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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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g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0
Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵  * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α  * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

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Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0
2 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
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Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

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d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

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v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

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