## Overview

The purpose of this example is to verify the concrete column design in RAM Structural System for the Australian standard AS3600-09. This example is based on the simple column and beam frame shown in Figures 1 through 3. The column under consideration is 500 mm square and is at the first level of a simple four-storey beam-column space frame.

*Figure 1 - Elevation view of beam-column frame with column under consideration circled.*

*Figure 2 - Three-dimensional view in RAM Structural System of beam-column frame with column under consideration circled.*

*Figure 3 - Column section showing reinforcing used in design example.*

## Criteria and Properties

- Design according to AS3600-09
- (4) storeys, 3500 mm storey-to-storey height typical
- 500 mm x 500 mm cross section
- f'
_{c}= 35 MPa, f_{y}= 460 MPa - Vertical reinforcing: (16) N16, typical
- Shear reinforcing: N10@350, typical
- Bar cover = 30 mm
- Slenderness reduction factor k = 1.0
- Column braced in both principal column directions at each storey
- Column resists axial load only

Various sections of the column design report in RAM Structural System are shown throughout this example to compare the verification calculations to results produced in RAM Structural System. The figure below shows the section of the report that summarises the properties and geometry used by the program.

*Figure 4 - Column design report in RAM Structural System showing properties and geometry used by the program.*

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## Loading

Dead Load

- Self-weight of 200 mm slab = 24 kN/m
^{3}x 200 mm = 4.80 kPa - Super-imposed dead load (SIDL) = 1 kPa

Live Load

- LL = 5 kPa (unreducible)
- Tributary area to column of 8x8 m

Column load per storey:

G = 8 x 8 x (4.8 + 1) = 371.2 kN

Q = 8 x 8 x 5 = 320 kN

Combined load:

1.35G = 1.35 x 371.2 = 501.1 kN

1.2G + 1.5Q = 1.2 x 371.2 + 1.5 x 320.0 = 925.4 kN

Controlling load combination is 1.2G + 1.5Q

Axial design force at the ground level:

N* = 4 x 925.4 = 3701.8 kN

## Design Methodology

The design procedure provided in Clause 10.2.1 is utilised. Accordingly, Clause 10.3.1 is used to classify the column as either short or slender, which will dictate the clauses to be used for design. As noted in *Criteria and Properties* in this document, the column is braced in each principal direction. Thus, use Clause 10.3.1(a), which states that a column is short if L_{e}/r is less than the greater of:

- 25
- alpha
_{c}(38 – f^{’}_{c}/15)(1 + M^{*}_{1}/M^{*}_{2})

where

alpha_{c }= sqrt(2.25 - 2.5N*/0.6Nuo) for N^{*}/0.6N_{uo} ≥ 0.15

alpha_{c }= sqrt(1/(3.5N*/0.6Nuo)) for N^{*}/0.6N_{uo} < 0.15

In this example there is no moment acting within the column. Per 10.3.1(b), when the absolute value of M^{*}_{1}/M^{*}_{2} is less than or equal to 0.05DN^{*}, the ratio shall be taken as zero. Thus the value of 25 will control the check of L_{e}/r.

L_{ux} = 3.5 m

k_{x} = 0.9 (restrained by beams as directed in Clause 10.3.1(b))

L_{ex} = k_{x}L_{ux} = 3.15 m

r = 0.3D = 0.3 x 500 mm = 0.15 m (per Clause 10.5.2)

L_{e}/r = 3.15/0.15 = 21 ≤ 25

Therefore the column is short and per Clause 10.2.1(a), the design shall be in accordance with Clauses 10.3, 10.6, and 10.7.

## Evaluation of Axial and Flexural Strength

Although there are no moments acting in the column, all types of columns are required to be designed for the minimum bending moment prescribed by Clause 10.1.2.

M*x = N* x 0.05 x D = 3701.8 x 0.05 x 0.5 = 92.5 kN

The column is evaluated for biaxial bending and compression per Clause 10.6.4:

(M*x/phiMux)^alpha_n + (M*y/phiMuy)^alpha_n <= 1.0

where

alpha_n = 0.7 + (1.7N*)/(0.6Nuo), 1 <= alpha_n <= 2

Determine the squash load N_{uo} per Clause 10.6.2.2:

alpha_n = 1 - 0.003f'c = 1 - 0.003 x 35 = 0.895, 0.72 <= alpha_1 <= 0.85

alpha_n = 0.85

Nuo = alpha_1 f'c Ac + fsyAs

= 0.85 x 35 x (500^2 - pi x 16^2/4 x 16) + 460 x pi x 16^2/4 x 16

= 8821.6 kN

alpha_n = 0.7 + 1.7 x (3701.8 / (0.6x8821.6)) = 1.89

*Figure 5 - Column design report in RAM showing combined loads and squash load calculation.*

Next, the design strength in bending under the design axial force N^{*}, phiM_{ux }and phiM_{uy}, must be determined. These values are calculated by considering the equilibrium of the cross section under a given axial load, with the assumption that the extreme fiber in compression is strained to the ultimate value for concrete of 0.003. Given these parameters, the location of the neutral axis of the section can be solved for such that the sum of the compression and tension resultants equilibrates the applied axial force. Summing the moments about the neutral axis then gives the design bending strength, phiM_{u}, coincident with N*.

The section strains and force resultants for the case of this analysis where N^{*} = phiN_{u} = 3,701.8 kN are shown in the diagram below.

*Figure 6 - Section strains and concrete and steel force resultants for the column cross section under an applied axial load of 6,169 kN.*

Resolving the forces and moments in this diagram gives:

Nu = 6,169 kN (as expected, equilibrating the applied forces)

Mu = 464.4 kN-m

Thus,

phiNu = 3,701.8 kN

phiMu = 278.6 kN-m

More generally, a plot of Nu vs. Mu can be generated by varying the load Nu, solving for the neutral axis location that achieves equilibrium at the state of ultimate concrete strain, and determining the corresponding moment Mu.

*Figure 7 - Plot of axial capacity versus flexural capacity for column.*

*Note: The spreadsheet used to generate the curve shown in Figure 7 can be downloaded at the following link:*

Now finalise the check for biaxial bending and compression per Clause 10.6.4:

(M*x/phiMux)^alpha_n + (M*y/phiMuy)^alpha_n <= 1.0

The minimum design moments are applied about both axes simultaneously,

2*(92.6/278.6)^1.89 = 0.249 < 1.0

Therefore design section passes biaxial bending checks.

*Figure 8 - Column design report in RAM Structural System showing phiMu.*

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*Figure 9 - Design results dialog in RAM Structural System showing reinforcing selected at storey in question and results of biaxial interaction evaluation.*