Check minimum wall thickness of pipe, Fig. 45:t > D -*- = 6.625 in /
36 ksi
f. = 0.083 in (0.21 cm)
V 8£ V^8 x 29,000 ksi v 't = 0.280 in > 0.083 in o.k.Check minimum cross-sectional area of steel pipe:
A 8 = <n(&-RD^(I?-Of)= -^[(6.625 in)
2- (6.065 in)
2
] = 5.6 in
2
(36.1 cm
2
)
A 0 = TTfl
2
= -^A
2 =
T-
x
(
6- 065 in
)
2 = 28- 9 in2
O
86- 5 cm2
4 4 )
- 5 cm2
- 9 in2
-V- ^.
5
;6i
"2
^ + ^^0.^2 = o.i6>4% O.k.
c 5.6 in2
+ 28.9 in
2- Analyze the selected column
In the absence of reinforcing bars:
Fmy = Fy +C 2 K^f-
ASA 0
Em = E + c,Ec-j-
Aswhere E 0 = w^1 -^5 V^, C 2 = 0.85, C 3 = 0.4.
The modulus of elasticity of the concrete
£C=145L5 VI? = 3267 ksiThe modified yield stress for composite design is
Fmy = 36 ksi + 0.85 x 3.5 ksi x^28 '^9 ."fFIGURE 45
outside diameter Dinside diameter Dr