1. Transposition...

Hi..

I have the following problem that i cannot seem to solve..I have to rearrange & solve for "S" in the following equation...

1*10^6=2{((1.8/S)^2*1/pi)^-0.256 - (20*10^-6)}/2*10^-6 * S^2.513 * pi^1.256 * ( -0.513)

The answer for S is 2.13.. This answer is no good to me alone, i must see how this is done

Regards....

2. Originally Posted by Bailey
Hi..

I have the following problem that i cannot seem to solve..I have to rearrange & solve for "S" in the following equation...

1*10^6=2{((1.8/S)^2*1/pi)^-0.256 - (20*10^-6)}/2*10^-6 * S^2.513 * pi^1.256 * ( -0.513)

The answer for S is 2.13.. This answer is no good to me alone, i must see how this is done

Regards....
Really unclear what that formula is. Gonna do this one step at a time so I don't get lost. Learn to use Latex and post up the real formula if this is wrong.

$\displaystyle 10^6 = \frac{2((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256} \cdot (-0.513)}$ ?

=> $\displaystyle 5 \cdot 10^5 = \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256} \cdot (-0.513)}$

=> $\displaystyle -2.565 \cdot 10^5 = \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513} \cdot \pi^{1.256}}$

=> $\displaystyle -2.565 \cdot 10^5 \cdot \pi^{1.256}= \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{2 \cdot 10^{-6} \cdot S^{2.513}}$

=> $\displaystyle -2.565 \cdot 10^5 \cdot \pi^{1.256} \cdot 2 \cdot 10^{-6}= \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{((\frac{1.8}{S})^2 \cdot \frac{1}{\pi})^{0.256} - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24}{\pi S^2})^{0.256} - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24^{0.256}}{(\pi S^2)^{0.256}}) - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24^{0.256}}{\pi^{0.256} S^{0.512}}) - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24^{0.256}}{\pi^{0.256} S^{0.512}}) - (20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24^{0.256}}{\pi^{0.256} S^{0.512}})}{S^{2.513}} - \frac{(20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{(\frac{3.24^{0.256}}{\pi^{0.256} S^{0.512}})}{S^{2.513}} - \frac{(20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle -0.513 \cdot \pi^{1.256} = \frac{3.24^{0.256}}{\pi^{0.256} S^{3.025}} - \frac{(20 \cdot 10^{-6})}{S^{2.513}}$

=> $\displaystyle S^{3.025} \cdot -0.513 \cdot \pi^{1.256} = \frac{3.24^{0.256}}{\pi^{0.256}} - S^{0.512} (20 \cdot 10^{-6})$

Now it's time to use some approximations...

=> $\displaystyle -2.16041709548325134 \cdot S^{3.025} = 1.00792717675284593 - 0.00002 \cdot S^{0.512}$

=> $\displaystyle -2.16041709548325134 \cdot S^{3.025} + 0.00002 \cdot S^{0.512} - 1.00792717675284593 = 0$

To give a rough idea of the answer...

$\displaystyle 2.16 S^3 = -1$

=> $\displaystyle S^3 = \frac{-1}{2.16}$.

=> $\displaystyle S = -0.7735981390$...

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3. Hi

I have attached a word doc containing the formula.. The formula that i am having trouble rearranging is used in fracture mechanics. S being sigma which is stress.
I have an answer of S = 2.13...

Thanks for the help....

Bailey..

4. $\displaystyle 10^6 = \frac{2((\frac{1.8}{\sigma})^2 \cdot \frac{1}{\pi})^{-0.256} - (20 \cdot 10^{-6})^{-0.256}}{2 \cdot 10^{-6} \cdot \sigma^{2.513} \cdot \pi^{1.256} \cdot (-0.513)}$

=> $\displaystyle 10^6 \cdot 2 \cdot 10^{-6} \cdot \pi^{1.256} \cdot (-0.513) = \frac{2((\frac{1.8}{\sigma})^2 \cdot \frac{1}{\pi})^{-0.256} - (20 \cdot 10^{-6})^{-0.256}}{\sigma^{2.513}}$

=> $\displaystyle -0.513 \pi^{1.256}= \frac{((\frac{1.8}{\sigma})^2 \cdot \frac{1}{\pi})^{-0.256} - (20 \cdot 10^{-6})^{-0.256}}{\sigma^{2.513}}$

=> $\displaystyle -0.513 \pi^{1.256}= \frac{1.8^{-0.512}}{\sigma^{-0.512} \pi^{-0.256}} - 0.00002^{-0.256}$

Taking approximations...

$\displaystyle -2.160417096\sigma^{2.513} - 0.9921351694 \sigma^{0.512} + 15.95645167 = 0$

=>$\displaystyle 2.160417096\sigma^{2.513} + 0.9921351694 \sigma^{0.512} -15.95645167 = 0$

Solve for $\displaystyle \sigma$ using Maple...

$\displaystyle \sigma = 2.132815069$ to 9 decimal places...

5. Thanks for the help Deadstar,, you're a good'un !!

Really appreciate it...