durabilty and fire resistance

8/12/2019 Durabilty and Fire Resistance

1/6

ProjectConcrete Design

Job ref

Part of structureMinimum concrete cover C nom

Sheet no ref

1 / 1 Drawing ref.

see Ref Calc by

M Omar, A Sharriff Date

01/01/2011 Check by Date

Ref Calculations Output

Task

Note 1Table B2

EN 19924.4.1.2 (3)

4.4.1.2 (5)

4.4.1.2 (6)

4.4.1.2 (7)

4.4.1.2 (8)

4.4.1.3

Note 1:Manual forthe designofconcretebuildingstructurestoEurocode2

Assuming the minimum concrete cover for the verification of durability and fire resistance As fol lows:

a) Concrete beam/slab: Fire resistance class: R90Exposure conditions: XC1

Support conditions: simple supported Max reinforcement diameter used: 20mm

b) Concrete column: Fire resistance class: R120Exposure conditions: XC3

f i = 0.7 Max reinforcement diameter used: 16mm

- Durability Find for the environment specified above:

1) The upper limit to the water/cement ratio =2) The lower limit to the cement content =

In addition to the above requirements for durability in any given environment are:3) good compaction4) adequate curing5) good detailing6) The lower limit to the thickness of the cover c min to the reinforcement

The greater value for c min satisfying the requirements for both bond and environmental

conditions shall be used.

c min = max {c min ,b; c min,dur + c dur, - c dur,st - c dur,add ; 14 mm} (4.2)

where: c min,b minimum cover due to bond requirement =

c min,dur minimum cover due to environmental conditions =

c dur, additive safety element =

c dur,st reduction of minimum cover for use of stainless steel =

c dur,add reduction of minimum cover for use of additional protection =

c d e v = 14mm cmin= a) 20mm b) 25mm

c n o m = c v = c m i n + c d e v =

- Fire resistance

Minimum cover to centreline of main reinforcement a =

Minimum dimensions with or thicknessinitial /final design initial /final design

a) Concrete slab/beam: slab: 200;25 / 200;25 beam: 200;39 / 250;46

b) Concrete column: final design bmin = 350 and a= 57mm or

initial = 500mm / no dim for a bmin = 450 and a= 51 mm

a) b)0.7 0.45

40kg/m 3340kg/m

a) b)20 1615 250 00 0

0 0

34 39


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Job ref

Part of structureSLS of deflection, cracking and vibration

Sheet no ref

1 / 1 Drawing ref.

see Ref Calc by




Table 4.2

Isrtuct emanual

Table 5.

Concretecentre

Figure 8

Table 11

Concrete slab= 7 m ( simply supported)

Concrete beam =6 m ( continuous)

Deflection Initial Design

With f yk being 500Mpa The span/effective Depth ratio is

One way simply supported slab = 20

Continuous beam= 18

Effective depth :

Concrete slab = 7/20 = 0.35m

Concrete beam= 6/18 = 0.33

Where use :

3/2 for

( )for For the final design

For concrete slab A s,req /bd2 = 0.5

For concrete beam A s,req /bd 2 = 01.5

Therefore the concrete slab span/effective depth ratio = 18.5

& concrete beam span/effective depth ratio = 26 or 30

Crack width

Assuming 2= 0.3 & = 1.25 ratio Gk/Qk= 2.8

From figure 6 su= 243Ratio of redistributed moment to elastic moment =1

Required reinforcement/ provided reinforcement = 1

Therefore use su

Concrete slab

Maximium spacing = 200mm

Concrete beam

Maximum bar diameter = 16mm

350mm

333mm

18.5

26 or 30

200mm

16mm


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Job ref

Part of structure

Foundation designSheet no ref

1 / 1 Drawing ref.

see Ref Calc by Date



Concise

EC2

Design BasisThe limit states of stability strength and serviceability needs to be considered.

Design ActionsDead load Slab = 25 x 3.75 x 2.375 x 0.2 = 44.5 kN

Roof = 0.6 x 3.75 x 2.375 = 5.34 kN

Beam = 25 x 0.4 x 0.3 x ( 3.75 + 2.375) = 18.375 kN

Column = 25 x 0.3 x 0.3 x 1.303 = 2.93 kN

Sum of permanent load (Gk) = 44.5 + 5.34 + 18.375 + 2.39 + 0.2 = 70.805 kN

live load

Imposed load = 0.6 kN/m 2 x 3.75 x 2.375 = 5.34kN

Horizontal load ( wind action )

Assuming half of the wind directly hitting the wal l acting on the foundation

Aref = (4.5 x 2.38 )/ 2 = 5.355

Fwe = 1 x (-0.323 x 5.4) = -1.74 K N

Fwi = 1 x ( -0.04 x 5.44) = -0.216 kN

Therefore the total force acting on foundation = -1.74 (- 0.216) = - 1.52 kN

Axial load

ULS

NED = 1.35 Gk + 1.5 Qk = 1.35 x 88.94 + 1.5 x 4.725 = 131.63 kN

HORIZONTAL LOAD (ULS)

NED = 1.35 x 0 + 1.5 x -1.52 = -2.28 kN

Design Moment

From column calculation Design Moment ( Med = M 02 + M 2 = 37.2+ 113=150.2 kNm)

Therefore determining ULS ( Geo) Assuming 50% permanent action & 50 % variable action

MED = 29.95 KNm

Member size of foundation

To satisfy eccentricity

e = Med/ Ned = 29.95 / 131.63 = 0.228

To achieve e d/6 try d = b = 1.36m

Try 1.5m x 1.5m x 0.8m deep e < d/6

Ground bearing pressure:

70.805 kN

5.34kN

- 1.52 kN

131.63 kN

-2.28 kN


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Job ref

Part of structure

Wind load calculation

Sheet no ref

1 / 1 Drawing ref.

see Ref Calc by




Exp 4.1

Istruct E

Table B2

Annex C

Table C1

Size of foundation = 131.63 / 120 = 1.097 m 2 ( Area of foundation)Therefore the size of foundation is> 1.097 Thus a moment will add up to the ground pressureHence try 2m x 2 m x 0.8m

Checking ground bearing pressure

N found = 25 x 2 x 2 x 0.8 = 80 kN

Structural resistance = N ED, found = 1.35 x 80 = 108kN

P = N ED+ N ED, found = 373 kN + 108 = 481 kN

2 = 481 / 2 x 2 29.95 x 6 /( 2 x 2 2) = 120.25 22.46

= 142.85 < 168 & = 97.65

Flexibility of foundation

d = 2 & d column = 0.3 , h = 0.8 ,Vmax = 0.8

Therefore V = 2/2 0.3/2 = 0.85> 0.8 Therefore foundation is flexible.

Cover

Cover

cnom = c min + c dev

where cmin = max[cmin,b, cmin,dur]

wherecmin,b = diameter of bar. Assume 20 mm main bars

cmin,dur = minimum cover due to environmental conditions = 40mm Assuming primarily XC3 / XC4, secondarily XF1, cmin,dur = 35 mm

cdev = allowance in design for deviation = 10 mm try c nom = 40 + 10 =50

= 50 mm to main bars

d = 800- 50mm = 750 20/2 = 740mm

Yield Strength of reinforcement

Fywd, eff = 250 + 0.25 Fywd = 250 + 0.25 x 740 = 435 N/mm 2

Fywd = k x fyk/ = 1.05 x 500/1.15 = 457 N/mm 2

K=(fe/fy) k 1.05 for steel grade A

Concrete compressive strength

Fck = 30 N/mm 2 , fcd = = 1 x 30 / 1.5 = 20 N/mm 2

Flexure design

Design moment at face of column

Mean GBP = (120.25 + 142.25)/ 2 = 131.25 kN/ m 2

M = w x v 2/2 = 131.25 x 0.85 2/2 = 47.41 knm

K = M ED / bd 2fck = 47.41x10 6/ ( 1000 x 740 2 x 30)= 0.003

Limit z/d to 0.95

ok

OK

740mm

435 N/mm 2

20 N/mm 2

z/d = 0.97


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Job ref

Part of structure


Sheet no ref

1 / 1 Drawing ref.

see Ref Calc by




Table 6.1

Z = 0.95 x 740 = 703mm

Klim > Kf ; Therefore NO compression reinforcement needed

Reinforcement required

As = M / ( 0.87z Fyk) = 47.71 x 10 6 / 0.87 x 703 x 500 = 156 mm 2

As, min = 0.26 fcm, btd/fyk

Fctm = 3.08 N/mm 2

Bt = 1

Asmin = 0.26 x 3.08 x 1000 x 740/500 = 1185 > 156 & 0.0013 x 1000 x 740 = 962 > 156

Try H20 B : As, prov = 7 x 320 = 2240

Asreq/As,prov = 1185/2240 = 0.53 : Therefore aggregate needs to be limited to 0 20mm

Punching shear design

Ved =Ned = 373KN

A1 = 300 2 = 0.09m 2

Ved, red = Ved - = A1 x ( GBP Self weight of foundation)

=0.09 x ( 120.25 ( 25 x 0.8) = 9kN

Ved,red= 373 9 = 364 kN

Shear stress at column face

Ved,max = Ved,red/ud

D = mean effective depth of slab (dy + dz)/2

Dy = h Cnom 0.5 = 800 50 0.5 x 20 = 740

Dz = h Cnom 1.5 = 800 50 1.5 x 20 = 720

D = ( 740 + 720 ) / 2 = 730mm

Vo = 300 x 4 = 1200

K = 0.7

W1 = c1 2/2 + c1c2 + 4c2d + 16 d 2+2 dc1

W1 = 300 2/2 + 300 x 300 + 4 x 300 x 730 + 16 x 730 2 + 2 x 730 x 300 = 10913419mm 2

U2 = 2 x 730 = 1460

U1 = 1460 x 2 = 9174mm

= 1 + k Med/ved x u1/w1

156 mm 2

UseAsmin

364 kN


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Job ref

Part of structure


Sheet no ref

1 / 1 Drawing ref.

see Ref Calc by




Bs EN 1992(6.50)

= 1 + 0.6 x 47.41/373 x 10 3 x 9173 / 10913418 = 1 N/mm 2 Ved,max = 1 x 373 x 10 3 / 1200 x 730 = 0.43 N/mm 2

Shear resistance of strutVrd,max = 0.5v, fcdFcd = 30/1.5 = 20V = 0.6 ( 1 FCK/250) = 0.6 ( 1 30/250) = 0.530.5 x 0.53 x 20= 5.3 N/mm 2

Vrd, max = 5.3 N/mm 2 > Ved, max = 0.43 N/mm 2

Maximum shear strength Vrd

Vrd = Crdc k ( 100 , fck) 1/3

Crdc = 0,12

K = 1 + 200/d = 1 + 200/730 = 1.52Vrdc = 0.12 x 1.52 ( 100 x 0.003 x 30 ) 1/3 = 0.36N/mm 2

As.bwd = 2240/ 1000 x740 = 0.003

Vrdc= Vmin x d/a

Vrdc = 0.035 fck x k 3/2 x 2d/aVrdc = 0.035 x 30 3/2 x 1.52 3/2 x 2 = 0.36 N/mm 2

Uout, eff = x Ved / ( Vrdc x d)

= 1.52 x 373 x 10 3/ 0.36 x 730 = 2157mm

Distance from column face = Uout eef 4h / 2

= 2157mm 4 x 800 / 2 = 0.36N/mm 2

Shear stress at Uo 2

Ved,red (Uo 2) = 0.5d

A0.5 = ( 0.5d + dcol/2)2

x = (730 x 0.5 + 300/2)2

x = 0.83Ved,red = Ved - Ved

Ved = 0.83 x ( 120.25 25 x 0.8 ) = 83.2kN

Ved,red = 373 83.2 = 289.6

Perimeter at 0.5 d = 0.5 x 730 = 365mm

Uo 5 = 365 x 2 x + 300 x 4 = 3493.4mm

Ved at U2 = x Ved /( Uo 5 x d )

1.52 x 113 x 10 3 / ( 3493 x 730 ) = 0.06 N/mm 2 < 1.52N/mm 2 = Ved

Therefore No shear reinforcement required

ok

0.36N/mm2

2157mm

0.36N/mm 2

durabilty and fire resistance

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