1
$\begingroup$

I’ve got a reasonably strong grasp of Newtonian mechanics, but as far as material science goes I don’t know which of the many material properties I need to use to solve my problem:

I’m trying to mount a heavy object at the end of a lever arm. I’ve got a 1 cm (0.4 in) diameter steel rod, and I’m planning to embed half of it in a concrete wall to hold it in place, and the other half will extend 2 m (6.6 ft) parallel to the ground with a rope holding 50 kg (120 lbs) sitting in a notch at the end. I’d like to buy a thinner rod if possible, but picturing this in my head it seems with about 980 Newton-meters (eh, 1330 ft-lbs?) of torque I’m likely to bend the rod.

Which material science properties do I use to figure out how to minimize elastic deformation and avoid plastic deformation entirely?

Note: This question was originally posted to Physics SE, and was moved here on request from @Gert :)

$\endgroup$
  • $\begingroup$ start here: meracalculator.com/engineering/deflection-solid-round-beams.php $\endgroup$ – Tiger Guy Aug 3 at 15:22
  • $\begingroup$ You can not do much to the properties of the rod because your selection is allready done. All you can do is check that the rod does not deform too much or swap it to another rod with a different material. Or use some other dimensions or hanging configuration. $\endgroup$ – joojaa Aug 3 at 15:24
  • $\begingroup$ @joojaa As mentioned above, I'm considering using a thinner rod, so I'm really trying to minimize rod thickness while retaining a certain comfortable margin of error $\endgroup$ – TheEnvironmentalist Aug 3 at 16:02
  • $\begingroup$ @TigerGuy There seems to be something wrong with that tool. It predicts deflection for a 2 m steel rod of 50 cm diameter with only 50 kg on the end at 58.5 m. Clearly a difficult feat with a 2 m rod $\endgroup$ – TheEnvironmentalist Aug 3 at 16:08
  • $\begingroup$ @TheEnvironmentalist you are sure right! I suspect a factor error, they do give the equations. $\endgroup$ – Tiger Guy Aug 3 at 16:18
5
$\begingroup$

Material Properties

For the linear-elastic analysis you describe, the key pieces of material information are the elastic modulus (E) to calculate deflection and the yield stress (Fy) to check if the material remains elastic under the given loading. For steel, E is commonly around 200 GPa. The Fy can vary significantly depending on the grade of steel, so you'll want to check for your particular product, but between 250 MPa and 350 MPa is a reasonable assumption for mild steel.

The E and Fy values will often be directly available but the actual stress-strain curve is helpful for developing understanding. (The image below is taken from the linked Wikipedia article.)

stress strain curve

This sort of plot can be generated by tension-testing to failure a small coupon of metal. The initial linear portion of the plot is where the material is still elastic. The slope of this line is the elastic modulus (aka Young's Modulus or Modulus of Elasticity). In this region, the strains are recoverable and if the load were removed the section would return to its original shape. The yield point is where you see the curve deviate from that initial linear portion. Beyond the yield point, the behavior is inelastic and permanent deformations occur. Some materials do not have a well defined yield point and Fy will be estimated by an offset method. Many resources are available online or in introductory Material Science textbooks if you're interested in learning more about stress-strain curves.

 

Analysis Process

Provided your cantilever beam is kept in the elastic region and the deflections are small, simple equations can be used to analyze beam response. It's important to recognize that for inelastic behavior or large deflections it would be necessary to employ a different methodology than the one presented below.

  1. Choose a material and cross section. This establishes E, Fy, and the moment of inertia, I, for the cross section.
  2. Calculate the maximum applied bending stress in the section and compare to Fy. Adjust the section as needed to ensure the beam remains elastic.
  3. Check deflection. Again, adjust section as needed in order to limit deflection.

 

1. Material and Cross Section

Assume E = 200 GPa, Fy = 250 MPa

Moment of Inertia for your 1 cm diameter cross section:

$$ I = \frac{\pi r^4}{4} = 4.91 x 10^{-10} \,m^4$$

Note that for beams in uniaxial bending, a solid circular cross section will be one of the least efficient sections. You want to concentrate material where bending stresses will be highest, which in this case is the lower edge of the beam cross section. (This is why I-shapes are so popular for beams in buildings and bridges.) For a given area of steel, a circular tube will offer improved efficiency over a solid bar and a rectangular tube with the long leg placed vertically would be even better.

 

2. Calculate Maximum Bending Stress and Check Against Fy

For a cantilever beam, the maximum moment will occur at the fixed end. Your problem statement indicates a 50kg mass and 2m cantilever length. Convert kg mass to N force by multiplying by acceleration due to gravity ($9.81 m/s^2$)

$$ M_{max} = PL = (0.49kN)(2m) = 0.98 \, kNm$$

For bending stress in a cross section via Euler-Bernoulli beam theory:

$$\sigma = \frac{My}{I}$$

And take $y = (bar\, radius)$ since the maximum stress will be at the edge of the bar. Thus,

$$\sigma_{max} = \frac{(0.98 kNm)(0.005m)}{4.91 x 10^{-10} \, m^4} = 9992 \,MPa$$

This far exceeds our assumed Fy of 250 MPa, so the 0.01 m diameter bar would definitely not remain elastic. You'll need to choose a section with a higher moment of inertia. I'd suggest setting up a spreadsheet to play around with various cross sectional geometries.

 

Check Deflection

For an elastic beam experiencing small deflections, the following equation predicts deflection at the free end of an end-loaded cantilever.

$$\delta_{max} = \frac{PL^3}{3EI}$$

Now, as discussed above, you'll need to change the section from the 1 cm diameter bar, but for purposes of illustration...

$$\delta_{max} = \frac{(0.98 \,kNm)(2 \,m)^3}{3(200\,GPa)(4.91 x 10^{-10} \,m^4)} = 13.3 \,m $$

Granted, this isn't physically possible for a 2 m long beam, but in practice the key thing to note is simply that this beam doesn't meet the requirements to even utilize this equation. The beam isn't elastic and the deflection is definitely not small. The theoretical max deflection decreases fairly rapidly with increasing bar diameter - as user NMech noted, a 9 cm diameter bar experiences 2 mm end deflection. However, I'd expect you're not that keen on using a 9 cm diameter bar. If you really need a single 2 m long cantilever end-loaded beam, it'd be recommended to explore more efficient cross sections than a circular bar.

 

Bottom Line

While a 1 cm diameter steel bar is not adequate for your load configuration, hopefully this information can give you a good start on selecting a suitable section.

Also note that the above neglected the self-weight of the beam. If you'd like to include self-weight in your analysis, it can be easily done by employing the principle of superposition (valid only for linear-elastic analyses). Essentially just add up the load effects from a cantilever beam with a uniformly distributed load (self-weight) and a cantilever beam with your point load at the free end. To that end, standard beam equations such these come in very handy for simple analyses.

| improve this answer | |
$\endgroup$
2
$\begingroup$

Although the deflections equation @kamran reported, are indeed what most textbooks give, the problem is that they only hold true for relatively small deflections.

I also agreewith @alephzero that there are a few errors in the calculations. e.g. I calculate $Ix = 0.0491 cm^4$.

By my calculations, you would need a steel rod with almost 11cm diameter to support a load of 50kg at the end with a deflection less than a 1 mm.

Even if you doubled the deflection to 2 mm, the steel rod would have to be just over 9 cm.

In this particular case, you are better off considering an I beam profile.

| improve this answer | |
$\endgroup$
0
$\begingroup$

I cm diameter bar at 2metres cantilever does not support its own weight within acceptable deflection limit, let alone 50kg load.

Assuming the wight of the bar is 200*8 gr=1600 gr.

$I_x= \pi d^4/64= 3.14*1/64=0.5cm^4 \quad \delta= Pl^3/3EI$ $\delta= 1.6*100^3/(3*0.5 *2039432)=$6 cm> L/180 6mm$

And this is deflection at half span, deflection at the 2 meters is much more.

You need to pick a bar much more substantial.

| improve this answer | |
$\endgroup$
  • $\begingroup$ What are these equations you're using? $\endgroup$ – TheEnvironmentalist Aug 3 at 20:34
  • $\begingroup$ These are the first semester material of basic statics. There is a wealth of material available if you search for basic statics. $\endgroup$ – kamran Aug 3 at 20:37
  • 1
    $\begingroup$ Just some extra help. These equations can be found searching for beam deflection, and second moment of inertia. And can be considered truly basic for enginners. If you want less material you should consider changing the geometry or using a hollow tube suficient compatible (less self weight). $\endgroup$ – Leafk Aug 3 at 21:29
  • $\begingroup$ @kamran to be honest this could also be solved without calculation. Just lifting a beam once will get you the idea quickly. $\endgroup$ – joojaa Aug 3 at 21:29
  • $\begingroup$ @kamran 3.14/64 isn't even close to 0.5, you seem to have specified E to 6 significant figures which is nonsensical, and you haven't bothered to solve the actual structure specified. IMO this is answer is worthless. $\endgroup$ – alephzero Aug 3 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.