Exam 1 Solution FL11

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MatS 2001 - Exam #1 Solution

October 7, 2011

Question 1 True/False (15 Points)

Circle True (T) or False (F) (Do not guess: +1 points for correct answer, 0 points if no answer isprovided, -1 point for incorrect answer).

a) T F The strain hardening exponent in the relationship T K T

n is 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 internal crack willcause fatigue failure before the external 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 GPa.

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 decreases 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 Plastic deformation results in a permanent change in shape.

l) T F Crack propagation in brittle failure becomes easier as the crack length increases.

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

n) T F For a brittle material, yield stress and tensile strength can be the same

o) T F True stress is always smaller than Engineering stress.



Question 2 Creep (15 Points)

The hook shown to the right will s

The hook operates continuously

Larson – Miller parameter is sho

L-M T 38.316 1.796logday

t

common log, base 10.

a) (10) If the useful life of the hfactor of 2), is to be 10 yearindicated portion of the hoo

L-M 583

From grapdesign stre

, or

7.14 i

F A

d

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upport a container of molten steel and must c

t a temperature of 583 C. The relationship b

n below:

310 x

when T is in Kelvin and rupture time is i

ook, with an applied stress ½ the maximum cs of continuous operation, what must be the d

?

3

2 2

+273 38.316 1.796 log 3650 10

stress 2,000 psi, or with safess = 1,000 psi

40,000 440 in , 40

1,000 psi

.

x

lbd

arry a load of 40,000 lb.

tween stress and

days. Note: “log” is

alculated stress (safetyiameter (in inches) of the

38.28

y factor,

50.93



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b) (5) Calculations by a “new” engineer stated that the diameter should be 3.5” (not the answer to part“a”). If a hook of this size was used, what would be its useful life (in years)?

3

2

L-M 535+273 38.316 1.796 log t 10

40,000( ) stress = 4,157 psi

3.54

Design stress = Working stress x safety factor = 4,157 x 2 = 8300 psi

From the graph: L-M 32.8 or,

32.8

log t

days

days

x

Applied working

432.8 138.316 9.7829 10

.856 1.796

1, or years = 0.0027days

x

t

Question 3 Poisson’s Ratio (5 Points)

A cylindrical specimen of an alloy, 8 mm in diameter, is stressed elastically in tension. A force of 15,700

N produces a reduction in specimen diameter of 5x10-3mm. Compute Poisson’s ratio for this material if

its modulus of elasticity is 140 GP



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Question 4 Fracture/Fatigue (35 Points)

An aluminum rod is used as a support in a critical application. The rod is subjected to stress cyclingbetween 50 MPa compression and 320 MPa tension. The material from which the rod is made can have

no flaw larger than 0.10 mm. Two alloys are being considered:

Yield Strength [MPa] Elastic Modulus [GPa] KIC

[MPa – m1/2

]

Al 7075 495 71.0 24

Al 2024 345 72.4 50

Both alloys follow the Paris Equation: da/dN =A (ΔK ) m where A=1.0x10-13

, m=3.1, and Y=1.73. These

values give crack growth in meters/cycle when ΔK is in MPa-m1/2

.

a) (5) Will either alloy fail upon initial loading? If so, which one(s) and why? If not, why not?

b) (5) What is the critical crack length for fatigue failure for each alloy?

c) (5) Without actually calculating cycles to failure, which rod will fail first in continuous operation?Why?

Alloy 7075 will fail first



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d) (5) What was the length of the rod when supporting the 35,000 N load?

e) (5) What is the length of the rod when the 35,000 N load is removed?

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