Deflection Curve of a Compound Spur Gear Shaft

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[TD="class: alt1, bgcolor: #E3E3E3"]Hi All!

I would really appreciate your help on something that has been bothering me for the past week.

I am uncertain on how to derive the deflection curve equation for a certain shaft. Loads are acting on both ends of the shaft, with the reaction forces being provided by two separate bearings that are located towards the middle of the shaft.

Is it correct to assume that by finding the Bending moment equations for each section of the shaft and then double integrating I will find deflection for each section of the shaft? I have attached a picture of my most recent attempt so far.

I am fairly certain that this is the right approach, however I am not sure as to whether or not I need to somehow combine all these moment equations into one equation so that I have a smooth deflection curve as opposed to a disjointed one.

Thank you for your help.[/TD]
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[QUOTE=pepster;6024][TABLE="class: tborder, width: 100%, align: center"]
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[TD="class: alt1, bgcolor: #E3E3E3"]Hi All!

Is it correct to assume that by finding the Bending moment equations for each section of the shaft and then double integrating I will find deflection for each section of the shaft? I have attached a picture of my most recent attempt so far.

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Not the way you are doing it. For one thing, even if you wrote the equations correctly, you have a mess to unravel the the constants .[BTW, your equations are written incorrectly. When you change domains you have to introduce the "step function" U(x)= 1 for x>0,=0 for x<0 ]
That is why engineers don't do it that way.
In this particular case, you do it by first solving R1, R2 and Fb (Ra=0)with initial conditions y1=y2=0.at R1 and R2
From static equilibrium you get the values of R1 and R2.
solve for M(x) ,Y'(x),Y(x) referred to the R1 position coordinate.
Now having that solution , you solve the opposite R1R2RA system (Rb=0) and add the two solutions

If you still insist on doing it mathematically, then
-R1x^2/2+C1 is the slope*EI
-R1x^3/6+C1x is the deflection*EI which satisfies y1=0
Now if you go to the second zone , x>30
M=-Fb(x-30)*U(x-30)
slope*EI==Fb(x-30)^2/2+C2
Matching slopes at x=30, etc.

The graphical way by integrating the M/EI diagram twice (area under curve) takes care of initial conditions and in my opinion is far easier to get reliable answers.
Of course you could go to the handbooks and easily get the answers.

Thanks zeke, you're a legend!
I had completely overlooked to match the slopes @30 and @130 to their respective equations which was giving me incorrect values for my coefficients.
Cheers,
pepster