Parallel RBD's (Reliabilty Block Diagram) help - Reliability & MTBF

[Estoril-5]

Hello guys,

im new to this forum, i need help.

a quick intro, i am from the UK, i am a mechanical engineer by degree and work in Through Life Management for an Engineering Co.

I am currently working on RBD's for reliability calcs etc etc.

Series calcs are easy to work out

Parrallel calcs are tough, i have the MTBF for both items in the parrallel RBD but dont know how to get the overall MTBF for both.

Can anyone help?

[Al Rosen]

Originally Posted by Estoril-5

Hello guys,

im new to this forum, i need help.

a quick intro, i am from the UK, i am a mechanical engineer by degree and work in Through Life Management for an Engineering Co.

I am currently working on RBD's for reliability calcs etc etc.

Series calcs are easy to work out

Parrallel calcs are tough, i have the MTBF for both items in the parrallel RBD but dont know how to get the overall MTBF for both.

Can anyone help?

Can you define RBD for some of us who are not familiar with the term?

[Tim Folkerts]

I believe he is talking about "Reliabilty Block Diagrams" which are used to symbolize combinations of part/systems for reliabilty calculations.

For two items in series, then the system still works if both are working. If R1 = 0.95 is the reilabilty of the first item and R2 = 0.75 is the reliabilty of the second, the reliabilty of the series combination is
R = R1*R2 = 0.7125
I.e. it is less reliable than either part separately.

If the two are in parallel, then the system still works if either are working. The equation is
R = 1- (1-R1)*(1-R2) = 0.9875
I.e. it is more reliable than either part separately.


Tim F

[Al Rosen]

Originally Posted by Tim Folkerts

I believe he is talking about "Reliabilty Block Diagrams" which are used to symbolize combinations of part/systems for reliabilty calculations.

For two items in series, then the system still works if both are working. If R1 = 0.95 is the reilabilty of the first item and R2 = 0.75 is the reliabilty of the second, the reliabilty of the series combination is
R = R1*R2 = 0.7125
I.e. it is less reliable than either part separately.

If the two are in parallel, then the system still works if either are working. The equation is
R = 1- (1-R1)*(1-R2) = 0.9875
I.e. it is more reliable than either part separately.


Tim F

Thanks Tim.

[Michael Walmsley]

Hope this helps.

Some of the math symbols may come out funky. Am using MathType.

[Estoril-5]

Excellent Guys!

exacyly what i wanted to hear

Now - how do i convert the reliability into an MTBF (mean time between failure?)

OR can i calculate the average MTBF of the parrallel RBD by using their respective MTBF's?

Cheers

[Estoril-5]

Any idea on how to turn reliability figures into MTBF's?

Cheers

[Michael Walmsley]

Rsys = 1-(1-R1)*(1-R2)

Let R1=.9
R2=.8
assuming a simple paralell system with (2) blocks of non-identical reliability:
let R1 = .9 , R2 = .8 , then
Rsys = 1-(1-.9)*(1-.8)
= 1-[(.1)*(.2)] =.98

now suppose
t=1000 hours

.98 = e^(-t/@)
ln (.98) = - (t/@)

-0.020202707 = -(1000/@)

@ = 1000/.0202
@= 49504.9505 hrs