Mrt IM Rigid shaft bearing TLC ID Solutionsolution 1here giv

Mr(t) IM Rigid shaft bearing TLC) ID

Solution

solution:

1)here given that gear ratio as

G=na/nb=Tb/Ta

Tb=number of teeth on gear b

Ta=number of teeth on gear a

G=1/3

2)here motor and first are mounted on rigid shaft hence it will not induced torsional vibration in system,hence whole inertia act on load shaft

3)hence inertia on load shaft at left end is

I1=Im+Ia+Ib/G^2

I1=Im+Ia+9Ib

and inertia on right side of shaft is

I2=Id/G^2=9Id

4)in this on load shaft this inertia acting and same torque from motor transfer to gear b hence resultant torque on system is Tr= (Tl-Tm)

5)stiffness of load shaft is given by

Kt2=Tr/m

m=angle turn by load shaft

Kt2=(Tl-Tm)/m

6)here degree of freedom of system is given by without torsion is

number of link=3

p1=lower pair=2 at beairing

p2=1 at gear contact

DOF=3(n-1)-2P1-P2=6-4-1=1

when torsion of shaft produces the system produces one more DOF and system resulatn DOF

DOF1=DOF+1=1+1=2

5)hence when opposite torque acting on two ends of shaft then it is two degree of freedom system

hence equation of motion is

for left end is

I1*m1\'\'+Kt2*m1=-Tm

in matrix form as

[Kt2-I1w1^2]m=Tm

for right end

I2*m2\'\'+Kt2m=Tl

in matrrix form

[Kt2-I2w2^2]m=Tl

8)here natural frequencies of system for 2 DOF system is as follows

w1^2*w2^2=.5{[(Kt1+Kt2)I2+Kt2*I1/I1*I2]+-[(Kt1+Kt2)I2+Kt2*I1/I1*I2]^2-4[(Kt1+Kt2)Kt2-Kt2^2/I1*i2]^.5]^.5}

on putting Kt1=0 and Kt2=[Tl-Tm/m]

we get finally

(w1*w2)^2=1.1834*[Tl-Tm/m][1/Id]

w1w2=0

9)where mode shape are given by

first mode shape

(m1/m2)₁=[Kt2/(-I1w1^2+(Kt1+Kt2))]

(m1/m2)₁=Kt2/-I1w1^2+Kt2

second mode shape

(m1/m2)₂=Kt2/-I2w2^2+Kt2

10) if inertia of gear are consider the we have to consider three rotor system,with two rotor at end of shaft for motor,load and at gear position we have consider another rotor and it will vary natural frequency of system and node shape