good to go
yes what is your question?
i haveit posted now @surry99
For part A) do you have information about the 3 suppliers they refer to?
here they are @surry99
ok thanks...which textbook are you using?
its' a old textbook not even hard back it's a paperback about soil testing.
any author?
chen b.
Hint 1) to put 1 yd^3 of solids into the dam, how many yd^3 of soil do you need? 2) Then for each supplier , taking into account the void ration, how much soil do you need from each supplier to provide 1 yd^3 of solids? 3) Once 1) and 2) are done, you should be able to get the cost from each supplier
itwoul d be $5.28 with void .90
yes...and how much do you save?
@Ash90 This is how I would approach the problem. If we're not sure how to proceed, start with definitions. The definition of void ratio is void ratio= e = \(\dfrac{V_v}{V_s} = \dfrac{V_v}{V_t-V_v} = \dfrac{n}{1-n}\) where n = ratio of voids to total volume = \(\dfrac{V_v}{V_t} = \dfrac{V_v}{V_s+V_v}=\dfrac{e}{1+e}\) We will be working with n because it is the ratio of voids to the total volume that we can measure. We have three suppliers, and as engineers, we need to evaluate every one of them in a fair and scientific way. We do not have the costs of compacting different fills to e=0.8, and for now, we assume cost is the same or borne by contractor. Supplier A: e = 0.9, cost = $5.28 / cy \(e=0.9,~ n=\dfrac{0.9}{1+0.9}=0.4736\) \(e=0.8,~ n=\dfrac{0.8}{1+0.8}=0.4444\) Thus after compaction, we have lost 0.4736-0.4444=0.0292 of the total volume. The pre-compaction volume required for each compacted cubic yard is therefore 1/(1-0.0292)=1/0.9708. Actual cost per cubic yard of compacted fill = $5.28/0.9708=$5.44 The same calculations must be calculated for each of the other two suppliers, and the savings deduced after choosing the lowest cost supplier. Supplier B: e = 2.0, cost = $3.91 / cy Supplier C: e = 1.6, cost = $5.19 / cy Once you have completed part A, you can calculate the actual volume required to provide a compacted (e=0.8) volume of 5 million cubic yards in the same way as in part A, using fill with e=1.2.
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