Is hydraulic hose too strong?

A basic hose assembly consists of fittings, a tube, reinforcement, and a cover. The reinforcement must provide enough support to keep the tube intact and support all the forces created by the pressurized fluid. The tube is a leak-free barrier that contains the fluid and is supported by the reinforcement. The cover serves the same purpose as our skin — to hold everything inside and provide protection from the ambient environment.

The hose segment must have fittings on both ends to complete the package. The fittings anchor the hose segment and complete the sealing system. The hose reinforcement reacts to the hydrostatic separating forces acting on the end fittings to keep the hose assembly together. In short, we can only speak about a hose assembly when we conduct the force analysis to organize the hose segment reinforcement design.

As an engineer new to the industry, I accepted the conventional approach to hose reinforcement design. Obviously, the approach has worked well. There was no reason to question the analysis. Then a concern came up. What would happen if we used a different tube wall thickness? The mean diameter of the reinforcement would be different, but the hydrostatic forces and the fitting’s nipple diameter would be the same. To a seasoned hydraulics guy, this didn’t make sense. What is going on? Maybe hose strength should be based on the nipple diameter instead of the mean diameter of the reinforcement.

Evans hose reinforcement analysis

Figure 1. This rendition of Evans’ free body diagram of a section of hydraulic hose shows that horizontal and vertical forces from hydrostatic pressure act at the mean diameter (Dm) of the hose reinforcement.

Colin Evans’ book, Hose Technology, is a respected engineering resource found in many hose manufacturing engineering libraries. Similar reinforcement engineering design approaches are repeated in several other resources. In Chapter 7, Evans presents a free body diagram, along with an analysis of the forces for the creation of reliable hose designs and manufactured product. Figure 1 is an enhanced rendition of Evans’ free body diagram model showing the forces H and V as acting directly on each braid strand at diameter, D_m.

 

Evans defines an axial and a radial force and designates them as vectors H and V:
H = P x D_m² × π ÷ 4, and
V = P × D_m × L ÷ 2.

Where:
H is the horizontal (axial) force vector from fluid pressure,
P is fluid pressure (not burst pressure, P_b),
D_mis mean diameter of the reinforcement,
V is the vertical (radial) force vector from fluid pressure, and
Lis the pitch (lead length) of the spiral.
Because these are vectors, we can calculate the magnitude and direction of the resultant vector, R.

The Evans analysis defines forces based on the reinforcement geometry. This analysis has the virtue of calculating a braid construction that has been used for many years and has been used to develop a large body of hose length designs.

Evans concludes with the (hoop force) burst formula for braided or spiral hose:

P_b = (2 × N × R × sinϴ) ÷ (D_m × L)

Where:
Nis the number of reinforcement ends,
P_b is hydrostatic pressure at burst. This value is measured and verified by experiment,
R is the breaking (tensile) strength of the ligament, and
ϴ is the braid angle: arctan [ 2L÷ (π D_m ) ].

Questions raised
Here is the fly in the ointment: The horizontal force calculation is based on the mean diameter of the reinforcement. So what? Well, consider we have a hose tube with a nominal (tube) inside diameter of 0.25 in., and further assume a pressure of 1000 psi. If we choose a tube wall thickness of 0.030 in. or 0.050 in., how much reinforcement do we need to keep the tube from failing? Assume the pitch is such that the reinforcement angle is 54° 44´and further assume the hydrostatic pressure acts at the mean diameter of the reinforcement.

Evans’ solution allows us to estimate both the burst value and the force the reinforcement must support; that value is N × R. This is true for any pressure from zero to burst pressure. Why is that? First, we’ll rearrange the Evans solution:

P × D_m × L ÷ (2 sin ϴ) = N × R

Now we find that the product of N and R for the tube with the 0.030-in. thick wall is about 130.7 lb. For the tube with the 0.050-in. wall, the product is about 166.6 lb. Why should the thicker, stronger tube need stronger reinforcement? To find out, let’s take a closer look at the free body diagram of a hose assembly that includes the end fitting.