四分の一円の断面定数
O
y dy
dA=r cos dy
y=r sin
dz
2
z=rsino
dA=r coso dz
Y
Iz =
小
Ly = √
A =
·
πT
2
πC
πr2
[ dA = √² (r cos 0)² do = [12² (0 + ½ sin 20)]²
[±²(0 = **²
2
4
TC
Gz
=
= √ y dA = S
²r sin 0 (r cos 0)² do = [²(cos 0)³]}=}³
==
33
Gy
= √ z dA = √²
2
=
1/3
0
2
³r sin & (r cos )² = = = = (COS Q)²)
これらより,重心の位置(yo, zo)は,次のように
なる.
Gy
4r
yo =
=
=
Zo
A 3π
A 3π
Gz 4r
+5
π