RC DESIGN PROJECT EXAMPLE

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Sheet No. 4/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

Bending Moments z-z axis

Stiffness, K = I/L = bh 3 /12L

Column; Roof to 1st Floor, L = 3.20m ; K = (150x300³)/(12x3200) = 1.05 x 105
0.95 x 105
1st Floor to G. Floor, L = 3.56m ; K = (150x300³)/(12x3556) = 2.25 x 105
2.88 x 105
G. Floor to Footing, L = 1.50m ; K = (150x300³)/(12x1500) = 2.88 x 105
1.08 x 105
Beam; Roof level, L = 3.96m ; K = (150x450³)/(12x3960) =
22.61 kNm
1st Floor level, L = 3.96m ; K = (150x450³)/(12x3960) =

G. Floor level, L = 3.96m ; K = (150x325³)/(12x3960) =
Roof to 1st Floor
Fixed end moment, wL2/12
Δ M = M JH = 17.3x3.96²/12 =
17.3 kN/m

Moment in column

M = ΔM x K c / ( K c + K JH /2 )

KJH/2 = 22.61x1.05/(1.05+(2.88/2)) = 9.53 kNm

Kc 3.96m

1st Floor to G. Floor Fixed end moment, wL2/12
Δ M = M JH = 23.3x3.96²/12 = 30.45 kNm
Kcu
23.3 kN/m Moment in upper column
M = ΔM x K cu / ( K cu + K cl +K JH /2)
= 30.45x1.05/(1.05+0.95+(2.88/2)) = 9.29 kNm

KJH/2 Moment in lower column
Kcl 3.96m M = ΔM x K cl / ( K cu + K cl +K JH /2)

= 30.45x0.95/(1.05+0.95+(2.88/2)) = 8.41 kNm
Fixed end moment, wL2/12
G. Floor to Footing
Δ M = M JH = 12.5x3.96²/12 = 16.34 kNm
Kcu
12.5 kN/m Moment in upper column

M = ΔM x K cu / ( K cu + K cl +K JH /2)

= 16.34x0.95/(0.95+2.25+(1.08/2)) = 4.15 kNm

KBC/2 Moment in lower column
Kcl 3.96m M = ΔM x K cl / ( K cu + K cl +K JH /2)

= 16.34x2.25/(0.95+2.25+(1.08/2)) = 9.83 kNm

Bending Moments y -y axis

Stiffness, K = I/L = hb 3 /12L

Column; Roof to 1st Floor, L = 3.20m ; K = (300x150³)/(12x3200) = 0.26 x 105
0.24 x 105
1st Floor to G. Floor, L = 3.56m ; K = (300x150³)/(12x3556) = 0.56 x 105
2.88 x 105
G. Floor to Footing, L = 1.50m ; K = (300x150³)/(12x1500) = 2.88 x 105
1.08 x 105
Beam; Roof level, L = 3.96m ; K = (150x450³)/(12x3960) =
11.11 kNm
1st Floor level, L = 3.96m ; K = (150x450³)/(12x3960) =

G. Floor level, L = 3.96m ; K = (150x325³)/(12x3960) =
Roof to 1st Floor
Fixed end moment, wL2/12
Δ M = M JH = 8.5x3.96²/12 =
8.5 kN/m

Moment in column

M = ΔM x K c / ( K c + K JH /2 )

KJH/2 = 11.11x0.26/(0.26+(2.88/2)) = 1.7 kNm
Kc 3.96m

1st Floor to G. Floor Fixed end moment, wL2/12
Kcu Δ M = M JH = 23.3x3.96²/12 = 30.45 kNm
23.3 kN/m
Moment in upper column
KJH/2 M = ΔM x K cu / ( K cu + K cl +K JH /2)
Kcl 3.96m = 30.45x0.26/(0.26+0.24+(2.88/2)) = 4.08 kNm

G. Floor to Footing Moment in lower column
Kcu M = ΔM x K cl / ( K cu + K cl +K JH /2)
12.5 kN/m
= 30.45x0.24/(0.26+0.24+(2.88/2)) = 3.77 kNm
KBC/2 Fixed end moment, wL2/12
Kcl 3.96m
Δ M = M JH = 12.5x3.96²/12 = 16.34 kNm

Moment in upper column

M = ΔM x K cu / ( K cu + K cl +K JH /2)

= 16.34x0.24/(0.24+0.56+(1.08/2)) = 2.93 kNm

Moment in lower column
M = ΔM x K cl / ( K cu + K cl +K JH /2)

= 16.34x0.56/(0.24+0.56+(1.08/2)) = 6.83 kNm

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Sheet No. 5/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

5.8.3.2(2) Determination of short or slender column GL J/2 (Roof to 1st Floor).
5.8.3.2(1)
5.8.3.1(1) Roof to 1st Floor, L = 3200 mm

Dimension and size

Column: b x h = 150 x 300 mm ; Clear height , Lz = 2750 mm ;Ly= 2750 mm

Beam :

Main beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Sec. beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Moment of inertia, I = bh 3 /12

Column; I zz = 150 x 300 3 /12 = 0.34 x 109 mm4
0.08 x 109 mm4
I yy = 300 x 150 3 /12 =

Main beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4
Sec beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4

Stiffness, K = EI / L

Column; K zz = 0.34 x 109/ 2750 = 1.23 x 105 mm4
= 0.08 x 109/ 2750 = 0.31 x 105 mm4
K yy = 1.14 x 109/ 3960 = 2.88 x 105 mm4
= 1.14 x 109/ 3960 = 2.88 x 105 mm4
Main Beam: Kmb1
0.213
Secondary Beam: Ksb1 0.213
0.100
Relative Column Stiffness, k = K col / 2(ΣK beam ) 0.100

z-z : Top end, k2 = 1.23 / 2 (2.88) = 0.213 > 0.1 use k2 =
0.213 > 0.1 use k1 =
z-z : Bottom end, k1 = 1.23 / 2 (2.88) = 0.053 < 0.1 use k2 =
0.053 < 0.1 use k1 =
y-y : Top end, k2 = 0.31 / 2 (2.88) =

y-y : Bottom end, k1 = 0.31 / 2 (2.88) =

Effective length of column (Lo),

√{(L o = L 1+ k1 ).( 1 + k2 )}
2 0.45 + k 1 0.45 + k 2

√{( ).( )}L ozz=
2750 1+ 0.213 1+ 0.213
2 0.45 + 0.213 0.45 + 0.213

= 1816.7 mm

√{(L oyy = 0.1 )}0.1
2750 1+ 0.45 + 0.1 ).( 1 +
2 0.45 + 0.1

= 1625 mm

Radius of gyration, i = √ (I/A) = 86.6 mm
= 43.3 mm
i zz = √(0.34 x 10 9 / 45000)
i yy = √ (0.08 x 10 9 / 45000)

Slenderness Ratio, λ = L o / i 1625 / 43.3 = 37.528
λ zz = L ozz / i zz = 1816.7 / 86.6 = 20.978 ; λ yy = L oyy / i yy =

Design data M NEd

f ck = 30 N/mm2 Mo2
f yk = 500 N/mm2 150
N Ed = 54.68 kN

M o2 = = 9.53 kNm

M o1 = -9.29 kNm 300 Mo1

Slenderness limit, λ lim = 20.A.B.C / (n 1/2 )

A = 1/(1 + 0.2 φ ef ) = 0.7 (φef not known)
(ω not known)
B = (1 + 2ω) 1/2 = 1.1

C = 1.7 - r m =1.7- -0.97 = 2.7 r m = (M o1 / M o2 ) = -9.29/ 9.53 = -0.97
f cd = f ck / γ c = 30/ 1.5 = 20.00 N/mm2
n = N ed / (A c f cd )
= 54.684 x 103
20.00 x 45000

= 0.0608

λ lim = 20 x 0.7 x 1.1 x 2.7 /( 0.06 )1/2 = 167.1
λ zz
λ yy < λ lim } SHORT COLUMN
< λ lim

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Sheet No. 6/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

Table 4.2 SPECIFICATION NEd
Table 4.4 Classification : Short braced column

4.4.1.3 Material: f ck = 30 N/mm2 Mo2
4.4.1.1(2) Concrete,

5.8.8.2 Reinforcement, f yk = 500 N/mm2 Mz
5.2(7)
Size (mm), b x h = 150 x 300 My
&
6.1(4) Effective length, L oz = 1.8167 m 150

Design Effective length, L oy = 1.625 m Mo1
Chart
Slenderness Ratio, λ z = 20.978 N Ed =
Mz =
λ y = 37.528 My = 54.68 kN
9.53 kNm
Assumed Φ link = 6 mm 300 1.70 kNm

Assumed Φ bar = 12 mm

CONSIDERATION TO DETERMINE NOMINAL COVER, Cnom

1. Bond requirement, C min,b = 20 mm

2. Durabality requirement, C min,dur = 15 mm

3. Fire resistance requirement refer to axis distance, a = 30 mm EN 1992-1-2

So, C minfire resistance = a - f bar /2 = 30 - 12/2 = 24 mm Table 5.2a

Allowance in design for deviation, DCdev = 10 mm Use;

So, C nom = C min + DCdev = 20 + 10 = 30 mm Cnom = 30 mm

DESIGN BENDING MOMENT

The imperfection moment,

M imp = N Ed .e i = N Ed . θi L 0 /2 ; e i = Max{ L o /400, h/30, 20}

e iz = Max{ 1816.74208144796/400, 300/30, e20iy} = Max{ 1625/400, 150/30, 20}

= Max{ 4.54185520361991,102,02m0}m = = Max{ 4.0625,5, 20}2=0 mm

M imp,z =54.684 x 20/1000 = 1 kNm M imp,y =54.684 x 20/1000 = 1 kNm

The design moment including the effect of imperfection, 2.8 kNm
M Edz9=.53 + 1.09368 = 10.6 kNm ; M Edy =1.7 + 1.09368 =

VERIFY BIAXIAL BENDING

ez = M Edz / N Ed = 10.6 x 106 / 54.684 x 103 = 194 mm
x 106 / 54.684 x 103 = 51 mm
ey = M Edy / N Ed = 2.8

(ey/h) / (ez/b) = (51 / 300)/ (194 / 150) = 0.13 < 0.2 }Not Consider
(ez/b) / (ey/h) = (194 / 150)/ (51 / 300) =
7.61 > 0.2 Biaxial Bending
λ z / λ y = 20.978/ 37.528
λ y / λ z = 37.528/ 20.978 5.8.9(4)
REINFORCEMENT DESIGN
h' = h - C nom - Φ link - 0.5 Φ bar = = 0.559 < 2 } Biaxial Bending Consider
b' = b - C nom - Φ link - 0.5 Φ bar = = 1.7889 < 2 Biaxial Bending

300 - 30 - 6 - 0.5(25) = 251.5 mm
150 - 30 - 6 - 0.5(25) = 101.5 mm

M Edz / h' = 10.6 x 106/ 251.5 = }42.1 M Edz > M Edy ==> use M'z
M Edy / b' = 2.8 x 106/ 101.5 =
27.6 h' b'

M'z = M Edz + β (h'/b') M Edy ; M'y = M Edy + β (b'/h') M Edz

From Table 3.22 BS8110, β = 0.54 <== β = 1 - NEd/bhfck ≥ 0.3

M'z = 10.6 + (0.54) (251.5/101.5) (2.8) = 14.3 kNm

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6+ 0.5(12) = 42.0 mm

d 2 /h= 42/300 = 0.14 use 0.15

N/bhfck = 54.684x10³/(150x300x30) = 0.04
M/bh2fck = 14.3x10⁶ / (300x150²x30) = 0.07

A s f yk / bhf ck = 0.16 <== Refer Graph below

A s = 0.16 bhf ck / f yk = 0.16x150x300x30 / 500 = 432 mm2

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Sheet No. 7/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

As,min = 0.1NEd/fyd = 0.1NEd / 0.87fyk = 0.1x54.684x10³ / (0.87x500) = 12.571 mm2
@ 0.002 Ac = 0.002 bh =
0.002x150x300 = 90 mm2 use 4H 12
( 452 mm2)

9.5.2(2) A s,max =0.04 Ac = 0.04 bh = 0.04x150x300 = 1800 mm2 z

9.5.2(3) Minimum size of Link = ¼ large of main bar =¼ (12) = 3 ===> use H 6
9.5.3
Maximum spacing = 20x12 = 240 mm@ 400 mm @ 150
So, s = 150 mm

At section 300 mm below and above beam and at lapped joint, Sv,max =0.6 x 150 z
H6 -150
= 90 mm use

CHECK BAR ARRANGEMENT DATA:

Column size, b x h = 150 x 300 mm Force, N Ed = 54.68 kN

Materials, f ck = 30 N/mm2 M Edz = 10.6 kNm

f yk = 500 N/mm2 M Edy = 2.8 kNm

Steel area;

All, A s = 452 mm2 4H 12 Link : H6 -300

z-z: A sz = 452 mm2 4H 12 Cnom = 30 mm

y-y: A sy = 452 mm2 4H 12

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6 + 0.5(12) = 42.0 mm

d 2z /h = 42/300 = 0.14 d 2y /b= 42/150 = 0.28

N/bhf ck = 54.684x10^3 / (150x300x30) = 0.04

A sz f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdz /bh 2 f ck = 0.07 =>M Rdz = 0.07 x 150 x 300² x 30 = 28350000 = 28 kNm

A sy f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdy /bh 2 f ck = 0.07 =>M Rdy = 0.07 x 300 x 150² x 30 = 14175000 = 14 kNm

N Rd =0.567 f ck A c + 0.87 f yk A s 962070 = 962 kN
= (0.567 x 30 x 150 x 300) + (0.87 x 500 x 452) =

N Ed / N Rd = 54.684 / 962 = 0.0568

N Ed / N Rd 0.1 0.7 1.0 Interpolation ==> N Ed / N Rd = 0.06

1.0 1.5 2.0 so, = 1

(M Edz / MRdz ) + (M Edy / MRdy ) ≤ 1.0

(10.6 / 28) 1 + (2.8 / 14) 1 = 0.38 + 0.20 = 0.58 < 1 ==> O.K

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 8/13

Reference Calculation Output

5.8.3.2(2) Determination of short or slender column GL J/2 (1st Floor to Ground Floor).
5.8.3.2(1)
5.8.3.1(1) 1st Floor to Ground Floor, L = 3560 mm

Dimension and size

Column: b x h = 150 x 300 mm ; Clear height , Lz = 3110 mm ;Ly= 3110 mm

Beam :

Main beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Sec. beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Moment of inertia, I = bh 3 /12

Column; I zz = 150 x 300 3 /12 = 0.34 x 109 mm4
0.08 x 109 mm4
I yy = 300 x 150 3 /12 =

Main beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4
Sec beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4

Stiffness, K = EI / L

Column; K zz = 0.34 x 109/ 3110 = 1.09 x 105 mm4
= 0.08 x 109/ 3110 = 0.27 x 105 mm4
K yy = 1.14 x 109/ 3960 = 2.88 x 105 mm4
= 1.14 x 109/ 3960 = 2.88 x 105 mm4
Main Beam: Kmb1
0.189
Secondary Beam: Ksb1 0.189
0.100
Relative Column Stiffness, k = K col / 2(ΣK beam ) 0.100

z-z : Top end, k2 = 1.09 / 2 (2.88) = 0.189 > 0.1 use k2 =
0.189 > 0.1 use k1 =
z-z : Bottom end, k1 = 1.09 / 2 (2.88) = 0.047 < 0.1 use k2 =
0.047 < 0.1 use k1 =
y-y : Top end, k2 = 0.27 / 2 (2.88) =

y-y : Bottom end, k1 = 0.27 / 2 (2.88) =

Effective length of column (Lo),

√{(L o = L 1+ k1 ).( 1 + k2 )}
2 0.45 + k 1 0.45 + k 2

√{( ).( )}L ozz=
3110 1+ 0.189 1+ 0.189
2 0.45 + 0.189 0.45 + 0.189

= 2014.9 mm

√{(L oyy = 0.1 )}0.1
3110 1+ 0.45 + 0.1 ).( 1 +
2 0.45 + 0.1

= 1837.7 mm

Radius of gyration, i = √ (I/A) = 86.6 mm
= 43.3 mm
i zz = √(0.34 x 10 9 / 45000)
i yy = √ (0.08 x 10 9 / 45000)

Slenderness Ratio, λ = L o / i
λ zz = L ozz / i zz = 2014.9 / 86.6 = 23.27 ; λ yy = L oyy / i yy = 1837.7 / 43.3 = 42.44

Design data M NEd

f ck = 30 N/mm2 Mo2
150
f yk = 500 N/mm2

N Ed = 150.95 kN

M o2 = = 8.41 kNm

M o1 = -4.15 kNm 300 Mo1

Slenderness limit, λ lim = 20.A.B.C / (n 1/2 )

A = 1/(1 + 0.2 φ ef ) = 0.7 (φef not known)
(ω not known)
B = (1 + 2ω) 1/2 = 1.1

C = 1.7 - r m =1.7- -0.49 = 2.2 r m = (M o1 / M o2 ) = -4.15/ 8.41 = -0.49
f cd = f ck / γ c = 30/ 1.5 = 20.00 N/mm2
n = N ed / (A c f cd )
= 150.95 x 103
20.00 x 45000

= 0.1677

λ lim = 20 x 0.7 x 1.1 x 2.2 /( 0.17 )1/2 = 82.5
λ zz
λ yy < λ lim } SHORT COLUMN
< λ lim

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 9/13

Reference Calculation Output

Table 4.2 SPECIFICATION NEd
Table 4.4 Classification : Short braced column

4.4.1.3 Material: f ck = 30 N/mm2 Mo2
4.4.1.1(2) Concrete,

5.8.8.2 Reinforcement, f yk = 500 N/mm2 Mz
5.2(7)
Size (mm), b x h = 150 x 300 My
&
6.1(4) Effective length, L oz = 2.015 m 150

Design Effective length, L oy = 1.838 m Mo1
Chart 23.27
Slenderness Ratio, λ z = 42.44 N Ed = 151.00 kN
M z = 8.41 kNm
λy = 6 mm M y = 3.77 kNm
12 mm
Assumed Φ link = 300

Assumed Φ bar =

CONSIDERATION TO DETERMINE NOMINAL COVER, Cnom

1. Bond requirement, C min,b = 20 mm

2. Durabality requirement, C min,dur = 15 mm

3. Fire resistance requirement refer to axis distance, a = 30 mm EN 1992-1-2

So, C minfire resistance = a - f bar /2 = 30 - 12/2 = 24 mm Table 5.2a

Allowance in design for deviation, DCdev = 10 mm Use;

So, C nom = C min + DCdev = 20 + 10 = 30 mm Cnom = 30 mm

DESIGN BENDING MOMENT

The imperfection moment,

M imp = N Ed .e i = N Ed . θi L 0 /2 ; e i = Max{ L o /400, h/30, 20}

e iz = Max{ 2015/400, 300/30, 20} e iy = Max{ 1838/400, 150/30, 20}

= Max{ 5.0375,10, 20} = 20 mm = Max{ 4.595,5, 20} =20 mm

M imp,z =151 x 20/1000 = 3 kNm M imp,y =151 x 20/1000 = 3 kNm

The design moment including the effect of imperfection, 6.8 kNm
M Edz = 8.41 + 3.02 = 11.4 kNm ; M Edy = 3.77 + 3.02 =

VERIFY BIAXIAL BENDING

ez = M Edz / N Ed = 11.4 x 106 / 151 x 103 = 75 mm
x 106 / 151 x 103 = 45 mm
ey = M Edy / N Ed = 6.8

(ey/h) / (ez/b) = (45 / 300)/ (75 / 150) = 0.30 > 0.2 }Not Consider
(ez/b) / (ey/h) = (75 / 150)/ (45 / 300) =
3.33 > 0.2 Biaxial Bending

5.8.9(4)

λ z / λ y = 23.266/ 42.44 = 0.5482 < 2 } Biaxial Bending Consider
= 1.8241 < 2 Biaxial Bending
λy/ λz = 42.44/ 23.266

REINFORCEMENT DESIGN

h' = h - C nom - Φ link - 0.5 Φ bar = 300 - 30 - 6 - 0.5(25) = 251.5 mm
150 - 30 - 6 - 0.5(25) = 101.5 mm
b' = b - C nom - Φ link - 0.5 Φ bar =

M Edz / h' = 11.4 x 106/ 251.5 = }45.3 M Edz < M Edy ==> use M'y
M Edy / b' = 6.8 x 106/ 101.5 =
67.0 h' b'

M'z = M Edz + β (h'/b') M Edy ; M'y = M Edy + β (b'/h') M Edz

From Table 3.22 BS8110, β = 0.54 <== β = 1 - NEd/bhfck ≥ 0.3

M'y = 6.8 + (0.54) (101.5/251.5) (11.4) = 9.3 kNm

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6+ 0.5(12) = 42.0 mm

d 2 /h= 42/150 = 0.28 use 0.25

N/bhfck = 151x10³/(150x300x30) = 0.11
M/bh2fck = 9.3x10⁶ / (300x150²x30) = 0.05

A s f yk / bhf ck = 0.02 <== Refer Graph below

A s = 0.02 bhf ck / f yk = 0.02x150x300x30 / 500 = 54 mm2

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 10/13

Reference Calculation Output

As,min = 0.1NEd/fyd = 0.1NEd / 0.87fyk = 0.1x151x10³ / (0.87x500) = 34.713 mm2
@ 0.002 Ac = 0.002 bh =
0.002x150x300 = 90 mm2 use 4H 12
( 452 mm2)

9.5.2(2) A s,max =0.04 Ac = 0.04 bh = 0.04x150x300 = 1800 mm2 y

9.5.2(3) Minimum size of Link = ¼ large of main bar =¼ (12) = 3 ===> use H 6
9.5.3
Maximum spacing = 20x12 = 240 mm@ 400 mm @ 150
So, s = 150 mm

At section 300 mm below and above beam and at lapped joint, Sv,max =0.6 x 150 y
H6 -150
= 90 mm use

CHECK BAR ARRANGEMENT DATA:

Column size, b x h = 150 x 300 mm Force, N Ed = 151.00 kN

Materials, f ck = 30 N/mm2 M Edz = 11.4 kNm

f yk = 500 N/mm2 M Edy = 6.8 kNm

Steel area;

All, A s = 452 mm2 4H 12 Link : H6 -300

z-z: A sz = 452 mm2 4H 12 Cnom = 30 mm

y-y: A sy = 452 mm2 4H 12

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6 + 0.5(12) = 42.0 mm

d 2z /h = 42/300 = 0.14 d 2y /b= 42/150 = 0.28

N/bhf ck = 151x10^3 / (150x300x30) = 0.11

A sz f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdz /bh 2 f ck = 0.07 =>M Rdz = 0.07 x 150 x 300² x 30 = 28350000 = 28 kNm

A sy f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdy /bh 2 f ck = 0.07 =>M Rdy = 0.07 x 300 x 150² x 30 = 14175000 = 14 kNm

N Rd =0.567 f ck A c + 0.87 f yk A s 962070 = 962 kN
= (0.567 x 30 x 150 x 300) + (0.87 x 500 x 452) =

N Ed / N Rd = 151 / 962 = 0.157

N Ed / N Rd 0.1 0.7 1.0 Interpolation ==> N Ed / N Rd = 0.16

1.0 1.5 2.0 so, = 1.05

(M Edz / MRdz ) + (M Edy / MRdy ) ≤ 1.0

(11.4 / 28) 1.05 + (6.8 / 14) 1.05 = 0.39 + 0.47 = 0.86 < 1 ==> O.K

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 11/13

Reference Calculation Output

5.8.3.2(2) Determination of short or slender column GL J/2 (Ground Floor to Footing).
5.8.3.2(1)
5.8.3.1(1) Ground Floor to Footing, L = 1500 mm

Dimension and size

Column: b x h = 150 x 300 mm ; Clear height , Lz = 1050 mm ;Ly= 1050 mm

Beam :

Main beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Sec. beam, b x h = 150 x 450 mm ; Length , L1 = 3960 mm ;

Moment of inertia, I = bh 3 /12

Column; I zz = 150 x 300 3 /12 = 0.34 x 109 mm4
0.08 x 109 mm4
I yy = 300 x 150 3 /12 =

Main beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4
Sec beam; I = 150 x 450 3 /12 = 1.14 x 109 mm4

Stiffness, K = EI / L

Column; K zz = 0.34 x 109/ 1050 = 3.21 x 105 mm4
= 0.08 x 109/ 1050 = 0.80 x 105 mm4
K yy = 1.14 x 109/ 3960 = 2.88 x 105 mm4
= 1.14 x 109/ 3960 = 2.88 x 105 mm4
Main Beam: Kmb1
0.559
Secondary Beam: Ksb1 0.559
0.140
Relative Column Stiffness, k = K col / 2(ΣK beam ) 0.140

z-z : Top end, k2 = 3.21 / 2 (2.88) = 0.559 > 0.1 use k2 =
0.559 > 0.1 use k1 =
z-z : Bottom end, k1 = 3.21 / 2 (2.88) = 0.140 > 0.1 use k2 =
0.140 > 0.1 use k1 =
y-y : Top end, k2 = 0.80 / 2 (2.88) =

y-y : Bottom end, k1 = 0.80 / 2 (2.88) =

Effective length of column (Lo),

√{(L o = L 1+ k1 ).( 1 + k2 )}
2 0.45 + k 1 0.45 + k 2

√{( ).( )}L ozz=
1050 1+ 0.559 1+ 0.559
2 0.45 + 0.559 0.45 + 0.559

= 815.86 mm

√{(L oyy = ).(0.14 )}0.14
1050 1+ 1+
2 0.45 + 0.14 0.45 + 0.14

= 649.58 mm

Radius of gyration, i = √ (I/A) = 86.6 mm
= 43.3 mm
i zz = √(0.34 x 10 9 / 45000)
i yy = √ (0.08 x 10 9 / 45000)

Slenderness Ratio, λ = L o / i ; λ yy = L oyy / i yy = 649.58 / 43.3 = 15.001
λ zz = L ozz / i zz = 815.86 / 86.6 = 9.42

Design data M NEd

f ck = 30 N/mm2 Mo2
150
f yk = 500 N/mm2

N Ed = 202.14 kN

M o2 = 9.83 kNm

M o1 = -4.915 kNm 300 Mo1

Slenderness limit, λ lim = 20.A.B.C / (n 1/2 )

A = 1/(1 + 0.2 φ ef ) = 0.7 (φef not known)
(ω not known)
B = (1 + 2ω) 1/2 = 1.1

C = 1.7 - r m =1.7- -0.5 = 2.2 r m = (M o1 / M o2 ) =-4.915/ 9.83 = -0.50
f cd = f ck / γ c = 30/ 1.5 = 20.00 N/mm2
n = N ed / (A c f cd )
= 202.14 x 103
20.00 x 45000

= 0.2246

λ lim = 20 x 0.7 x 1.1 x 2.2 /( 0.22 )1/2 = 71.5
λ zz
λ yy < λ lim } SHORT COLUMN
< λ lim

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 12/13

Reference Calculation Output

Table 4.2 SPECIFICATION NEd
Table 4.4 Classification : Short braced column

4.4.1.3 Material: f ck = 30 N/mm2 Mo2
4.4.1.1(2) Concrete,

5.8.8.2 Reinforcement, f yk = 500 N/mm2 Mz
5.2(7)
Size (mm), b x h = 150 x 300 My
&
6.1(4) Effective length, L oz = 0.816 m 150

Design Effective length, L oy = 0.65 m Mo1
Chart
Slenderness Ratio, λ z = 9.42 N Ed = 202.14 kN
M z = 9.83 kNm
λ y = 15.001 M y = 6.83 kNm

Assumed Φ link = 6 mm 300

Assumed Φ bar = 12 mm

CONSIDERATION TO DETERMINE NOMINAL COVER, Cnom

1. Bond requirement, C min,b = 20 mm

2. Durabality requirement, C min,dur = 15 mm

3. Fire resistance requirement refer to axis distance, a = 30 mm EN 1992-1-2

So, C minfire resistance = a - f bar /2 = 30 - 12/2 = 24 mm Table 5.2a

Allowance in design for deviation, DCdev = 10 mm Use;

So, C nom = C min + DCdev = 20 + 10 = 30 mm Cnom = 30 mm

DESIGN BENDING MOMENT

The imperfection moment,

M imp = N Ed .e i = N Ed . θi L 0 /2 ; e i = Max{ L o /400, h/30, 20}

e iz = Max{ 816/400, 300/30, 20} e iy = Max{ 650/400, 150/30, 20}

= Max{ 2.04,10, 20} = 20 mm = Max{ 1.625,5, 20} =20 mm

M imp,z =202.14 x 20/1000 = 4 kNm M imp,y =202.14 x 20/1000 = 4 kNm

The design moment including the effect of imperfection, 10.9 kNm
M Edz =9.83 + 4.0428 = 13.9 kNm ; M Edy =6.83 + 4.0428 =

VERIFY BIAXIAL BENDING

ez = M Edz / N Ed = 13.9 x 106 / 202.14 x 103 = 69 mm
x 106 / 202.14 x 103 = 54 mm
ey = M Edy / N Ed = 10.9

(ey/h) / (ez/b) = (54 / 300)/ (69 / 150) = 0.39 > 0.2 }Not Consider
(ez/b) / (ey/h) = (69 / 150)/ (54 / 300) =
2.56 > 0.2 Biaxial Bending

5.8.9(4)

λ z / λ y = 9.4207/ 15.001 = 0.628 < 2 } Biaxial Bending Consider
λ y / λ z = 15.001/ 9.4207 = 1.5924 < 2 Biaxial Bending

REINFORCEMENT DESIGN 300 - 30 - 6 - 0.5(25) = 251.5 mm
h' = h - C nom - Φ link - 0.5 Φ bar = 150 - 30 - 6 - 0.5(25) = 101.5 mm
b' = b - C nom - Φ link - 0.5 Φ bar =

M Edz / h' = 13.9 x 106/ 251.5 = }55.3 M Edz < M Edy ==> use M'y
M Edy / b' = 10.9 x 106/ 101.5 =
107.4 h' b'

M'z = M Edz + β (h'/b') M Edy ; M'y = M Edy + β (b'/h') M Edz

From Table 3.22 BS8110, β = 0.54 <== β = 1 - NEd/bhfck ≥ 0.3

M'y = 10.9 + (0.54) (101.5/251.5) (13.9) = 13.9 kNm

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6+ 0.5(12) = 42.0 mm

d 2 /h= 42/150 = 0.28 use 0.25

N/bhfck = 202.14x10³/(150x300x30) = 0.15
M/bh2fck = 13.9x10⁶ / (300x150²x30) = 0.07

A s f yk / bhf ck = 0.08 <== Refer Graph below

A s = 0.08 bhf ck / f yk = 0.08x150x300x30 / 500 = 216 mm2

created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU Muhammad bin Hamidun Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN
02DKA20F1241 13/13

Reference Calculation Output

As,min = 0.1NEd/fyd = 0.1NEd / 0.87fyk = 0.1x202.14x10³ / (0.87x500) = 46.469 mm2
@ 0.002 Ac = 0.002 bh =
0.002x150x300 = 90 mm2 use 4H 12
( 452 mm2)

9.5.2(2) A s,max =0.04 Ac = 0.04 bh = 0.04x150x300 = 1800 mm2 y

9.5.2(3) Minimum size of Link = ¼ large of main bar =¼ (12) = 3 ===> use H 6
9.5.3
Maximum spacing = 20x12 = 240 mm@ 400 mm @ 150
So, s = 150 mm

At section 300 mm below and above beam and at lapped joint, Sv,max =0.6 x 150 y
H6 -150
= 90 mm use

CHECK BAR ARRANGEMENT DATA:

Column size, b x h = 150 x 300 mm Force, N Ed = 202.14 kN

Materials, f ck = 30 N/mm2 M Edz = 13.9 kNm

f yk = 500 N/mm2 M Edy = 10.9 kNm

Steel area;

All, A s = 452 mm2 4H 12 Link : H6 -300

z-z: A sz = 452 mm2 4H 12 Cnom = 30 mm

y-y: A sy = 452 mm2 4H 12

d 2 =C nom + Φ link + 0.5 Φ bar = 30 + 6 + 0.5(12) = 42.0 mm

d 2z /h = 42/300 = 0.14 d 2y /b= 42/150 = 0.28

N/bhf ck = 202.14x10^3 / (150x300x30) = 0.15

A sz f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdz /bh 2 f ck = 0.09 =>M Rdz = 0.09 x 150 x 300² x 30 = 36450000 = 36 kNm

A sy f yk / bhf ck = 452 x 500 / (150 x 300 x 30) = 0.17
M Rdy /bh 2 f ck = 0.09 =>M Rdy = 0.09 x 300 x 150² x 30 = 18225000 = 18 kNm

N Rd =0.567 f ck A c + 0.87 f yk A s 962070 = 962 kN
= (0.567 x 30 x 150 x 300) + (0.87 x 500 x 452) =

N Ed / N Rd = 202.14 / 962 = 0.2101

N Ed / N Rd 0.1 0.7 1.0 Interpolation ==> N Ed / N Rd = 0.21

1.0 1.5 2.0 so, = 1.09

(M Edz / MRdz ) + (M Edy / MRdy ) ≤ 1.0

(13.9 / 36) 1.09 + (10.9 / 18) 1.09 = 0.35 + 0.58 = 0.93 < 1 ==> O.K

created by Mohd Rizalman