Week 12 Challenge

1. given data

isolated footing

column size= 300*400mm

axial load = 1500KN

Mu = 150KN.m

soil bearing capacity= 200KN/Sq.m

garde of steel (fy) - 415N/sq.m

grade of concrete(fck) = 25 N/sq.m

area = p/soil bearing capacity

= 1500/200

= 7.5 sq.m

size of the footing = 3.2m*3.2m

Qmax = P/A+M/Z

1500/10.24+150/3.2*3.2^2/6

=173.95KN/Sq.m

Qmin = 1500/10.24-150/3.2*3.2^2/6

= 119.01KN/Sq.m

Moment= W(L^2)/2

W= Qmax=173.95KN/Sq.m = 0.173 N/sq.mm=

L= (3.2-0.3)/2 = 1.45m

M= (173.95*1.45^2)/2

182.866 KN.M

Mu= 0.138*fckBd^2

182.866*10^6= 0.138*25*3200*d^2

d= 128.70mm

D = 400mm

d= D-75-12-12= 400-75-12-12 = 301mm

Ast = Mu/Bd^2

= 182.866*10^6/1000*301^2

= 2.01

by using SP16

Pt= 0.647

required Ast= 647Sq.mm

2. ans.

given data

two columns

C1, 400 mm x 400 mm carries service load of 800 kN

C2, 300 mm x 500 mm carrying a service load of 1200 kN

columns are at 3.0 m c/c distance

safe bearing capacity of soil is 200 kN/m² at a depth of 1.5 m below the ground level

M 20 and Fe 415 for columns and footing