Unit 1 Engineering Physics

12 CHAPTER 1. PROPERTIES OF MATTER

Worked out Example 1.3.1

a load of 500 N (112 lbf) is applied. Assume

that the deformation is totally elastic.

6.7For a bronze alloy, the stress at which plastic

deformation begins is 275 MPa (40,000 psi),

and the modulus of elasticity is 115 GPa

(16.7! 106 psi).

(a)What is the maximum load that may be

applied to a specimen with a cross-sectional

area of 325 mm^2 (0.5 in.^2 ) without plastic de-

formation?

(b)If the original specimen length is 115 mm

(4.5 in.), what is the maximum length to which

it may be stretched without causing plastic

deformation?

6.8A cylindrical rod of copper (E"110 GPa,

16! 106 psi) having a yield strength of 240

MPa (35,000 psi) is to be subjected to a load

of 6660 N (1500 lbf). If the length of the rod

is 380 mm (15.0 in.), what must be the di-

ameter to allow an elongation of 0.50 mm

(0.020 in.)?

6.9Compute the elastic moduli for the follow-

ing metal alloys, whose stress–strain behav-

iors may be observed in the “Tensile Tests”

module of Virtual Materials Science and En-

gineering (VMSE):(a) titanium,(b) tem-

pered steel,(c) aluminum, and (d) carbon

steel. How do these values compare with

those presented in Table 6.1 for the same

metals?

Questions and Problems • 189

6.10Consider a cylindrical specimen of a steel

alloy (Figure 6.21) 10.0 mm (0.39 in.) in di-

ameter and 75 mm (3.0 in.) long that is pulled

in tension. Determine its elongation when a

load of 20,000 N (4,500 lbf) is applied.

6.11Figure 6.22 shows, for a gray cast iron, the ten-

sile engineering stress–strain curve in the elas-

tic region. Determine (a)the tangent modulus

at 10.3 MPa (1500 psi) and (b)the secant

modulus taken to 6.9 MPa (1000 psi).

6.12As noted in Section 3.15, for single crystals of

some substances, the physical properties are

anisotropic; that is, they are dependent on

crystallographic direction. One such property

is the modulus of elasticity. For cubic single

crystals, the modulus of elasticity in a general

[uvw] direction,Euvw,is described by the re-

lationship

where and are the moduli of elas-

ticity in [100] and [111] directions, respec-

tively;!,",and #are the cosines of the angles

between [uvw] and the respective [100], [010],

and [001] directions. Verify that the val-

ues for aluminum, copper, and iron in Table

3.3 are correct.

6.13In Section 2.6 it was noted that the net bond-

ing energy ENbetween two isolated positive

E 81109

E! 100 " E! 111 "

1 a^2 b^2 #b^2 g^2 #g^2 a^22

1

Euvw"

1

E! 100 "$^3 a

1

E! 100 "$

1

E! 111 "b

Figure 6.21 Te n s i l e

stress–strain

behavior for a steel

alloy.

0.002

Strain

Stress (MPa)

0.00^0 0.04 0.08 0.12 0.16 0.20

200

100

600

500

400

300

Strain

0.000^0 0.004 0.006

400

300

200

100

500

Stress (MPa)

Metal Alloys

JWCL187_ch06_150-196.qxd 11/5/09 9:36 AM Page 189

Figure 1.11: Tensile stress-strain behaviour of a

steel alloy.(Picture courtesy :[ 1 ])

A cylindrical specimen of a steel

alloy 10.0 mm in diameter and

75 mm long that is pulled in

tension. From the given stress-

strain diagram (Figure1.11)of

the steel alloy, determine its

elongation when a load of 20,000

N is applied [ 1 ].

Solution:

The stress when a load of 20,000

N is applied is

‡=

F

A 0

=

F

fi

(^1) d
0
2
(^22)

=

20000 N

fi

(^110). 0 ◊ 10 ≠ (^3) m
2
22 =255MPa
Referring to Figure1.11, at this
stress level we are in the elastic
region on the stress-strain curve, which corresponds to a strain of 0.0012. Therefore
the corresponding elongation is
l=‘l 0 =(0.0012)(75 mm) = 0.090 mm

1.4 Types of Elastic Moduli

Depending on the type of load applied, i.e., tensile, compressive or shear, there are three
kinds of elastic moduli: Young’s Modulus, Bulk Modulus, and Shear modulus.

1.4.1 Young’s Modulus

A tensile load, applied as shown in Figure1.2(a) or in Figure1.12, produces an elongation
and positive linear strain.
F= applied external force.
A 0 = Area (before deformation) on which the force is applied
Then linear stress,‡=F/A 0
l 0 = length before deformation andl= length after deformation
Change in length (elongation) =l=l≠l 0
Linear (or longitudinal) strain,‘= (change in length)/(original length) =l/l 0

YoungÕs Modulus (Y) =

linear stress

linear strain

=

‡

‘

=

F/A 0

l/l 0

=

l 0 F

A 0 l

(1.3)

12