While the underlying equations governing structural mechanics may be complex, they can be made much more manageable with a few simplifying assumptions.
Most introductory “mechanics of materials” or “deformable body mechanics” rely on the major assumption that the materials described are homogeneous, isotropic, and linear-elastic (HILE). When any of these assumptions are violated, some of the simplicity they afforded is also lost.
When a structural section made up of multiple materials is analyzed, engineers typically transform the section properties of one material into an equivalent property of another by means of the modular ratio, which is the ratio of the Young’s moduli (moduli of elasticity) of the two materials.
This simple act of “transforming” sections by multiplying or dividing some property of the section avoids the need to apportion loads across various portions of the section, thereby keeping familiar equations in play.
Why do We Assume HILE Materials?
The general differential form of the equations that describe tensile and flexural members are incredibly complex, as they must necessarily be to accommodate the wide variety of materials that could conceivably be used in structural or structural-mechanical applications.
These equations allow for materials with properties that vary over space (non-homogenous materials), materials with properties that depend on the loading direction (think orthotropic materials like grainy wood, or even possibly anisotropic materials with more complex variation by direction), as well as materials with any arbitrary relationship between stress and strain.
However, this generality comes at a cost: analytical solutions to these general-form equations elude our current understanding and likely don’t exist at all.
In order to reduce these fancy equations to something usable, numerous simplifying assumptions are necessary, usually relating to the material properties themselves.
HILE Materials
HILE materials are homogenous (uniform over space), isotropic (properties don’t vary by loading direction), and linear-elastic (stress directly correlates to strain, at least over typical usage ranges).
Highly refined metals, like aluminum and steel, are great examples of materials that largely abide by these assumptions.
However, stick more than one of these metals together in one composite structural member, and you get variation in material properties over space, violating the homogeneity assumption.
As previously mentioned, whenever we violate an assumption, there is a price to be paid.
Composite Sections of Two Materials – The Hard Way
Without the benefit of the modular ratio, a composite (load-sharing) member composed of two different materials presents a bit of a problem.
Say we have a pair of 1″ x 2″ metal bars, glued together sufficiently along their lengths to ensure a no-slip condition (deflection compatibility) to form one overall 2″ x 2″ section, half of aluminum (E.a=10,000 ksi), and half of steel (E.s = 29,000 ksi). This composite bar is 100″ long, and is subjected to 200,000 lbs of axial tensile load. How much does the bar elongate?
The strain/deflection compatibility of the two metal bars making up our section tells us that they must both elongate the same amount, or delta.aluminum = delta.steel. The only way this can occur is if the portion of the total load each bar carries is proportional to its axial stiffness, as these “springs” are in parallel, adding their stiffnesses together.
The formula for axial elongation of a HILE rod is delta = F*L / E*A. We can abstract this as if it were a Hookean spring, where F = k*delta, or k (the spring constant or stiffness) = F/delta. Back on our axial elongation, a little rearrangement of the equation gives that F/delta = E*A / L, so the stiffness of an axial rod is E*A / L, which is applicable to both the aluminum and steel portions.
For the overall system, F.total / delta.total = k.total, and k.total is just the sum of the stiffnesses of each bar. Flipping that equation for ease, we get that delta.total (our goal) = F.total / k.total. k.total is just the sum of the two portions, so k.total = E.steel*A.steel/L.steel + E.aluminum*A.aluminum/L.aluminum.
Substituting in, that’s k.total = 29,000 ksi * 2 in² / 100 in + 10,000 ksi * 2 in² / 100 in, or k.total = 780,000 lbs/in.
Finally, returning back to the delta.total = F.total/k.total, that gives us that delta.total = 200,000 lbs / 780,000 lbs/in => delta.total = 0.256″.
How does the Modular Ratio Simplify Analyses?
If reading through all those steps made your eyes glaze over, you’re not alone.
Enforcing compatibility, backing out Hookean spring stiffnesses, and tracking whether various elements are in parallel or series is not for the faint of heart, and generally well above the level of what’s expected in an intro mechanics class. Luckily, there’s a much easier way, thanks to the modular ratio.
Axial Modular Ratio Manipulations
Let’s grab that axial example from above, where k.total = E.steel*A.steel / L + E.aluminum*A.aluminum / L (L.steel = L.aluminum = L). Everything is the same for the stiffness of the steel or the aluminum portions, except the area and the Young’s Modulus.
What if we pretended the entire area was aluminum? As aluminum is less stiff, this would be a conservative simplifying approach, but it might be too conservative. We could just add together the area of steel and aluminum directly, and get a total area of “aluminum”, then use that in an easy delta = F*L / E*A equation.
With the numbers from above, that’d give delta = 200,000 lbs * 100″ / 10,000,000 psi * 4 in² = 0.5″, almost double what we had calculated above!
If we multiply and divide the steel term of k.total by E.aluminum, we can get E.steel*E.aluminum*A.steel / L*E.aluminum, which doesn’t seem super helpful. But let’s define the modular ratio, n, as n = E.steel / E.aluminum, so that simplifies down to k.steel = n * E.aluminum*A.steel / L. This implies we can just multiply the area of the steel by this Modular Ratio, and then pretend it’s aluminum.
Crunching the numbers, delta = F * L / E.aluminum*(A.aluminum + n*A.steel) = 200,000 lbs * 100″ / 10,000,000psi * (2 in² + 2.9*2 in²) => delta = 0.257″, exactly what we got before, but with way less hassle!
Bending Modular Ratio Manipulations
We can extend the same concept to bending, and get a pretty similar result!
Bending deflections are functions of applied loading, end constraints, and Young’s Modulus and Moment of Inertia (second moment of area) of the section. For composite beams, we need to scale the moment of inertia, which corresponds with the area in axial extension.
If you’re looking to compute the section properties by hand, just scale the section parallel to the neutral axis by the modular ratio before you find your centroid in that direction, and calculate away. For an ordinary beam, this corresponds to the width-wise direction.
NOTE: You’ll need to perform this transformation scaling in both the strong axis and weak axis direction independently, returning to the original cross-section between.
For example, a rectangular section with an overall 10″ height and 4″ width, where the top 2″ are steel and the rest is aluminum will need to be scaled to look like a t-shape for strong-axis bending properties, reverted to the original shape, then have the steel section stretched vertically to compute the weak-axis bending properties.
Conclusion
Hopefully, this helped you understand the modular ratio and its uses a bit better, and I’m aiming to get a corresponding video up soon to help you visualize this topic better!
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