Deflection of beam check... 4

[Bert2]

Hi,

Its been a while since ive needed to calculate the defelection of a beam, in doing so i have came up with two quite different answers, using Macauleys and the direct deflection equation. calcs attached, any pointers in my errors would be soundly appreciated.

Thanks!

 

[prex]

C₁ has negative sign. Also there are problems with signs in the integrations, but they cancel out.
If you take this simple problem in symbolic form, recognizing, as an example, that R_A=R_B=wL/2, it will be much simpler to avoid trivial mistakes.
Look also in the first site below under Beams -> Single beam -> Simply supported -> Unif.load (or here) to check your result.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[rb1957]

yes, C1 negative.

also deflection units (N/mm*mm^4/(N/mm^2*mm^4) = mm
and index ... i expect something closer to E-17.

but good on you for trying 1st principles ...

 

[Bert2]

Thanks prex for comments i noticed my C1 error and now corrected.

also took a look at the link - nice tool thanks, it was showing a maximum deflection of 83x10^3 mm defelction . ..?

@ rb1957 - also got myself muddled up on my units thanks.

however just using the single equation i get -8.34x10^-15 mm.

just one last question is my Ixx value ok it seems quite large.....?

thanks bert.

 

[rb1957]

it's a pretty darn'd big beam, 2m deep, 1"thick, 2' wide ...

quick calc, 0.8*78^3/12 + 2*24*.8*39.4^2 = 91247 in4 = 3.8E10 mm4

 

[desertfox]

Hi Bert,
Firstly your Ixx value is correct. I calculate it to be 3.61*10^10 mm^4.
Regarding the deflection, I agree with your value on the standard equation ie 8.34mm*10^-12.
I am going to look at the Macauley method in more detail. It's good to get the old grey matter working, however, I think if it were me I would have just gone with the standard equation.
Regards,
desertfox

 

[desertfox]

Hi Bert2

Try this link look for case 4 page 10:-

http://www.freestudy.co.uk/statics/beams/beam tut3.pdf

 

[SNORGY]

I typically start with EI*d^4y/dx^4 = -w, integrate four times, solve for the boundary conditions, and get the answer that Bert2 ultimately got. For a uniformly distributed beam with simple supports, you would end up at 5wL^4/384EI anyway...so one could start there.

Seems the only thing at issue is the computation of I.

Regards,

SNORGY.

 

[Bert2]

@ rb1957 - yeah the beam is large; which in turn i guess gives a large Ixx value.


@Desterfox - Thanks for the confirmation, await your response to the macauley's.

To conclude the beam has a very small amount of deflection at 8.34mm*10^-12.?

Thanks for the help guys!

 

[leisure17]

Have you considered it might buckle first?

 

[rb1957]

20mm thick caps, deflecting nothing ... doesn't sound like a buckling issue

 

[prex]

The middle deflection is 8.3 mm.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[racookpe1978]

it's a pretty darn'd big beam, 2m deep, 1"thick, 2' wide ...

quick calc, 0.8*78^3/12 + 2*24*.8*39.4^2 = 91247 in4 = 3.8E10 mm4

Seems like a bigger problem (if in real life) will be the depth of the member: At nearly 2 meters depth and 15.7 meters long (51-1/2 feet) combined a relatively "thin" but very wide caps, won't the failure mechanism be the web twisting? (Unless you could always ensure you have a perfectly symmetric and even load exactly centered on the mid-plane of the beam every day with no induced moment being ever transferred to the beam.....)

Seems several intermediate stiffeners are essential if he wants little deflection.

 

[Bert2]

@ prex - could you elaborate how you came to the conclusion on 8.3mm deflection / what method did you use? - thanks.

@ racookpe1978 - i can see your point but with the small amount of deflection i dont see this as a problem.

bert.

 

[desertfox]

Hi Bert,
I initially agreed with your calculation on the deflection and I got the same value, however we both made the same error, which was that we mixed our units. The 'I' value that we calculated we both used in the standard equation and the units being mm^4. However the rest of the units were in metres. If you change the units of 'I' to metres m^4 you will find the deflection comes out at 8.34mm as stated by prex in his earlier post.
I also calculated the deflection by direct integration and got 8.34mm deflection again. I cannot post my solution yet but will do on Friday when I return home.
Still looking at Macauleys method and will post later.
Regards,
desertfox

 

[Bert2]

Ahh of course mm^4 for 'Ixx' thanks desertfox !

 

[desertfox]

Hi Bert2

Your welcome!
I'm believe Macauley's method is applicable for the for the standard case of beam loading which is the type your beam falls into.
Use the direct intergration method its much easier.

desertfox

 

[desertfox]

sorry I meant not applicable

 

[msquared48]

Other than refreshing your memory, which I will admit we all need to do at times, you are taking the long hard road to the solution.

The AISC steel book, and many other handbooks, have the solution for the deflection for any point "x" along a uniformly loaded simply supported beam, to include the maximum point of deflection, the center.

Mike McCann
MMC Engineering

http://mmcengineering.tripod.com

 

[desertfox]

Hi Bert2

My solution as promised.

desertfox