SME
Statics · Problem type
Radius of gyration
Use this for “Find the radius of gyration of a 20 × 100 mm bar about its strong axis.”
Diagram
A = 2,000 mm²
Ī = 1,667,000 mm⁴ about the centroidal axis
k
=
√(
I
A
):
the
area’s
spread
from
the
axis
This is an example. Type your own values to start a new problem.
A (mm²)
=
b (mm)
×
h (mm)
Ī (mm⁴)
=
b (mm)
×
(
h (mm)
)
3
12
k (mm)
=
Ī (mm⁴)
A (mm²)
Formulas
A
=
b
×
h
2,000 = 20 × 100
Ī
=
b
×
h
3
12
1.6667 ×
10
6
=
20
×
100
3
12
k
=
√(
Ī
A
)
28.8675
=
√(
1.6667 ×
10
6
2,000
)
Assumptions
•
k is about the horizontal centroidal axis: the distance at which all the area would give the same Ī.
•
For another section, type its Ī and A instead of b and h.
•
k feeds a column’s slenderness ratio L ÷ k.
Step-by-step
WE KNOW
b = 20 mm, h = 100 mm
FIND
area (A), moment of inertia about the centroidal axis (Ī), radius of gyration (k)
Step 1 · Find area (A)
Use A = b × h
A rectangle’s area is its width times its height.
A
=
20
×
100
A
=
2,000
mm²
Step 2 · Find moment of inertia about the centroidal axis (Ī)
Use Ī = b × h³ ÷ 12
A
rectangle
about
its
own
centroidal
axis:
Ī
=
∫
y
2
dA
over
the
area,
which
comes
to
bh
³
12
.
Ī
=
20
×
100
3
12
Ī
=
20
×
1,000,000
12
Ī
=
20,000,000
12
Ī
=
1.66667 ×
10
6
mm⁴
Step 3 · Find radius of gyration (k)
Use k = √(Ī ÷ A)
Divide
Ī
by
the
area,
then
take
the
square
root.
k
=
√(
1.66667 ×
10
6
2,000
)
k
=
√
833.335
k
=
28.8675
mm
Check
Put the numbers back in. Both sides should match.
2,000 = 20 × 100 ✓
1.66667 ×
10
6
=
20
×
100
3
12
✓
28.8675
=
√(
1.66667 ×
10
6
2,000
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.