Converting belt tension to radial compression force

[MechEgr]

Imagine you are getting dressed in the morning and fastening a belt around your waste. You pull on the loose end of the belt, developing tension T throughout the length of the belt. Using this information, how can you determine the amount of compression the belt is applying to your waste?

I apologize if this topic has already been covered. I could not find it in the forums, but if it has, please direct me to the correct thread.

Thanks.

 

[Jano L.]

Suppose the belt is circular and it is pulled by force T.

One can calculate the force perpendicular to the belt element delimited by angle sector of $\Delta \varphi$ radians (draw a picture).

There are two forces pushing the band of the belt towards the belly, each acting on one end of the band and having equal magnitude $T\Delta \varphi/2$. Their sum is $T\Delta \varphi$.

This is the total force pushing the band of the belt towards the belly.

With information on the radius and width of the belt, one can then calculate the pressure.

 

[jbriggs444]

Rather than compute the tension on a small belt element of size Δ, one can consider the tension on the front 180 degrees.

The net force on this semicircular segment is 2T. If you consider a vertical slice through yourself from hip to hip, the total force applied across this slice must therefore be 2T. The length of this slice is the diameter D of the circle formed by your belt.

Accordingly, the pressure within the bulk of your body must be given by 2T/D. This is the same as the pressure exerted by the belt.

[This "pressure" is per unit length, not per unit area]

Divide by the width of the belt to get pressure per unit area.

On the [questionable] assumption that your waist is circular and that the belt has length L and width W and just barely reaches around, this would mean that the pressure exerted by the belt on your body is given by 2piT/(L*W)

 

[MechEgr]

Jbriggs,

I follow your logic all the way until your very last point. I don't quite understand where the pi term came from in your equation. I understand that the force exerted over half of the cross-sectional area should be 2T/(L*W). It seems that somehow when you convert that to pressure applied circumferentially, you get an extra "pi" term from somewhere?

Thanks for your help.

 

[jbriggs444]

The pressure per unit distance at the centerline of your body must be given by 2T/D, you agree?

Given that an equilibrium exists, the pressure per unit distance around the circumference of your body must therefore also be given by 2T/D, you agree?

The length L of the belt is given by pi * D. It follows that D = L/pi.

This means that 2T/D can be rewrtten as 2T/(L/pi) = 2piT/L

Factoring in the width of the belt, the pressure per unit area is 2piT/(L*W)