Design of Shank for Biological Purpose - Analysis and Formulations

This article discusses the design of a shank for biological purpose, which is to be used as a bone to connect joints of socket and foot through a pin joint. It covers the analysis and formulations required for the design of the shank, including the assumptions made by Euler’s theory of buckling, the types of columns, and the methodology adopted. The article also provides a solution to the given problem, including the critical buckling load of the shank and the mass of the shank.

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FEBRUARY 25, 2018
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INTRODUCTION
Basic objective of this assessment is to design a shank for biological purpose. The shank which
is to be designed must be used as a bone to connect joints of socket and foot through a pin joint.
Considering the objective the shank is to be designed on the model of column. The column is an
axially loaded vertical member which are to take compressive load.
There are three types of column:
1. Long column.
2. Short column.
3. Intermediate Column.
Generally long column fails in buckling (bending), short column fails in crushing and
intermediate column may fails in simultaneous action of bucking and crushing.
Before doing any analysis on the shank design some assumptions made by Euler’s theory of
buckling must be considered:-
1. Material must be isotropic, homogeneous and elastic.
2. The column is straight.
3. The magnitude of young’s modulus is constant in tension and compression.
There for Euler’s buckling load is given by
Pc= n2 π2 EI
¿2
Where,
n= number of buckling loops. ‘n’ depends upon the number of joints/supports given to the
column. (Here the shank has two joints therefore n is taken as 1 because only single node is
formed during buckling)
E= Young’s modulus of elasticity.
I= Minimum Moment of inertia of cross section. It is taken minimum because buckling will start
about axis which has minimum moment of inertia.
le= effective length/equivalent length. It is the length between the two points where bending is
zero.
Pe= Euler’s critical load/ Euler’s crippling load.

NOTE:
1. For the long column the buckling load capacity is very small therefore it is easy to buckle
column hence long column fails in buckling.
2. For short column ‘Pe’ is very large therefore it is difficult to buckle short column, hence
before buckling short column gets crushed.
3. For the intermediate column both ‘Pe’ and crushing load ‘Pc’ is considerable therefore
failure is due to simultaneous action of buckling and crushing.
Analysis of the hollow circular shank provided be eccentric loading:
During eccentric loading a core is formed. The core is that area which if any eccentric load is
passed then there is only compression in the column.
Core diameter for hollow shank/column is given by
e= Di2+ Do2
8 Do
where,
Di= internal diameter of the shank.
Do= external diameter of the shank.
L=lenth of column
As for equivalent length is concerned:
Column Condition Equivalent/effective length ‘le’
1. Both ends hinged/pin jointed L
2. One end hinged other end is fixed L/√ 2
3. Both the ends are fixed L/2
4. One end is fixed other is free 2L
Methodology adopted:-
The steps required to evaluate and analyze the given problem are as follows:-
1. Assumptions done in order to simplify the problems.
2. Analyzing the given data.
3. Interpret the sketch of the shank and the point where the load is acting..
4. Obtaining the formulations.

5. Obtaining the maximum and minimum stress component.
6. Comparison of the maximum stress with the yield limit of the shank material.
7. Obtaining the strain and comparing the obtained value with the desired value of strain
8. Calculating the critical buckling load of the shank.
9. Calculating the mass of the shank.
Solution to the given problem:-
Do = outer diamter =4cm
e= 2cm eccentricity
Di =internal dia = 3.5cm
Weight of body =50 kg
.·. Load p=50× 9.81=490.5 N
Moment due to eccentricity loading
M= P¿e = 490.5 ×0.02=9.8 Nm
I yy= π
64 ( D0
4 −Di
4
)= π
64 ( 0.044−0.0354 )
I yy=5.20 ×10−8 m4
FOS =3
(1) σ max,min= p
A ± M
I yy
. y
A= π
4 ( DO2−Di2 ) = π
4 ( 0.042−0.0352 )
A= 2.945 ×10−4 m2
Y= DO
2 = 4
2 =2 cm=0.02 m
σ mx , min= 490.5
2.945 ×10−4 ± 9.8× 0.02
5.20 ×10−8
DO
DI
P

σ mx, min=1.6655± 3.769 Mpa
σ max=1.6655+3.769=5.4545 Mpa
σ min=1.6655−3.769=−2.1035 Mpa
FOS= σ yield
σmax
σ yield=FOS ×σ max=5.4545 ×3=16.3735 Mpa
Material given for the shank is Aluminum Alloy
EAL= modules of elasticity of Aluminum =70Gpa (kudela, 2010)
By the hooks law (Gere, 2009)
σ =E ∈∈=Strain
σ =σ max=5.4594 Mpa
%ε= σ
E = 5.4545
70 ×103 ×100 %=0.0078 % whichis under thelimit
-tolerance
(2) Critical buckling load Pc= π 2 EI
¿2 CITATION Jon06 \l 16393 (Jones, 2006)
le = equivalent length of shank =25cm=0.25m
pc= π 2 ×70 ×109 × 5.20× 10−8
( 0.25 ) 2
Pc=575805.7603N
PC=574.8057KN
(3) Density of Aluminum alloy is f=2.8g/cm2(Ashby, 2005)
F= 2.8× 10−3
( 10−2 )
2 =28 kg /m3

F= massofshank
volumeofshank = m
v
.·. v= πh
4 ( D0
2−Di
2
)
.·. v= π × 0.25
4 ( 0.042−0.0352 ) =7.3631× 10−5
m2
M=f× v=28 ×7.3631 ×10−5
=2.0617gm Answer.
References
Ashby, M. F., 2005. Materials Selection in Mechanical Design. 3 ed. Burlington: Butterworth-
herneman linacre house.
Gere, J. M. G. B. J., 2009. Mechanics of Materials. 7 ed. Toronto: Cenegage learning.
Jones, R. M., 2006. Buckling of Bars, Plates, and Shells. 2 ed. Blacksburg: Bull Ridge.
kudela, J. R. L., 2010. Wood Structure and Properties. Zvolen: IUFRO Rearch Group.