Page 65 - Shigley's Mechanical Engineering Design

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40 Mechanical Engineering Design
Difficulties arise when using the gauge length to evaluate the true strain in the
plastic range, since necking causes the strain to be nonuniform. A more satisfactory
relation can be obtained by using the area at the neck. Assuming that the change in vol-
ume of the material is small, Al = A 0 l 0 . Thus, l/l 0 = A 0 /A, and the true strain is
given by

l A 0
ε = ln = ln (2–17)
l 0 A
Returning to Fig. 2–6b, if point i is to the left of point u, that is, P i < P u , then the
new yield strength is

P i
m (2–18)
S = = σ 0 ε i P i ≤ P u
y
A
i
Because of the reduced area, that is, because A < A 0 , the ultimate strength also

i
changes, and is
P u

S = (c)
u
A
i
Since P u = S u A 0 , we find, with Eq. (2–14), that
S u A 0 S u
(2–19)
S = = ε i ≤ ε u
u
A 0 (1 − W) 1 − W
which is valid only when point i is to the left of point u.
For points to the right of u, the yield strength is approaching the ultimate strength,
and, with small loss in accuracy,
. . m
S = S = σ 0 ε i ε i >ε u (2–20)
y
u
A little thought will reveal that a bar will have the same ultimate load in tension after
being strain-strengthened in tension as it had before. The new strength is of interest
to us not because the static ultimate load increases, but—since fatigue strengths
are correlated with the local ultimate strengths—because the fatigue strength im-
proves. Also the yield strength increases, giving a larger range of sustainable elastic
loading.

EXAMPLE 2–1 An annealed AISI 1018 steel (see Table A–22) has S y = 32.0 kpsi, S u = 49.5 kpsi,
σ f = 91.1 kpsi, σ 0 = 90 kpsi, m = 0.25, and ε f = 1.05 in/in. Find the new values of
the strengths if the material is given 15 percent cold work.
Solution From Eq. (2–16), we find the true strain corresponding to the ultimate strength to be

ε u = m = 0.25

The ratio A 0 /A i is, from Eq. (2–13),

A 0 1 1
= = = 1.176
A i 1 − W 1 − 0.15