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 19P
To determine

(a)

To plot:

Grain size distribution curve.

Expert Solution
Check Mark

Answer to Problem 19P

MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 17, Problem 19P , additional homework tip  1

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  99
  10  2  87
  20  0.85  77
  40  0.425  65
  60  0.25  32
  80  0.17  23
  100  0.15  14
  200  0.075  7
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 19P , 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 19P

  D10=0.10mm

  D30=0.22mm

  D60=0.35mm

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  99
  10  2  87
  20  0.85  77
  40  0.425  65
  60  0.25  32
  80  0.17  23
  100  0.15  14
  200  0.075  7
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 19P , additional homework tip  3

Conclusion:

The values of D10, D30 and D60 from the sieve distribution graph are 0.10mm, 0.22mm and 0.35mm.

To determine

(c)

The value of uniformity coefficient

Expert Solution
Check Mark

Answer to Problem 19P

  Cu=3.50

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  99
  10  2  87
  20  0.85  77
  40  0.425  65
  60  0.25  32
  80  0.17  23
  100  0.15  14
  200  0.075  7
PanPan--

Concept used:

  Cu=D60D10

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

Calculation:

  Cu=D 60D 10=0.350.10=3.50

Conclusion:

The value of coefficient of uniformityis 3.50.

To determine

(d)

The value of coefficient of gradation

Expert Solution
Check Mark

Answer to Problem 19P

  Cc=1.38

Explanation of Solution

Given information:

Sieve No.Sieve No.Percent finer
  4  4.75  99
  10  2  87
  20  0.85  77
  40  0.425  65
  60  0.25  32
  80  0.17  23
  100  0.15  14
  200  0.075  7
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= ( 0.22 )20.35×0.10=1.38

Conclusion:

The value of coefficient of uniformity is 1.38.

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