LUKISAN STRUKTUR

12

6.0 REKA BENTUK RASUK DISOKONG MUDAH

Ref. Calculation Output

SPECIFICATION W kN/m

L h

b

Effective span, L = 8.25m

Characteristic actions :

Permanent, gk = 15 kN/m

Variable, qk = 10 kN/m

Design life = 50 years

Fire resistance = R60

Exposure classes = XC1

Materials:

Characteristic strength of concrete, fck = 20 N/mm2

(Table F.1 EN

206)

Characteristic strength of steel, fyk = 500 N/mm2

Characteristic strength of link, fyk = 500 N/mm2

Unit weight of reinforced concrete = 25 kN/m3

(Table A.1 EN

1991-1)

Assumed:

bar 1 = 20mm

bar 2 = 12mm

link = 8mm

(Excluding

selfweight)

(Table 2.1 EN 1990)

SIZE Overall depth, h = L /13 = 8250/13 = 20mm

Width, b = 0.4h = 0.4x 635 = 254mm

Use : b x h

= 250 x 650 mm

Table 4.2

Table 4.4N

4.4.1.2

4.4.1.3

4.4.1.1(2)

DURABILITY, FIRE & BOND REQUIREMENTS

Min. conc. Cover regard to bond, C min,b = 20mm

Min. conc. Cover regard to durability, C min, dur = 15mm

Min. required axis distance for R60 fire resistance,

asd = 30 + 10 = 40mm

Min. concrete cover regard to fire,

Cmin = asd - link - bar/2

= 40-8-0.5(20) = 22mm

Allowance in design for deviation, Cdev = 10 mm

Nominal cover,

Cnom = Cmin + Cdev = 20 + 10 = 30mm

Table 5.5 EN 1992-

1-2

Use :

Cnom = 35mm

13

Ref Calculation Output

Table A

1.2B

:EN 1990

LOADING & ANALYSIS

Beam selfweight = (0.25 x 0.65) x 25 = 4.06 kN/m

Permanent load (excluding selfweight) = 15.00 kN/m

Characteristic permanent action, gk = 19.06 kN/m

Characteristic variable action, qk = 10.00 kN/m

Design action,

Wd= 1.35gk + 1.5 qk = 1.35(19.06) + 1.5(10.00)

= 40.73 kN/m

Wd= 40.73 kN/m

L=8.25m

Shear force,

V = wd L/2

= 40.73(8.25)/2

= 168.0 kN

M

BENDINGMOMENT

M = wdL2/8

= 40.73(8.25)2/8

=346.6 kNm

MAIN REINFORCEMENT Effective Depth,

D = H Cnom - link - bar = 587mm

D = Cnom - link - bar/2 = 49mm

Design Bending Moment, Med = 346.6 Knm

K = M/Bd2fck

= 346.6x106 / (250 X 587

2 X 20)

= 0.201

Redistribution = 0% Redistribution ratio, = 1.0

Kbal = 0.454 ( - k1)/ k2 0.182( - k1)/k2]2

= 0.363 ( - k1) 0.116 ( -k1)2

= 0.167

K > Kbal Compression reinforcement is required

z = d [ 0.5 + 0.25 kbal/1.134 ]

= 0.82d

d

b

Using : EC2

K1 = 0.44

K2 = 1.25

14


= 0.82(587)

= 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 yielded

fsc = 0.87 fyk

Area of compression steel

As = ( K Kbal) fckbd2

/ 0.87fyk (d - d)

= ( 0.201 0.167) x (20 x 250 x 5872)

0.87 x 500 x (587 -49)

= 250 mm2

Area of tension steel

As = kbalfckbd2 / 0.87fyk Z bal + As

= 0.167 x (20 x 250 x 5872) + 252

0.87 x 500 x 481.8

= 1623 mm2

Minimum and maximum reinforcement area,

As, min = 0.26(fctm/fyk) bd

= 0.26 (2.21/500) bd

= 0.0011 bd use= 0.0013bd

= 0.0013 x 250 x 587 = 119mm2

As,max = 0.04AC = 0.04bh

= 0.04 x 250 x 650 = 6500mm2

Use : 3H 12

(339 mm2)

Use : 6H 20

(1885 mm2)

SHEAR REINFORCEMENT

Design shear force, VEd =168.0 Kn

Concrete strut capacity

VRd, max = 0.36b wdfck ( 1- fck/250) / (cot + tan )

= (0.36 x 250 x 587 x 20 (1-20/250)

Cot +tan

=338Kn = 22 deg cot = 2.5

=486Kn = 45 deg cot = 1.0

VED < VRD,max cot = 2.5

VED < VRD, max cot = 1.0

Therefore angle < 22o

= 0.5 sin -1

[VED / 0.18bwdfck(1-fck/250)]

= 0.5 sin-1

( 168.0 x 103

)

(0.18 x 250 x 587 x 20 (1- 20/250)

15


=0.5 sin

-1 (0.35)

= 10.1o

Use: = 22.0 o tan = 0.40 cot = 2.48

9.2.2 (6)

9.2.2 (5)

Shear links

ASW/S =VED/0.78fykd cot

=168.0 x 103 /(0.78x 500x 587x 2.48)

=0.297

Try link: H8 ASW = 101 mm2

Spacing s = 101/0.30

=339 mm

Max. spacing , s max = 0.75d =0.75x 587 =440mm

Minimum links

ASW/S = 0.08fck1/2

bw/fyk

=0.08 x(20)

x 250/ 500

=0.179

Try link: H8 ASW = 101 mm2

Spacing s = 101/0.18

=562mm > 0.75d =440mm

Shear resistance of minimum links

V min = (ASW/s)( 0.78fykd cot )

=(101/425) x (0.78 x 587 x 500 x 2.5)

=134kN

Use :

H8-325

Use :

H8-425

Links arrangement 168kN

X = (16-134)/40.7

=0.83m

134kN

134kN

168kN

16


Additional longitudinal reinforcement

Additional tensile force,

F td =0.5 vEd cot

= 0.5 x 168 x 2.48

= 208 Kn

MED,max/Z = 346.6 x 106 / 481.8

= 719 kN > F td

Additional longitudinal reinforcement,

As = F td/0.87fyk

= 207.94 x 103

/ 0.87 x 500

= 478 mm2

Use : 2H 20

(628 mm2)

DEFLECTION Percentage of required tension reinforcement

= As,req/bd

= 1624 / 250 x 587

= 0.011

Reference reinforcement ratio,

o = ( fck)1/2

x 10-3

= (20)1/2

x 10-3

= 0.0045

Percentage of required compression reinforcement

= As,req / bd

= 252 / 250 x 587

= 0.002

Factor for structural system , K = 1.0

> o (use equation 2)

1/d = K [11 + 1.5 fck o/ + 3.2 fck(o/ 1)3/2

] -> (1)

1/d = K [11 + 1.5fck o/ + 1/12 fck/] ->(2)

= 1.0 (11 + 3.21 + 0.23)

=14.4

Therefore basic span-effective depth ratio, l/d =14.44

Modification factor for span greater than 7m

= 7/span = 7/8 =0.85

17


Modification factor for steel are provided,

As,prov / As,req =1885 / 1624 =1.16 < 1.5

Therefore allowable span-effective depth ratio

(i/d) allowable =14.44 x 0.85 x 1.16 =14.22

Actual span effective depth

(i/d)actual =8250 /857 =14.1 < (i/d)allowable

Table 7.1 N CRACKING

Limiting crack width, w max = 0.3mm

Steel stress,

fs = fyk/1.15 x [Gk + 0.3 Qk/(1.35Gk + 1.5 Qk)](1/)

= (500/1.15) x [19.1 + 0.3(10) / 1.35 (19.1) + 1.5 (10)](1.0)

= 236 N/mm2

Max allowable bar spacing = 150 mm

Bar spacing

s = [250-2(35)-2(8)-(20)]/2

=72 mm < 150mm

18


DETAILING

19

7.0 REKA BENTUK RASUK SELANJAR

Ref Calculations Output

SPECIFICATION

Dimension:

Span, L = 8.00 m

Width, B = 3.00 m

Slab thickness = 110 mm

Beam size = 225 x 500 mm

Charasteristic loads:

Finishes, etc = 1.5 kN/m2

Variable, qk = 3.0 kN/m2

Materials:

Unit weight of concrete = 25 kN/m3

Charasteristic strength of concrete, fcl = 25 N/m2

Charasteristic strength of steel, fyk = 500 N/m2

Charasteristic strength of link, fyk = 500 N/m2

Nominal concrete cover = 30 mm

Assumed : bar 1 = 20 mm

bar 2 = 16 mm

link = 6 mm

Design beam 1a/A-D

Beam 1a/A-D

w kN/m w kN/m w kN/m

A 8.0 m B 8.0 m C 8.0 m D

20


EFFECTIVE FLANGE WIDTH

Effective flange width,

b eff = beff i +bw b

beff i= 0.2bi + 0.1lo 0.2lo

Span A-B, C-D

l =0.85l = 0.85 x 8000 = 6800 mm

b eff,1 = 960 mm

21


Action on beam 1a/A-D : Distribution of action from slabs to beam are as follows :

Panel 1-1a/A-D :

LY/LX = 8.00/3.00 =2.7 > 2.0, one-way slab

Shear coefficient, v = 0.50

Panel 1a-2/A_D:

LY/LX = 8.00/3.00 =2.7 > 2.0, one-way slab

Shear coefficient, v = 0.5

Characteristic permanent action:

Wo = 0.225 (0.50 0.11) x 25 = 2.19 kN/m

W1 = vnLx = 0.50 x 4.25 x 3.00 = 6.38 kN/m

W2 = vnLx = 0.50 x 4.25 x 3.00 = 6.38 kN/m

Gk = 14.95kN/m

Characteristic variable action :

W1 = vnLx = 0.50 x 3.00 x 3.00 = 4.50 kN/m

W1 = vnLx = 0.50 x 3.00 x 3.00 = 4.50 kN/m

Qk = 9.00 kN/m

Design load,

Wd = 1.35Gk + 1.5Qk

= 1.35(14.95) + 1.5(9.00) = 33.68 kN/m

ANALYSIS

Continuous beam with equal spans, uniform action and

Qk < Gk use moment and shear coefficient

From table 3.14 BS8110 Part 1

Wd kN/m Wd kN/m Wd kN/m

L m L m L m

A B C

D

0.11 FL 0.11 FL

0.09 FL 0.07 FL 0.09

FL

0.45 F 0.55 F 0.6 F

0.6 F 0.55F 0.45 F

F = WD L kn = 33.67 x 8.0 = 269 kN

Shear Force

Bending Moment

22


9.2.1.1

MAIN REINFORCEMENT

Span A-B & C-D

Effective depth, d = h Cnom - link 0.5 bar = 454.0 mm

Bending moment,

M = 0.09 FL

= 0.09 X 269 X 8.0 = 194 kNm

Mf = 0.567 fckbhf (d - 0.5h)

= 0.567 x 25 x 2145 x 110 ( 454.0 55 )

= 1334 kNm

M < Mf neutral axis within the flange

K = M/fck bd2

= 194 x106 / (25 x 2145 x 454.0

2)

= 0.018

z = d [ 0.5 + 0.25 K/1.134)] = 0.98d

Area of tension reinforcement

As = M/0.87 fyk z

= 194.0 x 106 / (0.87 x 500 x 0.95 x 454.0)

= 1034 mm2

Minimum and maximum reinforcement area,

As min = 0.26 (fctm /fyk ) bd

= 0.26 ( 2.56 / 500) bd

= 0.0013bd use= 0.0013bd

= 0.0013 x 225 x 454.0 = 136 mm2

As max = 0.04 Ac = 0.04 bh

= 0.04 x 225 x 500 = 4500 mm2

Support B & C

Effective depth, d = h Cnom - link - bar = 444.0 mm

Bending moment,

M = 0.11 FL

= 0.11 X 269 X 8.0 =237kNm

K = M/ bd2 fck

= 237.1 x 106 / ( 225 x 444

2 x 25)

= 0.214

Redistribution = 0% redistribution ratio, = 1.0

K bal = 0.454( - k1)/k2 0.182[ - k1)/ k2]2

= 0.363 ( - k1) 0.116 ( - k1)2

= 0.167

Use : 5H 20

(1571 mm2)

Use : EC2

K1 = 0.44

K2 = 1.25

23


K > kbal

Compression reinforcement is required

d = c + link + 0.5 bar =44mm z =d (0.5 + 0.25- kbal /1.134)

= 0.82d =0.82 x 444 =364.4mm

X = (d-z)/0.4 =199.0mm

d/x =44/199.0 = 0.22 < 0.38

the compression steel will have have yielded

area of compression steel

As = (k- kbal) fckbd2/0.87fyk (d- d )

= (0.214-0.17)x (25 x 225 x 4442)

0.87 x 500 x (444-44)

=299mm2

Area of tension steel

As = (k- kbal) fckbd2/0.87fyk z bal + As

= 0.167x (25 x 225 x 4442)

0.87 x 500 x 364 + 299

= 1466mm2

Asmin = 0.0013 bd =136mm2

Asmax = 0.04 bh = 4500mm2

Span B-C

Effective depth , d =h-cnom link- bar =444.0 mm

Bending moment,

M =0.07FL

=0.007 X 269 X8.0 =151 kNm

Mf =0.567 fckbhr(d-0.5h)

=0.567 x 25 x 1345 x 110 (454.0-55)

=837 kNm

M < Mf neutral axis within the flange

K =m/fckbd2

=151 x 106/ (25 x1345 x454.0

2)

=0.022

Z = d(0.5+ 0.25-k/1.134) = 0.98 d

Use : 5H 20

(1571 mm2)

Use : 2H 16

(402mm2)

24


Area of tension renforcement

As = M/0.87fyk =

= 150,9X106/(0.87 X 500 X 0.95 X 454.0)

=804mm2

Asmin =0.0013 bd =136mm2

Asmax =0.04 bh = 4500mm2

Use : 3H 20

(943mm2)

6.2.3

9.2.2 (6)

6.2.3 (7)

SHEAR REINFORCEMENT

Design shear force, VEd =168.0 Kn

Concrete strut capacity

VRd, max = 0.36b wdfck ( 1- fck/250) / (cot + tan )

= (0.36 x 250 x 587 x 20 (1-20/250)

Cot +tan

=338Kn = 22 deg cot = 2.5

=486Kn = 45 deg cot = 1.0

Support A & D

VED =0.45 F = 0.45 X 269 =121 kNm

< VRD,max cot = 2.5

< VRD, max cot = 1.0

Therefore angle < 22o

Shear link

Asw /s = ved /0.78fyk dcot

= 121.2 x 103/ (0.78 x 500 x 454 2.48)

=0.277

Try link : H6 Asw = 57mm2

Spacing, s =57/0.2777

= 204mm < 0.5d = 341 mm



f td = 0.5 ved cot

= 0.5 x 121 x 2.48

= 150 kN

MEd max/z = 194.0 X 106/431.3

=450 kN > F td

As req =F td/0.87fyk =150 x 10

3 /(0.87x 500)

= 345mm2

Use :H6-200

Use :2H-200

25

Ref. Calculations Output

9.2.2 (6)

6.2.3 (7)

9.2.2 (6)

6.2.3 (7)

Support B & C

VEd = 0.60 F = 0.60 x 269 = 162 kN

< VRd, max cot = 2.5


Therefore angle < 22

Shear links

Asw / s = VEd / 0.78 fyk d cot

= 161.6 x 103 / (0.78 x 500 x 454 x 2.48)

= 0.369

Try link : H6 Asw = 57 mm2

Spacing, s = 57 / 0.369

= 153 mm < 0.75d = 341 mm



Ftd = 0.5 VEd cot

= 0.5 x 162 x 2.48 = 200kN

As req = Ftd / 0.87fyk

= 200 x 103 / (0.87 x 500) = 460 mm

2

Support B & C

VEd = 0.55 F = 0.55 x 269 = 148 kN



Therefore angle < 22

Shear links

Asw / s = VEd / 0.78 fyk d cot

= 148.2 x 103 / (0.78 x 500 x 454 x 2.48)

= 0.338


Spacing, s = 57 / 0.338

= 167 mm < 0.75d = 341 mm



Ftd = 0.5 VEd cot

= 0.5 x 148 x 2.48 = 183 kN

As req = Ftd / 0.87fyk

= 183 x 103 / (0.87 x 500) = 422 mm

2

Use : H6 150

Use : 3H 16

( 603 mm2)

Use : H6 150

Use : 3H 16

( 603 mm2)

26


9.2.2 (5)

9.2.2 (6)

6.2.4

Minimum links

Asw / s = 0.08fck1/2

b w / fyk = 0.08 x (25)

x 225 / 500

= 0.180


Spacing, s = 57 / 0.18

= 314 mm

< 0.75d = 0.75 x 454 = 341 mm

Shear resistance of minimum links

Vmin = (Asw / s) (0.78dfyk cot )

= (57 / 300) x (0.78 x 454 x 500 x 2.5)

= 83 kN

Link arrangement

121 148 162 83

83 162 148 121

x1 x2 x3 | | | | |

x1 = (121 83) / 33.67 = 1.14 m

x2 = (162 83) / 33.67 = 2.34 m x3 = (148 83) / 33.67 = 1.94 m

Transverse steel in the flange

The longitudinal shear stresses are the greatest over a distance x

measured from the point of zero moment.

x = 0.5(L/2) = (L/4) = 8000 / 4 = 2000 mm

The change in moment over distance x from zero moment,

M = (wL/2) (L/4) (wL/4) (L/8) = 3wL2/32

= 3 x 33.7 x 8.02 / 32 = 202.04 kNm

The change in longitudinal force,

Fd = [ M / (d 0.5hf) ] x [ (b bw) / 2b ]

= 202.04 x 103

x

(2145 225)

(454 55) (2 x 2145)

= 227 kN

Use :

H6 - 300

27


Longitudinal shear stress

VED = FO / (hfx)

= 227 x 103 / (110 x 2000)

= 1.03 N/mm2

VED > 0.27fck = 0.27 x 2.03 = 0.55 N/mm2

Transverse shear reinforcement is required

Concrete strut capacity in the flange

vED max = 0.4fck (1-fck / 250) / (cot + tan )

= ( 0.40 x 25 ( 1- 25 / 250)

( cot + tan )

= 3.59 N/mm2 = 27 deg cot = 2.0

= 4.50 N/mm2

= 45 deg cot = 1.0

vED < vED, max cot = 2.0

vED < vED, max cot = 1.0

therefore angle < 27o

= 0.5 sin -1

[ vED / 0.2 fck(1 - fck / 250)]

= 0.5 sin -1

{ 1.03 }

{0.20 x 25 (1 25/250)}

= 0.5 sin -1

{0.23}

= 6.62o

Use : = 26.5o tan = 0.50 cot = 2.0

Transverse shear reinforcement

Asf / sf = vED hf / 0.87 fykcot

= 1.03 x 110 / (0.87 x 500 x 2.0)

= 0.13

Try : H10 Asf = 79 mm2

Spacing, sf = 79/0.13 = 605mm

28


Minimum transverse steel area,

As,min = 0.26( fctm / fyk) bhf

= 0.26 (2.56 / 500)bhf = 0.0013 bhf Use = 0.0013bhf

= 0.0013 x 1000 x 110.0

= 147 mm2/m

Use:

H10-400

(196 mm2/m)

DEFLECTION

Check at span A-B

Percentage of required tension reinforcement,

= As,req / bd

= 1034 / 225 x 454 = 0.010

Reference reinforcement ratio,

o = (fck)1/2

x 10-3

= (25)1/2

x 10-3

= 0.005

percentage of required compression reinforcement,

= As,req / bd

= 0 /225 x 454 = 0

Factor for structural system, K = 1.3

> o use equation (2)

1/d = K [ 11 + 1.5 fck o/ + 3.2 fck(o/ 1)3/2

] (1)

1/d = K [11 + 1.5fck o / + 1/12 fck / ] (2)

= 1.3( 11 + 3.71 + 0.00)

= 19.1

Therefore basic span-effective depth ratio, 1/d = 19.12

Modification factor for steel area provided,

=As,prov/ As,req

= 1571/1034

= 1.52 > 1.50

29


Therefore allowable span effective depth ratio,

(1/d)allowable = 19.12 x 0.88 x 1.50 = 25.23

Actual span effective depth

(1/d)actual = 8000/454.0 = 17.6 < (1/d)allowable

OK!

CRACKING

Limiting crack width, wmax = 0.3 mm

Steel stress,

fs = fyk / 1.15 x Gk + 0.3Qk / (1.35Gk + 1.5Qk)(1/)

= (500/1.15) x [(14.9 + (0.3 x 9.0)) / 33.7] x 1.0

= 262 N/mm2

Max allowable bar spacing = 150 mm

Bar spacing

s = [ 225 2(30) 2(6) (20)]/2

= 66.5 mm < 150mm

OK!

30


DETAILING

31

ANGGARAN BAHAN


Simply supported beam

Konkrit

Isipadu konkrit = Panjang (P) x Lebar (L) x Tinggi (H)

= 8.25 x 0.25 x 0.6

= 1.24m3

Tetulang

Panjang 1 batang tetulang = 12 m

Jenis tetulang 3H-20 (2), 3H-12

Panjang tetulang 3H-20 = 6 x 8.25

= 49.5 m

Bilangan tetulang 3H-20 = 4.125

Panjang tetulang 3H-12 = 3 x 8.25

=24.75 m


Continuous beam

Konkrit

Isipadu konkrit = Panjang (P) x Lebar (L) x Tinggi (H)

= 24 x 0.225 x 0.45

= 2.43 m3

Tetulang

Panjang 1 batang tetulang = 12 m

Jenis tetulang 3H-20, 2H-20 (4), 2H-25 , 1H-20 (2)

Panjang tetulang 3H-20 = 3 x 8

= 24 m

Bilangan tetulang 3H-20 = 2

Panjang tetulang 2H-20 = 4 x 8 (2)

= 64 m

32



= 16 m



= 16 m


33

6.0 KESIMPULAN

Kesimpulannya rasuk merupakan satu komponen dalam struktur bangunan

yang sangat penting. Tugasnya adalah untuk menerima dan menanggung beban dari

bumbung, lantai dan sebagainya. Penggunaan rasuk sangat la luas seperti di dalam

pembinaan ianya digunakan untung membuat jambatan, bangunan dan sebagainya.

Rasuk terdiri daripada pelbagai jenis antaranya ialah rasuk sokong mudah, rasuk

julur, rasuk selanjar dan sebagainya. Dalam pembinaan rasuk ianya haruslah di reka

bentuk terlebih dahulu mengikut spesifikasi yang telah ditetapkan dan ianya perlu

melepasi ujian kiraan supaya tiada sebarang kegagalan terjadi.

Terdapat pelbagai jenis sistem rasuk yang biasa ditemui dalam bidang

kejuruteraan, antaranya adalah rasuk disokong mudah, rasuk julur, rasuk terbina

dalam, rasuk juntaian dan rasuk berterusan. Setiap keadaan ini mempunyai kekuatan

yang berlainan bersesuaian dengan setiap kegunaannya. Apabila menerima bebanan,

setiap rasuk akan mengalami pesongan di mana rasuk tersebut akan melentur

mengikut arah daya yang dikenakan. Setiap rasuk mempunyai satu tahap di mana ia

akan melentur pada satu kedudukan yang maksimum apabila beban dikenakan.

LUKISAN STRUKTUR

Documents

knm wd

fyk d d

knm variable

knm design action

max cot

knm design life

cot tan

h cnom link