Week 12 Challenge
1)
Aim: To design an Isolated Footing Design using STAAD Foundation for the given data:-
Size of the column = 300 mm x 450 mm
Axial load = 1500 kN
Moment = 150 kNm ( Reversible)
SBC of soil = 200 kN/m2 at a depth of 1 m from ground level
Grade of concrete = M25
Grade of steel = Fe 415
Design is also to be verified through hand/manual calculations.
Procedure:
Step 1: Open Staad Foundation and click on General Mode to start the design of the combined footing.
Step 2: A new workspace window will open. Click on enable back up and Enter the support coordinates for the Isolated Footing. Let the footing coordinates be at the origin.
Step 3: Now click on column dimension and add dimensions.
Step 4: Now add load using the main navigator. Go to the Load and Factors Option on the Main Navigator panel. Choose the Create New Load Option to create the load cases for Dead Load and Live Load.
Add the given load condition of Fy = -1500 kN and Mx = 150 kN-m as shown below.
Step 5: Now generate load cases according to the Indian Standards.
Step 6: Now create Job information for analysis and design.
Step 7: Now fill data for the design of footing as given in question.
Step 8: Now click on Design option to analyse and design the footing.
Click on YES to start the Design Process. Design of the Footing has been done successfully.
Results: Following design result has been obtained
- Dimension of footing = 3 m x 3 m
- Over all depth of footing D = 650 m
- Depth of footing is checked for BM, Single shear & Double shear and found safe
- No tension steel is needed as there is no negative pressure.
- Development length = 648 mm
Now, for validation of design of footing using manual calculation.
Footing should be symmetric with respect to column as the moment is reversible (Mx & My are same) ,that is square footing is to be designed.
Service load = 1500 kN
Considering Self weight of footing & weight of backfill soil = 10% of service load
P = 1.1 x 1500= 1650 kN
Pu = 1.5 x 1650 kN = 2475 kN
Area required = 1.1P /(SBC) = (1.1 x 1500)/200 = 8.25 m2
Acal = 8.25 m2
Therefore L = B = sqrt(A) = sqrt(8.25)= 2.87 m
Now Provide square footing of size 3.0 m x 3.0 m
Aprovided = 9 m2 > Area required which is 8.25 m2 (OK)
NET UPWARD SOIL PRESSURE
σ(max, min)= Pd/Aprovided ± Mu/Z
=(1.5 x 1.1x1500)/9 ± ( 1.5x150/(B x L^2/6) ( B = L = 3 M)
= 275 ± 50 lN/m2
σ(max)= 325 kN/m2 < 370 kN/m2 ( Gross factored Bearing capacity)
σ(min) = 225 kN-m2 < 340 kN/m2
Eccentricity, e = M/P = 150/1500 = 0.1 m = 100 mm
L/6 = 3000/6 =500 mm
e < L/6 ,
Therefore no negative soil pressure will develop, hence compression steel is not required.
DEPTH OF FOOTING ON THE BASIS OF B.M.
The maximum bending moment at the face of column (M)
= (Pox(B-b)^2)/8 = (275 x 3(3-0.3)^2)/8 = 751.78 KN-m
Mu = 0.138 x fck x bd^2
751.78 x 10^6 = 0.138 x 20 x3000 xd^2
d = 301 mm, Let d = 305 mm
Adopt d = 600 mm due to shear consideration
Add effective cover = 50 mm
D = 650 mm
Depth of footing from one way shear-
d = Pu(L-a)/[2(P+τc x L^2)]
= 1.5 x 1.1 x 1500 (3-0.3) /2 (1.5 x 1.1 x 1500 + 350 x 9)
= 6682.5 / 11250 = 0.594 m = 594 mm < 600 ( OK)
Depth required for two way shear ( punching shear)
P/L^2[L^2-(a+d)^2] = 4(a+d) d x τp
[1.5 x 1.1 x 1500/9 ] [(9 - (0.3+0.6)^2] = 4 x 0.9 x 0.6 x τp
τp = 1043.98 KN/m2 = 1.044 N/mm2
Permissible τp = 0.25sqrt(fck)
= 0.25 sqrt(20)
= 0.25 x 4.472 = 1.118 N/mm2 > 1.044 N/mm2
Hence OK.
Reinforcement calculation
Ast =(0.5 fck)/fy [1-{1-(4.6 x Mu /fck B d^2)
Ast = 5081 m2
p = Ast/(B x d)x 100
= (5081/(3000 x 650) x100 = 0.26% > 0.12% which is min reqirement,
Hence OK
xu/d = 1.2-[1.44-(6.6Mu)/(fck x b x d^2)]^(1/2)
= 1.2 -[1.44-(6.6 x 751780 x 1000/20 x 3000 x 600 x600)^0.5
= 0.0998
Lever Arm Z = d(1-0.416 xu/d)
= 600( 1-0.416x0.0998)
= 575.089 mm
Ast = Mu/(0.87 x fy x Z)
= 751780 x 1000/0.87 x 415x 575.089 = 3620.66 mm2
STEP-5 : DEVELOPMENT LENGTH
As per IS 456: 2000,Cl 26.2.1.1
σs = 415 N/(mm^2)
τbd for deformed bar and M20 Grade of concrete = 1.92 N/mm^2
Ld = (ϕ×σs)/ 4 τbd
415x ϕ /4 x 1.92 = 54 ϕ = 54 x 12 =
Ld = 648 mm for 12 mm dia bar.
Result: Isolated Footing for the given data has been designed succesfully using STAAD Foundation and summary of the design is as follow:
- Dimension of footing: 3 mx 3 m
- Over all depth of footing D = 650 mm
- Depth of footing is checked for BM, Single shear & Double shear and found safe
- No tension steel is needed as there is no negative pressure.
- Development length = 648 mm.
2)
Aim: To design Combined footing for given data.
Given Data:
- Two columns C1 & C2 having size 400 mmx 400 mm & 300 mmx 500 mm respectively.
- Service load on C1 is 800 KN & on C2 is 1200 KN
- Column C1 is flushed with property line
- C/C distance between columns = 3.0 m
- SBC of soil = 200 KN/m2 at a depth of 1.5 m
- Grade of Concrete : M20
- Steel grade : Fe 415
Procedure:
Step 1: Open Staad Foundation file. Click on General mode and enter supports coordinates for column placement.
Step 2: Add column Dimensions and Pedestal and Anchor Bolt details.
Step 3: Add Load cases. Go to Create new load case -> Add column reaction -> Assign load for columns.
For column c1 service load = 800 KN.
For column c2 service load = 1200 KN.
Step 4: Now create load combination, select indian load combinations.
Step 5: Now create a job. Select column edit current job -> Click strip footing setup and Click create from selected nodes
Step 6: Now go to current job and add design parameters. Add cover and reinforcement then Add cover and soil. Add footing geometry and Add sliding and overturning factors.
Step 7: Go to design and Click design save ok.
Result: Design is successful with footing dimension as 9.2m x 7.5 m and thickness 0.5 m.
C-1 column is flushed with property line that is left over hang is 0.
Main steel top 8mm dia bar @ 50 mm c/c
Main steel bottom 8 mm dia bar @ 55 mm c/c
Secondory steel bottom 12 mm dia 60 mm c/c and top 8 mm dia 80 mm c/c
Design is safe against sliding, single shear, punching shear, soil pressure, overturning.
Hence, design of Combined Footing has been done succesfully.