ch02- Strain Mekanik Bahan

8/11/2019 ch02- Strain Mekanik Bahan

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2005 Pearson Education South Asia Pte Ltd

2. Strain

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CHAPTER OBJECTIVES

Define concept of

normal strain

Define concept of

shear strain

Determine normaland shear strain in

engineering

applications


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CHAPTER OUTLINE

1. Deformation2. Strain


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Deformation

Occurs when a force is applied to a body

Can be highly visible or practically unnoticeable

Can also occur when temperature of a body is

changed

Is not uniform throughout a bodys volume, thus

change in geometry of any line segment within

body may vary along its length

2.1 DEFORMATION


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To simplify study of deformation

Assume lines to be very short and located in

neighborhood of a point, and

Take into account the orientation of the line

segment at the point

2.1 DEFORMATION


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Normal strain

Defined as the elongation or contraction of a linesegment per unit of length

Consider lineABin figure below

After deformation, schanges to s

2.2 STRAIN


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Normal strain

Defining average normal strainusing avg(epsilon)

As s 0, s 0

2.2 STRAIN

avg=s s

s

=

s s

s

lim

BA along n


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Normal strain

If normal strain is known, use the equation to

obtain approx. final length of a shortline segment

in direction of nafter deformation.

Hence, when is positive, initial line will elongate,if is negative, the line contracts

2.2 STRAIN

s (1 + )s



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2.2 STRAIN

Units

Normal strain is a dimensionless quantity, asits a ratio of two lengths

But common practice to state it in terms ofmeters/meter (m/m)

is small for most engineering applications, sois normally expressed as micrometers permeter (m/m)where1m = 106

Also expressed as a percentage,e.g.,0.001 m/m = 0.1 %



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2.2 STRAIN

Shear strain

Defined as the change in anglethat occurs

between two line segments that were originally

perpendicularto one another

This angle is denoted by (gamma) andmeasured in radians (rad).



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2.2 STRAIN

Shear strain

Consider line segmentsABandACoriginating

from same pointAin a body, and directed along

the perpendicular nand taxes

After deformation, lines become curves, such thatangle between them atAis



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2.2 STRAIN

Shear strain

Hence, shear strain at pointAassociated with n

and taxes is

If is smaller than /2, shear strain is positive,otherwise, shear strain is negative

nt=

2

lim

BA along n

C A along t



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Cartesian strain components

Using above definitions of normal and shear strain,

we show how they describe the deformation of the

body

2.2 STRAIN

Divide body into smallelements with

undeformed dimensions

ofx,yandz



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Since element is very small, deformed shape of

element is a parallelepiped

Approx. lengths of sides of parallelepiped are

(1 + x) x (1 + y)y (1 + z)z

2.2 STRAIN



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Approx. angles between the sides are

2.2 STRAIN

2 xy

2 yz

2 xz

Normal strains cause a change in its volume

Shear strains cause a change in its shape

To summarize, state of strain at a point requires

specifying 3 normal strains; x, y, zand 3 shear

strains of xy,yz,xz



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Small strain analysis

Most engineering design involves applications

for which only small deformationsare allowed

Well assume that deformations that take place

within a body are almost infinitesimal, so normalstrainsoccurring within material are very small

compared to 1, i.e.,



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Small strain analysis

This assumption is widely applied in practical

engineering problems, and is referred to as

small strain analysis

E.g., it can be used to approximate sin = , cos= and tan = , provided is small

2.2 STRAIN


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2 St i



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EXAMPLE 2.1 (SOLN)

(a) Since normal strain reported at each point along

the rod, a differential segment dz, located at

positionzhas a deformed length:

dz = [1 + 40(103)z1/2] dz

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EXAMPLE 2.1 (SOLN)

(a) Sum total of these segments along axis yields

deformed length of the rod, i.e.,z = 0 [1 + 40(103)z1/2] dz

= z+ 40(103)(z3/2)|0= 0.20239 m

0.2 m

0.2 m

Displacement of end of rod is

B= 0.20239 m 0.2 m = 2.39 mm

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EXAMPLE 2.1 (SOLN)

(b) Assume rod or line segment has originallength of 200 mm and a change in length of

2.39 mm. Hence,

avg=s s

s =2.39 mm

200 mm = 0.0119 mm/mm

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EXAMPLE 2.3

Plate is deformed as shown in figure. In this

deformed shape, horizontal lines on the on plate

remain horizontal and do not change their length.

Determine

(a) average normal strainalong sideAB,

(b) average shear strain

in the plate relative toxandyaxes

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EXAMPLE 2.3 (SOLN)

(a) LineAB, coincident withyaxis, becomes line

AB after deformation. Length of lineAB is

AB = (250 2)2+ (3)2 = 248.018 mm

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EXAMPLE 2.3 (SOLN)

(a) Therefore, average normal strain forABis,

= 7.93(103) mm/mm

(AB)avg= AB

AB AB 248.018 mm 250 mm250 mm

=

Negative sign means

strain causes a

contraction of AB.

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EXAMPLE 2.3 (SOLN)

(b) Due to displacement ofBtoB, angle BACreferenced fromx,yaxes changes to .

Since xy= /2 , thus

xy= tan13 mm

250 mm 2 mm = 0.0121 rad( )

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CHAPTER REVIEW

Loads cause bodies to deform, thus points inthe body will undergo displacements or

changes in position

Normal strainis a measure of elongation orcontraction of small line segment in the body

Shear strainis a measure of the change inangle that occurs between two small line

segments that are originally perpendicular to

each other

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CHAPTER REVIEW

State of strain at a point is described by sixstrain components:

a) Three normal strains: x, y, z

b) Three shear strains: xy, xz, yz

c) These components depend upon the orientation ofthe line segments and their location in the body

Strain is a geometrical quantity measuredby experimental techniques. Stress in body

is then determined from material property

relations

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CHAPTER REVIEW

Most engineering materials undergo smalldeformations, so normal strain

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ch02- Strain Mekanik Bahan