Week 12 Challenge
SOLUTION 1.
ISOLATED FOOTING
REFER MODEL WEEK 12
OPEN STAAD FOUNDATION
GO TO FOUNDATION PLAN
LINEAR GRID SET UP
ENTER THE FOLLOWING INPUT
GO TO COLUMN POSITION.
ENTER THE FOLLOWING INPUT
GO TO COLUMN DIMENSION .
ENTER THE FOLLOWING INPUT
SHAPE =RECTANGULAR
COLUMN LENGTH = 0.45 M
COLUMN WIDTH = 0.3 M
GO TO PEDESTAL AND ANCHOR BOLT
SELECT NO
GO TO LOADS AND FACTORS
CREATE NEW LOAD CASE
LOAD CASE NO = 1
LOAD TITLE =DL
LOAD CASE TYPE =PRIMARY
LOADING TYPE =DEAD LOAD
ADD A COLUMN REACTION LOAD
UNIT =KN
FORCES
FY =-1500 KN
MOMENTS
UNIT =KNM
MZ= 150 KNM
CREATE NEW LOAD COMBINATIONS
LOAD CASE 101
LOAD COMBINATION TITLE =1DL
COMBINATION TYPE=SERVICE
DEAD LOAD
FACTOR =1
ADD LOAD COMBINATION
LOAD CASE 102
LOAD COMBINATION TITLE =1.5DL
COMBINATION TYPE=SERVICE
DEAD LOAD
FACTOR =1.5
ADD LOAD COMBINATION
CREATE JOB
JOB NAME = ISOLATED
JOB TYPE =ISOLATED
DESIGN CODE=INDIAN
DEFAULT UNIT TYPE= SI
SUPPORT ASSIGNMENT= ASSIGN TO ALL SUPPORTS
CREATE JOB
ISOLATED FOOTING
DESIGN PARAMETERS
CONCRETE AND REINFORCEMENT
COVER AND SOIL
FOOTING GEOMETRY
DESIGN THE ISOLATED FOOTING
RESULTS;
MANUAL CACULATION OF ISOLATED FOOTING
DETERMINE THE SIZE OF FOOTING.
SINCE THE GIVEN MOMENT IS REVERSIBLE,WE NEED TO HAVE A FOOTING THAT IS SYMMETRIC WITH RESPECT TO THE COLUMN.
ASSUME THE WEIGHT OF FOOTING AS 10 PERCENT OF THE COLUMN LOAD
LET US THE ASSUME E<L/6 WITH E=150*10^6/(1500*10^3)*1.1=90.90 MM
HENCE L>6*90.90=545.45 MM
FOR DETERMINATION OF FOOTING SIZE WE NEED WORKING LOADS AND MOMENTS
ASSUMING THE LOAD FACTOR IS 1.5
P=1500/1.5=1000 KN
M=150/1.5=100 KNM
HENCE
P/BL+M/(BL^2/6)< SBC
1000*1.1/BL+100*6/BL^2<200
FROM THIS WE GET
200BL^2-1100L-600=0
L^3-6.6L-3.60=0
SOLVING THIS
ADOPT FOOTING OF SIZE = 3.3M*3.3 M
CALCULATE THE NET PRESSURE BELOW THE FOUNDATION
FACTORED NET SOIL PRESSURE,Q1=Q@U,MAX=P@U/BL+6*M@U/BL^2
Q1=Q@U,MAX=1500/(3.3*3.3)+150*6/(3.3*3.3^2)=137.741+25.043=
=162.78 KN/M^2
Q@2=Q@U,MIN=162.78-25.043=112.698 KN/M^2
DETERMINE THE DEPTH OF FOOTING.
EVEN THOUGH THE DEPTH HAS TO BE FIXED BASED ON ONE WAY SHEAR
TO SIMPLY THE CALCULATION
LET US FIRST DETERMINE THE DEPTH DUE TO BENDING CONSIDERATION
CANTILEVER SPAN L@C=(L-C@2)/2=(3300-450)/2=1425 MM
THEN SOIL PRESSURE AT THE FACE OF THE COLUMN=
Q@3=Q@1-(Q@1-Q@2)*L@C/L=162.78-(162.78-112.698)*1.425/3.3
=141.15 KN/M^2=0.141 N/MM^2
THE MAXIMUM BENDING MOMENT AT THE FACE OF THE COLUMN
M@C=Q@3BL^2/2-(Q@3-Q@1)/2*BL^2*(2/3)=
M@C=0.141*3300*1425^2/2-(0.1627-0.141/2)*(3300*1425^2*2/3)
M@C=424.25*10^6 NMM
D=SQRT((M@C/0.138*F@CK*B))=SQRT((424.25*10^6/0.138*25*3300))=193.93 MM
OVERALL DEPTH =750 MM
d=750-8-50=692 MM
CHECK FOR ONE WAY SHEAR
THE CRITICAL SECTION IS AT A DISTANCE d=692 MM FROM THE FACE OF THE COLUMN
SHEAR SPAN=L@S=(L-C@2)/2-d=1425-692=733 mm
THE SOIL PRESSURE AT THIS LOCATION
Q@4=Q@1-(Q@1-Q@2)*L@S/L=162.78-(162.78-112.698)*0.733/3.3
=160.77 KN/M^2=0.1607 MPA
THE FACTORED SHEAR FORCE AT THIS PLANE IS
V@U=(Q@1+Q@4)/2*BL=(0.1627+0.1607/2)*3300*733*10^-3=391.22 KN
NOMINAL SHEAR STRESS=TAU@V=V@U/BD=391.22*10^3/(3300*692)=0.171 N/MM^2
FROM TABLE OF 19 OF IS 456
P@T=0.14 PERCENT
CHECK FOR PUNCHING SHEAR
THE CRITICAL SECTION IS AT A DISTANCE D/2=692/2=346 MM FROM THE FACE OF FOOTING
THE SOIL PRESSURE AT THIS LOCATION
Q@5=Q@1-(Q@1-Q@2)/L*L@PS=162.78-(162.78-112.698)/3.3*1.079=146.404 KN/M^2
AVERAGE PRESSURE=(162.78+146.404)/2=154.59 KN/M^2
PUNCHING SHEAR FORCE=V@U2=Q(LB-(C@1+D)*(C@2+d)=0.1545(3300*3300-(300+692)*(450+692))
=1507.47*10^3 N
NOMINAL SHEAR STRESS=TAU@V2=V@U2/B@O*D
B@0=2*(300+692)+(450+692)=4268 MM
NOMINAL SHEAR STRESS=TAU@V2=1507.47*10^3/(4268*692)=0.510 N/MM^2
LIMITING SHEAR STRESS=TAU@C=K@S*(0.25*SQRT(F@CK))
K@S=0.5+C1/C2<1=0.5+300/450=1.16 HENCE K@S=1
LIMITING SHEAR STRESS=TAU@C=1*(0.25*SQRT(25))=1.25 N/MM^2>0.510N/MM^2
HENCE IT IS SAFE IN PUNCHING SHEAR
DESIGN THE FLEXURAL REINFORCEMENT
R=M@U/B*D^2=424.25*10^6/3300*692^2=0.268
PROVIDE=0.14*3300*692/100=3197.04 MM^2
USING 16 MM DIA REQUIRED NUMBER=3197.04/201.06=16 NO
CORRESPONDING SPACING =(3300-2*75-16)/16=195 MM
PROVIDE 16 MM DIA AT 175 MM C/C
SOLUTION 2.
REFER MODEL COMBINED FOOTING 1
COMBINED FOOTING.
OPEN STAAD FOUNDATION
GO TO FOUNDATION PLAN
LINEAR GRID SET UP
ENTER THE FOLLOWING INPUT
GO TO COLUMN POSITION.
ENTER THE FOLLOWING INPUT
GO TO COLUMN DIMENSION .
ENTER THE FOLLOWING INPUT
COLUMN 1
SHAPE =RECTANGULAR
COLUMN LENGTH = 0.4 M
COLUMN WIDTH = 0.4 M
COLUMN 2
SHAPE =RECTANGULAR
COLUMN LENGTH = 0.5 M
COLUMN WIDTH = 0.3 M
GO TO PEDESTAL AND ANCHOR BOLT
SELECT YES
SUPPORT 1
PEDESTAL HEIGHT =1.5 M
PEDESTAL LENGTH =0.4 M
PEDESTAL WIDTH=0.4 M
SUPPORT 2
PEDESTAL HEIGHT =1.5 M
PEDESTAL LENGTH =0.5 M
PEDESTAL WIDTH=0.3 M
GO TO LOADS AND FACTORS
CREATE NEW LOAD CASE
LOAD CASE NO = 1
LOAD TITLE =DL
LOAD CASE TYPE =PRIMARY
LOADING TYPE =DEAD LOAD
ADD A COLUMN REACTION LOAD
UNIT =KN
FORCES
FY =-800 KN
ADD
ASSIGN TO SUPPORT 1
ADD A COLUMN REACTION LOAD
UNIT =KN
FORCES
FY =-1200 KN
ADD
ASSIGN TO SUPPORT 2
CREATE NEW LOAD COMBINATIONS
LOAD CASE 101
LOAD COMBINATION TITLE =1DL
COMBINATION TYPE=SERVICE
DEAD LOAD
FACTOR =1
ADD LOAD COMBINATION
LOAD CASE 102
LOAD COMBINATION TITLE =1.5DL
COMBINATION TYPE=SERVICE
DEAD LOAD
FACTOR =1.5
ADD LOAD COMBINATION
CREATE JOB
JOB NAME = COMBINED
JOB TYPE =COMBINED
DESIGN CODE=INDIAN
DEFAULT UNIT TYPE= SI
SUPPORT ASSIGNMENT= ASSIGN TO ALL SUPPORTS
CREATE JOB
COMBINED FOOTING
DESIGN PARAMETERS
CONCRETE AND REINFORCEMENT
COVER AND SOIL
FOOTING GEOMETRY
HERE WE NEED TO FIXED LEFT OVER HANG =YES
MINIMUNM LEFT OVERHANG LENGTH =0
DESIGN THE COMBINED FOOTING
RESULTS;
