Deformation - Orthopaedic Biomaterials in Research and Practice - page 40

Table 2.5 Polar moments of inertia

Polar moment
of inertia ( K ) Shape

Square 0.141 a 4

a

Triangle

( 3/80) a 4

a

1
2

Solid rod π r 4

r

1
2

ick-walled tube π( r 4 - r ´ 4 )

r ´

1
2

π r 3 t

t in-walled tube

2
3

in-walled tube,
open

π rt 3

PROBLEM 2.6
If a 16-mm-diameter thin-walled tube (wall thickness, t = 1.0 mm) were
to be used as an IM device, how much would its torsional resistance be
reduced if it were slit parallel to its axis?

ANSWER:
From Table 2.5:

1
2

3 4

K ( intact ) = (. )( )() 3148 1 = 804 mm

2
3

3 4

K () slit = (. )( )( ) 31481 68 = . mm

Therefore, the slit produces a 98% reduction in torsional stiffness.
Fortunately, IM devices function primarily in bending; a similar calcula-
tion shows no reduction if the slit is positioned on the level of the neutral
axis, and only modest reductions, depending on slit width, if the slit is
located at other positions.

Secondary properties from the stress-strain curve

In addition to the relationship between stress and strain, there are other, sec-
ondary properties that may be derived from the stress-strain curve. Strength