Lecture # 10: Hydrostatic Skeletons
Body Plan Evolution
1 cell
cellular sheet
cellular bilayer
one way gut
endoderm
ectoderm
bilayered canister
mouth
anus
cephalization
mesoderm
Body Plan Evolution
1 cell
cellular sheet
cellular bilayer
one way gut
endoderm
ectoderm
bilayered canister
mouth
anus
cephalization
mesoderm
coelom
ectoderm
endoderm
gut
mesoderm
coelom
pseudocoelom
ectoderm
mesoderm
endoderm
gut
How does stress in a worm depend on geometry?
Consider a hollow spherical animal….
P
r
P
P
d
slice in half
P= internal pressure
r = radius
d= thickness
what is stress in wall?
Define tension, T, as force/length
then T = s x d
s = force/area
=rp/2
= (p r2 p) / (2 p r d)
= (r p) / (2 d)
T=½rp
disk area ~ p r2
rim area ~ 2 p r d
LaPlace’s Law: Tension in wall of sphere is
proportional to radius and pressure.
Consider a cylindrical animal….
1) longitudinal slice
2) slice in half
Equivalent to spherical case,
Thus longitudinal tension, TLis
same as in sphere of equal
radius:
3) cap with
hemisphere
TL = ½ r p
Consider a cylindrical animal….
d
1) transverse wedge
sc = force/area
= (2 r d p) / (2 d d)
=rp/d
Again, TC = sc x d
TC = r p
sc
rim area
=2 d d
r
slice area
=2rd
d
Circumferential or ‘hoop stress’
is twice than longitudinal stress.
TC = 2 x TL
Implications of LaPlace’s Law:
1) Small worm withstand greater pressure than large worms.
P
P
2) Large worms should have thicker walls.
P
P
Pierre-Simon Laplace
1749-1827
3) Square cross sections should be rare.
tension
is infinite
L
Consider a helical worm:
L
Volume = p r2 L
Solve for volume in terms of q (helical angle):
D = L cos q
r = D sin q /(2 p )
V=
D3
q cos q
4p
sin2
Solve for dV/dq:
Maximum volume at q = 54.73o
muscle action
V = d3 sin2 q cos q
4p
Permissible
Morpho-space
elliptical
profile
circular
section
Ontogenetic scaling
of burrowing forces in the
earthworm Lumbricus terrestris
• Kim Quillin
• J Exp Biol 203, 2757-2770
(2000)