+0  
 
0
949
7
avatar+1832 

Any help with these two queistions? 

 

 

 

 

 Mar 13, 2016

Best Answer 

 #1
avatar+1037 
+15

Solution to #10


\(\text {Calculate the compression load in terms of the cross-sectional area.}\\ \displaystyle {3^2*\pi = 28.27 {in}^2} \Rightarrow \dfrac{150 {~kilopounds}}{28.27 {in}^2} = 5.305 {~ksi}\\\ \text {Calculate the longitudinal compression }\\ \dfrac {5.305}{8000}*12 in = 0.0079575{in}\\ \)

 

\(\text {Calculate the expansion in the radial direction.}\\ \dfrac{5.305}{8000}(0.35)*6{in} = 0.00139256 {in} \)

 

\(\text {Final length } (12 - 0.00796) = 11.99204in\\ \text {Final diameter } ( 6.0 + 6.00139) = 6.00139in\)

 

 

 

Source:

Formulae Esoterica

Processus algorythmic

Twelfth edition

Morgan le Fay

 Mar 13, 2016
 #1
avatar+1037 
+15
Best Answer

Solution to #10


\(\text {Calculate the compression load in terms of the cross-sectional area.}\\ \displaystyle {3^2*\pi = 28.27 {in}^2} \Rightarrow \dfrac{150 {~kilopounds}}{28.27 {in}^2} = 5.305 {~ksi}\\\ \text {Calculate the longitudinal compression }\\ \dfrac {5.305}{8000}*12 in = 0.0079575{in}\\ \)

 

\(\text {Calculate the expansion in the radial direction.}\\ \dfrac{5.305}{8000}(0.35)*6{in} = 0.00139256 {in} \)

 

\(\text {Final length } (12 - 0.00796) = 11.99204in\\ \text {Final diameter } ( 6.0 + 6.00139) = 6.00139in\)

 

 

 

Source:

Formulae Esoterica

Processus algorythmic

Twelfth edition

Morgan le Fay

Nauseated Mar 13, 2016
 #2
avatar+1832 
0

Immpressive Nauseated 

I totally get it 

 

Any idea about Q #11 ? 

 Mar 13, 2016
 #3
avatar+1832 
0

Any help about question 11 ? 

 Mar 13, 2016
 #4
avatar+1037 
+10

Solution to #11

 

To calculate the normalized strain for isotropic, homogenized compounds, exhibiting linear elastic properties, in cubical form, where the stresses are orthogonal in three dimensions, use the generalized Hooke’s law formula

 

\(\varepsilon_x=\dfrac{\sigma_x-\nu(\sigma_y-\sigma_z)}{E}\\\\ \dfrac{-15 -0.49(-15-15)}{1000} = -0.0003 in/in\)

 

\(\varepsilon_x=\varepsilon_y=\varepsilon_z=-0.0003 in/in\\ \Delta_x =\Delta_y=\Delta_z =(-0.0003in/in) *4in =-0.0012in \)

 

\({Final~ side~ length~ } 4-0.0012= 3.9988\\ {Final~ volume =} 3.9988^3\\ \Delta Volume ~4^3-3.9988^3=0.057583 \)

 

 

 

 

Source:

Formulae Esoterica

Processus algorythmic

Twelfth edition -Volume II

Morgan le Fay

Translated from ancient Babylonian

 Mar 14, 2016
 #5
avatar+121064 
+5

Nice work, Nauseated......I looked at this one...but...I found myself, here, only after a short few minutes :

 

 

I also asked for some help from that damned Sisyphus.....

 

But.....these days....it's useless......he has "zero" on his mind and is a complete rockhead......

 

 

 

cool cool cool

 Mar 14, 2016
 #6
avatar+1037 
+5

While translating the ancient Babylonian, I was in a “causality loop.” This isn’t as bad as an infinite loop, but you could be in it for eons, and never know until you escaped.

 

Sisyphus is definitely fixated on his bolder -- well more than LancelotLink is on his banana daiquiris.  I have wondered if his obsession would change if he knew the contents. The roman Zero is valuable –Archimedes probably though it one of the “holy grails” in his world. 

 

You description of Sisyphus is very close to some of the engineering techs I work with: useless rock-heads, with zero on their minds –but very focused on it. 

 Mar 14, 2016
 #7
avatar+1832 
0

Amazing Nauseated I got it.

Thank you. 

 Apr 9, 2016

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