Slab Design Me

8/11/2019 Slab Design Me

1/79

Design of slab

Given Data:-

1. No of storey = 52. Plinth height = 0.6 m

3. Length along shorter direction = 5.60 m

4. Length along longer direction = 6.00 m

5. Cover for mild exposure = 15.00 mm

6. Bars used = 10.00 mm (assummed)

7. Bars used in edge strip of slab = 8.00 mm (assummed)

8. Characteristic Strength of concrete, fck= 20.00 N/mm2

9. Characteristic Strength ofsteel, fy= 415.00 N/mm2

Loading

i) Loading on slab

a) Self wt due to dead load = 24000 N/m2

b) Loading due to 25 mm flooring

c) 10 mm thick ceiling plaster 250 N/m2

d) 10 mm thick turfelting 250 N/m2

ii) Load due to partition wall on slab = 1000 N/m2

iii) Live load on floor slab = 4000 N/m2

iv) Live load on roof = 1500 N/m2

v) Basic wind pressure = 1500 N/m2

Ref Step

1. Thickness of slab and durability consideration

lx = 5.6 m

Cover for mild exposure= 15.00 mm

Basic value of span / eff depth = 26

d= 215.3846 mm

Provided d = 110 mmOverall depth = 130 mm

2. Calculation of design load

Dead Load = 3620 N/m2

Live load = 1500 N/m2

Design load = (Factored Load) 1.5 (D.L + L.L) = 7680 N/m2

3. Calculation of moment

ly/lx = 1.071

Case -I - Two adjacent edges discontinous

x (+ve moment at mid span) = 0.0386x (-ve moment at continous edge) = 0.0513

y (+ve moment at mid span) = 0.035

y (-ve moment at continous edge) = 0.047

Longer direction:

8429.57 N-m (+ve moment)

11319.71 N-m (-ve moment)

Shorter direction:

Calculation

IS 456-

2000, Cl

23.2.1

IS 456-

2000,Annex D-

1.1, Table

26

IS 456-

2000,

Table 18


2/79



Case -II - Interior Panels

x (+ve moment at mid span) = 0.027

x (-ve moment at continous edge) = 0.0355

y(+ve moment at mid span) = 0.024


Longer direction:



Shorter direction:



Case -III - One short edge discontinous


x (-ve moment at continous edge) = 0.0413y (+ve moment at mid span) = 0.028


Longer direction:



Shorter direction:



Case -III - One long edge discontinous

x (+ve moment at mid span) = 0.0316x (-ve moment at continous edge) = 0.0420



Longer direction:



Shorter direction:



Maxm moment = 12345.70 N-m

Check depth for maxm B.M

Mmax= 0.138fckbd2

d = 66.88 < 110 mm OK

4. Check for shear

Maxm design shear, V = 0.5wlx= 21504 N

Shear Strss, c = V/bd = 0.195 N/mm2

Table C of

SP 16

Table 12.3

of 'RCC

Design' by

Ver hese &


3/79

< Safe minm shear stress, 0.28 N/mm2

5. Area of steel


Middle strip (For shorter direction)

Width of middle strip = 4.5 m

Steel at midspan

M = 8429.57 N-m d = 110 mm

M/bd2

= 0.697

% steel = 0.203%

Steel provided > 0.12% which is minimum

Ast = 223.3 mm2

Spacing of 10 mm dia bar= 351.72 mm

Provided spacing : 330 mm c/c < 3d or 450 mm

Steel at support

M = 11319.71 N-m d = 110 mm

M/bd2

= 0.936

% steel = 0.288%


Ast = 316.25 mm2



Middle strip (For longer direction)

Steel at midspan


M = 9296.61 N-m d = 100.00 mm

M/bd2

= 0.930

% steel = 0.274%Steel provided > 0.12% which is minimum

Ast = 273.5 mm2



Steel at support

M = 12345.70 N-m d = 100 mm

M/bd2

= 1.235

% steel = 0.371%


Ast = 371 mm2



Edge strip (For shorter direction)

Width of edge strip = 0.75 m

Minm reinforcement @ 0.12% of cross sectional area is provided

Ast = 156 mm2

(Considering 1 m in width)

IS 456-

2000,

Annex D-

SP 16

Table 2

IS 456

Table 19


4/79

No of 8 mm dia bar used = 3 nos

Edge strip (For longer direction)



Ast = 156 mm2



Torsional reinforcement

Area of steel (at corner where two adjacent edges discontinus) =

= 3/4 th Amount of reinforcement required for maxm +ve midspan BM =

167.475 mm2

Reinforcement is to be provided for a distance of lx/5 = 1.12 m

in 4 layers at mutually perpendicular direction both at top & bottom

Spacing of reinforcement of 8 mm dia = 300.1372 mm

Provided spacing : 300 mm c/c

Area of corner steel = 83.7375 mm2



6. Check of deflection


% of steel provided along lx direction = 0.203%

Modification factor = 1.6

Allowable value of span / eff depth = 41.6

d= 134.62 mm

> provided i.e. 110 mm, OK7. Check cracking

Steel > 0.12%, minm required, OK.

Spacing of steel < 3d, OK.

Diameter of steel < 200/8 i.e. 25 mm, OK.

.

IS 456-

2000,

Annex D-

1.8 & D-1.9

Amount of reinforcement at corner where one edge is continous & other

discontinous is to be provided equal to half of that provided earlier.

IS 456-

2000, Cl

23.2.1 &

Fig. 3

IS 456-

2000, Cl

25.5.2.1


5/79


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Design of slab

Given Data:-

1. Length along shorter direction = 5.60 m

2. Length along longer direction = 6.00 m

3. Cover for mild exposure = 15.00 mm

4. Bars used = 10.00 mm (assummed)

5. Bars used in edge strip of slab = 8.00 mm (assummed)

Loadingi) Loading on slab

a) Self wt due to dead load = 24000 N/m2

b) Loading due to 25 mm flooring 750 N/m2

c) 10 mm thick ceiling plaster 250 N/m2

d) 10 mm thick turfelting 250 N/m2

ii) Load due to partition wall on slab = 1000 N/m2

iii) Live load on floor slab = 4000 N/m2

iv) Live load on roof = 1500 N/m2

Ref Step1. Thickness of slab and durability consideration

lx = 5.6 m

Cover for mild exposure= 15.00 mm


d= 215.3846 mm

Provided d = 140 mm (assummed)

Overall depth = 160 mm

2. Calculation of design load

Dead Load = 5840 N/m2

Live load = 4000 N/m

2

Design load = (Factored Load) 1.5 (D.L + L.L) = 14760 N/m2

3. Calculation of moment

ly/lx = 1.071

Case -I - Two adjacent edges discontinousx (+ve moment at mid span) = 0.0386




Longer direction:

16200.58 N-m (+ve moment)21600.77 N-m (-ve moment)

Shorter direction:



Case -II - Interior Panels



Calculation

IS 456-

2000, Cl

23.2.1

IS 456-

2000,Table 18

IS 456-

2000,

Annex D-

1.1, Table

26


14/79



Longer direction:



Shorter direction:



Case -III - One short edge discontinous





Longer direction:



Shorter direction:14349.08 N-m (+ve moment)


Case -III - One long edge discontinous





Longer direction:



Shorter direction:



Maxm moment = 23822.56 N-m

Check depth for maxm B.M

Mmax= 0.138fckbd2

d = 92.91 < 140 mm OK

4. Check for shear

Maxm design shear, V = 0.5wlx= 41328 NShear Strss, c = V/bd = 0.295 N/mm

2

< Safe minm shear stress, 0.28 N/mm2

5. Area of steel


Middle strip (For shorter direction)


Table C of

SP 16

Table 12.3

of 'RCCDesign' by

Verghese &

IS 456

Table 19


15/79

Steel at midspan

M = 16200.58 N-m d = 140 mm

M/bd2

= 0.827 % steel = 0.241%


Ast = 337.40 mm2



Steel at support

M = 21600.77 N-m d = 140 mm

M/bd2

= 1.102 % steel = 0.328%


Ast = 458.53006 mm2



Middle strip (For longer direction)

Steel at midspan


M = 17866.92 N-m d = 130.00 mm

M/bd2

= 1.057 % steel = 0.313%


Ast = 407.14531 mm2



Steel at support

M = 23822.56 N-m d = 130 mm

M/bd2

= 1.410 % steel = 0.429%


Ast = 557.12436 mm2



Edge strip (For shorter direction)



Ast = 192 mm2



Edge strip (For longer direction)Width of edge strip = 0.7 m


Ast = 192 mm2



Torsional reinforcement

Area of steel (at corner where two adjacent edges discontinus) =

IS 456-

2000,

Annex D-

1.7

IS 456-


16/79

= 3/4 th Amount of reinforcement required for maxm +ve midspan BM =

253.0488 mm2

Reinforcement is to be provided for a distance of lx/5 = 1.12 m

in 4 layers at mutually perpendicular direction both at top & bottom



Area of corner steel = 126.5244 mm2



6. Check of deflection


% of steel provided along lx direction = 0.241%

Modification factor = 1.4

Allowable value of span / eff depth = 36.4

d= 153.85 mm> provided i.e. 140 mm, OK

7. Check cracking

Steel > 0.12%, minm required, OK.

Spacing of steel < 3d, OK.

Diameter of steel < 200/8 i.e. 25 mm, OK.

IS 456-

2000, Cl

25.5.2.1

2000,

Annex D-

1.8 & D-1.9

Amount of reinforcement at corner where one edge is continous & other

discontinous is to be provided equal to half of that provided earlier.

IS 456-

2000, Cl

23.2.1 &

Fig. 3


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Assumption

1. No axial load is coming on beam

3. In case of horizontal loading, point of inflexon occurs at mid span.

4. Maximum +ve B.M. occurs at the centre and maxmimum -ve B.M. occurs at the support.

5. Stiffness of column (I/L) is assummed to be same at each storey.

Ref Step

1. Calculation of load on beam

In larger direction

Overall depth of beam = 500 mm (assummed)

Width of beam = 300 mm (assummed)

Diameter of bar provided in longer direction = 25 mm

Diameter of bar provided in shorter direction = 25 mm

Diameter of 2 lgged stirrup = 8 mm

Spacing of 2 lgged stirrup = 200 mmCover of reinforcement = 25 mm

Height of each storey, h = 3.9 m

Dead Load

For weight of wall coming on beam, 25 to 30% should be deducted for opening

30% of weight of wall per meter on floor = 300 N/m2

Hence, dead load of floor slab on beam = 5540 N/m2

Dead load on beam from slab = 99276.8 N

(Bothside of beam is considered)

Eqivalent slab load, Wsl= 16546.13 N/m

Self weight of beam = 3600.00 N/m

Hence, total D.L., Wd(factored)= 30219.20 N/m

Live load

Live load of floor slab on beam = 4000 N/m2

L.L. on beam from slab = 71680.00 N


Eqivalent live load on beam per meter = 11946.67 N/m

Hence, total L.L., Wl(factored)= 17920.00 N/m

In shorter direction

Overall depth of beam = 450 mm (assummed)

Width of beam = 300 mm (assummed)Dead Load

Hence, dead load of floor slab on beam = 5540 N/m2

Dead load on beam from slab = 86867.2 N


Eqivalent slab load, Wsl= 15512 N/m

Self weight of beam = 3240.00 N/m

Hence, total D.L., Wd(factored) = 28128.00 N/m

2. Point of inflexon i.e. the point of zero moment occur at 1/10th span span frm the support in vertical loading.

Calculation

IS 456-

2000,

Cl 24.5


26/79

Live load

L.L. on beam from slab = 62720 N


Eqivalent live load on beam per meter = 11200 N/m

Hence, total L.L., Wl(factored) = 16800.00 N/m

2. Calculation of B.M. and S.F. on beam in longer direction.

Wd = 30219.20 N/m

Wl = 17920.00 N/m

Span Moment

B.M. near middle of end span = 155169.6 N-m

B.M. at middle of interior span = 99088.8 N-m

Support Moment

B.M. at support next to the end support = -180469.12 N-m

B.M. at other interior support = -162337.60 N-m

Shear forces

At end support = 120910.08 NAt support next to the end support (outer side) = 173301.12 N

At support next to the end support (inner side) = 164235.36 N

At other interior support = 155169.60 N

Hence, maxm +ve B.M. = 155169.6 N-m

Maxm -ve B.M. = -180469.12 N-m

Maxm S.F. = 173301.12 N

3. Calculation of B.M. and S.F. on beam in shorter direction.

Wd = 28128 N/m

Wl = 16800 N/m

Span Moment

B.M. near middle of end span = 126192.64 N-m

B.M. at middle of interior span = 80657.92 N-m

Support Moment

B.M. at support next to the end support = -146748.07 N-m

B.M. at other interior support = -132046.51 N-m

Shear forces

At end support = 105342.72 N

At support next to the end support (outer side) = 150958.08 N

At support next to the end support (inner side) = 143082.24 N

At other interior support = 135206.40 N

Hence, maxm +ve B.M. = 126192.64 N-m

Maxm -ve B.M. = -146748.07 N-m

Maxm S.F. = 150958.08 N

4. Analysis of frame with lateral loads (Considering wind load on longer side of the frame)

(Considering wind on larger side of the frame)

Total surface area = 482.4 Sqm

IS 456-

2000,

Cl

22.5.1,

Table

12

IS 456-

2000,

Cl

22.5.1,

Table

12

IS 456-

2000,

Cl

22.5.1

&

22.5.2,

Table

13

IS 456-2000,

Cl

22.5.1

&

22.5.2,

Table

13


27/79

Nos of bay: 4

Effective surface area (after 20% reduction for opening) = 385.92 Sqm

Basic wind pressure = 1500 N/m2

Total wind load = 578880 N

Wind load/bay = 144720 N

Wind load/bay/each storey = 28944 N

Hence, P = 28944 N

(assummed to be acting on junction)

Calculation of column shear

Assumption

From Portal method, we have, shear in each interior column =

2 X Shear in each exterior column of same storey (x) = 2x

5 storied building is being designed

At Gr floor

x + 2x + 2x + x = 2 x P/2 + 4 x P =

or, 6x = 5P = 144720 N

x = 24120 N

At 1st floor

x + 2x + 2x + x = 1 x P/2 + 4 x P =or, 6x = 4.5P = 130248 N

x = 21708 N

At 2nd floor

x + 2x + 2x + x = 1 x P/2 + 3 x P =

or, 6x = 3.5P = 101304 N

x = 16884 N

At 3rd floor

x + 2x + 2x + x = 1 x P/2 + 2 x P =

or, 6x = 2.5P = 72360 N

x = 12060 N

At 4th floor

x + 2x + 2x + x = 1 x P/2 + 1 x P =

or, 6x = 1.5P = 43416 N

x = 7236 N

At roof

x + 2x + 2x + x = 1 x P/2 =

or, 6x = 0.5P = 14472 N

x = 2412 N

Calculation of column moments

Assumption

Moments at centre is zero. So, point of contraflexure lies at midspan i.e. at mid height

End moment in column = Shear in column X h/2, where h is the storey height.

Moment (end) in exterior columnsAt Gr floor - 1st Floor

M (exterior) = 42330.6 N-m

At 1st floor - 2nd Floor


At 2nd floor - 3rd Floor


At 3rd floor - 4th Floor



28/79

At 4th floor - roof


Moment in interior columns

= 2 X Moment in exterior column of same storey.

Calculation of beam moments

At roof

M = 4703.4 N-m

At 4th floor

M = 18813.6 N-m

At 3rd floor

M = 37627.2 N-m

At 2nd floor

M = 56440.8 N-m

At 1st floor

M = 75254.4 N-m

At Gr floor

M = 49566.6 N-mAxial load in interior column = 0, due to wind load

5. Analysis of frame with lateral loads (Considering wind load on shorter side of the frame)

Total surface area = 337.68 Sqm

Nos of bay: 3

Effective surface area (after 20% reduction for opening) = 270.144 Sqm

Basic wind pressure = 1500 N/m2

Total wind load = 405216 N

Wind load/bay = 135072 N

Wind load/bay/each storey = 27014.4 N

Hence, P = 27014.4 N

(assummed to be acting on junction)

Calculation of column shear

Assumption

From Portal method, we have, shear in each interior column =

2 X Shear in each exterior column of same storey (x) = 2x

5 storied building is being designed

At Gr floor

x + 2x + 2x + 2x + x = 2 x P/2 + 4 x P =

or, 8x = 5P = 135072 N

x = 16884 N

At 1st floor

x + 2x + 2x + 2x + x = 1 x P/2 + 4 x P =or, 8x = 4.5P = 121564.8 N

x = 15195.6 N

At 2nd floor

x + 2x + 2x + 2x + x = 1 x P/2 + 3 x P =

or, 8x = 3.5P = 94550.4 N

x = 11818.8 N

At 3rd floor

x + 2x + 2x + 2x + x = 1 x P/2 + 2 x P =

Beam moment acts in opposite direction to that of column at that joint so that sum of column

moment is equal to the sum of beam moment.


29/79

or, 8x = 2.5P = 67536 N

x = 8442 N

At 4th floor

x + 2x + 2x + 2x + x = 1 x P/2 + 1 x P =

or, 8x = 1.5P = 40521.6 N

x = 5065.2 N

At roof

x + 2x + 2x + 2x + x = 1 x P/2 =or, 8x = 0.5P = 13507.2 N

x = 1688.4 N

Calculation of column moments

Assumption

Moments at centre is zero. So, point of contraflexure lies at midspan i.e. at mid height

End moment in column = Shear in column X h/2, where h is the storey height.

Moment (end) in exterior columns

At Gr floor - 1st Floor


At 1st floor - 2nd Floor

M (exterior) = 23046.66 N-mAt 2nd floor - 3rd Floor


At 3rd floor - 4th Floor


At 4th floor - roof


Moment in interior columns

= 2 X Moment in exterior column of same storey.

Calculation of beam moments

At roof

M = 3292.38 N-m

At 4th floor

M = 13169.52 N-m

At 3rd floor

M = 26339.04 N-m

At 2nd floor

M = 39508.56 N-m

At 1st floor

M = 52678.08 N-m

At Gr floor

M = 36385.02 N-mAxial load in interior column = 0, due to wind load

6. Determination of design moment

Beam (interior) in larger direction (at 3rd floor)

+ve moment due to vertical load = 155169.6 N-m

-ve moment due to vertical load = 180469.12 N-m

End(-ve) moment due to wind load: 26339.04 N-m

Design +ve moment at mid span = 155169.6 N-m

Beam moment acts in opposite direction to that of column at that joint so that sum of column

moment is equal to the sum of beam moment.


30/79

Design -ve moment at support = 206808.16 N-m

Axial load on interior column =

Beam (interior) in shorter direction (at 3rd floor)

+ve moment due to vertical load = 126192.64 N-m

-ve moment due to vertical load = 146748.07 N-m

End(-ve) moment due to wind load: 37627.2 N-m

Design +ve moment at mid span = 126192.64 N-mDesign -ve moment at support = 184375.27 N-m

Axial load on interior column =

Maxm S.F. = 173301.12 N

7. Design of flexural reinforcement in longer direction.For an under-reinforced section,Mmax= 0.138fckbd

Effective depth, d = 433 mm

Effective depth, d (provided) = 462.5 mm > reqd

At mid span

Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd

2

= 2.418 Pt= 0.804%

Ast(reqd) = 1205.77 mm2

No of 25 mm bars provided at bottom of mid span = 3

At support

Pt/100 = Ast/bd = (fck/2fy)*[1-(1-4.598R/fck)] where R = Mu/bd2= 3.223

Pt= 1.183%

Ast(reqd) = 1774.65 mm2

No of 25 mm bars provided at top of support = 4

8. Design of flexural reinforcement in shorter direction.

For an under-reinforced section,Mmax= 0.138fckbd2

Effective depth, d = 390 mm

Effective depth, d (provided) = 412.5 mm > reqd

At mid span


Pt= 0.826%

Ast(reqd) = 1115.74 mm2

No of 25 mm bars provided at bottom of mid span = 3

At support


Pt= 1.417%

Ast(reqd) = 1913.23 mm2

No of 25 mm bars provided at top of support = 4

8. Design of shear reinforcement.

Design shear force, V = 173301.12 N


31/79

Shear stress, v= V/bd = 1.155 N/mm2

Design shear stress of cocrete, c= *0.85 (0.8 fck)(15) -1+/6

where = max(1,0.8fck/0.89pt)

Pt= 1.183%

= 1.96

c= 0.661 N/mm2

< v

Hence, shear reinforcement necessary.Shear reinforcement provided for shear force, Vs = V - cbd

Vs = 74198.4 N

Using vertical stirrups of 2 legged 8 mm c/c (Fe 250 grade)

Asv= 100.53 mm2

Sv= 0.87 fyAsvd/Vs Sv= 226.25 mm

Provided spacing = 200 mm


32/79


33/79


34/79

Assumption

1. Typical interior column is considered. (Project work is consulted)

3. In case of horizontal loading, point of inflexon occurs at mid span.

4. Maximum +ve B.M. occurs at the centre and maxmimum -ve B.M. occurs at the support.5. Stiffness of column (I/L) is assummed to be same at each storey.

Ref Step

1. Calculation of load on column

(a) Column(interior, next to end support) at 3rd floor, Total no: 6

At 3rd floor, S.F. of beam at next to end support (longer direction)

Outer Side: 173301.12 N

Inner Side: 164235.36 N

Avg S.F. at c.g. of column = 168768.24 N

Load = 337536.48 N

S.F. of beam at next to end support (shorter direction)Outer Side: 150958.08 N

Inner Side: 143082.24 N

Avg S.F. at c.g. of column = 147020.16 N

Load = 294040.32 N

Load coming on column from roof slab and beam at roof level

Roof slab = 258048 N

Roof beam (longer side) = 23976 N

Roof beam (Shorter side) = 19353.6 N

Total: 301377.6 N

Load coming on column from 4th floor (assummed as same from 3rd floor)

i.e. 631576.8 N

Deduction of live load @ 10% for each floor from top:

From position of the column, total deduction: 20.00%

Total deduction = 40320 N

Design axial load on column : 1524.21 KN

(b) Column(exterior) at 3rd floor in longer side, Total no: 6

S.F. of beam at end support (longer direction): 120910.08 N

Load = 241820.16 N

S.F. of beam at end support (shorter direction): 105342.72 N

Load = 210685.44 N



Roof beam (longer side) = 11988 NRoof beam (Shorter side) = 19353.6 N

Total: 160365.6 N


i.e. 452505.6 N



Total deduction = 38707.2 N


2. Point of inflexon i.e. the point of zero moment occur at 1/10th span span frm the support in vertical

loading.

Calculation


35/79

Column(exterior) at 3rd floor in shorter side, Total no: 4


Load = 241820.16 N

S.F. of beam at end support (shorter direction): 105342.72 N

Load = 210685.44 N



Roof beam (longer side) = 23976 NRoof beam (Shorter side) = 9676.8 N

Total: 162676.8 N


i.e. 452505.6 N





(d) Column(corner) at 3rd floor, Total no: 4


Load = 241820.16 NS.F. of beam at end support (shorter direction): 105342.72 N

Load = 210685.44 N


Roof slab = 64512 N

Roof beam (longer side) = 11988 N

Roof beam (Shorter side) = 9676.8 N

Total: 86176.8 N


i.e. 452505.6 N





2. Calculation of bending moment on column

B.M. due to dead load & live load = 0

B.M. due to wind load (on larger side) = 47034 N-m

B.M. due to wind load (on shorter side) = 32923.8 N-m

Design B.M.= 47034 N-m

3. Provided section of column, size & no. of reinforcement:

400 mm X 400 mm

Main bar: 4 nos - 25 mm

Tie: 8 mm Cover: 40 mm

M25 fck= 25.00 N/mm2

4. Short Column or Slender Column ?

Unsupported length, ly = 3400 mm (for buckling about y-axis)

Unsupported length, lx 3450 mm (for buckling about x-axis)

Lateral dimension, Dx = 400 mm

Lateral dimension, Dy = 400 mm

Grade of concrete:


36/79


37/79

ley = kyx ly = 4488 mm

Buckling with respect to major (local x-) axis:

jt(Ic/ hs)= 1.09E+06 mm3

jt(Ib/ lb)= 8.14E+05 mm3

1= 2 = 0.47

Referring to Fig. 27 of the Code, ky= 1.39

lex = kxx lx = 4795.5 mm

Alternative: Effective Lengths by Formulas

Buckling with respect to minor (local y-) axis:

1= 2= 1.05

ky = 1.32

ley = 4471.26 mm

Buckling with respect to major (local x-) axis:1= 2= 1.34

kx = 1.40

lex = 4841.70 mm

ley/Dy = 11.22 < 12

lex/Dx = 11.99 < 12

The column should be designed as a short column.

5.

Minimum Eccentricities

ex,min= 20.233 (> 20.0 mm) < 0.5Dx= 20.00 mm

ey,min= 20.133 (> 20.0 mm) < 0.5Dy= 20.00 mm

Design load (factored) = 1524.21 KN

Primary moments for design

Mux = 47.03 KN-M

Muy = 32.92 KN-M

Corresponding (primary) eccentricities:

ex= 30.86 > ex,min= 20.233 mm

ey= 21.60 > ey,min= 20.133 mm

The primary eccentricities should not be less than the minimum eccentricities.

Primary moments for design:

Mux = 47.03 KN-M

IS 456

2000 Cl.

25.4

Ratio of the unsupported length (l) to the least lateral dimension (d) of a column i.e. l/d 60

Note 2: Considering the effective lengths given by the Code charts, the slenderness ratios of

the column are obtained as follows:

Note 1: The effective lengths predicted by the two different methods are fairly close.

2000

Annex E

Fig 27

IS 456

2000 Cl.

25.3.1

As the column is unbraced and bent in double

curvature, Mu= M2+ Mafor unbraced columns, where

M2is the higher of the two end moments


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Muy = 32.92 KN-M

Additional moments

Without modification factor, additional eccentricities

eax= Dx(lex/Dx)2/2000 29.30 mm

eay= Dy(ley/Dy)2/2000 24.99 mm

Assuming modification factors kax= kay 0.5 (to be verified later),

additional moments:Max= Pu(kax eax) 22.33 KN-M

May= Pu(kay eay) 19.05 KN-M

Total factored moments

Mux = Mux+ Max= 69.37 KN-M

Muy = Muy+ May= 51.97 KN-M

Trial section as slender column

Designing for a resultant uniaxial moment with respect to the minor axis,

99.68 KN-M

Pu= 1524.21 KN

0.38

0.06

Assuming 25 for main bars, 8 ties and 40 mm clear cover, d' = 60.5 mm

d'/D = 0.15 0.15

with "Reinforcement Distributed Equally on Four Sides" and referring chart of SP 16

p/fck= 0.04 preqd = 1.00 As,reqd= 1600 mm2

As,provided= 1963.50 mm2

> 1600.00 mm2

Hence, OK

pprovided= 1.23% p/fck= 0.049

Check additional moments

Assuming a clear cover of 40mm d' = 60.5 mm

d'/Dx= 0.15 0.15

d'/Dy= 0.15 0.15

Corresponding to p/fck= 0.049

For d'/Dx= 0.15 Pub,x/fckbD = 0.11

Pub,x= 440.00 KN

For d'/Dy= 0.15 Pub,y/fckbD = 0.11

Pub,y= 440.00 KN

Puz

= 0.45fck

Ag+ (0.75f

y 0.45f

ck)A

s P

uz=

2389.05 KN Modification factors:

The actual (revised) total moments are obtained as:

Mux = Mux+ Max= 66.85 KN-M

Referring to Charts of SP :16, the ultimate loads Pub,x & Pub,yat balanced failure can be

determined by considering the stress level fyd= 0.87fy

0.444

0.444

Pu/fckbD =

Mu/fckbD2=

Mu1.15(Mux2+Muy

2)


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Muy = Muy+ May= 49.83 KN-M

Check safety under biaxial bending

Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to

Pu/fckbD = 0.38 & p/fck= 0.049

For d'/Dx= 0.15 Mux1/fckbD2= 0.09

For d'/Dy= 0.15 Muy1/fckbD2= 0.09

Mux1= 144.00 KN-M > Mux = 66.85 KN-MMuy1= 144.00 KN-M > Muy = 49.83 KN-M

Pu/Puz= 0.638 which lies between 0.2 & 0.8

n= 1.73

6. Transverse reinforcement

Minimum diameter, t = 6.25 mm

Maximum spacing,st= 300 mm

Provided 8 ties @ 300 c/c

Trial section as short column

Designing for a resultant uniaxial moment with respect to the minor axis,

66.02 KN-M

Pu= 1524.21 KN

0.38

0.04

Assuming 25 for main bars, 8 ties and 40 mm clear cover, d' = 60.5 mm

d'/D = 0.15 0.15

with "Reinforcement Distributed Equally on Four Sides" and referring chart of SP 16p/fck= 0.01 preqd = 0.8 As,reqd= 1280 mm

2

As,provided= 1963.50 mm2

> 1280.00 mm2

Hence, OK

pprovided= 1.23% p/fck= 0.049

Check safety under biaxial bending

Referring to the design Charts in SP : 16, uniaxial moment capacities corresponding to

Pu/fckbD = 0.38 & p/fck= 0.049

For d'/Dx= 0.15 Mux1/fckbD2= 0.07

For d'/Dy= 0.15 Muy1/fckbD2= 0.07

Mux1= 112.00 KN-M > Mux = 47.03 KN-M

Muy1= 112.00 KN-M > Muy = 32.92 KN-M

Pu/Puz= 0.638 which lies between 0.2 & 0.8

n= 1.73

Transverse reinforcement

< 1 Hence, safe0.42

0.34 < 1 Hence, safe

IS 456

2000 Cl.26.5.3.2

Mu1.15(Mux2+Muy

2)

Pu/fckbD =

Mu/fckbD2=


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Assumption

Bearing capacity of soil = 120 KN/m2

Depth of soil tested for bearing capacity = 1.5 m

Characteristic strength of concrete, fck= 20 N/mm2

Clear cover: 75 mm

Diameter of bar used = 25 mm

Ref Step

1. Calculation of load on footing (under the column designed).

Load on column at 3rd floor = 1524.211 KN

Includes load from roof = 301.378 KN

-do- from 4th floor = 631.577 KN

-do- from 3rd floor = 631.577 KN

Add: Load from 2nd floor = 631.577 KN

Add: Load from 1st floor = 631.577 KN

Total: 2787.365 KN

Deduction of live load for 1st & 2nd floor = 20%

Total deduction = 40.32 KN

Total load after deduction = 2747.045 KNSelf wt of column @ 6 KN/m

180.9 KN (factored)

Total axial load on footing = 2927.945 KN

Self wt of footing (assummed)= 10.00%

292.794 KN

Design axial load = 3220.739 KN (factored)

Design bending moment = 84.661 KN-M

(due to wind load on larger side on G.L.)

2. Design of isolated footing, eccentrically loaded

Size of footingResultant eccentricity of loading at footing base,

e = 26.29 mm

Assuming e < L/6 (i.e., L > 158 mm

where, L = Dimension along the plane of section of moment.

B = Bradth

q1 & q2= Foundation pressure

Now, soil pressure = 180 KN/m2

BL2 17.893 L 2.82204 0

BL2 17.893 L 2.82204 0

L2 4.473 L 0.70551 0

L = 4.626 m

Assume B = 4.00 m

L = 4.75 m

Projection (in the short direction) = 2175.00 mm

Calculation

Assuming an enhanced soil pressure under ultimate loads is equivalent to considering

allowable pressures at the serviceability limit state. A load factor of 1.5 is assumed here.


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Projection (in the long direction) = 1800.00 mm

Maxm soil pressure, q1= 175.14 KN/m2

Minm soil pressure, q2= 163.88 KN/m2

Thickness of footing based on shear

(a) One-way shear

Assume: d = 975 mmDepth at edge = 250 mm

172.30 KN/m2

0.172 N/mm2

The section at critical plane is trapezoidal.

Width at top = 3300 mm (1:2 slope is maintained)

Eff depth at critical plane, d' = 650.00 mm

Depth of N.A = k d' = 183.95 mm k = 0.283

Width of the section at N.A, b' = 3710.35 mm

Shear force at critical plane,V' = 861.82 KN

Shear stress, =V'/(b'd') = 0.357 N/mm2 > allowable c = 0.36

(for M 20 concrete with nominal pt = 0.25) N/mm2

(b) Two-way shear

Calculated shear stress, v= ksc ks= 1

c = 1.12 N/mm2

Width of footing at critical plane, ao(x-x dirn) = 1375 mm

Width of footing at critical plane, bo(y-y dirn) = 1375 mm

Depth of footing at critical plane, do= 812.5 mm

Pressure under column axis, q = 169.51 KN/m2

Punching shear force, F = 2900.25 KN

Punching shear stress, v= 0.6490 < c= 1.12 N/mm2

(c) Depth from bending

Section Y-Y

Moment about y-y axis, Myy = 1603.80 KN-M

Section X-X

Moment about y-y axis, Myy = 1304.40 KN-M

Width of equivalent rectangle for trapejoidal shaped footing, beqv= b + (B-b)/8

beqv= 943.75 mm

For an under-reinforced section,Mmax= 0.138fckbd2

deff= 784.68 mm < d (assummed) = 975 mm

D 1070 mm

effective depth (long span) = 982.5 mm

The critical section is located d away from the column face. Intensity of pressure

contributing to the factored one-way shear is

IS 456

2000

Table 19

The critical section is located d/2 from the periphery of the column all around. The

average pressure contributing to the factored two-way shear is

IS 456

2000 Cl

31.6.3.1

Hence, one-way shear governs the footing slab thickness and d 975 mm. Assuming a

clear cover of 75 mm and a bar diameter of 25 mm,


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effective depth (short span) = 957.5 mm

3. Check maximum soil pressure (at factored loads)

Assumption

Unit weights of concrete = 24.00 kN/m3

unit weights of soil = 18.00 kN/m3

qmax-gross

= 186.754 kN/m3

> 180 kN/m3

Hence O.K.

4. Design of flexural reinforcement

(a) long span


Pt= 0.099% < Pt(assummed for one way shear) , 0.25 %

Pt(reqd)= 0.250%

Ast(reqd) = 11667.19 mm2

No of 25 mm bars provided = 24Corresponding spacing = 190

Development length required = 1175 mm

Length available from face of the column = 2175 mm > Reqd length

Hence, OK.

Provided 24 no 25 mm bars @ 190 mm c/c along long span.

(a) Short span


Pt= 0.101% < Pt(assummed for one way shear) , 0.25 %

Pt(reqd)= 0.250%

Ast(reqd) = 9575.00 mm2

No of 25 mm bars provided = 20

Corresponding spacing = 225

Development length required = 1175 mm

Length available from face of the column = 1800 mm > Reqd length

Hence, OK.

Provided 20 no 25 mm bars @ 225 mm c/c along long span.

Placing of steel

(a) Reinforcements in long direction placed uniformly across the full width.

5. Transfer of forces at column base

Capacity =0.45 fck x 2 x Area = 2880 KN < 2927.945

Hence, dowels are needed.

At least 4 no bars equal to 0.5% of the area of supported column are extended to the footing.

Reinf in central brand width/ Total reinf in short direction = 2/(1), where is the ratio of

the long side to the short side.

The critical sections for moment are located at the faces of the column in both directions

(XX and YY)

IS 456

2000 Cl

34.3.1

(b) Reinforcement in short direction will be placed in a central band equal to the width of

the footing and portion of the reinforcement determined for central band as per ratiobelow,


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Building Construction Tips

1. Anti-Termite Treatment:

a) Heptachlor 0.5 %

b) Chlordane 1.0%

c) Chloropyrifos 1.0%

2. Hydrated Lime & Clay Pozzolana:

3. Use of Flyash:

To prevent attack of termites, any of the following chemicals may be used by mixing with water to

make a solution that can be sprayed @ 5 liters per Sq Meter.

May be used for all masonry mortar and plaster. 20 to 30% cement can be saved by using Lime

Pozzolana. Lime Pozzolana mortar is cheaper than cement sand mortar. It also improves workability.

Qty of cement can be saved up to 20%, if Flyash is used in conjunction with Cement in mortars and

plasters.


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Cubical Content with floor wise structural cost

Ground Floor

Qty Unit Mark

1. Total Area: 176.53 Sq M A

2. Floor to Floor Height: 3 M B

3. Plinth Height: 0.6 M C

4. Depth of Foundation: 1.2 M D

5. Number of Floors 2 E

6. Ht of Parapet 1 M F

Cubical Content: [A X {B X E + C + (D/2) + F/2}] = 1359.28

Rate: 3,250.00 per Cum =

ABSTRACT

1. Total Structural Cost: = 4,417,663.25

2. Internal S & P works: 15% of Structural Cost = 662,649.49

3. Internal Electrical Works: 20% of Structural Cost = 883,532.65

4. Anti Termite: 1% of Structural Cost = 44,176.63

5. Quality Control: 1% of Structural Cost = 44,176.63

6. Foundation Treatment: 6% of Structural Cost = 265,059.80

7. External Electrification: 6% of Structural Cost = 265,059.80Sub Total 6,582,318.24

Add: Contingency: 3% = 197,469.55

Add: Administrative Charge: 8.5% = 559,497.05

7,339,284.84

4,417,663.25


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Cu M


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Spacing of Stirrup

Beam: Simply Supported Beam

Dimension: Wide: 200 mm

Eff Depth: 500 mm

Loading: F= 25000 N/m

(Uniformly distributed load including its own weight)

Eff Span: l = 6.00 mDetermination the length over which vertical stirrups to be provided

cbc = 5 N/mm2

c = 0.35 N/mm2

c max= 1.6 N/mm2

sv = 140 N/mm2

fy = 250 N/mm2

Maxm Shear Force, V = 75000 N

Shear Stress, v= 0.75 N/mm2 > 0.35 N/mm

2 Shear reinf

< 1.6 N/mm2

Distance (x) from the end of the beam where permissible shear stress is developed is given by

x = 1.6 m

Shear to be resisted by vertical stirrups, Vs = V - c . b . d 40000 N

Shear Reinf: Dia of reinf: 6 mm (2 legged stirrup)

Asv 56.57 mm2

Spacing, Sv at support: Asv . sv. D / Vs = 99.00 mm

Maxm Spacing: 0.75 d = 375 mm

Asv/bsv 0.4/fy where Asv 31.68 mm2

Sv Asvfy/0.4 b 176.78 mm

Hence, 6 mm dia 2 legged stirrups near support at 95 mm c/c is adopted.


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necessary


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Design of Beam

Data given:

Section of the Beam: Rectangular Grade of Concrete: M 30

Grade of Steel: Fe 250

b= 250 mm

d= 350 mm Analysis as underreinforced beaml= 4 mm

Ref Step

1 Depth of N.A. for balanced failure

SP 16

For M 30, Es= 200000 N/mm2

For Fe 250, fy = 250 N/mm2

x/d = 0.531

Balanced Percentage of Steel

IS 875 Assumed unit wt of RCC: 25 kN/m3

(Uniformly distributed load including its own weight)

Dead Load = 0.2 x 1 x 25 = 5 kN/m2

Live Load= 300 kg/m2

= 3 kN/m2

2 Factored Load, Moment and Shear

Factoted Load, w = 1.5 ( D.L + L. L)

= 12 kN/m2

Factored (design) moment, Mu = wL2/8

= 24 kN-m

Design Shear, Vu : wL/2 = 24 kN

3 Load for serviceability condition, ws

ws= 1.0 (D.L + L.L) = 8 kN/m2

Calculation

IS 456

Table 12


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Design of Doubly Reinforced Beam

Data given:

Section of the Beam: Rectangular fck = 30 N/mm2

fy = 415 N/mm2

b= 280 mmd= 510 mm Analysis as underreinforced beam

d' = 50 mm

Asc= 402 mm2

Ast= 2455 mm2

Determination of ultimate moment capacity

Ref Step

Method I Solution by Strain Compatibility

1. Choose a suitable NA depth

xu/d for Fe 415 = 0.48

x = 244.8 mm

x= 240 mm (adopted)

2. Check strain and stress in steel

Ec= 0.0035

Est= 0.0039

Yeild strain of Fe 415 = 0.0038 Steel yields

3. Total Tension in steel

T = 0.87 fyAst

T = 886.38 kN

4. Check strain in compression steel

Esc= 0.0035 (1-d'/x) 0.0028

Corresponding stress in steel = 350 N/mm2

Cs= 140.7 kN

5. Compression in concrete and its point of action

Calculation

Compression steel doesn't yield.


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Cc= 725.76 kN

Center of Cc= 0.42 x = 100.8 mm

6. Total Compression

C = 866.46 kN

7. Check T = C

T = 886.38 kN C = 866.46 kN

8. Ultimate moment i.e. moment of Csand Ccabout tension steel

Mu = 361.70 kNm

Method II Solution by formulae

1. Mu1for concrete failure as singly reinforced

= 301.51 kNm

2. Balanced steel for Fe 415, fck= 30

p = 1.43%

Ast

=2042.04

mm2

3. Compression in steel

d'/d = 0.10

fsc= 353 N/mm2

Cs= 141.91 kN

4. Value of Mu2due to compression steel

Mu2= 65.28 kNm

5. Total Muconsidering compression failure

Muc= Mu1+ Mu2 = 367 kNm

6. Additional tension steel available

Ast2= Ast- Ast1 = 412.96 mm2


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