Only
Content
Design of Rectangular Beam
The values are Displayed in Blue color are the Input values
1 Section Forces (From STAAD)
Beam Number
=
118
Load Case
Bending Moment at Support
Bending Moment at Mid Span
Max Shear Force at support, Vu
Max Shear Force at mid span, Vu
=
=
=
=
=
101
299
299
857
810
Torsional Moment at support, Tu
Torsional Moment at mid span, Tu
=
=
757 KN.m
757 KN.m
KN.m
KN.m
KN
KN
2 Section Geometry
0.0035
0.45 fck
3/7xu
Asc
x
A
xu
0.42 xu
d
D
N
0.002
4/7xu
c=0.36 fck b xu
Ast
Es
T
b
0.87fy/Es+0.002
Strain diagram
Stress diagram
3 Design Datas
1. Length of Beam
2. Width of Support
3. Depth of beam, D
=
=
=
8 m
230 mm
1000 mm
4. Width of Beam, B
=
300 mm
5. Compressive strength of Concrete, Fck
=
25 N/mm2
6. Yield Strength of steel, Fy
=
415 N/mm2
7. Clear cover to the reinforcement, C
8. Diameter of Bar
9. Diameter of Shear reinforcement
=
=
=
4 Calculation of Design Forces
At support
Equivalent bending moment, Me1
where
Mu = bending moment at the cross-section
Mt =
Tu(1+D/b/1.7)
Mt
30 mm
20 mm
8 mm
= Mu + Mt
As per Clause 41.4.2,
IS:456:2000
= 757000000((1 + (1000/300)/1.7)
= 1929.61 KN.m
1929.61>299
Provide additional tensile
reinforcement for torsion at support
At mid Span
Me1
=
2228.6 KN.m
As per Clause
41.4.2.1,
IS:456:2000
Equivalent bending moment, Me1
= Mu + Mt
where
Mu = bending moment at the cross-section
Mt =
Tu(1+D/b/1.7)
Mt
= 757000000((1 + (1000/300)/1.7)
= 1929.61 KN.m
1929.61>299
Provide additional tensile
reinforcement for torsion at mid
span
Me1
5 Effective depth required
Overall Depth, Dprovided
Effective depth provided, dprovided
=
2228.6 KN.m
=
1000 mm
= (D-CC-1/2 Ø of bar)
= (1000-30-1/2 x 20)
=
960 mm
Mu.lim
= (0.36 x Fck x (xu.max/d) x
(1-0.42(xu.max/d)) x bd2
xu.max/d
Mu.lim
=
=
d,reqd
= (Mu.lim/Qb)1/2
0.48
3.45 bd2
(IS:456:2000,
Annex G, clause
G.1.1.c)
= (2228610000 / ( 3.45 x 300 )) 1/2
=
1467.4 mm
1467.39 > 960
Revise the depth or Provide
doubly rein. beam
Mu.lim
=
953.86 KN.m
Mu>Mu.lim
Doubly
6 Calculation of Longitudinal reinforcement
I) Longitudinal Reinforcement for Singly Reinforced Section
a) At Support
=
Mu/bd2
pt
0
=
0
=
0
Ast
=
0 mm2
Ast.min
=
589.88 mm2
Ast
=
589.88 mm2
=
0 Nos
No of bars
b) At mid span
=
Mu/bd2
0
=
0
=
0
Ast
=
0 mm2
Ast.min
=
589.88 mm2
Ast
=
589.88 mm2
=
0 Nos
pt
No of bars
II) Longitudinal Reinforcement for doubly Reinforced Section
As per ANNEX -G,
Clause G.1.2,
IS:456:2000
a) At Support (Tension reinforcement)
Total tension steel,Ast
= Ast1+Ast2
Mu.lim/bd2
pt
= (953856000)/(300x960x960)
=
3.45
=
1.188
Ast1 = 3421.44 mm2
As2
= Mu-Mu.lim/0.87fy(d-d')
=
3837.7 mm2
Ast =
7259.1 mm2
No of bars
=
24 Nos
b) At support (Compression reinforcement)
Asc = 0.87fyAst2/(fsc-0.447fck)
fsc
= 0.87 fy
d'
=
40 mm
fsc
=
361.05 N/mm2
=
3960.3 mm2
No of bars =
13 Nos
Asc
d) At mid span (Tension reinforcement)
Mu/bd2
= (953856000)/(300x960x960)
pt
Ast
=
3.45
=
1.19
= 3421.44 mm2
As2
= Mu-Mu.lim/0.87fy(d-d')
=
3837.7 mm2
Ast =
7259.1 mm2
=
24 Nos
No of bars
e) At mid span (Compression reinforcement)
Asc = 0.87fyAst2/(fsc-0.447fck)
fsc
= 0.87 fy
d'
=
40 mm
fsc
=
361.05 N/mm2
=
3960.3 mm2
No of bars =
13 Nos
Asc
7 Check for shear capacity
Equivalent Shear at support, Ve
= Vu +1.6(Tu/b)
= 857+(1.6x757/300)
=
Equivalent Shear at mid span, Ve
861.04 KN
= Vu +1.6(Tu/b)
= 810+(1.6x757/300)
Nominal shear stress at support, tve
=
814.04 KN
=
Ve/bd
As per clause 40.1,
IS:456:2000
= 861040/(300x960)
=
pt =
2.976 N/mm2
2.5205
ß = 0.87fck/6.89pt
=
Allowable shear stress at support, tc
=
7.9567
0.429 N/mm2
2.976>0.429
Provide shear reinforcement
tc.max =
4.15
N/mm2
2.976<4.15
O.K
Minimum reinforcement
Asv/sv b = 0.4/0.87fy
sv =
0 mm
Shear reinforcement
Nominal shear stress at mid span, tve
=
47 mm
min spacing
300 mm
Vus = Vu-tc bd
=
sv =
spacing
733.45 KN
47 mm
Ve/bd
= 814040/(300x960)
=
pt =
2.827 N/mm2
1.3751
ß = 0.87fck/6.89pt
=
Allowable shear stress at mid span, tc
=
4.3408
0.549 N/mm2
2.827>0.549
Provide shear reinforcement
tc.max =
4.15
N/mm2
2.827<4.15
O.K
Minimum reinforcement
Asv/sv b = 0.4/0.87fy
sv =
0 mm
Shear reinforcement
Vus = Vu-tc bd
= 655.93 KN
sv =
53 mm
Longitudinal reinforcement
Beam Dimensions
B, mm
D, mm
Reinforcement, Area
Beam type
At support
At mid span
top
Top
Bottom
Bottom
spacing
53 mm
min spacing
300 mm
300
1000
Doubly
7259.1 3960.3 3960.3 7259.1
Bar dia
25
20
16
25
25
0
0
20
15
2
3
5
0
0
0
0
7359.4
628
Nos
602.88 2453.1
Shear reinforcement at support
Bar dia
Spacing
8
47
mm
mm No of Stirrups Legged
=
2
=
2
Shear reinforcement at support
Bar dia
8
Spacing
53
mm
mm No of Stirrups Legged
8 Check for deflection
l/d, Actual
l/d, Max
pt,
=
8.33
= B.V x M.F
= 100Ast/bd
fs
=
2.52
= 0.58x fyx(Ast required/Ast provided)
fs
=
Modification Factor
l/d, Max
712.27 N/mm2
0.4
=
= 0.4 X 26
=
10.4
10.4>8.33
O.K
The values are Displayed in Blue color are the Input values
1 Section Forces (From STAAD)
Beam Number
=
118
Load Case
Bending Moment at Support
Bending Moment at Mid Span
Max Shear Force at support, Vu
Max Shear Force at mid span, Vu
=
=
=
=
=
101
299
299
857
810
Torsional Moment at support, Tu
Torsional Moment at mid span, Tu
=
=
757 KN.m
757 KN.m
KN.m
KN.m
KN
KN
2 Section Geometry
0.0035
0.45 fck
3/7xu
Asc
x
A
xu
0.42 xu
d
D
N
0.002
4/7xu
c=0.36 fck b xu
Ast
Es
T
b
0.87fy/Es+0.002
Strain diagram
Stress diagram
3 Design Datas
1. Length of Beam
2. Width of Support
3. Depth of beam, D
=
=
=
8 m
230 mm
1000 mm
4. Width of Beam, B
=
300 mm
5. Compressive strength of Concrete, Fck
=
25 N/mm2
6. Yield Strength of steel, Fy
=
415 N/mm2
7. Clear cover to the reinforcement, C
8. Diameter of Bar
9. Diameter of Shear reinforcement
=
=
=
4 Calculation of Design Forces
At support
Equivalent bending moment, Me1
where
Mu = bending moment at the cross-section
Mt =
Tu(1+D/b/1.7)
Mt
30 mm
20 mm
8 mm
= Mu + Mt
As per Clause 41.4.2,
IS:456:2000
= 757000000((1 + (1000/300)/1.7)
= 1929.61 KN.m
1929.61>299
Provide additional tensile
reinforcement for torsion at support
At mid Span
Me1
=
2228.6 KN.m
As per Clause
41.4.2.1,
IS:456:2000
Equivalent bending moment, Me1
= Mu + Mt
where
Mu = bending moment at the cross-section
Mt =
Tu(1+D/b/1.7)
Mt
= 757000000((1 + (1000/300)/1.7)
= 1929.61 KN.m
1929.61>299
Provide additional tensile
reinforcement for torsion at mid
span
Me1
5 Effective depth required
Overall Depth, Dprovided
Effective depth provided, dprovided
=
2228.6 KN.m
=
1000 mm
= (D-CC-1/2 Ø of bar)
= (1000-30-1/2 x 20)
=
960 mm
Mu.lim
= (0.36 x Fck x (xu.max/d) x
(1-0.42(xu.max/d)) x bd2
xu.max/d
Mu.lim
=
=
d,reqd
= (Mu.lim/Qb)1/2
0.48
3.45 bd2
(IS:456:2000,
Annex G, clause
G.1.1.c)
= (2228610000 / ( 3.45 x 300 )) 1/2
=
1467.4 mm
1467.39 > 960
Revise the depth or Provide
doubly rein. beam
Mu.lim
=
953.86 KN.m
Mu>Mu.lim
Doubly
6 Calculation of Longitudinal reinforcement
I) Longitudinal Reinforcement for Singly Reinforced Section
a) At Support
=
Mu/bd2
pt
0
=
0
=
0
Ast
=
0 mm2
Ast.min
=
589.88 mm2
Ast
=
589.88 mm2
=
0 Nos
No of bars
b) At mid span
=
Mu/bd2
0
=
0
=
0
Ast
=
0 mm2
Ast.min
=
589.88 mm2
Ast
=
589.88 mm2
=
0 Nos
pt
No of bars
II) Longitudinal Reinforcement for doubly Reinforced Section
As per ANNEX -G,
Clause G.1.2,
IS:456:2000
a) At Support (Tension reinforcement)
Total tension steel,Ast
= Ast1+Ast2
Mu.lim/bd2
pt
= (953856000)/(300x960x960)
=
3.45
=
1.188
Ast1 = 3421.44 mm2
As2
= Mu-Mu.lim/0.87fy(d-d')
=
3837.7 mm2
Ast =
7259.1 mm2
No of bars
=
24 Nos
b) At support (Compression reinforcement)
Asc = 0.87fyAst2/(fsc-0.447fck)
fsc
= 0.87 fy
d'
=
40 mm
fsc
=
361.05 N/mm2
=
3960.3 mm2
No of bars =
13 Nos
Asc
d) At mid span (Tension reinforcement)
Mu/bd2
= (953856000)/(300x960x960)
pt
Ast
=
3.45
=
1.19
= 3421.44 mm2
As2
= Mu-Mu.lim/0.87fy(d-d')
=
3837.7 mm2
Ast =
7259.1 mm2
=
24 Nos
No of bars
e) At mid span (Compression reinforcement)
Asc = 0.87fyAst2/(fsc-0.447fck)
fsc
= 0.87 fy
d'
=
40 mm
fsc
=
361.05 N/mm2
=
3960.3 mm2
No of bars =
13 Nos
Asc
7 Check for shear capacity
Equivalent Shear at support, Ve
= Vu +1.6(Tu/b)
= 857+(1.6x757/300)
=
Equivalent Shear at mid span, Ve
861.04 KN
= Vu +1.6(Tu/b)
= 810+(1.6x757/300)
Nominal shear stress at support, tve
=
814.04 KN
=
Ve/bd
As per clause 40.1,
IS:456:2000
= 861040/(300x960)
=
pt =
2.976 N/mm2
2.5205
ß = 0.87fck/6.89pt
=
Allowable shear stress at support, tc
=
7.9567
0.429 N/mm2
2.976>0.429
Provide shear reinforcement
tc.max =
4.15
N/mm2
2.976<4.15
O.K
Minimum reinforcement
Asv/sv b = 0.4/0.87fy
sv =
0 mm
Shear reinforcement
Nominal shear stress at mid span, tve
=
47 mm
min spacing
300 mm
Vus = Vu-tc bd
=
sv =
spacing
733.45 KN
47 mm
Ve/bd
= 814040/(300x960)
=
pt =
2.827 N/mm2
1.3751
ß = 0.87fck/6.89pt
=
Allowable shear stress at mid span, tc
=
4.3408
0.549 N/mm2
2.827>0.549
Provide shear reinforcement
tc.max =
4.15
N/mm2
2.827<4.15
O.K
Minimum reinforcement
Asv/sv b = 0.4/0.87fy
sv =
0 mm
Shear reinforcement
Vus = Vu-tc bd
= 655.93 KN
sv =
53 mm
Longitudinal reinforcement
Beam Dimensions
B, mm
D, mm
Reinforcement, Area
Beam type
At support
At mid span
top
Top
Bottom
Bottom
spacing
53 mm
min spacing
300 mm
300
1000
Doubly
7259.1 3960.3 3960.3 7259.1
Bar dia
25
20
16
25
25
0
0
20
15
2
3
5
0
0
0
0
7359.4
628
Nos
602.88 2453.1
Shear reinforcement at support
Bar dia
Spacing
8
47
mm
mm No of Stirrups Legged
=
2
=
2
Shear reinforcement at support
Bar dia
8
Spacing
53
mm
mm No of Stirrups Legged
8 Check for deflection
l/d, Actual
l/d, Max
pt,
=
8.33
= B.V x M.F
= 100Ast/bd
fs
=
2.52
= 0.58x fyx(Ast required/Ast provided)
fs
=
Modification Factor
l/d, Max
712.27 N/mm2
0.4
=
= 0.4 X 26
=
10.4
10.4>8.33
O.K
