Axles and shafts of the 1½-Deck-Bus

The model has a mass of 16 kg, which is spreaded onto 6 wheels with a radius R = 50 mm.
The mass corresponds with a weight of 160 N, from which every front wheel carries 35 N and every rear wheel 22,5 N (measured with a spring balance).
By means of 2 driven wheels the model shall be accelerated on a flat bottom from zero speed to a velocity of 1 m/s (3,6 km/h) within 1 s (without consideration of the road resistance and the inertia of the drive components).


Front wheels
The stub axles of the front wheels are burdened with a swelling bending.
The average distance between fixing and force action is l = 13 mm. From this the static bending stress can be calculated to
s_b = M_b / W = F · l / W = 35 N · 13 mm / 6,283 mm³ = 72 N/mm².
The max. acceptable stress for swelling bending according to the fatigue strength diagram is s_sch = 255 N/mm², from which the half shall be used for the static load:
s_{b zul} = 127,5 N/mm² > 72 N/mm²      o.k.

Trailing wheels
Valid are the conditions of the front wheels with the difference, that the weight is 22,5 N and the average distance fixing and force acting is l = 20 mm. The static bending stress can be calculated to
s_b = M_b / W = F · l / W = 22,5 N · 20 mm / 6,283 mm³ = 72 N/mm² < 127,5 N/mm²       o.k.

Drive wheels
The shafts of the drive wheels are loaded with alternating bending and swelling torsion.
The average distance between fixing and force action is also l = 20 mm; from this the static bending stress also can be calculated to
s_b = M_b / W = F · l / W = 22,5 N · 20 mm / 6,283 mm³ = 72 N/mm².
Additional during acceleration of the model the drive generates a max. torsional stress of
t_{t max} = M_{t max} / W_p = F · R / W_p = 0,5 · m · a · R / W_p = 0,5 · m · v · R / t · W_p = 32 N/mm²
with an assumed average of t_t = 16 N/mm². The resulting equivalent stress with this can be calculated to
s_v = Ö[s_b² + 3(0,7 · t_t) ²] = 75 N/mm².
In this case dominates the bending stress; thus as the limit the acceptable stress for alternating bending according to the fatigue strength diagram s_w = 170 N/mm² is defined, from which the half for the static resp. average load shall be used:
s_{b zul} = 85 N/mm² > 75 N/mm²      o.k.

Maximum torque of the drive
According to the above calculation the stub axles and shafts are designed correct.
But the data for the acceleration of the model are only mathematical values, and the real torque on the drive shafts is to be examined further.

For the drive a motor with the following data is used:

Idle:	Rated voltage U = 6 V, rotation speed n = 93 1/s (5600 1/min), current I = 0,5 A
Nominal load:	Rated voltage U = 6 V, rotation speed n = 75 1/s (4500 1/min), current I = 1,7 A

with an armature resistance of R = 0,9078 Ohm and a motor constant c = 9,457 · 10^-3 Vs (see "Characteristics of DC-Motors").

However the motor is not operated with 6 V but with 8,4 V. Thereby the locked rotor torque increases from approx. 6 Ncm to max. approx. 8,5 Ncm (see "Characteristics of DC-Motors").

Between motor and drive shafts works a gear reduction 20:1; with this the max. locked rotor torque at the differential can be calculated to 1,7 Nm (without consideration of the mechanical losses), which is spreaded to M_t max = 0,85 Nm on each drive shaft. As average value also here the half shall be assumed:
M_t = 0,425 Nm.
The corresponding torsional stress
t_t = M_t / W_p = 34 N/mm²
results into an equivalent stress of
s_v = Ö[s_b² + 3(0,7 · t_t) ²] = 83 N/mm²,
which meets the assumed limit:
s_{b zul} = 85 N/mm² > 83 N/mm²      o.k.