Question

Design a spiral column to carry a factored axial load of 375 ksi. Use f’c = 3 ksi and fy = 40 ksi. Use a reinforcement ratio p = 0.02

Answer 1

Design Loads, P_u = 375 k

ACI Provisions for Spiral Column (ACI 10.3.5.1)

$P_{u}=\phi P_{n(max)} = 0.85\phi [A_{st} f_{y}+0.85f'_{c}(A_{g}-A_{st})]$

fot Spiral Column,
0.70
, also $\rho=0.02\Rightarrow A_{st}=0.02A_{g}$

$375 = 0.85\times 0.70 [0.02A_{g} \times 40+0.85\times 3(A_{g}-0.02A_{g})]$

or,

so, for Circular Cross section , Diameter

  $D=\sqrt{\frac{4\times 191.04}{\pi }}=15.56inch$

provide 16 inch Diameter Circular cross section, so area of steel,so A_g,(prov)   $A_{g,(prov))}=\frac{\pi\times 16^{2}}{4 }=201.06inch^{2}$

so for area of steel,

$375 = 0.85\times 0.70 [40A_{st} +0.85\times 3(201.06-A_{st})]$

A_st = 3.14 inch² ,so provide 8 numbers of #6 bar

so, A_st,(prov) = 8x0.44 = 3.52 inch²

$\rho _{prov} = \frac{A_{st(prov)}}{A_{g(prov)}}\times 100 = \frac{3.52}{201.06}\times 100=1.75$

Consider clear cover of 1.5'' and Pitch of Spiral (S) = 1.5''

#[email protected]" -8 nos #6 13' 16"

so,

$A_{Core}=\frac{\pi\times 13^{2}}{4 }=132.73inch^{2}$

$\rho_{s}=\frac{4A_{sp}}{DS} =\frac{4\times 0.11}{13\times 1.5}=0.0225$

Check Spiral Ratio Against ACI requirements

$\rho_{s}=\frac{0.45f'_{c}}{f_{y}}[\frac{A_{g}}{A_{Core}}-1]=\frac{0.45\times 3}{40}[\frac{201.06}{132.73}-1]=0.0173$

Since,0.0225>0.0173,the column satisfy the minimum spiral reinforcement requirement.