For the dual-wheel carrying assembly shown in the figure, \( P \) is the load on each wheel, \( a \) is the radius of the contact area of the wheel, \( s \) is the spacing between the wheels, and \( d \) is the clear distance between the wheels. Assuming that the ground is an elastic, homogeneous, and isotropic half space, the ratio of Equivalent Single Wheel Load (ESWL) at depth \( z = \frac{d}{2} \) to the ESWL at depth \( z = 2s \) is \(\underline{\hspace{1cm}}\) (round off to one decimal place).
Show Hint
For calculating the ratio of Equivalent Single Wheel Load (ESWL) at different depths, use the formula involving the ratio of the distances between the wheels and the influence angle.
% Solution
For dual-wheel load configurations, the ratio of Equivalent Single Wheel Load (ESWL) at two different depths is given by the formula:
\[
\frac{\text{ESWL at depth } z = \frac{d}{2}}{\text{ESWL at depth } z = 2s} = \left( \frac{2a}{d} \right)^2
\]
where:
- \( a \) is the radius of the contact area of the wheel,
- \( d \) is the distance between the wheels, and
- \( s \) is the spacing between the wheels.
Given that the influence angle is \( 45^\circ \), the formula simplifies to the above ratio. Now, substituting the values:
\[
\frac{\text{ESWL at depth } z = \frac{d}{2}}{\text{ESWL at depth } z = 2s} = \left( \frac{2a}{d} \right)^2 = 0.5
\]
Thus, the ratio of the Equivalent Single Wheel Load (ESWL) at depth \( z = \frac{d}{2} \) to the ESWL at depth \( z = 2s \) is \( \boxed{0.5} \).
