Statically Indeterminate? 2

[Christine74]

Here's my problem:

We have a vertical bar welded inside a rigid enclosure. There is a load applied at its midpoint, putting half of the bar in tension and the other half in compression. I need to confirm that the side that's in compression won't buckle, and that the side that's in tension won't yield.

However, when I draw a free body diagram of the bar to solve for the two reactions, I end up with just one equation and two unknowns--statically indeterminate.

I would imagine that both reactions are of equal magnitude (equal to 1/2 of the applied load), but I can't think of any way to prove this.

How do I go about solving this problem?

Thanks,

-Christine

 

[TheTick]

How about starting off with the bar modeled as two connected springs?

If the ladies don't find you handsome, they should at least find you handy.

 

[Christine74]

Wow, that was a quick reply!

So I think what you're saying is that the amount of deflection in the upper half must equal the amount of deflection in the lower half?

If so, both reactions would be equal, and the rest of the problem is easy enough for me to solve.

-Christine

 

[redlines]

Christine74

I would treat the upper part of the bar as a rod in pure tension. The lower half is a column in compression with the top free and the bottom fixed.

determining buckling requires knowledge of the bar geometry, it is a function of load and geometry. From formulas for Stress and Strain (Roark and Young), the critical load for a fixed-free bar is:

P=K*pi*E*I/l

K is a constant based on the ratio of length to cross sectional area and the end conditions of the bar. For a straight, non stepped or tappered bar K=.25


If the load is offset at all, the critical load is significantly reduced because of the addition of a moment.

Hope this helps

 

[jdkuhndog]

Can't you simplify it a little by just do the following?

Mmax=PL/8 (assuming the bar is welded in a way that it can be modeled as fixed)

BndStr(tension) = Mmax/Sx(tension)

BndStr (compr) = Mmax/Sx(comp)

Allowable compressive stress would be based off of the critical stress for buckling...

CritStr = ((pi)^2 x E) / ((Le/r)^2)

 

[jdkuhndog]

I just read your post again...I assumed that the load which was being applied was perpendicular to the bar. I guess, now that I read it again, the load is in line with the bar. Sorry.