in this

chapter that given thenine components of

stress at a point in a

material with three

arbitrary mutually-perpendicular planes, it

is possible to determine

the stress components

on any other three mutually-perpendicular planes. This is termed as

transformation of stresses and is a very useful tool in the analysis ofstructures.

Thus, the state of stress at any given location in a structure can be fully

described by specifying the three vectors (with three components each) onany arbitrary three mutually perpendicular faces (specified by the

directions of their outward normals). We conventionally express this inthe form of a matrix. This matrix

73has the three stress vectors as its three

rows.

73 This matrix is known as stress tensor. Just as a vector has a direction

associated with it, a tensorhas two directions associated with it. Whereas each

element in a vector has an associated coordinate direction, each element of atensor has two associated directions, one, the direction of the normal to the plane

on which it acts, and the other, the direction of the stress component itself.

We had further seen in Sec. 1.7 that the shear stresses on complementary

planes are equal in magnitude, so that instead of nine stress components

only six are required. This reduces the number of distinct stress

components to six.

We, in this chapter, present a framework for analysing the structure withcombined stresses (and strains). The strategy begins by realizing that it is

possible to transform the stresses (and strains) obtained in one set of axes

to any arbitrary set of axes, then to determine stresses in those axes where

the stresses picture is simpler. Once that is done, we can compare thesestresses with the maximum prescribed stresses for the material of thestructure.

Example 6.1 Wire over a pulley

Fig 6.3 shows a steel wire of diameter 1

mm that goes over a pulley of diameter500 mm and carries a load of 100 N as

shown. Determine the stress distribution

in the wire cross-section atAA.

Solution:

The section atAA is subjected to bending

as well as direct pull by the load of 100

N. The bending stresses are evaluated

by noting that the radius of curvature of the bent wire is (500 + 1)/2 mm.

Using this value of in Eq. 5.6, we get

The maximum value of this stress is at y = 0.0005 mm. Thus, the

maximum bending stress in the wire is 399.2 MPa, compressive on the

inside, and tensile on the outside as shown.

The stress due to the direct load is tensile everywhere equal to 100 Ndivided by the re of the wire, or 100 N/( 2 m2) = 127.3 kN.The maximum compressive stress on the inside, thus, is (399.2 127.3)

or 171.9 MPa, and the maximum tensile stress on the outside is (399.2 +

Fig. 6.3 Wire over a pulley

Fig. 6.2 Stress components


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127.3) or 426.5 MPa. This level of tensile strength is quite high and

probably exceeds the tensile strength of most steel wires74

.

Example 6.2 A spirally-wound cylindrical vessel75

Consider a spirally wound cylindrical vessel of diameter 40 mm (Fig.

6.4a), much in the fashion of the cardboard tube discussed in Example1.6. The vessel is

made from a thin

copper strip of width

30 mm wound

spirally. The edgesof the strip are brazed

together. The

diameter of the

cylinder is 40 mmand the wall

thickness is 2 mm.

What is the shearstrength of the brazed

joint if the internal pressure that this cylinder can withstand is 3MPa.

74Suppose we increase the wire diameter to 2 mm. Would the stresses

decrease? Let us find out. Since the radius of curvature of the wire over

the pulley has changed little (from 0.2505 m to 0.251 m) the bending

stresses increases due to change in the value ofy (from 0.0005 m to 0.001

m). Thus, the maximum value of bending stress changes to 798.4 MPa.Since the area of cross-section of the wire increases by a factor of 4, the

direct stress becomes one-fourth of the earlier value, or 31.8 MPa. Thus,

the maximum stress in the wire now is 830.2 MPa tensile! A veryinteresting result, indeed. Increasing the wire diameter result in the

increase in stress level, and the wire is sure to fail now.

We need to decrease the wire diameter to make it safer (if we cannot

increase the pulley diameter).

75 This problem can be solved in a straight forward manner by using the Mohr

circle approach introduced in Sec 6.4

Solution:

The hoop stress for a thin closed cylinder was evaluated in Sec. 1.8 as

pr/2t ( ), whereas the longitudinal stress zz wasevaluated aspr/t( ).As was discussed in Example 1.6, the angle of the spiral can be

obtained by imagining that as the brass strip is wound up, the strip shouldadvance in the axial direction a distance equal to the width of the strip as

we go around the tube through one circumference, D as shown in Fig.

6.4c. This angle is, therefore,

.Following the strategy adopted in Example 1.6 let us consider an

infinitesimal elementABCat the seam of the cylinder where the side ABis along the circumferential () direction,BCis along thez-direction, and

AC coincides with the seam (Fig. 6.4b). Clearly, is the angle evaluated above. Let the area of the face A be taken as unity. The area

of faceACis then Atan, and that of face ACis Acosec. Equilibrium

of the element requires:

On simplification we get

and

For the values ofzz, and calculated above, the values of nn and nt

are obtained as 23.2 MPa and = 7.5 MPa, respectively. Thus, the strengthof the brazed material in shear is just 7.5 MPa.

6.2 Plain stresses

As mentioned earlier, we require a total of six stress components to

specify the state of stress at a location in a structure. But in all theexamples we have dealt with so far not all components are present. This

fortunately is the situation in many problems of interest. It is seen that

Fig. 6.4 Stresses in a closed cylinder


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many of the stress components are either absent or are so small (in

comparison to the other components) that they can be neglected.

Let us first consider the simpler case of plane stresses. A structure (or a

structural element) is said to be under the condition ofplane stress if all

the stress vectors are withinone plane. This implies that

there are no stress

components in one the

coordinate directionperpendicular to that plane.The lamina of Fig. 6.5 is

under the state of plane

stress since there are no

components in the z-

coordinate direction76

.

Further, the use of equivalence of shear stresses on complementary planes

(i.e., xy = yx) reduces to number of distinct stress components in the case

of plane stress to three: xx, yy and xy.

6.3 Transformation of plane stressesWe will now show that given the three stress componentsxx, yy and xy at

a point on two mutually perpendicular planesx andy in the case of plane

stresses, it is possible to determine the stresses on any other plane at thatpoint. We have already done such an exercise for the simpler cases of

uniaxial stress (in Sec. 1.6) and for the case of plane stress without the

shear stress component xy ( in example 6.2). We follow exactly the same

approach here.

76It is interesting to note that the shear stress components on thez-plane, zx and

zy, which are notin thez-direction, must also vanish in the case of plane stress.

Why?

Consider the material element shown under plane stress in Fig. 6.6a. The

stresses acting on the x- and y- planes are shown. The equivalence of

shear stresses on complementary planes requires xy= yx. As the material

is under plane stress there are no z-components of stresses. Let us now

attempt to determine the stresses on planes x and y, rotted through anarbitrary angle , counter-clockwise as shown in Fig. 6.6b. Let us, for this

purpose isolate an infinitesimal wedge of material as shown in Fig. 6.6c.

Here the facesAB (of length x) andBC(of length y) are aligned alongthex- andy- directions. The face ACis aligned along the x-direction at

angle , and hence is ay-plane. Its length lis clearly equal to x/cosory/sin. Fig. 6.6c shows the forces on each face (of a unit length in the z-

direction). The equilibrium of the element gives:

(a)

() () (b)

Replacing lwith x/cosor y/sinwhere appropriate, we can simplify

these two equations to:

( ) (c)

(d)

We could, in a similar fashion, draw up a wedge of material whichexposes the x-face, and obtain a similar equation for the stress

component xx:

(e)

Eqs. a-c are conventionally cast in the following form77

which lead

directly to a graphical construction known as Mohr circle:

77 We have used sin2and cos2relations in Eqs. a-c for the purpose.

Fig. 6.5 Plane stress

(a) (b) (c)Fig. 6.6 Transformation of plane stress


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. / . / (6.1)

. / (6.2)

. / . / (6.3)The significance of this result is that it shows that it is possible to

calculate the stress on any arbitrary plane xor ygiven the stresses ontwo mutually perpendicular planesx andy.

In 1882, Otto Mohr, a French engineer devised a graphical procedure toimplement the above transformation which is quite useful. It is described

below.

6.4 Mohr circle

A Mohr circle is a special construction which helps in determination of

stress vectors on variously oriented planes. It is a plot of the components

of stress vectors on differently oriented planes at a location. The tensile

stresses are plotted along the horizontal axis while the shear stresses areplotted along the vertical axis.

Consider the state of stress given in Fig. 6.7.

1. The plane Xis plotted with coordinates (xx, xy) and plane Yisplotted with coordinates (yy, +xy). Note the difference in the

sign of the shear stress for the two planes.

For the purpose of understanding this sign convention, let us firstintroduce the concept ofx-like axes andy-like axes. The convention laysdown that a positive y-like direction is 90

oclock-wisefrom the positive x-

like direction. Thus, all axes marked as A in Fig. 6.8 are x-like, while all

axes marked asB arey-like.

The convention for plotting of shear stresses is the following:

A positive shear stress on an x-like face78

is plotted downward

while a positive stress on a y-like face is plotted upwards.Reverse for the negative shear stresses.

2. Join the lineXY. Let it intersect the axis at point C. With Cascentre and CXor CYas radius draw a circle as shown. This is theMohr circle. It should be easy to see from the construction of Fig.

6.8 that the co-ordinates of the centre Care [(xx+ yy)/2, 0], and

the radius of the circle is . /

78Anx-like face is a face with its outward normal along thex-like axis.Fig. 6.7 Mohr circle

Fig. 6.8 A isx-like whileB isy-like in each set


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3. To obtain stress components on a set of planes represented byoutward normals inx- andydirections obtained by rotating the

x- andy- axes clockwise through an angle , rotate the shaded

triangles clockwise through an angle 2(twice the angle throughwhich the axes are rotated. The points X and Y so obtained

represent the stress components on x- andy- planes. This isshown in Fig. 6.10.

The horizontal coordinate of point X is seen to be . Noting that . /, . /, and , we see that this is equal to xx asgiven by Eq. 6.1.

Similarly, the horizontal coordinate of point Y is seen to be . Noting that . /,

.

/, and

, we see that this is equal to yy

as given by Eq. 6.3.

The vertical coordinate of point X is seen as = . / , which is thenegative of as given by Eq. 6.2.

Thus we see that the coordinates of the points XandYgive the

components of the stresses on the rotated planes following the

sign convention stated above79

.

Let us summarize the graphical construction of Mohr circle to determine

the stress components on arbitrary planes:

Plot the stress components onx- andy- planes as pointsXand Yon the-plane using the convention that a positive shear stress on anx-like

face is plotted downward while a positive stress on a y-like face is

plotted upwards. Reverse for the negative shear stresses.

Join points Xand Y to locate the centre Cof the circle on the -axis, and draw the Mohr circle with CX(or CY) as radius.

To determine the stress components on planes which are at aclockwise angle from the original faces, rotate the diameter

YCXthrough an angle 2to obtain pointsXand Y.

The coordinates of pointsXand Ygive the stress components onthe new planes with normals in the X and Y directions. Thesigns of the shear stress components are to be interpreted using

the convention given above.

6.5 Principal planes, principal stresses andMaximum shear stresses

Consider a general plane stress situation as depicted in Fig. 6.10a. The

corresponding Mohr circle is drawn as Fig. 6.10b. The stress components

on thex-plane (xx, xy) are plotted as point Xwith coordinates (xx, xy).

The negative sign with the shear stress is because f the fact that this is an

x-like plane. They-plane is plotted as point Ywith coordinates (yy, +xy)

with the positive sign with the shear stress because this is a y-like plane.

The centre is located, as before, at [(xx+ yy)/2, 0], and the radius of the

circle is

.

/

.

79Namely, a positive shear stress on anx-like face

79is plotted downward while a

positive stress on ay-like face is plotted upwards. Reverse for the negative shearstresses.

(a) (b)

Fig. 6.9 Using Mohr circle to obtain stress components in arbitrary directions


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As we move along the circle from point Xand arrive at another point X,

say angle clockwise from point X, the magnitude of the vertical

coordinate decreases till we reach the point P1. The point X represents a

plane whose outwards normal is inclined at an angle /2 (half the angle

on the Mohr circle). Thus, the (positive) shear stress on this x-like planedecreases till it vanishes at point P1. This point P1 represents a plane

which has a special significance and is termed as a principal plane which

is defined as a plane that has no shear stress component acting on it. Thispoint is at an angle P from the pointXon the Mohr circle, and represents

a plane which is at angle P/2 from the planeXon the physical diagramshown as Fig. 6.10c. The point P2 on the Mohr circle is also a principal

plane. There are two such planes, 90o

apart. Note that the tensile stress 1

is the maximum possible tensile stress and 2 is the minimum possibletensile stress in any plane at the given point

80. These stresses are termed

as the twoprincipal stresses at this point.

80 The fact that principal stress 1 is the maximum tensile stress on any plane is

the reason why principal stresses play such an important role in mechanics of

It is easy to show that the sum of principal stresses 1and 2 is equal to

the sum of the two original tensile stresses81

xx and yy. One can verify

from Fig. 6.10b that

,or

. / . / (6.4)and . / . / (6.5)Planes S1 and S2 represent to other significant planes. These are the

planes with maximum values of shear stress. The maximum value of theshear stress is

. / (6.6)

These planes of maximum shear also carry tensile stress equal to

.

/in each case.

Given below are a few examples on the transformation of stresses which

illustrate some significant results.

Example 6.3 Uniaxial tension

Consider a two-force member such as the cross-linkab of the step ladder

shown in Fig. 6.11a. The member experiences only an axial tension.

Away from the two ends, the material is under uniform tensile stress, sayxx (St. Vennts principle: Section 1.4). Draw the Mohr circle for the

state of stress within this link, and determine the principal stresses and the

maximum shear stress (and the planes they act on).

solids. We shall see that brittle materials fail when the tensile stress at any point

in any direction exceeds the ultimate tensile strength of the material. Thus, to see

if a brittle structural element is safe, we just need to ensure that the maximum

principal stress is less than the ultimate tensile strength.81 In fact, one can establish easily from the Mohr circle that the sum of tensile

stresses on any two mutually perpendicular planes is the same.

Fig. 6.10 Principal planes and planes of maximum shear


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Solution:

Fig. 6.11b shows a small element of this link with the x- and y- faces

exposed, and all the stresses acting on it.

The stress components on the x-plane are (o, 0). This is plotted as the

point X in Fig. 6.11c. Point Y has the coordinates (0, 0), the stress

components on the y-plane. The centre C is midway point on the -axis

with coordinates (o/2, 0). The Mohr circle is drawn with the centre at C

and radius equal to length CX, that is, o/2.

We note from the Mohr circle thatx- andy-planes are the principal planes

themselves (no shear stress), and the maximum shear points S1 and S2 are

90o

clockwise from the Xand Ypoints on the Mohr circle. This implies

that physically these are 45o

from the x- and y-planes. The stress

components on these planes are shown in Fig. 6.11d.

Each of these planes has a tensile stress as well as a shear stress82

, all of

magnitude equal to o/2.

82 It is interesting to note that the planeS1 is anx-like plane. Positive shear stress

on anx-like plane is plotted downwards. Since the shear stress on this plane is

Figs. 6.11e and f illustrate a very interesting point. The first one shows a

rod made of ductile material after failure when subjected to a pure tensile

load, while the second one shows the fracture of a brittle material under

tensile load. The cup-and-cone failure of the ductile material indicates

that this material has failed in shear, because shear is the maximum on aplane inclined at 45

oto the axis. It is quite a general property of the

ductile materials: these are weaker in shear than in tension.

The failure of the brittle material (Fig. 6.11f) perpendicular to the axis

indicates it has failed along the principal plane P1, i.e., it has failed in

tension.Brittle materials are weaker in tension than in shear.

Example 6.4 Simple axial compression load

Two-force compressively loaded structures are very commonly

encountered. The vertical columns of the elevated water tank of Fig.

6.12a have compressive stresses at section ab. Draw the Mohr circle forthe state of stress within this column, and determine the principal stresses

and the maximum shear stress (and the planes they act on).

plotted upwards, it must be negative, or in the negative S2-direction, as the

outward normal to this plane is in the positive-direction.

Fig. 6.11 Uniaxial stresses


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Solution:

Fig. 6.12b shows a small element of this column with the x- andy- faces

exposed, and all the stresses acting on it.

The stress components on they-plne re (o, 0). This is plotted as pointYin Fig. 6.11c. PointXhas the coordinates (0, 0), the stress components

on the x-plane. The centre C is the midway point on the -axis with

coordintes (o/2, 0). The Mohr circle is drawn with the centre at Cand

radius equal to length CX, that is, o/2.

Thex- andy-planes are the principal planes themselves (no shear stress),

and the maximum shear points S1 and S2 are 90o

clockwise from the Xand Ypoints on the Mohr circle. This again implies that, physically, these

are 45o

from thex- andy-planes. The stress components on these planes

are shown in Fig. 6.12d. Each of these planes has a tensile stress as wellas a shear stress, all of magnitude equal to o/2. Note that the plane S1 is

an x-like plane. Positive shear stress on an x-like plane is plotteddownwards. Since the shear stress on this plane is plotted upwards, it

must be negative, or in the negative S2-direction, as the outward normalto this plane is in the positive-direction

Figs. 6.12e illustrates the mode of failure in a brittle material. Since most

brittle material are weak in tension (but not in compression), and since

there are no tensile stresses in this case (the whole of the Mohr circle

lying in the negative space), the material fails in shear. The failure lines

are largely inclined to the vertical axis.

Example 6.5 Hydrostatic Loading

Recall from the elementary fluid statics that in a fluid at rest there are no

shear stresses and the pressure at a point is invariant with the direction ofthe plane. Let us consider a structural element subject to compressive

stress o in bothx- andy- directions and with no shear stress xy. Draw the

Mohr circle and discuss the state of stress.

Solution:

In the absence of shear stress xy, both the planes X and Y have thecoordintes ( o, 0) and are plotted at the same location on the stress

graph. Clearly, the Mohr circle here is reduced to a single point.Consequently, there are no shear stresses on any plane, and the

compressive stress is oin all directions.

Example 6.6 A shaft under pure torsion

Consider a circular shaft subjected to pure torsion as shown in Fig. 6.14a.

It was shown in Chapter 3 that it is convenient to work in the polar

coordinates r, and z, and that the only component of stress in thesecoordinate planes is z (or z). Determine the principal axes and the

principal stresses in the shaft.

Fig. 6.12 Simple uniaxial compressive loading

Fig.6.13 Hydrostatic loading


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Solution:

As was seen in Chapter 3, the maximum shear stress occurs in an element

at the surface of the shaft where , whereR is the radius of theshaft, T is the twisting torque, and Izz is the polar moment of the shaft( Let this be denoted by o as shown in Fig.6.14b.

We note that the stress components on thez-plne re (0, o). Since thisis anx-like plane, the shear stress will be plotted upwards. PointZin Fig.

6.14c represents this plane. Similarly, the - plane which is a y-like plane

nd hs the sme stress components (0, o) is plotted downwards as point

. The Mohr circle is drawn with centre at origin and with radius equal too. Clearly, points P1 and P2 represent the principal planes, there beingno shear stress on them. These are 90

oaway clockwise from the Z- and

- points, respectively on the Mohr circle, or 45o clockwise fron z- and -

axes in the physical plane as shown in Fig. 6.14c.

Figs. 6.14 d and e show the failure of brittle and ductile materials,

respectively. As was discussed in Example 6.4 above, a brittle material is

weaker in tension than in shear. Therefore it breaks83

along the principal

plane P1. A ductile material, on the other hand is weaker in shear, and

therefore it fractures along the plane carrying the maximum shear, i.e., the

z-plane itself.

Example 6.7

Fig. 6.15 shows the stress components on an aircraft panel loaded in its

own plane. Determine the orientation of principal planes and the

principal stresses. Determine also the maximum shear stresses acting at

this point of the panel.

Solution:

The coordinates of pointXrepresenting thex-plane on the stress diagram

are (100, -40), this being an x-like plane. Similarly, the coordinates ofpoint Yare (30,+40). The coordinates of the centre Care [(110 + 50)/2,

83 Looking at the fracture line of the brittle shaft in Fig. 6.14d, can you determine

if the torque this shaft was subjected to was positive or negative?

Fig. 6.14 Stresses in a circular shaft

Fig. 6.15


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0], or (80, 0). The radius of the Mohr circle is . / , or. / 50, all in MPa. The principal stresses are:

- and - - -

For the first principal plane, angle 2=

.

Thus, the normal to the first principal plane inclined at an angle of

53.13o /2, or 26.6

oclockwise to the x-axis. The normal to the second

principal plane is likewise inclined at 26.6o

clockwise to they-axis.

The maximum shear stress at this location is equal to the radius of the

Mohr circle, i.e., 50 MPa.

Example 6.8 A torsional pendulum

Consider the torsional pendulum of

length 1 m shown in Fig. 6.16. The steelwire has a diameter of 1 mm. Can the

steel wire withstand a torsionalamplitude of 60

o, if the weight of the disc

is 50 N. Assume that the wire fails when

the shear stress exceeds 100 MPa, and a

factor of safety of 1.6 is desired.

Solution:

There are two sources of stresses in the wire: the direct tension, and the

shear stresses due to torsion. The direct tension is uniform across the

section of the wire:

,

-

The shear stresses due to twisting given by Eq. 3.2 vary with the radius

with the maximum value occurring at the outer surface.

Fig. 6.16a shows the stress components on an element at the surface of

the wire. Clearly, this is a case of combined stresses. To determine the

maximum shear stress in the material, we need to construct a Mohr circle.

The Mohr circle for this is as shown in Fig. 6.17b. The point Z with

coordinates (63.7 MPa, 41.8MPa) represents the stresses on the z-plane.

As this is an x-like plane, the positive shear stress is plotted downwards.The point with coordinates (0, 41.8 MPa) represents the -plane. The

centre of the circle is at , -, or at (31.8 MPa, 0). Theradius of the circle is

( )

, or 52.5 MPa. The

maximum shear on the wire, therefore, is 52.5 MPa and occurs on a plane

that makes an angle clockwise from thez-axis.This results in a factor of safety of 100 MPa/52.5 MPa = 1.9, well below

the prescribed factor. Hence the wire is safe.

Example 6.9 Combined stresses in a cylindrical pressure vessel

Consider a thin-walled cylindrical pressure vessel of radius R with wallthickness tas shown in Fig. 6.18a. We had determined in Sec. 1.8 that in

the polar coordinate directions r, and z, there are no shear stresses84

,

and that the tensile stresses were given by Eqs. 1.8-1.10 as

,

, and

84 This was established by using symmetry arguments

Fig. 6.16 Torsional Pendulum

(a) (b) (c)Fig. 6.17 Stresses in the wire of torsional pendulum


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Since there are no

shear stresses on

these planes, these

are the principal

planes. Determinethe magnitude and

direction(s) of the

maximum shearstress.

Solution:

Fig. 6.18 b shows a

two dimensional

element of the thin-

walled cylindricalvessel with stresses

shown. Fig. 6.18c shows the corresponding Mohr circle with centre at, - or (3pr/4t, 0), and the radius is or pr/4t.Thus the maximum stresses are equal to pr/4t, positive on the S2 plane

located 45o

anti-clockwise from thez- plane, and negative on the S1 plane

located 45

o

clockwise from thez- plane.

6.6 General 3-D stress

Let us now consider the general case of stress where the stresses are

confined to a plane. This meansthat at a point there are z-components of the stresses as

well. Before considering the

general transformation, let us

consider the case when the new

plane in which we seek todetermine the stresses has a

normal within the x-y plane asshown in Fig. 6.19.

We can determine the stresscomponent on the inclined plane

(with normal in x-direction) by

writing the equilibrium equations. It can be easily verified that xx and

xy are given by the same expressions as for the case of plane stress, i.e.,

Eqs. 6.1 and Eq. 6.2. This happens because in the two equilibrium

equations the contributions of zx and zy on the +z- and z- face cancel

each other. Thus, the construction of Mohr circle for the case of planestress can be used even in this general 3-D case for all planes with

normals in the x-y plane. However, the determination of the principal

planes would not be correct, since the usual Mohr circle would ensure thatthere is no shear stress xy o r yx on these planes. xz will, of course,

occur. This will be zero on these planes only ifxz and yz vanish. In canbe shown that thez-component of shear stress is given by

(6.7)Elaborate schemes for transforming 3-D stress have been evolved. Suffice

it to say here that:

The results given by Eqs. 6.1 and 6.3 and the construction of theMohr circle described above are valid for any orientation ofx-andy- axes within the x-y plane even ifyz, yz and xz are non-

zero.

However, if yz and xz are non-zero, the shear stress on the x-plane is given by Eq. 6.7 above, and the Mohr circle cannot beused for determining the principal planes and principal stresses.

6.7 Displacement and strain

Consider the deformation of a small element ABCD (Fig. 6.20) whichundergoes deformation under the action of a generalized load to acquire a

displaced location ABCDas shown. Its sides AB and AD of originallengths x and y are originally parallel to the coordinate axes. Let us

denote the x- and y- components of displacements of point A of the

element by u and v, respectively. Then the displacements of pointsB and

D can be written as . / and . /,respectively as shown. It is easy to verify that the length of the linesABand AD, that were originally x and y, respectively are now and , respectively. It is easy to see that the linear strainin thex-andydirections are given by:

Fig. 6.19 General 3-D stress

Fig. 6.18 Cylindrical pressure vessel


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and (6.8)It can also beverified, that the

line AB is

inclined to thevertical at an angle

equal to , theminus signindicating that thisrotation is

clockwise.

Similarly, the line

ADis inclined to

the horizontal atan angle equal to

. Thus, theangle between the

two lines is

reduced from a value of /2 by an amount equal to the sum of the two.This reduction in the angle between the two lines perpendicular to each

other to begin with is termed as the shear strain . Thus, (6.9)

We can, similarly, define the rotation of the element (about the z-axis

perpendicular to the paper) as . /.This determination of stress component assumes that the strains are an

order less than one.

6.8 Transformation of plane strains

We have related above the strains in the x- and y- directions to thegradients of the displacements in a material body. We turn next to the

question if we can relate the strains in any other direction to the strains in

the x-and y- directions, just as we related the stresses on an arbitrarilydirected plane to the stresses given in thex-andydirections.

Consider a plane strain situation depicted in Fig. 6.21. Let u and v and

represent, as before, the displacements inx-andydirections. Consider a

line AB of length lalong the x-

axis inclined at an angle to the

x- axis within the material. Letthis line move the locationAB

after displacement. For

convenience of analysis we movethe line AB parallel to itself to

the location AB. This isequivalent to subtracting from the

displacement of the point B the

displacement (u, v) of the pointA.We drop a perpendicularBD from

point B to line AB. For smalldisplacements, length AD length AB, so that DB represents the

elongation of the line AB. To determine this elongation we resort to the

construction of Fig. 6.21 whereinBCis parallel to thex-axis and CB isparallel to the y-axis. The coordinates of pointB are lcos nd lsin.

Clearly, the lengthBC, the additionalx-displacement of point B (over that

of pointA

) is , and the lengthCB,

the additionaly-displacement of point B (over that of point A) is . The elongation DB of line AB then is . The longitudinal strain xxat this point is obtainedby dividing this length by the original length l of line AB. Using the

values ofBCand CB evaluated above, we determine the strain as

. / Recognizing the partial derivatives as the strain components xx, xy and yy

we can write this as

, or onrearranging, . / . / (6.10)

We can, in a similar fashion, obtain the following equations:

Fig. 6.20 Deformation of a small element

Fig. 6.21 Strain in an arbitrary direction.


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. / , and(6.11)

. / . / (6.12)Eqs. 6.106.12 are seen to be identical to the Mohr circle equations, Eqs.

6.16.3 except that the stresses and are replaced by the strains and , and the shear stress is replaced by the shear strain .The significance of these results is that they show that it is possible tocalculate the strains on any arbitrary plane xor ygiven the strains ontwo mutually perpendicular planesx andy, just as it was possible to do so

for the stresses. Thus, the three strain components , describe completely the state of strain at any point. The identical nature

of the two sets of transformation equations suggests that it should be

possible to have a graphical construction for strains as the Mohr circle for

stress described above. The only significant change will be that the axes

for such a construction should be and in place ofand , the axes for

the Mohr circle for stresses. This construction is conveniently termed astheMohr circle for strain and is shown in Fig. 6.22.

We summarize below the procedure for the construction of Mohr circle

to determine the strain components on arbitrary planes:

Plot the strain components onx- andy- planes as pointsXand Yon the-/2plane using the convention that a positive shear strain on anx-like

face is plotted downward while a positive shear strain on ay-like face isplotted upwards. Reverse for the negative shear strains.

Join points Xand Y to locate the centre Cof the circle on the -axis, and draw the Mohr circle with CX(or CY) as radius.

To determine the strain components on planes which are at aclockwise angle from the original faces, rotate the diameter

YCXthrough an angle 2to obtain pointsXand Y.

The coordinates of pointsXand Ygive the strain components onthe new planes with normals in the X and Y directions. The

signs of the shear strain components are to be interpreted using

the convention given above.

Example 6.10 Distortion in computer graphics

An error in a computer-graphics software distorts the images appearing

on the monitor. The horizontal lines are stretched 1% and the vertical

lines are compressed 5% (See Fig. 6.23). Show that little rectangles on

screens may be distortedinto parallelograms.

Determine the maximumangular distortions of any

rectangles, and also the

orientation of these.

Solution:

Even though the image

on the screen is not a

material object, there isnothing in our discussion of the previous section which should restrict the

consideration of deformations and strain from being applied to it. We canthus, look at the problem as one of transformation of strains. In the

coordinate system of Fig. 6.23, the given data can be interpreted as: xx =

0.01, yy= 0.05 nd xx = 0.

The direction of maximum angular distortion is the direction of maximum

shear strain and that can be determined by the construction of the Mohrcircle of strain, as shown in Fig. 6.23. Point X on the diagram with

coordinates (+ 0.01, 0) represent the strains on the x-plane, while point Y

(a) (b)

Fig. 6.22 Using Mohr circle to obtain strain components in arbitrary directions

Fig. 6.23 Distortion in computer graphics


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with coordintes ( 0.05, 0) represent the strins on the y-plane. We

determine the centre of the circle on -axis midway between the two

points and draw the Mohr circle with radius [0.01(0.05)]/2 = 0.03.

The maximum shear strain occurs at planes S1 and S2 located 45o

clockwise from the x- and y- planes, respectively. As is clear from theMohr circle, the maximum angular strain S1S2 equal to two times the

radius of the circle, i.e., 0.06. This is the maximum distortion in the

image, appearing as a decrease in angle between two lines inclined at 45o

to thex- andy- directions.

6.9 Relation among elastic properties of a material

We had in Chapter 2 introduced three elastic properties of materials:

elastic modulus E, Poisson ratio , and shear modulus G. We can now

show that the three are related, so that there are only two independent

elastic properties. Consider a two-dimensional state of stress shown in

Fig. 6.24a where an element is subjected to pure shear stress o. The shear

strain is seen as . We draw its Mohr circle as Fig. 6.24b. Thisshows that the principal planes are inclined at 45

oto the x- and y- axes.

The orientation of this element is shown as Fig. 6.24c. The element has

tensile stress in the principal direction 1, and a compressive stress in the

principal direction 2, both of magnitude o. We next determine the

principal strains 1 and 2 as and . We next draw the Mohrcircle of strain for this state of strain as shown in Fig. 6.24e. Clearly, the

planes inclined at 45o

to the principal directions 1 and 2. These are the

originalx- andy- directions. The shear strain xy is seen from this Mohr

circle is equal to twice the radius, i.e., 2 1, or

. But this is

the same shear strain xy as obtained in Fig. 6.24a. Thus, . This leads to our desired result: (6.13)

Thus, there are only two independent elastic properties.

6.10 Strain gauges

Resistance strain gauges are based on the principle that when a wire is

stretched its resistance increases. Fig. 6.25 shows the layout of a common

type of strain gage. The gauge is laid outon a backing sheet which is bonded to the

structural member whose strain is to be

measured. The gauge consists of a long

wire been folded such that its entirelength is aligned in one direction, but is

confined to a small area. The long length

of the resistance wire increases thechange in resistance for a given strain

increasing the sensitivity of the gauge.The compact size of the gauge ensures

that the gauge responds to the strain in a

small region making it possible to measure the variation of strains withlocation. The two leads are soldered to the two tabs shown and

connected to a resistance measuring device.85

The gauge measures the

85 Such as a Wheatstone bridge circuit.

Fig. 6.25 Resistance strain gauge

Fig. 6.24


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linear strain on the member in the direction of its axis, which is the

direction along which the length of the wire is aligned.

Strain gauges find wide applications in measurement of forces and

acceleration. Many sensors for vibrations and aerodynamic forces use

strain gauges to pickup linear strains and

then convert them into

desired quantities. Fig.

6.26 shows one such

application. This is theschematic of an

experimental set up to

determine the

aerodynamic forces

that excite a model of human vocal chords. Here a strain gauge is appliednear the base of cantilevered reed. The oscillating aerodynamic forces on

the model vocal chord result in the vibrations of the reed. The resulting

oscillating strains are picked up by the strain gauge which is calibrated togive the variations of displacement with time. The nature and magnitude

of the forces can be inferred from the measured rate of growth of the

amplitude ofvibrations.

Though a straingauge measures only

the linear strains, it

can be used

judiciously with

Mohr circle tomeasure shear strain

as well. Fig. 6.27a

shows a shaft

subjected to a puretorque. This will

result in shear strains

as shown for an

infinitesimal element

at the surface of theshaft in Fig. 6.27b.

We determine next the principal directions and the principal strains

through the Mohr circle construction of Fig. 6.27c. We note that these

directions are at 90o

in the Mohr circle plane from the axial direction, or

inclined at 45o

to the axis, and the principle strains are equal to o/2,

tensile on P1 plane and compressive on P2 plane. Thus, the strain gaugeshould be fixed such that it is inclined at 45

oto the axis of the shaft.

Notice that the strain gauge shown in Fig. 6.27a is inclined 45o

counter-

clockwise, and therefore, it will measure linear strain on the principalplane P1. Thus, the linear strain measured by this strain gauge will be

o /2, or the maximum shear strain in the shaft will be twice the linearstrain measured by the strain gauge mounted as shown.

Example 3.11 Measuring pressure in a soft-drink can

We attempt to measure the pressure inside a soft-drink can by pasting a

strain gauge along the axial direction when it is sealed. The can is openedto release the pressure. The resulting strain (as compared to when the can

was pressurized) can be related to the inside pressure of the can.

However, there was an error in fixing the strain gauge and we end uphaving its axis off by an angle of 10

ofrom the axial direction. If the

aluminium can has a radius of 3 cm and a wall thickness of 0.07 mm,what was the pressure inside the can if the measured strain was 202 ,

compressive?

Solution:

This method of measuring the internal pressure of a can is quiteinnovative. When we stick a gauge to a can with pressure (and hence,

with the wall under stresses and tensile strain), the gauge will read zero to

begin with. As the can is popped open the strains on the walls will be

Fig. 6.27 Use of strain gauge to measure shear strain.

(a) Fixing the strain gauge at 45o to a shaft, (b) Shear

stresses on an element at the surface, (c) Mohr circle,

and (d) Princi al directions and stresses

Fig. 6.26 Schematic of a set up to pickup vibrations of

model vocal chords using strain gauge

(a) (b)

Fig. 6.28 Mohr circle for strains for a can


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relieved, but in the process, the gauge will be experience a compressive

strain, exactly of the same magnitude as the tensile strain in the can along

that axis of the strain when the can was pressurized. We will, therefore,

determine the tensile strain (in the can material) along the axis of the

strain gauge when the gage is subjected to an internal excess pressure p,and equate it to the (negative) of the gauge reading when the pressure has

been relieved.

Fig. 6.28a shows the schematic of the strain gauge stuck on the outside of

the can, while Fig. 6.28b shows the corresponding Mohr circle for strains

with internal excess pressurep. For a can of radius rand wall thickness t,

and With these stresses,

0 1, and 0 1

The centre Cof the Mohr circle is located at , or at

. The

radius of the circle is

, or

The axis of the gauge is

inclined 10o

clockwise to the z-axis. Thus, point G representing this axison Mohr circle is located at 20

oclockwise from the point Z as shown.

The linear stress at the gauge location is, therefore, OC Rcos20o, or

, -.The gauge is cemented on the can when the material has this strain. As

the pressure is relieved, this strain in the material disappears but the gaugeis subjected to a strain negative of this value. The measured vlue is 202

. We can, thus, evaluate the can pressure p by equating , - to +202106. Using the value of as 1/3 and ofE

as 70 GPa for aluminium, this excess pressure is determined as 1.77105

Pa, or about 1.7 times the atmospheric pressure.

6.10 Strain Rosettes

Even though the strain gauges respond only the axial strain along the

length of its sensing element, we can use a number of them in

combinations to determine the complete state of stress at a point. A strainrosette (Fig. 6.29) has three strain gauges on the same backing paper. If

the required strain components are xx, yy, and xy, the rosette is cemented

to the structure such that gauge a is aligned

with the x-axis. The linear strains measuredby gauges b and c (aligned at angles and,

respectively to the gauge a) are then given in

terms of the desired strains xx, yy, and xy byEq. 6.10 as

. / . / , and . / . /

These together with the fact that measured stress a is equal to xx give

three equations for the three desired strains components.

To give better accuracy in the determination of stresses, there are two

common types of rosettes available: the 45o-rosette with = 45

oand=

90o, and the the 60o-rosette with = 60o and = 120o as shown in Fig.6.30.

The relevant equations for the 45o-rosette are:

. / . / ,(6.14a)

. / , and (6.14b) . / . / (6.14c)

Solving the three equations, we get:

, , and (6.15)

Fig. 6.29 Strain rosette


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We can similarly obtain the strain components from the measurement ona 60

o-rosette as

, , and (6.16)Example 3.12

6.11 Criteria for Failure

We had seen in Sec. 2.6 that when a specimen of a material is subjected toa tension test it fails. A brittle material fails at a higher level of stress in

compression than in tension. We had also seen that in a tension test while

a ductile material showed a cup-and-cone failure, a brittle material

showed no such failure. As explained in Example 6.3, the cup-and-cone

failure of a ductile specimen suggested that the material failed in shear,86

and the flat failure of a ductile material was suggestive of a failure in

tension.

But all this is only for a tensile loading. What about failure in a puretorsion condition? When we subject a bar to a pure torsion, like the

loadings discussed in Chapter 3 and in Example 6.6, a ductile specimenfractures perpendicular to the axis (see Fig. 6.14f) indicating a failure in

86 Because in a uniaxial loading the maximum shear occurs at 45o to the axis.

shear, while a brittle specimen fails at an angle (see Fig. 6.14e) indicating

a failure in tension.87

A number of theories have been proposed to establish failure criteria for

structural elements in a general state of stress, i.e., when both tensile

stresses and shear stresses are present.

One simple theory known as maximum normal stress criterion postulatesthat the failure occurs when the

maximum tensile (or compressive) stress

in any direction exceeds the tensilestrength of the material. Since the

maximum tensile stresses, positive ornegative, are the principal stresses, the

failure is predicted when any of the

numerical value of the principal stresses

exceeds the yield strength of the material.In Fig. 6.31, thus, if the state of stress

falls within the shaded region the

structure is safe. This theory worries only

about the maximum value of the principal

stress and neglects the effect of the other

two principal stresses as long as they arebelow the yield strength. Nevertheless,

the results on brittle materials with aboutequal strength in tensile and compressive

loadings agree well with this theory.

Note that the yield strength is replaced by

the failure strength for brittle materials.

For brittle materials with vastly differenttensile and compressive strengths, the

Mohr criterion of failure gives fairly good prediction. Here, two Mohr

circles are drawn: one for the uniaxial tension test with the failure

strength (Y)T in tension as one principal stress and 0 as the other, and theother circle for the uniaxial compression test with the failure strength(Y)C in compression as one principal stress and 0 as the other, as in Fig.

87 Because in this case of pure torsion the principal planes where maximum

tensile stresses occurs are inclined at 45o to the axis.

(a) (b)

Fig. 6.30 (a) A 45o-rosette (b) A 60o-rosette

Fig. 6.31 Maximum principal

stress criterion

Fig. 6.32 Mohr criterion for

brittle materials strong in

compression


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6.32. The shaded area of the diagram is the safe zone. A structure

ruptures if its state of stress falls outside the shaded region.

The Maximum shear stress criterion or

Tresca criterion gives fairly good (but

conservative) results for ductilematerials. It postulates that the failure of

a ductile material occurs when the

maximumshear stress at a point exceeds

the yield value of the shear stress.

Value of the maximum shear stress at a

point is related to the values of the twoprincipal stresses 1 and2 at that point.

There are two distinct cases: one, when

the sign of the two principal stresses are the same, i.e., both of them are

either tensile or compressive, and the second, when one of them iscompressive and the other tensile.

Mohr circle for the first case is shown asFig. 6.33. Both 1 and 2 are positive in

this figure. Fig. shows the Mohr circle

for such a case. Here the maximum shearin the material does not occur in the

plane within which 1 and 2 lie, but in a

plane formed by the direction ofprincipal stress 1 and the third direction

(where the principal stress is 0). Themaximum shear stress thus, is 1/2. In a

uniaxial test the material yields at 1 =Y. Thus, 1 (or 2 in case it is larger than1) should be equal to or greater than Yfor yielding. Thus, lines AB and BC in

Fig. 6.34 represent the limiting case.

Any state of stress within the shadedsquare OABC is safe from yielding,while the points outside represent

yielding.

Mohr circle for the other case with the

two principal stresses having opposite signs is shown in Fig. 6.35. The

maximum shear stress in this case is given by || || or (1

2). Thus, the line CD in Fig. 6.33 represents the combination of principal

stresses 1 and 2 for which the maximum shear stress is equal to Y/2.

The boundary shaded area of Fig. 6.34 represents the combinations of1and 2 for which the maximum shear stress (on any plane) is Y /2. The

shaded area, thus, represents the safe zone. The material is expected toyield if the state of its stress falls outside this area. We can, thus, write

the maximum shear stress criterion as:

A structure fails whenever either

|| || or

| |

Let us next consider the specimen subject to pure torsion, a case that wasdiscussed as Example 6.6. We had seen there that the two principal

stresses 1 and 2 were of equal magnitude but of opposite signs. This is

the case represented by point S of Fig. 6.34. Clearly, then, Tresca criteria

suggests that the maximum shear stress that a specimen may be subjectedto before yielding is that which results in 1 = 2 = Y /2. The Mohr

circle for this case, drawn as Fig. 6.14c shows that these values of

principal stresses are obtained when the maximum shear stress in the

torsion test is itself equal to Y /2. But in actual torsion test on mostductile materials, the measured yield shear strength Y is about 0.58Y, a

good 15% higher than predicted by this theory.

Based on this and the fact that most ductile materials are able to sustainvery large hydrostatic

88loads, Von Mises proposed that it is the distortion

of a material, as opposed to its volumetric strain that causes it to yield.The proposed measure is the energy of distortion. We had seen in Eq.

2.17 that the strain energy per unit volume of the material is measured by:

( )From this total strain energy density we subtract the strain energy due to

change in volume to obtain the distortion energy.

If we deal in principal planes and principal stresses and strains, thisequation reduces to:

(6.17)

88 A hydrostatic load is one where a material is subjected to a uniform load in all

directions. See Example 6.5.

Fig. 6.34 Maximum shear stress

criterion

Fig. 6.33 Mohr circle for plane

stress

Fig. 6.35


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Using Eqs. 2.2-2.4, we can write the principalstrains in terms of principal

stresses: , -, etc. Using these strains, weget the total energy density as:

, -(6.18)

Let us next consider the energy of the volumetric strain. For this purposewe resolve the given stresses into hydrostatic stresses and deformation

stresses as in Fig. 6.36. Here we replace the average of the threeprincipal stresses, , by representing the hydrostaticstress. With the three principal stresses equal to , Eq. 6.19 gives thestrain energy of volumetric change as

(6.19)The energy of distortion (per unit volume) is obtained by subtracting uhfrom the total strain energy:

This can be recast as;

, -(6.20)

The theory proposes that a material yields when this distortion energy

(per unit volume) exceeds the distortion energy per unit volume in auniaxial tension specimen stressed to its yield strength.

In uniaxial tension test, 1 = Y, and 2 = 3 = 0.

Thus, the distortion energy at yield point is .

Von Mises yield criterion, thus, predicts yielding when

, - 2 (6.21)The ellipse of Fig. 6.37 is a plot of this for the plane stress condition. A

point inside the shaded portion is safe from yielding. Note that it is lessconservative that the maximum shear stress criterion, and is seen to give

good results in most cases.

Let us consider the torsion test again. At the yield point when = Y, the

principal stresses 1 and 2 are both

equal to Y (by Mohr circle of Example6.6), and therefore, by Eq. 6.21: , or , almostexactly the result obtained for most

ductile material (point S in Fig. 6.36).

Fig. 6.36 Resolving stresses into hydrostatic stresses and deformation stresses

Fig. 6.37 Von Mises yield criterion

Chapter 6 Combined Stresses...(SOM-201)

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Transcript of Chapter 6 Combined Stresses...(SOM-201)