Reaction Loads on Bent Hose 1

[Twoballcane]

Hello All,

OK I got one that I'm stumped on that came accross my desk. I have a hose that is 4in in diameter, 50in in lenght, and has a 3.5ft bend radius. Once 250psi is applied, what is the reaction forces at each end of the hose to keep it in a 3.5ft bend radius? I'm calculating the load to be F=PA, so A=(4)(50)=200in^2 (area down center of hose), thus F=250(200)=50,000lbF. The reaction loads would be 25,000lbF on each side of the hose. Wow that is huge! I'm doing somthing wrong.

Thank you in advance for your help!

Tobalcane
"If you avoid failure, you also avoid success."

 

[seymours2571]

If I understand correclty the force to maintain the 3.5ft bent radius in the 4" hose would be a function of the stiffness of the hose (imagine having the hose empty at 0 psig internal pressure). From my understanding, as the pressure increases in the hose the stiffness of the hose increases.

However, having said that I am not sure of the right approach to this problem. I guess if you knew roughly the material properties of the hose you could perform a crude FEA. You might only be able to simulate the reactions at the end from the bend configuration, giving you the increase in reaction force to maintain the 3.5 ft radius at 250 psi.

This is a tough one.

 

[Twoballcane]

Thanks seymours for responding. The hose is your typical fire man hose. I can probaly do a FEA, but I like to start with hand calcs to back up the FEA.

Tobalcane
"If you avoid failure, you also avoid success."

 

[seymours2571]

In Roarks 6th Ed. Table 28 Case 5 it depicts a toridial shell under uniform internal pressure. It is not clear to me if the formulas will help determine the reaction loads at the ends or not, but it might be a good place to start.

I found several other cases that might be applicable as well in Roarks, however I am not sure if the equations take into account any stress stiffening effects of the hose while under pressure.

Good luck on this one. Post back if you find a solution.

 

[dhengr]

Twoballcane:

I think you’re doing something wrong too. Your calcs. don’t make any sense to me. Your “A” is actually a circumferential length not an area, and thus your F=PA has dubius meaning. But, then you did make a full/closed circle of your hose, and that may be a start in the right direction for an analysis of the problem. Assuming you are not near a nozzle with impulse/momentum loads involved. What I see is an elastic toroidal shell, under internal pressure which I can maintain with end caps at 12 & 3 o’clock. And, what reactions need be applied to hold the 3.5' radius when I remove the rest of the torus.

Seymours2571 points us further in a right direction. His thought, my words: the hose has no stiffness at zero pressure, and much stiffness at 250psi; primarily a function of the material properties of the hose under pressure, that is elasticity, hose mean diameter, stiffness, tensile strength, etc. You are effectively dealing with a 50" canti. beam or single leaf spring and you want to conform it to a 3.5' radius. The reaction and the bending end load are equal, they are the shear in a canti. steel beam.

I haven’t done either of these problems in the last week or two, so I gotta think about them for a while. Instead of making Seymours and me do all your thinkin for ya, why don’t you take that hose out behind the firehouse, pressurize it, bend it around a 3.5' radius and measure the reactions with a spring scale? This would be the easy way, maybe the hose manuf’er. could inform you. I suspect you would have some trouble finding the hose properties we need for this analysis.

 

[dhengr]

I’ll bet one of the hydraulic hose manuf’ers. would have some insight on your problem. There hose run in a 90° arc from one hard fitting to another would experience exactly the same reactions or shear force in the hose, right above the hose end fitting. And, I’ll bet they care about that stress.

 

[FeX32]

You would need to know the flow rate. This in turn allows you to calculate the inertial force that the bend creates in order to force the flow in another direction.
Pressure alone tells you, just that, only pressure.


Fe

 

[Compositepro]

The pressure is contained by the fiber reinforcement in the hose wall. The reaction of the hose to pressure depend very much on the fiber orientations. A braided hose is relatively flexible because of the approximately +/-45 degree fibers. This allows the tube walls to expand and contract in the length direction while flexing and still contain the axial and hoop stresses do to pressure. When you flex the hose the fibers on the outside of the bend may become +/-40 while on the inside they are +/-50 degrees while the tension in each yarn stays fairly constant.

The fire hoses I've seen are usually 0-90 construction and not very flexible under pressure. When these are flexed they will tend to kink on the inside of the bend. This kind of hose behaves like a 4" diameter column with an axial compression strength of 250 psi. When you flex it, one side of the tube wall will act as a hinge point while the other side will buckle. The length of the hose is not very relevant.

 

[zekeman]

The forces on the ends would be the vectorial resultant of the time rate of change of momentum,
vector M'Vf-M'Vi
where
M'= mass rate of flow
vf final velocity vector
Vi initial velocity vector
All other forces are internal; the force at each end constraint is independent of the radius or the length of arc and is
F=M'V^2
Looks like
2*p*hose nozzle area
If the nozzle diameter is say 1/2 inch, the force is
2*250*.785*.25=93 lb

 

[Ron]

Absent the momentum of the water upon initially fast charging the hose, it would take an external force to change the bend radius, not an internal one. Further, the hose is trying to straighten out when pressurized, not curl up, so if the end locations don't change, where's the force to change the radius?

If you carefully lay the hose out with a 3.5' bend radius and fill it with water until the hose is expanded, but not pressurized to any great degree, then you add pressure to 250 psi, the only locational change of the hose has to come from lengthening of the hose. If the ends are restrained, then the hose will creep outward based on the geometry of the friction forces of the hose to contact surface. If no contact surface, it will depend on material variations in the hose fabric as to which way it will twist and turn.

I think your reaction at the radial chord would be 1/2(250)(12.56)=1570 lb., assuming I'm understanding the problem somewhere near to what it is!!

 

[GregLocock]

Looks like a hoop stress problem to me, the net outward 'pressure' on a 180 degree hoop is

250psi*pi*r^2*2/(pi*2*R*(2*r))
which might cancel down to
250psi*r/(2*R)

That seems to be going the right way, a bigger diameter hose will see more straightening pressure, as will a more tightly curved one.

Now just work out the resultants for a smaller slice theta, draw an FBd, and hey presto. FEA indeed.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[CarlB]

I see the restraint force you're looking for as being essentially the "thrust" which develops at a change in direction of a pipe. Which us Civil engineers address (below ground) with thrust blocks or restrained joints. The thrust is primarily a function of the pressure and change in direction. Flowrate is usually insignificant.

Principle is explained at

http://www.lmnoeng.com/Force/ForceBend.htm

 

[Tmoose]

I believe High pressure hoses jerk (straighten?) noticeably when pressure is applied. No flow necessary.

http://3.bp.blogspot.com/_nxvcU9yjv2s/SbP_Rv4_TAI/AAAAAAAABTU/pSTKE7OwUSA/s400/IMG_2946.png

At least one pump or hydraulic book suggests using a "T" as an elbow with leg capped, to reduce reaction force.

Do to an unfortunate layout some pumps at at a Polaroid plant (long since closed) had some piping that was over-reacting to to inherent pulsating reaction forces tweaking the elbows. I was looking forward to clamping a strut across the elbows' bends to stiffen them up, detune the system and reduce the force of the excitation, but never got the chance.

Dan T

 

[Twoballcane]

Thanks all for responding!! CarlB, I like the calculator that you posted and it looks like what I'm looking for my problem. Did not even think it could have been a fluids phenomena, I've been doing so much structural and dynamic for the past few months.

Tobalcane
"If you avoid failure, you also avoid success."

 

[bvanhiel]

Twoballcane,

I believe that CarlB's calculation will give you the force required to keep a rigid pipe elbow in place, not the force required to keep your hose from straightening itself.

I don't know the right answer, but my approach would be to think about the volume change that the hose would undergo when you bend it. Multiply the volume change times the pressure to get work. You can then figure out the F*d to create an equivilent amount of work.

The key assumption is that you know how the volume will change with bending. As others have pointed out it has a lot to do with the directions of the fibers in your hose. I think you can bracket your answer by assuming neural axis is either along the centerline or on the outer arc.

-b

 

[Twoballcane]

Thanks bvanhiel for responding,

The reaction loads at each end of the hose was one of the things I was looking for, the other was forces to keep it bent. Now that I slapped my forehead and realized that this is more of a fluids problem, I started looking into Impulse-Momentum Principles. The hose we are using are more like a firemen’s hose, so I am looking into ways to support and keep the bend radius in place. In doing this, I needed the loads or reaction loads to quantify what we need.


Tobalcane
"If you avoid failure, you also avoid success."

 

[zekeman]

The Carlb solution F of the resultant force is not an external force, but the tensile force in the pipe wall. The only reason for additional support is to minimize the additional stress of the centrifugal force of the stream moving around the curve, but if the hose has sufficient tensile capability, that support is unnecessary.
My answer to the OP is that there is no external support necessary to sustain the curvature in the hose following the
onset of flow and it will maintain the curve it started, within the tensile capability of the hose.

 

[rb1957]

ok zeke, try this ... get your garden hose, lay it on the ground in an arc, try the water on ... what happens.

to 2ball, are you only supporting the hose at the ends, and not along it's length ?

is the hose then "just" redirecting a flow ? someone posted above already that the reaction would be due to the change in momentum. by my calc 50" as an arc, 42" radius, is about 68deg,

 

[Twoballcane]

Zekeman, thanks for responding, the hose is not a stiff pipe, more like a firmen's hose.

rb1957, we will be supporting the ends and need-be the length of the hose. We will be testing in the next few weeks to see how the hose will act under pressure. I just needed a way to quantify the loads we may see for constraining purposes. Thanks again rb for the link! Also, I'm dusting off my fluids analysis for this job LOL.


Tobalcane
"If you avoid failure, you also avoid success."

 

[zekeman]

If you are constraining the 2 ends then I vote for CarlB and you really don't have to support it anywhere else providing that the tensile stresses are sustainable.
But if the "free" end is at the open nozzle then the force there is
M'V=W*V^2/g
M' mass flow rate
W' weight flow rate

Rb,
I'm aware of the rocket forces ala Newton and have seen gyrating hoses.

For a short hose (no pressure drop) the 250 psi and a nozzle size of 1", I got 200 Lbs of force at the open nozzle (corrected error in previous post); for a 2" nozzle this goes to 800 lbs. The force at the other end should be 250* .785*4^2= 3140 lb.
I believe that this result is not dependent on the curvature of the hose, nor should it change the existing curvature, but I haven't tested this conclusion (with a real hose as R1957 suggests I should).