MSE 235-Section 01&02
ASSIGNMENT 5
MECHANICAL PROPERTIES OF MATERIALS
1)Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10 mm. A tensile force of 1500 N
produces an elastic reduction in diameter (d) of 6.7x10-4 mm. Compute the modulus of elasticity, E (in GPa) for this alloy,
given that Poisson’s ratio () is 0.35.
1500 N
Since the deformation is elastic we can use Hooke’s Law and Poisson Ratio Equation:
∆𝑑
−6.4𝑥10−4
−6.4𝑥10−4
𝜀𝑥
𝑑
10
10
𝜗=− =− 𝑜 →𝜗=−
→ 0.35 = −
→ 𝜀𝑧 ≅ 1.83𝑥10−4
∆𝑙
𝜀𝑧
𝜀𝑧
𝜀𝑧
𝑙𝑜
𝜎 = 𝐸. 𝜀𝑧 (𝐻𝑜𝑜𝑘𝑒 ′ 𝑠𝐿𝑎𝑤)
→
𝐹
= 𝐸. 𝜀𝑧 →
𝐴𝑜
1500𝑁
𝑁
= 𝐸. (1.83𝑥10−4 ) → 𝐸 ≅ 104.4𝑥109 𝑃𝑎 ( 2 ) ≅ 104.4 𝐺𝑃𝑎
−3
10𝑥10 𝑚 2
𝑚
𝜋(
)
2
Type equation here.
2) A cylindrical metal specimen having an original diameter (do) of 12.8 mm and gauge length of 50.80 mm is pulled in tension
until fracture occurs. The diameter at the point of fracture (df) is 8.13 mm and fractured gauge length is 74.17 mm. Calculate the
ductility in terms of % reduction in area (RA) and % elongation.
lo = 50.80
mm
%𝑅𝐴 =
𝑑𝑜2 − 𝑑𝑓2
%𝐸𝐿 =
𝑑𝑜2
𝑥100 =
(12.8)2 − (8.13)2
𝑥100 ≅ 59.7%
(12.8)2
𝑙𝑓 − 𝑙𝑜
74.17 − 50.80
𝑥100 =
𝑥100 ≅ 46%
𝑙𝑜
50.80
3)A cylindrical metal specimen 12.7 mm in diameter (do) and 254 mm long (lo) is to be subjected to a tensile stress of 28 MPa. It
is known that at this stress level the resulting deformation will be totally elastic.
If the elongation (l) must be less than 0.080 mm, which of the metals given in table are suitable candidates? Why?
Since the deformation is elastic we can use Hooke′ sLaw
𝜎 = 𝐸. 𝜀𝑧 (𝐻𝑜𝑜𝑘𝑒 ′ 𝑠𝐿𝑎𝑤) → 𝜎 = 𝐸.
In the question it stated that ∆𝑙 < 0.08 𝑚𝑚 →
𝜎.𝑙𝑜
𝐸
∆𝑙
𝜎. 𝑙𝑜
𝜎. 𝑙𝑜
→ 𝑤ℎ𝑒𝑟𝑒 ∆𝑙 =
𝑜𝑟 𝐸 =
𝑙𝑜
𝐸
∆𝑙
< 0.08 𝑜𝑟
𝜎.𝑙𝑜
0.08 𝑚𝑚
<𝐸→
(28𝑥106 𝑃𝑎 )(254 𝑚𝑚)
(0.08 𝑚𝑚)
<𝐸
88.9𝑥109 𝑃𝑎 < 𝐸
→ 𝐸𝑥𝑐𝑒𝑝𝑡 𝐴𝑙 (𝐸𝐴𝑙 = 69𝑥109 𝑃𝑎) and Mg (𝐸𝑀𝑔 = 45𝑥109 𝑃𝑎) all of the materials given in Table are canditates.
4) ) A cylindrical specimen of an aluminum alloy having a diameter of 12.8 mm and a length of 50.8 mm is pulled in
tension. By using the stress-strain diagram shown:
a. Compute the modulus of elasticity
b. Determine the yield strength according to 0.2% rule.
c. Determine the ultimate tensile strength of this alloy.
d. This material is subjected to a force of 40,000N. Calculate the increase in length (l) of the bar when load is
on the bar.
(a)Elastic Modulus is the slope of the linear section of Stress-strain diagram
v
  0.012
E=
∆σ
200 − 0
→
≅ 62.500 MPa ≅ 62.5 GPa
∆ε
0.0032 − 0
(b) Green dashed line is drawn at 0.002 (or 0.2%) parallel to
elastic linear line. The intersection point of dashed green line
with stress-strain curve gives the yield strength.
𝜎𝑦𝑖𝑒𝑙𝑑 ≅ 290 𝑀𝑃𝑎
(c) Ultimate tensile strength is the maximum stress level in
stress-strain diagram. From the blue dashed line:
𝜎𝑈𝑇𝑆 ≅ 380 𝑀𝑃𝑎
∆𝑙
(d) We should first find corresponding strain (𝜀 = ):
𝑙𝑜
𝜎=
𝐹
→ 𝜎=
𝐴𝑜
40000𝑁
2
𝜋(
12.8𝑥10−3 𝑚
)
2
≅ 310.85 𝑀𝑃𝑎
From stress strain curve: 𝜀 ≅ 0.012
𝜀=
∆𝑙
∆𝑙
→ 0.012 =
→ ∆𝑙 = 0.6096 𝑚𝑚
𝑙𝑜
50.8
ASSIGNMENT 6
DISLOCATIONS and STRENGTHENING MECHANISMS
1)Why does slip in metals usually take place on the densest-packed plane and in the closest-packed direction?
Slip is favored on close packed planes since a lower shear stress for atomic displacement is required.
Moreover, slip occurs in close packed directions since less energy is required to move atoms in these directions
o
2)A single crystal of cadmium is oriented for a tensile test such that its slip plane normal makes an angle of 65 with
o
o
o
the tensile axis. Three possible slip directions makes an angle of 30 , 48 , and 78 with the same tensile axis.
a) Which of these three slip directions is most favored?
b) If plastic deformation begins at a tensile stress of 1.55 MPa, determine the critical
resolved shear stress for cadmium.
(a)Slip favors in a slip system at which resolved shear stress (R = CosCos) is max.
𝜏𝑅 = 𝜎𝐶𝑜𝑠65𝐶𝑜𝑠30 ≅ 0.37𝜎
=30 is most favored
o
𝜏𝑅 = 𝜎𝐶𝑜𝑠65𝐶𝑜𝑠48 ≅ 0.28𝜎
.
𝜏𝑅 = 𝜎𝐶𝑜𝑠65𝐶𝑜𝑠78 ≅ 0.09𝜎
(𝑏) 𝜏𝐶𝑅𝑆𝑆 = 𝜎𝑦 (𝐶𝑜𝑠∅𝐶𝑜𝑠)𝑚𝑎𝑥
τCRSS = 1.55 MPa Cos45Cos45 = 0.775 MPa
3) The critical resolved shear stress for copper is 0.48 MPa. Determine the max. possible yield strength for a single
crystal of copper pulled in tension.
𝜏𝐶𝑅𝑆𝑆 = 𝜎𝑦 (𝐶𝑜𝑠∅𝐶𝑜𝑠)𝑚𝑎𝑥 → 0.48 = 𝜎𝑦 𝐶𝑜𝑠45𝐶𝑜𝑠45 → 𝜎𝑦 = 0.96 𝑀𝑃𝑎
-2
4) The yield point for an iron that has an average grain diameter of 1x10 mm is 230 MPa. At a grain diameter of
-3
6x10 mm, the yield point increases to 275 MPa. At what grain diameter will the yield point be 310 MPa.
𝜎𝑦 = 𝜎𝑜 +
𝑘𝑦
√𝑑
Hall Petch Equation (defines the relationship between yield strength and average grain size, d)
− 230 = −𝜎𝑜 −
275 = 𝜎𝑜 +
𝑘𝑦
√1𝑥10−2 𝑚𝑚
𝑘𝑦
√6𝑥10−3 𝑚𝑚
2.9𝑘𝑦 = 45 → 𝑘𝑦 = 15.5
275 = 𝜎𝑜 +
15.5
√6𝑥10−3 𝑚𝑚
→ 𝜎𝑜 = 74.9
310 𝑀𝑃𝑎 = 74.9 +
15.5
√𝑑
→ 𝑑 = 4.35𝑥10−3 𝑚𝑚
ASSIGNMENT 7
FAILURE-I
1)Briefly explain why BCC and HCP metal alloys may experience a ductile-to-brittle transition with decreasing temperature,
whereas FCC alloys do not experience such a transition.
It is related to number of slip systems. Since the number of
slip systems in FCC metals at a specific temperature is
higher than that of BCC and HCP, FCC metals do not
experience ductile to brittle transition.
2)What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature 2.5x10 -4
mm and a crack length of 2.5x10-2 mm when a tensile stress of 170 MPa is applied?
a 1/2
Maximum stress at the crack tip is, σm = 2σo ( )
ρt
2a = 2.5x10−2 mm → a = 1.25x10−2 mm
1
1.25x10−2 mm 2
σm = 2(170 MPa) (
) → σm ≅ 2404,2 MPa
2.5x10−4 mm
3)A large plate is fabricated from a steel alloy that has a plain strain fracture toughness of 82.4 MPam. If, during service use,
the plate is exposed to a tensile stress of 345 MPa, determine the minimum length of internal crack that will lead to fracture.
Assume a value of 1.0 for Y.
𝐾𝐼𝐶 = 𝑌𝜎√𝜋𝑎 → 82.4 𝑀𝑃𝑎√𝑚 = (1.0)(345 𝑀𝑃𝑎)√𝜋(𝑎) → 𝑎 = 0.018 𝑚 → 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑐𝑟𝑎𝑐𝑘 𝑙𝑒𝑛𝑔𝑡ℎ = 2𝑎 = 0.036 𝑚 = 36 𝑚𝑚
4)Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPam. It has
been determined that fracture results at a stress of 300 MPa when the maximum internal crack length is 4.0 mm. For the same
component and alloy, will fracture occur at a stress level of 260 MPa when the maximum internal crack length is 6.0 mm? Why
or why not?
Fracture occurs if calculated fracture toughness for given stress level and crack length exceeds the theoretical fracture
toughness of the material. For computation we need to find parameter ‘’Y’’ first:
= 300 MPa; 2a= 4.0 mm  a= 2.0 mm
𝐾𝐼𝐶 = 𝑌𝜎√𝜋𝑎 → 40 𝑀𝑃𝑎√𝑚 = 𝑌(300 𝑀𝑃𝑎)√𝜋(2𝑥10−3 𝑚) → 𝑌 = 1.68
 when =260 MPa and 2a=6.0 mm (a=3.0 mm)  𝐾𝐼𝐶 = 1.68(260 𝑀𝑃𝑎)√𝜋(3𝑥10−3 𝑚) = 42.4𝑀𝑃𝑎√𝑚 > 40 𝑀𝑃𝑎√𝑚
 FRACTURE OCCURS
ASSIGNMENT 8
FAILURE-II
1)The fatigue data for a brass alloy under compressive and tensile stress cycle are given as follows:
Stress Amplitude(MPa)
170
148
130
114
92
80
74
Cycles to Failure
3.7x104
1.0x105
3.0x105
1.0x106
1.0x107
1.0x108
1.0x109
Make an S-N (stress amplitude versus logarithm cycles to failure) plot using
these data and answer the followings
(a) Is there an endurance limit for the alloy?
(b) Determine the fatigue strength at 7x107 cycles
(c) Calculate the maximum tensile stress at 7x107 cycles if the
reversed cyclic stress is applied to material
(d) Determine the fatigue life for 150 MPa.
(a) There is no endurance limit.
7
(b) From the figure at 7x10 cycles fatigue strength is  90
MPa
(c)
In reversed cycle σmax (tensile) = σmin (compression)
Stress amplitude, S =
σmax − (−σmin )
2
At 7x10 cycles stress amplitude is  90 MPa
7
90 𝑀𝑃𝑎 =
2𝜎𝑚𝑎𝑥
𝐹𝑚𝑎𝑥
→ 𝜎𝑚𝑎𝑥 (
) = 90 𝑀𝑃𝑎
2
𝐴𝑜
(d) From the figure at 150 MPa; Nf  90x10 = 9x10
3
4
2) A metal component is subjected to a fluctuating load cycle ranging from +60 MPa to +10 MPa.
(a) Draw a schematic stress vs time curve and identify the type of cyclic stress
(b) Calculate stress values given below and show them on stress vs time curves drawn in part (a)
i-The mean stress
ii-The stress range
iii-The stress amplitude
+max
(b)
REPEATED STRESS CYCLE
60 MPa
(𝑖) σm =
a
r
m
10 MPa
σmax +σmin
2
→ σm =
60+10
2
= 35 MPa
(ii) σr = σmax + (−σmin ) → σm = 60 + (−10) = 50MPa
(iii)S = σa =
σmax −σmin
2
=
60−10
2
= 25 MPa
-min
3) How can you determine whether a machine component is failed due to fatigue or not? Which precautions should be taken to increase the
fatigue life of a component?
Fatigue failure can be determined by examining the fracture surface:
Fracture surface contains crack initiation point, crack propogation part which is
composed of beackmarks, and a final failure region.
Crack initation point
beachmarks
Precautions that should be taken to increase fatigue life:
1-Sharp changes in cross-section in cross-section should be avoided,
2-Polish the surface of the material,
3-Create compressive stresses on the surface by shot peening or carburizing,
4-Avoid corrosive environment if possible.
Final failure region
4)Steady state creep data are given for a copper alloy pipe taken at 200oC:
Steady State Creep Rate, s (hr-1)
2.5x10-3
2.4x10-2
Stress,  (MPa)
55
69
𝑄𝑐
𝜀𝑠̇ = 𝐾2 𝜎 𝑛 𝑒 (−𝑅𝑇)
2.4𝑥10−2 𝐾2 69𝑛 𝑒
2.5𝑥10
−3 =
𝐾2 55𝑛 𝑒
140.000 𝐽/𝑚𝑜𝑙𝑒
(−8.314 𝐽/𝑚𝑜𝑙𝐾(523𝐾))
140.000 𝐽/𝑚𝑜𝑙𝑒
(−8.314 𝐽/𝑚𝑜𝑙𝐾(523𝐾))
2.4𝑥10−2 = 𝐾2 691.8 𝑒
→ 𝑛 =1.8
140.000 𝐽/𝑚𝑜𝑙𝑒
)
8.314 𝐽/𝑚𝑜𝑙𝐾(523𝐾)
(−
→ 𝐾2 ≅ 1.12𝑥109
o
At 250 C and 48 MPa, the steady state creep rate is:
𝜀𝑠̇ = 1.12𝑥109 (481.8 )𝑒
140.000 𝐽/𝑚𝑜𝑙𝑒
(−
)
8.314 𝐽/𝑚𝑜𝑙𝐾(523𝐾)
→ 𝜀𝑠̇ = 1.25𝑥10−2 ℎ𝑟 −1
If activation energy for creep is 140 kJ/mole, compute the steady
state creep rate of the pipe at a temperature of 250 oC and a stress
level of 48 MPa.