Determining hose strength is a key factor in successful hydraulic-system design, especially if questions arise about whether the hose is stronger than it needs to be. Thus, it’s a good idea to calculate the hydrostatic forces that must be supported by reinforcement. For an in-depth discussion about calculating hose strength, see “Is Hydraulic Hose Too Strong?” published in the November 2012 issue of Hydraulics & Pneumatics, or click here for the online version of the article.

Based on the calculations for hose strength, we can now discuss reinforcement, which is the product of the tensile force in each strand and the number of strands:

R × N = P × (1.36 × DN2),

where R = tension force vector associated with elongation (applied at 54.74°, the neutral angle), N = minimum number of ligaments needed to cover the hose tube, P = hydrostatic pressure of the fluid, and DN = nipple diameter when equilibrium exists between the hydrostatic forces and the reinforcement (with reinforcement strands aligned at the neutral angle).

Reinforcement ligaments support the fluid’s hydrostatic force by tension. The ligaments align themselves so that they’re equal to, and opposite in direction to, the hydrostatic force. Ultimately, the reinforcement moves to the neutral angle. Tensile strength of the ligaments and the number of them weigh heavily in the strength of hydraulic hose.

Hose reinforcement consists of either braided or spiraled ligaments of wire or yarn wound symmetrically about the hose tube. Specifically, total tension is distributed among the total number of ligaments in the construction. Therefore, the tensile force per ligament is:

R = (P × 1.36 × DN2)/N,

which becomes the required quantity for equal load sharing.

How does hose reinforcement share the load from hydrostatic pressure in a hose assembly? Each ligament in the construction is a spring that develops tension in proportion to how much it’s stretched (∆L). This raises the question: How can ligament elasticity and parallel construction provide effective hose reinforcement?

Reinforcement is elastic

Visualize hose reinforcement as a collection of parallel springs that align themselves against the hydrostatic forces. Each hose reinforcement strand of wire or yarn is a spring that exhibits a tensile-versus-elongation result (Figure 1). This type of curve is common for both wire and yarn, and its slope, R/ε, is the spring rate, K, or yarn tensile modulus, YTM, respectively. The 1,500-Denier yarn in Figure 1 has a breaking strength, Rmax, of 87 lb and elongation of about 3.7%. So for 1,500-Denier yarn reinforcement, yarn tensile modulus is:

YTM = 87 lb/0.037 = 2,351 lb.

Therefore, for a 10-in.-long strand of yarn, YTM is 235 lb/in. Just as with a spring, YTM and K change with increasing or decreasing length.

The curves shown in Figure 1 represent four para-aramid yarns. However, metallic wire tensile tests conducted by manufacturers generate similar data. In fact, metallic-wire properties are more easily found because the metals are homogeneous, isotropic, and elastic (and not undergoing a chemical reaction). Tensile tests with heavily work-hardened metallic wire produce a straight-line curve similar to that in Figure 1. The slope of the line defines the force-to-elongation relationship between tension and elongation. The slope is a function of the modulus of elasticity, and spring rate is given by:

K = R/∆L.

So for wire reinforcement the spring rate is change in force divided by strain:

K = (E × A)/L0,

and for yarn:

YTM = R (L0 × ∆L).

Consequently, for wire, the tensile load carried by a single strand is estimated by:

R = (E × A)/L0 × ∆L.

The point is that the wire spring rate is sensitive to the original ligament length because of the wire’s fixed modulus of elasticity and cross-sectional area. Therefore, whether yarn or wire, the original ligament length is the only design variable that impacts load sharing.

Equal load sharing in reinforcement requires each strand to carry an equal portion of the load created by the fluid pressure. However, this only happens if each strand carries the same tension, and the ligaments must start out at the same length.

One consequence of the spring-rate concept is that the shorter the original ligament length, the stiffer the spring. This means shorter ligaments become loaded quicker, making them first to break. Ideally, all ligaments should load at the same rate; to accomplish this, all ligaments must be the same length.