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DIA [1.3K]
3 years ago
12

A sample of soil has a volume of 0.45 ft^3 and a weight of 53.3 lb. After being dried inan oven, it has a weight of 45.1 lb. It

has a specific gravity of solids of 2.70. Compute its moisture content and degree of saturation before it was placed in the oven.
Engineering
1 answer:
topjm [15]3 years ago
7 0

Answer:

a) the moisture content before it was placed in the oven is 18.18%

b) degree of saturation for soil is 72.19%

Explanation:

Given the data in the question;

Moisture Content = [(Weight of soil before dry - dry weight) / dry weight] × 100

so we substitute

Moisture content = [(53.3 - 45.1) / 45.1 ] × 100

= (8.2/45.1) × 100

= 18.18%

Therefore the moisture content before it was placed in the oven is 18.18%

Dry Unit Weight = dry weight / volume

Dry Unit Weight = 45.1 lb / 0.45 ft³

Dry Unit Weight = 100.22 lb/ft³

we know that;

dry unit weight = (Specific gravity × unit weight of water) / (1 + e)

we also know that; unit weight of water is 62.43 lbf/ft³

so we substitute

e = (2.70×62.43 / 100.22) - 1

e = 1.68 - 1

e = 0.68

so void ratio e = 0.68

Now we determine the degree of saturation using the equation;

degree of saturation = (Moisture content × specific gravity) / void ratio

we substitute

degree of saturation = ( 18.18% × 2.7) / 0.68

= 0.49086 / 0.68

= 0.7219 ≈ 72.19%

Therefore degree of saturation for soil is 72.19%

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Answer:

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Explanation:

(a)

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Now substituting E for 10\times 10^{6} psi then

\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi

(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

But the outer diameter is given as 2 in hence

(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}

d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in

As already mentioned, 2t=d_o - d_i hence t=0.5(d_o - d_i)

t=0.5(2-1.785)=0.1075 in

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(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor
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Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

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stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

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Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

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Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

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