Calculate the bulk modulus and Derive the mean standard deviation

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The discussion focuses on calculating the bulk modulus and shear modulus of a material with a Young's modulus of 250 GN/m² and a Poisson's ratio of 0.32, yielding a bulk modulus of approximately 231.48 GPa and a shear modulus of about 94.7 GPa. Additionally, the mean standard deviation of ultimate tensile strength results from 10 samples is calculated using the standard deviation formula, resulting in approximately 29.90 N/mm², which converts to 0.0299 GPa. Participants clarify unit conversions and the relationship between Newtons per square millimeter and megapascals, emphasizing the importance of understanding SI prefixes. The discussion highlights the need for accuracy in calculations and unit conversions to avoid discrepancies in results.
Tiberious
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(apologies for re-posting, I am unable to find my original thread.) A material which has a Young's modulus of elasticity of 250 GN m-2 and a poisons ratio of 0.32, calculate:

(a) the bulk modulus of the material 


Relevant equation,
K=E/3(1-2v)
Inputting our known values,
K=250/3(1-2(0.32))
Answer:
= 231.48 Gpa
(b) the shear modulus of the material. 

Relevant equation,
G=E/2(1-v)
Inputting our known values,
G=250/2(1+(0.32))
Answer:
= 94.6^.9^. Gpa ≈94.7 GpaThe ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows. (a) Determine the mean standard deviation of these results.
(b) Express the values found in (a) in GPa.

As we are referring to a 'sample' we apply the below formula for Standard Deviation:

s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )

Calculating the mean value x ̅

(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10

x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2

(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 645.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56

Determining the sum of the above

∑▒237.16+31.36 +1062.76 +3180.95 +645.16 +1489.96 +275.56 +817.96 +129.96 +179.56

= 8,050.39 〖N mm〗^(-2)

Divide by N-1

(1/9)∙8050.39= 894.487 N 〖mm〗^(-2) ~ 894.5 N 〖mm〗^(-2)Determining the sample standard deviation σ

σ= √((894.487))=29.90 N 〖mm〗^(-2)=0.0299 MPa

0.0299 MPa=2.99∙〖10〗^(-5) GPa
 
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I get a Newton per square mm equal to 1 megaPascal or 0.001 GPa
 
Which part do you attain the 0.001 Gap in ? Also, would you be able to point out in which part of the above I have gone wrong. Evidentally there must be a fatal flaw as the answers are so different to what you have calculated.
 
I was just stating the equivalent units. there are 1000 mm in a meter, so 1 million square mm in a square meter. If there is 1 Newton per mm2, then there is 1 million Newton's per m2, which is 1 million Pascals, or 0.001 GigaPascals. For example 29 N(mm)^-2 is equal to 29 MPa. You may want to reference SI prefixes .
Here is one link that I like. https://www. unc. edu/~rowlett/units/prefixes.html
(revised link here - http://www.ibiblio.org/units/prefixes.html )
 
Last edited:
So, this should be correct.

Determining the sample standard deviation σ

σ= √((894.487))=29.90 N 〖mm〗^(-2)=29.90 MPa

Express the values found in (a) in GPa.

29.9 MPa=0.0299 GPa
 

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