13. A contractor needs 375 m³ of aggregate base for a highway construction project. It will be compactedto a dry unit weight of 18.9 kN/m³. This material is available in a stockpile at a local material supply yardat a water content of 7.75%, but is sold by the metric ton and not by cubic meters (ton 3 1000kg).a. How many tons of aggregate should the contractor purchase? Round your answer to thenearest whole number.b. A few weeks later, an intense rainstorm increased the water content of the stockpile to 14%. Ifthe contractor orders the same quantity for an identical section of the highway, how many cubic meters.of compacted aggregate base will he produce?

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13. A contractor needs 375 m³ of aggrega to a dry unit weight of 18.9 kN/m³. This n at a water content of 7.75%. but is sold b a. How many tons of aggregate s nearest whole number. h. A few weeks later an intense

13. A contractor needs 375 m³ of aggrega to a dry unit weight of 18.9 kN/m³. This n at a water content of 7.75%. but is sold b a. How many tons of aggregate s nearest whole number. h. A few weeks later an intense

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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5.24 Three aggregates are to be mixed together in the following ratio: Aggregate A 35% Aggregate B 40% Aggregate C 25% For each aggregate, the percent passing a set of five sieves is shown in the fol- lowing table: Sieve Size (mm) % Passing Agg. A % Passing Agg. B % Passing Agg. C 9.5 85 50 40 4.75 70 35 30 0.6 35 20 0.3 25 13 1 0.15 17 7 Determine the percent passing each sieve for the blended aggregate.
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Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket ½ ft3 Weight of empty bucket Weight of bucket filled with dry rodded coarse aggregate 20.3 lb Trial 1 = 69.6 lb Trial 2 = 68.2 lb Trial 3 = 71.6 lb a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the per- cent voids between aggregate particles for each trial.
Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained:Volume of bucket = 14 LWeight of empty bucket = 9.21 kgWeight of bucket filled with dry rodded coarse aggregate:Trial 1 = 34.75 kgTrial 2 = 34.06 kgMeasureSampleA B CWet Mass (g) 521.0 522.4 523.4Dry Mass (g) 491.6 491.7 492.1Absorption (%) 2.5 2.4 2.3Table P 5 . 9 determine the amount of water required to increase the moisture content of aggregate to reach absorption?5.9 Three samples of fine aggregate have the properties shown in Table Trial 3 = 35.74 kga. Calculate the average dry-rodded unit weightb. If the bulk dry specific gravity of the aggregate is 2.620, calculate thepercent voids between aggregate particles for each trial.
Three aggregates are to be mixed together in the following ratio: Aggregate A 35% Aggregate B 40% Aggregate C 25% For each aggregate, the percent passing a set of five sieves is shown in the following table: Determine the percent passing each sieve for the blended aggregate?
Referring to the aggregate gradations A, B, and C in Figure P5.26, answer the following questions: a. What is the maximum size of each gradation? b. What is the nominal maximum size of each gradation? c. Classify each gradation as dense, open, or gap indicating the reason for each classification. 100 A 80- 60 40 0.075 0.30 1.18 2.36 4.75 9.5 12.5 19.0 25.0 0.15 0.60 Sieve sizes (mm) FIGURE P5.26 Percent passing 20
Please help, thanks. Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = 0.3 cuft Weight of empty bucket is 19.8 lb Weight of bucket filled with dry rodded coarse aggregate = 57.6 lb If the bulk dry specific gravity of the aggregate is 2.63 and unit weight of water = to 62.4lb/ft^3, calculate the percent voids in the aggregate (one decimal only). Input magnitude only.
Three aggregates are to be mixed together in the following ratio:Aggregate A: 35%Aggregate B: 40%Aggregate C: 25%For each aggregate, the percent passing a set of five sieves is shown in Table PDetermine the percent passing each sieve for the blended aggregate.
Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: (Note: Write answer with 2 decimal places) Volume of bucket = 1/2 cu.ft Weight of empty bucket = 20.3 Ib Weight of bucket filled with dry rodded coarse aggregate: Trial 1 = 76.6 Ib Trial 2 = 75.1 Ib Trial 3 = 78.8 Ib a. Calculate the average dry-rodded unit weight b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the percent voids between aggregate particles for each trial. {Note: Bulk Dry specific gravity must be multiplied to the unit weight of water to obtain percent voids)
Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained: Volume of bucket = 0.3 cuft Weight of empty bucket is 18.1 lb Weight of bucket filled with dry rodded coarse aggregate = 58.7 lb Calculate the dry-rodded unit weight in lb/ft^3 (one decimal only). Input magnitude only.
Would it be reduced to half or almost half? Im so confused