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K05 Clevises & Yoke Ends

What Limits the Load of a Clevis Joint, and How Do You Check Pin Shear?

Type: Engineering reference Reading time: about 5 min Updated: 2026-11

What are the places it can fail?

Along the load path, from the rod to the bracket. Look at the complete load path: a rod end with enough radial capacity can still sit in an inadequate assembly if the pin bends, the clevis spreads, or the threads are not engaged far enough (K06-Q1-06).

  • The pin. Cut across (shear), pressed on the hole walls (bearing pressure), or bent.
  • The lugs and fork arms. Cracking or deformation around the pin holes under repeated shock. It is one of the results of a wrong clevis.
  • The thread. A documented loader-cylinder case: a generic clevis with no rated load and a mild steel pin. The thread loosened, the pin bore elongated and the rod end cracked at the thread root.
  • The rod. A long rod can buckle under push (K03-Q1-28).

How do you check the pin in shear?

A pin in a clevis is held at both fork arms and loaded by the part between them, so it is cut in two places. That is double shear. The average stress is the force, shared over two cross-sections:

τ = 2P ÷ (π × d²)

P is the force across the pin and d is the pin diameter. Atlas of Engineering uses this expression in a worked example.

Example: a 36 kN design load and a 20 mm solid pin (the same figures as in that worked example).

  • Area of the pin: π × 20² ÷ 4 = 314 mm², and two of them: 628 mm².
  • τ = 36,000 ÷ 628 = 57.3 MPa, which agrees with 2 × 36,000 ÷ (π × 400). The peak shear in a solid round pin is about 4/3 of the average (76 MPa here), so keep that in mind when you compare with an allowable value.

In this example the load already includes the safety factor. Compare the result with the shear strength of the pin material, divided by the factor you choose (Q1-06).

What about bearing pressure on the holes?

The pin also presses on the walls of the holes. The pressure is the force on that part divided by the pin diameter times the thickness of the part. In the same example, each fork arm is 12 mm thick and takes half the load: 18,000 ÷ (20 × 12) = 75 MPa. The central lug is 18 mm thick and takes all of it: 36,000 ÷ (20 × 18) = 100 MPa. A thin lug has higher pressure for the same pin, which is one reason lug thickness matters.

Why does the gap matter?

Because the pin bends. Move the lug away from the fork arms on each side, from a gap of 1 mm to 6 mm. The average shear stays at 57.3 MPa, but the pin's peak bending stress goes from about 264 MPa to about 378 MPa. The moment at the centre of the pin is M = P ÷ 8 × (b + 4g + 2t), where b is the lug thickness, g the gap on each side and t the arm thickness: 207,000 N·mm for the 1 mm gap and 297,000 N·mm for the 6 mm gap. The bending stress is then 32 × M ÷ (π × d³).

Take 300 MPa as an illustrative bending screening value for this example; not a material property or a code limit. On that screen the 6 mm gap fails while the 1 mm gap passes. The lesson holds without that number: a clevis that is loose side to side on its pin, or a lug that is much narrower than the fork opening, bends the pin more. Use spacers or a matching lug width (Q1-04).

What should you do?

Find the greatest force across the joint, push and pull, including shock (K06-Q1-02). Check the pin in double shear and in bending, with the actual gaps, and check bearing pressure on each hole. Then check the thread and the clevis body against the maker's rating; ask for it, because a generic part with no rated load is a risk. Last, choose the pin material and fit (Q1-06, K03-Q1-14).

What load does a pedal or vehicle control link put on a clevis?

On a pedal or lever, the load on the clevis is a person's force multiplied by the lever ratio, with a shock case on top (a foot slammed down, a pedal hit by a dropped tool).

An illustration with assumed numbers, not a design value. A 1,000 N peak foot force (a panic or abuse case can be higher) on a pedal with a 4:1 ratio gives 4,000 N in the pushrod. For an 8 mm pin in double shear, τ = 2 × 4,000 ÷ (π × 8²) ≈ 39.8 MPa on average, and about 53 MPa at the peak of a solid round pin (4/3 of the average). Bearing pressure on a 6 mm lug is 4,000 ÷ (8 × 6) ≈ 83 MPa. These are the same checks as above, and the gap check applies as well, because a pedal clevis often sits in a looser bracket than a machined cylinder mount.

SYZ's bolt shear capacity calculator screens an inch bolt in double shear; for a metric pin, use the formula above. The clevis rod ends page covers the part itself.

Do not read a rating from this example. The load that governs a pedal or pushrod joint, the safety factor and the acceptable parts are set by the vehicle maker, because a brake or clutch pushrod is a safety part.

FAQ

Is shear the only check on the pin?
No. Check bearing pressure on the holes and pin bending as well, with the real gaps.
Does this give me a load rating for my clevis?
No. The numbers are an example. A rating comes from the maker's data for your part.
Single shear or double shear?
A pin through a fork and a single lug is in double shear. A pin through an eye and one ear is in single shear, and carries the force over one cross-section.

Related reading

On this page
  1. What are the places it can fail?
  2. How do you check the pin in shear?
  3. What about bearing pressure on the holes?
  4. Why does the gap matter?
  5. What should you do?
  6. What load does a pedal or vehicle control link put on a clevis?
  7. FAQ
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