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Beam - Curved Spring; Bending stress, force & deflection Calculation 3

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see3p0

see3p0

Mechanical
Jan 28, 2011
14
I am designing a bent spring ;


A simplified version is illustrated in the above link. It consists of two straight sections and a curved section in the middle.

Multiple springs will hold an object in place.

I am looking for some formulae/ method to calculate the deflection, force and bending stress in the beam.

Essentially the spring force at each location.

I had a relatively similar problem last year and posted here;


Desertfox offered a 'strain energy method'

I am wondering how this might apply to my current problem or maybe someone could suggest a calculation method for this problem.

Thank you
 
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You should have already done this back in school. Look in your Shigley book.

Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
 
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i'd do section cuts normal to the CL.

as above, you can use curved beam equations, but i'd be surprised if your geometry really is a curved beam problem.

you're doing all this 'cause you're interested in the deflection of the end, ie the stiffness of the part ?
i mean, from a strength perspective, the critical section is at the edge of the fixed support.

this assumption (a fixed support) is probably going to be the biggest variable in matching with the real world.
 
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looking at it a 2nd time, is it going to be significantly different from a cantilever, 8mm long, 0.2mm thk ?
 
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If you can neglect the effect of the curved part and replace it with a simple angle , it's a very simple problem that doesn't require to go for strain energy: the equations for a cantilever are directly applicable.
Simply decompose the load into a transverse and an axial force and the sloped part is directly calculated; then apply the load and the moment to the horizontal portion, calculate it directly and then sum the deflections for the two portions.

prex
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Strain energy
From sketch small curvature neglected, all strain energy in 2 sections

E= integral of M^2/2EI ds=(Mx +My)^2/2EIds

My=Fy*x 0<x< 8
Mx=Fx*xsin@ x<6
Mx=Fx*6*sin@ x>6
Fx= fictitious x force, zero in this case but neede for the x deflection
s=x/cos@ for x< 6mm
ds =dx/cos@ for x<6
ds=dx for x>6

y deflection=partial derivative of E WR to Fy
x deflection =partial derivative of E WR to Fx , then set Fx=0
E=INT[(Fy*x+Fx*x*sin@)^2*dx/(2EIcos@)] limits 0>x<6
+INT [(Fy*x+Fx*6*sin@)^2)*dx/(2EI] limits 6>x<8
ydefl=part of E WRT Fy=INT[(Fy*x^2)*dx/(EIcos@)] limits 0>x<6
+INT [Fy*x^2)*dx/(EI] limits 6>x<8
xdefl=INT[sin@*Fy*x^2/EI*dx] limits 0 to 6
+INT[6*x*sin@Fy*dx/EI limits 6 to 8

 
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the straight pice near the support is easy enough.

for the sloping piece, i think you need to convert the load into it's components (one component compressing the spring, the other bending it)

like prex suggested above.
 
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Since you don’t have time to show a sketch of your problem, you really shouldn’t expect others to waste much of their time guessing what you are trying to do, and then trying to help you.
 
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I get Hooke's Law to be F = 1.58688 D, the spring stiffness is 1.58688 N/mm. This is about 9.06129 lbf/in imperial, typical of small Bowen Springs used underneath Dogs in our setting tools.

Work as attached, Curved Beam Theory slightly bastardized for our application.

Regards,
Cockroach
 
 http://files.engineering.com/getfile.aspx?folder=9d45d7f2-ea64-4154-abde-dce926c9d329&file=Sturni_Engg_Ltd_-_16_May_2012.pdf
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dhengr - what's wrong with the sketch? compared wit the usual we get round here it is rather good.

In my opinion to within engineering accuracy the beam effectively behaves like a straight one 8 mm long, in displacement and stress .

The reason is that at each x the moment is the same, and the difference in section properties is irrelevantly small at the tip, and non existent at the important bit.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
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GregLocock:
I tried to find an attachment which showed a sketch, and had no luck until I saw Cockroach’s sketch. Otherwise, I agree with your statements on the matter.
 
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"In my opinion to within engineering accuracy the beam effectively behaves like a straight one 8 mm long, in displacement and stress .
The reason is that at each x the moment is the same, and the difference in section properties is irrelevantly small at the tip, and non existent at the important bit. "


You are almost right for the angle shown, but for steeper angles the solution is not within engineering accuracy.
I got
y=F*(296+216/cos@)/3EI
in mm units
where you see the dependence on @.


 
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I really appreciate all the input folks.

Prex, RB1957 & GregLocock thank you, Zekeman thank you for the details and Cockroach thank you so much for the neat upload!

I need to digest this and maybe get back with a question or two when I get a chance to take this information in.

Have a nice weekend guys
 
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