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DATA:-

Density of concrete gconc. =

Density of wall =

Density of floor finish = N/mper mm thick

Breadth of beam = mm

Depth of beam = mm

Height of wall hwall = m

Self-weight of beam wbeam =

Dead load from beam Gk =

Design ultimate load from beam 1.4Gk =

Design service load from beam 1.0Gk =

Dead load from slab gk =

Live load qk =

Design ultimate load n = 1.4gk + 1.6qk =

Design service load n = 1.0gk

=

SHEAR FORCE COEFFICIENTvsx = bvxnlx

vsy = bvynlx

Note:- vs = vsx when l = ly; vs = vsy when l = lx

LOAD CASE 1 MOMENT DISTRIBUTION:-

kN/m kN/m kN/

A B Cm m mA

AB BC CD

SPAN AB:- SPAN BC:-

kNm kN/m kNm kNm

A m B B

kN/m2

7.80

25.0 kN/m3

gwall 19.0 kN/m3

gfloor finish 23.6

b 225

d 450

3.50

2.53

1.50

BS 8110 -1

5.30 kN/m2

kN/m

2.53 kN/m

3.54 kN/m

2.53 kN/m

1.50 kN/m2

2.00 kN/m2

Equ. 19

Equ. 20

Fig. 3.10

8.16 7.80

3.350 3.35 3.35

Joint B C

Member BA CB

DF's 1.000 0.500 0.500 0.500 0.500 0.

FEM's 0.0 11.4 -7.3 7.3 -7.3

Balance 0.00 -2.07 -2.07 0.00 0.00

Carry-over -1.04 0.00 0.00 -1.04 0.00

Balance 1.04 0.00 0.00 0.52 0.52

Carry-over 0.00 0.52 0.26 0.00 0.00

Balance 0.00 -0.39 -0.39 0.00 0.00 -

Carry-over -0.19 0.00 0.00 -0.19 -0.06

Balance 0.19 0.00 0.00 0.13 0.13

Carry-over 0.00 0.10 0.06 0.00 0.02

Balance 0.00 -0.08 -0.08 -0.01 -0.01 -

Carry-over -0.04 0.00 0.00 -0.04 0.00

Balance 0.04 0.00 0.00 0.02 0.02

Total 0.00 9.52 -9.52 6.69 -6.69

0.00 8.16 9.52 9.52 7.80

3.350 3.350

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MAB = MBC =

kN kN kN

Check : SFY = Check : SFY =

x = x =

SPAN EF:- SPAN FG:-

kNm kN/m kNm kNm

E m F F

MEF = MFG =

kN kN kN

Check : SFY = Check : SFY =

x = x =

LOAD CASE 2 MOMENT DISTRIBUTION:-

kN/m kN/m kN/

A B C

m m m

A

AB BC CD

SPAN AB:- SPAN BC:-

kNm kN/m kNm kNm

A m B B

MAB = MBC =

kN kN kN

Check : SFY = Check : SFY =

x = x =

SPAN EF:- SPAN FG:-

kNm kN/m kNm kNm

E m F F

MEF = MFG =

kN kN kN

Check : SFY = Check : SFY =

7.1811 kNm 2.8912

VA = 10.826 VB = 16.51 VB = 13.92

1.33 1.78

0.00 OK

6.88 7.80 8.74 8.74 8.16

3.1574 kNm 5.8285

3.350 3.350

15.419 VE = 12.519 VF = 13.628 VF =

1.60 1.89

0.00 OK

7.808.16 3.74

3.350 3.35 3.35

Joint B C

Member BA CB

DF's 1.000 0.500 0.500 0.500 0.500 0.

FEM's 0.0 11.4 -3.5 3.5 -7.3

Balance 0.00 -3.98 -3.98 1.90 1.90 -

Carry-over -1.99 0.00 0.95 -1.99 -0.95

Balance 1.99 -0.48 -0.48 1.47 1.47 -

Carry-over -0.24 0.99 0.73 -0.24 -0.48

Balance 0.24 -0.86 -0.86 0.36 0.36 -

Carry-over -0.43 0.12 0.18 -0.43 -0.30

Balance 0.43 -0.15 -0.15 0.37 0.37 -

Carry-over -0.07 0.22 0.18 -0.07 -0.08

Balance 0.07 -0.20 -0.20 0.08 0.08 -

Carry-over -0.10 0.04 0.04 -0.10 -0.07Balance 0.10 -0.04 -0.04 0.08 0.08 -

Total 0.00 7.11 -7.11 4.92 -4.92

0.00 8.16 7.11 7.11 3.74

3.350 3.350

8.1678 kNm -0.7164

VA = 11.546 VB = 15.791 VB = 6.9133

1.41 1.85

0.00 OK

5.07 7.80 6.55 6.55 3.84

5.1552 kNm 1.7403

3.350 3.350

7.9751 VE = 12.632 VF = 13.515 VF =

0.00 OK

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x = x =

LOAD CASE 3 MOMENT DISTRIBUTION:-

kN/m kN/m kN/

A B C

m m m

A

AB BC CD

SPAN AB:- SPAN BC:-

kNm kN/m kNm kNm

A m B B

MAB = MBC =

kN kN kN

Check : SFY = Check : SFY =x = x =

SPAN EF:- SPAN FG:-

kNm kN/m kNm kNm

E m F F

MEF = MFG =

kN kN kN

Check : SFY = Check : SFY =

x = x =

DESIGN BENDING MOMENTS AND SHEAR FORCES

SPAN MOMENTS

Load case 1

Load case 2

Load case 3

1.62 2.08

3.743.84 7.80

3.350 3.35 3.35

Joint B C

Member BA CB

DF's 1.000 0.500 0.500 0.500 0.500 0.

FEM's 0.0 5.4 -7.3 7.3 -3.5

Balance 0.00 0.96 0.96 -1.90 -1.90

Carry-over 0.48 0.00 -0.95 0.48 0.95 -

Balance -0.48 0.48 0.48 -0.71 -0.71

Carry-over 0.24 -0.24 -0.36 0.24 0.48 -

Balance -0.24 0.30 0.30 -0.36 -0.36

Carry-over 0.15 -0.12 -0.18 0.15 0.21 -

Balance -0.15 0.15 0.15 -0.18 -0.18

Carry-over 0.07 -0.07 -0.09 0.07 0.11 -

Balance -0.07 0.08 0.08 -0.09 -0.09Carry-over 0.04 -0.04 -0.05 0.04 0.06 -

Balance -0.04 0.04 0.04 -0.05 -0.05

Total 0.00 6.92 -6.92 4.98 -4.98

0.00 3.84 6.92 6.92 7.80

3.350 3.350

2.4802 kNm 5.0208

VA = 4.3631 VB = 8.4934 VB = 13.652

1.14 1.750.00 OK

5.15 3.74 6.22 6.22 8.16

3.350 3.350

-0.431 kNm 6.739

VE = 5.9419 VF = 6.5779 VF = 14.543

1.59 1.78

0.00 OK

Span AB Span BC Span CD Span DE Span EF S

MAB MBC MCD MDE MEF

7.18 2.89 3.84 3.74 3.16

8.17 -0.72 5.70 -0.08 5.16

-0.02 5.60 -0.432.48 5.02

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SUPPORT MOMENTS

Load case 1

Load case 2

Load case 3

SHEAR FORCES

Load case 1

Load case 2

Load case 3

DESIGN SPAN MOMENTS

DESIGN SUPPORT MOMENTS

DESIGN SHEAR FORCES

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1: Tension reinforcement size, f = mm

Layer 2: Tension reinforcement size, f = mm

Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =Steel compressive stress, fsc = N/mm

2

d'/x = = mm

MAB =

BS 8110-1

Cl. 3.4.4.4

If K

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final z = mm

Depth to the neutral axis x=(d - z)/0.45 = mm

Provide 0 - 12 mm = mm2

Provide 0 - 25 mm = mm2

Total = mm

Provide 0 - 12 mm = mm2

Provide 2 - 12 mm = mm2

Total = mm

check for minimum tension reinforcement

Rectangular section

As = 0.13*Ac/100 = mm

check for minimum compression reinforcement

Rectangular section

As = 0.2*Ac/100 = mm

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Layer 1:Compression reinforcement size, f = mm

Layer 2:Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mmSteel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MB =

BS 8110-1

Cl. 3.4.4.4

If K

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Provide 2 - 12 mm = mm2

Total = mm

check for minimum tension reinforcement

Rectangular section

As = 0.13*Ac/100 = mm

check for minimum compression reinforcement

Rectangular section

As = 0.2*Ac/100 = mm

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MBC =

BS 8110-1

Cl. 3.4.4.4

If K

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Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Layer 1:Compression reinforcement size, f = mm

Layer 2:Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MC =

BS 8110-1

Cl. 3.4.4.4

If K

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MCD =

BS 8110-1

Cl. 3.4.4.4

If K

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As = 0.2*Ac/100 = mm

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Layer 1:Compression reinforcement size, f = mm

Layer 2:Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

ME =

BS 8110-1

Cl. 3.4.4.4If K

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Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MEF =

BS 8110-1

Cl. 3.4.4.4If K

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z = 0.95d = mm

final z = mm

Depth to the neutral axis x=(d - z)/0.45 = mm

Provide 0 - 12 mm = mm2

Provide 0 - 12 mm = mm2

Total = mm

Provide 0 - 12 mm = mm2

Provide 2 - 12 mm = mm2

Total = mm

check for minimum tension reinforcement

Rectangular section

As = 0.13*Ac/100 = mm

check for minimum compression reinforcement

Rectangular section

As = 0.2*Ac/100 = mm

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =Steel compressive stress, fsc = N/mm

2

d'/x = = mm

MFG =

BS 8110-1

Cl. 3.4.4.4

If K

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Rectangular section

As = 0.13*Ac/100 = mm

check for minimum compression reinforcement

Rectangular section

As = 0.2*Ac/100 = mm

DESIGN FOR MOMENT - RECTANGULAR BEAM

Layer 1:Tension reinforcement size, f = mm

Layer 2:Tension reinforcement size, f = mm

Layer 1:Compression reinforcement size, f = mm

Layer 2:Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MG =

BS 8110-1

Cl. 3.4.4.4

If K

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Compression reinforcement size, f = mm

Shear reinforcement size, f = mm

Reinforcement cover, c = mm

Characteristic concrete cube strength, fcu = N/mm

Effective depth of tension reinforcement, d = mm

Characteristic strength of reinforcement, fy = N/mm2

Depth to compression reinforcement, d' = mm

Steel compressive strain, esc =

Steel compressive stress, fsc = N/mm2

d'/x = = mm

MGH =

BS 8110-1

Cl. 3.4.4.4

If K

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BS 8110-1

Cl. 3.4.4.4

If K

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Provide 0 - 16 mm = mm2

Provide 2 - 12 mm = mm2

Total = mm

check for minimum tension reinforcement

Rectangular section

As = 0.13*Ac/100 = mm

check for minimum compression reinforcement

Rectangular section

As = 0.2*Ac/100 = mm

DESIGN FOR SHEAR - SUPPORT A

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT B [LHS]

Effective depth = bv = mm

fyv = N/mm As =fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

As = M/0.87fyz = 93.24 mm2

0.00

226.19 226.19

Cl. 3.12.5.3

Table 3.27c

406.00 mm 225

131.63

202.5

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VAB = 11.55

Design shear stress, v = V/bvd = 0.1263.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

200.00 8

-30.53

100.53

404.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VBA = 16.51

Design shear stress, v = V/bvd = 0.182

3.58

OK

Cl. 3.4.5.4 100As/bvd = 0.2488

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(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT B [RHS]

Effective depth = bv = mm

fyv =N/mm

As =fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT C [LHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

a e . . 0.629

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.369 N/mm

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

250.00 8

-29.55

100.53

404.00 mm 225

410.0 226.2 mm

2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VBC = 13.92

Design shear stress, v = V/bvd = 0.153

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2488

0.629

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.369 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VCB = 12.49

Design shear stress, v = V/bvd = 0.137

3.58

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v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT C [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/b

vd)

1/3=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT D [LHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3(400/d)1/4/gm = 0.368 N/mmCl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VCD = 12.87

Design shear stress, v = V/bvd = 0.141

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1 20.0

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kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mmAs = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT D [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm

2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT E [LHS]

Effective depth = bv = mm

. . . .

Equ. 3.Design shear force, VDC = 13.33

Design shear stress, v = V/bvd = 0.146

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4 = 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 80.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VDE = 13.27

Design shear stress, v = V/bvd = 0.145

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

7/28/2019 Roof Beam Design

20/68

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT E [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

410.0 226.2 mm

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VED = 12.95

Design shear stress, v = V/bvd = 0.142

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5Table 3.7.

v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) abovev < 0.8sqrt(fcu) or 5N/mm

2Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VEF = 12.63

Design shear stress, v = V/bvd = 0.1383.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

7/28/2019 Roof Beam Design

21/68

DESIGN FOR SHEAR - SUPPORT F [LHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu

) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT F [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu =N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VFE = 13.63

Design shear stress, v = V/bvd = 0.149

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1Cl. 3.4.5.2

Equ. 3.

20.0Design shear force, VFG = 15.42

Design shear stress, v = V/bvd = 0.169

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

7/28/2019 Roof Beam Design

22/68

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT G [LHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT G [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2Equ. 3.

20.0

Design shear force, VGF = 12.79

Design shear stress, v = V/bvd = 0.140

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5Table 3.7.

v < (vc + 0.4) Provide nominal links in areas where v > vc250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

406.00 mm 225

410.0 226.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VGH = 3.15

Design shear stress, v = V/bvd = 0.034

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.2476

0.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

7/28/2019 Roof Beam Design

23/68

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT H [LHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mmProvide 0 legs 8 mm links = mm

2

DESIGN FOR SHEAR - SUPPORT H [RHS]

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd)1/3

=

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.00

0.00

404.00 mm 225

410.0 402.1 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VHG = 7.79

Design shear stress, v = V/bvd = 0.086

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.4424

0.762

(400/d)1/4

= 1.000

(fcu/25)1/3 = 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.447 N/mm

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 8

0.000.00

404.00 mm 225

410.0 402.1 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VHI = 21.90

Design shear stress, v = V/bvd = 0.241

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.4424

0.762

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.447 N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

7/28/2019 Roof Beam Design

24/68

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

DESIGN FOR SHEAR - SUPPORT I

Effective depth = bv = mm

fyv = N/mm As =

fcu = N/mm

kN

N/mm2

Lesser of 0.8sqrt(fcu) or 5.0 N/mm2

= N/mm2

v < lesser 0.8sqrt(fcu) or 5N/mm2

(100As/bvd) =

For members with shear reinforcement.

(1) Check if shear reinforcement is required

sv = mm bar size = mm

Nominal As = 0.4bvsv/0.87fyv = mm

Provide 2 legs 8 mm links = mm2

(2) Check if shear reinforcement is required

sv = mm bar size = mm

As = bvsv(v-vc)/0.87fyv = mm

Provide 0 legs 10 mm links = mm2

CHECK DEFLECTIONSPAN AB

bw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

200.00 8

-26.00

100.53

406.00 mm 225

410.0 226.2 mm

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VI = 16.10

Design shear stress, v = V/bvd = 0.176

3.58

OK

Cl. 3.4.5.4Table 3.8.

100As/bvd = 0.24760.628

(400/d)1/4

= 1.000

(fcu/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.368 N/mm

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

63.08

100.53

Cl. 3.4.5.5

Table 3.7.v > (vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (2)

0.00 10

0.00

0.00

BS 8110 - 1

Cl. 3.4.6225 59.4 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.2202

fs = 5*fy*As req/8*As prov = 67.256 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2

52

8.3 DEFLECTION OK

7/28/2019 Roof Beam Design

25/68

SPAN BCbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN CDbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN DEbw = mm As req =

b = mm As prov ==

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN EFbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

BS 8110 - 1

Cl. 3.4.6225 36.5 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.1354

fs = 5*fy*As req/8*As prov = 41.343 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2

52

8.3 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6225 41.4 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.1536

fs = 5*fy*As req/8*As prov = 46.906 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2

52

8.3 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6225 40.7 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.1509

fs = 5*fy*As req/8*As prov = 46.083 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2

52

8.3 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6225 37.5 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.139

f = 5*f *A /8*A = 42.452

7/28/2019 Roof Beam Design

26/68

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN FGbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN GHbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

SPAN HIbw = mm As req =

b = mm As prov =

=

M/bd2

=

The service stress is

Allowable span/effective depth ratio =

Actual span/effective depth ratio =

CHECK CRACKING

.

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2

52

8.3 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6

225 51.8 mm2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.2031

fs = 5*fy*As req/8*As prov = 58.67 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2.00

52

8.7 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6225 0.3 mm

2

225 226.2 mm2

bw/b 1

The basic span-to-effective depth ratio = 26

0.001

fs = 5*fy*As req/8*As prov = 0.2999 N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2))) = 2.00

52

6.5 DEFLECTION OK

BS 8110 - 1

Cl. 3.4.6225 93.2 mm

2

225 226.2 mm2

= 2.00

bw/b 1

The basic span-to-effective depth ratio = 26

0.3459

fs = 5*fy*As req/8*As prov N/mm2

Modification factor = 0.55 + ((477-fs)/(120*(0.9+M/bd2)))

= 105.63

52

9.2 DEFLECTION OK

BS 8110 - 1

Cl.Tension Reinf.

Clear distance btw bars in

tension.

Clear dist. btw the face of the beam

nearest longi. bars in tension.

7/28/2019 Roof Beam Design

27/68

(vc + 0.4) Provide nominal links in areas where v > vc, go to Item No. (2) above

v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (

N/mm2

Cl. 3.4.5.5

Table 3.7.v < (vc + 0.4) Provide nominal links in areas where v > vc

250.00 8

(400/d)1/4

= 1.000

(fcu

/25)1/3

= 0.928

vc = 0.79{100As/(bvd)}1/3

(400/d)1/4

/gm = 0.510

3.58

OK

Cl. 3.4.5.4

Table 3.8.100As/bvd = 0.6555

0.869

225

410.0 603.2 mm2

BS 8110 - 1

Cl. 3.4.5.2

Equ. 3.

20.0

Design shear force, VBA = 53.96

Design shear stress, v = V/bvd = 0.586

150.00 8

7.26

100.53

409.00 mm

a e . . v < 0.8sqrt(fcu) or 5N/mm2

Provide nominal links in areas where v > vc, go to Item No. (

7/28/2019 Roof Beam Design

63/68

CHECK CRACKING