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Steel Design
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Use this for “A lap splice has 6 Group A bolts, ⅞ in across, threads clear of the shear plane (Fₙᵥ = 68 ksi). Find the design shear strength φRₙ.”
Diagram
Ab
=
πd²
4
=
π
×
0.75
2
4
=
0.442
in².
One bolt: φrₙ = 0.75FₙᵥAb = 0.75 × 54 × 0.442 = 17.9 kips. φRₙ = n × φrₙ = 4 × 17.9 = 71.6 kips, every bolt cut once at the shear plane.
This is an example. Type your own values to start a new problem.
Ab (in²)
=
π
×
(
d (in)
)
2
÷
4
φrₙ (kip)
=
0.75
×
Fₙᵥ (ksi)
×
Ab (in²)
φRₙ (kip)
=
n
×
φrₙ (kip)
Formulas
Ab
=
π
×
d
2
4
0.4418
=
π
×
0.75
2
4
φrₙ = 0.75 × Fₙᵥ × Ab
17.8924 = 0.75 × 54 × 0.4418
φRₙ
=
n
×
φrₙ
71.5694 = 4 × 17.8924
Assumptions
•
AISC 360-22, LRFD: φ = 0.75 on the bolt’s nominal shear stress Fₙᵥ times its area.
•
One shear plane (a lap splice); standard holes; every bolt carries an equal share.
•
Group A bolts: Fₙᵥ = 54 ksi with threads in the shear plane, 68 ksi with them excluded.
•
Bearing and tear-out at the holes are checked separately; they may govern instead.
Step-by-step
WE KNOW
d = 0.75 in, Fₙᵥ = 54 ksi, n = 4
FIND
bolt area (Ab), strength of one bolt (φrₙ), strength of the group (φRₙ)
Step 1 · Find bolt area (Ab)
Use Ab = π × d² ÷ 4
The
shank
is
round,
so
its
cross-section
is
the
circle
of
diameter
d
.
Ab
=
π
×
0.75
2
4
Ab
=
π
×
0.5625
4
Ab
=
1.767
4
Ab = 0.4417865 in²
Step 2 · Find strength of one bolt (φrₙ)
Use φrₙ = 0.75 × Fₙᵥ × Ab
Each bolt is cut once, across its area Ab, at the stress Fₙᵥ; φ = 0.75.
φrₙ = 0.75 × 54 × 0.4417865
φrₙ = 40.5 × 0.4418
φrₙ = 17.89235 kip
Step 3 · Find strength of the group (φRₙ)
Use φRₙ = n × φrₙ
The
bolts
share
the
load
equally,
so
the
group
carries
n
times
one
bolt.
φRₙ = 4 × 17.89235
φRₙ = 71.5694 kip
Check
Put the numbers back in. Both sides should match.
0.4417865
=
π
×
0.75
2
4
✓
17.89235 = 0.75 × 54 × 0.4417865 ✓
71.5694 = 4 × 17.89235 ✓
Topic check
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