# Indeterminately Loaded Beam

1. Nov 15, 2008

### bakoo

Hey

This may seem like a homework question, but its actually extending my knowledge...

I attended a lab, and it was for beam bending, here is the setup...

Beam of length was clamped at either end, think this is called "Indeterminately Loaded Beam"? Two loads are then applied to it, one at 25% of the way, and one at 75% of the way.

The delfection was measured, and using a quarter bridge strain guage the strain was took.

Here is my problem, i want to know how to model this with theory, but i cant seem to figure out how to do it?

This is just to expand my understanding of loading, rather than any homework, but if it comes up in the future I want to know how to do my calculations.

Thanks for any help or light you can shed on this for me.

Last edited: Nov 15, 2008
2. Nov 15, 2008

### CivilEnergy

Can you clarify how it is clamped. Is it supported on both ends or is set up as a cantilever? If it is set up in cantilever configuration use this equation to model it.

deflection = Load/(6 * Modulus of Elasticity * Moment of Inertia) * (2 * l^3 - 3 * l^2*x + x^3)

l = length of beam from clamping point to end
x = distance from end that load is placed

For multiple loads you could use superposition principles to figure out the total deflection.

3. Nov 15, 2008

### bakoo

Its clamped on both ends

/|......L1........L2........|\
/|____V______V_____|\
/|............................|\

The V is where the L load is applied... For clarification __v_____v___ is the beam and |\ is the wall/clamp/fixing as in a cantilever beam. The .............. are just to format, so ignor

Strain gauage is possitioned between L1 and the wall/clamp

Thanks

4. Nov 15, 2008

### CivilEnergy

It should still work with superposition. In calculating the stress you would figure out three simultaneous equations for the beam, by breaking loose all the redundant fixtures, until the beam is determinate and replacing them with unknown reactions.

This beam set up is the same as a cantilever with a moment and an upward force acting to the free end.

You solve for the unknown reactions (moment and upward force) in the simultaneous equations. Now that you have solved for the unknown reactions plug them back into the free body diagram and solve for the internal beam stresses.

The part I am not to sure about is how to account for the fixed end theorem for both ends of the beam to get actual results instead of theoretical.

Hope that helps

5. Nov 16, 2008

### bakoo

Hey

i will use superpossition to try and solve this, i have the real results so if the theory is close they i will know i have it right.

Thanks again for the advice

6. Nov 16, 2008

### bakoo

I have found the name of the beam configuration.

Its related to as a "Fixed end beam" and i think you apply superpossition to it from simply supported and cantilever, but still working on this to find how its done?