Bulkhead under vaccum

[drodrig]

Hi,

I wrote some time ago asking for and end-cap to close a vacuum volume. I learnt a lot from that post and now I am bringing a similar case.

https://www.eng-tips.com/viewthread.cfm?qid=416816

After some calculations I wanted to present the new case here. Checking the results there are some results that make no sense and it is because this formula:

(From the Timoshenko book, Theory of plates and shells)

The value of "k" comes from this table and these cases:
("a" is the biggest and "b" the smallest radius of the bulkhead)

In the formula, the value of "k" grows when the ratio "a/b" grows. Which means that the maximum stress grows too.

Let's suppose two scenarios of the case 10 (plate clamped in the outer perimeter

Scenario A
Outer radius: 1150mm
Inner radius: 680 mm
ratio a/b: 1.70
value k: 0.365 (case 10, interpolating)
area: 2.7e6 mm2

Scenario B
Outer radius: 680mm
Inner radius: 25 mm
ratio a/b: 27.2
value k: 0.75 (case 10, extrapolating, it is asymptotic)
area: 1.4e6 mm2 (almost half)

How is it possible that the scenario B has higher stress (higher "h" value) when the area is almost half?
Because of the longer (clamped) perimeter of the scenario A?

thanks
regards,

 

[desertfox]

Hi drodrig

What is “h” defined as? I can see it in the formula and your post but nowhere in the diagrams.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[HTURKAK]

I am not sure why you decided that the scenario B has higher stress..

My points are

- By definiton, ( h ) is plate thickness ,

- q is uniformly distributed load ( refer to fig.34)

- P is the total load applied to the inner boundary of the plate.

- Suggest you to look also ROARK’s Formulas for Stress and Strain ( chp. 11 Circular plates ).

- If you have a software ,in general FEM is the way to go..







Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[drodrig]

Here is the definition of "h":

I get to the conclusion of the higher stress if we have the same thickness (h) and pressure (P)

The relation a/b (outer/inner diameter of the bulkhead) and not the absolute dimensions drive the equation.

What about the area? should it play a roll?

cheers,

 

[drodrig]

I also checked Roars, but found this nice (easy) equation in the Timoshenko

Next step is running some simulations, but I like first calculating "by hand" to get an idea

thanks

 

[HTURKAK]

The case 10 is circular plate under the uniformly distributed load , clamped at perimeter and with hole at center . In this case first formula SHALL BE USED..

σ max=(k*q*a^2)/h^2

..And lets check Scenario A
Outer radius: 1150mm
Inner radius: 680 mm
ratio a/b: 1.70
value k: 0.365
area: 2.7e6 mm2
assume UDL q=1.0 mPa , h=20 mm

σ max=0.365*1*1150**2/400 =1206 mPa



And check Scenario B
Outer radius: 680mm
Inner radius: 25 mm
ratio a/b: 27.2
value k: 0.75
area: 1.4e6 mm2

assume again UDL q=1.0 mPa , h=20 mm

σ max=0.75*1*680**2/400 =867 mPa

Apparently the scenario B does not have higher stress ...

Probably it could be better if you post more info. if this is a real case..








Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[drodrig]

HTURKAK,

You are absolutely right. I got confused with the capital P (single load) and the pressure.

I'm still puzzled because according to the formulas:

P = q*a^2

shouldn't a Pi be included in the formula?

the integrated load (P) should be equal to the pressure (q) times the area (Pi*a^2)

a is the radius

What about the inner radius? is this included in the K value?

I'm fixing my Excel and then I'll present the case I'm working on

thanks

 

[HTURKAK]

Dear drodrig (Mechanical)(OP),

You are in confusion probably for the reason of Figure 35.

- Refer to TABLE 3 again..

- The cases 2,3,4,5,7 and 10 are for uniformly distributed load and the first formula is applicable .( σ max=(k*q*a^2)/h^2

- The cases 1,6,8 and 9 are for concentrated load along the interior edge and the second formula is applicable .( σ max=(k*P)/h^2

- P is concentrated load and the unit is ( say kN ) . If Q is the uniformly distributed load along the inner edge , P =2ΠQ

- Regarding the Fig 35, Refer to Fig. 34, in which the plate is supported along the outer edge and carries a uniformly distributed load. Consider this plate cut by the cylindrical surface of radius b and perpendicular to the plate, you may find that along this section shearing force acting Q = Πqb^2/2Πb = qb/2 and the moment acting will be Mr = (q/16)* (3+ µ){a^2-b^2).

- The values at TABLE 3 are based on Poisson ratio μ = 0.3

I hope this respond answers your question..










Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB

 

[drodrig]

Again: you are totally right.

I can now see it.

All the numbers are now made

 

[drodrig]

Now I would like to explain the case I am working on.

We have a chamber which has a section (we call it "insert") that goes inwards. Here an sketch:

I want to calculate the wall thicknesses A, B and C, so we divide the problem in 3 sections to support an overpressure of 3 bars (200 N/m2 pressure difference)

- Starting with A.
It is clamped in the outer perimeter (bolted on a flange). Aluminium 5083 (tensile strength 145 MPa)

a, Outer radius: 1152 mm
b, Inner radius: 680 mm
a/b: 1.69
k: 0.365

h, Thickness: (k * q * a^2/sigma)*0.5 = (0.365 * 200,000 N/m2 * 1.152^2 / 145,000,000 N/m2)^0.5 = 25.85mm

So the plate must be at least 25.85 mm. I'll add a safety factor of 3 later.

- For the thickness B. I'll talk later, since there is buckling involved

- For the thickness C, same than A. I assume the outer perimeter clamped (since it will be screwed or welded to the insert-cylinder)
With the new values of k I get a thickness of 21.87 mm

Are these numbers and assumptions right?

I've been playing a bit with xcalcs and I don't get the same numbers

thanks,

 

[EdStainless]

How much thicker will your flanges at the corners of this be?
If they are not a lot more robust then then will flex considerably and you may need to reevaluate your flexure case.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[HTURKAK]

Short answer ;

- PLATE ( C ); Pls look section 16. Uniformly Loaded Circular Plates.

In this case , the edge neither clamped nor simple supported . Take the average of clamped and free case ( relevant formulas ,63 thru 66 and 69,70,71)

- PLATE ( A) ; This is also in-between case and superposition of center loading and uniform loading .

Center loading = ( the load is the reaction of plate C ) Average of case CASE 1 and CASE 6

Uniform loading = Average of case CASE 5 and CASE 7



My opinion..











Not to know is bad;
not to wish to know is worse.

NIGERIAN PROVERB