Mechanical Modeling of Material Behavior - Continuum Mechanics of Anisotropic Materials - page 18

Fig. 1.11 The standard linear
solid

F

η

k

k R

element under the same loading, there is no instantaneous deflection, but a creeping
deflection begins under the constant applied stress and proceeds asymptotically to a
rest value. There is only one lumped three-parameter element of interest, the
standard linear solid (SLS, Fig. 1.11 ).

Example 1.8.2

Using the Fig. 1.11 as a guide, derive the differential equation of the governing
force-deflection relationship of the standard linear solid.

Solution : Let F L and F R denote the force in the two branches, left and right, of the
standard linear solid; the total force F is then given by F

F R . Let x denote
the overall deflection of the standard linear solid element; the deflections in both
branches must be equal; the horizontal cross-bars in spring-dashpot models are not
allowed to rotate. The total deflection in the right branch is the sum, x , of the
deflection of the dashpot, x D , and the deflection of the spring, x S ; thus x

¼ F L þ

x S .
The equations describing the behavior of the three constituent elements are
F L ¼

¼ x D þ

(d x D /d t ), respectively. Note that the force in the
two elements on the right branch must be the same. These equations are combined
in the following manner. First, note that from F R ¼

kx,F R ¼ k R x S and F R ¼

k R x S it follows that (d x S /d t )

¼ (1/ k R )(d F R /d t ), from F R ¼ (d x D /d t ) it follows that (d x D /d t ) ¼ (1/ ) F R , and

from x ¼ x D þ x S it follows that (d x /d t ) ¼ (d x D /d t ) þ (d x S /d t ). Combining these

results it follows that

(d x = d t Þ¼ð 1 =Þ F R þð 1 = k R )(d F R = d t Þ: