Plant Stems with Radial Density Gradients

Coconut Palm http://en.wikipedia.org/wiki/

Image:Palmtree_Curacao.jpg

Palm: Density Gradient Vascular bundles: Honeycomb

Ground tissue (Parenchyma): Foam

Peripheral Stem Tissue

Rich, 1987

6 µm Kuo-Huang et al., 200410 µm

2 Young Old

Palm Stem: Density Gradient

Rich, PM (1987) Bot.Gazette 148, 42-50. 3

Palm Stem: Density at Breast Height

Densities of common woods

A single mature palm has a similar range of density as nearly all species of wood combined

4Rich, PM (1987) Bot.Gazette 148, 42-50.

Palm Stem: Density Gradient

r ro

ρ − ρmin

ρmax − ρmin

Iriartea gigantea: ρmin ≈ 0

ρ* ⎛ r ⎞ 2

= ⎠⎟ρmax ⎝⎜ ro

ro is the outer radius 5

ρ

ρ

6Rich, PM (1987) Bot.Gazette 148, 42-50.

E* = C *

ρmax

⎛ ⎝⎜

⎞ ⎠⎟

2.5

σ * = C *

ρmax

⎛ ⎝⎜

⎞ ⎠⎟

2Iriartea gigantea

Along Grain

Youn

g’s

Mod

ulus

(MP

a)

Dry Density (kg/m3)

Dry Density (kg/m3)

Mod

ulus

of R

uptu

re (M

Pa)

 

 

 

 

Palm Properties • Prismatic cells in palm deform axially (like

wood loaded along the grain)

• If Es was constant, would expect: E* = Es (ρ* ρs )

)2.5 • But measure: E* = C (ρ* ρmax

• Similarly with strength

7

Palm Properties

•  Es = 0.1-3.0 GPa in low density palm tissue from Washingtonia robusta (Rueggeberg et al., 2008)

•  Estimate in dense tissue (E* = 30 GPa; ρ*= 1000 kg/m3) Es = 45 GPa

•  Large variation in Es due to additional secondary layers in cell walls of denser tissue and increased alignment of cellulose microfibrils in those layers

8

Palm: Mechanical Efficiency Bending Stiffness

9

ρ =rro

⎛⎝⎜

⎞⎠⎟

n

ρmax

E = Cρ

ρmax

⎛⎝⎜

⎞⎠⎟

m

= Crro

⎛⎝⎜

⎞⎠⎟

mn

EI( )gradient =Cπro

4

mn + 4

EI( )gradientEI( )uniform

=4

mn + 4n + 2

2⎛⎝⎜

⎞⎠⎟m

Iriartea gigantea: n = 2, m = 2.5

(EI)gradient/(EI)uniform = 2.5

Palm: Mechanical Efficiency Bending Stress Distribution

10

σ (y) = Eε = Eκ y

σ (r,θ) = Crro

⎛⎝⎜

⎞⎠⎟

mn

κ r cosθ ∝ rmn+1

I. gigantea: n =2, m = 2.5

σ ∝ r6

Palm: Mechanical Efficiency Bending Strength Distribution

11

σ * ∝ρ

ρmax

⎛⎝⎜

⎞⎠⎟

q

∝rro

⎛⎝⎜

⎞⎠⎟

nq

Iriartea gigantea: n = 2, q = 2

σ * ∝ r4

Palm bending stress, strength

Stress distribution Strength distribution

Figure sources

Sources for all figures in: Cellular Materials in Nature and Medicine (2010)

Circular sections with radial density gradients: Palm Stems

• Palms can grow up to 20-40m - largest stresses from hurricane winds

• Unlike trees, palms do not have a cambium layer at the periphery, with dividing cells to allow increasein diameter as palm grows in height

• Instead, diameter of palm roughly constant as it grows in height

• Increasing stress resisted by cell walls increasing in thickness

• Add additional layers of secondary cell wall

• Produces radial density gradient

– Density higher at periphery and at base of stem

– A single stem can have densities from 100-1000 kg/m3, nearly spanning the density range of allwoods (balsa ∼200 kg/m3 → lignum vitae ∼ 1300 kg/m3)

• Specimen of palm taken from different radial positions tested in bending (Paul Rich, 1980s)

• Found Eaxial∗ = C ′ ρ∗ 2.46

• Might expect Eaxial∗ ∝ ρ - vascular bundles honeycomb-like

• But additional cell wall layers change Es: data Es=0.1-3 GPa

• Also: lower density palm has more ground tissue (parenchyma) with E∝ ρ if at high turgor, butE∝ ρ2 if at low turgor. (bending specimens dry)

• Modulus of rupture σ∗ = C ′′ ρ∗2.05

• Radial density gradient increases flexual rigidity

• Compare (EI) with density gradient to (EI) of section of same mass+radius but uniform density

• For Iriartea gigantea:

( ρ∗

ρmax

)=( r n

r0

)r0 = outer radius

n = 2

E = C = Cρmax r0

(EI)gradient =

∫ r0 ρ m 2πr r2dr= C∫0 r

(ρmax

)2

0

= C0

( rdr

r

)mn

πr30

( ρ )m ( r )mn∫r2 2πrdr = J = 2I

Cπ(EI)gradient =

rmn0

∫ r0

rmn+3dr0

Cπ rmn+4

= 0

rmn0 mn+ 4

Cπr4= 0

mn+ 4

Equivalent mass, r0, uniform density ρ̄:

ρ̄ 1∫ r0 ( r n 1 2��π rn+2

0 2= 2πr dr = =

ρ πr2 2 nmax 0 0 r0

)��πr0 r0 n+ 2 n+ 2

(EI)

= Cuniform maxdensit

( ρ̄

ρy

)m πr404

2 m πr4= C

(n+ 2

)0

4

(EI)

( )gradient Cπr4 4 n+= 0 2 m 4 n+ 2 m

=mn+ 4 C πr4EI 0

(2

uniform

)mn+ 4

(2

)(EI)

( )gradientI.gigantea m = 2.5 n = 2 = 2.5EI

uniform

Stress and Strength distribution

ρ

ρmax=( r n

r0Stress σ(y) = E ε

)= E K y

= C( ρ

ρmax

)mK y

= C( rr0

)mn

K r

K=curvature at thecross-section

y=distance from neutral axis

E = C( ρ m

ρmax

)

σ(r) ∝ rmn+1 I. gigantea m=2.5 n=2 σ(r) ∝ r6

Strength σ∗(r) = C ′( ρ

ρmax

)q= C

( r nq

r0

σ∗r

)∝ rnq I. gigantea q=2 n=2 σ∗(r) ∝ r4

Figure: if max normal stress at r = r0 is σ = σ∗ then bending stress distribution closely follows strengthdistribution!

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3.054 / 3.36 Cellular Solids: Structure, Properties and ApplicationsSpring 2015

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