Moment Curvature Plot: Explaining the Significance of the Slope

[Civilian2]

Hi, I was really hoping someone could help me...

I have a beam, simply supported and I place 2 point loads on it (equidistant from either end). I measure a range of deflections (of the beam) using gauges as I increase the loads (always having the same load at each point). I measure these deflections at each point load and at the centre of the beam.

By using a BMD (or just by experience) we know that the moment will be constant inbetween the two point loads (i.e over the centre of the beam).

By knowing the difference in deflection between the point at the centre of the beam (max deflection) and the point under one of the loads, I can calculate the curvature of that constant moment section of the beam.

I can then plot the Moment verus the curvature (x-axis). I get a straight line.

Can someone explain to me in layman terms what I get? As in, what the slope of the graph represents?

I get a straight line for this plot. Looking at it, I think I'm looking at a similar thing to the elastic section of a stress strain curve, so I'm thinking that slope of the moment curvature plot is the elastic modulus of the beam. I'm unsure about this though.

Anyone able to explain? The significance of this slope?

Cheers in advance.

 

[Civilian2]

Right, well I've figured it out.

So for those of you who care... its actually an extremely simple connection... the hours I spent completely unwarranted.

I in bending theory you find that

Curvature, or, 1/r = M/EI
where r is the radiius of curvature and M is the moment.

The slope of a moment vs curvature plot is M/(1/r)
Which is M * r

Since 1/r = M/EI then r = EI/M
therefore M*r = M * EI/M
which equals EI

which is the elastic modulus of the steel times the second moment of area.

Although I can't separate the two, EI give me the flexural rigidity of the beam which allows me to calculate strain.

So you see, this was one of those things that took me a lot more time than it was worth lol.

 

[IITian]

If you know the dimensions of the beam, you may well be able to separate the two

 

[FredGarvin]

Whoa. Thanks for digging this one up. Just a little over two years old now...