1. Callister 6.7 For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0!10 6 psi). (a) What is the maximum load that may be applied to a specimen with a

cross-sectional area of 130 mm2 (0.2 in.2) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum

length to which it may be stretched without causing plastic deformation?

2. Callister 6.24 Figure below shows the tensile engineering stress-strain behavior for a steel alloy. (a) What is the modulus of elasticity? (b) What is the proportional limit? (c) What is the yield strength at a strain off-set of 0.002 ? (d) What is the tensile strength?

3. Callister 6.45 A steel alloy specimen having a rectangular cross section of dimensions 19mm!3.2mm( 3/4 in. !1/8 in.) has the stress-strain behavior shown in Figure 6.21. If this specimen 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?

4. Callister 7.1 To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105 mm-2 . Suppose that all the dislocations in 1000 mm3 (cm3) were somehow removed and linked end to end. How far (in miles) would this chain extend? Now suppose that the density is increased to 109 mm-2 by cold working. What would be the chain length of dislocations in 1000 mm3 of material?

5. Core 8.06

T!

y!

895 MPa

825 MPa

!

E 10%

6. Core 8.08

(a) 3 (b) 3 (c) 3 (d) 1 (e) 2 and 3

7. Core 8.12

For A 302, y

!302=310 MPa, A 302= =

! y

F

"25.0

!

1x107N

0.25 " 310 "106N /m

2= 0.129 2

m

For A542, y

!542=585 MPa, A 542= =

! y 25.0 "

F

!

1x107N

0.25 " 585 "106N /m

2= 0.0683 2

m

87.7

542302=!!

Fe""" 33

/10 mkg! kgmmkgmAlM 3233302

302 1015.10)129.0()/1087.7()10( !=!!!=!!= " kgmmkgmAlM 3233302

542 1035.5)068.0()/1087.7()10( !=!!!=!!= " Weight saving = kg3

1015.10 ! - kg31035.5 ! = kg3

108.4 ! Thus we conclude that, because the A542 steel is son much stronger, much less of it is needed in order to support the tensile load required by this particular application. However, the downside to using the A542 is that it will be more brittle than the A302. Hence, if there are stress concentrators in either steel, the A542 will be more susceptible to potentially catastrophic cracking. The Type A 542 is more brittle.