Files 2 handouts-ch02

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2. Strain

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CHAPTER OBJECTIVES• Define concept of

normal strain• Define concept of

shear strain• Determine normal

and 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 body’s 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 line

segment per unit of length• Consider line AB in figure below• After deformation, Δs changes to Δs’

2.2 STRAIN

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Normal strain• Defining average normal strain using εavg (epsilon)

• As Δs → 0, Δs’ → 0

2.2 STRAIN

εavg = Δs − Δs’Δs

ε = Δs − Δs’Δs

limB→A along n

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Normal strain• If normal strain ε is known, use the equation to

obtain approx. final length of a short line segment in direction of n after deformation.

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

2.2 STRAIN

Δs’ ≈ (1 + ε) Δs

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2.2 STRAINUnits • Normal strain is a dimensionless quantity, as

it’s a ratio of two lengths• But common practice to state it in terms of

meters/meter (m/m)• ε is small for most engineering applications, so

is normally expressed as micrometers per meter (μm/m) where 1 μm = 10−6

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

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2.2 STRAINShear strain• Defined as the change in angle that occurs

between two line segments that were originally perpendicular to one another

• This angle is denoted by γ (gamma) and measured in radians (rad).

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2.2 STRAINShear strain• Consider line segments AB and AC originating

from same point A in a body, and directed along the perpendicular n and t axes

• After deformation, lines become curves, such that angle between them at A is θ’

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2.2 STRAINShear strain• Hence, shear strain at point A associated with n

and t axes is

• If θ’ is smaller than π/2, shear strain is positive,otherwise, shear strain is negative

γnt =π2

limB→A along nC →A along t

θ’−

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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 small elements with undeformed dimensions of Δx, Δy and Δz

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Cartesian strain components• Since element is very small, deformed shape of

element is a parallelepiped

• Approx. lengths of sides of parallelepiped are

(1 + εx) Δx (1 + εy)Δy (1 + εz)Δz

2.2 STRAIN

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Cartesian strain components• Approx. angles between the sides are

2.2 STRAIN

π2 − γxy

π2 − γyz

π2 − γxz

• 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; εx, εy, εz and 3 shear strains of γxy, γyz, γxz

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Small strain analysis• Most engineering design involves applications

for which only small deformations are allowed• We’ll assume that deformations that take place

within a body are almost infinitesimal, so normal strains occurring within material are very smallcompared to 1, i.e., ε << 1.

2.2 STRAIN

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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 θ = θ, cos θ = θ and tan θ = θ, provided θ is small

2.2 STRAIN

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EXAMPLE 2.1

Rod below is subjected to temperature increase along its axis, creating a normal strain of εz = 40(10−3)z1/2, where z is given in meters.Determine (a) displacement of end B of rod due to temperature increase, (b) average normal strain in the rod.

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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 position z has a deformed length:

dz’ = [1 + 40(10−3)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(10−3)z1/2] dz

= z + 40(10−3)(⅔ z3/2)|0= 0.20239 m

0.2 m

0.2 m

Displacement of end of rod isΔB = 0.20239 m − 0.2 m = 2.39 mm ↓

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EXAMPLE 2.1 (SOLN)

(b) Assume rod or “line segment” has original length of 200 mm and a change in length of 2.39 mm. Hence,

εavg = Δs’ − ΔsΔs

= 2.39 mm200 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 strain

along side AB, (b) average shear strain

in the plate relative to x and y axes

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EXAMPLE 2.3 (SOLN)

(a) Line AB, coincident with y axis, becomes line AB’ after deformation. Length of line AB’ isAB’ = √ (250 − 2)2 + (3)2 = 248.018 mm

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EXAMPLE 2.3 (SOLN)

(a) Therefore, average normal strain for AB is,

= −7.93(10−3) mm/mm

(εAB)avg = ABAB’ − AB 248.018 mm − 250 mm

250 mm=

Negative sign means strain causes a contraction of AB.

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EXAMPLE 2.3 (SOLN)

(b) Due to displacement of B to B’, angle BAC referenced from x, y axes changes to θ’. Since γxy = π/2 − θ’, thus

γxy = tan−1 3 mm250 mm − 2 mm = 0.0121 rad( )

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CHAPTER REVIEW

• Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position

• Normal strain is a measure of elongation or contraction of small line segment in the body

• Shear strain is a measure of the change in angle 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 six strain components:

a) Three normal strains: εx, εy, εz

b) Three shear strains: γxy, γxz, γyz

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

• Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations

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CHAPTER REVIEW

• Most engineering materials undergo small deformations, so normal strain ε << 1. This assumption of “small strain analysis” allows us to simplify calculations for normal strain, since first-order approximations can be made about their size