15-1. Moment of inertia of area（1〜12）
1
Cross-sectional areaA
b*h=
b=
h=
Moment of inertia of area I
b*h
3
/12 =
Moment of inertiaZ
I/e=b*h
2
/6 =
Moment of inertia of area k
2
I/A=h
2
/12 =
（ｋ=0.289*h）
2
Cross-sectional area A
b*h=
b=
h=
Moment of inertia of area I
b
3
*h
3
／(6*（b
2
+h
2
）)
=
Moment of inertia Z=I/e
b
2
*h
2
／(6*（b
2
+h
2
）
0.5
)
=
Moment of inertia of area k
2
=I/A
b
2
*h
2
／(6*（b
2
+h
2
）)
=
3
Cross-sectional area A
b*h/2
=
b=
h=
Moment of inertia of area I
b*h
3
/36
=
Moment of inertia Z=I/e
e
1
=2*h/3
=
e
2
=h/3
=
Z
1
=b*h
2
/24
=
Z
2
=b*h
2
/12
=
Moment of inertia of area k
2
=I/A
h
2
/18
=
（ｋ=0.236*h）
4
Cross-sectional area A
3*3
0.5
*b
2
／2
=
b=
Moment of inertia of area I
5*3
0.5
*b
4
／16
=0.5413*b
4
=
Moment of inertia Z=I/e
5*b
3
/8
=0.625*b
3
=
Moment of inertia of area k
2
=I/A
5*b
2
/24
=
（k=0.456*b）
5
Cross-sectional area A
3*3
0.5
*b
2
／2
=
b=
Moment of inertia of area I
5*3
0.5
*b
4
／16
=0.5413*b
4
=
Moment of inertia Z=I/e
5*3
0.5
*b
3
／16
=0.5413*b
3
=
Moment of inertia of area k
2
=I/A
5*b
2
/24
=
（k=0.456*b）
6
Cross-sectional area A
h*（b+b
1
/2 ）
=
b=
b1=
h=
Moment of inertia of area I
(6*b
2
+6*b*b
1
+b
1
2
)*h
3
／(36*(2*b+b
1
))
=
Moment of inertia Z=I/e
e
1
=（3*b+2*b
1
）*h／(3*(2*b+b
1
）)
=
Z
1
=(6*b
2
+6*b*b
1
+b
1
2
)*h
2
／(12*(3*b+2*b
1
))
=
Moment of inertia of area k
2
=I/A
（6*b
2
+6*b*b
1
+b
1
2
）*h
2
／(18*（2*b+b
1
）
2
)
=
7
Cross-sectional area A
2.8284*r
2
=
r=
Moment of inertia of area I
(1+2*2
0.5
)*r
4
／6
=0.6381*r
4
=
Moment of inertia Z=I/e
0.6906*r
3
=
Moment of inertia of area k
2
=I/A
0.2256*r
2
=
（k=0.475*r）
8
regular polygon
n= Number of edges
a= Side length
r
2
= Radius of circumscribed circle
r
1
= Radius of inscribed circle
The axis shall pass through the center
Cross-sectional area A
n*a*r
1
/2
=
n=
a=
r
2
=
r
1
=
Moment of inertia of area I
A*（6*r
2
2
-a
2
）/24
=
Moment of inertia Z=I/e
I／[r
2
*cos(π/n)]≒A*r
2
/4
=
Moment of inertia of area k
2
=I/A
（6*r
2
2
-a
2
）/24
=
9
d=2r
Cross-sectional area A
d
2
*π/4 =π*r
2
=
d=
Moment of inertia of area I
d
4
*π/64
=
Moment of inertia Z=I/e
d
3
*π/32
=
Moment of inertia of area k
2
=I/A
d
2
/16
=
（k=0.5*r）
10
Cross-sectional area A
（d
2
2
-d
1
2
）*π/4
=
d
1
=
d
2
=
Moment of inertia of area I
（d
2
4
-d
1
4
）*π/64 =
Moment of inertia Z=I/e
[(d
2
4
-d
1
4
)／d
2
]*π/32 =
Moment of inertia of area k
2
=I/A
（d
2
2
+d
1
2
）/16 =
11
Cross-sectional area A
d
2
*π/8
（ｒ
2
*π/2） =
r=
Moment of inertia of area I
（9*π
2
-64）*r
4
／(72*π)
=0.1098*r
4
=
Moment of inertia Z=I/e
e=0.5756*r
=
Z=0.1908*r
3
=
（On the other sideZ
2
To）
Z
2
=0.2587*r
3
=
Moment of inertia of area k
2
=I/A
（9*π
2
-64）*r
2
／(36*π
2
)
=0.0697*r
2
=
（k=0.264*r）
12
Cross-sectional area A
π*a*b
=
a=
b=
Moment of inertia of area I
a
3
*b*π/4
=
Moment of inertia Z=I/e
a
2
*b*π/4
=
Moment of inertia of area k
2
=I/A
a
2
/4
=
References: "Handbook of Mechanical Designs Must-Have for Engineers" (Revised Edition by Saburo Kano) Applied Structural Mechanics Beams and Columns Page 132 Refer to A, I, Z, k of various cross sections.。