Design of Footings: 1) Design of footing for C1 Column:
Load(w) = 695 KN Fck = 20 N/mm 2 Fy = 415 N/mm2 Size of column = 230 * 280 Safe bearing capacity of soil(q) = 160 KN/m 2 W = Weight of column = 695 KN W1= Weight of footing / self weight = 10% of 695 KN = 10/100 * 695 = W1= 69.5 KN So, Total load on footing = W + W1 = 965 + 69.5 = 764.5 KN Required Area of footing = 764.5/160 Area (A) = 4.77 m 2 Assume, we are take square footing So, Area = B * B 4.77 = B2 B = 2.18 ≈ 2.2 m So, Size of footing = 2.2 * 2.2 Net upward pressure (P) = W/A = 695/2.2 *2.2 P = 143.59 KN/m 2
Calculating of Moment Moment (M) = P4 /8 * B * (B-b) (B-b)2 = 143.59/8 * 2.2 (2.2 – 0.28) 2 M = 145.56 KN-m Depth of footing =√Mu/Ru * B * fck
D = √145.56 * 106 /0.138*20*2.2*1000 /0.138*20*2.2*1000 D = 154.83 ≈ 155 mm Check for shear:Take, d = 2 * 155 [For shear check] D = 300 mm For one way shear:Vu = Pu B [B-b/2 – d] = 143.59 * 2.2 [2.2 – 0.28/2 – 0.3] Vu = 208.49 KN
Τu = Vu/Bd = 208.49 *103 /2200 * 300 Τu = 0.315 Pt = 0.414 * fck/fy * x 4max /d * 100 Pt = 0.957 = 0.75 – 0.55 = 1.0 – 0.60 So, Tc = 0.59
Tc’= k * Tc Tc’= 1.20 * 0.59 Tc’= 0.708 > Tu
Calculating of Moment Moment (M) = P4 /8 * B * (B-b) (B-b)2 = 143.59/8 * 2.2 (2.2 – 0.28) 2 M = 145.56 KN-m Depth of footing =√Mu/Ru * B * fck
D = √145.56 * 106 /0.138*20*2.2*1000 /0.138*20*2.2*1000 D = 154.83 ≈ 155 mm Check for shear:Take, d = 2 * 155 [For shear check] D = 300 mm For one way shear:Vu = Pu B [B-b/2 – d] = 143.59 * 2.2 [2.2 – 0.28/2 – 0.3] Vu = 208.49 KN
Τu = Vu/Bd = 208.49 *103 /2200 * 300 Τu = 0.315 Pt = 0.414 * fck/fy * x 4max /d * 100 Pt = 0.957 = 0.75 – 0.55 = 1.0 – 0.60 So, Tc = 0.59
Tc’= k * Tc Tc’= 1.20 * 0.59 Tc’= 0.708 > Tu
Hence safe O.K Check for 2-way shear:-
Vu’ = Pu [B2 – (b + d)2] = 143.59 [(2.2) 2 – (0.28 + 0.300) 2]
Vu’ = 646.67 KN Τu’ = Vu’/4(b0+d)*d = 646.68 * 10 3 /4(280 + 300) 300) 300
Tu’ = 0.929 N/mm2 Tc’ = 0.25√fy Tc’ = 5.09 N/mm2 Ks = 0.5 + 1 Ks = 1.5 > 1 Adopt, K = 1
Now, Tc’ = Tc’ * Ks Tc’ = 5.09 N/mm2 Tc’>Tv’
Hence safe O.K. Adopt, d = 300 Provide 12 mm bars D= deff + 50 + ø + ø/2
S = 378 ≈ 370 mm deff = 302 mm
d’eff = d - ø d’eff = 290 mm – 4.6 Mu/fck Bd2] Bd’ Ast = 0.5 fck/fy [ 1 - √1 – 4.6
= 0.5 * 20/415 [1 - √1- 4.6*145.56 * 10 6 /20* 2200 * (290)2] 2200 * 290 Ast = 1460.24 mm 2
Number of bars = 1460.69/(π/4) * 122 Number of bars = 12.59 ≈ 13 bars Spacing:-
Spacing = (π/4) * 122 * 1000/ 1460.69 Spacing = 77.446 ≈ 80 mm Provide 12 mm ø bars @ 80 mm c/c.
2.) Design of footing for C2 column:-
Load (W) = 850 KN Fck = 20 N/mm2 Fy = 415 N/mm2 Size of column = 230 * 350 S.B.C(q) = 160 KN/m 2 Weight of column = W = 850 KN Weight of footing = W 1 = 10% of 850 W1 = 85 KN Total load on footing(W) = 935 KN Area = Load/S.B.C = 935/160 Area = 5.84 Shape of footing square Area = B * B 5.84 = B2
B = 2.41 ≈ 2.5 m Size of footing = 2.5 * 2.5 Net upward pressure(P) = W/2.5 * 2.5 P = 136 KN/m2 Calculation for moment :-
Moment (M) = P4 /8 * B (B-b2) = 136/8 * 2.5 (2.5 – 0.35)2 M = 196.45 KN-m
Depth of footing (d) = √Mu/Ru B fck
= √196.45 * 10 6 /0.138 * 20 * 2.5 * 1000 = 168.73 mm ≈ 170 mm d = 170 mm Take, d = 2d [for footing criteria] d = 340 mm Check for shear:For one way shear :Vu = Pu B [B-b/2 – d] = 136 * 2.5 [ 2.5 – 0.35/2 – 0.340] Vu = 249.9 KN Tu = 0.294 Pt = 0.414 * fck/fy * x 4max / D * 100 = 0.414 * 20 /415 * 0.48 * 100 Pt = 0.957 Tc = 0.59
Tc’ = K * Tc = 1.40 * 0.59
Tc’ = 0.708 > Tu Hence safe O.K Check for 2-way shear:-
Vu’ = Pu [B2 – (b + d)2] = 136 [(2.5)2 – ( 0.35 - 0.340) 2]
Vu’ = 785.25 KN Tu’ = Vu’/4(b0 + d) d = 785.25 * 10 3 /4(350 + 340) 340
Tu’ = 0.836 N/mm2
Tu’ = 0.25√fy Tc’ = 5.09 N/mm2 Ks= D.S + 1 KS = 1.5 > 1 Adopt, K = 1
Tc’ = Tc’(old) * K Tc’ = 5.09 N/mm2 Tc’ > Tv’ Hence safe O.K Adopt, d = 340 mm Provide 12 mm ø bars. D = deff + 50 + ø + ø/2 = 160 +50 + 12 + 6
D = 408 mm ≈ 410 mm deff = 342 mm
d’eff = d – ø = 330 mm Ast = 0.5 fck/fy [1 - √1 – 4.6 Mu/Fck Bd 2] B * d’ = 0.5 * 20/415 [ 1 - √1 – 4.6 * 196.45 * 10 6 / 20 * 2500 * 3302] 2500 * 330 Ast = 1724.42 mm2
Number of bars = 1724.42/(π/4) * 122 Number of bars = 15.25 ≈ 16 Spacing = 65.58 ≈ 70 mm Provide 12mm ø bars @ 70 mm c/c spacing.
3.) Design of footing for C3 column:-
Load(W) = 540 KN Fck = 20 N/mm2 Fy = 415 N/mm2 Size of column = 230 * 220 mm S.B.C (q) = 160 Kn/m 2 W = Weight of column = 540 KN W1= Weight of footing/ Self weight = 10% of 540 W1= 54 KN Total load on footing = 594 KN Area = 594/160 Area = 3.713 m 2 Assume, we are take square footing. So, Area = B * B B2= 3.713
B = 1.927 m ≈ 2m So, Size of footing = 2 * 2 m Net upward pressure(P) = W/A P = 135 KN/m2 Calculation of moment:M= P4 /8 * B * (B-b)2 M = 135/8 * 2 * (2 – 0.22)2
M = 106.93 KN-m
Depth of footing (d) = √Mu/Ru*B*fck = √106.93 * 106 /0.138 * 20 * 2000 d = 139.18 mm ≈ 140 mm Take,
d = 139.18 mm ≈ 140 mm d = 280 mm Check for shear:-
For one way shear:Vu = Pu B[B- b/2 – d] = 135 * 2 [2 – 0.22/2 – 0.28] Vu = 164.7 KN Tu = Vu/Bd Tu = 164.7 * 10 3 /2000 * 280 Tu = 0.29 Pt = 0.414 fck/fy * x 4max /d * 100 Pt = 0.957 0.75 – 0.55 1.00 – 0.60 So,
Tc = 0.59
Tc’ = K * Tc = 1.20 * 0.59
Tc’ = 0.708 > Tu Hence safe O.K Check for 2-way shear:-
Vu’ = Pu [B2 – (b + d)2] = 135 [(2)2 – (0.22 + 0.28)2]
Vu’ = 506.25 KN Tu’ = Vu’/4 (b0 + d) d = 506.25 * 10 3 /4(220 + 280) 280
Tu’ = 0.904 N/mm2 Tc’ = 0.25√fy Tc’ = 05.09 N/mm2 Ks = 0.5 + 1 Ks = 1.5 > 1 So, K=1
Tc’ = Tc * K Tc’ = 5.09 * 1 Tc’ = 5.09 N/mm2 Tc’ > Tv’ Hence safe O.K Take, d = 280 mm Provide 12 mm bars D = deff + 50 + ø + ø/2
D = 348 ≈ 350 mm So, deff = 282 mm
d’eff = deff – ø
d’eff = 270 mm Ast = 0.5 fck/fy [ 1 - √1- 4.6 Mu/fck Bd 2] B* d Ast = 0.5 20/415 [1 - √1 – 4.6 * 106.93 * 10 6 / 20* 2000 * (270)2] 2000 * 270 Ast = 1148.10 mm2
Number of bars = 1148.10/(π/4) * 122 Number of bars = 10.15 ≈ 11 bars Spacing:-
Spacing = (π/4) * 122 /1148.10 * 1000 Spacing = 98.50 ≈ 100 mm So, Provide 12 mm ø bar @ 100 mm c/c .
Design of Columns Column C1:- Use – M20 Concrete Steel – Fe-415 Height of the column l= 3.6 m It is restrained in position and direction at both ends. Both ends are fixed Left = 0.65 l = 0.65 x 3.6 = 2.43 m. Arial loads carried by column C 1= Load by beam B 1= 3*75*2 = 225 KN * 2 = 450 KN Load by beam B 2= 3 * 75 = 225 KN Total Arial load carried by column = B 1 + B2 P4 = 450 + 225 = 675 KN Assuming 1% steel, we get Ac= 99%Ag and Asc =1% Ag 675 * 1000 = 0.4 fck A c + 0.67 fy Asc 675 * 1000 = 0.4 * 20 * 99/100 Ag + 0.67 * 415 * 1/100 Ag Ag = 63081.16 mm 2 Using a rectangular column size = 230 * 280 mm Provide 230 * 280 mm Column
Check for slenderness of column Left/b = 2.43 * 1000/260 = 9.346 < 12 Therefore it is a short column Minimum Eccentricity Emin = l/500 + D/30 = 3600/500 + 260/500 + 260/30 = 14.87 Emin /D = 14.87/230 = 0.03174 < 0.05 Hence design as arially loaded short column is satisfactory. Longitudinal reinforcement: Asc= 1/100 * Ag required Asc= 1/100 * 63081.16 Asc= 631 mm2
Use 6 bars of 12 mm diameter: Lateral Ties: Diameter of lateral ties should not be less than a) Ø/4= 12/4 = 3 mm b) 5mm Hence use 6mm bars spacing of 8mm bars It shall be minimum of a) Least lateral dimension = 260 mm b) 16 * ø = 16 * 12 = 190 mm c) 300 mm Provide ø 8mm lateral ties at 190 mm c/c Total load transfer to the footing C 1 Total load = Arial load = Arial load on column C 1 + 3 * self weight of column Total load on footing C 1 = 675 + 3 * 0.23 * 0.28 * 25 * 3.6 = 675 + 20 = 695 KN
Design of Column Column C2 Use – M20 Grade concrete Use – Steel – Fe415 Height of column l = 3.6 m Both ends are fixed Left = 0.65 l = 0.65 * 3.6 Left = 2.43 m Arial load carried by column C 2 = load by B2 + Load by B3 + 2 * Load by B 1 Load by B 1 = 2 * 3 * 75 Load by B 1 = 450 KN Load by B 2 = 3 *75 = 225 KN Load by B 3 = 150 KN Total Arial load carried by column C 2 P4 = 450 + 225 + 150 P4 = 825 KN Assuming 1 % steel, we get Ac = 99% Ag & Asc= 1% Ag 825 * 1000 = 0.4 * 20 * 99/100 Ag + 0.67 * 415 * 1/ 100 Ag Ag = 77099.2 mm 2 Therefore using a rectangular column size = 230 mm * 350 mm Provide 230 mm * 350 mm column Check for slenderness of column Left/b = 2.43 * 1000/230 = 10.56 < 12
Therefore it is a short column Minimum Eccentricity Emin = l/500 + D/30 = 3600/500 + 300/30 = 14.8 Emin /D = 14/230 = 0.049 < 0.05 Hence satisfactory Longitudinal reinforcement Asc = 1/100 * Ag Asc = 1/100 * 77099.2 Asc = 771 mm2 Use 8 bars of 12mm diameter
Lateral Ties: Diameter of lateral ties should not be less than a) ø/4 = 12/4 = 3 mm b) 5mm Hence provide 8 mm bars Spacing:
It shall be minimum of a) Least lateral dimension = 300 mm b) 16 * ø = 16 * 12 = 192 mm c) 300 mm Therefore provide ø 8 mm bar at 190 mm c/c Total load transferred to the footing Load transferred = Arial load on column + 3 * self weight + c + column C 2 Load on footing C 2 = 825 + 3 * 0.3 * .3 * 3.6 * 25 Load on footing C 2 = 825 + 25 = 850 KN
Design of Column Column C3:
Use – M20 concrete Steel – Fe-415 Height of column l = 3.6 m Fixed at both ends Left = 0.65 l = 0.65 * 3.6 Left = 2.43 m Arial load carried by column C 3 = Load by B3 * 2 + Load by B 2 Total Arial load carried by column P4 = 2 * 3 * 50 + 225 Total Arial load P4 =525 KN Assuming 1 % steel, we get Ac = 99% Ag & Asc = 1% Ag 525 *1000 = 7.92 Ag + 2.7805 Ag Ag = 49063.12 mm 2 Therefore using a rectangular column size = 230 mm * 220 mm Provide 230 mm * 220 mm column Check for slenderness of column Left/b = 2430/230 = 10.56 <12 Therefore it is a short column Longitudinal reinforcement Asc = 1/100 * Ag
Asc = 1/100 * 49063.12 Asc = 490.63 mm 2 Provide 6 bars of 12 mm diameter Lateral Ties:
Diameter of lateral ties should not be less than a) ø/4 = 12/4 = 3 mm b) 5 mm Use 8 mm diameter bars. Spacing: It shall be minimum of a) Least lateral dimension = 230 b) 16 * ø = 16 * 12 = 192 ≈ 190 mm c) 300 mm Therefore provide 8 mm lateral ties at 190 mm c/c Total load transfer to the footing Total load transferred = Arial load on column + w + o + column * 3 [That is three floors there] = 525 KN + 15 KN Load on footing C 3 = 540 KN
Design of Beam: Design of beam B1
1.) Given Data:a.) Size – 230 * 450 mm b.) Fck = 25 N/mm 2 c.) Fy = 415 N/mm2 d.) L0 = Clear Span = 3.2 m e.) Ls = Width of support = 200 mm 2.) Loud calculation:Self weight of beam = 230 * 450 * 1 * 25 Self we of slab = ½ * 3.20 * 2.12 * 25 Finishing load = 1 Live load = 4 Weight of brick wall = 2 * 3.6 * 18 = 12.96 KN Applied load = 105.34 KN Factored load = 1.5 * 105.34 KN = 150 KN Bending Moment:BMapplied = Wu * L02 /24 = 150 * 3.42 /24 = 72.25 Kn – m Left => L0 + d = 3.2 + 0.450 = 3.650 Left => L0 + Ls = 3.2 + 0.1 + 0.1 = 3.4
Moment of resistance for Balance section:MR = 0.138 fck bd 2
for beam
= 0.138 * 25 * 230 * (450 – 30 – 16/2) 2 = 134.69 * 106 N-mm MRb= 134.69 KN – m Since MRB > BMapplied. Hence beam design as a simplify. Shear Force:Check for depth:dreq = √mu/Q.B => √72.25 * 106 /138 * 25 * 230 dreq = 301.75 mm So that is safe d req < dprov Area of Steel:Ast = 0.5 fck/fy [ 1- √1 – 4.6 Mu/fck bd 2] bd = 0.5 * 25/415 [1 - √1 – 4.6 * 72.25 * 10 6 /25 * 23 * 4122] * 230 * 412 Ast = 533.7 mm2 Number of bars:Numbers of bars = Ast/Ast 1 = 533.7/(π/4) * 162 = 2.443 ≈ 3 bars Check for deflection:Span/d eff = 20 * mf 1 3400/412 = 20 * mf 1 mf 1 => fs = 0.58 fy * area/ A provided
0.58 * 415 * 419.219/3 * (π/4) * 162
100906.01/603.185 = 167.288
Tension steel y. = Ast/bd * 100 = 0.6 %
3400/412 = 20 * 1.8
8.252 ≤ 36 Hence it is safe.
Check for development length:Ld ≤ 1.3 M1 /Vu + L0 Ld = 0.87 fy ø/4 * T bd = 0.87 * 415 * 16/4 * 1.4 = 1031.57 mm Vu = Wu * Leff /2 = 150 * 3.4/2 = 255 KN Ld ≤ 1.3 M1 /Vu + L0
1031.57 ≤ 1.3 * 72.25 * 106 /235.5877 * 103 + 3200 1031.57 ≤ 3598.68 mm
Hence it is safe. Design of stirrups:-
Vu = 235.5877 KN Zv = Vv/bd = 235.5877 * 10 3 /230 * 412 = 2.486 N/mm2 Vuc = Zc * b * d = p0T0O0
Zc => % Steel Ast/bd * 100 = 3 * (π/4) * 162 /230 * 412 * 100 = 0.6% Grade of concrete M-25 Zc = 0.52 Since Zv > Zc < Tmax Sheer P/8 should be provide. Spacing:-
Asv/b * Sv ≥ 0.4/ 0.87 fg 2 * (π/4) * 82/230 * Sv = 0.4/0.87 * 415 Sv = 300 mm Provided 8 mm ø @ 300 c/c
Design of beam B2 1.) Size of Beam :- 230 * 530 2.) Fck = 25 N/mm2 3.) Fy = 415 N/mm2 4.) Effective length = 4.43
1.) Load Calculations: Self weight of beam = 230 * 530 * 1 * 25 = 3.05 KN Self weight of slab = ½ * 4.23 * 1.6 * 25 = 84.6 KN Finishing = 1 KN Live load = 4 KN Self weight of brick wall = 0.1 * 3.6 * 18 +> 6.48 Total load = 92.65 KN + 6.48 => 99.13
Featured load => 99.13 * 1.5 => 148 ≈ 150 KN 2.) M0 = W * L2 /24 = 150 * 4.432 /24 = 122.6 KN – m Mu = 122.6 KN – m M0 of resistance for balance reaction:-
MR = 0.138 fck bd2 = 0.138 * 25 * 230 * (530 – 30 – 16/2)2 = 192.077 * 106 N- mm [MRB < 192.077 KN – m] Since M0 of resistance > applied M 0 Design of singly reinforced:-
Area of steel:-
Ast = 0.5 fck/fy [1 - √1 – 4.6 Mu/fck * b * d 2] * b * d = 0.5 * 25/415 * [1 - √1 – 4.6 * 122.6 * 106 / 25 * 230 * 4922] * 230 * 492 = 777.4092 mm2 Number of bars:-
Numbers of bars = Ast/Ast 1 => 777.4092/ (π/4) * 162 => 3.55 ≈ 4 Bars Check for depth:-
d = √M0/Q.B => √122.6 * 106 / 0.138 * fck * B d = 393 mm So depth is fine. Check for deflection:-
Span/D eff = 20* mf 1 4.43 * 1000/492 = 20 * mf 1 Mf 1 => fs = 0.58 fy * Area/ A provided
0.58 * 415 * 714.77/ 4 * (π/4) * 162
214.02
Tension steel = Ast/b*d * 100 = 43 %
4.43 * 1000/492 ≈ 20 * 1.4 [9 ≤ 28] Safe is deflection Check for development length:-
Ld ≤ 1.3 M1 /V0 + L0 Ld = 0.87 fy ø/ 4 Tbd => 0.87 * 415 * 16/ 4 *1.4 => 1031.51 m Vv = Wv * Leff /2 => 1.5 * 99.13 * 4.43/2 = 329.36 KN Ld ≤ 1.3 M1 /Vv + L0
1031.57 ≤ 1.3 122.6 * 106 / 329.36 * 103 + 4430 1031.57 ≤ 3982 + 4430 [1031.57 ≤ 4914.44 mm] Safe Spacing:-
Asv/b * Sv ≥ 0.4/0.87 fy 2 * (π/4) * S2 /230 * Sv = 0.4/0.57 * 415 Sv = 394 mm ≈ 400 mm [Sv = 400 mm] [Provided 8 mm @ 400 mm c/c.]
Design of Beam B3 Given data:-
Size = 230 * 450 mm Fck = 25 || fy = 415 L0 = Clear Span = 1.5 m Ls = Width of support = 200 mm Load Calculation:-
Weight of beam = 0.230 * 0.450 * 1 * 25 = 2.6 Weight of slab = ½ * 1.5 * 2.75 * 25 = 51.56 Finishing load = 1 KN Live load = 4 KN Self weight of way
0.1 * 3.6 * 18 = 6.48 KN
Total load = 65.64
Factored load = 1.5 * 65.64 = 100 KN
B.M= BMapplied = Wu * L02 /24 = 100 * 512 2 /24 = 15
= 9.6 ≈ 10 KM – m Leff => L0 + d = 1.5 + 0.012 = 1.512 Leff => L0 + Ls = 1.5 + 0.2 = 1.7 Moment of resistance for Balance section:-
MR = 0.138 fck * bd 2 = 0.138 * 25 * (230 *(450 – 30 – 12/2)) 2 = 136.002 KN – m
Hence MRB > BMapplied, hence beam design as singly reinforcement. Check for depth:-
dreq = √Mu/Q.b
= √10 * 106 /0.138 * 25 * 230 = 112.26 m Hence that is safe. Area of steel:-
Ast = 0.5 fck/fy [1 - √1 – 4.6 Mu/fck bd 2] bd = 0.5 * 25/415 [1- √1 – 4.6 * 10 * 10 6 / 25 * 230 * 4142] * 230 * 414 = 72.33 mm2 Numbers of bars:-
Numbers of bars = Ast/Ast 1 = 72.339/(π/4) * 122
= 0.503 ≈ 1 bar So have to have at least 2 bar. Check for deflection:-
Span/d eff ≈ 20 mf 1
512/414 ≈ 20 * 1.2 5.6 ≈ 24 Hence it is safe. Check for development length:-
Ld ≤ 1.3 * M1 /Vv + L0 Ld = 0.87 fy * ø/ 4 * Tbd = 0.87 * 415 * 12/4 * 1.4 = 773.678 mm V0 = W0 * Leff /2 = 100.25 * 1.512/2
100 * 1.512/2 => 100 * 1.512/2
75.6 KN
Ld ≤ 1.3 M1 /Vu + L0
773.678 ≤ 1.3 * 10 * 106 /75.6 * 103 + 1512 [773.678 ≤ 1683.9]
Design of Slab (S1) Given :
Room Size = 2.75X4.50m Type of Concrete Type of Steel
M.20 Fe. 415 HYSD bars.
Live load 4KN/m 2. Design Steps (1)
= = 1.636<2
Hence it is a two way Slab. Bearing = 200 mm. Assuming. Dia of bars = 12 mm in both the direction. Assuming 20 mm clear Cover.
= = 98.2mm. = 100mm. = d + cover + = 100+20+
Effective depth (dx)=
Overall depth (D)
= 126 mm.
Effective depth for longer Span
Dy = dx = 100-12 = 88mm. (2)
Effective Span Least of following two. I. Lx = clear span + bearing. = 2.75+0.2
overall depth. Taking max o the two we take 130 mm.
II.
= 2.95m Lx = clear Span + effective depth (dx) = 2.75+0.1 = 2.85 m There fore
Lx = 2.85m
Least of following two. i. Ly= Clear Span + bearing. = 4.5 + 0.2 = 4.7m. ii. Ly = Clear span + effective depth(dy) = 4.5+0.088 = 4.5888m Ly = 4.588m
(3)
Load Calculation
(i) (ii) (iii)
Self weight Live load Finishing load
= 0.126X1X25 = 3.15 KN/m. = 4X1 = 4 KN/m. = 1X1 = 1 KN/m. Total load W u = 8.15 KN/m.
Factored or Ultimate load = 1.5 X 8.15
Wu = 12.225 KN/m. (4)
Moment Calculation
= = 1.64.
Mux = dx Wu lx2. Muy = dy Wu lx2. Refering IS 456:2000 Page – 91, Table 26 ( corner Slabs are held down).
= 1.5 = 1.75
x
y
0.089
0.046
0.10
0.037
By linear Interpolation.
X = 0.089 +
x = 0.095. X = 0.046 -
x = 0.041. Mux Muy
= 0.095 X 12.225 X (2.85) 2 = 9.43 KN.m = 0.41 X 12.225 X (2.85) 2 = 4.07 KN.m
Design Constants
[ ] R = 0.36fck u
[ ] √ √ Dx required =
=
Dx required = 58.45mm Dx assumed = 100 mm Dx assumed > d x required. Design is sabe. (5)
Main Reinforcement Along shorter Span (Artx) Mux = 0.87fy Astx dx
9.43X10 6 = 36105 Astx – 7.49 Astx2
Astx = 277-11mm
2
Check for minimum Reinforcement.
Astmin = 0.12 % of bD. =
X 1000 X 100
= 120 mm2 Spacing of Barsd S=
= 408 mm.
But Maximum Spacing.
Least of. 1. 408 mm 2. 3d = 3x100 = 300 mm. 3. 300 mm. Providing 12 mm dia bars @ 300mm c/c along Shorter Span Main Reinforcement along longer Span. Muy = 0.87 fy Asty dy
= 4.07x106 = 31772.4Asty – 7.49 Asty2 2
Asty = 120 mm
Astmin = 120 mm2
OK
Spacing
= = 855.37 mm But Max. Spacing is
Ieast of 3d “ 3x88 = 264mm. i. ii. 855.37 mm. iii. 300 mm. Spacing = 264 mm.
Providing 12 mm dia bars @ 264 mm c/c along longer Span Torsional or Corner Steel
= x
Area of torsional steel =
=207.83
Spacing of 8mm dia bars.
=
= 241.85 mm 240 mm Provide 8mm dia bars @240mm c/c in a mesh
Mesh Size =
= = 0.57
Say 570
Check for deflection
Area of steel provided = = 376.99 377 mm2 % of steel provided
Pt =
=
= 0.377% Stress in Steel at Service load
Fs = 0.58fy
+ * *+
= 0.58X415 X
= 176.9 N/mm2. 190N/mm 2.
Reading out modification factor (k1) from (curve fig 4 of IS 456:2000) for = 190 N/mn 2. & pt = 0.377 Values.) Kt = 1.8
max
= 20 X kt = 20 x 1.8 = 36
Provided
=
max >
= 28.5
Provided
Hence, Ok.
TWOWAY SLAB REINFORCEMENT DETAIL. NO.
DIMENSION
2.75 X 4.5m SLAB-1 2.75 X 3.5m SLAB-2
SIZE
130mm D=100 Cover = 30 130mm D=100 Cover=30mm
MAIN STEEL
TORISIONAL REINBORE MENT.
12 @ 300 mm
12 @ 250 mm
8 @ 240 mm c/c for 570 mm.
10 @ 300 mm
8 @ 260mm
8 @ 300 mm c/c for 570 mm.
Design of slab (S2) Given Room Size Type of concrete Type of Steel Live Load
2.75x3.50 M-20 fe 415 HYSD bars 4kN/m2
Design steps :
i.
Hence it is a two way Bearing = 200 mm Assuming Dia of bars = 12 mm in both the direction Assuming 20 mm clear cover Effective depth (dx) =
= 98.21mm 100mm
Overll depth (D) = d + cover +
Take 130mm Effective depth for longer span. Dy = dx -
= 100 – 12 = 88mm.
Taking max. of the above two take overall depth = 130mm. ii.
Effective span Least of following two
i. ii.
Lx = clear span + bearing = 2.75+0.2 =2.95m Lx = clear span + effective depth (dx) = 2.75 + 0.1 = 2.85m Therefore Lx = 2.85m
Least of following two
i.
Ly = clear span + bearing
ii.
= 3.5 + 0.2 = 3.7 m Ly = clear span + effective depth (dy) = 3.5 + 0.088 = 3.588mm
= 3.588mm
iii.
Load Calculation
i. ii. iii.
Self weight Live Load Finishing load
= 0.13 x1x25 = 3.25KN/m = 4x1 = 4KN/m = 1x1 = 1KN/m Total load w = 8.25 KN/m.
Factored or Ulternate load Wu = 1.5 X 8.25 = 12.375 iv.
Moment Calculation
= 1.27 Mux Muy
= =
xwulx2 y wulx2
Rebering IS 456:2000 Page 96 table 26 ( Corner slabs are held down)
= 1.27 = 1.2 = 1.3
x 0.072 0.079
By linear interpolation x = 0.072 + x = 0.0769
y
0.059 0.055
y = 0.059 -
y = 0.0541 Mux = .0769x12.375x(2.85) 2 = 7.72 KN/m.
Muy = 0.051x12.375x(2.85) 2 = 5.4379 KN/m. Design constants (Ref Iscode 456 Page no. 96
[ Ru = 0.36 fck [ for M -20
dx required =
2.76
√ √ =
dx required = 52.28 dx assumed = 100mm dx assumed > dx requied Design is sabe
v.
Maain Reintorcement
Along shorter span. (Axtx fy) Mux - .87 fyAstx dx
7.72X10 6 = 36102 Astx – 7.49 Astx2 Astx = 224.25 mm2 Check for minimum reinforcement
Astin = .12% of bD =
= 120 mm2 Thus sabe. √√
]
Spacing of Bars
S= But Maximu spacing Least of i. 350.08mm ii. 3dx = 3x100=300m iii. 300mm
Provid 10 mm dia bars @ 300mm c/c along shorter span. Main Reinforcement along longer span. Muy = .87 fy Asty dy
5.4379x10 6 = 31772.4 Asty – 7.49 Asty2 5.4379x10 6 = 31772.4 Asty – 7.49 Asty2 Asty = 178.64 mm2 Ast min = 120mm2 Hence design safe. Spacing
= = 281.23mm But max spacing 18 Leas of i. ii. iii.
3d = 3x 88 = 264 mm 281 mm 300 m
Provide 8mm dia bars @ 260 c/c along longer span.
Torisional or corner steel Area of torisional steel = ¾ (mid span steel)
= 168.18mm2 Spacing of 8mm dia bars.
= = 298.72 300 mm Provide 8 mm dia bars @ 300mm c/c in a mesh Mesh size -
Say 570 mm
Check for deflection
Are of steel provided = = 314 mm2 % of steel provided Pt =
Stren in steel at service load Fs = .58 fy = .58 x 415 x = 171.90 190 N/mm2
Reading out modification factor (k1) born (Curve fig. 4 of IS 456 : 2000) for fs = I>on/mm 2 & pt = .314 value The value of fs 145 is tending like assumptiontes. Taning Kt = 1.8
= 20 1.8 = 36
= 28.5
>
Design of slab (S3) Given Room Size
3.20x1.20m
Type of concrete = M – 20 Type of steel – Fe – 415. HYSD bars 3.5 KN/m 2
Live load
Design Steps : -
1)
2) Assume Overall depth (100mm to 150 mm) Or D =
= 60 mm
Let D = 100mm
Effective depth d = 100 – clear cover = 10 – 15 – 5 = 80 mm.
3) Effective Span
Lest of the following (i) Lc + ls = 1.2 + 0.2 = 1.4 m (ii)
Lc + d = 1.2 + 0.08 = 1.28 m Eeffective span = left = 1.28 m.
4) Load Calculation
(i) (ii) (iii)
Dead load = 0.2x25
=
2.5KN/m 2 =
Finishing load
= 1 KN/m 2 7.5 KN/m 2
Total Load = Factored load per m = Wu
=
5) Factored Moment
Mu =
=
= 2.304 KN/m
6) Check for effective depth.
Dreq =
√ √
=
= 28.89mm. D assumed = 80mm. Hence design is safe
7) Area of main reinforcement.
Mu = 2.304x10 6 =0.87fu Astd
+ * + *
= 2.304x106 =0.87x415x80 Astd Ast = 81.49 mm 2
Check or minimum Reinforcement
Ast min = 0.12% of bD =
3.5 KN/m 2
Live load
= 120 mm2 Thus Ast = 120 mm 2
Spacing =
= 418.88mm. 415mm.
7.5x1.5 = 11.25 KN/m
8
415mm c/c.
Distribuition reinforcement Astmin = 0.12% bD =120mm.
Spacing =
= 418.88mm. 415mm. Check for Deflection
Area of steel provide = = 121.12 mm % of steel provide = pt = = = 0.15 + Fs = 0.58fy * = 0.58x415 x =238.47 240 N/mm2
Modification factor from curve (fig 4 OS 456) Fs = 240 & pt = 0.15 Kt = 2
max
= 20x kt = 20x2 =40
Provided max
=
>
Provided
2
ONE WAY SLAB REINFORCEMENT DETAIL N0.
SLAB-3 (rs-18)
DIMENSION
3.2X1.2M
SIZE
MAIN STEEL
100mm 8 @415mm 120mm 300mm
DISTRIBUTIONSTEEL
8 @415mm 300mm
Design of staircqse RISE = R = 150 mm TRADE = T = 250 mm Floor to floor height = 3.6m
= 12 Number of Rise = Hight of one flight =
Number of trade = 12 - 1 = 11 For 11 trade we need length = 11XT = 11X250 = 2750mm = 2.7 m. Land width = 3.97-2.75=1.20m centre to centre Effective span = 1.20+2.75+
(wall thickness) = 4.065
Thickness of waist slab
=
th to
= 300 + 0.240 mm We are take = t = 250 mm Overall depth D = 280 mm
2.75
1.2
0.23/2
Weight of waist slab = t
√ √ X 25
= .250
= 7.14 KN/m
= X .150 X 25
Weight of step
= X R X X 25
= 1.875 KN/m. Deqd load
= 1.875 7.14 = 9.025 KN/m.
With finishing is taken
= 1 (finishing) D L = 10 KN/m
In landing position
= 0.250 X 1 X 25 = 6.25KN/m.
With Arishing = 1 KN/m. Total
= 6.25 + 1 KN
Live laod
= 7.25 KN/m.
Factorey going portion of = 1.5 (10+3) = 19.5 KN/m.
Factored canding slab per meter width = 1.5 (7.7+3)
X 25
= 16.05 KN/m. 19.5 KN/m
16.05 KN/m
2.75
RA
= 1.315
RB
Taking moment about B
RA X 4.065
= 19.5 X 2.75(4.065= 144.25 + 13.87
RA X 4.065
= 198.12
RA
= 38.89
Shear force zero +
X
=
X
= 1.99
X
= 2m from A
Maxi. Bm
M
M
=
38.89X2 – 16.09 X
=
77.78 – 32. 1
=
45.68 KN/m
Reinforcement
) + 16.05
(√ )b.d X 250 X 1000 = 0.5X √
Ast = 0.5
Ast = 529.61 mm2
Distribution steel
Ast
=
X 1000X 250
300 mm2
Spacing =
Provided 8 mm bqd at 160 c/c
