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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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\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}

\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}

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