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Alex777 [14]
4 years ago
14

A soil had a liquid limit of 44, a plastic limit of 21, and a shrinkage limit of 14. In the summer, the in situ water content wa

s 18%. (a) Compute the plasticity index of the soil. (b) Calculate the liquidity index of the in situ soil. (c) In which consistency state was the in situ soil
Engineering
1 answer:
Marysya12 [62]4 years ago
4 0

Answer:

(a) 23

(b)-0.13

(c) In which consistency state was the in situ soil? Plastic limit = 1.1

Explanation:

(a) Compute the plasticity index of the soil.

The formula for plasticity index is given as

Plasticity Index = Liquid limit - Plastic limit

Liquid limit = 44

Plastic limit = 21

Plastics Index = 44 - 21

= 23

(b) Calculate the liquidity index of the in situ soil.

Liquidity Index = Water content - Plastic limit/Liquid limit - Plastic Limit

Water content = 18%

Plastic limit = 21

Liquid limit = 44

Liquidity Index =18 -21/44 - 21

- 3/23

= -0.1304347826

≈ -0.13

(c) In which consistency state was the in situ soil

The consistency state of a soil tells us how firm the soil is.

Consistency state/index = Liquid limit - Water content/ Plastic Index

= Liquid limit - Water content /Liquid limit - Plastic Limit

Water content of the in situ soil = 18%

=44 - 18/44 - 23

= 26/23

= 1.1304347826

= 1.1

Since the consistency index of this situ soil is above 1, it means the soil is at plastic limit.

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Please answer fast. With full step by step solution.​
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First, carry out the division:

<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)

Observe that

2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)

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(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)

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8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)

so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.

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<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)

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<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))

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\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n

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\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}

so that by substitution, we can write

\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}

Now condense <em>f(z)</em> into one series:

\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}

\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}

\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}

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A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

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length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

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<u>Part (A)</u> The stress on the steel wire;

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σ = 0.00063

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E = δ/σ

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E = 30,000,000 Psi

4 0
3 years ago
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