Bending spring rate bending beam 1

[tlee123]

I'm interested in the spring rate of a cantilevered beam (one end free, one end fixed). I've rearranged the formula from Roark's to be:

Spring Rate=3EI/(length^3) for a=0 (load at free end).

OK, now what I don't know is what happens when there is an axial compressive load in addition to the perpendicular load. I can't seem to find this case in the book. My intuition says that there may be a small decrease in the spring rate with an axial load. Does anyone have experience with this?

Thanks,

Tom

 

[Cockroach]

You are correct, since y=(-P L^3)/(3 E I) and recognizing that the Hooke Law is P = K y [P=transverse load, L=beam length, E=Young's Modulus, I=moment of inertia (by area)], then it follows K = (-3 E I)/L^3.

Adding an axial force would tend to buckle the Cantilever like a Euler Column. I would therefore consider this effect apart from transverse loading, it doesn't contribute to vertical deflection of the end beam. Where the axial load is visible is at the reaction of the support.

However I would think that the system spring rate would seem to behave stiffer with the addition of an axial load. I'm not confident on the exact mathematical relationship to the cantilever spring rate, my suspicion would be to combine the Euler Column and Cantilever spring rates like PARALLEL springs, since each effect experiences the same deflection. In that case, the spring rate behaves like resistors in a parallel electrical circut:

1/K = 1/K1 + 1/K2 or K = (K1 K2)/(K1 + K2)

for K1 = Cantilever, K2 = Euler spring rate(s). This would seem to suggest an increasing spring rate with axial load rather than decreasing.

This would be my best guess off the top of my head.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[GregLocock]

Ah, unless you are talking about the non linear effect when an axial force is applied to the end of a cantilever that has already been bent by a tip force.

Not too tricky to solve a particular case, Roark covers it.





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[tlee123]

Thanks for the quick replies, Cockroach and Greg. I was rushing when I wrote the original post and left out the details. Here are the numbers:

d=.05 in round wire
L=.6 in
E=3x10^7
A=2x10^-3
I=3x10^-7

K (axial)=AE/L=100,000 lb/in
K (bending)=3EI/L^3=125 lb/in

Since K(axial)>>K(bending) and if we use Cockroach's formula, K (effective) ~ K(axial). So even if the stiffness in bending increases, it's a very small effect.

Greg has a point and I will look into it. The axial loading of the bent wire will create a moment, tending to bend the wire more. This non-linear effect will therefore lower the effective bending stiffness.

What made me think of this original question is that I know if you axially load a beam close to the critical (buckling) load, you can grab the center of the beam with your hand and easily move the center back and forth; the tranverse stiffness has become very low. I've heard about this being done with 1/2" dia rods, which I thought sounded dangerous, and I verified this phenomena with loading of 1/32" dia wires.

Regards,

Tom

 

[tlee123]

Sorry, in my previous post the following line:

Since K(axial)>>K(bending) and if we use Cockroach's formula, K (effective) ~ K(axial).

Should read:

Since K(axial)>>K(bending) and if we use Cockroach's formula, K (effective) ~ K(bending).