Predicted deflection of a rectangular steel beam

[kiteboy]

Hi

Im am trying to find out a predicted deflection of a steel rectangular beam of uniform cross section (which is 12.7mm wide x 6.4mm deep). The beam is simply supported at both ends and has a span of 660mm. The load is to be placed in the middle of the beam.

I have the formula to be used to work out the deflection for the above case which is

d = C WL^3/EI

C = the constant of 1/48
W = the load
L= the span cubed
E = Youngs Modulus
I = 2nd Moment of Area

Ive worked out the I to be 277
E = 200kN/mm^2
W = 40
L= 0.66 cubed

Tried plugging in these numbers but they not comming out anywhere near the values I was expecting...about 4mm (I think)

Can anybody see anything wrong with what I am entering or any obvious mistakes

Thanks for any advice guys

 

[PhanthomJay]

Check your units and math. It's easy to get mixed up with the decimals when using metric units. Don't leave off the units in your givens or calculations! You should get the correct answer.

 

[kiteboy]

Thanks for the reply

I had a feeling it may be the units somewhere but I've never heared of some of these units in my life :) so not sure how to convert and to what

Should I convert them to inches or something...aint got a clue really

I know there are some online calculators that can be used but would rather get it right myself

Thanks again

 

[PhanthomJay]

Thanks for the reply

I had a feeling it may be the units somewhere but I've never heared of some of these units in my life :) so not sure how to convert and to what

Should I convert them to inches or something...aint got a clue really

I know there are some online calculators that can be used but would rather get it right myself

Thanks again

Well, you, like I, must be from the United States, where SI , in the field of structural engineering , is seldom used. I don't have a feel for the numbers, either. But leave it in metric units, don't convert to USA units, just use the applied load in Newtons, the length in meters, E in N/m^2 (paschals!), and I in m^4. The deflection will then have the units of meters. Multiply that result by 1000 to get the deflection in mm. You should note that 1000mm = 1m when converting .Don't take any shortcuts if you are not familiar with SI. I get a deflection of .004 m , or 4 mm.

 

[kiteboy]

Ok i think I have it at last

so the result I am getting is 0.00432 that's 4.32 mm...is that correct?

 

[PhanthomJay]

Yes, looks about right. Might want to round it off to 4.3 mm.

 

[kiteboy]

Amazing...thankyou

I have a similar issue and the results I am getting are not comming in as what they should be..yet again

Using the same beam and apparatus but this time the loads are two 20N loads placed symetrically on the beam 220mm apart and 220mm from a support

Using the same formula but this time with a constant of 23/648

Plugging the numbers in as before

23X40X0.66^3/ 648x 200 x 277

= a deflection of a about 7mm

I was expecting a deflection of 3.5 or 3.6 mm

Thanks again

 

[kiteboy]

Oh hang on does the load go into this formula as 20N as

which would give 3.6mm deflection??

 

[PhanthomJay]

Oh hang on does the load go into this formula as 20N as

which would give 3.6mm deflection??

Yes correct. The formula you referenced is for max deflection of a simply supported beam with 2 equal loads P symetrically placed at the one-third points from either end. P = 20N for this case. Are the deflection formulas being given to you, or are you looking them up in a table as I've been doing? Once you know the formula, its all algebra from there.