Unit 1 Engineering Physics

16 CHAPTER 1. PROPERTIES OF MATTER

should be at least four times this diameter; 60 mm is common. Gauge length

is used in ductility computations, as discussed in Section 6.6; the standard value is

50 mm (2.0 in.). The specimen is mounted by its ends into the holding grips of the

testing apparatus (Figure 6.3). The tensile testing machine is designed to elongate

the specimen at a constant rate and to continuously and simultaneously measure the

instantaneous applied load (with a load cell) and the resulting elongations (using an

extensometer). A stress–strain test typically takes several minutes to perform and is

destructive; that is, the test specimen is permanently deformed and usually fractured.

[The (a) chapter-opening photograph for this chapter is of a modern tensile-testing

apparatus.]

The output of such a tensile test is recorded (usually on a computer) as load

or force versus elongation. These load–deformation characteristics are dependent

on the specimen size. For example, it will require twice the load to produce the same

elongation if the cross-sectional area of the specimen is doubled. To minimize these

1214 in. 2

6.2 Concepts of Stress and Strain • 153

T

T

F

F

F

F

F

F

F

A 0

A 0

A 0

(a) (b)

(c) (d)

!

"

l l 0 l 0 l

Figure 6.2

A standard tensile

specimen with

circular cross

section.

Figure 6.1

(a) Schematic

illustration of how a

tensile load produces

an elongation and

positive linear strain.

Dashed lines

represent the shape

before deformation;

solid lines, after

deformation.

(b) Schematic

illustration of how a

compressive load

produces contraction

and a negative linear

strain. (c) Schematic

representation of

shear strain , where

!tan.

(d) Schematic

representation of

torsional

deformation (i.e.,

angle of twist )

produced by an

applied torque T.

f

ug

g

2"

Gauge length

Reduced section

2 "

"Diameter

"

1

4

3

4

(^38) Radius
0.505" Diameter
JWCL187_ch06_150-196.qxd 11/5/09 9:36 AM Page 153
d

A

A 0 > A

d 0 > d , Δd = (d 0 - d)

l 0 < l , Δl = (l - l 0 )

longitudinal strain = Δl/l 0

lateral strain = - Δd/d 0

d 0

z

x

Poissons ratio =

(d/d 0 )

l/l 0

=

l 0

d 0

d

l

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Figure 1.15:Axial(z)elongation (positive strain) and lateral(x)contractions(negative

strains)in response to an imposed tensile stress. Solid lines represent dimensions after

stress application; dashed lines, before.( Picture courtesy :[ 1 ])

the direction parallel to its length). This applied force produces an elongation of specimen

along its direction and contractions along all directions perpendicular to it. As a result,

the cylindrical specimen will have an increased length (l), a reduced diameter (d), and a

smaller area of cross-section (A), as shown in Figure1.15. Therefore, in addition to the

longitudinal strain[(l≠l 0 )/l 0 =l/l 0 ], we can also define a compressive or lateral strain

as follows:

lateral strain =

final diamater≠ original diameter

original diamater

=

d≠d 0

d 0

=≠

d 0 ≠d

d 0

= ≠

d

d 0

The Poisson’s Ratio(‹) is defined as

‹ = ≠

lateral strain

longitudinal strain

=

l 0

d 0

d

l

The negative sign ensures that Poisson ratio is positive, since the longitudinal and axial

strains have opposite signs for most materials. Many metals and other alloys, values

of Poisson’s ratio range between 0.25 and 0.35. Poisson ratio values of some common

materials are given in Table1.1.

For isotropic materials, shear and elastic moduli are related to each other and to

Poisson’s ratio according to

Y =2G(1 +‹)

16