Question

Can someone please assist me with the design of a 4.0m column, secured to a foundation at the bottom and fixed to 3 beams at the top. The loads are axial and at a magnitude of 1500kN total applied load. The design is based on Eurocode 2.

Code to be used when designing: Eurocode 2
concrete strength: C32
steel strength: 460
given dimensions for column: 300mm x 300mm

Answer 1

Solution:- the values given in the question are as follows:

dimension of column=300mm*300mm

fy=460 MPa

fc=32 MPa

load on column(P)=1500 KN

length of column(L)=4.0 m

Design of column according to limit state method:

factored load(Pu)=1.5*load on column=1.5*1500=2250 KN

factored load(Pu)=2250 KN

, for short column

where, r=radius of gyration=(I/A)^0.5

K=0.5, for both ends of column is fixed

r={(b*d^4/12)/(b*d)}^0.5=d/12^0.5=300/12^0.5

KLu/r=(0.5*4000)/(300/12^0.5) , [length of column(L)=4 m =4000 mm]

KLu/r=2000*12^0.5/300=23.09

KLu/r=23.0940 , soth column is short column

for short axialy loaded column

ultimate strength of column(Pu)=0.4fc*Ac+0.75fy*As

where, Ac=cross-section area of column , As=area of steel

gross area(Ag)=300*300=90000 mm^2

area of concrete(Ac)=90000-As

values put in equation-1 and find the area of steel(reinforcement)

2250*10^3=0.4*32*(90000-As)+0.75*460*As

2250000=1152000-12.8*As+345*As

2250000-1152000=332.2*As

1098000=332.2As

As==1098000/332.2=3305.23 mm^2

area of steel(As)=3305.23 mm^2

provide 23 mm dia bars

number of bars(n)=As/area of bar=3305.23/{(3.14/4)*23^2}=7.9598

provided 23 mm dia 8 bars(8#23mm)

area of steel provided(Aspro)=8*(3.14/4)*23^2=3322.12 3323 mm^2

area of steel provided(Aspro)=3323 mm^2

lateral ties:

diameter of lateral ties=/4=23/4=5.76 mm

provide 8 mm dia lateral ties

spacing of ties(s)=minimum of{least lateral dimension of column , 16* , 300 mm }

spacing of ties(s)=minimum of{300 mm , 16*23=368 mm , 300 mm}

provide spacing of ties is 300 mm centre to centre

[Ans]