2025-2026 PRACTICE PROBLEM SET
SOLUTIONS #1-3 MECH 321 (Properties and
Failure of Materials) Concordia University
Question 1
A specimen of ductile cast iron with a rectangular cross section of
dimensions 4.8 mm 15.9 mm (3/16 in. 5/8 in.) is deformed in tension.
Using the load-elongation data provided in the table on the right, complete
problems (a) through (f).
a) Plot the data as engineering stress versus engineering strain.
b) Compute the modulus of elasticity.
c) Determine the yield strength at a strain offset of 0.002.
d) Determine the tensile strength of this alloy.
e) Compute the modulus of resilience.
f) What is the ductility, in percent elongation?
Solution:
a) 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.
The data are plotted below in two adjacent plots: the plot on the left shows the entire stress-strain
curve, while the plot on the right extends just beyond the elastic region of deformation.
,b) The elastic modulus is the slope in the linear elastic region (Equation 6.10) as
c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain
curve at approximately 280 MPa (40,500 psi).
d) The tensile strength is approximately 410 MPa (59,500 psi), corresponding to the maximum
stress on the complete stress-strain plot.
, e) From Equation 6.14, the modulus of resilience is just
which, using data computed above, yields a value of
f) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-
hundred. The total fracture strain at fracture is 0.185; subtracting out the elastic strain (which is
about 0.001) leaves a plastic strain of 0.184. Thus, the ductility is about 18.4%EL.
Question 2
A steel alloy specimen having a rectangular cross
section of dimensions 19 mm × 3.2 mm has the
stress–strain behavior shown on the right. If this
specimen is subjected to a tensile force of 110,000 N
then:
a) Determine the elastic and plastic strain values.
b) If its original length is 610 mm, what will be its
final length after the load in part (a) is applied and
then released?
Solution:
a) Determining the applied stress:
𝐹𝐹 110000 𝑁𝑁
σ= 𝐹𝐹
= = =1809.2 𝑀𝑃𝑎
𝐴0 𝑏0𝑑0 (19×10−3 𝑚)(3.2×10−3 𝑚)
From the Figure, this point is in the plastic region so the specimen will experience both elastic and
plastic strains. The total strain at this point ε𝑡 is about 0.022. We are able to estimate the amount
of permanent strain recovery ε𝑒 from Hooke's law, as
σ
ε𝑒 =
𝐸
Since E = 207 GPa for steel (Table 6.1)
1809.2 MPa
ε𝑒 = 207×103𝑀𝑃𝑎 =0.0087
Plastic strain is just the difference between the total and elastic strains; that is
ε𝑝= ε𝑡 − ε𝑒 = = 0.022– 0.0087 = 0.0133
SOLUTIONS #1-3 MECH 321 (Properties and
Failure of Materials) Concordia University
Question 1
A specimen of ductile cast iron with a rectangular cross section of
dimensions 4.8 mm 15.9 mm (3/16 in. 5/8 in.) is deformed in tension.
Using the load-elongation data provided in the table on the right, complete
problems (a) through (f).
a) Plot the data as engineering stress versus engineering strain.
b) Compute the modulus of elasticity.
c) Determine the yield strength at a strain offset of 0.002.
d) Determine the tensile strength of this alloy.
e) Compute the modulus of resilience.
f) What is the ductility, in percent elongation?
Solution:
a) 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.
The data are plotted below in two adjacent plots: the plot on the left shows the entire stress-strain
curve, while the plot on the right extends just beyond the elastic region of deformation.
,b) The elastic modulus is the slope in the linear elastic region (Equation 6.10) as
c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the stress-strain
curve at approximately 280 MPa (40,500 psi).
d) The tensile strength is approximately 410 MPa (59,500 psi), corresponding to the maximum
stress on the complete stress-strain plot.
, e) From Equation 6.14, the modulus of resilience is just
which, using data computed above, yields a value of
f) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by one-
hundred. The total fracture strain at fracture is 0.185; subtracting out the elastic strain (which is
about 0.001) leaves a plastic strain of 0.184. Thus, the ductility is about 18.4%EL.
Question 2
A steel alloy specimen having a rectangular cross
section of dimensions 19 mm × 3.2 mm has the
stress–strain behavior shown on the right. If this
specimen is subjected to a tensile force of 110,000 N
then:
a) Determine the elastic and plastic strain values.
b) If its original length is 610 mm, what will be its
final length after the load in part (a) is applied and
then released?
Solution:
a) Determining the applied stress:
𝐹𝐹 110000 𝑁𝑁
σ= 𝐹𝐹
= = =1809.2 𝑀𝑃𝑎
𝐴0 𝑏0𝑑0 (19×10−3 𝑚)(3.2×10−3 𝑚)
From the Figure, this point is in the plastic region so the specimen will experience both elastic and
plastic strains. The total strain at this point ε𝑡 is about 0.022. We are able to estimate the amount
of permanent strain recovery ε𝑒 from Hooke's law, as
σ
ε𝑒 =
𝐸
Since E = 207 GPa for steel (Table 6.1)
1809.2 MPa
ε𝑒 = 207×103𝑀𝑃𝑎 =0.0087
Plastic strain is just the difference between the total and elastic strains; that is
ε𝑝= ε𝑡 − ε𝑒 = = 0.022– 0.0087 = 0.0133
