Page 40 - Civil Engineering Formulas

Radius of gyration d r = 4 2 D 2 + d 2 R 4 + r 2 = 2 2

Section modulus A πr 3 r = = 4 4 (approx) D 4 – d 4 D R 4 – r 4 R (approx) s is very small d m

= πd 3 32 0.1d 3 = π = 32 π = 4 = 0.8d m 2s when
I c I c

Moment of inertia A πr 4 r 2 = = 4 4 (approx) (D 4 – d 4 ) (R 4 – r 4 ) (R 2 + r 2 ) = 0.05 (D 4 – d 4 ) (approx)

= πd 4 64 0.05d 4 = π = 64 π = 4 A = 4
I I R

d m = (D + d) s = (D – d)
r
Section d d 1 2 1 2
r
D

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