Question

To obtain an undisturbed clay soil sample, an Area Ratio of 10 % needs to be achieved for a thin walled sampling tube. If the outer diameter of the tube is 50.8 mm, the inner diameter (in mm) is (rounded off to one decimal place).

Hint:

Area Ratio = (Do^2 - Di^2)/Di^2 x 100; solve for Di from the given Do and Ar.

Answer 1

Correct Answer: 48.4

The area ratio compares the cross section area of the tube's metal wall to the area of the soil sample the tube encloses. Working with the actual areas, rather than jumping straight to a diameter formula, makes the physical meaning clearer here.

The area of the full outer circle of the cutting edge is

\[ A_o = \frac{\pi}{4}D_o^2 = \frac{\pi}{4}(50.8)^2 = \frac{\pi}{4}(2580.64) = 2026.83 \ \text{mm}^2 \]

Let $A_i$ be the area enclosed by the inner edge, which is the area of the actual soil sample recovered. The wall area (metal displacing soil) is $A_o - A_i$. The area ratio is this wall area expressed as a percentage of the sample area:

\[ A_r = \frac{A_o - A_i}{A_i} \times 100 = 10 \]

so

\[ A_o - A_i = 0.10\,A_i \implies A_o = 1.10\,A_i \] \[ A_i = \frac{A_o}{1.10} = \frac{2026.83}{1.10} = 1842.57 \ \text{mm}^2 \]

Now convert this sample area back to a diameter using $A_i = \frac{\pi}{4}D_i^2$:

\[ D_i^2 = \frac{4 A_i}{\pi} = \frac{4 \times 1842.57}{\pi} = 2346.05 \] \[ D_i = \sqrt{2346.05} = 48.44 \ \text{mm} \]

Rounded to one decimal place, this gives the same result reached by working directly in diameters: a tube with this inner diameter keeps the sample disturbance within the 10% limit needed for good quality undisturbed clay samples.

\[ \boxed{D_i = 48.4 \ \text{mm}} \]