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22/8/2010
1
Determine design life
Determine preliminary size of beam
Estimate actions on
beam
Assess durability requirements and
determine concrete strength
Determine nominal cover for durability, fire and
bond
Analysis structure to obtain critical
moments and shear forces
Design of flexural
reinforcement
Design of shear
reinforcement
Verify deflection
Verify crackingProduce detail
drawing
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2
Step Task Standard
1 Determine design life, Exposure class & Fire resistance
EN 1990 Table 2.1
EN 1992-1-1: Table 4.1
EN 1992-1-2: Sec. 5.6
2 Determine material strengthBS 8500-1: Table A.3
EN 206-1: Table F1
3 Select size of beam EN 1992-1-1: Table 7.4N
EN 1992-1-2: Table 5.5
4 Calculate min. cover for durability , fire and bond requirements EN 1992-1-1: Sec. 4.4.1
5 Estimate actions on beam EN 1990 Table A1.1 and A1.2
6 Analyze structure to obtain critical moments and shear forces EN 1992-1-1: Sec. 5
7 Design flexural reinforcement EN 1992-1-1: Sec. 6.1
8 Design shear reinforcement EN 1992-1-1: Sec. 6.2
9 Check deflection EN 1992-1-1: Sec. 7.4
10 Check cracking EN 1992-1-1: Sec. 7.3
11 Detailing EN 1992-1-1: Sec. 8 & 9.2
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3
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4
Trial and error method:
Overall depth, h = L
13 to 18
Width, b = 0.3h – 0.6h
b
h
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5
EN 1992-1-2, gives a method for determining the minimum dimension of beams for fire resistance requirements
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6
One-way slab (Ly/Lx 2.0)
AB = CD
w = nLx/2 kN/m
AC = BDw = 0 kN/m
Two-way slab (Ly/Lx 2.0)
AB = CD
AC = BD
w = nLx/6 [(3 – (Lx/Ly)2] kN/m
w = nLx/3 kN/m
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7
w = vsx = vxnLx kN/mw
= v
sy
=
vynL
xkN
/m
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8
w = vsx = 0.35 nLx kN/m
w = vsx = 0.54 nLx kN/m
w =
vsy
= 0
.26
nL
xkN
/m
w =
vsy
= 0
.40
nL
xkN
/m
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9
22/8/2010
10
leff
Permanent action = 15 kN/m (NOT including self weight)Variable action = 10 kN/m
Design data
fck = 20 N/mm2
fyk = 500 N/mm2
Clear distance, ln = 8 m and supported on two masonry walls 250 mm thickExposure class = XC1Fire resistance = R60 (1 hour)Design life = 50 years
b
h
Effective Length
leff = ln + a1 + a2
= 8000 + 125 + 125 = 8250 mm
Use leff = 8.25 m
Beam size
Overall depth, h = L/13 = 8250/13= 635 mmWidth, b = 0.6h = 0.4 x 635 = 254 mm
Try beam size, b h = 250 650 mm
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11
Minimum concrete cover regard to bond, Cmin, b = 20 mm (Table 4.2)
Minimum concrete cover regard to durability, Cmin, dur = 15 mm (Table 4.4N)
Minimum required axis distance for R60 fire resistance, asd = a + 10 = 30 + 10 = 40 mm (Table 5.5: BS EN 1992-1-2: 2004)Minimum concrete cover regard to fire, Cmin, fire = asd - links - bar/2
= 40 – 8 – 20/2 = 22 mm
Allowance in design for deviation, Cdev = 10 mm
Nominal cover, Cnom = Cmin + Cdev = 22 + 10 = 32 mm
Use Cnom = 35 mm
Analysis
Beam self weight = 25.0 x (0.25 x 0.65) = 4.06 kN/mPermanent action (excluding selfweight) = 15 kN/mCharacteristics permanent action, Gk = 15 + 4.06 = 19.06 kN/m
Characteristics variable action, Qk = 10 kN/m
Design action, w = 1.35Gk + 1.5Qk = 40.73 kN/m
Vmax = 168.0 kNMmax = 346.6 kNm
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12
Main Reinforcement
Assumed 1 = 20 mm, 2 = 12 mm and links = 8 mmEffective depthd = 650 – 35 – 8 – 20 = 587 mm and d’ = 35 + 8 + 12/2 = 49 mm
K = M/fckbd2 = 0.201 Kbal = 0.167 (redistribution 0% and = 1.0)Compression reinforcement is required
z = 0.82d = 481.8 mm
x = (d – z)/0.4 = 263.1 mm
d’/x = 49/263.1= 0.19 0.38 The compression steel will have yield Use fsc = 0.87fyk
As’ = (K – Kbal)fckbd2/0.87fyk(d – d’)= (0.201 – 0.167)(20)(250)(587)2/(0.87)(500)(587 – 49)= 252 mm2
As = (Kbalfckbd2/0.87fykz) + As’= (0.167 x 20 x 250 x 5872/0.87 x 500 x 481.8) + 252= 1624 mm2
Check Area of ReinforcementAs, min = 0.26(fctm/fyk) bd
= 0.26 (2.21/500) bd = 0.0011 bd 0.0013 bd Use As, min = 0.0013 bd = 191 mm2
As, max = 0.04 Ac = 0.04 bh= 6500 mm2
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13
6H20 (As = 1885 mm2)
3H12 (As’ = 339 mm2)
650 mm
250 mm
Design shear force;VEd = 168 kN
Concrete strut capacity;
VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )= 0.36 250 587 20 (1 – 20/250)
(cot + tan )
for = 22, cot = 2.5 VRd, max = 338 kN = 45, cot = 1.0 VRd, max = 486 kN
Therefore, 22Use = 22; tan = 0.40; cot = 2.48
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14
Shear links;Asw/s = VEd / 0.78fykdcot
= 168 103 / (0.78 500 587 2.48) = 0.297
Try link: H8 Asw = 101 mm2
Spacing, s= 101/0.297 = 339 mm smax = 0.75d (440 mm)
Provide H8 – 325 mmMinimum links;Asw/s = 0.08fck
1/2bw / fyk
= 0.08 (20)1/2 250 / 500 = 0.179
Try link: H8 Asw = 101 mm2
Spacing, s= 101/0.179 = 562 mm smax = 0.75d (440 mm)
Provide H8 – 425 mm
H8-425H8-325 H8-325
(168 kN)
(134 kN)
(168 kN)
(134 kN)
Vmin = (Asw/s)(0.78d fykcot )
0.83 m 0.83 m6.58 m
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15
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot = 0.5 168 2.48= 208 kN
MEd, max / z = 346.6 106 / 481.8 = 719 kN 208 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 208 103 / 0.87 500= 478 mm2
Provide 2H20 ( 628 mm2)
Percentage of required tension reinforcement; = As, req / bd
=1624/ 250 587 = 0.011
Reference reinforcement ratio;o = (fck)
1/2 10-3 = (20)1/2 10-3 = 0.0045
Since o Use Eq. (7.16b) in EC 2 Cl. 7.4.2
Percentage of required compression reinforcement;’ = As’, req / bd
= 252 / 250 587 = 0.002
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For structural system, K = 1.0
= 1.0 (11 + 3.21 + 0.23) = 14.44
Basic span-effective depth ratio, l/d = 14.44
'
12
1
'5.111 ck
ock ffK
d
l
Modification factor for span greater than 7 m;7/span = 7/8.25 = 0.85
Modification factor for tension steel area provided;As, prov/As, req = 1885/1624 = 1.16 1.5
Allowable span-effective depth ratio;(l/d)allow = 14.44 0.85 1.16 = 14.22
Actual span-effective depth ratio8250/587 = 14.10 (l/d)allow = 14.22 OK
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Limiting crack width, wmax = 0.3 mm
Steel stress,
= (500/1.15) [(19.06 + 0.3 10.0) / 40.73]= 271 N/mm2
Maximum allowable bar spacing = 188.75 mm (from interpolation)
Actual bar spacing = [250 – 2(35) – 2(8) – 3(20)]/2= 52 mm 188.75 mm OK
1
)5.135.1(
3.0x
15.1 kk
kkyk
sQG
QGff
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18
250 600
250 450
250
600
250
600
250
450
250 600
Design datafck = 25 N/mm2
fyk = 500 N/mm2
LoadingVariable action = 3.0 kN/m2
Finishes etc. = 1.5 kN/m2
Brickwall (at 3 m height) = 2.6 kN/m2
Nominal concrete cover = 30 mm
Reinforcement1 = 25 mm2 = 12 mm links = 8 mm
Slab thickness, hf = 125 mm
h =
45
0 m
m
bw = 250 mm
beff
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19
b1 = 4000/2 – 250/2 = 1875 mmb2 = 2500/2 – 250/2 = 1125 mmlo = 6000 mm
beff, i = 0.2bi + 0.1lo 0.2lobeff, 1 = 0.2b1 + 0.1lo = 975 mm 0.2lo (1200 mm) and b1 (1875 mm)beff, 2 = 0.2b2 + 0.2lo = 825 mm 0.2lo (1200 mm) and b2 (1125 mm)
b = b1 + b2 + bw = 1875 + 1125 + 250 = 3250 mm
Effective flange width;beff = beff, i + bw b
= 975 + 825 + 250= 2050 mm b (3250 mm)
beff = 2050 mm
Loading on Slab
Slab self weight = 0.125 x 25 = 3.13 kN/m2
Finishes etc. = 1.50 kN/m2
Characteristics permanent load on slab, gk = 4.63 kN/m2
Characteristics variable load on slab, qk = 3.00 kN/m2
Design load, n = 1.35gk + 1.5qk = 10.75 kN/m2
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20
n =
10
.75
kN
/m
n =
10
.75
kN
/m
Loading on Beam
w1 = 0.54nLx = 0.54(10.75)(4.0) = 23.22 kN/mw2 = 0.50nLx = 0.50(10.75)(2.5) = 13.44 kN/mBeam selfweight, w3 = 0.25(0.45 – 0.125)(25) x 1.35 = 2.74 kN/mBrickwall, w4 = 2.6(3.0) x 1.35 = 10.53 kN/m
Total design action on beam B/1-2, w = w1 + w2 + w3 + w4
= 49.93 kN/m
6 m
w = 49.93 kN/m
Vmax = wL/2 = 149.79 kNMmax = wL2/8 = 224.68 kNm
PSF
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21
Design of the Main Reinforcement
Effective depth, d = 450 – 30 – 8 – 25 = 387 mm
Mf = 0.567fckbhf (d – 0.5hf)= 0.567(25)(2050)(125)(387 – 0.5 x 125) x 10-6
= 1178.7 kNm
M = 224.68 kNm Mf Compression reinforcement is not required
Design as similar to rectangular beam.
K = M/fckbd2 = 224.68 106 / (25 2050 3872) = 0.029
z = 0.97d 0.95d Use z = 0.95d
As = M/0.87fykz = 224.68 106 / (0.87 500 0.95 387) = 1405 mm2
Check Area of ReinforcementAs, min = 0.26(fctm/fyk) bd
= 0.26 (2.56/500) bd = 0.0013 bd 0.0013 bd Use As, min = 0.0013 bd = 126 mm2
As, max = 0.04 Ac = 0.04 bh= 4500 mm2
Provide 3H25 (As = 1473 mm2)
22/8/2010
22
h = 450 mm
bw = 250 mm
beff = 2050 mm
3H25
Shear design, deflection and cracking are more less the same as the previous
example
Self study !!!
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23
Shear force coefficient
• Table 3.5: BS 8110: Part 1
Elastic analysis
• Moment distribution method
• Theory of Structures I
Computer analysis
• Stiffness matrix system
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24
Shear Force Coefficient Method: BS 8110: 1997
Elastic Analysis
EN 1992-1-1: Cl. 5.1.3
Load Set 1: Alternate or adjacent spans loaded
Alternate span carrying the design variable and
permanent load (1.35Qk + 1.5Gk), other spans carrying only
the design permanent loads (1.35Gk)
Any two adjacent spans carrying the design variable and
permanent loads (1.35Qk + 1.5Gk), all other spans carrying
only the design permanent load (1.35Gk)
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25
Elastic Analysis
Alternate Span Loaded Adjacent Span Loaded
Elastic Analysis
Load set 2: All or alternate spans loaded
All span carrying the design variable and permanent
loads (1.35Qk + 1.5Gk)
Alternate span carrying the design variable and
permanent load (1.35Qk + 1.5Gk), other spans carrying
only the design permanent loads (1.35Gk)
UK NATIONAL ANNEX
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26
Elastic AnalysisAll Spans Loaded
Alternate Span Loaded
Moment Redistribution
Moment redistribution is the transfer of moments to the less
stressed sections as sections of peak moment yield on their
ultimate capacity being reached.
From a design viewpoint, this behavior can be taken
advantage of by attempting to effect a redistribute bending
moment diagram which achieves a reduction in the maximum
moment levels (and a corresponding increase in the lower
moments at other locations)
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27
Moment Redistribution
Such an adjustment in the moment diagram often leads to the
design of a more economical structure with better balanced
proportion, and less congestion of reinforcement at critical
sections. BS EN 1992-1-1: Cl. 5.5 permit the moment
redistribution, provided that;
- The resulting distribution remains in equilibrium with the loads
- The continuous beams are predominantly subject to flexure
- The ratio of adjacent spans be in the range of 0.5 to 2.0
Moment Redistribution
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28
Ly/Lx = 2.67 2.0One-way slab
Ly/Lx = 2.67 2.0One-way slab
Design datafck = 25 N/mm2
fyk = 500 N/mm2
Nominal cover = 30 mm
LoadingLive load = 3.0 kN/m2
Finishing, partition etc. = 1.5 kN/m2
Reinforcement1 = 20 mm2 = 16 mm links = 6 mm
Slab thickness, hf = 110 mm
500 mm
225 mm
beff
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29
b1 = 3000/2 – 225/2 = 1387.5 mmb2 = 3000/2 – 225/2 = 1387.5 mm
Span A-B and C-Dlo = 0.85 8000 = 6800 mmbeff, i = 0.2bi + 0.1lo 0.2lobeff, 1 = 0.2b1 + 0.1lo = 957.5 mm 0.2lo (1360 mm) and b1 (1387.5 mm)beff, 2 = 0.2b2 + 0.2lo = 957.5 mm 0.2lo (1360 mm) and b2 (1387.5 mm)
b = b1 + b2 + bw = 1387.5 + 1387.5 + 225 = 3000 mm
Effective flange width;beff = beff, i + bw b
= 957.5 + 957.5 + 225= 2140 mm b (3000 mm)
beff = 2140 mm
b1 = 3000/2 – 225/2 = 1387.5 mmb2 = 3000/2 – 225/2 = 1387.5 mm
Span B-Clo = 0.70 8000 = 5600 mmbeff, i = 0.2bi + 0.1lo 0.2lobeff, 1 = 0.2b1 + 0.1lo = 837.5 mm 0.2lo (1120 mm) and b1 (1387.5 mm)beff, 2 = 0.2b2 + 0.2lo = 837.5 mm 0.2lo (1120 mm) and b2 (1387.5 mm)
b = b1 + b2 + bw = 1387.5 + 1387.5 + 225 = 3000 mm
Effective flange width;beff = beff, i + bw b
= 837.5 + 837.5 + 225= 1900 mm b (3000 mm)
beff = 1900 mm
22/8/2010
30
Action on Slab, n kN/m2
Slab self weight = 0.110 x 25 = 2.75 kN/m2
Finishing, partition etc. = 1.50 kN/m2
Characteristics permanent action on slab, gk = 2.75 + 1.50 = 4.25 kN/m2
Characteristics variable action on slab, qk = 3.00 kN/m2
Action on Beam 1a/A-D, w kN/m
Beam self weight = 0.225(0.50 – 0.11)(25) = 2.19 kN/mSelf weight from slab = 0.5 x 4.25 x 3.0 x 2 = 12.75 kN/mTotal permanent action, gk = 2.19 + 12.75 = 14.94 kN/m
Variable action from slab, qk = 0.5 x 3.00 x 3 x 2 = 9.00 kN/m
Total design action, w = 1.35(14.94) + 1.5(9.00) = 33.67 kN/m
Bending Moment and Shear ForceFrom Table 3.5: BS 8110: Part 1
22/8/2010
31
0.45F
0.6F
0.55F
0.55F0.45F
0.6F
0.11FL 0.11FL
0.09FL 0.09FL
0.07FL
F = wL= 33.67 8= 269.36 kN
SFD
BMD
Effective Depth
d = 500 – 30 – 6 – 20 = 444 mm
Span A-B and C-D
Mf = 0.567fckbhf (d – 0.5hf)= 0.567(25)(2140)(110)(444 – 0.5 x 110) x 10-6
= 1298.0 kNm
M = 0.09FL = 193.9 kNm Mf NA lies in the flange
K = M/fckbd2 = 0.018z = 0.98d 0.95d Use z = 0.95d
As = M/0.87fykz = 193.9 x 106 / (0.87 x 500 x 0.95 x 444) = 1057 mm2
22/8/2010
32
Check Area of ReinforcementAs, min = 0.26(fctm/fyk) bd
= 0.26 (2.56/500) bd = 0.0013 bd 0.0013 bd Use As, min = 0.0013 bd = 126 mm2
As, max = 0.04 Ac = 0.04 bh= 4500 mm2
As, min As As, max
Provide 5H20 (As = 1571 mm2)h = 500 mm
bw = 225 mm
beff = 2140mm
2H20
3H20
Support B and C
M = 0.11FL = 237.0 kNm
K = M/fckbd2 = 0.214 Kbal (0.167) Compression reinforcement required
z = 0.82d d’ = 30 + 6 + (16/2) = 44 mmx = (d – z)/0.4 = 199 mmd’/x = 44/199 = 0.22 0.38, Use fsc = 0.87fyk
As’ = (K – Kbal)fckbd2/0.87fyk(d – d’)= (0.214 – 0.167)(25)(225)(444)2/(0.87)(500)(444 – 44)= 299 mm2
As = (0.167fckbd2/0.87fykz) + As’= (0.167 x 25 x 225 x 4442/0.87 x 500 x 0.82 x 444) + 299= 1466 mm2
22/8/2010
33
Check Area of ReinforcementAs, min = 0.013 bd = 136 mm2
As, max = 0.04 bh = 4500 mm2
Provide 2H16 (As’ = 402 mm2)Provide 5H20 (As = 1571 mm2)
h = 500 mm
bw = 225 mm
beff = 2140 mm
2H16
3H202H20
Span B-C
Mf = 0.567fckbhf (d – 0.5hf)= 0.567(25)(1900)(110)(444 – 0.5 x 110) x 10-6 = 1152 kNm
M = 0.07FL = 150.8 kNm Mf NA lies in the flange
K = M/fckbd2 = 0.016z = 0.98d 0.95d Use z = 0.95dAs = M/0.87fykz = 150.8 x 106 / (0.87 x 500 x 0.95 x 444)
= 822 mm2
As, min = 0.013 bd = 136 mm2
As, max = 0.04 bh = 4500 mm2
Provide 3H20 (As = 943 mm2)
h=
50
0 m
m
bw = 225 mm
beff = 1900 mm
3H20
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34
Concrete Strut Capacity
VRd, max = 0.36bwdfck(1 – fck/250) / (cot + tan )= 0.36 225 444 25 (1 – 25/250)
(cot + tan )
for = 22, cot = 2.5 VRd, max = 287 kN = 45, cot = 1.0 VRd, max = 414 kN
Support A & D
VEd = 0.45F = 0.45 269.36 = 121.2 kN VRd, max cot 2.5 VRd, max cot 1.0
Therefore, 22Use = 22; tan = 0.40; cot = 2.48
Shear links;
Asw/s = VEd / 0.78fykdcot = 121.2 103 / (0.78 500 444 2.48) = 0.277
Try link: H6 Asw = 57 mm2
Spacing, s= 57/0.277 = 204 mm smax = 0.75d (333 mm)
Provide H6 – 200 mm
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35
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot = 0.5 121 2.48= 150 kN
MEd, max / z = 194 106 / 431 = 450 kN 150 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 150 103 / 0.87 500= 345 mm2
Provide 2H16 ( As = 402 mm2)
Support B & C
VEd = 0.60F = 0.60 269.36 = 161.6 kN VRd, max cot 2.5 VRd, max cot 1.0
Therefore, 22Use = 22; tan = 0.40; cot = 2.48
Shear links;
Asw/s = VEd / 0.78fykdcot = 162 103 / (0.78 500 444 2.48) = 0.369
Try link: H6 Asw = 57 mm2
Spacing, s= 57/0.369 = 153 mm smax = 0.75d (333 mm)
Provide H6 – 150 mm
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36
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot = 0.5 162 2.48= 200 kN
MEd, max / z = 200 106 / 431 = 450 kN 200 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 200 103 / 0.87 500= 460 mm2
Provide 3H16 ( As = 603 mm2)
Support B & C
VEd = 0.55F = 0.55 269.36 = 148.1 kN VRd, max cot 2.5 VRd, max cot 1.0
Therefore, 22Use = 22; tan = 0.40; cot = 2.48
Shear links;
Asw/s = VEd / 0.78fykdcot = 148 103 / (0.78 500 444 2.48) = 0.338
Try link: H6 Asw = 57 mm2
Spacing, s= 57/0.338 = 167 mm smax = 0.75d (341 mm)
Provide H6 – 150 mm
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37
Additional longitudinal reinforcement;
Additional tensile force, Ftd = 0.5VEd cot = 0.5 148 2.48= 183 kN
MEd, max / z = 200 106 / 431 = 450 kN 183 kN
Additional tension reinforcement, As = Ftd / 0.87fyk
= 183 103 / 0.87 500= 422 mm2
Provide 3H16 ( As = 603 mm2)
Minimum links;Asw/s = 0.08fck
1/2bw / fyk
= 0.08 (25)1/2 225 / 500 = 0.180
Try link: H6 Asw = 57 mm2
Spacing, s= 57/0.180 = 314 mm smax = 0.75d (341 mm)
Provide H6 – 300 mm
Shear resistance of minimum links;Vmin = (Asw/s)(0.78d fykcot )
= (57/300)(0.78 444 500 2.48)= 83 kN
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38
121.2
161.8
148.1
148.1121.2
161.8
83
83
Transverse steel in the flange;
x = 0.5(0.85L/2) = 6800/4 = 1700 mm
Change of moment over distance x from zero moment:
M = 121.2 1.7 – 33.67 1.7 1.7/2 = 157.4 kNm
Change in longitudinal force;
= 157.4 103 (2140 – 225)
(444 – 55) (2 2140)
= 181.0 kN
f
wf
f
d
2/)(x
)2/( b
bb
hd
MF
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39
Longitudinal shear stress;
vEd = Ftd / (hf x)
= 181.0 103 / (110 1700)
= 0.97 N/mm2
Since vEd (0.97 N/mm2) 0.27fctk = 0.27 1.80 = 0.49 N/mm2
Transverse steel reinforcement is required
Concrete strut capacity in the flange;
vRd, max = 0.4fck (1 – fck/250) / (cot + tan )
= 0.4 25 (1 – 25/250)
(cot + tan )
for = 27, cot = 2.0 vRd, min = 3.59 N/mm2
= 45, cot = 1.0 vRd, max = 4.50 N/mm2
vEd (0.97 N/mm2) vRd, min cot = 2.0 (3.59 N/mm2)
and
vEd (0.97 N/mm2) VRd, max cot = 1.0 (4.50 N/mm2)
Therefore, use = 27 ; tan = 0.50 ; cot = 2.0
22/8/2010
40
Transverse shear reinforcement;
Asf / sf = vEdhf / 0.87fykcot
= 0.97 110 / (0.87 500 2.0)
= 0.12
Try H10: Asf = 79 mm2
Spacing, sf = 79/0.12 = 658 mm
Minimum transverse steel area;
As, min = 0.26 (fctm/fyk) bhf
= 0.26 (2.60/500) bhf
= 0.0013bhf 0.0013bhf
= 0.0013 1000 110 = 147 mm2
Provide H10 – 400 (As = 196 mm2/m)
Percentage of required tension reinforcement; = As, req / bd
=1057/ 225 444 = 0.010
Reference reinforcement ratio;o = (fck)
1/2 10-3 = (25)1/2 10-3 = 0.005
Since o Use Eq. (7.16b) in EC 2 Cl. 7.4.2
Percentage of required compression reinforcement;’ = 0
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41
For structural system, K = 1.3
= 1.3 (11 + 3.71 + 0.00) = 19.1
Basic span-effective depth ratio, l/d = 8000/444 = 18.01
'
12
1
'5.111 ck
ock ffK
d
l
Modification factor for span greater than 7 m;7/span = 7/8 = 0.875
Modification factor for tension steel area provided;As, prov/As, req = 1571/1057 = 1.48 1.5
Allowable span-effective depth ratio;(l/d)allow = 18.01 0.875 1.48 = 23.22
Actual span-effective depth ratio8000/444 = 18.01 (l/d)allow = 23.22 OK
22/8/2010
42
Limiting crack width, wmax = 0.3 mm
Steel stress,
= (500/1.15) [(14.94 + 0.3 9.0) / 33.7] 1.0= 262 N/mm2
Maximum allowable bar spacing = 150 mm (from interpolation)
Actual bar spacing = [225– 2(30) – 2(6) – 3(20)]/2= 46.5 mm 150 mm OK
1
)5.135.1(
3.0x
15.1 kk
kkyk
sQG
QGff
Ly/Lx = 2.67 2.0One-way slab
Ly/Lx = 2.67 2.0One-way slab
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Load
Minimum load, wmin = 1.35gk = 20.17 kN/mMaximum load, wmax = 1.35gk + 1.50qk = 33.67 kN/m
8 m 8 m 8 m
wmax = 33.67 kN/m
A B C D
8 m 8 m 8 m
wmax = 33.67 kN/m
A B C D
8 m 8 m 8 m
A B C D
CASE 1
CASE 2
CASE 3
wmax = 33.67 kN/m
wmin = 20.17 kN/m
wmin =20.17 kN/m wmin = 20.17 kN/m
wmax = 33.67 kN/m
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Case 1: All Span Maximum Load
End momentMAB = MBC = MCD = wL2/12 = 33.67(8)2/12 = 179.57 kNm
StiffnesskBA = kCD = 3EI/L = 3/8kBC = kCB = 4EI/L = 4/8
Distribution factorBA : BC = CD : CB = 0.43 : 0.57
0.43 0.57 0.57 0.43
-179.57 179.57 -179.57 179.57 -179.57 179.57
179.57 0.00 0.00 0.00 0.00 -179.57
89.78 0.00 0.00 -89.78
-38.60 -51.17 51.17 38.60
0.00 25.58 -25.58 0.00
-11.00 -14.58 14.58 11.00
0.00 7.29 7.29 0.00
-3.13 -4.15 4.15 3.13
0.00 2.07 2.07 0.00
-0.89 -1.18 1.18 0.89
0.00 0.59 -0.59 0.00
-0.25 -0.34 0.34 0.25
0.00 0.17 -0.17 0.00
-0.07 -0.10 0.07 0.10
0.00 0.05 0.05 0.00
-0.03 -0.03 0.03 0.03
0.00 215.38 --215.38 215.38 -215.38 0.00
A B C D
Moment Distribution
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Shear ForceCase 1
Total moment at B1 = 08VA – 33.67(8)(4) + 215.38 = 0VA = 107.7 kNVB1 = 33.67(8) – 107.7 = 161.6 kN
Total moment at C1 = 08VB2 – 33.67(8)(4) + 215.38 – 215.38 = 0VB2 = 134.68 kNVC1 = 33.67(8) – 134.68 = 134.68 kN
Total moment at D = 08VC2 – 33.67(8)(4) – 215.38 = 0VC2 = 161.6 kNVD = 33.67(8) – 161.6 = 107.7 kN
8 m
33.67 kN/m
VA VB1
215.38 kNm
8 m
33.67 kN/m
VB2 VC1
215.38 kNm215.38 kNm
8 m
33.67 kN/m
VC2 VD
215.38 kNm
VA = 107.7 kNVB = VB1 + VB2 = 296.3 kNVC = VC1 + VC2 = 296.3 kNVD = 107.7 kN
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Do the same for Case 2 and 3 loading condition
Self study !!!
Case 1
Case 3Case 2
Shear Force Diagram
161.7
134.7
113.1
161.7
134.7
113.1
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Case 1
Case 3
Case 2
Bending Moment Diagram
11.4
215.38 215.38
189.9
96.9
189.9
From the bending moment diagram in Example 4, reduced the moment at the support for 20%
Moment redistribution
Elastic analysis moment at support = 215.38 kNmReduced moment at support 20% = 0.8 x 215.38 = 172.30 kNm
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Shear ForceCase 1
Total moment at B1 = 08VA – 33.67(8)(4) + 172.30 = 0VA = 113.14 kNVB1 = 33.67(8) – 113.14 = 156.22 kN
Total moment at C1 = 08VB2 – 33.67(8)(4) + 172.30– 172.30 = 0VB2 = 134.68 kNVC1 = 33.67(8) – 134.68 = 134.68 kN
Total moment at D = 08VC2 – 33.67(8)(4) – 172.30 = 0VC2 = 156.22 kNVD = 33.67(8) – 156.22 = 113.14 kN
8 m
33.67 kN/m
VA VB1
172.30 kNm
8 m
33.67 kN/m
VB2 VC1
172.30 kNm172.30 kNm
8 m
33.67 kN/m
VC2 VD
172.30 kNm
VA = 113.1 kNVB = VB1 + VB2 = 290.9 kNVC = VC1 + VC2 = 290.9 kNVD = 113.1 kN
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Do the same for Case 2 and 3 loading condition
Self study !!!
Shear Force Diagram: Before Redistribution
Case 1
Case 3Case 2
161.7
134.7
113.1
161.7
134.7
113.1
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113.1
156.2
156.2
113.1
156.2
156.2
Case 1
Case 3Case 2
Shear Force Diagram: After Redistribution
Bending Moment Diagram: Before Redistribution
Case 1
Case 3
Case 211.4
215.38 215.38
189.9
96.9
189.9
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Case 1
Case 3
Case 2
Bending Moment Diagram: After Redistribution
189.9
11.3
189.9
172.2 172.2
96.7
Design the beam as in Example 3
The difference is the addition of value in determining Kbal
Self Study and Do Your Project. Then you’ll know !!!
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