A bolt that sheared off in a machine bracket. The bracket held a 200 N load on a lever 200 mm from the bolt. The bolt was M8, 8.8 grade. The lever was steel plate welded to a column. The customer thought the bolt was too small. It was the right size for single shear — but the load was applied as a lever, creating both shear and prying. This is about bolted bracket shear design.
The shear load
The bolt sees a shear force from the lever:
F_shear = F · L_lever / L_bolt
Wait — no. The load is applied at the end of the lever. The bolt holds the bracket to the column. The shear force on the bolt equals the applied load (200 N). But the lever moment also creates a prying force. The bracket is 50 mm tall, with the bolt at the top. The 200 N load at 200 mm from the bolt creates a moment of 40,000 N·mm. This moment pries the bracket away from the column at the bottom edge. The prying force on the bolt:
F_pry = M / h
Where h is the distance between the bolt and the bottom edge (50 mm). F_pry = 40,000 / 50 = 800 N. That’s 4x the applied load! The bolt sees 200 N shear + 800 N prying tension. The combined load is much higher than just the shear.
The bolt stress
An M8 bolt (stress area 36.6 mm²) at 8.8 grade (yield 640 MPa). The prying tension creates axial stress: σ = 800 / 36.6 = 21.9 MPa. The shear stress: τ = 200 / 36.6 = 5.5 MPa. The combined stress (von Mises): σ_von = √(σ² + 3τ²) = √(480 + 91) = √571 = 23.9 MPa. That’s way below yield. Why did it break?
Because the bracket was a cantilever that flexed. Under cyclic loading (the lever was cycled 100,000 times), the bracket flexed. The bolt bent slightly each cycle. The bending fatigue at the bolt thread root caused failure. The bolt wasn’t in pure shear — it was in bending fatigue.
What I changed
1. Added a second bolt below. Two bolts spaced 50 mm apart share the prying moment. Each bolt sees half the prying force (400 N). The bending stress drops by half. The bolt survives 10x more cycles.
2. Added a support gusset. I welded a triangular gusset under the bracket. The gusset carries the moment directly into the column. The bolts only hold the bracket in place — they don’t carry the moment. The prying force drops to near zero. The bolts are now in pure shear (200 N). No bending. No fatigue.
3. Used a reamed bolt (interference fit). For shear-critical joints, I use a reamer bolt (ground to h7 tolerance) in a reamed hole (H7). The bolt fits perfectly — no clearance. The load transfers through the bolt cross-section, not through friction. The shear strength doubles vs a standard clearance bolt.
Single vs double shear
| Configuration | Shear planes | Shear area for M8 | Max shear load (8.8) |
|---|---|---|---|
| Single shear (bracket to column) | 1 | 36.6 mm² | 13 kN |
| Double shear (bracket between two plates) | 2 | 73.2 mm² | 26 kN |
For a bracket that must carry heavy load, I design it as double shear — the bracket plate fits between two lugs on the column. The bolt sees two shear planes. The load capacity doubles. This is standard for pivot joints and lever brackets.
The friction grip
For non-critical joints, the friction between the bracket and column carries the shear (not the bolt). The bolt preload creates normal force. Friction force = μ × F_preload. For M8 at 8.8 grade, preload is 21,000 N. μ = 0.2 (steel on steel). Friction = 4,200 N. That’s the shear capacity. But friction joints slip under shock. For cyclic loads, use reamed bolts in double shear — don’t rely on friction.
The load I check: not just the applied shear, but the prying moment from lever arms. A 200 N load at 200 mm creates 800 N prying. Add a gusset to carry the moment, or use two bolts. For shear-critical joints, use reamed bolts in double shear. The sheared bolt wasn’t undersized — it was bending-fatigued from prying.
