Ask question

Ask question

All categories

3 years ago

5

An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil

ler (Sp. Cr. 2.748). The compacted specimen contains 6% asphalt binder (Sp. Gr. 1.088) by weight of total mix, and has a bulk density of 143.9 lb/ft^3 Ignoring absorption, compute the percent voids in total mix, percent voids in mineral aggregate, and the percent voids filled with asphalt.

1 answer:

den301095 [7]3 years ago

7 0

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112 which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler will be = 5 × 112/100

= 5.6%

Percentage of the asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air void = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) × 100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate = -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

You might be interested in

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

4 years ago

According to the rules of dimension how should written notes appear

Arturiano [62]

Explanation:

Leaders for notes should be straight, not curved, and point to the center of circular views of holes wherever possible. Leaders should slope at 45°, 30° or 60° with horizontal but may be made at any convenient angle except vertical or horizontal.

7 0

3 years ago

Consider a turbojet powered airplane flying at a standard altitude of 30,000ft at a velocity of 500 mph. The turbojet engine its

grandymaker [24]

Answer:

T = 20.42 N

Explanation:

given data

standard altitude = 30,000 ft

velocity Ca = 500 mph = 0.4 m/s

inlet areas Aa = 7 ft² = 0.65 m²

exit areas Aj = 4.5 ft² = 0.42 m²

velocity at exit Cj = 1600 ft/s = 487.68 m/s

pressure exit $\rho$ j = 640 lb/ft² = 0.3 bar

solution

we get here thrust of the turbojet that is express as

thrust of the turbojet T = Mg × Cj - Ma × Ca + ( $\rho$ j Aj - $\rho$ a Ag ) .............1

here Ma = Mg

Ma = $\rho$ a × Ca Aa = 0.042 kg/s

put value in equation 1 we get

T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )

T = 20.42 N

5 0

3 years ago

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-

Hitman42 [59]

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.

We have:

$r_{c}=1.9*10^{-4} [mm]$

$l_{c}=3.8*10^{-2} [mm]$

$\sigma_{c}=140 [MPa]$

We replace:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}$

We get:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]$

5 0

3 years ago

I study to get good grades because my parents want to send me to the college of my choice.” This is an a. Intrinsic motivational

sashaice [31]

c. Both of These

hope it helps

lub.....

5 0

3 years ago

Read 2 more answers