Shear stress in an annular plate

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Hoser415
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Hi Everyone,
I am having a problem finding the shear stresses in an annular plate. The problem is as follows. I have an annular plate with outer radius a, inner radius b. Outer and inner edges are fixed. I also have an annular line load at radius r. I am trying to calculate the shear stresses at a and b. Using Roarks equations to find the shear line forces Qa and Qb I get one of the forces as a positive number and one as a negative number...this makes no sense to me as both forces should be in the same direction as the applied force, at least that's what it seems to me. So am I calculating them correctly and just don't fully understand what is going on or is there some mistake in my math...I've checked the equations about ten times and can't find anything wrong. Any hlep would be great. Thanks.
 
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normal to the plate
 
According to Roark's: Case 1h- Outer edge fixed, inner edge fixed

Shear at the inner radius support:
[tex]Q_{b}=w\frac{C_{2}L_{6}-C_{5}L_{3}}{C_{2}C_{6}-C_{3}C_{5}}[/tex]

Shear at the outer radius support:
[tex]Q_{a}=Q_{b}\frac{b}{a}-\frac{wr_{o}}{a}[/tex]

Where:

[tex]C_{2}=\frac{1}{4}[1-(\frac{b}{a})^2(1+2ln\frac{a}{b})][/tex]

[tex]C_{3}=\frac{b}{4a}[[(\frac{b}{a})^2+1]ln\frac{a}{b}+(\frac{b}{a})^2-1][/tex]

[tex]C_{5}=\frac{1}{2}[1-(\frac{b}{a})^2][/tex]

[tex]L_{3}=\frac{r_{o}}{4a}[[(\frac{r_{o}}{a})^2+1]ln\frac{a}{r_{o}}+(\frac{r_{o}}{a})^2-1][/tex]

[tex]L_{6}=\frac{r_{o}}{4a}[(\frac{r_{o}}{a})^2-1+2ln\frac{a}{r_{o}}][/tex]

Whew! Hope I got all the latex right...

The only way i can see one being opposite in sign to the other is if you accidentally messed up the radius the load is acting at...
 

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That is exactly what I used, However here is my problem. w is a positive number, and for constants come out to a number that is greater than one. Therefore Qb is greater than w, which makes sense as it is on a smaller radius than w. However, this means that Qb*(b/a)-w*(ro/a) makes Qa a negative number at some point when ro approaches b. My question is how can that be...It doesn't seem like there should be shear in the opposite direction of the applied force on the outer edge. If I am wrong could someone please explain this to me, I am pretty sure my constants are right because I've checked them a bunch of times. Just in case anyone wants to check them I have
a=19.75 in
b=6in
ro=10.976
v=.33

I get c2=0.171948169
c3=0.029897023
c5=0.453853549
c6=0.112031464
L3=0.010799721
L6=0.067209158
 
Sorry I mean Qa gets negative at some point as r0 approaches a.

Also my w= 6162 lbs/in

giving me a Qb of ~7213 and a Qa of ~ -1233
 
Your numbers look good, I think the problem is that Qa and Qb are not the reactions, they are the calculated shear force at that location... In a shear force diagram for your system to be in equilibrium it must have a shear force that passes through 0 and into negative values. So the shear force at the inner edge is positive, and the shear force at the outer edge is negative.

However, like you said the reactions will both be in the opposite direction of the load, so just take the absolute value of the calculated values of Qa and Qb to get Ra and Rb.
 
I noticed that deflections could be pretty large depending on the thickness of your plate. Be sure to check that your maximum deflection is less than half of the thickness of the plate or the equtions being used are not valid.
 
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Thanks so much, that makes sense.
 
In order to calculate the maximum deflection I'm assuming you use the Ky values in the table for this particular case and use the formula y=Ky*(w*a^3)/D, but for this case I'm in between the k values given, do you know of another place that has a table of more k values or where I might find a formula to calculate a specific value (that is if they are calculated and not experimentally determined).
 
I would do what Roark's recommends to do for interpolation in Table 23. That is to adjust your problem to match the values listed in the table and then use those values to interpolate in between to get your specific condition. You could simply interpolate between the values given in the table, but they say that that could lead to a lesser degree of accuracy.