PH8151EPUnit1

1.3. ENGINEERING STRESS-STRAIN DIAGRAM

Ultimate Tensile Strength : After yielding, the stress necessary to continue plastic
deformation in metals increases to a maximum (the point 4 in Figure1.6). The
maximum ordinate in the stress- strain diagram is the ultimate tensile strength. At
this maximum stress, a small constriction orneckbegins to form at some point on
the specimen, and all subsequent deformation is confined at this neck.

Rupture (Fracture) Strength: If the stress is further increased beyond the tensile
strength, fracture ultimately occurs at the neck. The fracture strength corresponds
to the stress at fracture (the point 5 in Figure1.6).

Worked out Example 1.3.

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.

Strain

Stress (MPa)

0.00^0 0.04 0.08 0.12 0.16 0.

200

100

600

500

400

300

Strain

0.000^0 0.004 0.

400

300

200

100

500

Stress (MPa)

Metal Alloys

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

Figure 1.8: 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, is pulled in tension.

From the given stress-strain di-

agram (Figure 1.8) of the steel

alloy, determine its elongation

when a load of 20,000 N is ap-

plied [ 1 ].

Solution:

The stress when a load of 20,

N is applied is

‡=

F

A 0

=

F

fi

1

d 0

2

(^22)

=

20000 N

fi

1

10. 0 ◊ 10 ≠^3 m

2

22 =255MPa

Referring to Figure1.8, 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

1.3.1 True Stress and True Strain

From Figure1.5, the decline in the stress necessary to continue deformation past the
maximum, the point M, seems to indicate that the metal is becoming weaker. This is
not at all the case. The cross-sectional area is decreasing rapidly within the neck region,
where deformation is occurring. The stress, as computed from‡=F/A 0 , is on the basis
of the original cross- sectional areaA 0 before any deformation and does not take into
account this reduction in area at the neck.

True stress‡T is defined as the loadF divided by the instantaneous cross-sectional

PH8151 9 LICET