General Design Principles for DuPont Engineering Polymers - Module

Form of section Area A
d
d
R
R
1 1
b
B
b
1 1
y 1
y 2
R R0
1
R R0
1
y 1
y 2
2
y1 R 1 y 1
2
2
2
R 1 1
αα
1
R
R
R
t
1
R
α
2
2
2
2
α
α α
2
y 1
y 2
y 1
1
y 2
y 1
1
y 2
t
(1)
(2)
(3)
(1) Circular sector
(2) Very thin annulus
(3) Sector of thin annulus
A = 1 bd
2
A = 1 (B + b)d
2
Table 4.01. Properties of Sections (continued)
y 1 = 2 d
3
y 2 = 1 d
3
y 1 = d
y 2 = d
A = πR 2 y 1 = y 2 = R
A = π(R 2 – R 2 )
0
A = 1 π R 2
2
A = α R 2
A = 1 R 2 (2α
2
– sin 2α)
A = 2 πRt
A = 2 αRt
Distance from
centroid to extremities
of section y1, y2
2B + b
3(B + b)
B + 2b
3(B + b)
17
Moments of inertia I1 and I2
about principal
central axes 1 and 2
I 1 = 1 bd 3
36
I1 = d3 (B 2 + 4Bb + b 2 )
36(B + b)
I = 1 πR 4
4
y 1 = y 2 = R I = 1 π(R 4 – R 4 )
y 1 = 0.5756R
y 2 = 0.4244R
y1 = R 1 –
2 sin α
� 3α �
y2 = 2R
sin α
3α
y1 = R 1 –
4 sin3 α
� 6α – 3 sin 2α �
y 2 =
R
4 sin3 α
� – cos α
6α – 3 sin 2α
�
r 1 =
Radii of gyration r1 and r2
about principal
central axes
r 1 = 0.2358d
r = 1 R
2
d
6(B + b)
r = 4 0 √ (R2 + R 2 1 )
4
0
I 1 = 0.1098R 4
I2 = 1 πR 4
8
I1 = 1 R 4 α + sin α cos α
4 �
I2 = 1 R 4 �α– sin α cos α�
4
I 1 = R4
α – sin α cos α
4 �
+ 2 sin 3 α cos α
–
– 16 sin 2 α
9α
16 sin 6 α
9(α – sin α cos α
4
I2 =
R
(3 � – 3 sin � cos �
12
– 2 sin 3 � cos ��
r 1 = 0.2643R
r 2 = 1 R
2
y 1 = y 2 = R I = π R 3 t r = 0.707R
y 1 = R
1 – sin α
� α �
y1 = 2R sin α � – cos α
α
�
I1 =R 3t α + sin α cos α – 2 sin2 � α �
α
I 2 =R 3 t (α – sin α cos α)
�
�
r 1 =
√ 2(B2 + 4Bb + b 2 )
1 R√ 1 + sin α cos α 16 sin2 α
2 α 9α 2
r2 = 1 R 1 –
sin α cos α
2 √ α
r 2 = 1 R √ 1 + 2 sin3 α cos α
2 � – sin α cos α
–
64 sin6 α
9(2α – sin 2α) 2
r 2 = 1 R√ 1 – 2 sin3 α cos α
2 3(α – sin α cos α)
r 1 =
R √ α + sin α cos α – 2 sin2 α/α
2α
r 2 = R √
α – sin α cos α
2α