PH8151 Engineering Physics Chapter 1

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

12

1

4 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.

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.

f

ug

g

2"

Gauge length

Reduced section

2 "

"Diameter

"

1

4

3

4

Radius

3

8

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 ])

deformation is produced. Then shear stress,‡=F/A 0. The shear strain“is defined as

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

(1.5)

Shear moduli for some common materials are given in Table1.1.

Worked out Example 1.4.

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 (“)

=

‡

x/L

)shear stress =G

x

L

=(1. 5 ◊ 1010 Pa)

(5m)

(10000m)

=7. 5 ◊ 106 N/m^2

1.4.3 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

=≠

‡

‘

=≠

P

V/V

=≠

PV

V

(1.6)

14