# How to design axial strength load of the column

This is another useful presentation by S.L. Khan, the renowned civil engineer. In this construction video tutorial, Mr, Khan briefly explains how to design the axial load strength of the column by applying the following formula :-

ØPn = Øα[0.85 fc´ (Ag-Ast) + fy Ast]

ØPn stands for design axial load strength of the column

Øα stands for design capacity (for tie column α = 0.80, Ø = 0.65)
(for spiral column α = 0.85, Ø = 0.75)

fc´ stands for grade of concrete alias strength of concrete

fy stands for grade of steel alias strength of steel

Ag stands for gross sectional area of the column

Ast stands for area of steel in section

In this video, a solution is given to the following problem :-

Example : Determine the design axial load strength for the square tie column provided. Assume fc´ = 4ksi and fy = 60 ksi

The sizes of the columns are 14 x 14 inches. The primary longitudinal bars are 8#6

The size of the tie alias ring is #3@12” c/c (center to center)

The concrete cover is 1.5 inches

Based on the above dimensions, the solution is given as follow :-

ØPn = Øα[0.85 fc´ (Ag-Ast) + fy Ast]

Design capacity = α = 0.80, Ø = 0.65

Ag (gross cross sectional of the column) = 14 x 14 = 196 in2 (square inches)

Ast (area of steel and steel is round here) = πD^{2}/4

Since the main bars are 8 in numbers , multiply the formula with 8 i.e. πD^{2}/4 x 8

We know diameter is no. 6 bars i.e. 6/8 inches

So, π(6/8)^{2}/4 = 3.535 in^{2}

After putting all the values, we get the following :-

ØPn = 0.65(0.80)[0.85 x 4 (196 – 3.535) + 60 x 3.535] = 450.57 kips

To get more detail, go through the following video tutorial presented by the renowned civil engineer, SL Khan.

Video Source: SL Khan