CivilReinforced ConcretemediumCourseworkPE / Advanced

Nominal Axial Capacity of a Tied Column

A square tied column is $400\ \text{mm} \times 400\ \text{mm}$ and reinforced with a total steel area $A_{st} = 4000\ \text{mm}^2$ .

Material strengths: $f'_c = 30\ \text{MPa}$ , $f_y = 415\ \text{MPa}$ .

Find the nominal concentric axial capacity $P_o$ .

Given

Ag	160000 mm^2	Gross section (400 x 400)
Ast	4000 mm^2	Total longitudinal steel area
f'c	30 MPa	Concrete strength
fy	415 MPa	Steel yield strength

Hint 1

Concrete and steel both carry load — sum their contributions.

Hint 2

Careful with the concrete area: the steel occupies part of the gross section.

Hint 3

P_o = 0.85f'_c(A_g − A_st) + f_y·A_st.

Worked solution — try the problem first

$P_o = 0.85 f'_c (A_g - A_{st}) + f_y A_{st}$

Concrete contribution. Note the steel area is subtracted from the gross — the steel displaces concrete, and counting both would double-count that region:

$A_g - A_{st} = 160{,}000 - 4000 = 156{,}000\ \text{mm}^2$ $0.85(30)(156{,}000) = 25.5 \times 156{,}000 = 3{,}978{,}000\ \text{N}$

Steel contribution: $415 \times 4000 = 1{,}660{,}000\ \text{N}$

Total: $P_o = 3{,}978{,}000 + 1{,}660{,}000 = 5{,}638{,}000\ \text{N}$

$\boxed{P_o = 5638\ \text{kN}}$

In design you would not use $P_o$ directly. Code limits a tied column to $\phi P_{n,max} = 0.65 \times 0.80 \times P_o \approx 2932$ kN, where the $0.80$ accounts for accidental eccentricity and $0.65$ is the compression-controlled resistance factor.

Concepts:Reinforced concrete columnsAxial capacityComposite action

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