Page 134 - Mechanical Behavior of Materials

Section 4.3 Engineering Stress–Strain Properties 135

points during the test. After fracture, the broken halves were reassembled, and the following
measurements were made: (1) Marks originally 25.4 mm apart and on opposite sides of the
necked region were 38.6 mm apart due to the lengthwise stretching in the specimen. (2) Similar
marks originally 50.8 mm apart were 70.9 mm apart. (3) The final minimum diameter in the
necked region was 5.28 mm.

(a) Determine the following materials properties: elastic modulus, 0.2% offset yield
strength, ultimate tensile strength, percent elongation, and percent reduction in area.
(b) Assume that the test was interrupted upon reaching a strain ε = 0.0070, and the
specimen unloaded to zero force. Estimate the elastic strain recovered and the plastic
strain remaining. Also, what would be the new length of the original 50.8 mm gage
section?

Solution (a) For each force value, engineering stresses are first calculated from Eq. 4.1, as
given in the first column of Table E4.1(b). For example, for the first force value above zero,
P P 4(6670 N) N
σ = = 2 = 2 = 102.3 2 = 102.3MPa
A i πd /4 π(9.11 mm) mm
i
Plotting these versus the corresponding engineering strains gives Fig. E4.1(a). However, on this
graph, the yield region is too crowded to see the needed detail, so the data for the beginning of the
test are plotted with a sensitive strain scale in Fig. E4.1(b). The first four data points appear to lie
on a straight line in Fig. E4.1(b), so that a least squares line of the form y = mx is fitted to these,
giving the elastic modulus as E = 201,200 MPa (Ans.). A line is drawn on Fig. E4.1(b) parallel
to the slope E and through the plastic strain offset of ε po = 0.002. The intersection of this with
the stress–strain curve gives the yield strength as σ o = 264 MPa (Ans.). From Fig. E4.1(a), or
from the numerical values in Table E4.1(b), it is evident that the highest stress reached, and hence
the ultimate tensile strength, is σ u = 395 MPa (Ans.).
Noting that the last line of Table E4.1 corresponds to fracture, the stress and strain at
fracture are σ f = 266.8 MPa and ε f = 0.366. The latter gives a percent elongation at fracture
of ε f % = 100ε f = 36.6% (Ans.). The corresponding elongation after fracture can then be
estimated from Eq. 4.7:

σ f 266.8MPa
ε pf = ε f − = 0.366 − = 0.365, ε pf % = 36.5% Ans.
E 201,200 MPa
The length measurements made from the broken specimen give additional values for the
elongation after fracture:

L f − L i (38.6 − 25.4) mm
ε pf = = = 0.520, ε pf % = 52.0% (over 25.4mm)
L i 25.4mm
(70.9 − 50.8) mm
L f − L i
ε pf = = = 0.396, ε pf % = 39.6% (over 50.8mm)
L i 50.8mm
Ans.