Cantilever beam bending

[o_0]

I'm trying to calculate the tip deflection of a horizontal cantilever beam for various materials and lengths, assuming the only force is gravity.

According to the lecture notes, the formula is this:
\[y^{tip}_{gravity} = \frac{3 \rho (L^4)}{2 E t^2}g\]
(Just in case the picture doesn't show up, it's this: y^{tip}_{gravity} = \frac{3 \rho (L^4)}{2 E t^2}g)

Using matlab I get the following results:

	L = 10 microns	L = 100 microns
SiO2	0.0050	               50.3
Si	0.0018	               17.8
Al	0.0057	               56.7

However, the results for 100 microns don't seem to make sense, since the units would be meters. My code is below, I also got the values of the constants from the lecture notes. The only values I converted were the densities, from g/cm3 to g/m3 because everything else was given in terms of meters. Any assistance would be appreciated, thanks in advance.

% dimensions in meters
L1 = 10e-6;
L2 = 100e-6;

W = 5e-6;
T = 1e-6;

g = 9.8; % m/s^2
E_SiO2 = 73;
E_Si = 190;
E_Al = 70;
% Young's modulus N/m^2

% density (g/m^3)
p_SiO2 = 2.5*100*100*100;
p_Si = 2.3*100*100*100;
p_Al = 2.7*100*100*100;

% Tip deflection, only force is gravity

y_tip_SiO2_L1 = ((3*p_SiO2*(L1^4))/(2*E_SiO2*(T^2)))*g 

y_tip_SiO2_L2 = ((3*p_SiO2*(L2^4))/(2*E_SiO2*(T^2)))*g

y_tip_Si_L1 = ((3*p_Si*(L1^4))/(2*E_Si*(T^2)))*g 

y_tip_Si_L2 = ((3*p_Si*(L2^4))/(2*E_Si*(T^2)))*g 

y_tip_Al_L1 = ((3*p_Al*(L1^4))/(2*E_Al*(T^2)))*g 

y_tip_Al_L2 = ((3*p_Al*(L2^4))/(2*E_Al*(T^2)))*g

Also, for low pressure damping it wasn't specified what pressure to put in the formula, can I just put in atmospheric? Or 1 torr?

 

[BradtheRad]

Didn't pore through the calculations.

Sometimes a coefficient is contrived to counteract a large exponent somewhere. To keep things manageable.

Also, try checking your code by running the formula for a situation which has an expected outcome, and in units that match your prediction.

Seems like atmosphere should have very little damping effect. I would think it makes more difference what fluid is surrounding, rather than the pressure it's at.

 

[o_0]

Thanks,

The fluid is air, assumed to be nitrogen, formula was already simplified using this assumption but pressure is still in there. I'll just use 1 torr since rough vacuum seems like a suitably low pressure. I had already decided this but won't edit it since you already replied to it.

I don't really have any expected outcomes at my disposal, would you happen to know if that's the right formula? Or is there another one?

Maybe the huge results mean that the beam wouldn't be stable, it would just fall off by itself?

 

[BradtheRad]

In my mind I was thinking how a metal bar would behave in real life.

Say, an aluminum bar, 3 feet long and 1/2 in. square, might bend downward about 1/16 in. under its own weight.

Putting a 1 lb on the end, might cause another 1/16 inch downward.

This isn't the best engineering practice, but it will show up the difference between microns and meters.

Or a steel ruler. It's commonly found around the house. And it will show measureable deflection. Flip it over to be sure you're measuring real deflection rather than existing curvature.

There are probably examples in the textbook which are better to use than what I was thinking.

 

[o_0]

---

Thanks so much for your help. The textbook is unhelpful, we never use it. I figure it would be easier to use this website: **broken link removed**

I clicked on "cantilever" then picked the first one at the top, with Q and L. I put Young's modulus as 70 for Al, the moment of inertia I calculated as 4.1667e-25 from \[ \frac{1}{12} w (t^3)\]
I put 9.8 (gravity) for Q, 0.00001 for L (100 microns) but what is x? I probably shouldn't leave it at the default, 100. The lecture notes have a diagram on slide 19 (attached), but I still can't figure out what x is? If I leave x = 100, I get a totally useless graph for displacement (it tells me that for x = 100 microns the displacement is somewhere between 0 and 1 million, but much closer to 0).






I also tried x = 0.00001 but it gives identical graphs and tables. The table looks like this:

x Displacement Slope Moment Shear 
0.00000 0.00000 0.00000 0.00000 -0.00098 
0.00001 78539.37169 869513794773.75170 0.00000 -0.00088 
0.00002 293437.65250 1565766538190.37180 0.00000 -0.00078 
0.00003 616135.07092 2108009458178.43140 0.00000 -0.00069 
0.00004 1021431.82855 2515493782666.49900 0.00000 -0.00059 
0.00005 1487488.10010 2807470739583.14650 0.00000 -0.00049 
0.00006 1995824.03341 3003191556856.94300 0.00000 -0.00039 
0.00007 2531319.74944 3121907462416.45900 0.00000 -0.00029 
0.00008 3082215.34228 3182869684190.26460 0.00000 -0.00020 
0.00009 3640110.87911 3205329450106.92900 0.00000 -0.00010 
0.00010 4199966.40027 3208537988095.02440 0.00000 0.00000

In this table, x is in inches but I don't think it's the same as the x I was talking about before.

 

[BradtheRad]

X is any location on the beam where you choose to calculate how much is the deflection Y.

To conform your code to the theoretical formula, you might as well measure Y at the end (X=L). It's the easy way.

For any other location, you cannot take a simple proportion. The beam is not straight. So the formula gets complicated because it must account for the beam curving slightly.

 

[o_0]

I tried x = 100 (default), x = 1 and x = 0.0001. They all produce the same table posted previously.

I even tried your example of the 3 ft. long Al bar. I converted all values to be in terms of inches: (E = 69444.444, I = 0.005208, q=386.4, L=36, x=1) and got the following results, just as nonsensical as my values:

x Displacement Slope Moment Shear 
0.00000 0.00000 0.00000 250387.20000 -13910.40000 
3.60000 4194.59717 128996.16905 202813.63200 -12519.36000 
7.20000 15671.79264 232288.30442 160247.80800 -11128.32000 
10.80000 32906.27835 312732.40985 122689.72800 -9737.28000 
14.40000 54552.19423 373184.48907 90139.39200 -8346.24000 
18.00000 79443.12825 416500.54584 62596.80000 -6955.20000 
21.60000 106592.11636 445536.58389 40061.95200 -5564.16000 
25.20000 135191.64253 463148.60698 22534.84800 -4173.12000 
28.80000 164613.63874 472192.61883 10015.48800 -2782.08000 
32.40000 194409.48495 475524.62319 2503.87200 -1391.04000 
36.00000 224310.00918 476000.62382 0.00000 0.00000