T Beam Design
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T BEAM DESIGN STEP 1 : GIVEN DATA Grade of Concrete
fck
Grade of Steel fy gc Unit Weight of concrete 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 OF BREADTH OF THE FLANGE
2 20 N/mm 2 415 N/mm 2 25 KN/m
500 50.35
125.38 124.26 3 500 460
300 120 16 8
KN-m KN-m KN m mm mm mm mm mm mm
From IS 456:2000 Pg No 37 clause 23.1.2 Breadth of the flange bf = ((l/6)+bw+6df)
1520
mm
STEP 3 : CHECK FOR DEPTH From IS 456:2000 Pg No 70 xumax/d= From IS 456:2000 Pg No 96 clause G.1.1 c Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d2*fck 125380000
0.1728
0.7984
Required Effective depth
2 6000 d d2= d Req=
0.48
2 151465.1603 mm 390 mm
Required Overall depth Provided overall depth
D Req= D=
430 mm 500 mm
Provide an effective depth Provide an overall depth depth
dp= D pro=
460
Hence
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 Actual Bending moment of the beam M
50350000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 c Assume Xu max = X u = Df
xumax/d From IS 456:2000 Pg No 96 clause G.1.1 c Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d2*fck Mu lim
0.093913 0.89043 6432640000 Mu 1
0.260869565
537919488
N-mm
M < Mu1
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) 50350000
-166083 Ast 165530
2 0.91054276 Ast
331613 553.003998 Ast1=
2 182096.3328 mm
Ast2=
2 303.6672302 mm
Ast=
2 303.6672302 mm
Area of mainReinforcement
58
CASE 2 : NUTRAL AXIS LIES OUT SIDE THE 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 Mu flange =((0.45fck(bf-bw)df(d-.5D))
276696000
N-mm
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
15205.65019
mm2
WEB REINFORCEMENT Mu web =( M - Mu flange) Mu = Mu web b= bw
-226346000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) -226346000 -166083 Ast 224038.79
2 24.9726 Ast
390122 -57955.7868 Ast1=
2 7810.988769 mm
Ast2=
2 -1160.38636 mm
Asw=
2 -1160.38636 mm
Area of mainReinforcement
Total area of reinforcement = Asf +As w
Ast
14045.26383
mm2
FINAL AREA OF REINFORCEMENT =
Ast
303.6672302
mm2 59
STEP 4 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) Area of the main reinforcement =((3.14*d1^2)/4)
200.96
No of reinforcement
mm2
2
STEP 5 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS
t v=((V/b*d))
tv
2 0.900434783 N/mm
tc
0.220048718 2 1 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19
For
100 Ast/( b*d) 0.22005
IT IS SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
100.48
mm2
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300
mm
minimum spacing
300
mm
300
mm
Adopt spacing of stirups 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
mm 1210 mm
1214.557485
300
mm 60
STEP 5 : CALCULATION OF AREA OF REINFORCEMENT
AT SUPPORT : From IS 456:2000 Pg No 96
Actual Bending moment of the beam M
125380000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) 125380000
-166083 Ast 122716.31
288799
2 24.972625 Ast
43366.693 Ast1=
2 5782.317778 mm
Ast2=
2 868.2846321 mm
Ast=
2 868.2846321 mm
Area of mainReinforcement
STEP 6 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) rod dia Area of the main reinforcement =((3.14*d1^2)/4) No of reinforcement
16 200.96
mm2
5
61
STEP 7 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS
t v=((V/b*d))
tv
2 0.900434783 N/mm
tc
0.629191762 2 0.5 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19
For
100 Ast/( b*d) 0.62919
IT IS NOT SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
100.48
mm2
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300
mm
minimum spacing
300
mm
300
mm
Adopt spacing of stirups 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
mm 1210 mm
1214.557485
300
mm
62
fck
Grade of Steel fy gc Unit Weight of concrete 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 OF BREADTH OF THE FLANGE
2 20 N/mm 2 415 N/mm 2 25 KN/m
500 50.35
125.38 124.26 3 500 460
300 120 16 8
KN-m KN-m KN m mm mm mm mm mm mm
From IS 456:2000 Pg No 37 clause 23.1.2 Breadth of the flange bf = ((l/6)+bw+6df)
1520
mm
STEP 3 : CHECK FOR DEPTH From IS 456:2000 Pg No 70 xumax/d= From IS 456:2000 Pg No 96 clause G.1.1 c Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d2*fck 125380000
0.1728
0.7984
Required Effective depth
2 6000 d d2= d Req=
0.48
2 151465.1603 mm 390 mm
Required Overall depth Provided overall depth
D Req= D=
430 mm 500 mm
Provide an effective depth Provide an overall depth depth
dp= D pro=
460
Hence
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 Actual Bending moment of the beam M
50350000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 c Assume Xu max = X u = Df
xumax/d From IS 456:2000 Pg No 96 clause G.1.1 c Mu lim=0.36 * (xu max /d)((1-0.42 xu max/d))b*d2*fck Mu lim
0.093913 0.89043 6432640000 Mu 1
0.260869565
537919488
N-mm
M < Mu1
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) 50350000
-166083 Ast 165530
2 0.91054276 Ast
331613 553.003998 Ast1=
2 182096.3328 mm
Ast2=
2 303.6672302 mm
Ast=
2 303.6672302 mm
Area of mainReinforcement
58
CASE 2 : NUTRAL AXIS LIES OUT SIDE THE 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 Mu flange =((0.45fck(bf-bw)df(d-.5D))
276696000
N-mm
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
15205.65019
mm2
WEB REINFORCEMENT Mu web =( M - Mu flange) Mu = Mu web b= bw
-226346000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) -226346000 -166083 Ast 224038.79
2 24.9726 Ast
390122 -57955.7868 Ast1=
2 7810.988769 mm
Ast2=
2 -1160.38636 mm
Asw=
2 -1160.38636 mm
Area of mainReinforcement
Total area of reinforcement = Asf +As w
Ast
14045.26383
mm2
FINAL AREA OF REINFORCEMENT =
Ast
303.6672302
mm2 59
STEP 4 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) Area of the main reinforcement =((3.14*d1^2)/4)
200.96
No of reinforcement
mm2
2
STEP 5 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS
t v=((V/b*d))
tv
2 0.900434783 N/mm
tc
0.220048718 2 1 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19
For
100 Ast/( b*d) 0.22005
IT IS SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
100.48
mm2
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300
mm
minimum spacing
300
mm
300
mm
Adopt spacing of stirups 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
mm 1210 mm
1214.557485
300
mm 60
STEP 5 : CALCULATION OF AREA OF REINFORCEMENT
AT SUPPORT : From IS 456:2000 Pg No 96
Actual Bending moment of the beam M
125380000
N-mm
From IS 456:2000 Pg No 96 clause G.1.1 b Mu=0.87* fy*Ast*d((1-(Ast*fy)/(b*d*fck)) 125380000
-166083 Ast 122716.31
288799
2 24.972625 Ast
43366.693 Ast1=
2 5782.317778 mm
Ast2=
2 868.2846321 mm
Ast=
2 868.2846321 mm
Area of mainReinforcement
STEP 6 : CALCULATION OF NUMBER OF REINFORCEMENT: No of reinforcement =((Ast / ast)) rod dia Area of the main reinforcement =((3.14*d1^2)/4) No of reinforcement
16 200.96
mm2
5
61
STEP 7 : CHECK FOR SHEAR STRESS ( I ) NOMINAL SHEAR STRESS
t v=((V/b*d))
tv
2 0.900434783 N/mm
tc
0.629191762 2 0.5 N/mm
( ii ) PERMISSIBLE SHEAR STRESS
From IS 456:2000 Pg No 73 table 19
For
100 Ast/( b*d) 0.62919
IT IS NOT SAFE AGAINST SHEAR Assume 2 legged 8 mm diameter stirups Asv=(3.14*d2^2)/4)
100.48
mm2
Spcing of stirups Sv =((Asv*0.87*fy)/(0.4*bw)
300
mm
minimum spacing
300
mm
300
mm
Adopt spacing of stirups 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
mm 1210 mm
1214.557485
300
mm
62
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