SME
Steel Design · Problem type
Block shear
Use this for “A bolted end has Agv = 4.5 in², Aₙᵥ = 3.2 in² and Aₙₜ = 1.1 in², with Fᵧ = 50 ksi and Fᵤ = 65 ksi. Find its block shear strength.”
Diagram
Lengths are in the areas’ ratio: the holes take Agv − Aₙᵥ = 0.75 in² from the shear plane. φRₙ = 0.75 × min(0.6FᵤAₙᵥ + UbsFᵤAₙₜ, 0.6FᵧAgv + UbsFᵤAₙₜ) = 0.75 × min(121.8, 108.3) = 81.2 kips.
This is an example. Type your own values to start a new problem.
Rᵥᵣ (kip)
=
0.6
×
Fᵤ (ksi)
×
Aₙᵥ (in²)
Rᵥᵧ (kip)
=
0.6
×
Fᵧ (ksi)
×
Agv (in²)
φRₙ (kip)
=
0.75
×
(min(
Rᵥᵣ (kip)
,
Rᵥᵧ (kip)
)
+
Ubs
×
Fᵤ (ksi)
×
Aₙₜ (in²)
)
Rₜ
Tension rupture
worked out (not typed) · kip
Formulas
Rᵥᵣ = 0.6 × Fᵤ × Aₙᵥ
78.3 = 0.6 × 58 × 2.25
Rᵥᵧ = 0.6 × Fᵧ × Agv
64.8 = 0.6 × 36 × 3
Rₜ = Ubs × Fᵤ × Aₙₜ
43.5 = 1 × 58 × 0.75
φRₙ = 0.75 × (Rᵥᵧ + Rₜ)
81.225 = 0.75 × (64.8 + 43.5)
Assumptions
•
AISC 360-22 J4.3, LRFD with φ = 0.75: the end tears out along its shear and tension planes.
•
The shear plane gives the smaller of rupture on its net area and yielding on its gross area.
•
Ubs = 1 when the tension stress is uniform, as with one line of bolts; 0.5 when it is not.
•
Each area is the plane’s length, less any holes on it, times the plate thickness.
Step-by-step
WE KNOW
Agv = 3 in², Aₙᵥ = 2.25 in², Aₙₜ = 0.75 in², Fᵧ = 36 ksi, Fᵤ = 58 ksi, Ubs = 1
FIND
shear rupture (Rᵥᵣ), shear yielding (Rᵥᵧ), tension rupture (Rₜ), block shear strength (φRₙ)
Step 1 · Find shear rupture (Rᵥᵣ)
Use Rᵥᵣ = 0.6 × Fᵤ × Aₙᵥ
Shear rupture: the net shear plane, holes taken out, tears at 0.6Fᵤ.
Rᵥᵣ = 0.6 × 58 × 2.25
Rᵥᵣ = 34.8 × 2.25
Rᵥᵣ = 78.3 kip
Step 2 · Find shear yielding (Rᵥᵧ)
Use Rᵥᵧ = 0.6 × Fᵧ × Agv
Shear yielding: the whole shear plane, holes included, yields at 0.6Fᵧ.
Rᵥᵧ = 0.6 × 36 × 3
Rᵥᵧ = 21.6 × 3
Rᵥᵧ = 64.8 kip
Step 3 · Find tension rupture (Rₜ)
Use Rₜ = Ubs × Fᵤ × Aₙₜ
Tension rupture: the net tension plane tears at Fᵤ, scaled by Ubs.
Rₜ = 1 × 58 × 0.75
Rₜ = 43.5 kip
Step 4 · Find block shear strength (φRₙ)
Use φRₙ = 0.75 × (Rᵥᵧ + Rₜ)
The block tears along the weaker of the two shear limits plus the tension plane; φ = 0.75.
78.3 > 64.8: shear yielding governs
φRₙ = 0.75 × (64.8 + 43.5)
φRₙ = 0.75 × 108.3
φRₙ = 81.225 kip
Check
Put the numbers back in. Both sides should match.
78.3 = 0.6 × 58 × 2.25 ✓
64.8 = 0.6 × 36 × 3 ✓
43.5 = 1 × 58 × 0.75 ✓
81.225 = 0.75 × (64.8 + 43.5) ✓
Topic check
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