Supplementary material
Trees and shrubs differ biomechanically
Markku Larjavaara
University of Helsinki
Assuming that wood is linear-elastic, e.g. follows Hooke’s law [1],stems are slender cylinders, the maximum height, hmax(Maximum height in Figure 1 of the letter), not leading to elastic buckling is
Eq. 1
, whereE is the modulus of the elasticity of fresh wood, ρ is the fresh density of wood, g gravitational acceleration, and d stem diameter [2]. This is a useful approximation althoughreal plants have tapering stems, heterogeneous wood and branches and leaves that cause gravitational loads.In addition shrubs might need extra rigidity to bounce upright when suppressed by other plants or debris.
Assumingthat maximal strain is the quotient of the modulus of rupture of fresh wood, σ, and E. In a column bending to form a perfect semicircle, the neutral plane[1] is located in the middle, the maximal tension in the upper outer fibres, and equal stress but compression in the lower inner fibres. Because the length of a semicircle is proportional to the radius of the semicircle, the extremestrain is proportional to the quotient ofthis radius and stem diameter. Therefore,
Eq. 2.
, where hshrubmin(Minimum shrub height in Figure 1) is the minimum height enabling the top to bend to the ground.This is a rough approximation of minimum height as real shrubs have a tapering stem and wide crowns allowing shorter shrubs to touch the ground, but often are subject to forces such as trampling somewhere along the stem, requiring even greater bending.
To obtain theexact maximum possible shrub stem height,hshrubmax (red arrow in Figure 1) Equations 1 and 2 can be combined to obtain the positive h for which hshrubmin and hmax are equal
Eq. 3.
For example, with wood of the mature Pinus sylvestris tree, when σ is 4.6 x 107 Nm-2, E is 7.3 x 109 Nm-2,ρ is790 kgm-3[3], and g is 9.81 m s-2, the tallest possible cylindrical stem able resist elastic buckling and bend to the ground without breaking has hshrubmax of 7.5 m and d of 0.030m.
The toppling-over torque or momentis greatest at the base of tree stem as it is the product length of the lever and of a temporary force such as wind. Assuming that the maximum force that the tree needs to resist, F, acts on top of the rigid stem and is not dependent on stem dimensions, for example, caused by a given wind speed blowing on a given leaf area per tree, then the maximum height to resist these temporary forces, htreemax (Maximum tree height in Figure 1), is
Eq. 4.
[1].Therefore, reaching the maximum possible shrub stem height of 7.5 m with a tree strategy and a toppling-over force, F, of 100 N at the tree top would require a stem diameter of 0.055m (light blue arrow in Figure 1).With smaller topping-over forces, as in a wind protected rain forest understory, with small leaf area or with very flexible branches lowering the sail area, hmax andhtreemax might cross, and therefore hmaxlimit height even for trees not being able to bend to the ground, but a closer examination of these complex situations is not realistic based on Eq. 4 assuming unrealistically a rigid stem[4].
Assuming a fixed toppling-over force is useful when examining a theoretically potential range of stem dimensions for a plant to raise its given leaf area as in even-aged closed stands but could be misleading if thinking of actual crown sizes for given stem diameters. For example, if leaf area and therefore F is proportional to the square of d, then htreemax is proportional to d, and Max tree height in Figure 1 would be a straight line. Increasing wind speeds with increasing height would lower further htreemax for larger trees.
References
1. Niklas, K.J. and Spatz, H.C. (2012) Plant Physics,University of Chicago Press
2. McMahon, T. (1973) Size and shape in biology. Science 179, 1201–1204
3. Lavers, G. (1983) The strength properties of timber (third edition, revised by G.L. Moore). In Building Research Establishment Report, Her Majesty's Stationary Office
4. Lopez, D. et al. (2014) Drag reduction, from bending to pruning. Epl 108