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Wewaii [24]
2 years ago
12

The in situ moist unit weight of a soil is 17.3 kN/m3 and the moisture content is 16%. The specific gravity of soil solids is 2.

72. This soil is to be excavated and transported to a construction site for use in a compacted fill. If the specification calls for the soil to be compacted to a minimum dry unit weight of 18.1 kN/m3 at the same moisture content of 16%, how many cubic meters of soil from the excavation site are needed to produce 2000 m3 of compacted fill?
Engineering
1 answer:
Temka [501]2 years ago
3 0

Answer:

Explanation:

Given that,

Moist content w = 16%

The in situ moist unit weight of the soil : γ(in situ) = 17.3 kN/m³

Specific gravity of the soil

G(s) = 2.72

Minimum dry unit weight of the soil

γd(compacted) = 18.1 kN/m³

Moist content is same as above

w = 16%

Question

how many cubic meters of soil from the excavation site are needed to produce 2000 m³ of compacted fill?

Let determine the in situ dry unit weight γd(in-situ) using the relation

γd(in-situ) = γ(in-situ) / [1 + (w/100)]

γd(in-situ) = 17.3/ [1 + (18/100)]

γd(in-situ) = 17.3 / ( 1 + 0.18)

γd(in-situ) = 17.3 / 1.18

γd(in-situ) = 14.66 kN/m³

To determine the Volume of the soil to be excavated (Vex)

Let the Volume to be excavated = V

We can use the relation

V=V(fill) × γd(compacted) / γd(in situ)

Given that, V(fill) = 2000m³

V(fill) is the volume of the compacted fill

Therefore,

V=V(fill) × γd(compacted) / γd(in situ)

Vex = 2000 × 18.1 / 14.66

Vex = 2469.13 m³

So, the excavated volume of the soil is 2469.13 m³

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Answer:

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Now, from the conversation of mass,

A_2*V_2/u_2 = A_1*V_1/u_1

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4 0
2 years ago
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A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch
Sedbober [7]

Answer:

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Explanation:

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2 years ago
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Answer:

The change in length of the circular strut DC = 0.0028 in.

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We have the following parameters or information in the question given above:

=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.

The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC  is given in  the attached picture below.

According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.

Where FCD = p × [4 + 5] ÷ [sin Ф × 4].

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Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.

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The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.

We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.

(CV = vertical deflection of CD).

The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.  

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8 0
2 years ago
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