What data should the maker give you?
Two numbers, in addition to dimensions: an ultimate retainer strength and a ball push-out force. SYZ's weld-on ball joint guide lists both for each configuration it has produced, and the range across them is wide, from 20,000 to 162,000 lb for retainer strength and from 14,600 to 68,600 lb for push-out force. Ask for the same two values, and for the test method behind them, for any weld-on part you consider. Size from the housing data and the weld package, not from the pin diameter alone: the pin can be strong enough while the housing or the ball retention is not.
Which load do you size to?
The worst load that can really happen, in each direction. Shock is a separate case from the steady load: a jam, an impact, a hard stop or a sudden reversal produces a short spike that can be several times the working load. An implement arm that swings and bumps is a varying-direction case with shock. Write down the normal load, the peak load and the abnormal case, and size to the highest one that can occur.
How do you choose the safety factor?
By how the load varies, and on the right basis. Two kinds of strength figure are in use. A limit load is where permanent deformation starts, and an ultimate load is where the part breaks. The ultimate is usually, but not always, about 1.5 times the limit load. Typical lower limits for the static safety factor are 2 to 3 for a constant load in a constant direction, 3 to 5 for a fluctuating load in a constant direction, and 5 to 8 for loads in varying directions (the figures THK publishes for rod ends). Those factors are applied to the static capacity, which is a guide to mechanical strength. They are not applied to a breaking load. Mixing the bases is the error to avoid, so state which one you used (K01-Q1-02 sets out the same rule for rod ends).
For a weld-on housing that is rated by ultimate strength, one way to keep the basis honest is to convert. Treat the factor as applying to a limit-level figure, and estimate that figure as the ultimate divided by 1.5. Then the ultimate you need is the working load times the factor times 1.5. Example, with illustrative numbers: a worst-case working load of 6,000 lb (26.7 kN) that changes direction. A factor of 5 asks for an ultimate retainer strength of at least 6,000 × 5 × 1.5 = 45,000 lb, and a factor of 8 asks for 72,000 lb. The 1.5 is the usual ratio, not a guarantee, so if the maker gives both figures, use the maker's. An ultimate or proof load is a test threshold, not a load the part should work at.
Cyclic loads: why is a static safety factor not enough?
Because a welded joint can fail by fatigue at a load far below its static strength. Fatigue life depends on three things: the stress range in each cycle (not the peak), the number of cycles, and the detail of the weld. The weak spots in a welded socket are the weld toe, where the weld meets the base steel, and the weld root. Both are stress concentrations, and cracks start there (see Q1-07). To a first approximation the fatigue strength of an as-welded detail does not rise with the strength of the steel, so a stronger steel does not fix a weld that is fatiguing.
Structural welding standards deal with this by classing weld details into fatigue categories. Each category gives an allowable stress range for a given number of cycles. Look at the fatigue provisions of the welding code that governs your structure, and at the weld detail in the drawing. Which category applies depends on the weld type, the access and the loading direction, and that decision belongs to the designer and the welding engineer. A load that reverses direction, such as an implement arm pushing and pulling, gives the largest stress range for a given peak.
Is the weld itself strong enough?
The load path runs from the pin to the ball, the housing, the weld and the structure, and the weld is the link most often left unchecked. The simplest check is the shear stress on the throat of a fillet weld that runs all the way round the housing. The throat is 0.707 times the leg of the weld, the throat area is the throat times the weld length, and the stress is the load divided by that area.
Example, with illustrative geometry only: a housing 60 mm in diameter welded all round with a 6 mm fillet. The throat is 0.707 × 6 = 4.24 mm, the weld length is π × 60 = 188.5 mm and the throat area is about 800 mm². For the 6,000 lb (26.7 kN) working load the shear stress is 26,689 ÷ 800 = 33 MPa. If the load swings from zero to 6,000 lb every cycle, the stress range is 33 MPa. If it reverses fully, from +6,000 to −6,000 lb, the range is about 67 MPa. A larger leg lowers the stress in proportion: 8 mm gives 25 MPa. The 6 mm leg and the 60 mm diameter are only an example, not recommended sizes.
Compare the result with two limits from the applicable code: the allowable stress for the weld metal and base steel under static load, and the allowable stress range for the fatigue category of the detail at the number of cycles you expect. This calculation assumes the load is shared evenly around the weld. For a load across the housing axis, such as the normal working load on a lift-arm pin, the shear around a ring weld is not uniform: its peak is up to about twice the average, which the usual weld-group check absorbs in its allowable stress. In practice a side load or a bending moment concentrates the stress on one side, and the toe has a stress concentration that a nominal stress does not show, so treat the figure as a screening check and let a qualified person finish the design. Weld size, procedure and process are not set here (Q1-04).
Check the pin too
The pin passing through the ball is in double shear, and the shear stress is 2F ÷ (π × d²) (K01-Q1-02). For the 26.7 kN working load used above, a 25.4 mm pin sees 26 MPa and a 28.6 mm pin sees 21 MPa. A bending load on the pin and the lug it sits in is a separate check.
What about combined radial, axial and side loads?
Check each direction against the maker's data, and keep the load in line where you can. For a ball on a pin, the ball push-out force speaks to a load along the pin axis, which is a side load on the arm. For a ball stud, the equivalent is the pull-out along the stem (Q1-01 explains the two constructions). A load across the axis adds bending to the pin and the housing, and moves the weld stress to one side.
What do you give the maker?
The worst-case load in each direction, how often it reverses, the shock you expect, the number of cycles in the life you need, the pin size and the plate or tube it will be welded into, and what happens if it fails.
What changes on a self-built chassis or a race vehicle?
The load in a link is not the weight of the vehicle. It comes from the wheel or axle load, scaled by the suspension's lever (motion) ratio, by acceleration and braking, and by bumps and landings. The multiplier has to come from measurement or the chassis designer. Our motion ratio and link load calculator shows how geometry changes the force in a link.
An illustration with assumed numbers, for a link that carries a corner load directly, such as a pushrod or a shock link: a 3.0 kN static corner load and a peak of three times that gives 9.0 kN. A link standing 20° off the line of that force carries 9.0 ÷ cos 20° ≈ 9.6 kN along itself. The 3× peak is a load estimate, not a safety factor, and 9.6 kN is a load, not a capacity: the safety factor and basis from the sections above still apply on top, or the chassis designer's factor. Such a link works in tension and compression, which is the reversing case above, so the weld sees the larger stress range. In compression it must also resist buckling; see the tube buckling calculator.
The rest of this article carries over. For a link in a steering or suspension system, the chassis designer sizes it, and the welding must follow the vehicle maker's or event organiser's current rules. This article makes no claim that any part meets them.