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DaniilM [7]
2 years ago
5

An earth fill, when compacted will occupy a net volume of 187,000 cy. The borrow material that will be used to construct this fi

ll is a stiff clay. In this “bank” condition, the borrow material has a wet unit weight of 129 lb per cubic foot (cf), a water content (w %) of 16.5%, and an in-place void ratio (e) of 0.620. The fill will be constructed in layers of 6 inch depth, loose measure, and compacted to a dry unit weight (Yd ) of 114 lb per cf at a moisture content of 18.3%...
Engineering
1 answer:
salantis [7]2 years ago
5 0

The <em>required</em> volume of borrow pit excavation was gotten as;

<em><u>V_net,borr ≈ 192,054 cy</u></em>

The missing part of the question is;

Compute the required volume of borrow pit excavation.

  • We are given;

Net volume of earth fill; V_net,fill = 187,000 cy

<em>Wet unit weight</em> of borrow material; γ_wet = 129 lb/ft³

Water content; m = 16.5% = 0.165

<em>Dry unit weight</em> of fill yard; γ_d,fill = 114 lb/cf

  • Now, to get the dry unit weight of the borrowed material, we will use the formula;

γ_d = γ_wet/(1 + m)

Plugging in the relevant values, we have;

γ_d,borr = 129/(1 + 0.65)

γ_d,borr = 111 lb/cf

     Since the net volume is in cy units, then let us convert the dry volumes to lb/cy.

Thus, from conversion tables, we have;

γ_d,fill = 114 lb/cf × 27 cf/cy

γ_d,fill = 3078 lb/cy

Similarly;

γ_d,borr = 111 lb/cf × 27 cf/cy

γ_d,borr = 2997 lb/cy

Thus, to find the volume of the borrow pit excavation, we will use the formula;

V_net,fill × γ_d,fill = V_net,borr × γ_d,borr

Plugging in the relevant values gives us;

187,000cy × 3078 lb/cy = V_net,borr × 2997 lb/cy

Thus;

V_net,borr = (187000 × 3078)/2997

<em>V_net,borr ≈ 192,054 cy</em>

Read more at; brainly.com/question/15090365

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3.115× 10^{-3} meter

Explanation:

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A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
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Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

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Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

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Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
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Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

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Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

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