Question

ETCM 4330 To develop the Blue Creek mine site, a wash pit area needs to be developed on the east side of the overburden. The area is shown on th Assignment #2 g area has been leased on the south side of the site to stock pile e map. A road must be built to transport materials to the temporary working area using materials from the pit excavation. be 20' wide and 4000' long. The existing clayey-silt topsoil will be compacted in a 6" lift requiring 3 passes of avibratorytampingfootcompactar lob efficency is anticipated to be 75% from the gem bearing strata will be deposited on top of the compacted topsoil to a depth of 18" in 6" lifts. Each lift requires 2 passes of the vibratory tamping foot compactor. Job efficiency is anticipated to be 75% Pit excavation will be done with 2 hydraulic excavators having 4 CY buckets, 2 articulating 20 ton dump trucks, and one crawler tractor dozer with a 10 wide, 4' high universal blade What is the expected production for compacting the topsoil base? What is the expected production for compacting the gravel topping? How many BCY of Gravel will be required to construct the road? (assume only 5' depth of gravel to be removed) How many BCY of topsoil will be removed? 1. 2. 3. 4. 4000 N 3800 00, E2000加… 198.001/ 3000 N 2 N 19 19 2000 10.001- N 1000.00, E0.00 1000€- Temporary Work Area NI0.08, E800.00 100.00] (74,001 5000 6000 74.00) 8 srsa 1000 2000 3000 4000 Note: Elevations in brackets [xxx.xx]"

Please help, show work.

Answer 1

Let the point with co-ordinates are designed as follows Co-ordinates 0.0 1950.00 2900.00 3800.00 1000.00 800.00 5800.00 5800.00 2000 Elevation (ft) 100 95 95 98 100 The distance between two point is given by To find thelength AB 800 0 x -5800 y,-1950 AB 5800-800) +(1950-0) AB 5366.79ft

Similarly point BC-950ft CD-3905ft DE 3440.93ft EA 1280.62 ft (1) Surface area of the polygon is calculated by dividing the area into triangle from a common point Area of polygon ABCDE-area of triangle AABCA ACD AADE

Δ ABC For We have AB- 5366ft AB -5366.79ft BC-950ft BC-950ft AC 5780ft AC-5780.14ft So we have to calculate S AB +BC+AC 5366.79+950+5780.14 S- S-6048.465 So we can get area bv following formula B) (S-B-AC) .4 16048.465(6048.465-5366.79) (6048465-950) (6048.465-5780.14) A 2375000ft

Similarly remaining two triangle will be calculated ACD 7760000ft ADE-2120000f2 ABCDE 2375000+7760000+212000 ARCD 12255000ft Total area of polygon (2) An overburden of S' common earth must be removed volume of overburden be removed-surface area x height(depth) of overburden #12255000 x5-61 275000f' 2269444yd(lt0037yd) Hence total overburden to be removed -2269444BCY Depth of gold stata+83.00-74.00 9ft total gold strata to be removed- surface area x depth of gold strata V 12255000x9 V-110295000ft V 4085000 yd

8 Since total gold strata to be removed - 4085000 yd Depth of ges bearing strata is calculated as follows Average depth of bearing strata -average elevation of points -5" Average elevation - Since average elevation of gem-bearing strata -99.6-5-94.6ft Hence average depth of gen bearing strata 94.6-83.00-11.6ft Volume of gem-bearing material to be removed surface area x average depth 100+95+9598+110 - 99.6ft 12255000 x11.6 -142158000ft3 -5265111yd 3 Hence volume of gem bearing material to be removed -5265111yd