Ask Question
21 August, 15:37

4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, respectively. a. Find the load and shrinkage factors for this soil. b. How many trucks loads with a capacity of 5 lcy/truck would be required to haul 750,000 ccy of this soil to a project

+1
Answers (1)
  1. 21 August, 17:25
    0
    a. load factor = 0.893

    shrinkage factor = 0.848

    b. Number of Trucks loads = 113,585 Trucks loads

    Explanation:

    Here, we start by identifying the factors as given in the question.

    γn = 2800 lb/cy

    γloose = 2500 lb/cy

    and γcompacted = 3300 lb/cy

    a. Mathematically,

    Load factor = γloose/γn = 2500/2800 = 0.893

    Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848

    b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;

    ρlcy = 5

    Load factor * Shrinkage factor = ρloose/γn * γn/γcompacted = ρlcy/ρccy

    0.893 * 0.848 = 5/ρccy

    ρccy = 5 / (0.893 * 0.848) = 6.603

    The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers