Week 12 Challenge

1.

Given

Column dimensions = 300x450 mm

Axial load = 1500kN

Mu = 150kN-m

Assume that the moment is reversible

The safe bearing capacity of the soil = 200 kN/m2

Depth= 1 m from the ground level

M25 and Fe415 for the footing

Procedure :

Open Staad Foundation and click on general mode -> Foundation plan -> Linear grid setup

Go to column position -> Enter value or click on origin (0,0,0) -> column dimensions -> pedestal and anchor bolt -> put No

Go to loads and factors -> create new loadcase -> Add column reaction load -> Assign to column

Axial load = P = 1500 kN

So Pu = 1500*1.5 = 2250 kN

Mu = 150 kN

Go to generate load combinations -> choose Indian -> Generate it

Go to soil properties -> choose Bearing capacity factors -> create new job -> define all design parameters

2.

Aim : To design Combined Footing with given data

Given ,

Two columns , C1 = 400x400 mm

Service load = 800 kN

C2 = 300x500 mm

Service load = 1200 kN

The column C1 is flushed with the property line

The columns are at 3.0 m c/c distance

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

Use M20 and Fe415 for columns and footing

Procedure :

Open staad foundation -> Click on general mode -> Foundation plan -> Linear grid setup

Go to column position -> column

Dimension as given in question -> Add pedestal and anchor bolt

create new load case -> Add column reaction -> Assign load to columns

For column C1 , service load = 800kN

For column C2 ,service load = 1200kN

Generate load combination -> add indian load combinations

Create new job -> Combined footing -> choose indian -> select column edit current job -> click strip footing setup

click create from selected nodes

Go to design parameters -> add cover and reinforcement -> Add cover and soil -> Add footing geomety

Add sliding and overturning factors

Go to design -> click design save ok