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MSE 230 Assignment 5 Solutions Fall 2014 1. We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the

load necessary to produce an elongation of 7.6 mm. It is first necessary to compute the strain at

yielding from the yield strength and the elastic modulus, and then the strain experienced by the test

specimen. Then, if

ε(test) < ε(yield)

deformation is elastic, and the load may be computed using Equations (6.1) and (6.5). However, if

ε(test) > ε(yield)

computation of the load is not possible inasmuch as deformation is plastic and we have neither a

stress-strain plot nor a mathematical expression relating plastic stress and strain. We compute

these two strain values as

ε(test) = ∆llo

= 7.6 mm250 mm = 0.03

and

ε(yield) = σyE =

275 MPa103 x 103 MPa

= 0.0027

Therefore, computation of the load is not possible as already explained.

2. This problem asks that we assess the four alloys relative to the two criteria presented. The first

criterion is that the material not experience plastic deformation when the tensile load of 27,500 N is

applied; this means that the stress corresponding to this load not exceed the yield strength of the

material. Upon computing the stress

σ = F

Ao =

F

π⎝⎜⎛

⎠⎟⎞do

22 =

27500 N

π⎝⎜⎛

⎠⎟⎞10 x 10-3 m

22

= 350 x 106 N/m2 = 350 MPa

Of the alloys listed in the table, the Ti and steel alloys have yield strengths greater than 350 MPa.

Relative to the second criterion, it is necessary to calculate the change in diameter ∆d for

these two alloys. From Equation (6.8)

ν = - εxεz

= - ∆d/doσ/E

2

Now, solving for ∆d from this expression,

∆d = - νσdo

E

For the steel alloy

∆d = - (0.27)(350 MPa)(10 mm)

207 x 103 MPa = - 4.57 x 10-3 mm

Therefore, the steel is a candidate.

For the Ti alloy

∆d = - (0.36)(350 MPa)(10 mm)

107 x 103 MPa = - 11.8 x 10-3 mm

Therefore, the Ti alloy is not acceptable.

3.* This problem calls for us to make a stress-strain plot for a ductile cast iron, given its tensile load-

length data, and then to determine some of its mechanical characteristics.

(a) The data are plotted below on two plots: the first corresponds to the entire stress-strain curve,

while for the second, the curve extends just beyond the elastic region of deformation.

0.20.10.00

100

200

300

400

Strain

Stre

ss (M

Pa)

3

(b) The elastic modulus is the slope in the linear elastic region as

E = ∆σ∆ε =

100 MPa - 0 psi0.0005 - 0 = 200 x 103 MPa = 200 GPa

(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain

curve at approximately 280 MPa. 4.* (a) In order to compute the final length of the brass specimen when the load is released, it first

becomes necessary to compute the applied stress thus

σ = F

Ao =

F

π⎝⎜⎛

⎠⎟⎞do

22 =

6000 N

π⎝⎜⎛

⎠⎟⎞7.5 x 10-3 m

22 = 136 MPa

Upon locating this point on the stress-strain curve (Figure 6.12), we note that it is in the linear, elastic

region; therefore, when the load is released the specimen will return to its original length of 90 mm.

(b) In this portion of the problem we are asked to compute the final length, after load release, when

the load is increased to 16,500 N. Again, computing the stress

σ = 16500 N

π⎝⎜⎛

⎠⎟⎞7.5 x 10-3 m

22 = 373 MPa

0.0060.0050.0040.0030.0020.0010.0000

100

200

300

Strain

Stre

ss (M

Pa)

4

The point on the stress-strain curve corresponding to this stress is in the plastic region. We are able

to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic

region; this line intersects the strain axis at a strain of about 0.08 which is the amount of plastic strain. The final specimen length li may be determined from Equation (6.2) as

li = lo(1 + ε) = (90 mm)(1 + 0.08) = 97.20 mm

5. Step 1: Calculate σT: σT =

FAi

=15, 000Nπ 0.005m( )2

=191MPa

Step 2: Solve for εT: lnσT = lnK + nlnεT ln εT =

lnσT − lnKn

=ln 191( ) − ln 315( )

0.54

εT = 0.396 Step 3: Solve for l0: εT = ln

lil0⇒ 0.396 = ln 1.5

l0⇒ l0 =

1.5e0.396

=1.01m

6.* The Young’s modulus of the new tensile specimen will be higher. During the original tensile test the necked region experienced a high degree of molecular alignment relative to the original fiber. The alignment increasing the orientation of strong covalent bonds along the fiber axis thereby increasing the modulus along the fiber axis (see pages 592 and 594 in Callister). The strain to failure is lower because the polymer chains have already been stretched out thereby reducing their “stretchiness” (no, not a technical term).

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7.

The [001] and 001 [ ] are along the z-axis which is perpendicular to the [110] and the 1 1 0[ ] . Slip involves atomic planes sliding past one another. It is difficult to see how sliding can occur in the [110] and 1 1 0[ ] when the stress is applied in the [001] and 001 [ ].

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8*. a) τ = σ cos ø cosλ σ = the applied tensile stress τ = the resolved shear stress Slip initiates when the applied tensile stress is equal to the yield stress σy. The resolved shear stress when σ=σy is called the critical resolved shear stress or τCRSS. τCRSS = σy cos ø cosλ, a = [001], b = [111], c = [0 1 1]

cosø =[001] ⋅[111]

1 3=13

cosλ =[00 1 ] ⋅[01 1]

1 2=

12

τCRSS = (7.8 MPa) 13

" #

$ % 12

" #

$ %

= 3.2 MPa

b) The applied stress refers to the tensile stress applied during a mechanical test. The yield stress is the applied tensile stress needed to initiate plastic deformation (or slip). The shear stress is the component of the tensile stress in the slip direction and is calculated using τ=σcosøcosλ. The critical resolved shear stress is the shear stress that is sufficient to initiate slip. Note that the critical resolved shear stress is never more than 50% of the yield stress.