Luke J. answered • 12/31/20

BS Mechanical Engineering with Experience Tutoring Relevant Material

I find that it is good practice to extract your Givens (a list of values given from the prompt related to recognizable variables used in equations from your class/course) and Finds (a list of other recognizable variables to hone in on what you are actually solving for).

__Given__:

σ_{max} = 165 MPa = 1.65 ( 10^{8} ) Pa

E = 200 GPa = 2.00 ( 10^{11} ) Pa

b = 10 mm = 1.0 ( 10^{-2} ) m

t = 1.5 mm = 1.5 ( 10^{-3} ) m

__Find__:

R = ? m = ? mm smallest value, essentially at maximum stress

M = ? N * m = ? N * mm maximum value at max stress

__Solution__:

It should be noticed that 2 of the 3 pieces of Hooke's Law were given:

A stress σ and the modulus of Elasticity (a.k.a. Young's Modulus) E.

So:

σ = E * ε

It looks like I'm doing this "for fun", but the strain ε is actually needed to find R.

ε = E / σ_{max} ε = ( 2.0 (10^{11}) Pa ) / ( 1.65 ( 10^{8} ) Pa ) ε ≈ 1.21 ( 10^{-3} ) m / m

Because strain needs its usual units (even though it is technically unitless), I have added on the meter per meter units because it will be useful for unit cancellation in the next step.

A property of strain due to bending is this:

ε = - y / ρ where y is the height measured above the neutral axis and ρ is the radius of curvature (in this case, the radius to reach the neutral axis).

If you'd like to see the derivation for this, go here: http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?topic=me&chap_sec=04.1&page=theory

If I were to have an image at my disposal, I would depict the following:

The ring (essentially) has an inner and outer radius. The inner radius is R. The reason why the inner radius is R is because the steel strip is wrapped around a spool of radius R, the value that is to be found. The outer radius is R + t.

The radius of curvature, in this case, is the average between these two.

Thus, the radius of curvature ρ is:

ρ = R + t / 2

Inputting this into the equation yields:

ε = - y / ( R + t / 2 )

The variable y has restrictions. It can only take on values that are either within the material being analyzed or at the extreme surfaces of the material (where the analyzed material ends).

In this problem, y is restricted as such:

- t / 2 ≤ y ≤ t / 2

Now, depending on where you analyze the strain in the ring can alter the sign of the equation for strain due to bending. So, in this solution, the bottom of the ring will be observed (as if the ring were standing up like a car tire on a flat surface). This is so that no mismatching of sign convention will occur.

Interpreting the strain equation, it says that (looking at the bottom of the ring cross-section):

If you move above the neutral axis (legitimately upward, not in reference to the ring), you will encounter compressive strain because the fibers are being compressed together with the "smiley face" positive moment that the bottom of the ring is making.

If you move below the neutral axis (legitimately downward), you will encounter tensile strain because the fibers are being stretched. A good experiment to try that with is a sizeable piece of cardboard. Make a "smiley face" and see how the top of the cardboard looks scrunched and the bottom of the cardboard looks slightly stretched.

Since the stress that was given was for tensile stress, you should measure below the neutral axis. And because we are looking for maximum stress, you should take the extreme surface for your y-value. Thus:

y = - t / 2

Inputting into the strain equation:

ε = - ( - t / 2 ) / ( R + t / 2 ) ε = t / ( 2R + t )

The negatives cancelled and the 2 was distributed to the denominator terms.

Rearranging for R yields:

2R + t = t / ε 2R = ( t - t ε ) / ε R = t * ( 1 - ε ) / 2 ε

Using the values determined and given:

R = ( 1.5 (10-3 ) m ) * ( 1 - 1.21 ( 10^{-3} ) m / m ) / ( 2 * 1.21 ( 10^{-3} ) m / m )

**R ≈ 0.618 m ≈ 618 mm**

Next, the bending stress in a material thru a cross-section with second moment of area (a.k.a. moment of inertia) I, **positive** bending moment M, located at some height y above or below the neutral axis ( plus or minus height ) when observing the cross-section at the bottom of the material:

σ_{b} = - M * y / I

Even though the material is a ring, the main body is a cylinder with a hole in the middle. If you observe a cross-section of the ring, it will be a rectangle with length b and height t.

The moment of inertia of a rectangle about its Center of Mass x-axis (a.k.a. its x'-axis) is:

I = ( 1 / 12 ) * b * h^{3}

I = ( 1 ( 10^{-2} ) m ) * ( 1.5 ( 10^{-3} ) m )^{3} / 12

I ≈ 1.25 ( 10^{-12} ) m^{4}

Rearranging the bending stress equation for the internal bending moment M when you are at y = - t / 2 and at maximum tensile stress σ_{max}:

Note: I simplified the rearrangement that I noted above.

M = 2 * σ_{max} * I / t = 2 * ( 1.65 ( 10^{8} ) N / m^{2} ) * ( 1.25 ( 10^{-12} ) m^{4} ) / ( 1.5 ( 10^{-3} ) m )

**M ≈ 0.275 N * m ≈ 275 N * mm**

I hope this helps! Feel free to message me on this site for further explanation!