Challenge

Foundation Design using SAFE

1.Given:

square footing for a column size of 400X400

compression axial load for the load combination of ( 1.5 DL + 1.5 LL ) is 2000Kn

the safe soil bearing capacity is 150KN/M^2 at a depth of 2 meters below EGL

Assume M25 concrete grade

Solution:

step 1)

required data

unfactored load = 2000/1.5 = 1333.3 KN

Dead load breakup = 1333.3/2 = 666.7 KN

Live load breakup = 1333.3/2 = 666.7 KN

subgrade modulus of soil = safe bearing capcity X factored to ultimate / maximum settlement

= 150 X 3 / 0.05 = k = 9000 KN/m^3

Rebar yeild strength ( Fy ) = 500N/mm^2

Rebar ultimate strength ( Fu ) = Say 575 N/mm^2

Elastic modulus of M25 Concrete = 2500 n/mm^2

Step 2)

safe design

open the safe software

click on the new model

in that choose single footing option

and press ok

after that fill all information like this \

and press ok

Step 3)

define material property

from define

then same for rebar

step 4)

Slab Properties

define slab properties from define

step 5) load cases

go to load cases from define

step 6) load combination

go to define

and click on the load combination

step 7) Run the analysis

step 8) draw design strip

step 9)

check

safe bearing capacity

from reaction

punching shear design

step 10)

strip forces

flexural reinforcement

Result: our sqaure isometric footing done successfully with all checkn

2.AIM: to deisgn a pile foundation usig safe software

Introduction: design pile footing by using safe software and do all required check, and try to design economic and safe structure, pile footing is done where hard strata available on depth

Data required

unfactored load = 1175KN

dead load breakup = 1175/2 = 587.5 KN

Live load breakup = 1175/2 = 587.2

Point spring of pile = A X E / L

= 196349 X 25000/ 16000 = 306796 KN

k for SAFE analysis = 306.8 KN/m

Rebar yeild strength (FY) = 500N/mm^2

rebar ultimate strength = 575 N/mm^2

Elastic modulus of M25 concrete = 25000N/mm^2

number of piles required = unfactored load with self weight addition / indivisual pile capacity

= 1175 X 1.1/200 = 6.5

consider 8 piles

take center to center distance between piles = 3 X piles diameter = (1500 mm)

Edge distance from pile center = 1 X pile dimeter = (500mm)

Assume pile cap depth = 500mm

step 2)

safe design

open new model

and put all required information

step 3)

material properties

click on material property from define

add M25 grade concrete material

and 500fy reabar material

step 4)

define slab property

step 5)

draw slab

draw the slab by using quick draw tool

step 6) draw refrence point:

draw refrence point

where we loacate the pile by the help of draw refrence point and replicate option from edit

like that

step 7) add the pile

add the piles at the point

draw the pile by using quick draw area around pint

set dia as 500mm

and draw by cursor

like that

step 8) same draw column

by choosing rectangular slab

step 9) assign a spring

selcect all the pile point

now go to draw

then support data and point spring

step 10) add load

assign load to column

simply select the column point

go to assign

go to joint load

and then select garvity dead and live load are 587.5 kn

step 11) check design preference

check design prefrence before analysis from design option

step 12) run the analysis

step 13) check for rection

check base rection for unfactored load

when we design for steel we need to check for factored load

step 14) check for punching shear design

from disply option

step 15) design strip

to know the area of steel required

go to edit option

and click on add design strip

same do for Y direction and strip layer B

so strip is assign for X and Y direction

check width of strip

from display option

step 16) define load combination

from define option

step 16) again run the analysis

and check the r/f required

go to display option then slab design

and strip A press apply

and for strip B for y direction steel

Result: design of pile foundation is done successfully with given data

and also done all check satisfy

and get quantity of steel required

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