Biomedical Engineering Reference
In-Depth Information
Fig. 3.10 An illustration
for problem 3.4.1
Solution : Substitution of the given stress tensor T and the action-at-a-distance force
d into the stress equations of motion ( 3.39 ) yields the fact that all the components of
the acceleration are zero:
r€
x 1 ¼
c 1
c 1 þ
0
¼
0
; r€
x 2 ¼
c 4 þ
c 4 þ r
g
r
g
¼
0
; r€
x 3 ¼
0
:
Problems
3.4.1. Derive the stress equations of motion in two dimensions, ( 3.40 ), from the
less rigorous argument that consists of applying Newton's second law to the
differential element of an object. This element is of length d x 1 and of width
d x 2 (see Fig. 3.10 ). The stresses acting on the element are all referred to the
point ( x 1 , x 2 ) that is the lower left-hand corner of the element. Use the
stresses to calculate the forces acting on the various faces of the element.
In order to determine the stresses on opposite faces expand the stresses on
one face in a one term Taylor series about the first face. Please keep in mind
that, even though the stress tensor is symmetric, the shear stress T 12 on a face
with a normal in a positive direction is opposite in direction from the shear
stress T 12 on a face with a normal in a negative direction.
3.4.2. The components of a stress matrix are
2
4
3
5;
x 2 þ
x 1
x 2 Þ
A
½
B
ð
2 ABx 1 x 2
0
x 1 þ
x 2
x 1 Þ
T
¼
2 ABx 1 x 2
A
½
B
ð
0
x 1 þ
x 2 Þ
0
0
AB
ð
where A and B are constants. Does this stress matrix satisfy the stress
equations of motion? For what action-at-a-distance force field does it satisfy
the stress equations of motion? Assume that the acceleration of the object
is zero.
3.4.3. Using the equations of motion to determine the acceleration of an object for
which d
¼
5 e 1 þ
6 e 2 þ
7 e 3 , the density,
r
, is 2 and T is given by
2
4
3
5 :
3 x 1
x 3
3 x 1
T ¼
x 3
4 x 2
3 x 3
3 x 1
3 x 3
7 x 3
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