OBTAINED by ADDING a cumulative percentage of aggregates retained on each of the standard sieves ranging from 80 mm to 150 microns.Fine aggregate is always mixed with proportion to the coarse aggregate and the following formula is used for percentage proportion fine aggregate to coarse aggregate.\(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)where,R = proportion of fine aggregate to the combined aggregate by weightp1 = fineness modulus of fine aggregatep2 = fineness modulus of coarse aggregatep = desired fineness modulus for concrete mixSolution:The percentage of fine aggregate is given by:\(\frac x1\)= \(\frac{{F{M_{coarse}} - F{M_{combined}}}}{{F{M_{combined}} - F{M_{fine}}}} \times 100 = \;\frac{{7.6 - 6.4}}{{6.4 - 2.8}} \times 100 = 33.33\;\% \)Confusion Point:Concept 1Since in the Qs proportion of fine aggregate in coarse aggregate is asked and total percentage is not given. Then we will use LET,x % = FA and y % = CA and combined (x + y) %.So, x.FA + y.CA = (x + y).Combined,On Solving,As proportion of FA in CA = x / y % = (CA - Combined) / (Combined - FA) % \(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)Concept 2If given find amount of FA, then the formula will be\(\rm{R=\frac{p_2-p}{p_2-p_1}\times 100}\)

"> OBTAINED by ADDING a cumulative percentage of aggregates retained on each of the standard sieves ranging from 80 mm to 150 microns.Fine aggregate is always mixed with proportion to the coarse aggregate and the following formula is used for percentage proportion fine aggregate to coarse aggregate.\(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)where,R = proportion of fine aggregate to the combined aggregate by weightp1 = fineness modulus of fine aggregatep2 = fineness modulus of coarse aggregatep = desired fineness modulus for concrete mixSolution:The percentage of fine aggregate is given by:\(\frac x1\)= \(\frac{{F{M_{coarse}} - F{M_{combined}}}}{{F{M_{combined}} - F{M_{fine}}}} \times 100 = \;\frac{{7.6 - 6.4}}{{6.4 - 2.8}} \times 100 = 33.33\;\% \)Confusion Point:Concept 1Since in the Qs proportion of fine aggregate in coarse aggregate is asked and total percentage is not given. Then we will use LET,x % = FA and y % = CA and combined (x + y) %.So, x.FA + y.CA = (x + y).Combined,On Solving,As proportion of FA in CA = x / y % = (CA - Combined) / (Combined - FA) % \(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)Concept 2If given find amount of FA, then the formula will be\(\rm{R=\frac{p_2-p}{p_2-p_1}\times 100}\)

">

If in a concrete mix the fineness modulus of coarse aggregate is 7.6., the fineness modulus of fine aggregate is 2.8 and the economic value of the fineness modulus of combined aggregate is 6.4 then the proportion of fine aggregate in coarse aggregate is

Highway Engineering Stone Aggregates in Highway Engineering . 6 months ago

  2   0   0   0   0 tuteeHUB earn credit +10 pts

5 Star Rating 1 Rating

Concept:Fineness Modulus: It is basically an Index which defines the coarseness and fineness of the aggregate. It is OBTAINED by ADDING a cumulative percentage of aggregates retained on each of the standard sieves ranging from 80 mm to 150 microns.Fine aggregate is always mixed with proportion to the coarse aggregate and the following formula is used for percentage proportion fine aggregate to coarse aggregate.\(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)where,R = proportion of fine aggregate to the combined aggregate by weightp1 = fineness modulus of fine aggregatep2 = fineness modulus of coarse aggregatep = desired fineness modulus for concrete mixSolution:The percentage of fine aggregate is given by:\(\frac x1\)= \(\frac{{F{M_{coarse}} - F{M_{combined}}}}{{F{M_{combined}} - F{M_{fine}}}} \times 100 = \;\frac{{7.6 - 6.4}}{{6.4 - 2.8}} \times 100 = 33.33\;\% \)Confusion Point:Concept 1Since in the Qs proportion of fine aggregate in coarse aggregate is asked and total percentage is not given. Then we will use LET,x % = FA and y % = CA and combined (x + y) %.So, x.FA + y.CA = (x + y).Combined,On Solving,As proportion of FA in CA = x / y % = (CA - Combined) / (Combined - FA) % \(\rm{R=\frac{p_2-p}{p-p_1}\times 100}\)Concept 2If given find amount of FA, then the formula will be\(\rm{R=\frac{p_2-p}{p_2-p_1}\times 100}\)

Posted on 20 Nov 2024, this text provides information on Highway Engineering related to Stone Aggregates in Highway Engineering. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

Take Quiz To Earn Credits!

Turn Your Knowledge into Earnings.

tuteehub_quiz

Tuteehub forum answer Answers

Post Answer

No matter what stage you're at in your education or career, TuteeHub will help you reach the next level that you're aiming for. Simply,Choose a subject/topic and get started in self-paced practice sessions to improve your knowledge and scores.