Biomedical Engineering Reference
In-Depth Information
Fig. 8.5 Bending of the bar
F
z
L / 2
L / 2
Δ
z
8.3.2.3 Bending of the bar. Case 2
The bar of length
L
is loaded at the center by the force
F
z
. Both ends of the bar are
fixed to the supports (Fig.
8.5
). The bar section has the moment of inertia
I
. For this
case, the deflection
D
z
is:
L
3
F
Z
D
z
¼
I
;
(8.32)
48
E
and the rigidity
r
is:
48
E
I
r
¼
Fz
=D
z
¼
:
(8.33)
L
3
Substitution of equation (8.30) into equation (8.33) gives:
S
4
48
E
A
I
A
48
E
B
I
B
r
A
¼
¼
¼
S
r
B
(8.34)
L
A
S
3
L
B
We can conclude from equation (8.34) that a size reduction by
S
times gives a
rigidity reduction by
S
times.
8.3.2.4 Torsion of the bar
The bar of length
L
is fixed to the wall (Fig.
8.6
). The end of this bar is fixed
perpendicularly to another bar of length
L
1
. Both ends of bar
L
1
are loaded by the
forces
F
z
that form the force couple. This couple turns the bar
L
that has a polar
moment of inertia
I
P
and produces a displacement
D
z
of the ends of bar
L
1
. Bar
L
1
is
considered absolutely rigid. The angle
y
of the torsion of bar
L
can be calculated
using the formula:
T
L
y ¼
I
p
;
(8.35)
'
where
T
is the moment of the force couple
F
z
,
L
is the bar length,
'
is the shear
modulus, and
I
p
is the polar moment of inertia. We have:
T
¼
F
z
L
1
;
(8.36)
