1. A spiral ramp is driven in a limestone to a vertical depth of 600 ft with the

Question

1. A spiral ramp is driven in a limestone to a vertical depth of 600 ft with the following parameters Ramp cross-section = 18 x 18 feet. b. Ramp grade 12 % C. Inner diameter = 102 ft d. Outer diameter 120 ft. Determine a. The number of spirals of the ramp b. The length of the ramp at the mean diameter c. The bank volume of material to be excavated d. The loose volume of material to be excavated assuming a swell factor of 30% e. The tonnage of material assuming that SG 2.65 for limestone f. The number of shifts to complete construction, assuming that 7 linear feet of ramp are excavated in every shift

Explanation / Answer

1. Number of spirals = 600/depth of spiral per revolution =

Depth of spiral per revolution = Ramp grade * Revolution length

Revolution length = 3.14 * (D+d)/2 = 348.716 fts. ,

Thus depth of spiral per revolution = 0.12 * 348.716 = 41.846 fts.

Thus, no of spirals = 14.33 = 14

2. Ramp length = no .of spirals * revolution of length of ramp at mean dia. = 14.33 * 348.716 = 5000 fts.

3. Bank volume excavated = Cross sectional area * ramp length = 18*18*5000 = 16,20,000 cu.ft.

4. Loose volume to be excavated = Bank volume * (1+swell factor) = 21,06,000 cu.ft.

5. Tonnage of material = 2.65 * 0.0283168 * 21,06,000 / 1000 = 1,58,033 tons

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.