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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!
MIT OpenCourseWarehttp://ocw.mit.edu
3.054 / 3.36 Cellular Solids: Structure, Properties and ApplicationsSpring 2015
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