CHAPTER OBJECTIVES

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# CHAPTER OBJECTIVES - PowerPoint PPT Presentation

CHAPTER OBJECTIVES. To define the concept of normal strain To define the concept of shear strain To determine the normal and shear strain in engineering applications. CHAPTER OUTLINE. Deformation Strain. 2.1 DEFORMATION. Deformation Occurs when a force is applied to a body

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Presentation Transcript
CHAPTER OBJECTIVES
• To define the concept of normal strain
• To define the concept of shear strain
• To determine the normal and shear strain in engineering applications
CHAPTER OUTLINE
• Deformation
• Strain
2.1 DEFORMATION

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

Assumptions 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

Deformed body

Undeformed body

2.2 STRAIN

Normal strain

Defined as the elongation or contraction of a line segment per unit of length

Consider line AB in figure above

After deformation, Δs changes to Δs’

Δs’ − Δs

avg =

Δs

Deformed body

Undeformed body

lim

B→A along n

Δs − Δs’

=

Δs

2.2 STRAIN
• Defining average normal strain using avg (e = epsilon)

As Δs → 0, Δs’ → 0

Δs’ ≈(1 + )Δs

2.2 STRAIN
• If normal strain  is known, use the equation to obtain approx. final length of a short line segment in direction of n after deformation.

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

2.2 STRAIN

Units

• 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)
• is small for most engineering applications, so it is normally expressed as micrometers per meter (μm/m) where 1 μm = 10−6m
• Also expressed as a percentage, e.g.,0.001 m/m = 0.1 %
2.2 STRAIN

Shear 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).

Undeformed body

Deformed body

2.2 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’

n

t

2

’

lim

B→A along n

C→A along t

γnt =

Undeformed body

Deformed body

2.2 STRAIN

Shear strain

• Hence, shear strain at point A associated with n andt axes is

Shear strain positive if q’ < /2, and

Shear strain negative if q’ > /2.

Undeformed point

2.2 STRAIN

Cartesian strain components

Using above definitions of normal and shear strain, we show how to describe the deformation of the body

A point in the un-deformed body is regarded as a small element having dimensions of Δx, Δy and Δz

Undeformed point

Approx. lengths of sides of parallelepiped are

(1+z)Δz

(1+x)Δx

(1+y)Δy

Deformed point

2.2 STRAIN
• Since element is very small, deformed shape of element is a parallelepiped

(1 + x)Δx

(1 + y)Δy

(1 + z)Δz

(1+z)Δz

2

2

2

(1+x)Δx

− γxy

− γyz

− γxz

(1+y)Δy

Deformed point

2.2 STRAIN

Approx. angles between the sides are

Normal strains, e, cause a

change in its volume

Shear strains, g, cause a

change in its shape

To summarize, state of strain at a point requires

specifying 3 normal strains of x,y,zand

3 shear strains of γxy,γyz,γxz

2.2 STRAIN

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 small compared to 1, i.e.,  << 1.
• 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 sinθ= θ, cosθ = θ and tanθ = θ, provided θ is very small
EXAMPLE

The rigid beam is supported by

a pin at A and wires BD and CE.

If the load P on the beam is

displaced 10 mm downward,

determine the normal strain

developed in each wire.

dBD

dCE

A

10 mm

B

C

dBD = 4.28 mm

dCE = 7.14 mm

EXAMPLE (CONT.)

Free-body diagram

dBD = change in length of BD

dCE = change in length of CE

EXAMPLE (CONT.)

Normal strain developed in each wire

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
• State of strain at a point is described by six strain components:
• Three normal strains: x, y, z
• Three shear strains: γxy, γxz, γyz
• These components depend upon the orientation of the line segments and their location in the body
CHAPTER REVIEW
• Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations
• Most engineering materials undergo small deformations, so normal strain  << 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