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T BEAM DESIGN STEP 1 : GIVEN DATA 2
Grade of Concrete
fck
20 N/mm
Grade of Steel
fy
Unit Weight of concrete
gc
415 N/mm 2 25 KN/m
Size of the beam 300 Max +ve Bending moment of the beam M1 Max -ve Bending moment of the beam M2 Shear force of the beam F Length of the beam l Overall depth of the beam D Effective depth of the beam d Breadth of the web bw Depth of the flange df Diameter of Main Reinforcement d1 Diameter of distribution Reinforcement d2 STEP 2 : CALCULATION CALCULATION OF BREADTH OF THE FLANGE
2
500 50.35 KN-m
125.3 125.38 8 KN-m KN-m 124. 124.26 26 KN 3 m 500 mm 460 mm 300 mm 120 mm 16 mm 8 mm mm
From IS 456:2000 Pg No 37 clause 23.1.2 2 3.1.2 1520 mm
Breadth of the flange bf = ((l/6)+bw+6df)
STEP 3 : CHECK FOR DEPTH From IS 456:2000 Pg No 70 xumax/d=
0.48
From IS 456:2000 Pg No 96 clause G.1.1 G.1 .1 c 2
Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d *f ck 125380000
0.1728
0.7984
2
6000 d
2
2
151465.1603 mm
Required Effective depth
d= d Req=
Required Overall depth
D Req=
430 mm
Provided overall depth
D=
500 mm mm
Provide an effective depth
dp= D pro=
460
390 mm
Hence
Provide an overall depth depth
500 mm 57
STEP 4 : CALCULATION OF AREA OF REINFORCEMENT
AT MID SPAN: From IS 456:2000 Pg No 96 CASE 1 :
NUTRAL AXIS LIES WITH IN THE FLANGE 50350000 N-mm
Actual Bending moment of the beam M
From IS 456:2000 Pg No 96 clause G.1.1 c Assume Xu max = X u = Df
xumax/d
0.260869565
From IS 456:2000 Pg No 96 clause G.1.1 c 2
Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d *f ck Mu lim
0.093913
0.89043 6432640000 Mu 1
537919488 N-mm
M < Mu1
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* f y*Ast*d((1-(Ast*f y)/(b*d*f ck )) 50350000
-166083 Ast 165530
331613
2
0.91054276 Ast
2
553.003998 Ast1=
182096.3328 mm
Ast2=
303.6672302 mm
Ast=
303.6672302 mm
Area of mainReinforcement
2
2
58
CASE 2 : NUTRAL AXIS LIES OUT SIDE T HE FLANGE
From IS 456:2000 Pg No 96 clause 2.2 M > Mu 1 Assume Xu =X u max = ( df/0.43)
279.0697674
xumax/d
0.606673407
calculation of moment of resistance Mu 2 Mu 2= Mu flange + Mu web 276696000 N-mm
Mu flange =((0.45fck(bf-bw)df(d-.5D)) CALCULATION OF REINFORCEMENT FLANGE REINFORCEMENT M = Cuf + Cu d) Cuf =( 0.45fck(bf-bw)df
Asf =(( Cuf )/(0.87* fy)
Cuf
5490000
Cud
44860000
Asf
2
15205.65019 mm
WEB REINFORCEMENT -226346000 N-mm
Mu web =( M - Mu flange) Mu = Mu web b= bw
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* f y*Ast*d((1-(Ast*f y)/(b*d*f ck )) -226346000 -166083 Ast 224038.79
2
24.9726 Ast
2
390122 -57955.7868 Ast1=
7810.988769 mm
Ast2=
-1160.38636 mm
Asw=
-1160.38636 mm
Area of mainReinforcement
2
2
2
Total area of reinforcement = Asf +As w
Ast
14045.26383 mm
FINAL AREA OF REINFORCEMENT =
Ast
303.6672302 mm
2
59
STEP 4 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) 2
Area of the main reinforcem ent =((3.14*d1^2)/4)
200.96 mm
No of reinforcement
2
STEP 5 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS t
2
v=((V/b*d))
tv
0.900434783 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19 100 Ast/( b*d) For
0.22005
0.220048718 tc
2
1 N/mm
IT IS SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
2
100.48 mm
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300 mm
minimum spacing
300 mm
Adopt spacing of stirups
300 mm
Assume 2 legged 8 mm diameter stirups Vs = Vu-tc b d Spacing of stirups Sv =(0.87 fy Asv d)/Vs Adopt sapcing of stirups Sv FINAL SPACING OF STIRIUPS Sv
-13740 Kn 1214.557485 mm 1210 mm 300 mm 60
STEP 5 : CALCULATION OF AREA OF REINFORCEMENT
AT SUPPORT : From IS 456:2000 Pg No 96
125380000 N-mm
Actual Bending moment of the beam M
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* f y*Ast*d((1-(Ast*f y)/(b*d*f ck )) 125380000
-166083 Ast 122716.31
288799
2
24.972625 Ast
2
43366.693 Ast1=
5782.317778 mm
Ast2=
868.2846321 mm
Ast=
868.2846321 mm
Area of mainReinforcement
2
2
STEP 6 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) rod dia
16 2
Area of the main reinforcem ent =((3.14*d1^2)/4)
200.96 mm
No of reinforcement
5
61
STEP 7 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS t
2
v=((V/b*d))
tv
0.900434783 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19 100 Ast/( b*d) For
0.62919
0.629191762 2
tc
0.5 N/mm
IT IS NOT SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
2
100.48 mm
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300 mm
minimum spacing
300 mm
Adopt spacing of stirups
300 mm
Assume 2 legged 8 mm diameter stirups Vs = Vu-tc b d Spacing of stirups Sv =(0.87 fy Asv d)/Vs Adopt sapcing of stirups Sv