# Fixed end beam

1. Nov 16, 2008

### bakoo

Hey

I recently observed a lab where we had a fixed end beam of 1m, and applied two loads to it at 0.25m and 0.75m.

I was trying to figure out the theory of what i had just seen, but so far coming up with nothing that actually seems close to the actual results i logged down.

This is not homework, but i want to know it incase it surfaces in the future in work, test or even real life.

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

/| represents the wall/clamp/rigid fixture

L

____ represents beam

...... represents nothing as is used to format the diagram

I also want to try and model the stress /strain.

A quarter bridge strain guage was placed in-between L1 and the clamp/wall/rigid fixture

Any help or light to shed on this problem would be great.

Maybe this would be better placed in homework too even though in theory its not?

2. Nov 17, 2008

### redargon

what did you observe and how did it differ from what you tried to calculate? What theories have you applied? Are you interested in the stress measurements from the strain gauge set up, or the deflection that you saw? A little more info and I'm sure somebody could help.

3. Nov 21, 2008

### bakoo

Hey

What i observed was the bending of the beam to a deflection in (mm) and the stress of the beam increasing when more weight was applied.

I want to model this. Basics are that the beam was fixed support, 1m in lengh, 2 loads applied, 6kg at 400mm and 4kg at 850mm. B and D of beam was 22.2 and 6.34 respective to give second moment of area/intertia of 471.454.

I am interested in shear-force and bending, reactions at either end, shear stress, deflection and how the stress gauge reading. Stress gauge as located at 290mm to the mid-point of the strain gauge.

So far i have tried superpossition using standard bending moment, shearforce equations and have gone wrong, i think with superpossition.

I am now trying to apple the Macaulay Method but not 100% how to apply it to a fixed end beam with 2 loads.

Any help would be most useful

4. Nov 21, 2008

### kv_home

superposition is the way to go - straightforward problem - go back and start over, looking for algebra mistakes - if you have a library handy - Roarkes formulas for stress and strain will have a textbook examples of this beam (and most other) beam problems

5. Nov 21, 2008

### redargon

with the info you gave, i did a force equilibrium and a moment equilibrium ie Sum of forces=0 and sum of moments=0.

]----------[
a..............b

There are opposing forces in the vertical direction at a and b. So, L1+L2=Fa+Fb.
With that you can write Fa in terms of L1 and L2 and Fb. Fb can also be written in terms of L1,L2 and Fa. 2 equations, 2 unknowns

Taking moments about a (and using the convention that clockwise moments are positive), Ma=L1*(distance from L1 to a) + L2*(distance from L1 to a) + Fb*(distance from Fb to a)
then same for moments around b
Now you have 4 equations and 4 unknowns (Ma,Mb,Fa,Fb). So sub some of the equations into each other until you find the unknowns in terms of L1,L2 and the known lengths.

From this you should get the forces at a and b and use shear stress (tau)=F/A to find the shear stresses at each end. You have A (22.2*6.34).
Use the Forces to calculate the moments at the ends. That should be it.

I did some quick clacs and got shear stress at a to be 0.293MPa and at b to be 0.404MPa.

I don't understand exactly where the strain gauge is positioned and I'm still trying to figure out how you would calculate the deflection, but I hope that helps a little.

6. Nov 22, 2008

### bakoo

Thanks for that info, working back thorugh it now to see what i get.

Any ideas on how to find the strain in the strain guage?

Strain guage (quarter bridge) is located at 290mm.

7. Nov 22, 2008

### bakoo

I have formulated this equation from some example i have found, but i dont know how to solve this equation?

$$\therefore\: -MA*1 + RA \frac{1^2}{2} - 29.43 * 0.4 - 19.62 * 0.85$$

The example i am trying to follow is:

$$\therefore -\;MA * 20 + RA * \frac{20^2}{2}- \frac{20^3}{8} + \frac{10^3}{8} -6 * 5^2\;=\;0$$

=

$$\displaystyle \therefore\;\;\;\;\;\;10\;R_a\;-\;M_a\;=\;\frac{205}{4}$$

----

I dont know, and cant figure out how to achieve this result for the top formula? Calculus is not my strong point at all...

Last edited: Nov 22, 2008
8. Nov 24, 2008

### bakoo

Hey

Can you confirm that

Ma = L1 * (Distance from L1 to A) * L2 (Distance from L1 to A) * fb (distance from fb to a)

Is the *L2(DL1 TO A) meant to be *L2 (Distance from L2 to A)???

I have solved for Ra, but now i am lost on how to solve for Rb

Same goes for Ma and Mb, little unsure if i have the correct answer for Ma and dont know how to find Mb.

Is the convention Fa and Fb the same as Ra and Rb? R standing for reactions?