EMCH 315 Final Spring 2018.pdf

E MCH 315 (All Sections) Spring 2018

Practice Problems for Preparation for Final Exam
2
[1] A tensile specimen of initial length 100 mm and cross-sectional area 25 mm is
subjected to an axial load that produces the true stress - true strain curve shown.
The logarithmic plot of true stress - true strain is also shown for the region beyond the yield
point. (a) Describe the true stress-true strain behavior by determining the coefficients (σo
and n) in the equation: σt= σo εt (b) Determine the length of the specimen when the true
n


strain is 0.3. (c) Determine the cross-sectional area when the true strain is 0.06. (d)
Determine the tensile strength (true and engineering) of the specimen.




m
60 100




er as
True Stress (MPa)
50
True Stress (MPa)




co
40




eH w
30




o.
20
10
rs e
ou urc
0 10
0 0.1 0.2 0.3 0.4 0.5 0.6 0.01 0.1 1
True Strain True Strain
o
aC s


Figure 1 Figure 2
vi y re



(a) From figure 2 find slop of the straight line and true stress
when true strain is 1. Extrapolate the curve to t=1, that will
ed d
ar stu




give 0= 65 MPa
n = slope = (log65 – log30) / (log1 – log0.01) = 0.155
sh is




(b) t = 0.3, L0 = 100mm,
Th




Use t = ln(Li/L0) Li = L0 x et
Solve for Li = 135 mm

(c) Do same as in (b) but now use t = ln(A0/Ai), A0 = 25mm2,
solve for Ai = 23.5 mm2
(d) Use (TS)nom = 0nne-n = 37.9 MPa and (TS)t =0nn = 45.73
MPa


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, [2] A 440C stainless steel alloy has the following properties: density 0.28 lb/in3, elastic
modulus 29 Mpsi, yield strength 275 ksi, and work hardening exponent 0.04. What would
be the ultimate tensile strength (find both nominal and true)?
(Hint: use y as nom and find total strain (true and nominal))

Given: n = 0.04, E = 29 Mpsi, y = 275ksi = nom
Find (UTS)nom = 0nne-n and (UTS)t =0nn (we know n but do not
know 0), use parabolic law to find out 0 .
t =0tn .
But we do not know t and t;
Use the following constitutive equations for finding out t and t.




m
er as
t = nom (nom +1) and t = ln (nom +1)




co
eH w
Now we do not know nom, use Hooke’s law for finding out e (elastic




o.
strain) and add 0.002 (permanent strain by definition of yield
rs e
strength) to it to get total strain (nom) at y.
ou urc
e = nom /E = 0.009483 nom = 0.002+ 0.009483 = 0.011483
o
aC s


Now put all the numbers in the equations above:
vi y re



t = 278.158 ksi t = 0.011476 0 = ????? ksi
Now put all the numbers in (UTS)nom = 0nne-n and (UTS)t =0nn
ed d




(UTS)nom = ????ksi and (UTS)t =????? ksi
ar stu




[3] A steel alloy rod will experience cyclic axial loading during a single day according to
sh is




the stress condition schedule below. For this steel alloy, the tensile strength is 250 MPa and
Th




the S-N curve for fully reversed loading is given. (a) What is the fatigue/endurance limit of
the steel alloy? (b) Determine the number of days that the rod can endure the prescribed
load schedule before failing due to fatigue.
R = –1
Reversed stress amplitude (MPa )




300
# of max min
cycles stress stress 250

(MPa) (MPa) 200

150
1. 1100 50 -50
100
2. 1100 150 -50 88
3. 1100 200 0 50

0
1.E+03 1.E+04 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09
Number of cycles to failure
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