Week 12 Challenge

Ans 1 )

Aim:-

        To design Isolated Footing for a column 300 mm x 450 mm carrying axial load of 1500 kN and Mu of 150kNm using STAAD Foundation. Also Validate the design through hand calculation.

Procedure:-

Step 1:-

• Open STAAD Foundation Software, and Create a New General Foundation.
• Then Setup Units as SI Unit.

        

Step 2:-

• In Main Navigator Window, select Column Position and give the coordinates as (0,0,0).

        

• Click Column Dimension and give Length and Width as 0.45 and 0.3 m

        

Step 3:-

• Now go to Loads & Factors menu, then click Create New Load cases.
• Then Add New Load cases of Dead Load and Live Load as Fy = -1500 kN and Mu = 150 kNm.      

         

• Now Generate Load Combination as per Indain Standard as shown below.

          

        

Step 4:-

• Create a New Job in Job Setup menu.
• Set Job Name as Isolated Footing and Job Type Isolated.
• And set Design code as Indian (IS456:2000).
• Then select all Load Cases and Create Job.

        

Step 5:-

• Now go to Design Parameter.
• Click Concrete & Reinforcement, Give input as shown below.

        

• Then set Cover & Soil, set Soil bearing capacity as 200 kPa.
• And Minimum % of Contact area for Ultimate loads as 95%.

       

• Now Footing Geametry, Set Minimum and Maximum dimensions for Footing.

        

Step 6:-

• Select Design option and Software will analysis and design the Footing.

        

Result:-

• Thus the Footing design was done successfully using STAAD Foundation.

       

       

        

Validation of Design through Hand Calculation:-

Footing should be symmetric with respect to column as the moment is reversible (M_x & M_y are same, that is square footing is to be designed.

Service load = 1500kN

Considering Self weight of footing & weight of backfill soil = 10% of service load

P = 1.1 x 1500 = 1650 kN

P_u = 1.5 x 1650 = 2475 kN

Step 1:-

Area Calculation:

Area required = 1.1 x P/SBC = 1.1 x 1500 / 200 = 8.25 m².

Therefore L = B = √A = √8.25 = 2.87 m.

Provide square footing of size 3.0 m x 3.0 m

A_provided = 9m² > Area required which is 8.25 m² (OK)

Step 2:-

Net Upward Soil Pressure:

σ_{(max, min)} = P_d / A_Provided + M_u / Z

                = (1.5 x 1.1 x 1500 / 9) ± (1.5 x 150 / (B × L² / 6)     (B = L = 3 m)

                = 275 ± 50

σ_(max) =  325 kN/m² < 370 kN/m² (Gross Factored Bearing capacity) ( ok) 

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.

Step 3:-

Depth of Footing in the basis of Bending Moment:

The maximum bending moment at the face of column (M)

M = P₀ × (B−b)² / 8 = 275 × 3(3 − 0.3)² / 8 = 751.78 kNm

M_u = 0.138f_ckbd²

751.78 × 10⁶ = 0.138 x 20 x 3000 x d²

d = 301 mm.

Say 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 = P_u(L−a) / 2(P + τ_c × L²)

   = 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 / L² [L² − (a + d)²] = 4 (a + d) d × τ_p

[1.5 x 1.1 x 1500 / 9][(9 − (0.3 + 0.6)²] = 4 x 0.9 x 0.6τ_p

τ_p = 1043.98 kN/m² = 1.044 N/mm²

Permissible τ_p = 0.25 √fck

                     = 0.25 √20

                     = 0.25 x 4.472 = 1.118 N/mm² > 1.044 N/mm² (OK)

Step 4:-

Reinforcement Calculation:

A_st = 0.5 f_ck/ f_y [1 − {1 − (4.6 × M_u / f_ckBd²)

A_st = 5081 mm²

p = (A_st / B×d) × 100

   = (5081 / 3000 x 650) x 100 = 0.26% > 0.12% which is min requirement, hence OK

x_u / d = 1.2 − [1.44 − 6.6 x M_u / (f_ck × b × d²)]^1/2    

         = 1.2 − [1.44 − (6.6 x 751780 x 1000 / 20 x 3000 x 600 x 600)^0.5

         = 0.0998

Lever Arm Z = d (1 - 0.416 x_u / d)

                   = 600 ( 1 - 0.416 x 0.0998)

                   = 575.089 mm

A_st = M_u / (0.87 × f_y × Z)

      = 751780 X 1000 / (0.87 X 415 X 575.089) = 3620.66 mm²

Step 5:-

Development Length:

As per IS 456:2000, Clause 26.2.1.1

σ_s = 415 N/mm²

τ_bd for deformed bar and M20 Grade of concrete = 1.92 N/mm²

L_d = ϕ × σ_A / 4τ_bd

    = 415 x ϕ / (4 x 1.92) = 54ϕ

    = 54 x 12 = 648 mm for 12 mm dia bar

Ans 2.

    Aim    To design Combined Footing for a columns C1 400 mm x 400 mm and C2 300 mm x 500 mm carrying Service load of 800 kN and 1200kN using STAAD Foundation.

Procedure:-

Step 1:-

• Open STAAD Foundation Software, and Create a New General Foundation.
• Then Setup Units as SI Unit.

        

Step 2:-

• In Main Navigator Window, select Column Position and give the coordinates as (0,0,0) for Column C1 and give coordinate as (3,0,0) for Column C2.

        

Step 3:-

• Add and assign Load of -800 kN for C1 and -1200 kN for C2.

        

• Generate Load Combination as Indian Table as Shown brlow.

        

Step 4:-

• Next Create a New Job Setup and Name it as Combined Footing.
• Set Job Type as Combined as shown below.
• Then select the Column C1 and C2, click Strip footing setup and create.

        

Step 5:-

• Now Give Parameters for Design as shown below.

        

Step 6:-

• Then click Design option to Analyse the Footing.

Result:-

• Model and design of Combined footing was done successfully using STAAD Foundation.
• Results are attached below.