Problem 26 - An Introduction to the Mechanics of Solids in SI Units

Question

A cross section of a cilium (see Prob. 3.21) is shown in the figure. The dark areas are fibrils which are thought to be responsible for the cilium motion. The bending moment at the base of the cilium is estimated to be \(5 \times 10^{-7} \mathrm{~N} \cdot \mathrm{m}\), and an experimental value of the radius of curvature at the base is \(6 \mu \mathrm{m}\). Assuming that the bending forces are carried by the fibrils alone, estimate the elastic modulus of the fibrils. The total second moment \(I_{z z}\) of all the fibril cross-sectional areas is approximately \(4 \times\) \(10^{-8} \mathrm{~mm}^{4}\)

Step-by-Step Solution

Answer

The estimated elastic modulus of the fibrils is approximately \(2.083 \times 10^{6}\) Pa.

Step 1: Convert Units

We have to convert the units from µm and mm^4 to m and m^4 before substituting into our formula. We know, 1 µm = 1e-6 m and 1 mm^4 = 1e-12 m^4. Therefore, r = 6 µm = \(6 \times 10^{-6}\) m and I = \(4 \times 10^{-8}\) mm^4 = \(4 \times 10^{-20}\) m^4.

Step 2: Substitute Values into the Equation

Substitute values into the equation, E = M/(I*r). So, E = \(5 \times 10^{-7}\)/( \(4 \times 10^{-20}\)*\(6 \times 10^{-6}\) ).

Step 3: Calculating Elastic Modulus

After performing the operations, E = \(2.083 \times 10^{6}\) Pa. Therefore, the estimated elastic modulus of the fibrils is approximately \(2.083 \times 10^{6}\) Pa.

Key Concepts

Bending Moment

Understanding the bending moment is crucial when analyzing the structural integrity of any material or beam under flexural load. Essentially, it is the rotational force acting on a beam, causing it to bend. In the context of our exercise, the bending moment is estimated at the base of the cilium and given a value of \(5 \times 10^{-7} \mathrm{N} \cdot \mathrm{m}\). This value is a measure of the torque that the fibrils have to withstand due to the bending forces applied. To visualize this, imagine a diving board as it bends under the weight of a diver. The bending moment at the base, where the board is fixed, determines how much the board will flex and the stress experienced along the material's length.

When estimating the elastic modulus, we use this bending moment to determine how much the fibrils resist deformation. It is a snapshot of the forces at play within the cilium, which will lead us to understand the stiffness of the fibrils when we combine this information with the radius of curvature and the second moment of area, which we will explore further.

Radius of Curvature

The radius of curvature provides us with an understanding of how sharply a beam or structure is bending. It is the distance from the center of the curvature circle to the curve itself. In a bent cilium, the smaller the radius of curvature, the sharper the bend at the base. Our exercise provides an experimental value of the radius of curvature as \(6 \mu \mathrm{m}\).

The radius of curvature is pivotal because it factors into the calculation of the elastic modulus. A smaller radius indicates a tighter curve and, generally, a stiffer material to withstand such bending without breaking. Conversely, a larger radius means a gentler bend and potentially a more flexible material. Hence, calculating the material's stiffness—the elastic modulus—requires this geometric parameter to understand how the bending moment translates to actual deformation regarding curvature.

Second Moment of Area

The second moment of area, also known as the area moment of inertia, is a property of a cross-section that reflects how its area is distributed about an axis. The further the area is from the axis, the higher the second moment of area, which means greater resistance against bending and twisting. In our exercise, the total second moment of all the fibril cross-sectional areas (referred to as \(I_{zz}\)) is approximately \(4 \times 10^{-8} \mathrm{mm}^{4}\).

The second moment is instrumental in establishing the cilium's elastic modulus, specifically because it measures this distribution within the fibrils that are resisting the bending moment. Intuitively, if more material is located far from the neutral axis (the centerline of the bend), it will offer more resistance to bending and, therefore, display a higher second moment of area. It's comparable to how a wider beam can support more weight without sagging as much as a narrower one.

Unit Conversion

Unit conversion is essential when working through engineering and physics problems to ensure consistency and correctness in calculations. In our problem-solving steps, we start by converting the units from micrometers (\(\mu\mathrm{m}\)) and millimeters to the fourth power (\(\mathrm{mm}^{4}\)) to meters (\(\mathrm{m}\)) and meters to the fourth power (\(\mathrm{m}^{4}\)) respectively. This process is vital because it aligns the units with the standard metric units used in calculating the elastic modulus.

Unit conversion is generally straightforward but requires careful attention to prevent significant errors. For instance, there's a colossal difference between a radius expressed in millimeters and one in meters due to the exponential scaling. Misconversions could lead to results that are off by orders of magnitude, leading to incorrect conclusions about material properties. In our case, specific values are \(1 \mu\mathrm{m} = 1 \times 10^{-6}\) meters and \(1 \mathrm{mm}^{4} = 1 \times 10^{-12}\) meters to the fourth power, which illustrates the precision needed in scientific calculations.