Lect 1&2 Stress and Strain 1

7/29/2019 Lect 1&2 Stress and Strain 1

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Stress and Strain


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Force Equilibrium

Consider an element ofcontinuous (no voids) andcohesive (no cracks, breaks and defects) materialsubjected to a number of externally applied loads asshown in Fig. 2.1a). It is supposed that the memberis in equilibrium.


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Force Equilibrium

If we now cut this body, the applied forces can bethought of as being distributedover the cut area A asin Fig. 2.1b). Now if we look at infinitesimal regionsA, we assume the resultant force in this infinitesimalarea is F. In fact, Fis also a distributed force. WhenA is extremely small, we can say that the distributed

force Fis uniform.


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Force Equilibrium

if we look at the whole sectionedarea, we can say that the entirearea A is subject to an infinite

number of forces, where eachone (of magnitude F) acts overa small area of size A. Now, wecan define stress.


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Stress

Definition: Stress is the intensity of theinternal force on a specific plane passingthrough a point. Mathematically, stress can beexpressed as

Dividing the magnitude of internal force F by theacting area A, we obtain the stress. If we let A

approach zero, we obtain the stress at a point. Ingeneral, the stress could vary in the body, whichdepends on the position that we are concerning.The stress is one of most important concepts thatwe introduced in mechanics of solids


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Normal and Shear Stress

As we known, force is a vector that has both

magnitude and direction. But in the stressdefinition, we only consider the magnitudeof the force. Obviously, this may easilyconfuse us. Lets still take patch A as anexample.

As we can see, force Fis not perpendicularto the sectioned infinitesimal area A. If weonly take the magnitude of the force into

account, apparently, the stress may notreflect the real mechanical status at thispoint. In other words, we need to considerboth magnitude and direction ofthe force


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Normal and Shear Stress

Now lets resolve the force Fin normalandtangentialdirection of the acting area asFig.2.1b). The intensity of the force or forceper unit area acting normally to sectionA is

called Normal Stress, (sigma), and it isexpressed as


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Normal and Shear StressIf this stress pulls on the area it is referred

as Tensile Stress and defined asPositive. If it pushes on the area it iscalled Compressive Stress and definedas Negative.

The intensity or force per unit area actingtangentially to A is called Shear Stress,(tau), and it is expressed as


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Average Normal Stress

To begin, we only look at beams thatcarry tensile or compressive loadsand which are long and slender.

Such beams can then be assumed tocarry a constant stress, and Eq. (2.2)can be simplified to:

We call this eitherAverage Normal StressorUniform

Uniaxial Stress.


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Units of Stress

The units in the SI system is the Newtonper square meter or Pascal, i.e. : Pa =N/m2.

In engineering, Pa seems too small, so we

usually use: Kilo Pascal KPa (=Pa103) e.g. 20,000Pa=20kPa

Mega Pascal MPa (=Pa106) e.g.

20,000,000Pa=20MPaGiga Pascal GPa (=Pa109) e.g.

20,000,000,000Pa=20GPa


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Example 2.1: An 80 kg lamp is supported by a

single electrical copper cable ofdiameter d= 3.15 mm. What is thestress carried by the cable.

To determine the stress in thewire/cable as Eq. (2.4), we need the

cross sectional areaA of the cableand the applied internal force F:


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Allowable Stress

From Example 2.1, we may concern whetheror not 80kg would be too heavy, or say100.6MPa stress would be too high for thewire/cable, from the safety point of view.

Indeed, stress is one of most importantindicators of structural strength. When thestress (intensity of force) of an element

exceeds some level, the structure will fail.For convenience, we usually adopt allowableforce or allowable stress to measure thethreshold of safety in engineering


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Allowable Stress

Moreover, there are following severalreasons that we must take into account inengineering:

The load for design may be different fromthe actual load.

Size of structural member may not bevery precise due to manufacturing andassembly.

Various defects in material due tomanufacturing processing.


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Allowable Stress

One simple method to consider such uncertaintiesis to use a number called the Factor of Safety,F.S., which is a ratio of failure load Ffail(foundfrom experimental testing) divided by theallowable one Fallow

If the applied load is linearly related to the stress

developed in the member, as in the case of using = F /

A, then we can define the factor of safety as a ratio of thefailure stress failto the allowable stress

allow


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` Allowable Stress

Usually, the factor ofsafety is chosen to begreater than 1 in order to avoid the potentialfailure. This is dependent on the specific designcase. For nuclear power plant, the factor of safetyfor some of its components may be as high as 3.

For an aircraft design, the higher theS.F. (safer), the heavier the structure, therefore the

higher in the operational cost. So we need tobalance the safety and cost.

The value ofF.S. can be found in design codes and

engineering handbook. More often, we use Eq.(2.6) to compute the allowable stress:


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Example 2.2:

In Example 2.1, if the maximum allowable

stress for copper is Cu,allow=50MPa, pleasedetermine the minimum size of thewire/cable from the material strengthpoint of view.

Obviously, the lower the allowable stress, the bigger the

cable size. Stress is an indication of structural strength

and elemental size.


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Stress Analysis

In engineering, there are two significant problemsassociated with stress as follows.

Problem (1) Stress Analysis: for a specificstructure, we can determine the stress level.

With the stress level, we then justify the safetyand reliability of a structural

member, i.e. known sizeA and load F, to determinestress level: = F A

Problem (2) Engineering Design: Inversely, wecan design a structural member based on theallowable stress so that it can satisfy the safetyrequirements, i.e. known materials allowablestress allowand load F, to design the element

size: allow A F


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Deformation

Whenever a force is applied to a body, itsshape and size will change. These changesare referred as deformations. Thesedeformations can be thought of being

eitherpositive (elongation) ornegative (contraction) in sign asshown in Fig. 2.2.


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Deformation

It is however very hard to make a relativecomparison between bodies or structuresof different size and length as theirindividual deformations will be different.

This requires the development of theconcept ofStrain, which relates thebodys deformation to its initial length.


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Strain

Normal StrainThe elongation (+ve) or contraction

(ve) of a body per unit length is

termed Strain.


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Strain

Lets take the arbitrarily shaped body in Fig. 2.3 as an example.Consider the infinitesimal line segmentAB that is containedwithin the undeformed body as shown in Fig. 2.3(a). The lineAB lies along the n-axis and has an original length of S.After deformation, points A and B are displaced to A and Band in general the line becomes a curve having a length SThe change in length of the line is therefore S-S. Weconsequently define the generalized strain mathematically as


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Average Normal Strain

If the stress in the body is everywhere

constant, in other words, the deformationis uniform in the material (e.g. uniformuniaxial tension or compression) as shownin Fig. 2.2, the strain can be computed by

i.e. the change in length of the body over its original

length,


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Unit of StrainFrom Eqs. (2.8) and (2.9), we can notice that the

normal strain is a dimensionless quantity since itis a ratio of two lengths. Although this is thecase, it is common in practice to state it in termsof a ratio of length units. i.e. meters per meter(m/m)

Usually, for most engineering applications is verysmall, so measurements of strain are inmicrometers per meter (m/m) or (/m).Sometimes for experiment work, strain isexpressed as a percent, e.g. 0.001m/m = 0.1%.

A normal strain of 480m for a one-meter lengthis said


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Example 2.3

In Example 2.1, if it is measured that thecable was elongated by 1.35 mm due tothe weight of the light, what would its

strain be?


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STRESS-STRAIN RELATIONSHIP,

HOOKE'S LAW

Material Test and Stress-Strain Diagram

The material strength depends on its ability tosustain a load without undue deformation orfailure. The property is inherent in the materialitself and must be determined by experiment.

One of the most important tests to perform in thisregard is the tension or compression. To do

so, a bunch of standard specimen is made. The testis performed in universal test machine.


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Stress-strain RelationshipShown in Fig. 2.4 is the specimen and test result of

Stress-Strain Diagram. The Stress-Straindiagram consists of 4 stages during the wholeprocess, elastic, yielding, hardening and neckingstages respectively. From yielding stage, somepermanent plastic deformation occurs.


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Stress-strain RelationshipAbout 90% of engineering problems only concernthe elastic deformation in structural members andmechanical components. Only 10% of engineeringwork concerns plastic and other nonlinear stage(e.g. metal forming). In this subject, we are onlyinvolved in the linear elastic region, in which therelationship between the strain and stress is linear


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Hookes Law

The stress-strain linear relationship was discoveredby Robert Hook in 1676 and is known as Hooke'slaw. It is mathematically represented by Eq. (2.10),

where Eis terms as the Modulus of Elast ic i tyorYoung'sModuluswith units of N/m2 or Pa.

For most of engineering metal material, GPa is used, e.g.

mild steel is about 200GPa~210GPa.

Lect 1&2 Stress and Strain 1

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