# deflection calculation

• posted

Morning,

I am running a deflection calculation to size the amount of force needed to pre-bow a piece of rectanglur steel 2mm. I have run the calculation two ways and am getting different results. Can someone point me to where my error is? Thanks in advance

Here is one run

Y= P(L^3)

--------

3EI

P=7500 lbs L(length of moment arm) = 4.8inches E(Modulus of elasticity) = 205 Gpa = 29732737 psi I(Moment of inertia) = .781 in^4

Deflection = 7500 * ((4.8)^3)

--------------

3 * (29732737) * (.781in^4)

Y = 829,440

-------

69663802

Y = .011 in = .254 millimeter

Here is the other run

P=7500lbs = 3401 kg L = 4.8 in = 12.192 cm E = 205 Gpa = 2090418 kg force/cm^2 I = .781 in^4 = 1.9 cm^4

Y = (3401)*((12.192)^3)

-------------------------

3 (2090418) (1.9) Y = 6163558

-----------

11915382.6

Y = .51cm

Y = 5.1mm

• posted

I'd be concerned about the accuracy of any calculation which contains a unit of kilogram-force!

Convert everything to N and m.

P=33362N L=0.12192m E=205e9 N/m² I=3.251e-7 m^4

Plug that into your calculator and you get: Y=3.02e-4 m = 0.0119"

Actually, I think the problem isn't in the kg-f (even though that still makes me cringe). I think it's in your I conversion

I=0.781 in^4=0.781in^4 * (2.54cm / 1 in)^4 = 32.5 cm^4

But that alone doesn't seem to account for the discrepancy in your answer, so there's something else too...

-Paul

• posted

It looks like you have two different difficulties here. One is a matter of unit conversions that is preventing your from getting equivalent results when makeing this calculation in two different unit systems. If you make all your unit conversions correctly, this problem will definitely go away.

The calculation you have been making is for elastic deformation, and when the load is removed, the material will spring back to its original shape. In your original statement you said something about a need to "pre-bow" the material. This sounds almost like you want to permanently bend the material into the form. If a permanent bend is what you want, that will require that the material be bent far enough to yield the material. The trick is to bend it past where you want it such that it will spring back to the point where you want it to be. This will leave the part with locked up stresses due to the fact that it has been yielded, and then those stresses have been partially relieved when the load is removed. The elastic beam calculation above has very little relevance for this problem.

It is hard to be sure which problem is the one you are concerned with, but you need to be clear about the matter in your own mind.

• posted

I think Paul keyed in on the conversion problem. Thanks! We are actually counting on the spring back of the material after unclamping th weldment. Thanks for all the help guys!

• posted

I ran a quickie check on a beam program (Archon's)

No guarantees, but I found that a 7500 lb point load on the end of a

4.8 inch cantilever deflected it 0.0119 inch or 0.3mm

So I'd expect about 50000 lb to make that 2 mm deflection.

Brian W p.s. You d>Morning,

• posted

From: "bgal" snipped-for-privacy@vizient.com

I did a quick run of your data thru DTBeam ( at

free functional demo version is there ). I ran the English units. The deflection for your 7500 lb load is correct at 0.011906 inches. Thus... it's easy to conclude that some of your unit conversions for your metric input values are incorrect. Note also... it's perfectly ok to use kg as a "pseudo" unit of force as long as you are assuming that your gravitational acceleration is constant. There are countries where this is a common practice. For instance... it's quite common to see kg listed on canned food goods and people from many countries will quote their body weights in kg. And yes... I've known of companies ( non-US companies ) requesting that software be required to work with kg.

I also used DTBeam to automatically convert all of the above English units to metric units and reanalyze. Here are equivalent metric input values and the resulting deflection.

L =121.92 mm I = 325076.74339 mm^4 E = 20904.179318 kg/mm^2 = 205000.023 N/mm^2 Load = 3401.94219 kg

Resulting Deflection = 0.30242 mm

Dan :-)

• posted

bgal wrote: [reposting my reply from 12/23]

** Calculating with US customary units:

That should be Y = 0.0119 in or 0.303 mm (25.4 mm = 1 in).

** Recalculating with SI units:

Should be I = 0.781 in^4 or 32.5 cm^4.

It's a good idea to carry the units through in your calculations. Units should cancel out so the answer is in the proper units.

(33.361e3 N)*(0.122 m)^3 Y = ------------------------ = 3.03e-4 m or 0.303 mm 3*(205e9 N/m^2)*(3.25e-7 m^4)

or

(33.361e3 N)*(122 mm)^3 Y = ------------------------ = 0.303 mm 3*(205e3 N/mm^2)*(3.25e5 mm^4)

- CJF.

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