# SOLUTIONS by nikeborome

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```									                                        SOLUTIONS

Mechanics of Biomaterials ENGSCI 274SC 2003
1.

Constitutive
Law

2.   (i)   My / I           (ii) I   y 2 dA

3.

3wL                3wLx wx 2
 V ( x)           wx, M ( x)      
8                   8   2

4.
So the neutral axis is 4mm up from the base of the beam.

5.
1     1        1 3
v          2 PLx  6 Px 
2

EI                  

6.
z         zx

 yx
 xx
x

y

7.   (no shear stress anywhere)
Could evaluate max. shear stresses in the other two perpendicular planes, but
this was not expected (question could have been a bit clearer)

8.    homogeneous: material properties are the same everywhere
incompressible: volume remains constant

9.     12 means the (negative of the) transverse strain in the “2” direction divided by
the strain in the “1” (applied) direction

10. Types of biological materials that can be described by the linear elastic model:
hard tissue, e.g. bone, or soft tissue under low stress.
Biological materials that cannot be described well by the linear elastic model:
soft tissue at higher stress levels

11. The equations are
 2   0
The question actually implies that  1   0 , but this doesn’t make sense, since
you need a compressive stresses  1 ,  2 if the material is pushed up against the
constraint. Question a bit ambiguous here.

12.
       

           
E        

13. (i) A biological material which is fairly viscoelastic: Myocardium
One which is not very viscoelastic: Bone
(ii) Creep:                       Relaxation:

14. creep compliance: the strain response to a unit load
Useful to know because we can then obtain the strain response to any loading
(using the hereditary integral condtitutive law)

15. Let pi  p,      p o  0 and ri  ro  r (the average radius)
pi ri3  p o ro3 ri3 ro3  po  pi  1
 t r                     
ro3  ri3                  
2 ro3  ri3 r 3
p 3 ri3 ro3           p 3 3 
          ri  3   3                   r
ro  ri3
3
        2r      ro  ri3  2 
      
Now ro3  ri3  r  t / 2  r  t / 2  3r 2t  rt 3 / 4  3r 2t for small t
3             3

Thus we have
pr
t 
2t

16.   0 
1
 1   2 2   2   3 2   3   1 2
2
SECTION B

1.
(a)

the solution is
27           9
RA       P, M A   PL
32           64

(b)
(i)
Consider the equilibrium of a triangular-shaped piece of material:

F   x   :   AB   x OB cos   y OA sin    OB sin    OA cos
x

etc.
(ii)
We have
5
  0
x
4
7
  0
y
4
3
      0
4

(iii)
Adding the first two stress transformation equations gives
      x  y
x    y

so that the sum of the stresses in any coordinate is a constant.
For   0 o :  x   y  3 o
For   30o :  x   y  (5 / 4) o  (7 / 4) o  3 o
So they are the same.
(iv)
Here the principal stresses are simply  1  2 0 ,  2   0
(v)

0              2 0

(vi)
It’s non-linear (not a straight line) and it’s viscoelastic (if it was not viscoelastic, the
load/unload curves would coincide – the response is different on the way down
because the material’s response changes with time)
(vii)
Uniaxial Test – Young’s modulus, info about stress/strain in 1-d
Biaxial Test – young’s modulus, Poisson’s ratio, info about anisotropy
Shear Test – Shear modulus
Compression (confined) Test – Bulk Modulus
(viii)
The tensorial shear strain is half the value of the engineering shear strain.
2.
(a)
(i)
We have
 1  0.75 0  10 9
(ii)
The max. shear stresses are
0
 1 2 
4
    2 3
0
0
 13 
4
So, according to the maximum shear stress theory, failure will occur when
Y       0
   0  2 Y
24
Since the yield stress is  Y   0 / 3 , it does not fail.

(b)
Stress: 10 kPa . Strain:  (t )  0.011  e  t / 100  .
Creep Compliance is the response due to a unit stress, so
J (t ) 
0.01
1  e t / 100 
10  10    3

 10 6 1  e t / 100 
Then, with  (t )  t Pa , d / dt  1 ,
 (t )  10 6 t  100e t / 100 e t / 100  1
t           
 10 6   100 1  e t / 100   

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