DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE F-1

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 461 KN B = 300 mmMux 121 KNm L = 500 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

e = 121*10^3461*1.10

= 239 mm

Assuming e < L/6 (i.e. , L > 6 * 239 = 1434 mm ) and load factor for bearing capacity as 1

461*1.10 + 121 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 461 * 1.15 * L - 121 * 6 <= 0

Assuming B = 1.00 m => L>= 2.997 m or #VALUE! mB = 1.50 m => L>= 2.223 m or #VALUE! mB = 2.00 m => L>= -0.800 m or 1.814445 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2000 mm L = 2000 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 115.3 + 90.8BL BL2/6

= 206.0 KN/m2

qu,min = Pu + Mux = 115.3 - 90.8BL BL2/6

= 24.5 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 206.0 - 90.8 * {(750-d)/2}/1000

= 171.96875 + 0.045375 d

= 184 KN/m2 ( assuming d = 255 mm )

= 0.184 N/mm2

Vu1 = 0.184 * 2000 * (750-d)

= 275309 - 367.07875 d 181703.92

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2000 * d = 720 d 183600

Vu1 <= Vuc => 275309 - 367 d <= 720dVuc = Vu1 =

183600 > 181703.9188 provide, THICKNESS OF FOOTING IS SAFEd >= 255 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 115.3 KN/m2= 0.115 N/mm2

Vu2 = 0.115[2000 * 2000 - (300 + d)(500 + d) ]

Assuming d = 255 mm

Vu2 = 411812.125 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/500but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (500 + d)] * 2 * d

d = 404 mmVuc = 1624080 N

= 1624 KN

Vuc > Vu21624 411.812125 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 255 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 311 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 6 = 294 mm

Effective depth (short span) dy = 294 - 12 = 282 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 461 / (2 * 2 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1) + 121 * 6 / (2 * 2 ^2)

= 115.25 32.25 90.75

= 238.25 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a) Long Span

cantilever projection = 750 mmwidth = 2000 mmdx = 294 mmqu = 0.1379 N/mm2 at face of column

= 0.2060 N/mm2 at footing edge

Mux = (0.138 * 2000 * 750 ^2) + ( 0.206 -0.138 ) * (1/2) *2000 * 750 ^2 * 2/3

= 77589844 + 25523437.5= 103113281 103.11 kN-m

R = = 0.596471848 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.596/25] = 0.001411112 * 500

Pt = 0.141 %

Pt assumed for one-way shear= 0.25 > 0.141

Ast req = 1470 mm2

Using 12 Φ of bars, no. of bars required = 1470 / 113= 14

Spacing = 146.153846 mm

Provide 14Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 700 mm Hence , ok

(b) Short Span

cantilever projection = 750 mmwidth = 2000 mmdx = 282 mmqu = 0.1153 + 0.2060 ) /2

= 0.1606 N/mm2

Muy = 0.161 * 2000 * 850 /2

= 116051563 Nmm

R = = 0.729663765 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.73/25] 0.00173792 * 500

Pt = 0.174

Ast req = 980.177372 mm2

Ast min = 840 mm2 < 980.177372 mm2

Using 12 Φ of bars, no. of bars required = 980 / 113= 9

Spacing = 237.5 mm

Provide 9Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 800 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE F-1DESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 461 KN B = 300 mmMux 121 KNm L = 500 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2000 mm L = 2000 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 14 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 140 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 9 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 230 mm

qa

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 741 KN B = 300 mmMux 134 KNm L = 500 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

e = 134*10^3741*1.10

= 164 mm

Assuming e < L/6 (i.e. , L > 6 * 164 = 984 mm ) and load factor for bearing capacity as 1

741*1.10 + 134 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 741 * 1.15 * L - 134 * 6 <= 0

Assuming B = 1.00 m => L>= 4.054 m or #VALUE! mB = 1.50 m => L>= 2.910 m or #VALUE! mB = 2.00 m => L>= -0.692 m or 2.322544 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2250 mm L = 2250 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 146.4 + 70.6BL BL2/6

= 217.0 KN/m2

qu,min = Pu + Mux = 146.4 - 70.6BL BL2/6

= 75.8 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 217.0 - 70.6 * {(875-d)/2}/1125

= 189.505258 + 0.031370828 d

= 200 KN/m2 ( assuming d = 320 mm )

= 0.200 N/mm2

Vu1 = 0.2 * 2250 * (875-d)

= 392852 - 448.9738272 d 249180.38

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2250 * d = 810 d 259200

Vu1 <= Vuc => 392852 - 449 d <= 810dVuc = Vu1 =

259200 > 249180.3753 provide, THICKNESS OF FOOTING IS SAFEd >= 320 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 146.4 KN/m2= 0.146 N/mm2

Vu2 = 0.146[2250 * 2250 - (300 + d)(500 + d) ]

Assuming d = 320 mm

Vu2 = 664898.6 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/500but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (500 + d)] * 2 * d

d = 404 mmVuc = 1624080 N

= 1624 KN

Vuc > Vu21624 664.8986 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 320 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 378 mmProvide D = 400 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 400 - 50 - 8 = 342 mm

Effective depth (short span) dy = 342 - 16 = 326 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 741 / (2.25 * 2.25 ) + {( 24 * 0.4) + 18 * (1.5 - 0.4)} *1) + 134 * 6 / (2 * 2.25 ^2)

= 146.3703704 32.25 70.58436

= 249.2047325 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 875 mmwidth = 2250 mmdx = 342 mmqu = 0.1621 N/mm2 at face of column

= 0.2170 N/mm2 at footing edge

Mux = (0.162 * 2250 * 875 ^2) + ( 0.217 -0.162 ) * (1/2) *2250 * 875 ^2 * 2/3

= 139583205 + 31524005.49= 171107210 171.11 kN-m

R = = 0.650179961 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.65/25] = 0.001542342 * 500

Pt = 0.154 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.154

Ast req = 1923.75 mm2

Using 16 Φ of bars, no. of bars required = 1923.75 / 201= 10

Spacing = 238.888889 mm

Provide 10Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 825 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 875 mmwidth = 2250 mmdx = 326 mmqu = 0.1464 + 0.2170 ) /2

= 0.1817 N/mm2

Muy = 0.182 * 2250 * 975 /2

= 194279583 Nmm

R = = 0.81247395 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.812/25] 0.001943412 * 500

Pt = 0.194

Ast req = 1425.49478 mm2

Ast min = 1080 mm2 < 1425.49478 mm2

Using 16 Φ of bars, no. of bars required = 1425 / 201= 8

Spacing = 307.142857 mm

Provide 8Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 925 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 741 KN B = 300 mmMux 134 KNm L = 500 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2250 mm L = 2250 mm D = 400 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 10 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 230 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 8 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 300 mm

qa

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 996 KN B = 300 mmMux 148 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

e = 148*10^3996*1.10

= 135 mm

Assuming e < L/6 (i.e. , L > 6 * 135 = 810 mm ) and load factor for bearing capacity as 1

996*1.10 + 148 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 996 * 1.15 * L - 148 * 6 <= 0

Assuming B = 1.00 m => L>= 5.081 m or 3.308947 mB = 1.50 m => L>= 3.583 m or #VALUE! mB = 2.00 m => L>= -0.630 m or 2.820807 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2500 mm L = 2500 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 159.4 + 56.8BL BL2/6

= 216.2 KN/m2

qu,min = Pu + Mux = 159.4 - 56.8BL BL2/6

= 102.5 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 216.2 - 56.8 * {(1025-d)/2}/1250

= 192.89088 + 0.0227328 d

= 201 KN/m2 ( assuming d = 370 mm )

= 0.201 N/mm2

Vu1 = 0.201 * 2500 * (1025-d)

= 515836 - 503.25504 d 329631.64

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2500 * d = 900 d 333000

Vu1 <= Vuc => 515836 - 503 d <= 900dVuc = Vu1 =

333000 > 329631.6352 provide, THICKNESS OF FOOTING IS SAFEd >= 370 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 159.4 KN/m2= 0.159 N/mm2

Vu2 = 0.159[2500 * 2500 - (300 + d)(450 + d) ]

Assuming d = 370 mm

Vu2 = 906395.4 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 906.3954 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 370 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 428 mmProvide D = 450 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 450 - 50 - 8 = 392 mm

Effective depth (short span) dy = 392 - 16 = 376 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 996 / (2.5 * 2.5 ) + {( 24 * 0.45) + 18 * (1.5 - 0.45)} *1) + 148 * 6 / (2 * 2.5 ^2)

= 159.36 32.25 56.832

= 248.442 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 392 mmqu = 0.1696 N/mm2 at face of column

= 0.2162 N/mm2 at footing edge

Mux = (0.17 * 2500 * 1025 ^2) + ( 0.216 -0.17 ) * (1/2) *2500 * 1025 ^2 * 2/3

= 222719052 + 40801232= 263520284 263.52 kN-m

R = = 0.685964921 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.686/25] = 0.001630182 * 500

Pt = 0.163 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.163

Ast req = 2450 mm2

Using 16 Φ of bars, no. of bars required = 2450 / 201= 13

Spacing = 200 mm

Provide 13Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 975 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 376 mmqu = 0.1594 + 0.2162 ) /2

= 0.1878 N/mm2

Muy = 0.188 * 2500 * 1100 /2

= 284011200 Nmm

R = = 0.803562698 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.804/25] 0.001921212 * 500

Pt = 0.192

Ast req = 1805.93903 mm2

Ast min = 1350 mm2 < 1805.93903 mm2

Using 16 Φ of bars, no. of bars required = 1806 / 201= 9

Spacing = 300 mm

Provide 9Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1050 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 996 KN B = 300 mmMux 148 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2500 mm L = 2500 mm D = 450 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 13 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 200 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 9 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 300 mm

STAAD MODEL NODE NO. 87

qa

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 748 KN B = 300 mm

Mux 108 kN-m 135 kN-m L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

e = 108*10^3748*1.10

= 131 mm

Assuming e < L/6 (i.e. , L > 6 * 131 = 786 mm ) and load factor for bearing capacity as 1

748*1.10 + 108 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 748 * 1.15 * L - 108 * 6 <= 0

Assuming B = 1.00 m => L>= 3.948 m or 1.986187 mB = 1.50 m => L>= 2.809 m or #VALUE! mB = 2.00 m => L>= -0.582 m or 2.227435 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2250 mm L = 2250 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 147.8 + 56.9BL BL2/6

= 204.6 KN/m2

qu,min = Pu + Mux = 147.8 - 56.9BL BL2/6

= 90.9 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 204.6 - 56.9 * {(900-d)/2}/1125

= 181.88642 + 0.025283951 d

= 190 KN/m2 ( assuming d = 330 mm )

= 0.190 N/mm2

Vu1 = 0.19 * 2250 * (900-d)

= 385216 - 428.0177778 d 243970.13

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2250 * d = 810 d 267300

Vu1 <= Vuc => 385216 - 428 d <= 810dVuc = Vu1 =

267300 > 243970.1333 provide, THICKNESS OF FOOTING IS SAFEd >= 330 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 147.8 KN/m2= 0.148 N/mm2

Vu2 = 0.148[2250 * 2250 - (300 + d)(450 + d) ]

Assuming d = 330 mm

Vu2 = 676522.8 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 676.5228 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 330 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 388 mmProvide D = 400 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 400 - 50 - 8 = 342 mm

Effective depth (short span) dy = 342 - 16 = 326 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 748 / (2.25 * 2.25 ) + {( 24 * 0.4) + 18 * (1.5 - 0.4)} *1) + 108 * 6 / (2 * 2.25 ^2)

= 147.7530864 32.25 56.88889

= 236.8919753 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 900 mmwidth = 2250 mmdx = 342 mmqu = 0.1591 N/mm2 at face of column

= 0.2046 N/mm2 at footing edge

Mux = (0.159 * 2250 * 900 ^2) + ( 0.205 -0.159 ) * (1/2) *2250 * 900 ^2 * 2/3

= 145008000 + 27648000= 172656000 172.66 kN-m

R = = 0.656065114 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.656/25] = 0.001556762 * 500

Pt = 0.156 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.156

Ast req = 1923.75 mm2

Using 16 Φ of bars, no. of bars required = 1923.75 / 201= 10

Spacing = 238.888889 mm

Provide 10Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 850 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 900 mmwidth = 2250 mmdx = 326 mmqu = 0.1478 + 0.2046 ) /2

= 0.1762 N/mm2

Muy = 0.176 * 2250 * 975 /2

= 188435000 Nmm

R = = 0.788032001 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.788/25] 0.001882572 * 500

Pt = 0.188

Ast req = 1380.86269 mm2

Ast min = 1080 mm2 < 1380.86269 mm2

Using 16 Φ of bars, no. of bars required = 1381 / 201= 7

Spacing = 358.333333 mm

Provide 7Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 925 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 748 KN B = 300 mmMux 108 kN-m L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2250 mm L = 2250 mm D = 400 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 10 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 230 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 7 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 350 mm

STAAD MODEL NODE NO. 88

FULL MXz

qa

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 386 KN B = 300 mmMux 118 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 118*10^3 2000

386*1.10 2100

= 278 mm

Assuming e < L/6 (i.e. , L > 6 * 278 = 1668 mm ) and load factor for bearing capacity as 1

386*1.10 + 118 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 386 * 1.15 * L - 118 * 6 <= 0

Assuming B = 1.00 m => L>= 2.734 m or #VALUE! mB = 1.50 m => L>= 2.052 m or #VALUE! mB = 2.00 m => L>= -0.839 m or 1.688042 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2000 mm L = 2100 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 91.9 + 80.3BL BL2/6

= 172.2 KN/m2

qu,min = Pu + Mux = 91.9 - 80.3BL BL2/6

= 11.6 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 172.2 - 80.3 * {(825-d)/2}/1050

= 140.641399 + 0.038224814 d

= 150 KN/m2 ( assuming d = 250 mm )

= 0.150 N/mm2

Vu1 = 0.15 * 2000 * (825-d)

= 247826 - 300.3952057 d 172727.2

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2000 * d = 720 d 180000

Vu1 <= Vuc => 247826 - 300 d <= 720dVuc = Vu1 =

180000 > 172727.1986 provide, THICKNESS OF FOOTING IS SAFEd >= 250 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 91.9 KN/m2= 0.092 N/mm2

Vu2 = 0.092[2000 * 2100 - (300 + d)(450 + d) ]

Assuming d = 250 mm

Vu2 = 350980 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 350.98 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 250 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 308 mmProvide D = 320 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 320 - 50 - 8 = 262 mm

Effective depth (short span) dy = 262 - 16 = 246 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 386 / (2 * 2.1 ) + {( 24 * 0.32) + 18 * (1.5 - 0.32)} *1) + 118 * 6 / (2 * 2.1 ^2)

= 91.9047619 32.25 80.27211

= 204.4268707 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 825 mmwidth = 2000 mmdx = 262 mmqu = 0.1091 N/mm2 at face of column

= 0.1722 N/mm2 at footing edge

Mux = (0.109 * 2000 * 825 ^2) + ( 0.172 -0.109 ) * (1/2) *2000 * 825 ^2 * 2/3

= 74260222 + 28618440.23= 102878663 102.88 kN-m

R = = 0.749363838 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.749/25] = 0.001786632 * 500

Pt = 0.179 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.179

Ast req = 1310 mm2

Using 16 Φ of bars, no. of bars required = 1310 / 201= 7

Spacing = 333.333333 mm

Provide 7Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 775 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 825 mmwidth = 2100 mmdx = 246 mmqu = 0.0919 + 0.1722 ) /2

= 0.1320 N/mm2

Muy = 0.132 * 2100 * 850 /2

= 100169464 Nmm

R = = 0.788217081 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.788/25] 0.001883032 * 500

Pt = 0.188

Ast req = 972.771683 mm2

Ast min = 806.4 mm2 < 972.771683 mm2

Using 16 Φ of bars, no. of bars required = 973 / 201= 5

Spacing = 475 mm

Provide 5Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 800 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 386 KN B = 300 mmMux 118 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2000 mm L = 2100 mm D = 320 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 7 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 330 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 5 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 470 mm

STAAD MODEL NODE NO. 89

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 666 KN B = 300 mmMux 133 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 133*10^3 2200

666*1.10 2200

= 182 mm

Assuming e < L/6 (i.e. , L > 6 * 182 = 1092 mm ) and load factor for bearing capacity as 1.5

666*1.10 + 133 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 666 * 1.15 * L - 133 * 6 <= 0

Assuming B = 1.00 m => L>= 2.732 m or #VALUE! mB = 1.50 m => L>= 2.009 m or #VALUE! mB = 2.00 m => L>= -0.653 m or 1.629686 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2200 mm L = 2200 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 137.6 + 74.9BL BL2/6

= 212.5 KN/m2

qu,min = Pu + Mux = 137.6 - 74.9BL BL2/6

= 62.7 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 212.5 - 74.9 * {(875-d)/2}/1100

= 182.739823 + 0.034065296 d

= 193 KN/m2 ( assuming d = 310 mm )

= 0.193 N/mm2

Vu1 = 0.193 * 2200 * (875-d)

= 372103 - 425.2601427 d 240272.36

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2200 * d = 792 d 245520

Vu1 <= Vuc => 372103 - 425 d <= 792dVuc = Vu1 =

245520 > 240272.3557 provide, THICKNESS OF FOOTING IS SAFEd >= 310 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 137.6 KN/m2= 0.138 N/mm2

Vu2 = 0.138[2200 * 2200 - (300 + d)(450 + d) ]

Assuming d = 310 mm

Vu2 = 603943.2 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 603.9432 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 310 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 368 mmProvide D = 380 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 380 - 50 - 8 = 322 mm

Effective depth (short span) dy = 322 - 16 = 306 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 666 / (2.2 * 2.2 ) + {( 24 * 0.38) + 18 * (1.5 - 0.38)} *1.5) + 133 * 6 / (2 * 2.2 ^2)

= 137.6033058 32.25 74.94365

= 244.7969572 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 875 mmwidth = 2200 mmdx = 322 mmqu = 0.1529 N/mm2 at face of column

= 0.2125 N/mm2 at footing edge

Mux = (0.153 * 2200 * 875 ^2) + ( 0.213 -0.153 ) * (1/2) *2200 * 875 ^2 * 2/3

= 128797999 + 33470927.64= 162268927 162.27 kN-m

R = = 0.711378834 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.711/25] = 0.001692772 * 500

Pt = 0.169 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.169

Ast req = 1771 mm2

Using 16 Φ of bars, no. of bars required = 1771 / 201= 9

Spacing = 262.5 mm

Provide 9Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 825 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 875 mmwidth = 2200 mmdx = 306 mmqu = 0.1376 + 0.2125 ) /2

= 0.1751 N/mm2

Muy = 0.175 * 2200 * 950 /2

= 173805837 Nmm

R = = 0.843720931 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.844/25] 0.002021442 * 500

Pt = 0.202

Ast req = 1360.83254 mm2

Ast min = 1003.2 mm2 < 1360.83254 mm2

Using 16 Φ of bars, no. of bars required = 1361 / 201= 7

Spacing = 350 mm

Provide 7Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 900 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 666 KN B = 300 mmMux 133 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2200 mm L = 2200 mm D = 380 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 9 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 260 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 7 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 350 mm

STAAD MODEL NODE NO. 90

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 969 KN B = 300 mmMux 147 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 147*10^3 2500

969*1.10 2500

= 138 mm

Assuming e < L/6 (i.e. , L > 6 * 138 = 828 mm ) and load factor for bearing capacity as 1

969*1.10 + 147 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 969 * 1.15 * L - 147 * 6 <= 0

Assuming B = 1.00 m => L>= 4.973 m or 3.140052 mB = 1.50 m => L>= 3.512 m or #VALUE! mB = 2.00 m => L>= -0.637 m or 2.768881 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2500 mm L = 2500 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 155.0 + 56.4BL BL2/6

= 211.5 KN/m2

qu,min = Pu + Mux = 155.0 - 56.4BL BL2/6

= 98.6 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 211.5 - 56.4 * {(1025-d)/2}/1250

= 188.34432 + 0.0225792 d

= 197 KN/m2 ( assuming d = 370 mm )

= 0.197 N/mm2

Vu1 = 0.197 * 2500 * (1025-d)

= 504040 - 491.74656 d 322093.77

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2500 * d = 900 d 333000

Vu1 <= Vuc => 504040 - 492 d <= 900dVuc = Vu1 =

333000 > 322093.7728 provide, THICKNESS OF FOOTING IS SAFEd >= 370 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 155.0 KN/m2= 0.155 N/mm2

Vu2 = 0.155[2500 * 2500 - (300 + d)(450 + d) ]

Assuming d = 370 mm

Vu2 = 883593 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 883.593 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 370 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 428 mmProvide D = 450 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 450 - 50 - 8 = 392 mm

Effective depth (short span) dy = 392 - 16 = 376 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 969 / (2.5 * 2.5 ) + {( 24 * 0.45) + 18 * (1.5 - 0.45)} *1) + 147 * 6 / (2 * 2.5 ^2)

= 155.04 32.25 56.448

= 243.738 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 392 mmqu = 0.1652 N/mm2 at face of column

= 0.2115 N/mm2 at footing edge

Mux = (0.165 * 2500 * 1025 ^2) + ( 0.211 -0.165 ) * (1/2) *2500 * 1025 ^2 * 2/3

= 216954903 + 40525548= 257480451 257.48 kN-m

R = = 0.67024274 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.67/25] = 0.001591552 * 500

Pt = 0.159 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.159

Ast req = 2450 mm2

Using 16 Φ of bars, no. of bars required = 2450 / 201= 13

Spacing = 200 mm

Provide 13Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 975 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 376 mmqu = 0.1550 + 0.2115 ) /2

= 0.1833 N/mm2

Muy = 0.183 * 2500 * 1100 /2

= 277186800 Nmm

R = = 0.784254187 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.784/25] 0.001873182 * 500

Pt = 0.187

Ast req = 1760.78559 mm2

Ast min = 1350 mm2 < 1760.78559 mm2

Using 16 Φ of bars, no. of bars required = 1761 / 201= 9

Spacing = 300 mm

Provide 9Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1050 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 969 KN B = 300 mmMux 147 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2500 mm L = 2500 mm D = 450 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 13 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 200 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 9 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 300 mm

STAAD MODEL NODE NO. 110

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 980 KN B = 300 mmMux 150 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 150*10^3 2500

980*1.10 2500

= 139 mm

Assuming e < L/6 (i.e. , L > 6 * 139 = 834 mm ) and load factor for bearing capacity as 1.5

980*1.10 + 150 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 980 * 1.15 * L - 150 * 6 <= 0

Assuming B = 1.00 m => L>= 3.551 m or #VALUE! mB = 1.50 m => L>= 2.545 m or #VALUE! mB = 2.00 m => L>= -0.591 m or 2.028812 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2500 mm L = 2500 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 156.8 + 57.6BL BL2/6

= 214.4 KN/m2

qu,min = Pu + Mux = 156.8 - 57.6BL BL2/6

= 99.2 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 214.4 - 57.6 * {(1025-d)/2}/1250

= 190.784 + 0.02304 d

= 199 KN/m2 ( assuming d = 370 mm )

= 0.199 N/mm2

Vu1 = 0.199 * 2500 * (1025-d)

= 510729 - 498.272 d 326368.36

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2500 * d = 900 d 333000

Vu1 <= Vuc => 510729 - 498 d <= 900dVuc = Vu1 =

333000 > 326368.36 provide, THICKNESS OF FOOTING IS SAFEd >= 370 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 156.8 KN/m2= 0.157 N/mm2

Vu2 = 0.157[2500 * 2500 - (300 + d)(450 + d) ]

Assuming d = 370 mm

Vu2 = 894994.2 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 894.9942 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 370 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 428 mmProvide D = 450 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 450 - 50 - 8 = 392 mm

Effective depth (short span) dy = 392 - 16 = 376 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 980 / (2.5 * 2.5 ) + {( 24 * 0.45) + 18 * (1.5 - 0.45)} *1.5) + 150 * 6 / (2 * 2.5 ^2)

= 156.8 32.25 57.6

= 246.65 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 392 mmqu = 0.1672 N/mm2 at face of column

= 0.2144 N/mm2 at footing edge

Mux = (0.167 * 2500 * 1025 ^2) + ( 0.214 -0.167 ) * (1/2) *2500 * 1025 ^2 * 2/3

= 219538600 + 41352600= 260891200 260.89 kN-m

R = = 0.6791212 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.679/25] = 0.001613362 * 500

Pt = 0.161 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.161

Ast req = 2450 mm2

Using 16 Φ of bars, no. of bars required = 2450 / 201= 13

Spacing = 200 mm

Provide 13Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 975 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1025 mmwidth = 2500 mmdx = 376 mmqu = 0.1568 + 0.2144 ) /2

= 0.1856 N/mm2

Muy = 0.186 * 2500 * 1100 /2

= 280720000 Nmm

R = = 0.794250792 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.794/25] 0.001898032 * 500

Pt = 0.190

Ast req = 1784.15118 mm2

Ast min = 1350 mm2 < 1784.15118 mm2

Using 16 Φ of bars, no. of bars required = 1784 / 201= 9

Spacing = 300 mm

Provide 9Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1050 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 980 KN B = 300 mmMux 150 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2500 mm L = 2500 mm D = 450 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 13 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 200 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 9 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 300 mm

STAAD MODEL NODE NO. 91

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 1080 KN B = 300 mmMux 147 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 147*10^3 2600

1080*1.10 2600

= 124 mm

Assuming e < L/6 (i.e. , L > 6 * 124 = 744 mm ) and load factor for bearing capacity as 1.5

1080*1.10 + 147 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 1080 * 1.15 * L - 147 * 6 <= 0

Assuming B = 1.00 m => L>= 3.789 m or 1.980303 mB = 1.50 m => L>= 2.694 m or #VALUE! mB = 2.00 m => L>= -0.551 m or 2.134857 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2600 mm L = 2600 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 159.8 + 50.2BL BL2/6

= 209.9 KN/m2

qu,min = Pu + Mux = 159.8 - 50.2BL BL2/6

= 109.6 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 209.9 - 50.2 * {(1075-d)/2}/1300

= 189.197026 + 0.019300795 d

= 197 KN/m2 ( assuming d = 380 mm )

= 0.197 N/mm2

Vu1 = 0.197 * 2600 * (1075-d)

= 549305 - 510.981452 d 355132

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2600 * d = 936 d 355680

Vu1 <= Vuc => 549305 - 511 d <= 936dVuc = Vu1 =

355680 > 355132.0482 provide, THICKNESS OF FOOTING IS SAFEd >= 380 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 159.8 KN/m2= 0.16 N/mm2

Vu2 = 0.16[2600 * 2600 - (300 + d)(450 + d) ]

Assuming d = 380 mm

Vu2 = 991296 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 991.296 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 380 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 438 mmProvide D = 450 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 450 - 50 - 8 = 392 mm

Effective depth (short span) dy = 392 - 16 = 376 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 1080 / (2.6 * 2.6 ) + {( 24 * 0.45) + 18 * (1.5 - 0.45)} *1.5) + 147 * 6 / (2 * 2.6 ^2)

= 159.7633136 32.25 50.18207

= 242.1953801 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1075 mmwidth = 2600 mmdx = 392 mmqu = 0.1684 N/mm2 at face of column

= 0.2099 N/mm2 at footing edge

Mux = (0.168 * 2600 * 1075 ^2) + ( 0.21 -0.168 ) * (1/2) *2600 * 1075 ^2 * 2/3

= 253062544 + 41560682.89= 294623227 294.62 kN-m

R = = 0.737431187 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.737/25] = 0.00175712 * 500

Pt = 0.176 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.176

Ast req = 2548 mm2

Using 16 Φ of bars, no. of bars required = 2548 / 201= 13

Spacing = 208.333333 mm

Provide 13Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1025 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1075 mmwidth = 2600 mmdx = 376 mmqu = 0.1598 + 0.2099 ) /2

= 0.1849 N/mm2

Muy = 0.185 * 2600 * 1150 /2

= 317810836 Nmm

R = = 0.864608822 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.865/25] 0.002073742 * 500

Pt = 0.207

Ast req = 2027.2921 mm2

Ast min = 1404 mm2 < 2027.2921 mm2

Using 16 Φ of bars, no. of bars required = 2027 / 201= 11

Spacing = 250 mm

Provide 11Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1100 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 1080 KN B = 300 mmMux 147 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2600 mm L = 2600 mm D = 450 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 13 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 200 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 11 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 250 mm

STAAD MODEL NODE NO. 92

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 1333 KN B = 300 mmMux 150 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

e = 150*10^31333*1.10

= 102 mm

Assuming e < L/6 (i.e. , L > 6 * 102 = 612 mm ) and load factor for bearing capacity as 1

1333*1.10 + 150 <= 250 * 1 KN/m2

BL

250 * 1 * BL2 - 1333 * 1.15 * L - 150 * 6 <= 0

Assuming B = 1.00 m => L>= 6.425 m or 5.1687 mB = 1.50 m => L>= 4.450 m or 3.147663 mB = 2.00 m => L>= -0.521 m or 3.45377 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2800 mm L = 2800 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 170.0 + 41.0BL BL2/6

= 211.0 KN/m2

qu,min = Pu + Mux = 170.0 - 41.0BL BL2/6

= 129.0 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 211.0 - 41 * {(1175-d)/2}/1400

= 193.819307 + 0.014642337 d

= 200 KN/m2 ( assuming d = 430 mm )

= 0.200 N/mm2

Vu1 = 0.2 * 2800 * (1175-d)

= 658380 - 560.3234329 d 417440.92

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2800 * d = 1008 d 433440

Vu1 <= Vuc => 658380 - 560 d <= 1008dVuc = Vu1 =

433440 > 417440.9238 provide, THICKNESS OF FOOTING IS SAFEd >= 430 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 170.0 KN/m2= 0.17 N/mm2

Vu2 = 0.17[2800 * 2800 - (300 + d)(450 + d) ]

Assuming d = 430 mm

Vu2 = 1223592 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 1223.592 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 430 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 488 mmProvide D = 500 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 500 - 50 - 8 = 442 mm

Effective depth (short span) dy = 442 - 16 = 426 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 1333 / (2.8 * 2.8 ) + {( 24 * 0.5) + 18 * (1.5 - 0.5)} *1) + 150 * 6 / (2 * 2.8 ^2)

= 170.0255102 32.25 40.99854

= 243.2740525 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1175 mmwidth = 2800 mmdx = 442 mmqu = 0.1766 N/mm2 at face of column

= 0.2110 N/mm2 at footing edge

Mux = (0.177 * 2800 * 1175 ^2) + ( 0.211 -0.177 ) * (1/2) *2800 * 1175 ^2 * 2/3

= 341373871 + 44339496.4= 385713367 385.71 kN-m

R = = 0.705118517 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.705/25] = 0.001677342 * 500

Pt = 0.168 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.168

Ast req = 3094 mm2

Using 16 Φ of bars, no. of bars required = 3094 / 201= 16

Spacing = 180 mm

Provide 16Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1125 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1175 mmwidth = 2800 mmdx = 426 mmqu = 0.1700 + 0.2110 ) /2

= 0.1905 N/mm2

Muy = 0.191 * 2800 * 1250 /2

= 416772959 Nmm

R = = 0.820204795 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.82/25] 0.001962692 * 500

Pt = 0.196

Ast req = 2341.10163 mm2

Ast min = 1680 mm2 < 2341.10163 mm2

Using 16 Φ of bars, no. of bars required = 2341 / 201= 12

Spacing = 245.454545 mm

Provide 12Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1200 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 1333 KN B = 300 mmMux 150 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2800 mm L = 2800 mm D = 500 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 16 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 180 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 12 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 240 mm

STAAD MODEL NODE NO. 93

qa

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 1083 KN B = 300 mmMux 147 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 147*10^3 2600

1083*1.10 2600

= 123 mm

Assuming e < L/6 (i.e. , L > 6 * 123 = 738 mm ) and load factor for bearing capacity as 1.5

1083*1.10 + 147 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 1083 * 1.15 * L - 147 * 6 <= 0

Assuming B = 1.00 m => L>= 3.796 m or 2.001939 mB = 1.50 m => L>= 2.699 m or #VALUE! mB = 2.00 m => L>= -0.550 m or 2.138355 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2600 mm L = 2600 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 160.2 + 50.2BL BL2/6

= 210.4 KN/m2

qu,min = Pu + Mux = 160.2 - 50.2BL BL2/6

= 110.0 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 210.4 - 50.2 * {(1075-d)/2}/1300

= 189.640813 + 0.019300795 d

= 197 KN/m2 ( assuming d = 390 mm )

= 0.197 N/mm2

Vu1 = 0.197 * 2600 * (1075-d)

= 551085 - 512.6371188 d 351156.52

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2600 * d = 936 d 365040

Vu1 <= Vuc => 551085 - 513 d <= 936dVuc = Vu1 =

365040 > 351156.5237 provide, THICKNESS OF FOOTING IS SAFEd >= 390 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 160.2 KN/m2= 0.16 N/mm2

Vu2 = 0.16[2600 * 2600 - (300 + d)(450 + d) ]

Assuming d = 390 mm

Vu2 = 988864 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 988.864 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 390 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 448 mmProvide D = 450 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 450 - 50 - 8 = 392 mm

Effective depth (short span) dy = 392 - 16 = 376 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 1083 / (2.6 * 2.6 ) + {( 24 * 0.45) + 18 * (1.5 - 0.45)} *1.5) + 147 * 6 / (2 * 2.6 ^2)

= 160.2071006 32.25 50.18207

= 242.639167 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 1075 mmwidth = 2600 mmdx = 392 mmqu = 0.1689 N/mm2 at face of column

= 0.2104 N/mm2 at footing edge

Mux = (0.169 * 2600 * 1075 ^2) + ( 0.21 -0.169 ) * (1/2) *2600 * 1075 ^2 * 2/3

= 253729251 + 41560682.89= 295289934 295.29 kN-m

R = = 0.739099929 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.739/25] = 0.001761232 * 500

Pt = 0.176 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.176

Ast req = 2548 mm2

Using 16 Φ of bars, no. of bars required = 2548 / 201= 13

Spacing = 208.333333 mm

Provide 13Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1025 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 1075 mmwidth = 2600 mmdx = 376 mmqu = 0.1602 + 0.2104 ) /2

= 0.1853 N/mm2

Muy = 0.185 * 2600 * 1150 /2

= 318573817 Nmm

R = = 0.866684522 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.867/25] 0.002078952 * 500

Pt = 0.208

Ast req = 2032.37979 mm2

Ast min = 1404 mm2 < 2032.37979 mm2

Using 16 Φ of bars, no. of bars required = 2032 / 201= 11

Spacing = 250 mm

Provide 11Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 1100 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 1083 KN B = 300 mmMux 147 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2600 mm L = 2600 mm D = 450 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 13 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 200 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 11 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 250 mm

STAAD MODEL NODE NO. 94

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 551 KN B = 300 mmMux 129 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 129*10^3 2100

551*1.10 2100

= 213 mm

Assuming e < L/6 (i.e. , L > 6 * 213 = 1278 mm ) and load factor for bearing capacity as 1.5

551*1.10 + 129 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 551 * 1.15 * L - 129 * 6 <= 0

Assuming B = 1.00 m => L>= 2.456 m or #VALUE! mB = 1.50 m => L>= 1.830 m or #VALUE! mB = 2.00 m => L>= -0.689 m or 1.497351 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2100 mm L = 2100 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 124.9 + 83.6BL BL2/6

= 208.5 KN/m2

qu,min = Pu + Mux = 124.9 - 83.6BL BL2/6

= 41.4 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 208.5 - 83.6 * {(825-d)/2}/1050

= 175.686057 + 0.039798232 d

= 187 KN/m2 ( assuming d = 290 mm )

= 0.187 N/mm2

Vu1 = 0.187 * 2100 * (825-d)

= 324372 - 393.1778426 d 210350.43

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2100 * d = 756 d 219240

Vu1 <= Vuc => 324372 - 393 d <= 756dVuc = Vu1 =

219240 > 210350.4257 provide, THICKNESS OF FOOTING IS SAFEd >= 290 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 124.9 KN/m2= 0.125 N/mm2

Vu2 = 0.125[2100 * 2100 - (300 + d)(450 + d) ]

Assuming d = 290 mm

Vu2 = 496675 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 496.675 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 290 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 348 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 8 = 292 mm

Effective depth (short span) dy = 292 - 16 = 276 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 551 / (2.1 * 2.1 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1.5) + 129 * 6 / (2 * 2.1 ^2)

= 124.9433107 32.25 83.57629

= 240.7695983 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 825 mmwidth = 2100 mmdx = 292 mmqu = 0.1429 N/mm2 at face of column

= 0.2085 N/mm2 at footing edge

Mux = (0.143 * 2100 * 825 ^2) + ( 0.209 -0.143 ) * (1/2) *2100 * 825 ^2 * 2/3

= 102090443 + 31286260.93= 133376704 133.38 kN-m

R = = 0.744894868 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.745/25] = 0.001775572 * 500

Pt = 0.178 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.178

Ast req = 1533 mm2

Using 16 Φ of bars, no. of bars required = 1533 / 201= 8

Spacing = 285.714286 mm

Provide 8Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 775 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 825 mmwidth = 2100 mmdx = 276 mmqu = 0.1249 + 0.2085 ) /2

= 0.1667 N/mm2

Muy = 0.167 * 2100 * 900 /2

= 141805102 Nmm

R = = 0.886450313 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.886/25] 0.002128572 * 500

Pt = 0.213

Ast req = 1233.71631 mm2

Ast min = 882 mm2 < 1233.71631 mm2

Using 16 Φ of bars, no. of bars required = 1234 / 201= 7

Spacing = 333.333333 mm

Provide 7Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 850 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 551 KN B = 300 mmMux 129 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2100 mm L = 2100 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 8 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 280 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 7 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 330 mm

STAAD MODEL NODE NO. 95

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 430 KN B = 300 mmMux 138 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 138*10^3 2000

430*1.10 2000

= 292 mm

Assuming e < L/6 (i.e. , L > 6 * 292 = 1752 mm ) and load factor for bearing capacity as 1.5

430*1.10 + 138 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 430 * 1.15 * L - 138 * 6 <= 0

Assuming B = 1.00 m => L>= 2.245 m or #VALUE! mB = 1.50 m => L>= 1.704 m or #VALUE! mB = 2.00 m => L>= -0.782 m or 1.412345 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2000 mm L = 2000 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 107.5 + 103.5BL BL2/6

= 211.0 KN/m2

qu,min = Pu + Mux = 107.5 - 103.5BL BL2/6

= 4.0 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 211.0 - 103.5 * {(775-d)/2}/1000

= 170.89375 + 0.05175 d

= 185 KN/m2 ( assuming d = 270 mm )

= 0.185 N/mm2

Vu1 = 0.185 * 2000 * (775-d)

= 286543 - 369.7325 d 186715.23

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2000 * d = 720 d 194400

Vu1 <= Vuc => 286543 - 370 d <= 720dVuc = Vu1 =

194400 > 186715.225 provide, THICKNESS OF FOOTING IS SAFEd >= 270 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 107.5 KN/m2= 0.108 N/mm2

Vu2 = 0.108[2000 * 2000 - (300 + d)(450 + d) ]

Assuming d = 270 mm

Vu2 = 387676.8 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 387.6768 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 270 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 326 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 6 = 294 mm

Effective depth (short span) dy = 294 - 12 = 282 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 430 / (2 * 2 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1.5) + 138 * 6 / (2 * 2 ^2)

= 107.5 32.25 103.5

= 243.25 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 775 mmwidth = 2000 mmdx = 294 mmqu = 0.1308 N/mm2 at face of column

= 0.2110 N/mm2 at footing edge

Mux = (0.131 * 2000 * 775 ^2) + ( 0.211 -0.131 ) * (1/2) *2000 * 775 ^2 * 2/3

= 78554242 + 32118421.88= 110672664 110.67 kN-m

R = = 0.640200056 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.64/25] = 0.00151792 * 500

Pt = 0.152 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.152

Ast req = 1470 mm2

Using 12 Φ of bars, no. of bars required = 1470 / 113= 14

Spacing = 146.153846 mm

Provide 14Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 725 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 775 mmwidth = 2000 mmdx = 282 mmqu = 0.1075 + 0.2110 ) /2

= 0.1593 N/mm2

Muy = 0.159 * 2000 * 850 /2

= 115058125 Nmm

R = = 0.723417616 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.723/25] 0.001722482 * 500

Pt = 0.172

Ast req = 971.476204 mm2

Ast min = 840 mm2 < 971.476204 mm2

Using 12 Φ of bars, no. of bars required = 971 / 113= 9

Spacing = 237.5 mm

Provide 9Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 800 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 430 KN B = 300 mmMux 138 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2000 mm L = 2000 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 14 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 140 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 9 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 230 mm

STAAD MODEL NODE NO. 96

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 315 kN B = 300 mm

Mu(RDM) 111.6 kN-m 124 kN-m L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 111.6*10^3 2000

315*1.10 2000

= 322 mm

Assuming e < L/6 (i.e. , L > 6 * 322 = 1932 mm ) and load factor for bearing capacity as 1.5

315*1.10 + 111.6 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 315 * 1.15 * L - 111.6 * 6 <= 0

Assuming B = 1.00 m => L>= 1.876 m or #VALUE! mB = 1.50 m => L>= 1.442 m or #VALUE! mB = 2.00 m => L>= -0.742 m or 1.203708 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2000 mm L = 2000 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 78.8 + 83.7BL BL2/6

= 162.5 KN/m2

qu,min = Pu + Mux = 78.8 - 83.7BL BL2/6

= -5.0 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 162.5 - 83.7 * {(775-d)/2}/1000

= 130.01625 + 0.04185 d

= 140 KN/m2 ( assuming d = 240 mm )

= 0.140 N/mm2

Vu1 = 0.14 * 2000 * (775-d)

= 217093 - 280.1205 d 149864.08

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2000 * d = 720 d 172800

Vu1 <= Vuc => 217093 - 280 d <= 720dVuc = Vu1 =

172800 > 149864.08 provide, THICKNESS OF FOOTING IS SAFEd >= 240 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 78.8 KN/m2= 0.079 N/mm2

Vu2 = 0.079[2000 * 2000 - (300 + d)(450 + d) ]

Assuming d = 240 mm

Vu2 = 286564.6 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 286.5646 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 240 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 296 mmProvide D = 300 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 300 - 50 - 6 = 244 mm

Effective depth (short span) dy = 244 - 12 = 232 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 315 / (2 * 2 ) + {( 24 * 0.3) + 18 * (1.5 - 0.3)} *1.5) + 111.6 * 6 / (2 * 2 ^2)

= 78.75 32.25 83.7

= 194.7 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 775 mmwidth = 2000 mmdx = 244 mmqu = 0.0976 N/mm2 at face of column

= 0.1625 N/mm2 at footing edge

Mux = (0.098 * 2000 * 775 ^2) + ( 0.162 -0.098 ) * (1/2) *2000 * 775 ^2 * 2/3

= 58610489 + 25974028.13= 84584517 84.58 kN-m

R = = 0.710364462 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.71/25] = 0.001690272 * 500

Pt = 0.169 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.169

Ast req = 1220 mm2

Using 12 Φ of bars, no. of bars required = 1220 / 113= 11

Spacing = 190 mm

Provide 11Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 725 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 775 mmwidth = 2000 mmdx = 232 mmqu = 0.0788 + 0.1625 ) /2

= 0.1206 N/mm2

Muy = 0.121 * 2000 * 850 /2

= 87133500 Nmm

R = = 0.809429808 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.809/25] 0.001935832 * 500

Pt = 0.194

Ast req = 898.224056 mm2

Ast min = 720 mm2 < 898.224056 mm2

Using 12 Φ of bars, no. of bars required = 898 / 113= 8

Spacing = 271.428571 mm

Provide 8Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 800 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 315 kN B = 300 mmMu(RDM) 111.6 kN-m L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2000 mm L = 2000 mm D = 300 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 11 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 190 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 8 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 270 mm

STAAD MODEL NODE NO. 97

FULL MXz

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 498 KN B = 300 mmMux 139 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 139*10^3 2100

498*1.10 2100

= 254 mm

Assuming e < L/6 (i.e. , L > 6 * 254 = 1524 mm ) and load factor for bearing capacity as 1.5

498*1.10 + 139 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 498 * 1.15 * L - 139 * 6 <= 0

Assuming B = 1.00 m => L>= 2.391 m or #VALUE! mB = 1.50 m => L>= 1.798 m or #VALUE! mB = 2.00 m => L>= -0.751 m or 1.481162 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2100 mm L = 2100 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 112.9 + 90.1BL BL2/6

= 203.0 KN/m2

qu,min = Pu + Mux = 112.9 - 90.1BL BL2/6

= 22.9 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 203.0 - 90.1 * {(825-d)/2}/1050

= 167.601462 + 0.042883366 d

= 180 KN/m2 ( assuming d = 280 mm )

= 0.180 N/mm2

Vu1 = 0.18 * 2100 * (825-d)

= 311172 - 377.1784904 d 205562

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2100 * d = 756 d 211680

Vu1 <= Vuc => 311172 - 377 d <= 756dVuc = Vu1 =

211680 > 205562.0227 provide, THICKNESS OF FOOTING IS SAFEd >= 280 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 112.9 KN/m2= 0.113 N/mm2

Vu2 = 0.113[2100 * 2100 - (300 + d)(450 + d) ]

Assuming d = 280 mm

Vu2 = 450485.8 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 450.4858 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 280 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 338 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 8 = 292 mm

Effective depth (short span) dy = 292 - 16 = 276 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 498 / (2.1 * 2.1 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1.5) + 139 * 6 / (2 * 2.1 ^2)

= 112.9251701 32.25 90.05507

= 235.2302397 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 825 mmwidth = 2100 mmdx = 292 mmqu = 0.1322 N/mm2 at face of column

= 0.2030 N/mm2 at footing edge

Mux = (0.132 * 2100 * 825 ^2) + ( 0.203 -0.132 ) * (1/2) *2100 * 825 ^2 * 2/3

= 94493768 + 33711552.48= 128205321 128.21 kN-m

R = = 0.716013238 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.716/25] = 0.00170422 * 500

Pt = 0.170 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.170

Ast req = 1533 mm2

Using 16 Φ of bars, no. of bars required = 1533 / 201= 8

Spacing = 285.714286 mm

Provide 8Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 775 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 825 mmwidth = 2100 mmdx = 276 mmqu = 0.1129 + 0.2030 ) /2

= 0.1580 N/mm2

Muy = 0.158 * 2100 * 900 /2

= 134338776 Nmm

R = = 0.839776905 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.84/25] 0.002011582 * 500

Pt = 0.201

Ast req = 1165.90936 mm2

Ast min = 882 mm2 < 1165.90936 mm2

Using 16 Φ of bars, no. of bars required = 1166 / 201= 6

Spacing = 400 mm

Provide 6Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 850 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 498 KN B = 300 mmMux 139 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2100 mm L = 2100 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 8 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 280 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 6 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 400 mm

STAAD MODEL NODE NO. 98

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 582 KN B = 300 mmMux 138 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 138*10^3 2200

582*1.10 2200

= 216 mm

Assuming e < L/6 (i.e. , L > 6 * 216 = 1296 mm ) and load factor for bearing capacity as 1.5

582*1.10 + 138 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 582 * 1.15 * L - 138 * 6 <= 0

Assuming B = 1.00 m => L>= 2.567 m or #VALUE! mB = 1.50 m => L>= 1.909 m or #VALUE! mB = 2.00 m => L>= -0.707 m or 1.560889 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2200 mm L = 2200 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 120.2 + 77.8BL BL2/6

= 198.0 KN/m2

qu,min = Pu + Mux = 120.2 - 77.8BL BL2/6

= 42.5 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 198.0 - 77.8 * {(875-d)/2}/1100

= 167.081313 + 0.035345946 d

= 177 KN/m2 ( assuming d = 290 mm )

= 0.177 N/mm2

Vu1 = 0.177 * 2200 * (875-d)

= 341363 - 390.1296018 d 228225.42

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2200 * d = 792 d 229680

Vu1 <= Vuc => 341363 - 390 d <= 792dVuc = Vu1 =

229680 > 228225.4155 provide, THICKNESS OF FOOTING IS SAFEd >= 290 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 120.2 KN/m2= 0.12 N/mm2

Vu2 = 0.12[2200 * 2200 - (300 + d)(450 + d) ]

Assuming d = 290 mm

Vu2 = 528408 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 528.408 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 290 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 348 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 8 = 292 mm

Effective depth (short span) dy = 292 - 16 = 276 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 582 / (2.2 * 2.2 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1.5) + 138 * 6 / (2 * 2.2 ^2)

= 120.2479339 32.25 77.76108

= 230.2590158 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 875 mmwidth = 2200 mmdx = 292 mmqu = 0.1362 N/mm2 at face of column

= 0.1980 N/mm2 at footing edge

Mux = (0.136 * 2200 * 875 ^2) + ( 0.198 -0.136 ) * (1/2) *2200 * 875 ^2 * 2/3

= 114666868 + 34729233.19= 149396101 149.40 kN-m

R = = 0.796435997 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.796/25] = 0.001903472 * 500

Pt = 0.190 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.190

Ast req = 1606 mm2

Using 16 Φ of bars, no. of bars required = 1606 / 201= 8

Spacing = 300 mm

Provide 8Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 825 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 875 mmwidth = 2200 mmdx = 276 mmqu = 0.1202 + 0.1980 ) /2

= 0.1591 N/mm2

Muy = 0.159 * 2200 * 950 /2

= 157974793 Nmm

R = = 0.942642358 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.943/25] 0.002270212 * 500

Pt = 0.227

Ast req = 1378.47275 mm2

Ast min = 924 mm2 < 1378.47275 mm2

Using 16 Φ of bars, no. of bars required = 1378 / 201= 7

Spacing = 350 mm

Provide 7Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 900 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 582 KN B = 300 mmMux 138 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2200 mm L = 2200 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 8 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 300 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 7 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 350 mm

STAAD MODEL NODE NO. 99

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 477 KN B = 300 mmMux 136 KNm L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 136*10^3 2100

477*1.10 2100

= 259 mm

Assuming e < L/6 (i.e. , L > 6 * 259 = 1554 mm ) and load factor for bearing capacity as 1.5

477*1.10 + 136 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 477 * 1.15 * L - 136 * 6 <= 0

Assuming B = 1.00 m => L>= 2.332 m or #VALUE! mB = 1.50 m => L>= 1.758 m or #VALUE! mB = 2.00 m => L>= -0.750 m or 1.449964 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2100 mm L = 2100 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 108.2 + 88.1BL BL2/6

= 196.3 KN/m2

qu,min = Pu + Mux = 108.2 - 88.1BL BL2/6

= 20.1 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 196.3 - 88.1 * {(825-d)/2}/1050

= 161.659494 + 0.041957826 d

= 173 KN/m2 ( assuming d = 270 mm )

= 0.173 N/mm2

Vu1 = 0.173 * 2100 * (825-d)

= 299702 - 363.2750243 d 201617.74

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2100 * d = 756 d 204120

Vu1 <= Vuc => 299702 - 363 d <= 756dVuc = Vu1 =

204120 > 201617.7434 provide, THICKNESS OF FOOTING IS SAFEd >= 270 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 108.2 KN/m2= 0.108 N/mm2

Vu2 = 0.108[2100 * 2100 - (300 + d)(450 + d) ]

Assuming d = 270 mm

Vu2 = 431956.8 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 431.9568 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 270 mm

Assuming clear cover = 50 mm and bar diameter = 16 mm

D >= 328 mmProvide D = 350 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 350 - 50 - 8 = 292 mm

Effective depth (short span) dy = 292 - 16 = 276 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 477 / (2.1 * 2.1 ) + {( 24 * 0.35) + 18 * (1.5 - 0.35)} *1.5) + 136 * 6 / (2 * 2.1 ^2)

= 108.1632653 32.25 88.11144

= 228.5247004 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 825 mmwidth = 2100 mmdx = 292 mmqu = 0.1270 N/mm2 at face of column

= 0.1963 N/mm2 at footing edge

Mux = (0.127 * 2100 * 825 ^2) + ( 0.196 -0.127 ) * (1/2) *2100 * 825 ^2 * 2/3

= 90792994 + 32983965= 123776959 123.78 kN-m

R = = 0.691281302 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.691/25] = 0.001643262 * 500

Pt = 0.164 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.164

Ast req = 1533 mm2

Using 16 Φ of bars, no. of bars required = 1533 / 201= 8

Spacing = 285.714286 mm

Provide 8Nos OF 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 775 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 825 mmwidth = 2100 mmdx = 276 mmqu = 0.1082 + 0.1963 ) /2

= 0.1522 N/mm2

Muy = 0.152 * 2100 * 900 /2

= 129462245 Nmm

R = = 0.809292796 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.809/25] 0.001935492 * 500

Pt = 0.194

Ast req = 1121.80785 mm2

Ast min = 882 mm2 < 1121.80785 mm2

Using 16 Φ of bars, no. of bars required = 1122 / 201= 6

Spacing = 400 mm

Provide 6Nos 16Φ bars at uniform spacing in the long direction

Dvelopment length required == 752 mm< 850 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 477 KN B = 300 mmMux 136 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2100 mm L = 2100 mm D = 350 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 8 Nos OF 16 Φ bars at uniform spacing in the long direction

WITH SPACING 280 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 6 Nos 16 Φ bars at uniform spacing in the long direction

WITH SPACING 400 mm

STAAD MODEL NODE NO. 100

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 274 kN B = 300 mm

Mux 84 kN-m 120 kN-m L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 84*10^3 1800

274*1.10 1800

= 279 mm

Assuming e < L/6 (i.e. , L > 6 * 279 = 1674 mm ) and load factor for bearing capacity as 1.5

274*1.10 + 84 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 274 * 1.15 * L - 84 * 6 <= 0

Assuming B = 1.00 m => L>= 1.629 m or #VALUE! mB = 1.50 m => L>= 1.252 m or #VALUE! mB = 2.00 m => L>= -0.643 m or 1.044956 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 1800 mm L = 1800 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 84.6 + 86.4BL BL2/6

= 171.0 KN/m2

qu,min = Pu + Mux = 84.6 - 86.4BL BL2/6

= -1.9 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 171.0 - 86.4 * {(675-d)/2}/900

= 138.580247 + 0.048010974 d

= 150 KN/m2 ( assuming d = 230 mm )

= 0.150 N/mm2

Vu1 = 0.15 * 1800 * (675-d)

= 181792 - 269.3209877 d 119848.17

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 1800 * d = 648 d 149040

Vu1 <= Vuc => 181792 - 269 d <= 648dVuc = Vu1 =

149040 > 119848.1728 provide, THICKNESS OF FOOTING IS SAFEd >= 230 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 84.6 KN/m2= 0.085 N/mm2

Vu2 = 0.085[1800 * 1800 - (300 + d)(450 + d) ]

Assuming d = 230 mm

Vu2 = 244766 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 244.766 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 230 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 286 mmProvide D = 300 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 300 - 50 - 6 = 244 mm

Effective depth (short span) dy = 244 - 12 = 232 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 274 / (1.8 * 1.8 ) + {( 24 * 0.3) + 18 * (1.5 - 0.3)} *1.5) + 84 * 6 / (2 * 1.8 ^2)

= 84.56790123 32.25 86.41975

= 203.2376543 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 675 mmwidth = 1800 mmdx = 244 mmqu = 0.1062 N/mm2 at face of column

= 0.1710 N/mm2 at footing edge

Mux = (0.106 * 1800 * 675 ^2) + ( 0.171 -0.106 ) * (1/2) *1800 * 675 ^2 * 2/3

= 43537500 + 17718750= 61256250 61.26 kN-m

R = = 0.571607935 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.572/25] = 0.001350612 * 500

Pt = 0.135 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.135

Ast req = 1098 mm2

Using 12 Φ of bars, no. of bars required = 1098 / 113= 10

Spacing = 188.888889 mm

Provide 10Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 625 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 675 mmwidth = 1800 mmdx = 232 mmqu = 0.0846 + 0.1710 ) /2

= 0.1278 N/mm2

Muy = 0.128 * 1800 * 750 /2

= 64687500 Nmm

R = = 0.667685419 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.668/25] 0.001585272 * 500

Pt = 0.159

Ast req = 662.008929 mm2

Ast min = 648 mm2 < 662.008929 mm2

Using 12 Φ of bars, no. of bars required = 662 / 113= 6

Spacing = 340 mm

Provide 6Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 700 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 274 kN B = 300 mmMux 84 kN-m L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 1800 mm L = 1800 mm D = 300 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 10 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 180 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 6 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 340 mm

STAAD MODEL NODE NO. 101

FULL MXz

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 202 KN B = 300 mm

Mud 102.2 KNm 146 kN-m L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 102.2*10^3 2000

202*1.10 2000

= 460 mm

Assuming e < L/6 (i.e. , L > 6 * 460 = 2760 mm ) and load factor for bearing capacity as 1.5

202*1.10 + 102.2 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 202 * 1.15 * L - 102.2 * 6 <= 0

Assuming B = 1.00 m => L>= 1.609 m or #VALUE! mB = 1.50 m => L>= 1.260 m or #VALUE! mB = 2.00 m => L>= -0.768 m or 1.064399 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 2000 mm L = 2000 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 50.5 + 76.7BL BL2/6

= 127.2 KN/m2

qu,min = Pu + Mux = 50.5 - 76.7BL BL2/6

= -26.2 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 127.2 - 76.7 * {(775-d)/2}/1000

= 97.448125 + 0.038325 d

= 105 KN/m2 ( assuming d = 210 mm )

= 0.105 N/mm2

Vu1 = 0.105 * 2000 * (775-d)

= 163519 - 210.99275 d 119210.52

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 2000 * d = 720 d 151200

Vu1 <= Vuc => 163519 - 211 d <= 720dVuc = Vu1 =

151200 > 119210.5225 provide, THICKNESS OF FOOTING IS SAFEd >= 210 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 50.5 KN/m2= 0.051 N/mm2

Vu2 = 0.051[2000 * 2000 - (300 + d)(450 + d) ]

Assuming d = 210 mm

Vu2 = 186833.4 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 186.8334 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 210 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 266 mmProvide D = 275 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 275 - 50 - 6 = 219 mm

Effective depth (short span) dy = 219 - 12 = 207 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 202 / (2 * 2 ) + {( 24 * 0.275) + 18 * (1.5 - 0.275)} *1.5) + 102.2 * 6 / (2 * 2 ^2)

= 50.5 32.25 76.65

= 159.4 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 775 mmwidth = 2000 mmdx = 219 mmqu = 0.0677 N/mm2 at face of column

= 0.1272 N/mm2 at footing edge

Mux = (0.068 * 2000 * 775 ^2) + ( 0.127 -0.068 ) * (1/2) *2000 * 775 ^2 * 2/3

= 40690091 + 23786251.56= 64476343 64.48 kN-m

R = = 0.672174715 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.672/25] = 0.001596292 * 500

Pt = 0.160 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.160

Ast req = 1095 mm2

Using 12 Φ of bars, no. of bars required = 1095 / 113= 10

Spacing = 211.111111 mm

Provide 10Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 725 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 775 mmwidth = 2000 mmdx = 207 mmqu = 0.0505 + 0.1272 ) /2

= 0.0888 N/mm2

Muy = 0.089 * 2000 * 850 /2

= 64176062.5 Nmm

R = = 0.748863013 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.749/25] 0.001785392 * 500

Pt = 0.179

Ast req = 739.150754 mm2

Ast min = 660 mm2 < 739.150754 mm2

Using 12 Φ of bars, no. of bars required = 739 / 113= 7

Spacing = 316.666667 mm

Provide 7Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 800 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 202 KN B = 300 mmMud 102.2 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 2000 mm L = 2000 mm D = 275 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 10 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 210 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 7 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 310 mm

STAAD MODEL NODE NO. 102

FULL MXz

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODE

DESIGN OF ISOLATED FOOTING

Given: ColumnPu 205 KN B = 300 mm

Mux 98 KNm 140 kN-m L = 450 mm

250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25Fe = 500

Assuming the weight of the footing plus backfill to constitute about 10 % of Pu,resultant eccentricity of loading at footing base,

Footing result

e = 98*10^3 1800

205*1.10 1800

= 435 mm

Assuming e < L/6 (i.e. , L > 6 * 435 = 2610 mm ) and load factor for bearing capacity as 1.5

205*1.10 + 98 <= 250 * 1.5 KN/m2

BL

250 * 1.5 * BL2 - 205 * 1.15 * L - 98 * 6 <= 0

Assuming B = 1.00 m => L>= 1.588 m or #VALUE! mB = 1.50 m => L>= 1.242 m or #VALUE! mB = 2.00 m => L>= -0.748 m or 1.048443 m

An economical proportion of the base slab is generally one in which the projection beyond the face of column is approximately equal in both direction

Therefore provide, B = 1800 mm L = 1800 mm

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OKThickness of footing based on shear

Factored (net) soil pressure qu,max = Pu + Mux = 63.3 + 100.8BL BL2/6

= 164.1 KN/m2

qu,min = Pu + Mux = 63.3 - 100.8BL BL2/6

= -37.6 KN/m2

(a) One-way shear

The critical section is located d away from the column face, therefore the average pressure contributing to the factored one-way shear is

qu = 164.1 - 100.8 * {(675-d)/2}/900

= 126.286008 + 0.0560128029 d

= 139 KN/m2 ( assuming d = 220 mm )

= 0.139 N/mm2

Vu1 = 0.139 * 1800 * (675-d)

= 168410 - 249.49588477 d 113520.91

Assuming γc = 0.36 Mpa for M25 concrete with nominal Pt = 0.25(IS:456, Table 19)

Vuc = 0.36 * 1800 * d = 648 d 142560

Vu1 <= Vuc => 168410 - 249 d <= 648dVuc = Vu1 =

142560 > 113520.9053 provide, THICKNESS OF FOOTING IS SAFEd >= 220 mm PROVIDE DECIDED THICKNESS HERE

(b) Two-way shear

The critical section located d/2 from the periphery of the column all round. The average pressure contributing to the factored two-way shear is

qu = 63.3 KN/m2= 0.063 N/mm2

Vu2 = 0.063[1800 * 1800 - (300 + d)(450 + d) ]

Assuming d = 220 mm

Vu2 = 182170.8 N

For two-way shear resistance, limiting shear stress of concrete

γcz = where Ks= 0.5 + 300/450but limited to 1

γcz = 1.250 Mpa

Vuc = γcz * b*d= 1.25* [(300 + d) + (450 + d)] * 2 * d

d = 404 mmVuc = 1573580 N

= 1574 KN

Vuc > Vu21574 182.1708 provide, THICKNESS OF FOOTING IS SAFE

Hence one way shear governs the footing slab thickness and d >= 220 mm

Assuming clear cover = 50 mm and bar diameter = 12 mm

D >= 276 mmProvide D = 300 mm PROVIDE DECIDED THICKNESS FOR ONE & TWO WAY SHEAR

Effective depth (long span) dx = 300 - 50 - 6 = 244 mm

Effective depth (short span) dy = 244 - 12 = 232 mm

Check maximum soil pressure

Assuming, unit weight of concrete = 24 KN/m2unit weight of soil = 18 KN/m2

qmax-gross = 205 / (1.8 * 1.8 ) + {( 24 * 0.3) + 18 * (1.5 - 0.3)} *1.5) + 98 * 6 / (2 * 1.8 ^2)

= 63.27160494 32.25 100.823

= 196.3446502 250 KN/m2

OUTCOME ACTING PRESSURE IS LESS THEN SBC HENCE OK

Design of flexural reinforcement

(a)Long Span

cantilever projection = 675 mmwidth = 1800 mmdx = 244 mmqu = 0.0885 N/mm2 at face of column

= 0.1641 N/mm2 at footing edge

Mux = (0.088 * 1800 * 675 ^2) + ( 0.164 -0.088 ) * (1/2) *1800 * 675 ^2 * 2/3

= 36281250 + 20671875= 56953125 56.95 kN-m

R = = 0.5314536583 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.531/25] = 0.001253222 * 500

Pt = 0.125 %% OF STEEL PROVIDED % OF STEEL REQUIRED

Pt assumed for one-way shear= 0.25 > 0.125

Ast req = 1098 mm2

Using 12 Φ of bars, no. of bars required = 1098 / 113= 10

Spacing = 188.888889 mm

Provide 10Nos OF 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 625 mm Hence , MEANS DEVLOPMENT BAR IS OK

(b)Short Span

cantilever projection = 675 mmwidth = 1800 mmdx = 232 mmqu = 0.0633 + 0.1641 ) /2

= 0.1137 N/mm2

Muy = 0.114 * 1800 * 750 /2

= 57552083 Nmm

R = = 0.5940357393 MPa

Pt / 100 = 25[1 -√1 - 4.598 * 0.594/25] 0.001405182 * 500

Pt = 0.141

Ast req = 586.802629 mm2

Ast min = 648 mm2 < 586.802629 mm2

Using 12 Φ of bars, no. of bars required = 587 / 113= 6

Spacing = 340 mm

Provide 6Nos 12Φ bars at uniform spacing in the long direction

Dvelopment length required == 564 mm< 700 mm Hence , ok

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME EMO FOOTING CODEDESIGN OF ISOLATED FOOTINGINPUT DATA ColumnPu 205 KN B = 300 mmMux 98 KNm L = 450 mmqa 250 KN/m2 DEPTH OF FOUNDATION = 1.5 m Fck = 25

Fe = 500

final resultSIZE OF FOOTING

B = 1800 mm L = 1800 mm D = 300 mmBOTTOM REINFORCEMENT IN LONGER SIDE L =Provide 10 Nos OF 12 Φ bars at uniform spacing in the long direction

WITH SPACING 180 mmBOTTOM REINFORCEMENT IN SHORTER SIDE B =Provide 6 Nos 12 Φ bars at uniform spacing in the long direction

WITH SPACING 340 mm

STAAD MODEL NODE NO. 103

FULL MXz

qa

Lf

Bf

thickness Df

BL2/6

Ks (0.25)√Fck

kN/m2 <

Mu/bdx2

47 Φ (for M20 with Fe 415)

Mu/bdy2

47 Φ (for M20 with Fe 415)

DAMDUM INDUSTRIAL ESTATE BHUTANBUILDING NAME -: EMO

Footing SummaryL B D IN L DIRECTION IN B DIRECTION

DIA SPACING NO. DIA SPACING NO. FORMULA FOR NO. OF BARSNODE 84 2000 2000 350 12 140 14 12 230 9 15 461 116 121NODE 86 2250 2250 400 16 230 10 16 300 8 741 130 134NODE 87 2500 2500 440 16 210 12 16 260 10 996 144 148NODE 88 2250 2250 400 16 230 10 16 300 8 748 129 135NODE 89 2000 2100 320 16 330 7 16 470 5 386 112 118NODE 90 2200 2200 380 16 260 9 16 350 7 666 133NODE 110 2500 2500 450 16 200 13 16 300 9 969 967 140 147NODE 91 2500 2500 450 16 200 13 16 300 9 980 150NODE 92 2600 2600 450 16 200 13 16 250 11 1080 147NODE 93 2800 2800 500 16 180 16 16 240 12 1333 155 150NODE 94 2600 2600 450 16 200 13 16 250 11 1083 147NODE 95 2100 2100 350 16 280 8 16 330 7 551 129NODE 96 2000 2000 350 12 140 14 12 230 9 430 138NODE 97 1900 1900 300 12 180 11 12 250 8 315 111.6 124NODE 98 2100 2100 350 16 280 8 16 400 6 498 482 139NODE 99 2200 2200 350 16 300 8 16 350 7 582 138NODE 100 2100 2100 350 16 280 8 16 400 6 477 136NODE 101 1800 1800 300 12 180 10 12 340 6 274 84 120NODE 102 2000 2000 275 12 210 10 12 270 8 202 210 146NODE 103 1800 1800 300 12 180 10 12 280 7 205 140