Reaction Loads on Bent Hose
Reaction Loads on Bent Hose
(OP)
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!
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."





RE: Reaction Loads on Bent Hose
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.
RE: Reaction Loads on Bent Hose
Tobalcane
"If you avoid failure, you also avoid success."
RE: Reaction Loads on Bent Hose
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.
RE: Reaction Loads on Bent Hose
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,
RE: Reaction Loads on Bent Hose
RE: Reaction Loads on Bent Hose
Pressure alone tells you, just that, only pressure.
Fe
RE: Reaction Loads on Bent Hose
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.
RE: Reaction Loads on Bent Hose
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
RE: Reaction Loads on Bent Hose
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!!
RE: Reaction Loads on Bent Hose
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
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RE: Reaction Loads on Bent Hose
Principle is explained at http://www.lmnoeng.com/Force/ForceBend.htm
RE: Reaction Loads on Bent Hose
h
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
RE: Reaction Loads on Bent Hose
Tobalcane
"If you avoid failure, you also avoid success."
RE: Reaction Loads on Bent Hose
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
RE: Reaction Loads on Bent Hose
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."
RE: Reaction Loads on Bent Hose
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.
RE: Reaction Loads on Bent Hose
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,
RE: Reaction Loads on Bent 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."
RE: Reaction Loads on Bent Hose
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).
RE: Reaction Loads on Bent Hose
Is it valid to assume that your hose is a stiff beam? There is a separate set of equations to describe the behaviour hanging ropes and chains, which may be more accurate at describing your hose. I am not even going to try to describe this.
Will the hose deform elastically at maximum stress, or will it buckle?
RE: Reaction Loads on Bent Hose
Tobalcane
"If you avoid failure, you also avoid success."
RE: Reaction Loads on Bent Hose
To restrain the hose under static condition think of a bourdon tube springing back under changes of internal pressure. Such the internet for analysis of bent tubes under internal pressure, I am sure you'll come across some thesis and read the paragraphs dealing with longitudinal and radial strains that will allow you to determine the restraining forces.
RE: Reaction Loads on Bent Hose
RE: Reaction Loads on Bent Hose
Tobalcane
"If you avoid failure, you also avoid success."