14 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
TTFFFFFFFA 0A 0A 0(a) (b)(c) (d)!"l l 0 l 0 lFigure 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.fugg2"
Gauge lengthReduced 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
L
Δx
Figure 1.13: Shear deformation- dashed lines represent the shape before deformation;
solid lines, after deformation.(Picture courtesy :[ 1 ])
the tangent of the strain angle◊, the angle through which a line, which was originally
perpendicular to the tangential force, has turned.
Shear strain“=tan◊= x/L. For small strains,tan◊¥◊and hence,“¥◊¥ x/L.
Shear Modulus (G) =
shear stress
shear strain
=
‡
“
=
F/A 0
x/L=
F
◊A 0
(1.4)
Shear moduli for some common materials are given in Table1.1.Worked out Example 1.4.2
The distortion of the earth’s crust is an example of shear on a large scale. A particular
rock has a shear modulus of 1. 5 ◊ 1010 Pa. What shear stress is applied when a 10
km layer of rock is sheared a distance of 5 m?Solution:Shear Modulus (G) =shear stress
shear strain=
‡
“
=
shear stress
x/L)shear stress =G x
L=(1. 5 ◊ 1010 Pa)(5m)
(10000m)=7. 5 ◊ 106 N/m^21.4.3 Bulk Modulus..............................
The Bulk Modulus (K) of a substance is a measure of the ability of a substance to
withstand changes in volume when under compression on all sides. Its magnitude is equal
to the applied pressure divided by the bulk strain (or volume strain). Bulk strain is
the relative decrease of the volume. Bulk stress (or volume stress),‡=P, the applied
pressure. Bulk strain,‘= (change in volume)/(original volume) = V/V.Bulk Modulus K = ≠volume stress
volume strain