Question

I need the formula four component reliability when strength is exponentially distributed and stress is normally distributed. please drive this formula and show the steps so I may apply it.

77% ■ 1 1 :13 rent -Area@ど(e -C.1%).jo".忘jo) et(iseise)t

here are the formulas for exponential strength denoted by G and normal stress denoted by s.

what is the formula for component reliability for when the strength function is exponentially distributed and the stress function has a normal distribution. the answer should be equal to Rc=.6096( this is what our professor gave us to check our work)

Answer 1

2 2 Fs

$f_{g}(s)=\lambda_{s} e^{-\lambda_{s} s}$

we keep it is to derive the equation

$R=\int_{0}^{\infty} F_{S}\left[\int_{0}^{S} f_{g}(s) d s\right] d s$

it yields

2 exp 1-2 R=

2 exp

  $\frac{1}{^{\sigma} s \sqrt{2 \pi}}\int_{0}^{\infty} \exp \left[-\frac{1}{2}\left[\frac{s-N_{\mathrm{S}}}{\sigma_{\mathrm{s}}}\right]^{2}\right] \cdot e^{-\lambda \mathbf{S}} \mathrm{d} \mathbf{s}$

$=1-\phi\left[-\frac{N_{\mathrm{S}}}{\sigma_{\mathrm{S}}}\right]-\frac{1}{\sigma_{\mathrm{S}} \sqrt{2 \pi}} \mathrm{T}$

here

$\mathrm{T}=\int_{0}^{\infty} \exp \left[-\frac{1}{2 \sigma_{\mathrm{S}}^{2}}\left[\left(\mathrm{s}-N_{\mathrm{S}}+\lambda \sigma_{\mathrm{S}}^{2}\right)^{2}+2 N_{\mathrm{S}} \sigma_{\mathrm{S}}^{2} \lambda-\lambda^{2} \sigma_{\mathrm{S}}^{4}\right]\right] \mathrm{ds}$

For convenience, we let

$t=\left(s-N_{S}+\lambda \sigma_{S}^{2}\right) / \sigma_{S}$

then ơSdt ds :

The reliability assumes the form

$R=1-\Phi\left[-\frac{N_{S}}{\sigma_{S}}\right]-\frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \exp \left[-\frac{t^{2}}{2}\right]\cdot \exp \left[-\frac{1}{2}\left(2 N_{\mathrm{s}} \lambda-\lambda^{2} \sigma_{\mathrm{S}}^{2}\right)\right] \mathrm{d} \mathrm{t}$

$where\: \: \: \: \: x=\frac{N_{s}-\lambda \sigma_{s}^{2}}{\sigma_{s}}$

$R=1-\Phi\left[-\frac{N_{S}}{\sigma_{S}}\right]-\exp \left[-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s\right.\right.--\lambda_{s}^{2} \sigma_{S}^{2} ) ]\left[1-\Phi\left[-\frac{^{N} s^{-\lambda} s^{\sigma_{S}^{2}}}{\sigma_{S}}\right]\right]$

putting values in equation

$R=1-0-\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.--\lambda_{s}^{2} \sigma_{S}^{2} ) ]\left[1-0\right]$

$R=1-\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.-\lambda_{s}^{2} \sigma_{S}^{2} ) ]$

$R_{c}=1-R$

$R_{c}=\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.-\lambda_{s}^{2} \sigma_{S}^{2} ) ]$

putting values

R,, 0.60957

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Answer 2

2 2 Fs

$f_{g}(s)=\lambda_{s} e^{-\lambda_{s} s}$

we keep it is to derive the equation

$R=\int_{0}^{\infty} F_{S}\left[\int_{0}^{S} f_{g}(s) d s\right] d s$

it yields

2 exp 1-2 R=

2 exp

  

2 2

$=1-\phi\left[-\frac{N_{\mathrm{S}}}{\sigma_{\mathrm{S}}}\right]-\frac{1}{\sigma_{\mathrm{S}} \sqrt{2 \pi}} \mathrm{T}$

here

$\mathrm{T}=\int_{0}^{\infty} \exp \left[-\frac{1}{2 \sigma_{\mathrm{S}}^{2}}\left[\left(\mathrm{s}-N_{\mathrm{S}}+\lambda \sigma_{\mathrm{S}}^{2}\right)^{2}+2 N_{\mathrm{S}} \sigma_{\mathrm{S}}^{2} \lambda-\lambda^{2} \sigma_{\mathrm{S}}^{4}\right]\right] \mathrm{ds}$

For convenience, we let

$t=\left(s-N_{S}+\lambda \sigma_{S}^{2}\right) / \sigma_{S}$

then ơSdt ds :

The reliability assumes the form

$R=1-\Phi\left[-\frac{N_{S}}{\sigma_{S}}\right]-\frac{1}{\sqrt{2 \pi}} \int_{x}^{\infty} \exp \left[-\frac{t^{2}}{2}\right]\cdot \exp \left[-\frac{1}{2}\left(2 N_{\mathrm{s}} \lambda-\lambda^{2} \sigma_{\mathrm{S}}^{2}\right)\right] \mathrm{d} \mathrm{t}$

$where\: \: \: \: \: x=\frac{N_{s}-\lambda \sigma_{s}^{2}}{\sigma_{s}}$

$R=1-\Phi\left[-\frac{N_{S}}{\sigma_{S}}\right]-\exp \left[-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s\right.\right.--\lambda_{s}^{2} \sigma_{S}^{2} ) ]\left[1-\Phi\left[-\frac{^{N} s^{-\lambda} s^{\sigma_{S}^{2}}}{\sigma_{S}}\right]\right]$

putting values in equation

$R=1-0-\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.--\lambda_{s}^{2} \sigma_{S}^{2} ) ]\left[1-0\right]$

$R=1-\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.-\lambda_{s}^{2} \sigma_{S}^{2} ) ]$

$R_{c}=1-R$

$R_{c}=\exp [-\frac{1}{2}\left(2 N_{\mathrm{s}}{\lambda} s.\right.-\lambda_{s}^{2} \sigma_{S}^{2} ) ]$

putting values

R,, 0.60957

IF YOU LIKED THE ANSWER PLEASE PROVIDE POSITIVE FEEDBACK