Balancing a Rotating Shaft - Determining Masses & Positions

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Homework Statement

A.

A shaft 2m long rotates at 1500 revs min^-1 between bearings as shown. The bearings forces of 5kN and 3kN acting in the same plane as shown. A single mass is to be used to balance the shaft, so that the reactions are zero. The mass is to be placed at a radius of 200mm from the shaft centre, 180^o from the direction of the bearing reactions. Determine the size and position (a and b) of the mass to be used.

B.

The shaft in part A is to be balanced using two mases (m₁ and m₂) placed 0.5m and 1.5m from end A and 180^o from the direction of the bearing reactions, each on radius arms 100mm long. Calculate the sizes of m₁ and m₂.

Homework Equations

F=m*r*ω²

The Attempt at a Solution

A.
ω=1500 revs min^-1=1500*(2π/60)=50π rad s^-1

To balance the shaft the bearing reactions must equal to centrifugal force of the mass.

R_A+R_B=F
F=5kN+3kN=8kN
F=m*r*ω²
8000=m*0.2*50π²
4934.8022m=8000
m=1.6211kg

Calculating moment about R_B

R_A*2-8*b=0
10-8b=0
8b=10
b=1.25m

a+b=2m
a=2-1.25=075m

B.
F₁+F₂=8kN

Taking moment about R_B

R_A*2-F₁*1.5-F₂*0.5=0
10-1.5*F₁-0.5*F₂=0
1.5*F₁+0.5*F₂=10

F₁+F₂=8
1.5*F₁+0.5*F₂=10

Solving equation

F₁=8-F₂
1.5*(8-F₂)+0.5*F₂=10
12-1.5*F₂+0.5*F₂=10
F₂=2kN

F₁+2=8
F₁=6kN

F₁=m₁*r*ω²
6000=m₁*0.1*50π²
2467.4011m₁=6000
m₁=2.4317kg

F₂=m₂*r*ω²
2000=m₂*0.1*50π²
2467.4011m₂=2000
m₂=0.8106kg

Did I got it right?
I'm not sure where I went wrong, but I suspect the m₁ should be smaller then m₂ as R_A is bigger then R_B. Any tips?

 

[SammyS]

Homework Statement

A.
View attachment 86596
A shaft 2m long rotates at 1500 revs min^-1 between bearings as shown. The bearings forces of 5kN and 3kN acting in the same plane as shown. A single mass is to be used to balance the shaft, so that the reactions are zero. The mass is to be placed at a radius of 200mm from the shaft centre, 180^o from the direction of the bearing reactions. Determine the size and position (a and b) of the mass to be used.

B.
View attachment 86597

The shaft in part A is to be balanced using two mases (m₁ and m₂) placed 0.5m and 1.5m from end A and 180^o from the direction of the bearing reactions, each on radius arms 100mm long. Calculate the sizes of m₁ and m₂.

Homework Equations

F=m*r*ω²

The Attempt at a Solution

A.
ω=1500 revs min^-1=1500*(2π/60)=50π rad s^-1

To balance the shaft the bearing reactions must equal to centrifugal force of the mass.

R_A+R_B=F
F=5kN+3kN=8kN
F=m*r*ω²
8000=m*0.2*50π²
4934.8022m=8000
m=1.6211kg

Calculating moment about R_B

R_A*2-8*b=0
10-8b=0
8b=10
b=1.25m

a+b=2m
a=2-1.25=075m

B.
F₁+F₂=8kN

Taking moment about R_B

R_A*2-F₁*1.5-F₂*0.5=0
10-1.5*F₁-0.5*F₂=0
1.5*F₁+0.5*F₂=10

F₁+F₂=8
1.5*F₁+0.5*F₂=10

Solving equation

F₁=8-F₂
1.5*(8-F₂)+0.5*F₂=10
12-1.5*F₂+0.5*F₂=10
F₂=2kN

F₁+2=8
F₁=6kN

F₁=m₁*r*ω²
6000=m₁*0.1*50π²
2467.4011m₁=6000
m₁=2.4317kg

F₂=m₂*r*ω²
2000=m₂*0.1*50π²
2467.4011m₂=2000
m₂=0.8106kg

Did I got it right?
I'm not sure where I went wrong, but I suspect the m₁ should be smaller then m₂ as R_A is bigger then R_B. Any tips?

It looks right to me.

There is an easier way to do this.

Consider: At what location along the shaft, do the bearing reaction forces produce zero torque?

 

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Thanks SammyS for you reply. Can you expand a bit more on that?

 

[SammyS]

Thanks SammyS for you reply. Can you expand a bit more on that?

For part A:

Consider the moment about some point on the shaft, a distance $\ x\$ to the right of point A, the left end of the shaft. Find what value of $\ x\$ makes that moment zero.

$\displaystyle \ -(5x)+3(2-x)=0 \$