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Introduction to the Science of Engineering Materials - Midterm Exam | MATS 2001, Exams of Materials science

Material Type: Exam; Professor: Schott; Class: Introduction to the Science of Engineering Materials; Subject: Materials Science; University: University of Minnesota - Twin Cities; Term: Fall 2013;

Typology: Exams

2012/2013

Uploaded on 10/09/2013

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Exam 1 - MatS 2001
Introduction to the Science of Engineering Materials
Fall 2013 – October 4, 2013
Name (Print)
ID #
EXAM RULES:
All notes, books, notebooks, note cards, cell phone, etc. must be off the desk
and inside closed and zippered backpacks. Students should be separated by at
least one empty chair. Only simple scientific calculators (cannot store
formulae or text) are allowed.
SHOW YOUR WORK!!
Numerical answers with incorrect concepts, or solution procedures or no work shown will
receive no credit, even if the numerical answer is correct.
GOT QUESTIONS? ASK!!
Equations you may need and the periodic table are on the last page. If you need extra paper to
work a problem use the back of the page but please indicate this.
Constants
Avogadro's number NA = 6.023 x 1023 1eV = 1.602 x 10-19 J
atomic mass unit 1 amu = 1.66 x 10-27 kg speed of light c = 3.0 x 108 m/sec
Bohr radius ao = 0.529 Å electron charge e = 1.602 x 10-19 C
Boltzman constant k = 8.617 x 10-5 eV/K Planck constant h = 6.626 x 10-34 J sec
proton mass mp = 1.672 x 10-27 kg neutron mass mn = 1.675 x 10-27 kg
electron mass me = 9.11 x 10-31 kg gas constant, R = 1.9872 cal mol-1 K-1
gas constant, R = 8.314 J mol-1 K-1 1 Pa = 1.45 x 10-4 psi
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Exam 1 - MatS 2001

Introduction to the Science of Engineering Materials

Fall 2013 – October 4, 2013

Name (Print)

ID

EXAM RULES:

All notes, books, notebooks, note cards, cell phone, etc. must be off the desk

and inside closed and zippered backpacks. Students should be separated by at

least one empty chair. Only simple scientific calculators (cannot store

formulae or text) are allowed.

SHOW YOUR WORK!!

Numerical answers with incorrect concepts, or solution procedures or no work shown will

receive no credit, even if the numerical answer is correct.

GOT QUESTIONS? ASK!!

Equations you may need and the periodic table are on the last page. If you need extra paper to

work a problem use the back of the page but please indicate this.

Constants

Avogadro's number N A

= 6.023 x 10

1eV = 1.602 x 10

J

atomic mass unit 1 amu = 1.66 x 10

kg speed of light c = 3.0 x 10

m/sec

Bohr radius a o

= 0.529 Å electron charge e = 1.602 x 10

  • C

Boltzman constant k = 8.617 x 10

eV/K Planck constant h = 6.626 x 10

J sec

proton mass m p

= 1.672 x 10

kg neutron mass m n

= 1.675 x 10

kg

electron mass m e

= 9.11 x 10

kg gas constant, R = 1.9872 cal mol

K

gas constant, R = 8.314 J mol

K

1 Pa = 1.45 x 10

psi

Question 1 (15 points)

Circle True or False

This time, question 1 was graded as 1 point when right, 0 when wrong or not answered. (In the past, it

was graded with 1 point when right, 0 when not answered, -1 when wrong.) The average estimated would

be 8 or 9. There were students guessing the answer by choosing true for all of them.

a) T F The strain hardening exponent in the relationship

s T

= Ke T

n

is equal tois equal to

the true strain at the point where necking begins.

b) T F A stiff material has a low modulus of elasticity.

c) T F Poisson’s ratio is based on constant volume in the elastic region

d) T F If internal and surface cracks are initially the same size, the surface crack

will cause fatigue failure before the internal crack.

e) T F The Larson – Miller Parameter directly relates creep failure time to

applied stress.

f) T F Typical yield stresses for metals are in the range of 10-300 MPa.

g) T F The most important factor in determining a material’s resistance to fracture is

the number of cracks in it.

h) T F Crack growth rate in fatigue increases as the crack length increases.

i) T F Ceramics have higher KIC values than metals.

j) T F KIC is small for metals because they have some ductility.

k) T F Fracture can happen before the yield stress is reached.

l) T F Polymers are covalently bonded.

m) T F Elements on the left side of the periodic table tend to form ionic bonds with

elements from the right side.

n) T F Melting temperature increases as bond energy increases.

o) T F In metals the valence electrons are free to drift through the entire metal.

(a) (10 pts) Calculate the elastic modulus, (in GPa), the yield stress, yield strain,

tensile strength, and strain at necking. Show your work on the graph.

(2 points for each of the five items, elastic modulus, the yield stress, yield strain,

tensile strength, and strain at necking)

..

 

200 MPa 0 MPa

E

0 0

80

025 0 0

GPa

(If the method shown on the graph is wrong and the answer

is not close to 80 GPa, 0 point; if the method shown on the graph is right, but answer is not

close to 80GPa, 1 point. )

Yield stress is found using a line with slope = E and 0.2% offset. (If did not use the 0.

method, 0 point)

As shown on the graph: y = 250 MPa

Yield strain can be read directly from the graph: y = 0.

As an alternative,

.

,

   . y

250 MPa

0 002

80 000 M a

0

P

005

Tensile strength is the maximum of the curve = 450 MPa

Strain at necking is strain at tensile strength = 0.

Most people got 10 points for this part.

(b) (15 Pts) A 2 m long, 2mm diameter rod of this alloy is loaded with 128 kg. What

are the resulting stress, strain, length, and diameter of the rod while

loaded? When the load is removed, what is the length of the rod?

(3 points for each of the five items, stress, strain, length, diameter, length after load

is removed)

With 128 kg.,

2 2

(128 kg)(9.81 m)(4)

399.3 MPa

(2 mm) s

From the graph: total = 0.125 (Most people used Hooke’s law to do this, which is

not right. At this stress the material is plastically deformed. If instead Hooke’s

law was used, the strain, 0.005, is much too small (just check the plot) and 0

points were given for the strain part. For the length and diameter, if the 0.

strain was used and the calculation done correctly, 3 points for each correct

answer, but the total score of part (b) should not be more than 10 points. If the

answer is not correct, 0, 1, or 2 points were given for the answers, depending on

quality of the equations and calculations.)

0

l l ( 1  ) ( 2 0 m)(1+0.125) = 2.25 m.

2

2

0 0

0 0

2

mm 2 m

, or , or , D=1.89 mm

4 2.25 m

or

1 399.3 MPa 1.125 450 MPA

A 2.79 mm , D 1.88 mm

i

i i i

i

T

i i

T

A l D

Al A l A

l

F

plastic 0 plastic

plastic total elastic plastic

400 MPa

80 GPa

l l (1 ) (2.0 m)(1 0.120) 2.24 m

or  

     

Most people got 4-10 points for part b.

2

1/

1/

0.0026 m

600 MPa 0.0026 54.2 MPa-m

54.2 MPa-m

967 MPa

IC

c

IC

IC

c

K

a

Y
K
K

Y a

Question 4 (10 points)

The hook shown to the right will support a container of molten steel and must carry a

load of 40,000 lb. The hook operates continuously at a temperature of 560 C. The

relationship between stress and the Larson – Miller parameter is shown on the graph

below:

where, for this material:

 

3

L-M T 38.316 1.796 log 10

day

t x

T is in Kelvin and rupture time is in days.

If the useful life of the hook, with a safety factor of 1.6 for the load, is to be 7 years of

continuous operation, what must be the diameter (in inches) of the indicated portion of

the hook?

Most students got the L-M value but many students could not calculate the stress or forgot to consider the

safety factor

LM  (^)  560  (^273)  [ 38. 316  1. 796 log( 365 x 7 yrs )] x 10

 3

From graph, Stress» 5 , 200 psi. Applying a safety factor of 1. 6 , the working stress

is 2800 psi.

F

 A , or

40 , 000 lb

lb

in

2

 22. 86 in

2

, d

2

 

 29. 1 in

2

d  5. 39 in.

0

F
A

T

i

F
A

0

l

l

 ln

f

T

i

l

l

  E 

E  2 G  1  

  G  (^ )

n

T T

  K 

1/

C

E

a

  K^  Y^^ ^  a

da

dN

 A  K

 

m

   

   

2 /2 2 /

0

/

m m

c

f (^) m m m

a a

N

m AY  

    

 

1

exp

n C

d Q

K

dN RT

3

( log ) 10 r

LM T C t

  

%  1  exp  0. 25 ( E )

2

 