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4 Design for area of steel and shear for doubly reinforced beam by limit state design method
Ref IS 456-2000 Cl G-1.2 For sections with compression reinforcement
Given
fy fck b D clear cover clear cover cg cg d d/
on ten.face on com.face of tension of comp. for tension for comp.
face steel face steel face steel face steel
N/mm2
N/mm2 mm mm mm mm mm mm mm mm
500 30 500 1000 25 25 40 40 935 65
Check for section : doubly Ref IS 456-2000 Cl G 1.2
Ref IS 456-2000 Cl G-1.2
Mulim pt lim Ast1 Mu Mu- Mulim fsc Asc Ast2
SP 16 SP 16 req req
Table C Table E req obtained considered Cl G-1.2 Cl G-1.2
kN.m % mm2
kNm Cl G 1.2 value value N/mm mm2
mm2
1744.08 1.13 5293.04 2000 255.92 0.070 0.20 329 894.11 676.23
Total Ast pt Result Asc pc Result
req on Nos. dia prov reg req on Nos. dia prov reg
tension mm tesion compression mm comp.
face 4 28 steel face 0 12 steel
mm2 4 28 % mm2 3 16 %
5969.27 Ast prov- 4926.02 1.05 not ok 894.11 Ast prov- 603.19 0.13 not ok
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, C l 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
N/mm2
kN % N/mm2
N/mm2
N/mm2
30 700 1.05 1.50 0.67 3.5
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup Vus/d prov. Result
req req stirrup dia of stirrup sp assumed kN/cm
kN kN kN kN/cm N/mm2
mm legs mm Cl 40.4 a Cl 40.4a
700 313.23 386.78 4.14 415 12 2 200 4.083 Not ok
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm491.97 300 200 Hence ok
Side face reinforcement
Ref IS 456-2000 Cl 26.5.1.3
b D side face spc b/w
of reinf. bars not to
web req. / face no. dia of Ast prov. exceed
mm mm Cl 26.5.1.3 per face bar mm2
Cl 26.5.1.3
500 1000 250 2 10 157.08 300 mm
side face reinf. mm2/face prov.
tau_v tau_c,design for shear
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Check for span to depth ratio
Ref IS 456-2000 Cl 23.2.1
Type of fy span d pt req. pt prov. pc MFt MFc
beam N/mm2 mm mm % % %
Cont.Bea 500 12000 935 1.28 1.05 0.13 0.73 1.038707
l/d l/d Result
prov Cl 23.2.1 Cl 23.2.1
12.83 16.43 Okay
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Design of Beam (Terminal 2-bay frame)
1.) Given:-
Span = 12 m
D = 1000 mm
b = 500 mm
Clear Cover on Tens. Face = 25 mmClear Cover on Comp. Face = 25 mm
fck = 30 kN/mm2
Fy = 415 kN/mm2
Ru 4.14 kN/mm2
Effective cover on tens face d'= 65 mm
Effective cover on comp face dc= 65 mm
Effective depth, d = 935 mm
d'/d = 0.069518717
fsc = 329 kN/mm2
2.) Load combinations considered :-As per IS 1893:2000 part 1, the following loading cases are taken into consideration :-
1.) 1.5(DL+LL)
2.) 1.2(DL+LL)
3.) 1.2(DL+LL+EQ X)
4.) 1.2(DL+LL+EQ Y)
5.) 1.2(DL+LL-EQ X)
6.) 1.2(DL+LL-EQ Y)
7.) 1.5DL
8.) 1.5(DL+EQ X)
9.) 1.5(DL+EQ Y)
10.) 1.5(DL-EQ X)
11.) 1.5(DL-EQ Y)
12.) 0.9DL + 1.5EQ X
13.) 0.9DL + 1.5EQ Y
14.) 0.9DL - 1.5EQ X
15.) 0.9DL - 1.5EQ Y
For obtaining maximum moments, an envelope of all the combinations is taken in STAAD.
3.) Support Design :-
11
Roof
2
3
9
10
4
5
6
7
8
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i.) End supports reinforcement:-
a.) Parking to Floor 8 :-
Max negative moment = 1580 kNm
Mu lim= 1809.65 kNm
Ast1= 5615.911685 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 5615.911685 mm2
dia.
mm
5 28 tension
5 28 steel
Ast prov. = 6157.52 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2 kN % N/mm2 N/mm2 N/mm2
30 482.751 1.32 1.03 0.94 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/mm N/mm2
mm legs mm
482.751 439.45 43.30 0.46 415 12 2 300
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
b.) Basements, Roof, and floors 9 to 11 :-
Max negative moment = 1350 kNm
Mu lim= 1809.65 kNm
Ast1= 4637.370176 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 4637.370176 mm2
dia.
Result
tau_v > tau_c,desig
tau_v
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Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/mm N/mm2 mm legs mm
482.751 392.70 90.05 0.96 415 12 2 300
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
i.) Intermediate supports reinforcement:-
a.) Basements, Parking, and floors 1 to 5 :-
Max negative moment = 1260 kNm
Mu lim= 1809.65 kNm
Ast1= 4275.09577 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 4275.09577 mm2
dia.
mm
5 25 tension
4 25 steel
Ast prov. = 4417.86 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2
kN % N/mm2
N/mm2
N/mm2
30 445 0.94 0.95 0.79 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/mm N/mm2
mm legs mm
445 369.33 75.68 0.81 415 12 2 300
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
b ) Roof and floors 6 to 11
tau_v > tau_c,desigtau_v
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Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2 kN % N/mm2 N/mm2 N/mm2
30 445 0.73 0.95 0.7 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/cm N/mm2 mm legs mm
445 327.25 117.75 1.26 415 12 2 300
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
4.) Mid span Ast calculation :-
Max positive moment = 547.961 kNm
Result
tau_v > tau_c,desig
tau_v
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Design of Beam (Terminal 2-bay frame)
1.) Given:-
Span = 12 m
D = 1000 mm
b = 500 mm
Clear Cover on Tens. Face = 25 mm
Clear Cover on Comp. Face = 25 mm
fck = 30 kN/mm2
Fy = 415 kN/mm2
Ru 4.14 kN/mm2
Effective cover on tens face d'= 65 mm
Effective cover on comp face dc= 65 mm
Effective depth, d = 935 mm
d'/d = 0.069518717
fsc = 329 kN/mm2
2.) Load combinations considered :-As per IS 1893:2000 part 1, the following loading cases are taken into consideration :-
1.) 1.5(DL+LL)2.) 1.2(DL+LL)
3.) 1.2(DL+LL+EQ X)
4.) 1.2(DL+LL+EQ Y)
5.) 1.2(DL+LL-EQ X)
6.) 1.2(DL+LL-EQ Y)
7.) 1.5DL
8.) 1.5(DL+EQ X)
9.) 1.5(DL+EQ Y)
10.) 1.5(DL-EQ X)
11.) 1.5(DL-EQ Y)
12.) 0.9DL + 1.5EQ X
13.) 0.9DL + 1.5EQ Y
14.) 0.9DL - 1.5EQ X
15.) 0.9DL - 1.5EQ Y
For obtaining maximum moments, an envelope of all the combinations is taken in STAAD.
3.) Support Design :-
Roof
11
10
9
8
7
6
5
4
3
2
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Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2 kN % N/mm2 N/mm2 N/mm2
30 702.249 1.32 1.50 0.94 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/mm N/mm2
mm legs mm
702.249 439.45 262.80 2.81 415 12 2 250
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 250 Hence ok
i.) Intermediate supports reinforcement:-
a.) Basements, Parking, and floors 1 to 6 :-
Max negative moment = 1600 kNm
Mu lim= 1809.65 kNm
Ast1= 5705.044616 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 5705.044616 mm2
dia.
mm
5 28 tension
5 28 steel
Ast prov. = 6157.52 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steelAst prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2 kN % N/mm2 N/mm2 N/mm2
30 613.291 1.32 1.31 0.94 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assumekN kN kN kN/mm N/mm
2mm legs mm
613.291 439.45 173.84 1.86 415 12 2 300
Result
tau_v > tau_c,desig
tau_v
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Total Ast Required on tension face = 4156.626506 mm2
dia.
mm
3 28 tension
4 28 steel
Ast prov. = 4310.27 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc tc max
prov. Cl 40.1 Table 19 Table 20
kN/mm2 kN % N/mm2 N/mm2 N/mm2
30 646.151 0.92 1.38 0.78 4
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy assuming no. stirrup
req req stirrup dia of stirrup sp assume
kN kN kN kN/cm N/mm2
mm legs mm
646.151 364.65 281.50 3.01 415 12 2 250
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm408.34 300 250 Hence ok
4.) Mid span Ast calculation :-
5
4
3
6
No. of Bars Result
Result
tau_v > tau_c,desigtau_v
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Max positive moment = 808.482 kNm
Mu lim= 1809.65 kNm
Ast1= 2595.448362 mm2
Ast2= - mm
2
Asc= - mm2
Total Ast Required on tension face = 2595.448362 mm2
dia.
mm
3 25 tension
3 25 steel
Ast prov. = 2945.24 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression0 28 steel
Ast prov. = 0.00 Prov > req
No. of Bars Result
Mu Limiting is more than Mu induced, therefore, beam is designed as Singly
Reinforced Beam
No. of Bars Result
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Design of Beam (Terminal 2-bay frame)
1.) Given:-
Span = 12 m
D = 1000 mm
b = 500 mmClear Cover on Tens. Face = 25 mm
Clear Cover on Comp. Face = 25 mm
fck = 30 kN/mm2
Fy = 415 kN/mm2
Ru 4.14 kN/mm2
Effective cover on tens face d'= 65 mm
Effective cover on comp face dc= 65 mm
Effective depth, d = 935 mm
d'/d = 0.069518717
fsc = 329 kN/mm2
2.) Load combinations considered :-As per IS 1893:2000 part 1, the following loading cases are taken into consideration :-
1.) 1.5(DL+LL)
2.) 1.2(DL+LL)
3.) 1.2(DL+LL+EQ X)
4.) 1.2(DL+LL+EQ Y)
5.) 1.2(DL+LL-EQ X)
6.) 1.2(DL+LL-EQ Y)
7.) 1.5DL
8.) 1.5(DL+EQ X)
9.) 1.5(DL+EQ Y)
10.) 1.5(DL-EQ X)
11.) 1.5(DL-EQ Y)
12.) 0.9DL + 1.5EQ X
13.) 0.9DL + 1.5EQ Y
14.) 0.9DL - 1.5EQ X
15.) 0.9DL - 1.5EQ Y
For obtaining maximum moments, an envelope of all the combinations is taken in STAAD.
3.) Support Design :-
Roof
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i.) End supports reinforcement:-
a.) Roof and Ground floor to floor 11
Max negative moment = 1900 kNm
Mu lim= 1809.65 kNm
Ast1= 6686.075776 mm2
Ast2= 287.6484137 mm2
Asc= 315.6700905 mm2
Parking
B1
B2
Mu limiting is less than Mu induced, therefore, beam is designed as Doubly
Reinforced Beam
1
10
9
8
7
6
5
4
3
2
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Total Ast Required on tension face = 6973.72419 mm2
dia.
mm
6 28 tension
6 28 steel
Ast prov. = 7389.03 Prov > req
Total Ast Required on compression face = 315.6700905 mm2
dia.
mm
2 16 compression
0 16 steel
Ast prov. = 402.12 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1fck Vu pt tv tc
prov. Cl 40.1 Table 19
kN/mm2 kN % N/mm2 N/mm2
30 731.658 1.67 1.57 1.05
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy
req req
kN kN kN kN/mm N/mm2
731.658 490.88 240.78 2.58 415
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
b.) Basements :-
Max negative moment = 1650 kNm
Mu lim= 1809.65 kNm
Ast1= 5931.045941 mm2
Ast2= - mm2
No. of Bars Result
Mu Limiting is more than Mu induced, therefore, beam is designed as Singly
Reinforced Beam
No. of Bars Result
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Asc= - mm2
Total Ast Required on tension face = 5931.045941 mm2
dia.
mm
5 28 tension
5 28 steelAst prov. = 6157.52 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc
prov. Cl 40.1 Table 19
kN/mm2 kN % N/mm2 N/mm2
30 702.249 1.32 1.50 0.94
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy
req req
kN kN kN kN/mm N/mm2
702.249 439.45 262.80 2.81 415
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 250 Hence ok
i.) Intermediate supports reinforcement:-
a.) Basements, Parking, and floors 1 to 6 :-
Max negative moment = 1600 kNm
Mu lim= 1809.65 kNm
Mu Limiting is more than Mu induced, therefore, beam is designed as Singly
Reinforced Beam
No. of Bars Result
No. of Bars Result
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Ast1= 5705.044616 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 5705.044616 mm2
dia.
mm
5 28 tension
5 28 steel
Ast prov. = 6157.52 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc
prov. Cl 40.1 Table 19
kN/mm2 kN % N/mm2 N/mm2
30 613.291 1.32 1.31 0.94
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu tcb d Vus Vus/d fy
req req
kN kN kN kN/mm N/mm2
613.291 439.45 173.84 1.86 415
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 300 Hence ok
b.) Roof and floors 7 to 11 :-
Max negative moment = 1230 kNm
Mu lim= 1809.65 kNm
Ast1= 4156.626506 mm2
Mu Limiting is more than Mu induced, therefore, beam is designed as Singly
Reinforced Beam
No. of Bars Result
No. of Bars Result
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Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 4156.626506 mm2
dia.
mm
3 28 tension
4 28 steel
Ast prov. = 4310.27 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression
0 28 steel
Ast prov. = 0.00 Prov > req
Check for shear in beams (limit state design method)
Ref IS 456-2000 Cl 40.1, Cl 40.2.3, Table 19, Table 20 & Cl 40.2.1
fck Vu pt tv tc
prov. Cl 40.1 Table 19
kN/mm2 kN % N/mm2 N/mm2
30 646.151 0.92 1.38 0.78
Design for shear reinforcement (vertical stirrups)
Ref IS 456-2000 Cl 40.4a
Vu t
cb d Vus Vus/d fyreq req
kN kN kN kN/cm N/mm2
646.151 364.65 281.50 3.01 415
Check for minimum and maximum spacing of stirrup
Min stirrup Max stirrup stirrup Result
spacing mm spacing mm sp prov.
Cl 26.5.1.6 Cl 26.5.1.5 mm
408.34 300 250 Hence ok
4.) Mid span Ast calculation :-
No. of Bars Result
Roof
11
No. of Bars Result
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Max positive moment = 808.482 kNm
Mu lim= 1809.65 kNm
Ast1= 2595.448362 mm2
Ast2= - mm2
Asc= - mm2
Total Ast Required on tension face = 2595.448362 mm2
dia.
mm
3 25 tension
5
4
3
2
1
Parking
B1
B2
Mu Limiting is more than Mu induced, therefore, beam is designed as Singly
Reinforced Beam
No. of Bars Result
6
10
9
8
7
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3 25 steel
Ast prov. = 2945.24 Prov > req
Total Ast Required on compression face = 0 mm2
dia.
mm
0 28 compression0 28 steel
Ast prov. = 0.00 Prov > req
No. of Bars Result
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tc max
Table 20
N/mm2
4
assuming no. stirrup Vus/d prov. Result
stirrup dia of stirrup sp assumed kN/mm
mm legs mm Cl 40.4 a Cl 40.4a
12 2 300 2.722 Hence ok
Result
tau_v > tau_c,design for shear
tau_v
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tc max
Table 20
N/mm2
4
assuming no. stirrup Vus/d prov. Result
stirrup dia of stirrup sp assumed kN/mm
mm legs mm Cl 40.4 a Cl 40.4a
12 2 250 3.267 Hence ok
tau_v tau_c,design for shear
Result
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tc max
Table 20
N/mm2
4
assuming no. stirrup Vus/d prov. Result
stirrup dia of stirrup sp assumed kN/mm
mm legs mm Cl 40.4 a Cl 40.4a
12 2 300 2.722 Hence ok
Result
tau_v > tau_c,design for shear
tau_v
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tc max
Table 20
N/mm2
4
assuming no. stirrup Vus/d prov. Resultstirrup dia of stirrup sp assumed kN/cm
mm legs mm Cl 40.4 a Cl 40.4a
12 2 250 3.267 Hence ok
Result
tau_v > tau_c,design for shear
tau_v
