22/8/2010...22/8/2010 15 additional longitudinal reinforcement; additional tensile force, f td =...

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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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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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Trial and error method:

Overall depth, h = L

13 to 18

Width, b = 0.3h – 0.6h

b

h

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EN 1992-1-2, gives a method for determining the minimum dimension of beams for fire resistance requirements

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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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w = vsx = vxnLx kN/mw

= v

sy

=

vynL

xkN

/m

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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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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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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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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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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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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



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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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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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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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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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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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



As, max = 0.04 Ac = 0.04 bh= 4500 mm2

Provide 3H25 (As = 1473 mm2)

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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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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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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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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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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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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


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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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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


= 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


= 837.5 + 837.5 + 225= 1900 mm b (3000 mm)

beff = 1900 mm

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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

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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

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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

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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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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


Shear links;

Asw/s = VEd / 0.78fykdcot = 121.2 103 / (0.78 500 444 2.48) = 0.277




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MEd, max / z = 194 106 / 431 = 450 kN 150 kN


= 150 103 / 0.87 500= 345 mm2

Provide 2H16 ( As = 402 mm2)

Support B & C



Shear links;

Asw/s = VEd / 0.78fykdcot = 162 103 / (0.78 500 444 2.48) = 0.369




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MEd, max / z = 200 106 / 431 = 450 kN 200 kN


= 200 103 / 0.87 500= 460 mm2


Support B & C



Shear links;

Asw/s = VEd / 0.78fykdcot = 148 103 / (0.78 500 444 2.48) = 0.338




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MEd, max / z = 200 106 / 431 = 450 kN 183 kN


= 183 103 / 0.87 500= 422 mm2


Minimum links;Asw/s = 0.08fck

1/2bw / fyk

= 0.08 (25)1/2 225 / 500 = 0.180




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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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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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

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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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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

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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



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43

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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22/8/2010...22/8/2010 15 additional longitudinal reinforcement; additional tensile force, f td =...

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