Page 193 - Mechanical Behavior of Materials

194 Chapter 5 Stress–Strain Relationships and Behavior

where ˙ε = dε/dt is the strain rate. Substitution from Fig. 5.2 and P = c˙x yields the relationship
between η and c:

cL
η = (5.6)
A
Equations 5.3 and 5.6 also apply to the spring and dashpot elements in the transient creep model.
Before proceeding to the detailed discussion of elastic deformation, it is useful to further discuss
plastic and creep deformation models.

5.2.1 Plastic Deformation Models

As discussed in Chapter 2, the principal physical mechanism causing plastic deformation in metals
and ceramics is sliding (slip) between planes of atoms in the crystal grains of the material, occurring
in an incremental manner due to dislocation motion. The material’s resistance to plastic deformation
is roughly analogous to the friction of a block on a plane, as in the rheological model of Fig. 5.1(b).
For modeling stress–strain behavior, the block of mass m can be replaced by a massless
frictional slider, which is similar to a spring clip, as shown in Fig. 5.3(a). Two additional models,
which are combinations of linear springs and frictional sliders, are shown in (b) and (c). These
give improved representation of the behavior of real materials, by including a spring in series with
the slider, so that they exhibit elastic behavior prior to yielding at the slider yield strength σ o .In
addition, model (c) has a second linear spring connected parallel to the slider, so that its resistance
increases as deformation proceeds. Model (a) is said to have rigid, perfectly plastic behavior; model
(b) elastic, perfectly plastic behavior; and model (c) elastic, linear-hardening behavior.
Figure 5.3 gives each model’s response to three different strain inputs. The first of these is
simple monotonic straining—that is, straining in a single direction. For this situation, for models (a)
and (b), the stress remains at σ o beyond yielding.
For monotonic loading of model (c), the strain ε is the sum of strain ε 1 in spring E 1 and strain
ε 2 in the (E 2 ,σ o ) parallel combination:

σ
ε = ε 1 + ε 2 , ε 1 = (5.7)
E 1
The vertical bar is assumed not to rotate, so that both spring E 2 and slider σ o have the same strain.
Prior to yielding, the slider prevents motion, so that strain ε 2 is zero:
σ
ε 2 = 0, ε = (σ ≤ σ o ) (5.8)
E 1
Since there is no deflection in spring E 2 , its stress is zero, and all of the stress is carried by the
slider. Beyond yielding, the slider has a constant stress σ o , so that the stress in spring E 2 is (σ − σ o ).
Hence, the strain ε 2 and the overall strain ε are

σ − σ o σ σ − σ o
ε 2 = , ε = + (σ ≥ σ o ) (5.9)
E 2 E 1 E 2