Question

Based on the equal deflection criterion, the expression for the equivalent wheel load Pe is _________

(a) \(F_e \sqrt{P_e}=(F_1+F_2)\sqrt{P} \)

(b) \(F_e \sqrt{P_e}=(F_1-F_2) \sqrt{P}\)

(c) \(F_e \sqrt{P_e}=(F_1*F_2) \sqrt{P}\)

(d) \(F_e \sqrt{P_e}=(F_1/F_2) \sqrt{P}\)

The question was asked in an online interview.

This interesting question is from Flexible Pavement topic in portion Design of Flexible & Rigid Pavement of Geotechnical Engineering II

Answer 1

The correct option is (a) \(F_e \sqrt{P_e}=(F_1+F_2)\sqrt{P} \)

The explanation is: The equivalent wheel load Pe is given by,

\(F_e \sqrt{P_e}=(F_1+F_2)\sqrt{P},\)

Where, Pe=equivalent wheel load

P=wheel load of each of the dual tyres

Fe=settlement factor for equivalent wheel load

F1= settlement factor contributed by one tyre of the

F2= settlement factor contributed by other tyre.