2_foundation Master_gable Column(Baxi Gable)

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    Foundation Design for AZAD MOTORS Page: 1 of 2

    1.0 FOUNDATION DESIGN FOR GABLE COLUMNS OF 120 X 43 X6.5gable column

    4.309 x 4.309

    0.500 x 0.350 0.300

    20o

    2.50 2.000

    0.500 0.200

    2.000 x 2.000

    FIG. 1.

    1.1 INPUT DATA D+W D+L

    Total dead load on foundation t = 1.482 2.450Maximum Bending Moment at Top of foundation along X-axis t-m = 0.000 0.000

    Maximum Bending Moment at Top of foundation along Y-axis t-m = 9.030 2.554

    Maximum horizontal Force along x-AXIS t = 1.200 0.294

    Maximum horizontal Forcealong Y-axis t = 0.000 0.000

    Maximum Compressive force on main leg t = 20.825 2.450

    Maximum Tensile force on main leg t = 0.000 0.000

    1.2 PROPOSED SIZE OF FOUNDATION

    Total height of foundation m = 2.500 2.500

    Height of foundation above ground level m = 0.300 0.300

    Side of pedestal PLx m = 0.500 0.500

    Ply m = 0.350 0.350

    Thickness of base slab at end m = 0.200 0.200

    Maximum thickness of base slab m = 0.500 0.500

    Length of foundation along X-axis m = 2.000 2.000

    Length of foundation along Y-axis m = 2.000 2.000

    Angle of reposeo

    = 30 30

    Net Bearing capacity of soil (Assumed) t/m2

    = 12.2 12.2

    Depth at which bearing pressure is considered m = 15.000 15

    Density of soil t/m3

    = 1.76 1.76

    Density of concrete t/m3

    = 2.4 2.4

    1.3 CHECK FOR OVERTURNING

    Total BM along X-axis = 0.000 + 1.200 x 2.500 t-m = 3 0.735

    Total BM along Y-axis = 9.030 + 0.000 x 2.500 t-m = 9.03 2.554

    To resist the structure against overturning soil frutum formed from the edge of slab at an

    angle off (Angle of repose), half the frutum so formed will resist the overturning

    Side of soil frustum at bottom along conductor m = 2.000 2.000

    Side of soil frustum at bottom across conductor m = 2.000 2.000

    Side of soil frustum at top along conductor

    = 2.000 + 2 x{ 2.000 x tan ( 30 )} m = 4.309 4.309

    Side of soil frustum at top across conductor

    = 2.000 + 2 x{ 2.000 x tan ( 30 )} m = 4.309 4.309

    Area of frustum at bottom A1 = 2.000 x 2.000 m2

    = 4 4

    Area of frustum at top A2 = 4.309 x 4.309 m2

    = 18.5709376 18.5709376

    Height of frustum h m = 2.000 2.000

    Volume of frustum = h x( A1 + A2 + sqrt(A1 x A2 )} / 3

    = 2.000 x( 4 + 18.57094 + 8.618802 )/ 3 cum = 20.7931599 20.7931599

    Volume of half frustum = 20.79316 / 2 cum = 10.3965799 10.3965799

    Weight of half frustum = 10.39658 x 1.76 t = 18.2979807 18.2979807

    Resisting moment = 18.29798 x 2.000 / 2 t-m = 18.2979807 18.2979807

    Overturning moment along x-axis due to soil frustum t-m = 3 0.735

    Factor of safety = 18.29798 / 3 = 6.09932689 24.8952118Permissible factor of safet = 1.5 1.5

    Hence OK Hence OK

    Resisting moment = 18.29798 x 2.000 / 2 t-m = 18.2979807 18.2979807

    Overturning moment along y-axis t-m = 9.030 2.554

    Factor of safety = 18.29798 / 9.03 = 2.02635445 7.16444036

    Permissible factor of safet = 1.5 1.5

    Hence OK Hence OK

    1.4 CHECKING FOR SOIL PRESSURE

    Dead load of Base slab = 2.000 x 2.000 x 0.200 x 2.4 t = 1.920 1.920

    Dead load of Trap. Slab = 0.300 x( 4 + 0.7225 + 1.7 )/ 3 x 2.4 t = 1.541 1.260

    Dead load of pedestals = 1 x 0.500 x 0.350 x 2.0 x 2.4 t = 0.840 4.800

    Dead load of soil =( 2.000 x 2.000 - 4 x 0.500 x 0.350 )x 1.70 x 1.76 = 9.874 8.976

    Total dead load on foundation = 1.920 + 1.541 + 0.840 + 9.874 t = 14.175 16.956

    Load from columns t = 1.482 2.45

    Total vertical load = 14.175 + 1.482 t = 15.657 19.406Max. B. M. at base along x-axis t-m = 0.000 0.000

    Max. B. M. at base along y-axis t-m = 9.030 2.554

    Area of base = 2.000 x 2.000 m2

    = 4.000 4.000

    Modulus of section Zxx = 2.000 x 2.000 x 2.000 / 6 m3

    = 1.33333333 1.33333333

    Modulus of section Zyy = 2.000 x 2.000 x 2.000 / 6 m3

    = 1.33333333 1.33333333

    Pressure due to direct load = 15.657 / 4.000 t/m2

    = 3.914 4.852

    Pressure due to B.M. Mxx = 0.000 / 1.333333 t/m2

    = 0.000 0.000

    Pressure due to B.M. Myy = 9.030 / 1.333333 t/m2

    = 6.773 1.916

    Maximum bearing pressure = 3.914 + 0.000 + 6.773 t/m2

    = 10.687 6.767

    Pressure due to earth surcharge = 1.76 x( 2.20 - 15 ) t/m2

    = -22.528 #VALUE!

    Net soil pressure = 10.687 - -22.528 t/m2

    = 33.215 #VALUE!

    Net bearing capacity of soil t/m2

    = 11 11

    Unsafe #VALUE!1.5 DESIGN OF BASE SLAB

    Neglecting soil pressure, net pressure t/m2

    = 33.215 #VALUE!

    Consider width of foundation m = 1 1

    Upward load = 33.215 x 1 t/m = 33.21475 #VALUE!

    Cantilever along X-axis =( 2.000 - 0.850 )/ 2 m = 0.575 0.75

    Cantilever across Y-axis =( 2.000 - 0.350 )/ 2 m = 0.825 0.75

    Maximum b.m. along x-axis = w l l / 2

    = 33.21475 x 0.575 x 0.575 / 2 t-m = 5.49081336 #VALUE!

    Maximum b.m. along y-axis = 33.21475 x 0.825 x 0.825 / 2 t-m = 11.3033946 #VALUE!

    Maximum of above two t-m = 11.3033946 #VALUE!

    kg-m = 11303.3946 #VALUE!

    Main Column

    File: 144402682.xls.ms_office - Sheet: FOUND-Col Status: 5/9/2013

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    Foundation Design for AZAD MOTORS Page: 2 of 2

    1.51 Concrete Strength

    Grade of concrete = M- 20

    Grade of steel = Fe- 415

    Here c = 66.66667 kg/cm2

    t = 2300 kg/cm2

    m = 14

    n1 = n / d = m c /( m c + t )= 14 x 66.66667 /( 14 x 66.66667 + 2300 )= 0.28866

    a1 = 1 - n1 / 3 = 0.90378

    q = 0.5 x a1 x n1 x c

    = 0.5 x 0.90378 x 0.28866 x 66.66667 = 8.696166 kg/cm2

    1.52 CHECKING FOR DEPTH OF SLAB

    Effective depth required = Sqrt { BM /( q x B )}

    = Sqrt { 11303.39 / 8.696166 x 1 )} cm = 36.05 #VALUE!

    Overall depth of slab mm = 500 500

    Clear Cover mm = 50 50

    Dia of main Reinforcement along x-axis mm = 10 10

    Dia of main Reinforcement along y-axis mm = 10 10

    Effective depth provided = 500 - 50 - 5 mm = 445 445

    cm = 44.5 44.5

    d reqd. cm = 36.05 #VALUE!

    Hence OK #VALUE!1.53 Reinforcement Design along x-axis

    Area of steel reqd. = BM /( a1 x d x t )

    = 549081.3 /( 0.90378 x 44.5 x 2300 ) cm2

    / m = 5.94 #VALUE!

    Minimum steel reqd = 0.15 %

    = 0.15 x 44.5 x 100 / 100 cm2

    / m = 6.675 6.675

    Hence area of steel to be provided cm2

    / m = 6.675 #VALUE!

    Spacing of 10 mm dia bars = 78.53982 / 6.675 cm c/c = 11.7662646 #VALUE!

    mm c/c = 110 #VALUE!

    maximum ermissible s acin = 3 x d

    = 3 x 445 mm c/c = 1335 1335

    So provide 10 mm bars @ 110 mm c/c mm c/c 110 #VALUE!

    1.54 Reinforcement Design y-axis

    Area of steel reqd. = BM /( a1 x d x t )

    = 1130339 /( 0.90378 x 44.5 x 2300 ) cm2 / m = 12.22 #VALUE!Minimum steel reqd = 0.15 %

    = 0.15 x 44.5 x 100 / 100 cm2

    / m = 6.675 6.675

    Hence area of steel to be provided cm2

    / m = 12.2196364 #VALUE!

    Spacing of 10 mm dia bars = 78.53982 / 12.21964 cm c/c = 6.4273448 #VALUE!

    mm c/c = 60 #VALUE!

    maximum permissible spacing = 3 x d

    = 3 x 445 mm c/c = 1335 1335

    So provide 10 mm bars @ 60 mm c/c mm c/c 60 #VALUE!

    1.6 DESIGN OF PEDESTAL

    To accommodate the Anchor bolts & bearing plate, the size of pedestal is decided as 500 x 500

    1.61 Check in Compression

    Maximum Compressive force on main leg t = 5.206 0.613

    Maximum Tensile force on main leg t = 0.000 0.000

    Max BM along x-axis t-m = 0.000 0.000Max BM along y-axis t-m = 9.030 2.554

    Max compression per leg = 5 / 1 t = 5.21 0.61

    Maximum tension per leg = 0 / 1 t = 0.00 0.00

    Maximum comp. Stress in pedestal = 5206 /( 50 x 50 ) kg/cm2

    = 2.0825 0.245

    Permissible Comp. Stress kg/cm2

    = 53.3333333 53.3333333

    Hence OK Hence OK

    1.62 Reinforcement

    Minimum area of steel reqd. = 0.8 %

    = 0.008 x 50 x 50 cm2

    = 20 20

    Proposed dia of bars mm = 16 16

    No. of bars nos. = 5 10

    Area of steel proposed = 5 x 2.01 cm2

    = 20.11

    Hence OK Hence OK

    Considering dia of rings mm = 8 8Spacing of rings (Clause 26.5.3.2 c) 1) i) of IS:456-2000) mm = 500 500

    (Clause 26.5.3.2 c) 1) ii) of IS:456-2000) mm = 256 256

    (Clause 26.5.3.2 c) 1) iii) of IS:456-2000) mm = 300 300

    Adopt spacing of rings mm c/c = 250 250

    1.63 Check in Tension

    Maximum tension in leg t = 0.000 0.000

    Area of reinforcement provided cm2 = 20.11

    Actual tensile stress = 0 / kg/cm2

    = #VALUE! 0

    Permissible tensile stress kg/cm2

    = 2300 2300

    #VALUE! Hence OK

    -x-x-x-x-

    Bill of Quantities

    Base slab 2.000 x 2.000 x 0.200 Cum = 0.8 0.8

    Tapered Slab 2.000 x 2.000 x 0.18 Cum = 0.72 0.72

    Pedestals 0.500 x 0.500 x 2.000 x 1 Cum = 0.5 0.5

    Cum 2.02 2.02

    say Cum 2.5 2.5

    Jal Reinforcement Adopt

    Bar dia along x-axis mm 10 10 10

    Spacing of bars along x-axis 110 #VALUE! #VALUE!

    Bar dia along y-ayis mm 10 10 10

    Spacing of bars along y-ayis 60 #VALUE! #VALUE!

    Pedestals

    No. of Main bars 5 10

    Dia of main bars mm 16 16

    Dia of rings 8 8

    Spacing of rings mm 250 250

    -x-x-x-x-

    File: 144402682.xls.ms_office - Sheet: FOUND-Col Status: 5/9/2013