A compression spring in a valve that sagged. The valve was a pressure relief valve in a hydraulic system. The spring was supposed to hold the poppet closed at 100 bar. After six months, the valve cracked open at 80 bar. The spring had lost 20% of its free length. The maintenance team ordered the same spring from the same catalog number. The new spring also sagged. The spring design was wrong, not the material or the heat treat.
The spring rate is the easy part
The spring rate k for a round-wire helical compression spring is:
k = (G × d⁴) / (8 × D³ × Na)
Where G is the shear modulus of the wire material (79,000 N/mm² for spring steel), d is the wire diameter, D is the mean coil diameter, and Na is the number of active coils. For a spring with d = 3 mm, D = 20 mm, Na = 8, the rate is:
k = (79000 × 81) / (8 × 8000 × 8) = 6,399,000 / 512,000 = 12.5 N/mm
That spring compresses 8 mm per 100 N of load. The relief valve needed 15.9 N/mm to hold 100 bar on the poppet area. The installed spring was 12.5 N/mm. The valve was undersprung from day one. It cracked open at about 80 bar — close to what the spring math says.
But the rate wasn’t the failure. The sag was. The spring rate didn’t change with time — the spring just got shorter. The free length dropped from 80 mm to 64 mm. The preload dropped because the spring was shorter. The valve cracked open early.
Why springs sag
Spring sag is permanent set. The wire yields in torsion under the operating stress. If the design stress is too high for the material, the spring takes a permanent set the first few times it’s compressed. A properly designed spring is stressed below the material’s elastic limit at full compression — typically below 45% of the tensile strength for a hardened spring steel, with a safety factor built in.
The sagging spring was operating at 62% of the wire’s tensile strength at full stroke. That’s too high for a long-life spring. The design rule: the operating stress should be under 45% of tensile strength for a spring that cycles more than 100,000 times. The valve cycled about 30 times a day — 6,500 cycles a year. That’s a “static” spring by cycle count, so the 45% rule relaxes to 55%. But the spring was at 62%.
Checking the stress
The stress in a compression spring is torsion in the wire. The formula, with the Wahl correction factor K_w accounting for curvature:
τ = K_w × (8 × F × D) / (π × d³)
K_w = (4C – 1)/(4C – 4) + 0.615/C, where C = D/d is the spring index.
For the failing spring: F = 600 N at full stroke, D = 20, d = 3, C = 6.67. K_w = (26.7-1)/(26.7-4) + 0.615/6.67 = 1.19 + 0.09 = 1.28.
τ = 1.28 × (8 × 600 × 20) / (π × 27) = 1.28 × 96,000 / 84.8 = 1,449 N/mm²
The wire tensile strength (music wire, 3 mm) is about 2,330 N/mm². The stress ratio is 62%. Over the elastic limit for a spring that needs to hold its length. The spring set.
The fixed spring used d = 3.5 mm wire, D = 20 mm, Na = 7. The rate went up to 19.4 N/mm, the stress at 600 N dropped to 1,080 N/mm² (46% of tensile), and the spring held its length. The valve holds 100 bar. No sag in two years.
Buckling: the failure nobody checks
A compression spring that’s too slender buckles sideways when compressed. The rule: if the free length L0 divided by the mean diameter D is more than about 4 (with the ends not guided), the spring buckles. For the failing spring, L0/D = 80/20 = 4.0. It was right at the edge. The spring was installed inside a valve bore that guided it — so it didn’t buckle. But a spring of the same proportions installed on a rod without a guide would bow sideways and rub against the machine.
The fix for a slender spring: either guide it (as this one was) or reduce the free length, or increase the diameter. A spring with L0/D over 4 should be designed to run inside a bore or over a rod with a diameter clearance of about 10% of D. The guide prevents the lateral movement.
Ends and their effect on rate
Closed and ground ends change the active coil count. A spring with 8 total coils and closed-ground ends has 6 active coils (Na = Nt – 2). The rate formula uses active coils. If you count total coils instead, you get a rate that’s 25% too high. Most catalog springs are closed and ground. Read the catalog table carefully — the rate is quoted for the finished spring, not for the theoretical active coils.
Also, the end coils (closed, unground) don’t contribute to the rate but do contribute to the solid height. The solid height (all coils touching) determines whether the spring can reach full compression without coil binding. The failing spring had a solid height of 24 mm. The valve full stroke compressed it to 30 mm — no binding. Fine.
The design checklist
- Calculate the required rate from the load and deflection.
- Pick wire diameter and mean diameter to get the rate. Use the rate formula, not a guess.
- Check the stress at full stroke with the Wahl correction. Keep it under 45% of tensile for cyclic, 55% for static.
- Check the solid height — the spring must not coil-bind before full stroke.
- Check buckling — L0/D under 4, or provide a guide.
- Specify the end type (closed, closed-ground, plain) and the finish (shot-peened for cyclic springs).
The sagging valve spring failed all the checks except coil binding. It was undersprung, overstressed, and borderline slender. The catalog said “valve spring.” The catalog didn’t say what stress it would run at in this application.
A spring is a stress calculation, not a catalog pick. Check the rate, the stress with the Wahl factor, the solid height, and buckling before you install it. The relief valve that cracked open at 80 bar wasn’t a worn spring — it was a spring that was 20% undersprung and 17% overstressed from the day it was installed. The same catalog number will sag the same way. Do the math on the next one.
