Not quite.

You need to bend the pipe. This reuqires a fair bit more technical detail than can be offered on this forum. You need to understand the construction details of a pipe that is a composite, its orthotropic in the three planes that represents the bending action for which the spring constants must be derived. You also need to understand the difference between empty and prescribed crossectional loadings and thieir influences.

Note, an empty pipe cut along its axis, with overlap, will want to collapse radially to a smaller size. So, friction contact with the overlap should be included.

There are other factors that also need attention that sits in the domain of designer IP.

We don't know who you are, who you work for, and the particulars of your interests. Can you help us? ■

If idler spacing is 2m then the horizontal component of the idler load will be 2x(0.193-0.00341) = 0.3792 kN. The vertical component of the load will be 2x[(60+110)*(9.81/1000)] = 3.3354 kN. The resolved idler load = [(0.3792^2)+(3.3354^2)]^0.5 = 3.3569 kN. The direction of the resolved idler load is Theta=arcsin (0.3792/3.3569)= 6.49 degrees from the vertical (towards inside of horizontal curve).

Joe Dos Santos ■

If idler spacing is 2m then the horizontal component of the idler load will be 2x(0.193-0.00341) = 0.3792 kN. The vertical component of the load will be 2x[(60+110)*(9.81/1000)] = 3.3354 kN. The resolved idler load = [(0.3792^2)+(3.3354^2)]^0.5 = 3.3569 kN. The direction of the resolved idler load is Theta=arcsin (0.3792/3.3569)= 6.49 degrees from the vertical (towards inside of horizontal curve).

Joe Dos Santos ■

Dear Joe,

Are you advising the interested parties that the details you have given are sufficient to conclude the radial idler pressures?

I contend they are not and will result in a further increased radial force to physically bend the pipe into the chosen radii! This increased radial force will require increase the axial tension along the arc of contact beyond your prescribed conditions.

I do appologize that I did not include this significant component in my earlier posting. I also appologize for my sloppy eye sight in posting the T/R value which you rightly corrected. ■

Larry,

What I have posted is the math to determine radial load to the inside of the curve and, when vectorially added to the gravity loads, to determine the (transverse) resultant idler load and the direction of that load.

This has not adressed the crushing or crimping of the cross-section, which is dependent on the mechanics of the pipe belt in the rolled shape. This becomes accentuated (at a tight horizontal curve) at a material load discontinuity of an otherwise fully loaded conveyor.

This also does not address any travel resistance issues which are dependent on the Ky and Kx calculations for the pipe belt section and its idlers. This radial loads (along with the gravity loads) do figure adversely into the resistance calculations.

The calculations posted merely calculate the transverse idler loads at a point (along the pipe conveyor) of given radius, tension, and material and belt loads, regardless of how the tension value was acheived. This is statics and accounts for the magnitude of the transverse forces on the idlers regardless of the complexity of the internal stresses and strains in the belt itself.

Joe Dos Santos ■

Yes, your comments are obvious.

My comment was to highlight the less obvious, less it be obscured by our rhetoric. I also would refer to the bending force or bending moment as a significant reaction factor, even a dominant force with high modulus belts and tight radii in some cases, which acts opposite to the horizontal inward radial force ie. the pipe wishes to straighten out after being bent. ■

Larry,

Your comments caused me to think in order to explain why the equations that I presented are correct, that is, the idler loads are determined by statics regardless of the internal stresses and strains in the belt (not counting the belt's outward crowding into the restraining idler hoop as it tends to unroll).

Simplistically, idealistically, a single moment couple, applied at the ends of the long curve, produces the curvature without applying any additional (due to moment) loads on the idlers, so at best any bending resistance will only be felt at the ends where a an applied or restraining force causes that end moment breaking over each of the end idlers. More realistically, depending on the idler spacing, the belt curves and breaks over each successive idler. This makes the explanation a little more complicated but not the results.

Your argument would mean that the idler loads, caused by tension and curvature, are not equal throughout the curve. Statics requires that the external tension loads be equal and opposite the reactions which are the idler loads (according to Pr=T/R). Inspection of the geometry along the curve clearly demonstrates that they are equal.

This can be proven rigorously but who has the time and space?

What do you think?

Joe Dos Santos ■

DEAR SIR

ATTACHED PICTURE NOT AVAILABLE

PIPE DIA AND PANEL SPACING IS REQUIRED

Pl CHECK BELT WT SEEMS TO BE HIGHER SIDE WHAT IS BELT SPEC

Do you want to calculate radial load o rollers

A R SINGH ■

THe radial or horizontal component is: T/R = 135 kN / 700 m = 0.183 kN/m of axial length. Multiply this by the idler spacing and you get the reaction into the idler frame excluding the centrifugal force. ■

The radial load due to tension T/R = 135 (kN/m)/700 (m) = 0.193 (kN/m). The Centripetal Force MV^2/R = (170*3.75^2)/700 = -0.00341 (kN/m) (direction opposite the radial load due to tension). These add vectorially with the belt and material weight loads. The vector sum is multiplied by the idler spacing to get the idler loads. Direction of the load vector is also determined arcsin(theta)=Vertical Load/Total Load Vector. Because of the nature of the load application only the radial load due to tension and the centripetal load apply to crushing of the pipe cross-section. The material load helps to preserve the cross-section but not locally at any material flow discontinuity.

Joseph A. Dos Santos ■

Dear Sirs,

I have read the thread about the pipe belt rigidity, but I still don't understand how to calculate the normalforce on the idlers due to belt-crowling.

Main question:

How much force is exerted on each idler, when the belt is formed into a pipe shape?

The pipe-belt tends to crowl out, so the belt exerts a force on the idlers.

Previousilly I have used, the bending formula for beams to calculate the bending force, but the formulas can't be used when the curling of the beam is too high (in this case 360 degrees!).

My teacher says the problem should be solved in Ansys, because it is a non-lineair problem. I would like to know a easier way.

So is it possible to calculate the forces on the idler due to the belt rigidity, without using numerical methods? An estimation should be sufficient.

I hope u all can help me(maybe a drawing)

with kind regards,

S.S.kalidien

Tu Delft ■

Dear Sandjay,

I suppose you are familiar with the formulea @=M*l/(E*I)

For a 360 degrees bending you can write:

@=M*pi*D/(E*I)=2*pi() rad.

M = the bending moment

D = the diameter of the pipe

next step: @=M*pi*D/2/pi/E/I=M*D/(2*E*I)

where E*I is the factor of stiffness of the belt.

If you know the stiffness of the belt you are able to define the value for M: the bending moment which has a liniaair value in relation with the diameter of the pipe.

I hope I have been of help for you.

Kind regards

Johan Brands

The Netherlands

i.s.t@wanadoo.nl

tel. 06 22 515 981 ■