SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Bolt in single or double shear
Use this for “A 16 mm bolt holds 30 kN in double shear. Find the average shear stress in it.”
Diagram
A clevis: the middle plate pulls between two outer ones, so two cross-sections share P.
A
=
πd²
4
=
201.1
mm²
for
each
cut.
τ
=
P
2A
=
74.6
MPa.
This is an example. Type your own values to start a new problem.
A (mm²)
=
π
×
(
d (mm)
)
2
4
τ (MPa)
=
P (kN)
×
A (mm²)
Formulas
A
=
π
×
d
2
4
201.0619
=
π
×
16
2
4
τ
=
P
n
×
A
74.6039
=
30,000
2 × 201.0619
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
Single shear (a lap joint) cuts the bolt once, n = 1; double shear (a clevis) cuts it twice, n = 2.
•
τ is the average over each cut cross-section; bearing and bending of the bolt are left out.
•
The steps work in N, mm and MPa: 1 MPa is 1 N/mm².
Step-by-step
WE KNOW
P = 30 kN, n = 2, d = 16 mm
FIND
bolt area (A), average shear stress (τ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
P = 30 kN = 30,000 N (1 kN = 1000 N)
Step 2 · Find bolt area (A)
Use A = π × d² ÷ 4
The cross-section of a round bar is a circle: π times the diameter squared, over 4.
A
=
π
×
16
2
4
A
=
π
×
256
4
A
=
256π
4
A
=
201.0619
mm²
Step 3 · Find average shear stress (τ)
Use τ = P ÷ (n × A)
Each
shear
plane
cuts
one
bolt
cross-section,
so
the
load
spreads
over
n
areas.
τ
=
30,000
2 × 201.0619
τ
=
30,000
402.1238
τ
=
74.6039
MPa
Check
Put the numbers back in. Both sides should match.
201.0619
=
π
×
16
2
4
✓
74.6039
=
30,000
2 × 201.0619
✓
Topic check
See what you know: 5 questions on this topic
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