Chapter 17, Problem 18P

To determine

(a)

To plot: Grain size distribution curve.

Answer to Problem 18P

Explanation of Solution

Given information:

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

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:

Conclusion:

The grain distribution curve is plotted.

To determine

(b)

The values of $D_{10}$, $D_{30}$ and $D_{60}$.

Answer to Problem 18P

  $D_{10}=0.075\text{mm}$

  $D_{30}=0.180\text{mm}$

  $D_{60}=0.347\text{mm}$

Explanation of Solution

Given information:

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

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 $D_{10}$, $D_{30}$ and $D_{60}$ are determined.

Calculation:

Conclusion:

The values of $D_{10}$, $D_{30}$ and $D_{60}$ from the sieve distribution graph are $0.075\text{mm}$, $0.180\text{mm}$ and $0.347\text{mm}$.

To determine

(c)

The value of uniformity coefficient

Answer to Problem 18P

  $C_u=4.627$

Explanation of Solution

Given information:

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

Concept used:

  $C_u=\frac{D_{60}}{D_{10}}$

  $\begin{array}{c}{C}_u\to \text{Coefficient of uniformity}\\ D_{60}\to \text{Grain diameter at 60\% passing}\\ D_{10}\to \text{Grain diameter at 10\% passing}\end{array}$

Calculation:

  $\begin{array}{c}{C}_u=\frac{D_{60}}{D_{10}}\\ =\frac{0.347}{0.075}\\ =4.627\end{array}$

Conclusion:

The value of coefficient of uniformityis $4.627$.

To determine

(d)

The value of coefficient of gradation

Answer to Problem 18P

  $C_c=1.245$

Explanation of Solution

Given information:

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

Concept used:

  $C_c=\frac{D_{30}^2}{D_{60}\times D_{10}}$

  $\begin{array}{c}{C}_c\to \text{Coefficient of gradation}\\ D_{60}\to \text{Grain diameter at 60\% passing}\\ D_{30}\to \text{Grain diameter at 30\% passing}\\ D_{10}\to \text{Grain diameter at 10\% passing}\end{array}$

Calculation:

  $\begin{array}{c}{C}_c=\frac{D_{30}^2}{D_{60}\times D_{10}}\\ =\frac{{\left(0.180\right)}^2}{0.347\times 0.075}\\ =1.245\end{array}$

Conclusion:

The value of coefficient of uniformity is $1.245$.