Roof Beam Design
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Transcript of Roof Beam Design
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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
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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
-
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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
-
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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
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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
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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.
-
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(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