Fundamentals of Geotechnical Engineering (MindTap Course List)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
Question
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Chapter 3, Problem 3.21CTP

(a)

To determine

Find the number of truckloads required to construct the fill.

(a)

Expert Solution
Check Mark

Answer to Problem 3.21CTP

The number of truckloads required to construct the fill is 10,539trips_.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is wpit=15%.

The unit weight of the undisturbed soil in the borrow pit is (γ)pit=19.1kN/m3.

The specific gravity of the undisturbed soil is Gs=2.70.

Loading truck:

The capacity of the truck is Vtruck=4.80m3.

The net weight of the soil and water when the truck is loaded with is W=72.7kN.

Compacted fill:

The dry unit weight of the compacted fill is (γd)fill=17.3kN/m3.

The finished volume of the compacted fill is Vfill=38,500m3.

The moisture content of the fill is wfill=15%.

Calculation:

Find the weight of the solids (Ws)truck in each truck using the relation:

(Ws)truck=W1+wpit

Substitute 72.7 kN for W and 15% for wpit.

(Ws)truck=72.71+15100=72.71.15=63.2kN

Find the weight of the solids in the compacted fill (Ws)fill using the relation:

(Ws)fill=(γd)fillVfill

Substitute 17.3kN/m3 for (γd)fill and 38,500m3 for Vfill.

(Ws)fill=17.3×38,500=666,050kN

Find the number of truckloads required using the relation:

Numberoftruckloads=(Ws)fill(Ws)truck

Substitute 666,050 kN for (Ws)fill and 63.2 kN for (Ws)truck.

Numberoftruckloads=666,05063.2=10,539trips

Therefore, the number of truckloads required to construct the fill is 10,539trips_.

(b)

To determine

Find the volume of the pit that remains in the borrow area.

(b)

Expert Solution
Check Mark

Answer to Problem 3.21CTP

The volume of the pit that remains in the borrow area is 40,102.5m3_.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is wpit=15%.

The unit weight of the undisturbed soil in the borrow pit is (γ)pit=19.1kN/m3.

The specific gravity of the undisturbed soil is Gs=2.70.

Loading truck:

The capacity of the truck is Vtruck=4.80m3.

The net weight of the soil and water when the truck is loaded with is W=72.7kN.

Compacted fill:

The dry unit weight of the compacted fill is (γd)fill=17.3kN/m3.

The finished volume of the compacted fill is Vfill=38,500m3.

The moisture content of the fill is wfill=15%.

Calculation:

Find the weight of the soil (Wsoil) using the relation:

(Ws)fill=Wsoil1+wpit

Substitute 666,050 kN for (Ws)fill and 15% for wpit.

666,050=Wsoil1+15100Wsoil=666,050(1+0.15)=765,957.5kN

Find the volume of the pit (Vpit) using the relation:

Vpit=Wsoil(γ)pit

Substitute 765,957.5 kN for Wsoil and 19.1kN/m3 for (γ)pit.

Vpit=765,957.519.1=40,102.5m3

Therefore, the volume of the pit that remains in the borrow area is 40,102.5m3_.

(c)

To determine

Find the liters of water will have to be added per truck load.

(c)

Expert Solution
Check Mark

Answer to Problem 3.21CTP

The liters of water will have to be added per truck load is 193.3liters_.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is wpit=15%.

The unit weight of the undisturbed soil in the borrow pit is (γ)pit=19.1kN/m3.

The specific gravity of the undisturbed soil is Gs=2.70.

Loading truck:

The capacity of the truck is Vtruck=4.80m3.

The net weight of the soil and water when the truck is loaded with is W=72.7kN.

Compacted fill:

The dry unit weight of the compacted fill is (γd)fill=17.3kN/m3.

The finished volume of the compacted fill is Vfill=38,500m3.

The moisture content of the fill is wfill=15%.

Calculation:

Neglect the moisture lost by evaporation during excavation, haulage, and handling.

The moisture content increased from 15% to 18%.

Find the weight of the water (Ww)  to be added using the relation:

Ww=(Ws)truck(wfillwpit)

Substitute 63.2 kN for (Ws)truck, 18% for wfill, and 15% for wpit.

Ww=63.2×(1810015100)=1.896kN

Consider the unit weight of water as γw=9.81kN/m3.

Find the volume of water (Vw) to be added using the relation:

Vw=Wwγw

Substitute 1.896 kN for Ww and 9.81kN/m3 for γw.

Vw=1.8969.81=0.1933m3×1,000liter1m3=193.3liters

Therefore, the liters of water will have to be added per truck load is 193.3liters_.

(d)

To determine

Find the water content at saturation.

(d)

Expert Solution
Check Mark

Answer to Problem 3.21CTP

The water content at saturation is 19.7%_.

Explanation of Solution

Given information:

Borrow pit:

The water content of the undisturbed soil is wpit=15%.

The unit weight of the undisturbed soil in the borrow pit is (γ)pit=19.1kN/m3.

The specific gravity of the undisturbed soil is Gs=2.70.

Loading truck:

The capacity of the truck is Vtruck=4.80m3.

The net weight of the soil and water when the truck is loaded with is W=72.7kN.

Compacted fill:

The dry unit weight of the compacted fill is (γd)fill=17.3kN/m3.

The finished volume of the compacted fill is Vfill=38,500m3.

The moisture content of the fill is wfill=15%.

Calculation:

Find the void ratio (efill) in the compacted fill using the relation:

(γd)fill=Gsγw1+efill

Substitute 17.3kN/m3 for (γd)fill, 2.70 for Gs, and 9.81kN/m3 for γw.

17.3=2.70×9.811+efill1+efill=26.48717.3efill=0.531

Find the water content (w) using the relation:

Sefill=wGs

Substitute 1 for S, 0.531 for efill, and 2.70 for Gs.

1×0.531=w×2.70w=0.197=19.7%

Therefore, the water content at saturation is 19.7%_.

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