Biomedical Engineering Reference
In-Depth Information
phases of the heterogeneous media and their interaction. The effective stiffness,
K
eff
, properties can be used in the analysis of a structure composed of the compos-
ite material by
AE
K
=
total
combined
(5.22)
eff
L
total
EXAMPLE 5.7
A bioengineer is designing a new artificial ligament using a composite structure having a
total cross-sectional area of 4 cm
2
and a length of 10 cm. Also, each structure is made up
of material 1 (Young's modulus of
E
=
600 MPa) and material 2 (Young's modulus of
E
=
10 MPa), in proportions of 80% and 20%, respectively. In the first design, constituents are
arranged in parallel, while in design 2, constituents are arranged in series. A tensile axial
load of 1,000N is applied to each structure. For each design, determine the tensile force
and the tensile stress in each constituent along the cross-section, the strain and axial exten-
sion in each constituent, the total extension, the effective modulus, and the total effective
stiffness. Which design more closely
resembles a real tendon, which has a Young's modulus
of 400 MPa? Which one would you choose?
Solution:
Total axial force
=
1,000N
Total axial stress
=
total force/total area
=
1,000/0.0004 =2.5 MPa
(a) Material in parallels
From (5.19), total axial stress
=
φ
E
1
ε
+
(1
−
φ
)
E
2
ε
2.5[MPa]
0.0052
Hence,
ε =
=
0.8 * 600[MPa]
(1
0.8) * 10[MPa]
+−
Force in Material 1
=
0.0052 * 0.004 [m
2
] * 0.8 * 600 * 10
6
[Pa]
=
995.9N
Force in Material 2
=
1,000
−
995.9= 4.1N
ε
1
=
ε
2
=
ε
=
0.0052
Total
extension
= Material 1
extension
= Material 2
extension
= 0.0052 * 0.10[m] = 0.52
mm
From (5.20), effective modulus
E
=
0.8 * 600 [MPa]
+
(1
−
0.8)10[MPa]
=
482
MPa
6
0.004 * 482 * 10 [Pa]
From (5.22),
K
1,928 kN m
=
=
eff
0.1[m]
(b) Materials in series
Force in Material 1
=
Force in Material 2
=
1,000N
Stress in Material 1
=
Stress in Material 2
=
2.5 MPa
ε
1
=
σ
/
E
1
=
2.5[MPa]/600[MPa]
=
0.0042
ε
2
=
σ
/
E
2
=
2.5[MPa]/10[MPa]
=
0.25
Material 1
extension
=
0.0042 * 0.8 * 0.10 [m]
=
0.33 mm
Material 2
extension
=
0.25 * 0.2 * 0.10 [m]
=
5 mm
Total
extension
=
5.33 mm
Total strain
=
5.33/100 = 0.053
From (5.121), effective modulus
600[MPa] * 10[MPa]
E
46.9 MPa
=
=
0.8 * 10[MPa]
(1
0.8)600[MPa]
+−
From (5.22)
6
0.004 * 46.9 * 10 [Pa]
K
=
=
188.8 kN m
eff
0.1[m]
Since the effective modulus of the composite in parallel is close to the modulus of the
natural tendon, a composite with a parallel configuration needs to be selected.
