Deflection equation for continuous beam with unequal spans 4

[conradlovejoy]

I can find published formulas for the critical deflection of a continuous beam with a uniformly distributed load and equal spans (wl[sup]4[/sup]/185EI); however, I would like to know if anyone has ever taken the time to derive an equation for a two-span continuous, uniformly-loaded beam with unequal spans. I can get programs to give results but I'd like a usable equation that I can write into a spreadsheet I am building.

 

[conradlovejoy]

Additionally, running examples through Enercalc and using an iterative process, I can get very similar results to that of Enercalc by using wl[sup]4[/sup]/137EI where l is forced to be the longer of the two spans...but that seems like really ugly math and I wondered if anyone who is more of a calculus wiz has calc'd this out.

 

[rb1957]

it's simple enough to derive, though lengthy, from beam moments.

Could you use a conservative estimate (like deflection of longer span) ? ... like you have !

 

[jayrod12]

Generally I'm prone to just designing for the longest span as simply supported for deflection. Yes it's conservative, but its quick and easy.

 

[robyengIT]

page 3 case 1 of PDF file ???

 

[rb1957]

fantastic ! robyeng hits another home run !

using this I calc for equal spans (L), y = 1/192*qL^4/EI.

another day in paradise, or is paradise one day closer ?

 

[JStephen]

Neglect the middle support, calculate deflection due to the uniform load, then use point load design to calculate reaction at that point to make zero deflection, and it should be pretty straight-forward.

 

[Celt83]

user IDS has a spreadsheet that you may find useful:

https://newtonexcelbach.com/2017/02/22/conbeamu-4-13/

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[conradlovejoy]

Thanks for all the great responses, everyone!

 

[rb1957]

@JStephen ... yes, unit load method. So from reactions you can calculate the bending moment and then the displacement. Difficult to do for a generalised geometry.

since Roby answered the question (exactly), I'll give him the LPS.

another day in paradise, or is paradise one day closer ?

 

[Celt83]

To expand on JStephen's comment:
dx,udl = w x (L^3 - 2 L x^2 + x^3)/ 24 E I

da,point = P a^2 b^2 / 3 E I L

da,point = da,udl

P a^2 b^2 / 3 E I L = w a (L^3 - 2 L a^2 + a^3)/ 24 E I

where b = L-a

P a^2 (L-a)^2 / 3 E I L = w a (L^3 - 2 L a^2 + a^3)/ 24 E I

P (a^4 - 2 a^3 L + a^2 L^2) / 3 E I L = w (a L^3 - 2 a^3 L + a^4) / 24 E I

P = w (a L^4 - 2 a^3 L^2 + a^4 L) / 8 (a^4 - 2 a^3 L + a^2 L^2)

a=0.5 L, P = 0.625 L w
a=0.3 L, P = 0.720237 L w
a=0.6 L, P = 0.645834 L w

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[rb1957]

yes, and now the end reactions (which change with "a") then derive the moment distribution along the two spans, (you may be able to show peak moment on the longer span) and then the deflections ...
to arrive at the equation roby posted.

another day in paradise, or is paradise one day closer ?