ME 4733: Deformation and Fracture of
Engineering Materials
Spring 2002
Problem Set 1
1) Hertzberg, 1.3
A 200-mm-long rod with a diameter of 2.5mm is loaded with a 2000-N weight. If the
diameter decreases to 2.2 mm, compute the following:
(a) The final length of the rod.
(b) The true stress and true strain at this load.
(c) The engineering stress and strain at this load.
Solution:
(a) Initial state:
diameter do = 2.5mm , l0 = 200mm
Final state:
diameter d f = 2.2mm , l f
Assume this is a constant-volume process
d 2f
d02
π ⋅ l0 = π ⋅ l f
4
4
 d 2
 2.5  2
0
⇒ The final length of the rod is l f =   ⋅ l0 =   ⋅ 200mm = 258.26mm
 2.2 
 df 
(b)
(c)
2000N
P
=
= 526.4 MPa ;
2
(2.2mm) 2
df
3.14 ⋅
π⋅
4
4
lf
258.26
= 0.2556 .
The true strain is εtrue = ln = ln
l0
200
The true stress is σ true =
P
2000N
=
= 407.4 MPa ;
2
d0
(2.5mm) 2
π⋅
3.14 ⋅
4
4
l f − l0 258.26 − 200
=
= 0.2913.
The engineering strain is εtrue =
l0
200
The engineering stress is σ true =
Note:
(1) In tensile test, the true stress is always higher than engineering stress. How about in
compressive state?
(2) In tensile test, the true strain is always lower than engineering strain.
2) Hertzberg, 1.8
A 5-cm-long circular rod of 1080 as-rolled steel (diameter=1.28cm) is loaded to failure in
tension. What was the load necessary to break the sample? If 80% of the total elongation
was uniform in character prior to the onset of localized deformation, computer the true
stress at the point of incipient necking. ( refer to table 1.6a in Hertzburg textbook )
Solution:
(1.28cm) 2
The initial cross-sectional area of the rod is A0 = π
= 1.287cm2 .
4
The tensile strength, σ ts , is defined as the maximum load divided the initial crosssectional area, which is at the top point of the engineering stress-strain curve. For 1080
as-rolled steel σ ts = 965MPa (from Hertzburg, table 1.6a).
The load, P, necessary to break the sample should reach this top point, so
`
P = σ ts ⋅ A0 = 965 × 10 6 ⋅ 1.287 × 10−4 N = 124.2 kN
The initial length of the rod is l0 = 5cm . From Hertzburg, table 1.6a, the total
elongation is 12% for 1080 as-rolled steel. This total elongation consists mainly of two
parts, uniform elongation (80%) and localized necking (20%), elastic deformation is
small and can be neglected. At the point of incipient necking, the elongation is uniformly
distributed. After that, deformation contributes only (or mostly) to localized necking. So,
at the point of the incipient necking, the rod length is
l = (1 + 0.12 × 0.8) ⋅ l0 = 1.096 × 5cm = 5.48cm .
At the point of the incipient necking the rod arrives at the state to bear the maximum
load, which is P = 124.2 kN . The plastic deformation can be approximated as constantvolume process, and the rod has uniform elongation for the process before necking, so the
cross-sectional area at the point for incipient necking is
l
5
A = A0 0 = 1.287cm2 ×
= 1.174 cm2 .
l
5.48
The true stress at the point of incipient necking is
P
124.2 kN
σ true = =
= 1058MPa .
A 1.174 × 10−4 m 2
Note:
The elongation at the point of the incipient necking is the uniform part of the elongation
during necking.
3) Hertzberg, 1.11
A platform is suspended by two parallel rods, as shown in the sketch, with each rod being
1.28cm in diameter. Rod A is manufactured from 4340 steel [Q + T (650°)] ( E = 210GPa ,
σ ys = 855MPa ); rod B is made from 7075-T6 aluminum alloy ( E = 70GPa ,
σ ys = 505MPa ).
a) What uniform load can be applied to the platform before yielding will occur?
b) Which rod will be the first to yield? (Hint: Both rods experience the same elastic
strain)
Solution:
b) For rod A, the strain at the yielding point is
σ A 855 × 10 6 Pa
εysA = ysA =
= 4.071 × 10−3
210 × 10 9 Pa
E
For rod B, the strain at the yielding point is
σ ysB 505 × 10 6 Pa
B
εys = B =
= 7.214 × 10−3
9
70 × 10 Pa
E
The platform needs to be always kept horizontal, that means both rods experience
the same elastic strain. Since εysA < εysB , so rod A will be the first to yield.
b)
The strain the system can experience without yielding is
ε = min(εysA,εysB ) = 4.071 × 10−3
The load applied to the two rods is
P = ( E A + E B )εA
So,
(1.280 × 10−2 ) 2
P = (210 + 70) × 10 9 × 4.071 × 10−3 × π
N = 147 kN
4
Note:
Here we can see the elastic strain, for 4340 steel ε = min(εysA,εysB ) = 4.071 × 10−3 is much
smaller than plastic strain, with is in the order of 0.10. When strain is very small, there is
no difference between engineering strain and true strain.
4) Hertzberg, 1.12
A 100-mm-long rod of 1340 steel [Q + T (205°)] is subjected to a load of 50,000N. If the
diameter of the rod is 10 mm, what strain would the rod experience? What strain would
remain if the load were removed?
Solution:
The stress is
P 50000 N
σ= =
= 636.6 MPa
0.012
A
π
4
The yield strength of 1340 steel [Q + T (205°)] is 1590MPa. The current stress state is
below the yielding point, so the elastic theory can apply. Since Young’s modulus does
not change mush for different treatment, we can use the Young’s modulus for steel from
the last problem as E = 210GPa . Using Hooke’s law, the strain in the rod is
σ 636.6 × 10 6 Pa
ε= =
= 3.03 × 10−3
9
E
210 × 10 Pa
For elastic deformation the strain returns to zero when load removed.
Note:
Here because the elastic strain is small, there is no difference between engineering and
true strain. So in the problem, only “strain” is mentioned.