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Transverse Shear Formula Approximation

A problem was recently assigned in our Machine Design class. It asked us to evaluate the maximum shear stress for a cylindrical beam. The inner diameter was 17 mm while the outer diameter was 25 mm.

Two formulas are derived from the general shear stress formula, \tau=\frac{VQ}{bI}:

 \tau_{\text{max}}=\frac{4V}{3A} (Eq 4.15c) \tau_{\text{max}}\approx\frac{2V}{A} (Eq 4.15d)

Equation 2 is only valid if the wall thickness is less than 10%(?) of the inner(?) radius(?).

Of course, this puts us in a quandary. Should we still apply the thin-wall formula anyway? Does Equation 1 have any significance?

In fact, Equation 1 gives us an indicator of the direction of the error from using Equation 2. As the inner radius gets closer and closer to 0% of the outer radius, Equation 1 will be closer to the actual value of Q.

Q, by the way, is the first moment of area above the neutral axis. It is used to measure how far portions of area are from a certain axis. I-beams, with much of their mass away from the neutral axis, will have large Q values. Larger Q values will tend to resist shear stress. The textbook includes this formula: Q=\sum{\overline{y}dA}, where
\overline{y} refers to the distance to the centroid of the area segment dA.

So where does MATLAB fit in this puzzle? It turns out the solution can be closely approximated without using calculus with roughly a dozen lines of code and several seconds of computation. The actual value of Q is determined using an area integral, which is also written as a sum of many, many small areas (shown above). More accurately, any time a small "piece of area" is within the coordinates of the beam, the small area times its centroid is summed with every other piece of area.

By splitting the area from (-25mm,-25mm) to (25mm, 25mm) into 1,562,500 smaller pieces, each piece can be evaluated, one-by-one, and added to Q when our conditions apply: Anytime the radius of a point (from the origin) is less than 25mm and greater than 17mm, the vertical distance (centroid) is multiplied by the miniscule area and added to Q. Each little evaluated square is 0.04mm by 0.04mm.

The MATLAB code is listed here:

      OD = .025;
ID = .017;

step = OD / 1250;

Q = 0;
for y = 0 : step : OD/2 ;
for x = -OD/2 : step : OD/2 ;
R = sqrt(x^2 + y^2);
if and(R <= OD/2, R >= ID/2)
Q = Q + y*(step)^2;
endif;
end;
end;

I = (pi/64) * ((OD)^4 - (ID)^4);
b = OD - ID;

disp(Q/(I*b));

Personally, I use GNU Octave on a cheap, Toshiba netbook running Windows 7 Starter. MATLAB tends to have faster solving algorithms. In Octave, the calculation takes roughly 86.15 seconds. On my desktop at home, the computation in Octave takes about 10 seconds. If performance is an issue, the step size can be reduced. Changing this variable from 1250 to 600, the computation occurs in 21.18 seconds. Less than 0.05% relative error is present when the step size is reduced from 1250 to 600. That is, the X and Y dimensions are split into 1250 or 600 equally-spaced values.

• τmax = 5052.5 * V for a solid beam (Equation 1)
• τmax7578.8 * V for a thin-walled beam (Equation 2)
• τmax7401.4 * V using the MATLAB approximation shown above

For completeness, the graphs below show the accuracy of each formula as the inner radius changes. The x-axis shows the percentage (out of 100) of the circular beam's cross-section which is hollow. In other words, it shows the ratio of the inner diameter to the outer diameter.

If the inner Home