COLUMN SUPPORTING AN AXIAL LOAD AND UNIAXIAL BENDING EXAMPLE 3.9

An internal column sized 300 x 300 mm in a multi-storey building is subjected to an ultimate

axial load (NEd) of 1600 kN and bending moment, M of 60 kNm including the effect of

imperfections.

Design the column assuming fck = 30 N/mm2, fyk = 500 N/mm2 and nominal cover, c = 30 mm.

Solution:

1. Check the design moment, MEd.

Moment due to geometric imperfections, Mimp = ei. NEd

ei = max { lo/400, h/30, 20} mm

No need to calculate Mimp since it has already been considered.

2. Design of main reinforcement.

Use the design chart

a) Calculate d value.

Assume bar = 25 mm and link = 8 mm

d = h – c - link - bar/2

= 300 – 30 – 8 – 25/2 = 249.5 mm

b) Calculate d/h (to determine the chart to be used)

d/h = 249.5 / 300 = 0.83 0.85

c) Calculate N/bhfck and M/bh2fck

Eccentricity due

to imperfections

Minimum eccentricity

h d

COLUMN SUPPORTING AN AXIAL LOAD AND UNIAXIAL BENDING EXAMPLE 3.9

d) Determine the value of Asfyk / bhfck from the chart (point where the lines intersect).

From the chart,

Therefore, provide: 4H25 (1960 mm2)

e) Check the reinforcement provided with the minimum and maximum requirement.

As,max = 0.04 Ac = 0.04 (3002) = 3600 mm2

As,min (368) < As,prov (1960) < As,max (3600) OK!

0.074

0.59

Intersection

point

COLUMN SUPPORTING AN AXIAL LOAD AND UNIAXIAL BENDING EXAMPLE 3.9

3. Design links

a) Minimum link diameter = max { 6 mm, ¼ of the maximum longitudinal bar diameter)

i) 6 mm

ii) ¼ bar = ¼ (25) = 6.25 mm

Use: 8 mm

b) Maximum link spacing, scl,tmax

The least of:

i) 20 bar = 20 x 25 = 500 mm

ii) Lesser dimension of the column = 300 mm

iii) 400 mm

Provide spacing, scl,tmax = 300 mm

Therefore, for the links, provide H8 – 300 c/c spacing.

4. Sketch the cross-sectional detailing of this column.

300

300

4H25

H8-300