SME
Biomechanics · Problem type
Bending stress in a hollow bone
Use this for “A femur shaft is a tube 30 mm across with a 16 mm canal. What stress does 100 N·m of bending cause?”
Diagram
I
=
π(
rₒ
4
−
rᵢ
4
)
4
=
π(
15
4
−
8
4
)
4
I = 36,500 mm⁴
A
solid
rod
as
wide
has
I
=
39,800
mm⁴:
the
tube
keeps
91.9%
of
its
stiffness
with
71.6%
of
its
bone.
σ
=
Mrₒ
I
=
100,000
N·mm
×
15
mm
36,500
mm⁴
=
41
MPa
at
the
outer
edge,
zero
on
the
neutral
axis.
This is an example. Type your own values to start a new problem.
I (mm⁴)
=
π(
(
rₒ (mm)
)
4
−
(
rᵢ (mm)
)
4
)
4
σ (MPa)
=
M (N·m)
×
rₒ (mm)
I (mm⁴)
Isolid
Solid rod’s I
worked out (not typed) · mm⁴
kI
Stiffness kept
worked out (not typed) · %
kA
Bone kept
worked out (not typed) · %
Formulas
I
=
π
×
rₒ
4
−
rᵢ
4
4
36,544
=
π
×
15
4
−
8
4
4
σ
=
M
×
rₒ
I
41.05
=
100,000
×
15
36,544
Isolid
=
π
×
rₒ
4
4
39,761
=
π
×
15
4
4
kI
=
100
×
I
Isolid
91.9
=
100
×
36,544
39,761
kA
=
100
×
rₒ
2
−
rᵢ
2
rₒ
2
71.6
=
100
×
15
2
−
8
2
15
2
Formulas are worked in N·mm. The step-by-step shows the conversions.
Assumptions
•
The shaft is a round tube of elastic bone, bent like a straight beam.
•
Stress grows with distance from the neutral axis, so it is largest at the outer surface.
•
The solid rod for comparison is as wide as the bone, with no canal.
Step-by-step
WE KNOW
rₒ = 15 mm, rᵢ = 8 mm, M = 100 N·m
FIND
second moment (I), stress at the surface (σ), solid rod’s I (Isolid), stiffness kept (kI), bone kept (kA)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N·mm).
M = 100 N·m = 100,000 N·mm (1 N·m = 1000 N·mm)
Step 2 · Find second moment (I)
Use I = π × (rₒ⁴ − rᵢ⁴) ÷ 4
Take the canal’s share from the solid rod’s: the bone near the axis adds little stiffness.
I
=
π
×
15
4
−
8
4
4
I
=
π
×
50,625 − 4,096
4
I
=
π
×
46,529
4
I
=
46,529π
4
I
=
36,544
mm⁴
Step 3 · Find stress at the surface (σ)
Use σ = M × rₒ ÷ I
Bending
stress
is
My
I
,
and
the
outer
surface
is
y
=
rₒ
from
the
neutral
axis.
σ
=
100,000
×
15
36,544
σ
=
1,500,000
36,544
σ
=
41.05
MPa
Step 4 · Find solid rod’s I (Isolid)
Use Isolid = π × rₒ⁴ ÷ 4
A solid rod as wide as the bone, for comparison.
Isolid
=
π
×
15
4
4
Isolid
=
π
×
50,625
4
Isolid
=
50,625π
4
Isolid = 39,761 mm⁴
Step 5 · Find stiffness kept (kI)
Use kI = 100 × I ÷ Isolid
Bending
stiffness
goes
with
I
,
so
compare
the
tube’s
I
with
the
solid
rod’s.
kI
=
100
×
36,544
39,761
kI
=
3,654,400
39,761
kI = 91.9%
Step 6 · Find bone kept (kA)
Use kA = 100 × (rₒ² − rᵢ²) ÷ rₒ²
The ring’s area over the solid circle’s: the π cancels.
kA
=
100
×
15
2
−
8
2
15
2
kA
=
100
×
225 − 64
15
2
kA
=
100
×
161
15
2
kA
=
100
×
161
225
kA
=
16,100
225
kA = 71.6%
Check
Put the numbers back in. Both sides should match.
36,544
=
π
×
15
4
−
8
4
4
✓
41.05
=
100,000
×
15
36,544
✓
39,761
=
π
×
15
4
4
✓
91.9
=
100
×
36,544
39,761
✓
71.6
=
100
×
15
2
−
8
2
15
2
✓
Topic check
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