- #1

Tiberious

- 73

- 3

(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