Geo - Embankment Problem

[Road Guy]

An Earth Dam Fill is to be built using a fine silty clay @ 95% Standard Procotr.

The soil to be used is from a borrow pit where the soil is at its optimum moisture contentwith a moist unit weight = 112.1 #/CF.As the soil is hauled in 15 CY trucks is has an e=1.47

Height of Fill 20'

Crest Width of Fill 20'

Length of dam 450'

Upstream Slope 2:1

Downstream Slope 3:1

Soil Max dry density 106 #/CF

Soil optimum moisture w=18%

Specific Gravity of solids 2.68

1. The Volume of the dam per ft-length is?

A. 1200

B. 1400

C. 1600

D. 3000

E. 4000

2. What is the volume required to build a 10' section of the dam?

A. 2080

B. 9410

C. 14800

D. 20800

E. 34700

3. The total volume (CY) of borrow hauled to build the dam

A. 15700

B. 17800

C. 20800

D. 23800

E. 34700

4. The # of truckloads needed to construct the dam is.

A. 1050

B. 1110

C. 1560

D. 2180

E. 2310

5. If the specification for fill placement is changed to 100% of the STD Proctor

how much would the # of trucks needed change by?

A. + 10%

B. +5%

C.no change

D. - 5%

E. - 10%

6. Disregard the changes in question #5, But assume the moist unit weight of the

borrow soil remains at 112.1 #/CF. If the moisture content of the in place sooil is found to be 23%

the required volume to construct the dam would change by?

A. + 5.5%

B. + 4.5%

C.no change

D. - 4.5%

E. - 5.5%

 

[3gorgesdam]

Here is my answer,

1) 1400 cf per ft length

2) Assume the question is asking the volume of borrowed soil...

Here is my thought:

Assume 1 ct of filled, the dry density of the filled rdry=rmax*95%=106*.95=100.7

Ms=rdry*Vt=100.7*1=100.7

Msfill = Msborrow

rdryborrow = rwetborrow/(1+w)=112.1/(1+.18)=95

Vtborrow=Msborrow/rdryborrow=100.7/95=1.06

The total Volumn of 10' section =1400*10=14000

The total borrowed soil =14000*1.06=14840

3) 24733 CY

4)1648 Truck

Of course the rest is just repeating the above procedure.

 

[Road Guy]

1. 1400 CF

2. 14840 CF

3. 34,700 CY

4. 2310 trucks

5. 5% Increase

6. 4.5% increase

_________________________________

3.

Density (Dry:Truck)= Gs*62.4 / (1 + e) = 2.68*62.4/2.47=67.7 PCF

V(hauled) = ((1400*450)/27))*(100.7/67.7) = 34,700 CY

4. # trucks = 34,700 / 15 = 2310 trucks

5. truckloads = (1400)(450)(106pcf) / ((27)(15)(67.7)) = 2435

% change = (2435-2310) / (2310) = 5.4%

 

[3gorgesdam]

Thanks RoadGuy, I found my errors. Didn't even use e=1.47 to find a new r dry.

 

[Road Guy]

FYI, this is from the OLD NCEES sample exam(before they switched the test format tomultiple choice) its the only question I have from it, would love to get my hands on the full copy.