Topic 5 - Stress & Strain (1)

7/27/2019 Topic 5 - Stress & Strain (1)

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Chapter 5

STRESS AND STRAIN5.1 INTRODUCTION5.2 STRESS5.3 PLANE STRESS5.4

MORHS CIRCLE FOR PLANESTRESS

5.5 THIN-WALLED PRESSUREVESSELS

5.6 STRAIN5.7 PLANE STRAIN5.8 MOHRS CIRCLE FOR PLANE

STRAIN

5.9 STRAIN ROSETTE


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5.1 Introduction

The notion of stress originates from ourdesire to quantify internal or external forces

distributed, respectively, in a body or along its

boundary. Stresses are those forces distributed

over an infinitesimal unit area cut out of a body in

certain directions or over an infinitesimal unit

area on the bounding surface. The study of

stress-related problems, in general, is referred to

as kinetics.

Stresses may be related to strains in solids or

rates of deformation in fluids through constitutivehypotheses. These hypotheses relate the variation

of stress with respect to the variation of strain,

which we sometimes called stress-strain

relationship. These relationships are important in

the description of mechanical properties ofmaterial, such as hardening, hysteresis etc.

Strain, on the other hand, is a geometrical

concept and is associated with the geometrical


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change of a point in the material. The mechanics

of deformations and strains is referred to as

kinematics.

5.2 Stress

Stress is defined mathematically as

. (5.1)

Thus, the concept of stress is related to the

concept of point in the space. The most general

state of stress at a point may be represented by six

components,

as shown in the figure below:


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Fig. 5.1

5.3 Plane stress

Plane stress is a state of stress in which two faces

of the cubic element are free of stress as shown inFig. 5.2.

Fig. 5.2


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Thus a structural member is defined as in the

state of plane stress when

(5.2)

State of plane stress can occurs

1.In thin plate subjected to forces acting in themidplane of the plate.

2.On thefree-surface of a structural element ormachine component, i.e. at any point of the

surface not subjected to an external force, as

in the cantilever beam shown in Fig. 5.3

Fig. 5.3


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5.3.1 Transformation of plane stress

Consider the conditions for equilibrium of aprismatic element with faces perpendicular to the

x,y,x andy axes, as shown in Fig. 5.4

Equilibrium of force in thex axis yields

(5.3)

Fig. 5.4


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Equilibrium of force in they-axis yields

(5.4)

Solving (5.3) for and (5.4) for , we

have

(5.5)

(5.6)

Recalling now

,


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and

,

we can rewrite (5.5) as

. (5.7)

Similarly, we can rewrite (5.6) as

2cos2sin2'' xy

yx

yx+

=

(5.8)

To obtain , we replace in (5.7) with

that the y-axis forms with the x-axis.

Since and

we have


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(5.9)

Adding (5.7), (5.8) and (5.9) together,

(5.10)

Since , Eq. (5.10) tells us that thesum of the normal stresses exerted on a cubic

element of material is invariant with respect to

the orientation of that material.

5.3.2 Principal stresses

Eqs. (5.7) and (5.8) are in fact parametric

equations of a circle. By eliminating from Eq.

(5.7) and (5.8), we arrive at

.


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Setting

, (5.11)

we now have

(5.12)

This is shown in Fig. 5.5.

Fig. 5.5

This shows us that the Principal stresses occur

on the principal planes of stress with zero


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shearing stresses. The min. and max. stresses are

given respectively as

(5.13)

and the principal plane inclined at an angle

(5.14)

Notice that defines two angles separated by

90o

.

This is shown in Fig. 5.6.


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Fig. 5.6

From (5.12), if , shearing stress is

maximum, i.e.

(5.15)

and this occurs at


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(5.16)

Note that defines two angles separated by 90o

and offset from by 45o

. This is shown in Fig.

5.7.

Fig. 5.7

The corresponding normal stress is given by

(5.17)


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Example: For the state of plane stress shown,

determine (a) the principal planes, (b) the

principal stresses and (c) the maximum shearingstress and the corresponding normal stress.

Soln:

(a)Using Eq. (5.14), .(b)Using Eq. (5.13),


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(c)Using Eq. (5.15), .(d) Using (5.16), .

The corresponding normal stress is

computed using (5.17), which is


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Example: A single horizontal force P of 150lb

is applied to D of lever ABD. Determine (a) the

normal and shearing stresses on an element atpoint H having sides parallel to the x and y axes,

(b) the principal planes and principal stresses at

H.


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Solution

Evaluate the normal and shearing stresses at H.


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5.4 Mohrs circle for plane stress

With the physical significance of Mohrs circlefor plane stress established, it may be applied

with simple geometric considerations. Critical

values can be estimated graphically.

For a known state of plane stress ,

plot the points Xand Yand construct the circlecentered at C, as shown in Fig. 5.8

Fig. 5.8


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The principal stresses are obtained atA andB.

The direction of rotation ofOx to Oa is the same

as CXto CA.

Fig. 5.9

With Mohrs circle uniquely defined, the state of

stress at other orientations may be depicted. For

the state of stress at an angle w.r.t thexy axes,

construct a new diameterXY and an angle

w.r.t.XY.


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Fig. 5.10

Fig. 5.11

Normal and shear stresses are obtained from the

coordinatesXY.


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For centric axial loading on thin plate:

Fig. 5.12

For torsional loading:

Fig. 5.13


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Example: For the state of stress shown, (a)

construct Mohrs circle and determine (b) the

principal planes, (c) the principal stresses and (d)the maximum shearing stress and the

corresponding normal stress.

Soln:

(a)


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The Mohrs circle is thus plotted as shown:

(b) The principal planes are


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(C) The principal stresses are, from Mohrs circle

(d) The maximum shearing stress, from Mohrs

cicle, is . This occurs at

The corresponding normal stress, from Mohrs

circle, is .


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Example: For the state of stress shown,

determine (a) the principal planes and the

principal stresses, (b) the stress componentsexerted on the element obtained by rotating the

given element counterclockwise through 30o

.

Soln:


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(On x-axis, the normal stress is +ve and shear

clockwise. On y-axis, the normal stress is +ve

and shear anticlockwise. Thus XY).

(a)Principal planes and stresses


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(b) Stress components at 30o

counterclockwise


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From Mohrs circle


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5.5 Thin-walled pressure vessels

5.5.1 Cylindrical vessels

The plane stress analysis we studied previously

can be applied to the stress analysis of

thin-walled pressure vessels problems. Why?

Consider a cylindrical vessel of inner radius rand

wall thickness t, containing a fluid under pressure,

Fig. 5.14.

Fig. 5.14

Because the vessel is symmetry, no shearing

stress is exerted on element. The stresses on the

surface of the cylinder are therefore the principal

stresses. The principal stresses on the surface of


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the cylinder is termed

.

We want to determine the stresses exerted on a

small element of wall sides respectively parallel

and perpendicular to the axis of the cylinder, Fig.

5.15.

Fig. 5.15

5.5.1.1 Hoop stress

To obtain the magnitude of hoop stress, taking

force equilibrium in the z-axis, Fig. 5.15, yields


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Solving for gives

(5.18)

5.5.1.2 Longitudinal stress

To obtain the longitudinal stress, we cut a section

perpendicular to the axis of the cylinder.

Fig. 5.16


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Equilibrium of force in x-axis gives

Solving for yields

(5.19)

Remarks:

1.Notice that

2.The limit of validity of Eq. (5.18) and (5.19)must be such that

(5.20)

Notice that the rterm in (5.19) is in fact a meanvalue

(5.21)


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Using (5.21), we see that

.

We can expand the RHS using Binomial theorem

and get

Our argument runs like this:

(a)If , , which cannot be true.(b)If , if we take the first-order

approximation and this cannot be true also.

(c)The only possibility is that . Notethat the argument is insufficient


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since this can also means that .

This is the fundamental idea behind

approximation and analysis.

5.5.1.3 Mohrs circle

Returning now to our problem at hand, the hoop

and longitudinal stresses we found can be plotted

on a Mohrs circle as follows:

Point A corresponds to hoop stress and point B

corresponds to longitudinal stress.


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Fig. 5.17

The maximum in-plane shear stress can be

computed as

(5.22)

The maximum out-of-plane shear stress

corresponds to a 45o

rotation of the plane stress

element around a longitudinal axis, giving


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(5.23)

5.5.2 Spherical vessels

We now consider a spherical vessel of inner

radius rand thickness t, containing a fluid under

gage pressurep, Fig. 5.18.

Fig. 5.18

Because of symmetry, there is no shear stress on

the surface of the sphere. Therefore the stresses

exerted on the sphere must be the principal


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stresses and are of equal magnitude, i.e.

(5.24)

Consider a half-section of the sphere as shown in

Fig. 5.19. We want to find out the stress acting on

the thin wall.

Fig. 5.19

Equilibrium of force in the x-axis


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Therefore,

(5.25)Maximum out-of-plane shearing stress

(5.26)

and this can be shown on a Mohrs circle as

follows:

Fig. 5.20


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With this, we conclude our study on stresses,

plane stress. Next, we look at strain the concepts

associated with it.

5.6 Strain

Strain is a geometrical or kinematical concept

that describes the deformation of a body. Theterm deformation signifies the entire geometric

change by which the points in a body in the initial

state with all loads absent go to another

configuration as a result of the action of loads.

The aforementioned initial state we shall call theundeformed state, and the subsequent state

occurring in the presence of loads we call the

deformedstate. The deformation, so defined, will

be seen to include the following contributions for

each element of a body:

1.Rigid-body translation and rotation2. A dilatation contribution from changes in

geometry associated with the volume change

of the element


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3.A distortion contribution from the remainingchanges in geometry of the element, which

includes e.g. change in angularity betweenline segments. This contribution is

sometimes call deviatoric changes.

Before we define strain, it is important to define

displacement field vector, u. Let us consider the

position vector of a point P in the undeformed

state, . After deformation, point P moves to a

new point P with position vector . The

displacement field is thus defined as

(5.27)

The displacement field is obviously a function of

both and time, t.

Following this, strain is thus defined as

(5.28)


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The concept of strain is seen to be associated with

the concept of point. The most general state of

strain at a point can be represented by sixcomponents,

5.7 Plane strain

Under special conditions, three-dimensional

strain analysis can be analyzed as a

two-dimensional one. This is called plane strain

analysis.

A structural member is in a state of plane strain

when one normal strain and two shear strain

components are zeros, i.e.

(5.29)

In fact, to be strictly plane strain, we also require


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that .

Thus, in laymans term, we say plane strainoccurs when deformations of the material take

place in parallel planes and are the same in each

of those planes.

For example, a plate subjected along its edges to

a uniformly distributed load and restrained fromexpanding or contracting laterally by smooth,

rigid and fixed supports, Fig. 5.21.

Fig. 5.21

Another example is a long bar subjected to


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uniformly distributed transverse loads. State of

plane strain exists in any transverse section not

located too close to the ends of the bar.

Fig. 5.22

A thick-walled circular cylinder constrained at

both ends is also another example of plane strain.

5.7.1 Transformation of plane strain

Similar to the plane stress, it can be demonstrated

that plane strain is invariant under transformation

of coordinate system, i.e.

.

This shows that strain, like stress, is a tensor

quantity that obeys the tensors transformation


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law.

Performing the similar analysis on strain

(Cauchys Tetrahedron Lemma), we can obtain

(5.30)

(5.31)

(5.32)

5.8 Mohrs circle for plane strain

Equations (5.30), (5.31) and (5.32) are

parametric equations of a circle and thus can be

represented by Mohrs method.

Defining now

, (5.33)


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and

, (5.34)

we can plot the strain circle as shown in Fig. 5.23.

Fig. 5.23

The principal axes can be computed as


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, (5.35)

and the maximum and minimum strains can be

evaluated using the formula

(5.36)

The maximum in-plane shearing strain is given

by

(5.37)


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5.9 Strain rosetteThe strain gauge is the most common device for

measuring strain, Fig. 5.24

Fig. 5.24

A single gauge will yield only a normal strain inthe direction of the gauge. Thus in application,

we must use a cluster of gauges or strain rosette

to give pertinent information about the state of

strain at a point, Fig. 5.25.


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Fig. 2.25

If the rosette is aligned at 45o

with respect to each

other, and are obtained directly and is

obtained indirectly as

(5.38)

If the rosette is aligned at some angles

with respect to each other, normal and shearing

strains may be obtained from normal strains in

any three directions via the formula

(5.39)

(5.40)

(5.41)


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Example: A strain rosette aligned 45o

with

respect to each other recorded the following

readings, G1=0.002, G2=0.001 and G3=-0.004.What are the principal strains?

Solution:

We let G1 corresponds tox-axis, G2 to OB, G3 toy-axis

From formula

Notice that (dont get confused!)

To obtain the principal planes, we have


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To obtain the principal strains, we have

Thus .

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Topic 5 - Stress & Strain (1)

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