Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-07 GTBL042-Callister-v2 August 9, 2007 13:52

Questions and Problems • 239

Engineering Stress Engineering

(MPa) Strain

315 0.105

340 0.220

On the basis of this information, compute the

engineeringstress necessary to produce anen-

gineeringstrain of 0.28.

7.22Find the toughness (or energy to cause frac-

ture) for a metal that experiences both elas-

tic and plastic deformation. Assume Equation

7.5 for elastic deformation, that the modulus

of elasticity is 103 GPa (15× 106 psi), and that

elastic deformation terminates at a strain of

0.007. For plastic deformation, assume that the

relationship between stress and strain is de-

scribed by Equation 7.19, in which the values

forKandnare 1520 MPa (221,000 psi) and

0.15, respectively. Furthermore, plastic defor-

mation occurs between strain values of 0.007

and 0.60, at which point fracture occurs.

7.23Taking the logarithm of both sides of Equation

7.19 yields

logσT=logK+nlogT (7.32)

Thus, a plot of logσTversus logTin the plastic

region to the point of necking should yield a

straight line having a slope ofnand an inter-

cept (at logσT=0) of logK.

Using the appropriate data tabulated in

Problem 7.15, make a plot of logσTversus log

Tand determine the values ofnandK. It will

be necessary to convert engineering stresses

and strains to true stresses and strains using

Equations 7.18a and 7.18b.

Elastic Recovery After Plastic Deformation

7.24A steel alloy specimen having a rectangu-

lar cross section of dimensions 19 mm ×

3.2 mm (^34 in.×^18 in.) has the stress–strain

behavior shown in Figure 7.33. If this speci-

men is subjected to a tensile force of 110,000 N

(25,000 lbf) then

(a)Determine the elastic and plastic strain

values.

(b)If its original length is 610 mm (24.0 in.),

what will be its final length after the load

in part (a) is applied and then released?

Flexural Strength (Ceramics)

7.25A three-point bending test is performed on a

spinel (MgAl 2 O 4 ) specimen having a rectan-

gular cross section of heightd=3.8 mm (0.15

in.) and widthb=9 mm (0.35 in.); the distance

between support points is 25 mm (1.0 in.).

(a)Compute the flexural strength if the load

at fracture is 350 N (80 lbf).

(b)The point of maximum deflectionyoc-

curs at the center of the specimen and is

described by

y=

FL^3

48 EI

whereEis the modulus of elasticity and

Iis the cross-sectional moment of inertia.

Computeyat a load of 310 N (70 lbf).

7.26A three-point bending test was performed on

an aluminum oxide specimen having a circular

cross section of radius 5.0 mm (0.20 in.); the

specimen fractured at a load of 3000 N (675

lbf) when the distance between the support

points was 40 mm (1.6 in.). Another test is to

be performed on a specimen of this same ma-

terial, but one that has a square cross section of

15 mm (0.6 in.) length on each edge. At what

load would you expect this specimen to frac-

ture if the support point separation is main-

tained at 40 mm (1.6 in.)?

Influence of Porosity on the Mechanical Properties

of Ceramics

7.27The modulus of elasticity for spinel

(MgAl 2 O 4 ) having 5 vol% porosity is 240 GPa

(35× 106 psi).

(a)Compute the modulus of elasticity for the

nonporous material.

(b)Compute the modulus of elasticity for

15 vol% porosity.

7.28Using the data in Table 7.2, do the following:

(a)Determine the flexural strength for non-

porous MgO assuming a value of 3.75 for

nin Equation 7.22.

(b)Compute the volume fraction porosity at

which the flexural strength for MgO is

74 MPa (10,700 psi).

Stress–Strain Behavior (Polymers)

7.29From the stress–strain data for poly(methyl

methacrylate) shown in Figure 7.24, determine