MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months)
5th Edition
ISBN: 9781305581159
Author: Nicholas J. Garber; Lester A. Hoel
Publisher: Cengage Learning US
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Chapter 17, Problem 18P
To determine

(a)

To plot: Grain size distribution curve.

Expert Solution
Check Mark

Answer to Problem 18P

MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 17, Problem 18P , additional homework tip  1

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  100
  10  2  90
  20  0.85  80
  40  0.425  70
  60  0.25  40
  80  0.17  29
  100  0.15  19
  200  0.075  10
PanPan--

Concept used:

Plot the grain size distribution curve between the sieve number (mm) and the percent finer. The sieve number (mm) is plotted in x -axis and percent finer in y -axis.

Calculation:

MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 17, Problem 18P , additional homework tip  2

Conclusion:

The grain distribution curve is plotted.

To determine

(b)

The values of D10, D30 and D60.

Expert Solution
Check Mark

Answer to Problem 18P

  D10=0.075mm

  D30=0.180mm

  D60=0.347mm

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  100
  10  2  90
  20  0.85  80
  40  0.425  70
  60  0.25  40
  80  0.17  29
  100  0.15  19
  200  0.075  10
PanPan--

Concept used:

Plot the grain size distribution curve between the sieve number (mm) and the percent finer. The sieve number (mm) is plotted in x -axis and percent finer in y -axis. Then the corresponding values of D10, D30 and D60 are determined.

Calculation:

MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 17, Problem 18P , additional homework tip  3

Conclusion:

The values of D10, D30 and D60 from the sieve distribution graph are 0.075mm, 0.180mm and 0.347mm.

To determine

(c)

The value of uniformity coefficient

Expert Solution
Check Mark

Answer to Problem 18P

  Cu=4.627

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  100
  10  2  90
  20  0.85  80
  40  0.425  70
  60  0.25  40
  80  0.17  29
  100  0.15  19
  200  0.075  10
PanPan--

Concept used:

  Cu=D60D10

  CuCoefficient of uniformityD60Grain diameter at 60% passingD10Grain diameter at 10% passing

Calculation:

  Cu=D 60D 10=0.3470.075=4.627

Conclusion:

The value of coefficient of uniformityis 4.627.

To determine

(d)

The value of coefficient of gradation

Expert Solution
Check Mark

Answer to Problem 18P

  Cc=1.245

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  100
  10  2  90
  20  0.85  80
  40  0.425  70
  60  0.25  40
  80  0.17  29
  100  0.15  19
  200  0.075  10
PanPan--

Concept used:

  Cc=D302D60×D10

  CcCoefficient of gradationD60Grain diameter at 60% passingD30Grain diameter at 30% passingD10Grain diameter at 10% passing

Calculation:

  Cc=D 302D 60×D 10= ( 0.180 )20.347×0.075=1.245

Conclusion:

The value of coefficient of uniformity is 1.245.

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