Answer to Question #241959 in Mechanical Engineering for Paul

Question

A truck and trailer must be able to accelerate at 0.3 m/s2 up an incline of 1 in 20. The truck’s engine generates a torque of 900Nm. It has a moment of inertia of 30kg.m2 . The efficiency of the truck’s transmission is 83%. The mass of the truck is 12 tons, and that of the trailer is 9tons. The tractive resistance for both the truck and the trailer combined is 60N/ton. The truck has 12 wheels and the trailer has 8 wheels. The wheels on both vehicles are identical. Each wheel has a diameter of 1.2m, a mass of 75kg and a radius of gyration of 500mm. Determine two suitable overall gear ratios. 

Answer 1

Find the incline:

$\theta=\arctan\frac1{20}=2.86°.$

Find the traction force required to provide this acceleration up the hill:

$ma=F-f,\\\space\\ F=ma+f=(12000+9000)0.3+\\+60·(12+9)=\\=7560\text{ N}.$

Find the torque that corresponds to this propulsive force:

$T_F=Fr=7560·0.5=3780\text{ Nm}.$

The angular acceleration of truck's wheels is

$\alpha_W=\frac aR=\frac{0.3}{0.6}=0.5\text{ rad/s}^2.$

The angular acceleration of truck's engine parts ($\Omega$ - gear ratio) is

$\alpha_E=\frac{a}{\Omega r}=\frac{0.3}{0.5\Omega}=\frac{0.6}\Omega.$

Moment of inertial of wheels:

$I_W=\frac12Nm_WR^2=\frac12·20·75·0.6^2=270\text{ kg-m}^2.$

Find the transmission (output) torque that corresponds to T_F above:

$T_o=T_F+I_W\alpha_W+I_E\alpha_E=\\ =3915+\frac{18}\Omega.$

Find the gear ratio that would make it possible for a 900 Nm engine to output 7590 Nm of mechanical torque:

$\Omega=\frac{T_o}{T_E}=\frac{3915+18/\Omega}{900},\\\space\\ \Omega_1=4.35,\\\space\\ \Omega_2=-0.00459.$