fundamentals of stress analysis - lec 04

7/24/2019 Fundamentals of Stress Analysis - Lec 04

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MansouraMansoura UniversityUniversity -- Faculty ofFaculty of EngineeringEngineering

Production and Mechanical Design Engineering DepartmentProduction and Mechanical Design Engineering Department


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PRE5311- Fundamentals of Stress Analysis -Lecture No. (4) : Strain and material properties.

Dr. T. A. ENAB, Production & Mechanical Design Engineering Department, Faculty of Engineering, Mansoura University. 22


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IntroductionIntroduction::

They relate to changes in the size and shape of a body

The state of stress at a point can be determined if the stress components onmutually perpendicular planes are given

A similar operation applies to the state of strain to develop the transformation

relations that give 2D and 3D strains in inclined directions in terms of thestrains in the coordinate directions

The plane strain transformation equations are especially important in

experimental investigations




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DeformationDeformation::

pictured by the dashed lines (see Fig.)

o n s sp ace o , o , an so on, un a e po n s n e o yare displaced to new positions

Displacements ofany twopoints such as & are may be a consequence of:

deformation (straining)

rigid-body motion (translation and rotation)

or some combination

The body is said to be strainedif the relative positions of points in the body

.

If no straining has taken place, (rigid-body motion) the distance



0


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DeformationDeformation::

of points within the body with respect to an appropriate coordinate reference

. . .

In the 2D case the components of displacement of point A to A can be

represen e yuan v n exan ycoor na e rec ons, respec ve y.

In general, the components of displacement at a point, occurring in the x, y,an z rect ons, are enote yu, v, an w, respect ve y.

The displacement at everypoint within the body constitutes the displacement

field,u =u(x,y,z),v =v(x,y,z), andw=w(x,y,z).

In engineering structures mainlysmall displacementsare considered




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SuperpositionSuperposition::

displacement) to be determined is a linearfunction of the loads that produce

.

For the foregoing condition to exist, material must be linearly elastic.

Superposition cannot be applied to plastic deformations.

The motivation for superposition is the replacement of a complex loadconfiguration by two or more simpler loads.

In superposition situations, the total quantity owing to the combined

loads acting simultaneously on a member may be obtained by determiningseparately the quantity attributable to each load and combining the

individual results.




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SuperpositionSuperposition::

Normal stressescaused by axial forces and bendingsimultaneously may

e o a ne y superpos on, prov e a e com ne s resses o noexceed the proportional limit of the material

Shearing stresses caused by a torque and a vertical shear force acting

simultaneously in a beam may be treated by superposition.




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StrainStrain definitiondefinition::

becomeAB.

e eng o s x. PointsA andBhave each been displaced:

Aan amountu, andB,u+ u.

Stated differently, point has been displaced by an amount uin addition to

displacement of pointA, and the length xhas been increased by u.

Normal strain the unit chan e in len th is defined as:

If the deformation is distributeduniformlyover the original length, the normal

whereL,Loand are the final length, the original length, and the change



, .


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StrainStrain definitiondefinition::

.

Stressis used to measure the intensity of internal force.

Normal strain, , used to measure change in size.

Shear strain, , used to measure change in shape




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PlanePlane strainstrain::

remain in the same plane.

wo- mens ona v ews o an e emen w e ges o un t eng s su ec e oplane strain are shown in three parts in figure.

We note that this element has no normal strain z and no shearing strains

xzandyzin thexzandyzplanes, respectively.

Strain components x, , andxin the xy plane




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, .

deformation may be regarded as possessing the following features:

a c ange n eng exper ence y e s es g. aa relative rotation without accompanying changes of length (Fig. b)

Deformations of an element: (a) normal strain; (b) shearing strain

The two normal or longitudinal strains are:

Apositivesign is applied toelongation; anegativesign, tocontraction.




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,

shearing strain and denoted byxy:

e s ear s ra n s pos t ve w en e r g t ang e e ween wo pos ve ornegative) axesdecreases.

That is, if the angle between +x and + or x and decreases, we have

positivexy; otherwise the shear strain is negative.




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ThreeThree--dimensionaldimensional strainstrain::

, , , ,

essentially identical analysis leads to the following normal and shearing

Noting that components of shearing strain are similarly related:

The above ex ressions are the straindis lacement relations of continuum

mechanics .

,

strain rather than the matter of cause and effect.




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whereux=u,uy=v,xx=x, and so on.

The factor in Eq. facilitates the representation of the strain transformationequations in indicial notation.

The longitudinal strains are obtained when i=j; the shearing strains are found

whenijandij=ji.

It is apparent from Eqs. that:

Just as the state of stress at a oint is described b a nine-term arra so strain

Eq. represents nine strains composing the symmetric strain tensor (ij=ji):




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,

coordinate setis thus established in a deformedbody;xyzis, in this instance,

.

For deformation: the xyz set is established in the undeformed body. In this

case,xyz s re erre o as a agrang an coor na e sys em.

Strains are indicated asdimensionlessquantities

The normal and shearing strains are also frequently described in terms of

units such as inches per inch or micrometers per meter and radians or micro

radians, respectively.

The strains for engineering materials in ordinary use seldom exceed 0.002,

which is equivalent to 2000 106 or 2000.




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A 0.8-m by 0.6-m rectangle ABCD is drawn on a thin plate prior to

oa ng. u sequen o oa ng, e e orme geome ry s s own y edashed lines in Fig. Determine the components of plane strain at pointA.




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The straindisplacement relations are:

By setting x= 800 mm and y= 600 mm

The normal strains are calculated as follows:

In a like manner, we obtain the shearing strain:

The positive sign indicates that angle ADhas decreased




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LargeLarge strainsstrains::

consistent with the magnitude of deformations usually found in engineering

.

The more general large orfinite straindisplacement relationships are given

n erms o esquare o e e emen eng ns ea o e eng se .

Therefore, with reference o figure we write:

In which:

andAD=dx




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LargeLarge strainsstrains::

, -

Similarly, we have:

It can also be verified that thefinite shearing straindisplacement relationis:

In small displacement theory, the higher-order terms in the above Eqs. are

.

The expressions for three-dimensional state of strain may readily be

generalized from the preceding equations.




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fundamentals of stress analysis - lec 04

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