This exercise is about the appliance of the belt friction formula (here is an online belt friction calculation tool).
Task
The band brake shown should be operated right-turning and left-turning. What is the maximum holding torque in right-turning and left-turning rotation without the brake drum rotating?
Solution
A solution video will be published soon on this channel.
The first step is the free-cut of the operator lever. The force FS doesn't have a real impact here, but it is included for formal reasons.
The balance of forces in x-direction is
∑Fx=0=FAx−FS⋅cos45°
The balance of forces in y-direction is
∑Fy=0=FAy+FS⋅sin45°+FB−F
Both balances don't yield useful informations, but the balance of moments around A gives
∑M(A)=0=FB⋅r−F⋅l
FB=F⋅lr
Next, the relationship between the rope force FS and the braking force FB is noted using the rope friction formula (Euler-Eytelwein). This is done separately for the two different directions of rotation.
Right-turning
In the case of the left-turning moment, F B lies in the strained strand, F S in the loose strand. The following applies:
FB≤FS⋅eµ⋅α
FS≥FB⋅e−µ⋅α
In union with the equilibrium of moments around the drum pivot, it follows:
∑M=0=FS⋅r+ML−FB⋅r
ML=FB⋅r−FS⋅r
ML=F⋅lr⋅r−F⋅lr⋅e−µ⋅α⋅r
ML=F⋅l⋅(1−e−µ⋅α)
Left-turning
At the right-turning moment, FB lies in the loose strand, FS in the strained strand. The following applies:
FS≤FB⋅eµ⋅α
In union with the equilibrium of moments around the drum pivot, it follows:
∑M=0=FS⋅r−MR−FB⋅r
MR=FS⋅r−FB⋅r
MR=F⋅lr⋅eµ⋅α⋅r−F⋅lr⋅r
MR=F⋅l⋅(eµ⋅α−1)
