Answer:
A worn inner CV joint often makes a clunking noise during starts and stops.
Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 ×
= 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 ×
/ 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
number of pulses produced = 162 pulses
Explanation:
give data
radius = 50 mm
encoder produces = 256 pulses per revolution
linear displacement = 200 mm
solution
first we consider here roll shaft encoder on the flat surface without any slipping
we get here now circumference that is
circumference = 2 π r .........1
circumference = 2 × π × 50
circumference = 314.16 mm
so now we get number of pulses produced
number of pulses produced =
× No of pulses per revolution .................2
number of pulses produced =
× 256
number of pulses produced = 162 pulses