
#1
Aug310, 12:37 PM

P: 8

Does anyone know the spring constant for a clamped/fixed beam with length L that is uniformly loaded? I know the spring constant for a clamped beam that is centrally load by a point load is
k= 192EI/L^3 I know for a uniformly distributed is different. 



#2
Aug310, 01:00 PM

P: 5,462

No idea what you mean by spring constant.
The formula you have quoted is for the midspan deflection of a beam with fully contrained end supports and a central point load. 



#3
Aug310, 01:32 PM

P: 8

well if you work out the units to what i posted you will see that the k has units of N/m, aka stiffness (not deflection of units 'm')
That is the stiffness of the beam, or how much force it takes to deflect the beam 1 unit of length depending on what set of units are used. 



#4
Aug310, 02:08 PM

P: 5,462

Spring constants for beams
I suppose what you are looking for is
k= 384EI/L^3 Bearing in mind that this is to be multiplied by the total load =wL, as opposed to W in the case of the point load. 



#5
Aug310, 02:15 PM

P: 8

Cool! thanks.
not to make it seem like I don't believe you, because I do and that constant looks familiar. but I'm preparing a calculation document for a company. Do you happen to have the source of that info? 



#6
Aug310, 02:39 PM

P: 5,462

Feel free to solve the differential equation for yourself
[tex]EI\delta =  \frac{{wL{x^3}}}{{12}} + \frac{{M{x^2}}}{2} + \frac{{w{x^4}}}{{24}} + Ax + B[/tex] You have the usual boundary conditions ie the deflections and their first derivatives are zero at the supports and the first derivative is also zero at the centre. These lead to A = B = 0 and M (midspan) is given by [tex]M = \frac{{w{L^2}}}{{24}}[/tex] 



#7
Aug310, 08:37 PM

P: 8

I worked it out and you were spot on. Thanks a bunch!




#8
Aug310, 09:26 PM

P: 8

Here's another question:
I was doing this problem via FEM by hand and came out with another spring coefficient. It's in the attachment, check it out tell me what you think. the spring constant k is found in the very last equation and is the first term to the left of the equal sign that is being multiplied by U21. it's 24EI/L^3 



#9
Aug510, 02:20 AM

P: 5,462

Where did this come from?
It sems to be only part of the story. You have displayed the equivalent nodal force calculation for a beam element, but it is unfinished. If we remember that L_{e} = L/2 and substitute Then [tex]{U_{21}} =  \frac{w}{{24EI}}{\left( {\frac{L}{2}} \right)^4}[/tex] for unit load w = 1 and U_{21} = 1/k So [tex]k = \frac{1}{{{U_{21\left( {w = 1} \right)}}}} =  \frac{{384EI}}{{{L^4}}}[/tex] I apologise for the power error in my earlier rendition I copy & pasted your formula and forgot to change the power. 



#10
Aug510, 01:02 PM

P: 8

truth is, its a confidential document which is why only some of the story is shown. I went on the assumption that the information given would be enough for you to understand.
If you are gonna use unit load of 1 you still have to use the units. since you didnt, the units are wrong for the stiffness, k. your power was right before. but like a dummy i now see i overlooked where the 382 came from because i forgot to evaluate at L/2 



#11
Aug510, 02:11 PM

P: 5,462

The formulae are the same.
I am just used to working in deflection terms not spring constants. Glad we got there in the end. 


Register to reply 
Related Discussions  
spring constants  Introductory Physics Homework  7  
Spring constants  General Engineering  1  
Spring constants  Introductory Physics Homework  2  
Spring constants and such  Introductory Physics Homework  3  
spring constants  Introductory Physics Homework  6 