lecture # 10: hydrostatic skeletons

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Lecture # 10: Hydrostatic Skeletons

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Lecture # 10: Hydrostatic Skeletons. 1 cell. cellular sheet. cellular bilayer. one way gut. ecto- derm. endo- derm. mouth. anus. bilayered canister. Body Plan Evolution. cephalization. mesoderm. 1 cell. cellular sheet. cellular bilayer. one way gut. ecto- derm. endo- derm. - PowerPoint PPT Presentation

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Page 1: Lecture # 10: Hydrostatic Skeletons

Lecture # 10: Hydrostatic Skeletons

Page 2: Lecture # 10: Hydrostatic Skeletons

1 cell cellular sheetcellular bilayer

bilayered canister

ecto-derm

endo-derm

one way gut

mouthanus

cephalization

mesoderm

Body Plan Evolution

Page 3: Lecture # 10: Hydrostatic Skeletons
Page 4: Lecture # 10: Hydrostatic Skeletons
Page 5: Lecture # 10: Hydrostatic Skeletons

1 cell cellular sheetcellular bilayer

bilayered canister

ecto-derm

endo-derm

one way gut

mouthanus

cephalization

mesoderm

Body Plan Evolution

Page 6: Lecture # 10: Hydrostatic Skeletons
Page 7: Lecture # 10: Hydrostatic Skeletons
Page 8: Lecture # 10: Hydrostatic Skeletons

ectoderm

mesoderm

endoderm

gut

coelom

coelom

Page 9: Lecture # 10: Hydrostatic Skeletons

ectoderm

mesoderm

gut

pseudocoelom

endoderm

Page 10: Lecture # 10: Hydrostatic Skeletons
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Page 20: Lecture # 10: Hydrostatic Skeletons

Consider a hollow spherical animal….

slice in half

d

rP

P= internal pressurer = radius

d= thickness

PP

what is stress in wall?

= force/area = (r2 p) / (2 r d) = (r p) / (2 d)

disk area ~ r2

rim area ~ 2 r d

Define tension, T, as force/length

then T = x d = r p / 2

T = ½ r p

LaPlace’s Law: Tension in wall of sphere is proportional to radius and pressure.

How does stress in a worm depend on geometry?

Page 21: Lecture # 10: Hydrostatic Skeletons

Consider a cylindrical animal….

Equivalent to spherical case,Thus longitudinal tension, TLis same as in sphere of equal radius:

TL = ½ r p

1) longitudinal slice 2) slice in half

3) cap withhemisphere

Page 22: Lecture # 10: Hydrostatic Skeletons

Consider a cylindrical animal….

1) transverse wedge

c = force/area = (2 r p) / (2 d ) = r p / d

Again, TC = c x d

TC = r p

r

c

slice area=2r

rim area=2 d

d

Circumferential or ‘hoop stress’is twice than longitudinal stress.

TC = 2 x TL

Page 23: Lecture # 10: Hydrostatic Skeletons

Implications of LaPlace’s Law:

Pierre-Simon Laplace 1749-1827

1) Small worm withstand greater pressure than large worms.

2) Large worms should have thicker walls.

3) Square cross sections should be rare.

P P

P P

tension is infinite

Page 24: Lecture # 10: Hydrostatic Skeletons

Consider a helical worm:

Volume = r2 L

Solve for volume in terms of (helical angle):

D = L cos r = D sin /(2 )

V = D3 sin2 cos 4

Solve for dV/d

Maximum volume at = 54.73o

L

L

Page 25: Lecture # 10: Hydrostatic Skeletons

Permissible Morpho-space

V = d3 sin2 cos 4

ellipticalprofile

circularsection

muscle action

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Ontogenetic scaling of burrowing forces in the earthworm Lumbricus terrestris

• Kim Quillin • J Exp Biol 203, 2757-2770 (2000)