16 CHAPTER 1. PROPERTIES OF MATTERshould 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 these12
1
4 in.^26.2 Concepts of Stress and Strain • 153
TTFFFFFFFA 0A 0A 0(a) (b)(c) (d)!"l l 0 l 0 lFigure 6.
A standard tensile
specimen with
circular cross
section.Figure 6.
(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.fugg2"
Gauge lengthReduced section
2 ""Diameter"1
4
3
4Radius
3
80.505" DiameterJWCL187_ch06_150-196.qxd 11/5/09 9:36 AM Page 153
dAA 0 > Ad 0 > d , Δd = (d 0 - d)l 0 < l , Δl = (l - l 0 )longitudinal strain = Δl/l 0lateral strain = - Δd/d 0
d 0
zxPoisson s ratio = ( d/d 0 )
l/l 0=
l 0
d 0 d
l
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strains)in response to an imposed tensile stress. Solid lines represent dimensions after
stress application; dashed lines, before.( Picture courtesy :[ 1 ])specimen in Figure1.15having original lengthl 0 , diameterd 0 and cross-sectional areaA 0.
Assume that a longitudinal forceFis applied to the cylinder in the z-direction (i.e., along
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 0The Poisson’s Ratio(‹) is defined as‹ = ≠lateral strain
longitudinal strain=l 0
d 0 d
lThe 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 Table1.1.16