RC DESIGN PROJECT EXAMPLE

JABATAN KEJURUTERAAN AWAM
PROGRAM DIPLOMA KEJURUTERAAN AWAM

DCC50203
REINFORCED CONCRETE DESIGN (EUROCODE 2)

SESSION 1:2021/2022

REINFORCED CONCRETE DESIGN PROJECT :

CADANGAN 1 UNIT RUMAH BANGLO 2 TINGKAT DI ATAS LOT
2386/3, JLN TENGKU MUHAMMAD, MUKIM KUALA KUANTAN

DESIGN BY MATRIKS NO.
BIL. NAME 02DKA20F1234
1 Ahmad bin Albab 02DKA20F1241
2 Muhammad bin Hamidun

SUPERVISED BY
MOHD RIZALMAN BIN ABD AZIZ

A1 / AO SIZE OF

BASIC 2-STOREY BUILDING ARCHITECTURE PLAN

1. GROUND FLOOR PLAN
2. FIRST FLOOR PLAN
3. ROOF PLAN
4. FRONT ELEVATION
5. REAR ELEVATION

6. RIGHT SIDE ELEVATION
7. LEFT SIDE ELEVATION

8. SECTION X-X & Y-Y













PROPOSED STRUCTURAL PLAN
TO

DESIGN

- NAME & SIZE OF BEAM FOR ALL LEVEL
 ROOF BEAM LAYOUT PLAN
 FIRST FLOOR BEAM LAYOUT PLAN
 GROUND FLOOR BEAM LAYOUT PLAN

- SLAB WITH THICKNESS AND CONSIDERATION OF EVERY PANEL
AS 0NE-WAY OR TWO-WAY

- COLUMN LOCATION

- TRUSSES

12 3 45 67

7316

3962 1829

1829 3962 1677 1981 1829

GB6 (150x325)

J
3962
2762 1200 (150x325) GB5 (125x250)
(150x325) GB10 (150x325)
H1 (150x325)
H (125x250) GB4 (125x400)
(125x250)
G
1220
1829

3962 GB8 (125x350)

F (125x250)

2134 (125x250) GB3 (125x400)GB7 (150x400)GB9 (150x400) GB11 (150x400)

E

2743 GB14 (150x500)

1524D GB2 (125x400) (125x250)
C (125x250) GB1 (125x400)
B GB13 (150x500)
1829 GB15 (150x500) (150x250)
GB16 (150x500)
3962

GB12 (150x500)

A

Ground Beam Layout Plan
Scale 1:100

12 3 45 67

3962 7316
1829
1829

3962 1677 1981 1829

J FFB8 (150x450)

H1 3962 Slab
H 2762 1200 125mm thk
G
Slab
F 125mm thk
E
FFB10 (125x250) FFB7 (125x350)
D FFB10a (125x350)
C FFB14 (125x250) FFB6 (125x350) FFB11 (150x450) FFB14 (125x250)
B FFB12 (125x350)
Slab Slab 1220
100mm thk 100mm thk 1829

dp
75mm

3962 Canteliver Slab FFB5 (125x350)
150mm thk
Slab
150mm thk

Slab
100mm thk

FFB14 (125x250) FFB14 (125x250)

FFB4 (125x350)

2134 Slab dp Slab FFB13 (150x450)
100mm thk 75mm 100mm thk

2743 FFB17 (150x500) FFB9 (150x450) FFB3 (125x350) Slab
125mm thk
Slab
100mm thk FFB14 (125x250)

1524 FFB2 (125x350) FFB14a (150x450)
FFB14 (125x250)
dp
75mm
Slab
100mm thk

FFB1 (125x350)

1829 FFB15 (150x500) FFB20 (150x450)
FFB21 (150x500)
3962 FFB18 (150x500)
FFB19 (150x450)

A FFB16 (150x500)

1st Floor Beam Layout Plan
Scale 1:100

12 3 45 67

7316

3962 1829

1829 3962 1677 1981 1829

RB8 (150x450)

J
3962
2762 1200
H1 RB7 (125x350)
H RB6 (125x350) 1829
G
3962 RB9 (150x450)
RB5 (125x350)RB10 (125x250)
2134 RB11 (150x450) RB13 (150x450)
F
2743
RB4 (125x350)
1829 1524
E

RB3 (125x350)

D RB2 (125x350)
C RB1 (125x350)
B

3962

A

Roof Beam Layout Plan
Scale 1:100

12 3 45 67

1829 3962 7316 1829
3962 1829 1677 1981
R7
R8 R9 R10 R9 R8 R7

RB8 (150x450)3962 T10
2762 1200
J
701
1036

T9

1036

H1 RB7 (125x350) T8

1036
H RB6 (125x350)
T1 1829

1036

G 10363962 RB9 (150x450) RB5 (125x350) T1
RB10 (125x250) T1
1036 RB11 (150x450)

T1

1036

F RB4 (125x350) T1

2134 1036

RB13 (150x450) T2

915

E T3

916 RB3 (125x350) T4
T5
2743 761 T6 456

762

D RB2 (125x350)

1829 1524 1200

T7 R6

C 780 RB1 (125x350) R5 R4 R3

R1 R2 R1

B

3962

A

Roof Beam Trusses Layout Plan
Scale 1:100

12 3 45 67

3962 7316
1829
1829

3962 1677 1981 1829

J FFB8 (150x450)

H1 3962 Slab
H 2762 1200 125mm thk
G
Slab
F 125mm thk
E
FFB10 (125x250) FFB7 (125x350)
D FFB10a (125x350)
C FFB14 (125x250) FFB6 (125x350) FFB11 (150x450) FFB14 (125x250)
B FFB12 (125x350)
Slab Slab 555
A 100mm thk 100mm thk T8

dp 1200
75mm
T7
3962 Canteliver Slab FFB5 (125x350) 1007
150mm thk
Slab T7
150mm thk

Slab
100mm thk

FFB14 (125x250) FFB14 (125x25102)00

FFB4 (125x350) T6

2134 1200

Slab dp Slab FFB13 (150x450) T6
100mm thk 75mm 100mm thk

FFB9 (150x450) 1200

FFB17 (150x500) FFB3 (125x350) Slab T6
125mm thk 1200
Slab
2743 100mm thk T6

FFB2 (125x350) 1200

dp FFB14 (125x250)
75mm
Slab T6
100mm thk
1524 FFB14a (150x450) 1200
FFB14 (125x250) FFB1 (125x350) T5

FFB20 (150x450) 1334
FFB21 (150x500)
1829 T12 T4
1200
FFB15 (150x500) FFB18 (150x500)
FFB19 (150x450) T1
T11 1200

T9 T10 T2
1200
3962
T3

FFB16 (150x500) 1258

850

R1 R2 R3 R2 R1

1st Floor Beam Trusses Layout Plan
Scale 1:100

DESIGN SPECIFICATION
AND

ESTIMATION ACTION FOR ALL LEVEL

Project: Design by: Date: Check by:
Sheet No.
CADANGAN 1 UNIT RUMAH BANGLO 2
Output
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU

MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation

DESIGN SPECIFICATIONS
STRUCTURE
2 Storey Reinforced Concrete Building Under Residential Usage.

DESIGN CODES & LITERATURE
1. MS EN 1990: 2010 Eurocode 0 - Basis of structural design
2. MS EN 1991: 2010 Eurocode 1 - Action on structures
3. MS EN 1992: 2010 Eurocode 2 - Design of concrete structures
4. Uniform building by-law 1984 (G.N. 5178/85, Inter. Law Book Series
5. MS 1553: 2002 Code of practice on wind loading for building structure
6. Manual for the design of concrete building structures to Eurocode 2,
The Institution of Structural Engineers, UK, 2006
7. BS EN 1997-1: Geotechnical Design

Table 2.1 DESIGN WORKING LIFE
EN 1992 50 Years

Table 4.1 EXPOSURE CLASS
EN 1992 Superstructure:-
XC3 Moderate humidity: Concrete inside buildings with moderate or high air humidity,

External concrete sheltered from rain.
Substructure:-
XC4 Cyclic wet and dry : Concrete surface subject to water contact

UBBL FIRE RESISTANCE
0.5 Hour (Residential 2 Storey building)

EN 1992 MATERIALS = 30 N/mm²
Sec. 3.1 Characteristic strength of concrete, f ck = 500 N/mm²
Sec. 3.2 Characteristic strength of main reinforcement, f yk = 500 N/mm²
Characteristic strength of shear reinforcement, f ywk

FOUNDATION

Type of soil : Firm, reddish brown yellow mottled CLAY with some sand & gravel

Water Level : 6 m below ground level

SPT value : 7

Bearing Capacity = 100 kN/m²

Type of foundation : Pad Footing

EN 1991 ACTIONS

Permanent actions

Table A1 Unit weight of reinforced concrete = 25 kN/m³
= 10 kN/m³
Table A7 Unit weight of water = 2.6 kN/m²

Ref.[9] Table 2.13 Brickwall : 125 mm thk = 1.15 kN/m²
= 0.67 kN/m²
Floor finishes : = 0.05 kN/m²
= 1.00 kN/m²
Ref.[9] Table 2.14 Rendering, Screed (50 mm thk) = 0.10 kN/m²

Ref.[9] Table 2.14 Clay floor tiles = 0.67 kN/m²
= 0.20 kN/m²
Ref.[9] Table 2.14 Carpet = 0.30 kN/m²
= 1.00 kN/m²
Ref.[9] Table 2.14 Ceiling (Asbestos / Plaster) = 1.00 kN/m²
= 0.10 kN/m²
Ref.[9] Table 2.14 Services

Roof :

Roof tiles

Insulated panel

H.W. Purlins

Ceiling (Asbestos / Plaster)

H.W Truss

Services

Variable actions = 0.25 kN/m²
Roof = 1.50 kN/m²
All area usage (Residential usage) = 10.00 kN/m²
Weight of water = 33.50 m/s
Wind load : Basic wind speed

Project: Design by: Date: Check by:
Sheet No.
CADANGAN 1 UNIT RUMAH BANGLO 2
Output
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU

MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation

Plan ROOF BEAM
Roof

LOAD qk:-
gk:-
Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²

Ceiling(Asbestos/Plaster) = 0.30 kN/m²
H.W Truss
Services = 1.00 kN/m²
gk
= 1.00 kN/m²

= 0.10 kN/m²

= 3.27 kN/m² qk = 0.25 kN/m²

Plan First Floor Beam (Loading from Roof)
First Floor

LOAD

gk:- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°

Insulated panel = 0.20 kN/m²

H.W. Purlins = 0.30 kN/m²

Ceiling(Asbestos/Plaster) = 1.00 kN/m²

H.W Truss = 1.00 kN/m²

Services = 0.10 kN/m²

gk = 3.27 kN/m² qk = 0.25 kN/m²

First Floor Beam (Loading from Slab area)

gk (Slab with 100 mm thk):- qk:(Imposed load on slab):-
All area of slab
Slab 0.100 x 25 = 2.50 kN/m²
qk = 1.50 kN/m²
Rendering, Screed (50 mm thk) = 1.15 kN/m²

Clay floor tiles = 0.67 kN/m²

Ceiling(Asbestos/Plaster) = 1.00 kN/m²

Services = 0.10 kN/m²

gk = 5.42 kN/m²

Loading by = 2.6 kN/m²
brickwall (125 mm

Plan GROUND BEAM (NON-SUSPENDED SLAB)
Ground Floor qk:-

LOAD
gk:-

Selfweight only

Loading by = 2.6 kN/m²
brickwall (125 mm

SLAB DESIGN

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 1/3

Reference Calculation Output

DESIGN SLAB GRIDLINE 2-4/J-H
SPECIFICATION

Long span, l y = 3962 mm
Short span, l x = 3962 mm

ly /lx = 3962 / 3962 = 1.0 < 2.0 Two way slab
3 edge discontinuous (1 long edge continuous)

Characteristic Actions: = 2.92 kN/m2 (Excluding selfweight)

Permanent, g k = 1.50 kN/m2

Variable, q k

Design life = 50 Years (Table 2.1 EN 1990)

Fire resistance = 60 (Section 5.7 EN 1992-1-2)

Exposure classes = XC3

Materials : 30 N/mm2
500 N/mm2
(Table F.1 EN 206) Characteristic strength of concrete, f ck = =
3.1.2(3) = 25 kN/m3
Characteristic strength of steel, f yk =
(Table A.1 EN
1991-1) Unit weight of reinforced concrete

Assumed: f bar = 8 mm

CONSIDERATION TO DETERMINE NOMINAL COVER, Cnom

Table 4.2 1. Bond requirement, C min,b = 8 mm
Table 4.4N 15 mm
4.4.1.2 2. Durabality requirement, C min,dur =

4.4.1.3 3. Fire resistance requirement refer to axis distance,
4.4.1.1(2)
a= 15 mm Table 5.8 EN 1992-1-2
Table 7.4N
So, C minfire resistance = a - f bar /2

= 10 - 8/2 Compare 1,2&3, max

= 6 mm C min = 15 mm

Allowance in design for deviation, DCdev = 10 mm
25 mm
So, C nom = C min + DCdev

= 15 + 10 = C nom= 25 mm

SLAB THICKNESS

Minimum thickness for fire resistance = 80 mm Table 5.8 EN 1992-1-2
= 99 mm Try:
Estimated thickness considering deflection control,
h = 100 mm
L/h = 40 So, h = 3962/40

ACTIONS

Slab selfweight = 0.100 x 25 = 2.50 kN/m2

Permanent load (Excluding selfweight) = 2.92 kN/m2
= 5.42 kN/m2
Table A1.2B : EN Characteristic permanent action, g k = 1.50 kN/m2
1990 Characteristic variable action, q k = 9.57 kN/m2
Design action, n d = 1.35g k + 1.5q 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 2/3

Reference Calculation Output

ANALYSIS

Table 3.12 : BS ly /lx = 1.0 ==> Continuous edge bsx = 0.057 bsy = 0
==> Mid Span bsx = 0.043 bsy = 0.044
8110

Cont. Long Edge Support, M sx = bsx n l 2 = 0.057 x 9.57 x 3.962² = 8.56 kNm
x 0.00 kNm
6.46 kNm
Cont. Short Edge Support, M sy = bsy n l 2 = 0 x 9.57 x 3.962² = 6.61 kNm
x

Short span, M sx = bsx n l 2 = 0.043 x 9.57 x 3.962² =
x

Long span, M sy = bsy n l 2 = 0.044 x 9.57 x 3.962² =
x

MAIN REINFORCEMENT

Effective depth,

d x = h -C nom -0.5f bar = 100 - 25 - 8/2 =71.0 mm
d y = h -C nom -1.5f bar = 100 - 25 - 8(1.5) =63.0 mm

9.2.1.1 Minimum and maximum reinforcement area,
6.1
6.1 A s,min = 0.26(f ctm/f yk) bd = 0.26 x (2.9 500) x bd
6.1
= 0.0015 bd = 0.0015 x1000 x 71 = 107 mm2/m Use secondary bar:
= 4000 mm2/m H8 - 300
A s,max = 0.04Ac = 0.04 x1000 x 100
(168 mm2/m)

Long Support - Cont. Edge , M sx = 8.56 kNm/m

K = M / bd 2f ck = 8.56 x 106 / (1000 x 71 2 x 30)

= 0.057 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.95 d < 0.95d 293 mm2/m Use:
8.56 x 106 (0.87 x 500 x 0.95 x 71 ) =
A s = M / 0.87 f yk z = H8 - 150 Top.

(335 mm2/m)

Short span : - Midspan , M sx = 6.46 kNm/m

K = M / bd 2f ck = 6.46 x 106 / (1000 x 71 2 x 30)

= 0.043 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.96 d > 0.95d 220 mm2/m Use:
A s = M / 0.87 f yk z = 6.46 x 106 (0.87 x 500 x 0.95 x 71 ) =
H8 - 150 bot.

(335 mm2/m)

Long span : - Midspan , M sy = 6.61 kNm/m

K = M / bd 2f ck = 6.6 x 106 / (1000 2 x 30)

x 63

= 0.055 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.95 d < 0.95d Use:
6.6 x 106 (0.87 x 500 x 0.95 x 63 ) = 254 mm2/m
A s = M / 0.87 f yk z = H8 - 150 bot.

(335 mm2/m)

SHEAR

ly /lx = 1.0 ==> Continuous edge bvx = 0.45 bvy = - Table 3.13 : BS 8110
0.29 V Ed = 17.06 kN
==> Discontinuous edge bvx = 0.3 bvy =
17.06 kN
Maximum design shear force, VED = bvx n l x = 0.45 x 9.57 x 3.962 =

6.2.2 Design shear resistance,

V Rd,c = [ 0.12 k (100r 1 f ck)1/3 ] bd k = 11+((220000//d71)1)/21/2 <== 22..608 > 2.0 Use : 2.0
k = 1 + (200/d )1/2 <≤ =2.20.0; 0.0047 Use : 0.0047

r 1 = A sl/bd ≤ 0.02 ; r 1 = A sl/3b3d5 ≤(100.0020 x 71) =

V Rd,c = 0.12 x 2.0 x (100 x 0.0047 x 30) 1/3 x 1000 x 71
x 1000
= 41226 N = 41.2 kN

V min = [ 0.035k 3/2f 1/2 ] bd = 0.035 x 2.0 3/2 x 30 1/2 x 71
ck

= 38497 N = 38.5 kN

So, V Rd,c = 41.2 kN > V Ed Ok !

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 3/3

Reference Calculation Output
7.4
DEFLECTION
Table 7.4N
Percentage of required tension reinforcement,
7.3.3
9.3.1 r = A s,req / bd = 220 / 1000 x 71 = 0.0031

Reference reinforcement ratio,
= (30) 1/2x10-3
r o =(f ck) 1/2 x 10-3 = 0.0055

Factor for structural system, K = 1.3

r < ro Use equation (7.16a)

l  ro  ro 13 / 2 
= K 11  1.5 r r 
d  f ck  3.2 f ck   (7.16a)

= 1.3 (11 + 14.5 + 11.77) = 48.5

Modification factor for spans,pan less than 7 m

= 1.00

Modification factor for steel area provided,

= A s,prov/A s,req = 335 / 220 = 1.52 > 1.5
72.71
Therefore allowable span-effective depth ratio,
(l /d )allowble
(l /d )allowble = 48.5 x 1.00 x 1.50 =

Actual span-effective depth

(l /d )actual = 3962 / 71 = 55.8 < Ok !

CRACKING Ok!
Ok !
h = 100 mm < 200 mm Ok !

Main bar :

S max,slabs = 3h ≤ 400 mm = 300 mm

Max. bar spacing = 150 mm < S max,slabs

Secondary bar :

S max,slabs = 3.5h ≤ 450 mm = 350 mm

Max. bar spacing = 150 mm < S max,slabs

DETAILING Concrete cover = 25 mm
Thickness, h = 100 mm

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. 1/3
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

DESIGN SLAB GRIDLINE 4 - 6 /D - F
SPECIFICATION

Long span, l y = 4877 mm
Short span, l x = 3658 mm

ly /lx = 4877 / 3658 = 1.3 < 2.0 Two way slab
2 adjacent edges discontinuous

Characteristic Actions: = 2.92 kN/m2 (Excluding selfweight)

Permanent, g k = 1.50 kN/m2

Variable, q k

Design life = 50 Years (Table 2.1 EN 1990)

Fire resistance = 60 (Section 5.7 EN 1992-1-2)

Exposure classes = XC3

Materials : 30 N/mm2
500 N/mm2
(Table F.1 EN 206) Characteristic strength of concrete, f ck = =
3.1.2(3) = 25 kN/m3
Characteristic strength of steel, f yk =
(Table A.1 EN
1991-1) Unit weight of reinforced concrete

Assumed: f bar = 8 mm

CONSIDERATION TO DETERMINE NOMINAL COVER, Cnom

Table 4.2 1. Bond requirement, C min,b = 8 mm
Table 4.4N 15 mm
4.4.1.2 2. Durabality requirement, C min,dur =

4.4.1.3 3. Fire resistance requirement refer to axis distance,
4.4.1.1(2)
a= 15 mm Table 5.8 EN 1992-1-2

So, C minfire resistance = a - f bar /2

= 10 - 8/2 Compare 1,2&3, max

= 6 mm C min = 15 mm

Allowance in design for deviation, DCdev = 10 mm
25 mm
So, C nom = C min + DCdev

= 15 + 10 = C nom= 25 mm

SLAB THICKNESS

Minimum thickness for fire resistance = 80 mm Table 5.8 EN 1992-1-2
= 91 mm Try:
Table 7.4N Estimated thickness considering deflection control,
h = 100 mm
L/h = 40 So, h = 3658/40

ACTIONS

Slab selfweight = 0.100 x 25 = 2.50 kN/m2

Permanent load (Excluding selfweight) = 2.92 kN/m2
= 5.42 kN/m2
Table A1.2B : EN Characteristic permanent action, g k = 1.50 kN/m2
1990 Characteristic variable action, q k = 9.57 kN/m2
Design action, n d = 1.35g k + 1.5q k

created by Mohd Rizalman

Project: Design by: Date: Check by: 2/3
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 0.045 Sheet No.
MUHAMMAD, MUKIM KUALA KUANTAN 0.034

Reference Calculation Output

ANALYSIS

Table 3.12 : BS ly /lx = 1.3 ==> Continuous edge bsx = 0.069 bsy =
==> Mid Span bsx = 0.051 bsy =
8110

Cont. Long Edge Support, M sx = bsx n l 2 = 0.069 x 9.57 x 3.658² = 8.84 kNm
x 5.76 kNm
6.53 kNm
Cont. Short Edge Support, M sy = bsy n l 2 = 0.045 x 9.57 x 3.658² = 4.35 kNm
x

Short span, M sx = bsx n l 2 = 0.051 x 9.57 x 3.658² =
x

Long span, M sy = bsy n l 2 = 0.034 x 9.57 x 3.658² =
x

MAIN REINFORCEMENT

Effective depth,

d x = h -C nom -0.5f bar = 100 - 25 - 8/2 =71.0 mm
d y = h -C nom -1.5f bar = 100 - 25 - 8(1.5) =63.0 mm

9.2.1.1 Minimum and maximum reinforcement area,
6.1
6.1 A s,min = 0.26(f ctm/f yk) bd = 0.26 x (2.9 500) x bd
6.1
6.1 = 0.0015 bd = 0.0015 x1000 x 71 = 107 mm2/m Use secondary bar:
= 4000 mm2/m H8 - 300
6.2.2 A s,max = 0.04Ac = 0.04 x1000 x 100
(168 mm2/m)

Long Support - Cont. Edge , M sx = 8.84 kNm/m

K = M / bd 2f ck = 8.84 x 106 / (1000 x 71 2 x 30)

= 0.058 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.95 d < 0.95d Use:
8.84 x 106 (0.87 x 500 x 0.95 x 71 ) = 303 mm2/m
A s = M / 0.87 f yk z = H8 - 150 Top.

(335 mm2/m)

Short span : - Midspan , M sx = 6.53 kNm/m

K = M / bd 2f ck = 6.53 x 106 / (1000 x 71 2 x 30)

= 0.043 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.96 d > 0.95d 223 mm2/m Use:
A s = M / 0.87 f yk z = 6.53 x 106 (0.87 x 500 x 0.95 x 71 ) =
H8 - 200 bot.

(251 mm2/m)

Short support : - Cont. Edge , M sx = 5.76 kNm/m

K = M / bd 2f ck = 5.76 x 106 / (1000 x 71 2 x 30)

= 0.038 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.97 d > 0.95d 196 mm2/m Use:
A s = M / 0.87 f yk z = 5.76 x 106 (0.87 x 500 x 0.95 x 71 ) =
H8 - 250 Top.

(201 mm2/m)

Long span : - Midspan , M sy = 4.35 kNm/m

K = M / bd 2f ck = 4.4 x 106 / (1000 x 63 2 x 30)

= 0.037 < K bal = 0.167 Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.97 d > 0.95d 167 mm2/m Use:
4.4 x 106 (0.87 x 500 x 0.95 x 63 ) =
A s = M / 0.87 f yk z = H8 - 300 bot.

(168 mm2/m)

SHEAR

ly /lx = 1.3 ==> Continuous edge bvx = 0.5 bvy = 0.4 Table 3.13 : BS 8110
V Ed = 17.50 kN
==> Discontinuous edge bvx = 0.33 bvy = 0.26

Maximum design shear force, VED = bvx n l x = 0.5 x 9.57 x 3.658 = 17.50 kN

Design shear resistance,

V Rd,c = [ 0.12 k (100r 1 f ck)1/3 ] bd k = 11+((220000//d71)1)/21/2 <== 22..608 > 2.0 Use : 2.0
k = 1 + (200/d )1/2 <≤ =2.20.0; 0.0035 Use : 0.0035

r 1 = A sl/bd ≤ 0.02 ; r 1 = A sl/2b5d1 ≤(100.0020 x 71) =

V Rd,c = 0.12 x 2.0 x (100 x 0.0035 x 30) 1/3 x 1000 x 71
x 1000
= 37456 N = 37.5 kN

V min = [ 0.035k 3/2f 1/2 ] bd = 0.035 x 2.0 3/2 x 30 1/2 x 71
ck

= 38497 N = 38.5 kN

So, V Rd,c = 38.5 kN > V Ed Ok !

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. 3/3
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output
7.4
DEFLECTION
Table 7.4N
Percentage of required tension reinforcement,

r = A s,req / bd = 223 / 1000 x 71 = 0.0031

Reference reinforcement ratio,
= (30) 1/2x10-3
r o =(f ck) 1/2 x 10-3 = 0.0055

Factor for structural system, K = 1.3

r < ro Use equation (7.16a)

l  ro  ro 13 / 2 
= K 11  1.5 r r 
d  f ck  3.2 f ck   (7.16a)

= 1.3 (11 + 14.4 + 11.33) = 47.7

Modification factor for spans,pan less than 7 m

= 1.00

Modification factor for steel area provided,

= A s,prov/A s,req = 251 / 223 = 1.13 < 1.5
53.88
Therefore allowable span-effective depth ratio,
(l /d )allowble
(l /d )allowble = 47.7 x 1.00 x 1.13 =

Actual span-effective depth

(l /d )actual = 3658 / 71 = 51.5 < Ok !

CRACKING Ok!
Ok !
7.3.3 h = 100 mm < 200 mm Ok !
9.3.1
Main bar :

S max,slabs = 3h ≤ 400 mm = 300 mm

Max. bar spacing = 200 mm < S max,slabs

Secondary bar :

S max,slabs = 3.5h ≤ 450 mm = 350 mm

Max. bar spacing = 300 mm < S max,slabs

DETAILING Concrete cover = 25 mm
Thickness, h = 100 mm

created by Mohd Rizalman

BEAM DESIGN
SIMPLY SUPPORTED BEAM

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Output 1/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN
Calculation

DESIGN SIMPLY SUPPORTED BEAM RECTANGULAR SECTION

BEAM POSITION : RB 1 ( 125 x 350 )

LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

ESTIMATION OF LOADING FROM ROOF

gk (Roof):- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²
H.W Truss
Sgk = 0.30 kN/m²

= 1.00 kN/m² Sqk = 0.25 kN/m²
= 2.17 kN/m²
Action area from plan
gk (Ceiling level):- ESTIMATION OF LOADING FROM CEILING for beam RB1

Ceiling(Asbestos/Plaster) = 1.00 kN/m²
Services = 0.10 kN/m²
Sgk = 1.10 kN/m²

R1/R2 3.96 m
width = 0.78m

0.85m

30°

1.24m

gk (Roof) = 2.17 x 1.24 / (COS(30) 3.11 kN/m
1.36 kN/m
gk (Ceiling level) = 1.1 x 1.24 = 1.09 kN/m
0.1205.125xx00..33550 x 25 = 5.56 kN/m
Beam seflweight =
Sgk = 0.36 kN/m
qk (Roof) =
0.25 x 1.24 / (COS(30)) =

Design action, 1.35(5.56) + 1.5(0.36) = 8.05 kN/m
w d = 1.35g k + 1.5q k =

w d = 8.05 kN/m

L = 3.96 m

Shear Force,

V = w d L /2
= 15.9 kN

Bending Moment,
M M = w d L 2/8

= 15.8 kNm

Design life = 50 Years (Table 2.1 EN 1990)
Fire resistance = R30 (Sec. 5 EN 1992-1-2)
Exposure classes = XC3
Materials :
Table 4.1
Characteristic strength of concrete, f ck = 30 N/mm2 (Table A.3 BS 8500)
3.1.2(3) = 500 N/mm2
3.2.2(2) Characteristic strength of steel, f yk = 500 N/mm2
=
Characteristic strength of link, f yk ; f Compression = 25 kN/m3
12 mm ;
Unit weight of reinforced concrete (Table A.1 EN 1991-1)

Assumed: f Tension = 12 mm f link = 6 mm

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. 2/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN Output
Calculation
Table 4.2
Table 4.4N SIZE Use : b x h = 125 x 350 mm

4.4.1.2 DURABILITY, FIRE & BOND REQUIREMENTS

4.4.1.3 Min. conc. cover regard to bond, C min,b = 12 mm
4.4.1.1(2)
Min. conc. cover regard to durability, C min,dur = 15 mm
6.1 Min. required axis distance for = fire resistance,
Table 5.5 EN 1992-1-2

a sd = 20 + 10 = 30 mm

Min. concrete cover regard to fire,

C min = a sd - f link - f bar/2 = 30 - 6 - (12/2) = 18 mm
10 mm
Allowance in design for deviation, D0C.5dev =

Nominal cover, Use:
C nom = 25 mm
C nom = C min + DCdev = 15 + 10 = 25 mm

MAIN REINFORCEMENT d'
Effective depth,
= 350 - 25 - 6 - 12 = 307 mm dh
d = h - C nom - f link - f Tension = 25 + 6 + 12/2 = 37 mm
d' = Cnom + f link + f comp. /2

Bending, Moment,M= 15.8 kNm b
15.8 x 106
K= M = = 0.045 < K bal = 0.167 ==> Compression
bd 2f ck (125 x 307² x 30) 0.96d reinforcement is not
required
z = d [0.5 + 0.25 - K/1.134) ] = d [0.5 + 0.25 - 0.045/1.134) ] =

= 0.95 x 307 = 291.7 mm

Area of tension steel;

A s,req = M = 15.8 x 10⁶ = 124.5 mm2 Use : 2H 12
0.87f yk z 0.87 x 500 x 291.7 ( 226 mm2)

9.2.1.1 Minimum and maximum reinforcement area;

A s,min= 0.26 ( f ctm )bd = 0.26 ( 2.9 )bd = 0.0015 bd
f yk 500 57.6 mm2

A s,min= 0.0015 bd = 0.0015 x 125 x 307 =

As,max= 0.04 bh = 0.04 x 125 x 350 = 1750 mm2 Use : 2H 12 mm
( 226 mm2)

6.2 SHEAR REINFORCEMENT V Ed = 15.9 kN
6.2.3
V Rd, max = 0.36b wdf ck(1 - f ck/250)
(cot q + tan q )

= 0.36 x 125 x 307 x 30(1 - 30/250) = 364716
(cot q + tan q ) (cot q + tan q )

V Rd, max ( q =22 o ) = 364716 = 125.8 kN > }V Ed cot 22 o = 2.5
(2.5 + 0.4) tan 22 o = 0.4
cot 45 o = 1.0
V Rd, max ( q =45 o ) = 364716 = 182.4 kN > tan 45 o = 1.0
(1 + 1)

Therefore angle q < 22°

Use : θ = 22.0 o tan q = 0.4 cot q = 2.5

Shear links

9.2.2 (6) Max. spacing, s max = 0.75d = 0.75 x 307 = 230 mm
Try link :
A sw = V ed = 15.9 x 10³ = 0.05
s 0.78f ykd cot q 0.78 x 500 x 307 x 2.5 = 1131 mm

H6 so, A sw = 57 mm2 : Spacing, s = 57 Use : H6 - 225
0.05

6.2.3(7) Additional longitudinal reinforcement

Additional tensile force,

DF td =0.5V Ed cot q = 0.5 x 15.9 x 2.5 = 20 kN Use : 1H 12
( 113 mm2)
A s req. = ΔF td = 20x10³ = 46 mm2
0.87f yk (0.87 x 500) created by Mohd Rizalman

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Output 3/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN
7.4 Calculation

Table 7.4N DEFLECTION

7.3 Percentage of required tension reinforcement,
Table 7.1N
Table 7.3N r = A s,req = 124.5 = 0.003
DETAILING bd 125 x 307

Reference reinforcement ratio,

r o =(f ck) 1/2 x 10-3 = (30) 1/2x10-3 = 0.0055

Percentage of required compression reinforcement,

r ' = A s ' ,req = 0 = 0.000
bd 125 x 307

Factor for structural system, K = 1.0

r < ro Use equation --> (7.16a)

l  ro  3.2  ro 13 / 2 
= K 11  1.5 r r 
d  f ck f ck   (7.16a)

= 1.0 (11 + 13.9 + 10.00) = 34.87
Modification factor for :-
> 7 = 1.00
span less than 7 m 7 = 1.82 > 1.5
226
steel area provided, = A s,prov = 124.5 use: 1.5
A s,req

So, allowable, l = 34.87 x 1 x 1.5 = 52.305
Actual , d
l = 3962 = 12.9 < 52.31 Ok !
d 307

CRACKING 160 300
200 250
Limiting crack width, w max = 0.3 mm 240 200
280 150
Steel stress, 320 100
360 50
fs = f yk x Gk  0.3Qk 1

1.15 (1.35Gk  1.5Qk ) 

= 500 x 5.56 + (0.3x 0.4)) x 1.0
1.15 8.05

= 306 N/mm2

Max. allowable clear bar spacing = 100 mm s
Ok !
Clear spacing,

s = [ 125 - 2(25) - 2(6) - 2(12) ] / 1

= 39 mm < 100 mm

2H 12

H6 - 225

2H 12
3962

RB 1 ( 125 x 350 )

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Muhammad bin Hamidun Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1241 Output 1/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN
Calculation

DESIGN SIMPLY SUPPORTED BEAM RECTANGULAR SECTION

BEAM POSITION : RB 6 ( 150 x 350 )

LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

ESTIMATION OF LOADING FROM ROOF

gk (Roof):- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²
H.W Truss
Sgk = 0.30 kN/m²

= 1.00 kN/m² Sqk = 0.25 kN/m²
= 2.17 kN/m²
Action area from plan
ESTIMATION OF LOADING FROM CEILING for beam RB6

gk (Ceiling level):- = 1.00 kN/m²

Ceiling(Asbestos/Plaster) = 0.10 kN/m²
Services = 1.10 kN/m²
Sgk

T1 1.04 m
width =

30°

RB6 9.32 m

gk (Roof) = 2.17 x 1.04 x 9.32 / (COS(30)) / 9.32 = 2.61 kN/m
1.14 kN/m
gk (Ceiling level) = 1.1 x 1.04 = 1.31 kN/m
5.06 kN/m
Beam seflweight = 0.15 x 0.35 x 25 = 0.30 kN/m

Sgk = 7.28 kN/m

qk (Roof) = 0.25 x 1.04 x 9.32 / (COS(30)) / 9.32 =

Design action,

w d = 1.35g k + 1.5q k = 1.35(5.06) + 1.5(0.3) =

w d = 7.28 kN/m

L = 3.96 m

Shear Force,

V = w d L /2
= 14.4 kN

Bending Moment,
M M = w d L 2/8

= 14.3 kNm

Design life = 50 Years (Table 2.1 EN 1990)
Fire resistance = R30 (Sec. 5 EN 1992-1-2)
Exposure classes = XC3
Materials :
Table 4.1
Characteristic strength of concrete, f ck = 30 N/mm2 (Table A.3 BS 8500)
3.1.2(3) = 500 N/mm2
3.2.2(2) Characteristic strength of steel, f yk = 500 N/mm2
=
Characteristic strength of link, f yk ; f Compression = 25 kN/m3
12 mm ;
Unit weight of reinforced concrete (Table A.1 EN 1991-1)

Assumed: f Tension = 12 mm f link = 6 mm

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. 2/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN Output
Calculation
Table 4.2
Table 4.4N SIZE Use : b x h = 150 x 350 mm

4.4.1.2 DURABILITY, FIRE & BOND REQUIREMENTS

4.4.1.3 Min. conc. cover regard to bond, C min,b = 12 mm
4.4.1.1(2)
Min. conc. cover regard to durability, C min,dur = 15 mm
6.1 Min. required axis distance for = fire resistance,
Table 5.5 EN 1992-1-2

a sd = 20 + 10 = 30 mm

Min. concrete cover regard to fire,

C min = a sd - f link - f bar/2 = 30 - 6 - (12/2) = 18 mm
10 mm
Allowance in design for deviation, D0C.5dev =

Nominal cover, Use:
C nom = 25 mm
C nom = C min + DCdev = 15 + 10 = 25 mm

MAIN REINFORCEMENT d'
Effective depth,
= 350 - 25 - 6 - 12 = 307 mm dh
d = h - C nom - f link - f Tension = 25 + 6 + 12/2 = 37 mm
d' = Cnom + f link + f comp. /2

Bending, Moment,M= 14.3 kNm b
14.3 x 106
K= M = = 0.034 < K bal = 0.167 ==> Compression
bd 2f ck (150 x 307² x 30) 0.97d reinforcement is not
required
z = d [0.5 + 0.25 - K/1.134) ] = d [0.5 + 0.25 - 0.034/1.134) ] =

= 0.95 x 307 = 291.7 mm

Area of tension steel;

A s,req = M = 14.3 x 10⁶ = 112.7 mm2 Use : 2H 12
0.87f yk z 0.87 x 500 x 291.7 ( 226 mm2)

9.2.1.1 Minimum and maximum reinforcement area;

A s,min= 0.26 ( f ctm )bd = 0.26 ( 2.9 )bd = 0.0015 bd
f yk 500 69.1 mm2

A s,min= 0.0015 bd = 0.0015 x 150 x 307 =

As,max= 0.04 bh = 0.04 x 150 x 350 = 2100 mm2 Use : 2H 12 mm
( 226 mm2)

6.2 SHEAR REINFORCEMENT V Ed = 14.4 kN
6.2.3
V Rd, max = 0.36b wdf ck(1 - f ck/250)
(cot q + tan q )

= 0.36 x 150 x 307 x 30(1 - 30/250) = 437659.2
(cot q + tan q ) (cot q + tan q )

V Rd, max ( q =22 o ) = 437659.2 = 150.9 kN > }V Ed cot 22 o = 2.5
(2.5 + 0.4) tan 22 o = 0.4
cot 45 o = 1.0
V Rd, max ( q =45 o ) = 437659 = 218.8 kN > tan 45 o = 1.0
(1 + 1)

Therefore angle q < 22°

Use : θ = 22.0 o tan q = 0.4 cot q = 2.5

Shear links

9.2.2 (6) Max. spacing, s max = 0.75d = 0.75 x 307 = 230 mm
Try link :
A sw = V ed = 14.4 x 10³ = 0.05
s 0.78f ykd cot q 0.78 x 500 x 307 x 2.5 = 1131 mm

H6 so, A sw = 57 mm2 : Spacing, s = 57 Use : H6 - 225
0.05

6.2.3(7) Additional longitudinal reinforcement

Additional tensile force,

DF td =0.5V Ed cot q = 0.5 x 14.4 x 2.5 = 18 kN Use : 1H 12
( 113 mm2)
A s req. = ΔF td = 18x10³ = 41 mm2
0.87f yk (0.87 x 500) created by Mohd Rizalman

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Output 3/3

Reference MUHAMMAD, MUKIM KUALA KUANTAN
7.4 Calculation

Table 7.4N DEFLECTION

7.3 Percentage of required tension reinforcement,
Table 7.1N
Table 7.3N r = A s,req = 112.7 = 0.002
DETAILING bd 150 x 307

Reference reinforcement ratio,

r o =(f ck) 1/2 x 10-3 = (30) 1/2x10-3 = 0.0055

Percentage of required compression reinforcement,

r ' = A s ' ,req = 0 = 0.000
bd 150 x 307

Factor for structural system, K = 1.0

r < ro Use equation --> (7.16a)

l  ro  3.2  ro 13 / 2 
= K 11  1.5 r r 
d  f ck f ck   (7.16a)

= 1.0 (11 + 18.4 + 24.15) = 53.53
Modification factor for :-
> 7 = 1.00
span less than 7 m 7 = 2.01 > 1.5
226
steel area provided, = A s,prov = 112.7 use: 1.5
A s,req

So, allowable, l = 53.53 x 1 x 1.5 = 80.295
Actual , d
l = 3960 = 12.9 < 80.30 Ok !
d 307

CRACKING 160 300
200 250
Limiting crack width, w max = 0.3 mm 240 200
280 150
Steel stress, 320 100
360 50
fs = f yk x Gk  0.3Qk 1

1.15 (1.35Gk  1.5Qk ) 

= 500 x 5.06 + (0.3x 0.3)) x 1.0
1.15 7.28

= 308 N/mm2

Max. allowable clear bar spacing = 100 mm s
Ok !
Clear spacing,

s = [ 150 - 2(25) - 2(6) - 2(12) ] / 1

= 64 mm < 100 mm

2H 12

H6 - 225

2H 12

3960

RB 6 ( 150 x 350 )

BEAM DESIGN
CONTINOUS BEAM

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Muhammad bin Hamidun Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1241 Output 1/7

Reference MUHAMMAD, MUKIM KUALA KUANTAN
Calculation

DESIGN CONTINUOUS 2-SPAN SUPPORTED BEAM RECTANGULAR SECTION

BEAM POSITION : RB 4 ( 125 x 350 )

LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

ESTIMATION OF LOADING FROM ROOF

gk (Roof):- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²
H.W Truss
Sgk = 0.30 kN/m²

= 1.00 kN/m² Sqk = 0.25 kN/m²
= 2.17 kN/m²
Action area from plan
gk (Ceiling level):- ESTIMATION OF LOADING FROM CEILING for beam RB4

Ceiling(Asbestos/Plaster) = 1.00 kN/m²
Services = 0.10 kN/m²
Sgk = 1.10 kN/m²

T1 1.04 m
width =

30°

RB4 9.32 m

gk (Roof) = 2.17 x 1.04 x 9.32 / (COS(30)) / 9.32 = 2.61 kN/m

gk (Ceiling level) = 1.1 x 1.04 = 1.14 kN/m

Beam seflweight = 0.125 x 0.35 x 25 = 1.09 kN/m

Sgk = 4.84 kN/m

qk (Roof) = 0.25 x 1.04 x 9.32 / (COS(30)) / 9.32 = 0.30 kN/m

Σgk = 4.84 kN/m Σgk = 4.84 kN/m
Σqk = 0.30 kN/m Σqk = 0.30 kN/m

3.96m 3.66m

(2) '(4) '(6)

Design Action W(4-6)max = 1.35gk + 1.5qk
W(2-4)max = 1.35gk + 1.5qk = (1.35 x 4.84) + (1.5 x 0.3)
= (1.35 x 4.84) + (1.5 x 0.3) = 6.98 kN/m
= 6.98 kN/m
W(2-4)min = 1.35gk W(4-6)min = 1.35gk

= (1.35 x 4.84) = (1.35 x 4.84)
= 6.53 kN/m = 6.53 kN/m

Beam Stiffness, K K(4-6) = 3EI/L = 3 / 3.66 = 0.82
K(2-4) = 3EI/L = 3 / 3.96 = 0.76

Distribution Factor, FA 0.76 / (0.76 + 0.82) = 0.48
FA(2-4) = 1 0.82 / (0.76 + 0.82) = 0.52
FA(4-2) = K(4-2) /(K(4-2) + K(4-6) =
FA(4-6) = K(4-6) /(K(4-2) + K(4-6) = 6.98x3.96²/12 = 9.12 kNm
FA(6-4) = 1 6.53x3.96²/12 = 8.53 kNm
6.98x3.66²/12 = 7.79 kNm
Fixed End Moment, FEM 6.53x3.66²/12 = 7.29 kNm
FEM(2-4)max = W(2-4)max .L(2-4)2 / 12 =
FEM(2-4)min = W(2-4)min .L(2-4)2 / 12 =
FEM(4-6)max = W(4-6)max .L(C4-6)2 / 12 =
FEM(4-6)min = W(4-6)min .L(4-6)2 / 12 =

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 2/7

Reference Calculation Output

CASE 1; max,max. 6.98 kN/m
6.98 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

D 3.96m C 3.66m A
Span (m)
0.48 0.52 1.00
F.A 1.00 9.12 -7.79 7.79
FEM -9.12 -0.64 -0.69 -7.79
Balance 9.12 4.56 -3.90
Distribute -0.32 -0.35 0.00
Balance 12.72 -12.72

M on Support 0.00
(kNm)

12.72 kNm 6.98 kN/m
6.98 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.96R(2) + 12.72 - (6.98 x 3.96² / 2) = 0
3.96R(2) + 12.72 - 54.73 = 0 3.66 R(6) + 12.72 - (6.98 x 3.66² / 2) = 0 xy 0
3.96R(2) + -42.01 = 0 00 8.06
3.96R(2) = 42.01 3.66 R(6) + 12.72 - 46.75 = 0 0 10.61 -12.72
R(2) = 42.01 / 3.96 1.52 0 -12.72
R(2) = 10.61 kN 3.66 R(6) + -34.03 = 0 3.96 -17.03 6.19
3.96 16.25
3.66 R(6) = 34.03 6.29 0 0
7.62 -9.30 0
R(6) = 34.03 / 3.66 0
7.62 0
R(6) = 9.3 kN 00

∑Fy↑+ = 0 0 -3.54
R(2) + R(4) + R(6) - (6.98 x 3.96) - (6.98 x 3.66) = 0 1.52 -3.54
10.61 + R(4) + 9.3 - (27.6408) - (25.5468) = 0 0.76 -3.54 1.52m

R(4)-33.28 = 0
R(4) = 33.28 kN

Shear Force Diagram 6.29 3.54
7.62 3.54
16.25 6.95 3.54 1.33m

10.61

0 1.52m 1.33m
0

-9.30

-17.03

Bending Moment Diagram

-12.72

0 0
8.06 6.19

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 3/7

Reference Calculation Output

CASE 2; max,min. 6.53 kN/m
6.98 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

DCA

Span (m) 3.96m 3.66m

F.A 1.00 0.48 0.52 1.00

FEM -9.12 9.12 7.29 7.29

Balance 9.12 -7.88 -8.53 -7.29

Distribute 4.56 -3.64
Balance -0.44 -0.48

M on Support 0.00 5.37 -5.37 0.00
(kNm)

5.37 kNm 6.53 kN/m
6.98 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.96R(2) + 5.37 - (6.98 x 3.96² / 2) = 0 3.66 R(6) + 5.37 - (6.53 x 3.66² / 2) = 0
3.96R(2) + 5.37 - 54.73 = 0 3.66 R(6) + 5.37 - 43.74 = 0 xy 0
3.96R(2) + -49.36 = 0 3.66 R(6) + -38.37 = 0 00 11.12
3.96R(2) = 49.36 3.66 R(6) = 38.37 0 12.46 -5.37
R(2) = 49.36 / 3.96 1.79 0 -5.37
R(2) = 12.46 kN R(6) = 38.37 / 3.66 3.96 -15.18 8.41
R(6) = 10.48 kN 3.96 13.42
6.02 0 0
∑Fy↑+ = 0 7.62 -10.48 0
R(2) + R(4) + R(6) - (6.98 x 3.96) - (6.53 x 3.66) = 0 0
12.46 + R(4) + 10.48 - (27.6408) - (23.8998) = 0 7.62 0
00
R(4)-28.6 = 0
R(4) = 28.6 kN 0 -4.15
1.79 -4.15
Shear Force Diagram 0.89 -4.15 1.79m

12.46 13.42 1.60m 6.02 4.15
0 1.79m -15.18 7.62 4.15
6.82 4.15 1.60m

0

-10.48

Bending Moment Diagram

-5.37

0 0
11.12 8.41

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 4/7

Reference Calculation Output

CASE 3; min,max. 6.98 kN/m
6.53 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

DCA

Span (m) 3.96m 3.66m

F.A 1.00 0.48 0.52 1.00

FEM -8.53 8.53 -7.79 7.79

Balance 8.53 -0.36 -0.39 -7.79

Distribute 4.27 -3.90
Balance -0.18 -0.19

M on Support 0.00 12.27 -12.27 0.00
(kNm)

12.27 kNm

6.53 kN/m 6.98 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.66 R(6) + 12.27 - (6.98 x 3.66² / 2) = 0
3.96R(2) + 12.27 - (6.53 x 3.96² / 2) = 0 3.66 R(6) + 12.27 - 46.75 = 0 xy 0
3.66 R(6) + -34.48 = 0 00 7.4
3.96R(2) + 12.27 - 51.2 = 0 3.66 R(6) = 34.48 0 9.83 -12.27
1.51 0 -12.27
3.96R(2) + -38.93 = 0 R(6) = 34.48 / 3.66 3.96 -16.03 6.35
R(6) = 9.42 kN 3.96 16.13 0
3.96R(2) = 38.93 6.27 0 0
7.62 -9.42 0
R(2) = 38.93 / 3.96
7.62 0
R(2) = 9.83 kN 00

∑Fy↑+ = 0 0 -3.28
R(2) + R(4) + R(6) - (6.53 x 3.96) - (6.98 x 3.66) = 0 1.51 -3.28
9.83 + R(4) + 9.42 - (25.8588) - (25.5468) = 0 0.75 -3.28 1.51m

R(4)-32.16 = 0 1.35m 6.27 3.28
R(4) = 32.16 kN 0 7.62 3.28
6.95 3.28 1.35m
Shear Force Diagram -9.42

16.13
9.83

0 1.51m

-16.03

Bending Moment Diagram

-12.27

0 6.35 0

7.4 11.12 kNm Shear Maximum 12.46 kN
12.72 kNm 1. Support 2 = 17.03 kN
CONCLUSION 2. Support 4 = 10.48 kN
Moment Maximum 8.41 kNm 3. Support 6 =

1. Span 2-4 =
2. Support 4 =
3. Span 4-6 =

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 5/7

Reference Calculation Output
Table 4.1
3.1.2(3) DESIGN
3.2.2(2)
Materials : = 30 N/mm2 (Table A.3 BS 8500)
Table 4.2 Characteristic strength of concrete, f ck = 500 N/mm2
Table 4.4N Characteristic strength of steel, f yk = 500 N/mm2 (Table A.1 EN 1991-1)
Characteristic strength of link, f yk = 25 kN/m3
4.4.1.2 Unit weight of reinforced concrete
4.4.1.3
4.4.1.1(2) Assumed: f Tension = 12 ; f Compression = 12 mm ; f link = 6 mm

9.2.1.1 SIZE Use : b x h = 125 x 350 mm

6.1 DURABILITY, FIRE & BOND REQUIREMENTS

6.1 Min. conc. cover regard to bond, C min,b = 12 mm

6.1 Min. conc. cover regard to durability, C min,dur = 15 mm
#REF!
Table 5.5 EN 1992-1-2
a sd = 20 + 10 = 30 mm
Use:
Min. concrete cover regard to fire, C nom = 25 mm

C min = a sd - f link - f bar/2 = 30 - 6 - (12/2) = 18 mm
10 mm
Allowance in design for deviation, D0.C5 dev =

Nominal cover,

C nom = C min + DCdev = 15 + 10 = 25 mm

Effective depth, 350 - 25 - 6 - 12 = 307 mm d'
d = h - C nom - f link - f Tension = 25 + 6 + 12/2 = 37 mm
d' = Cnom + f link + f comp. /2 =

Minimum and maximum reinforcement area; dh
b
A s,min= 0.26 ( f ctm )bd = 0.26 ( 2.9 )bd = 0.0015 bd
f yk 500
65.6 mm2
A s,min= 0.0015 bd = 0.0015 x 125 x 350 = 1750 mm2

As,max= 0.04 bh = 0.04 x 125 x 350 =

MAIN REINFORCEMENT AT SPAN 2-4 Bending, Moment, M = 11.12 kNm

K = M / bd 2f ck
= 11.12 x 106 / (125 x 307 x 30) = 0.031 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.97d >0.95d

= 0.95 x 307 = 291.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = 11.1 x 106 / (0.87 x 500 x 291.7) = 88 mm2

MAIN REINFORCEMENT AT SUPPORT 4 Bending, Moment, M = -12.72 kNm

K = M / bd 2f ck
= -12.72 x 106 / (125 x 307 x 30) = -0.036 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 1.03d >0.95d

= 0.95 x 307 = 291.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = -12.7 x 106 / (0.87 x 500 x 291.7) -= 100 mm2

MAIN REINFORCEMENT AT SPAN 4-6 Bending, Moment, M = 8.41 kNm

K = M / bd 2f ck
= 8.41 x 106 / (125 x 307 x 30) = 0.024 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.98d >0.95d

= 0.95 x 307 = 291.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = 8.4 x 106 / (0.87 x 500 x 291.7) = 66 mm2
created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Mohd Rizalman
Muhammad bin Hamidun Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1241 Output 6/7

Reference MUHAMMAD, MUKIM KUALA KUANTAN
6.2 Calculation
6.2.3
SHEAR REINFORCEMENT V Ed = 17.0 kN

V Rd, max = 0.36b wdf ck(1 - f ck/250)
(cot q + tan q )

= 0.36 x 125 x 307 x 30(1 - 30/250) = 364716
(cot q + tan q ) (cot q + tan q )

V Rd, max ( q =22 o ) = 364716 }= 125.8 kN > cot 22 o = 2.5
(2.5 + 0.4) V Ed tan 22 o = 0.4
= 182.4 kN > cot 45 o = 1.0
V Rd, max ( q =45 o ) = 364716 tan 45 o = 1.0
(1 + 1)

Therefore angle q < 22°

Use : θ = 22.0 o tan q = 0.4 cot q = 2.5

Shear links

9.2.2 (6) Max. spacing, s max = 0.75d = 0.75 x 307 = 230 mm
Try link :
A sw = V ed = 17.0 x 10³ = 0.06
s 0.78f ykd cot q 0.78 x 500 x 307 x 2.5 = 943 mm

H6 so, A sw = 57 mm2 : Spacing, s = 57 Use : H6 - 225
0.06

6.2.3(7) Additional longitudinal reinforcement

Additional tensile force,

DF td =0.5V Ed cot q = 0.5 x 17.0308 x 2.5 = 21 kN Use : 1H 12
( 113 mm2)
A s req. = ΔF td = 21x10³ = 48 mm2
0.87f yk (0.87 x 500)

7.4 DEFLECTION

Percentage of required tension reinforcement,

r = A s,req = 88.0 = 0.002
bd 125 x 307

Reference reinforcement ratio,

r o =(f ck) 1/2 x 10-3 = (30) 1/2x10-3 = 0.0055

Percentage of required compression reinforcement,

r ' = A s ' ,req = 0 = 0.000
bd 125 x 307

Table 7.4N Factor for structural system, K = 1.3

r < ro Use equation --> (7.16a)

l  ro  3.2  ro 13 / 2 
= K 11  1.5 r r 
d  f ck f ck   (7.16a)

= 1.3 (11 + 19.6 + 28.68) = 77.09

Modification factor for :-

span less than 7 m > 7 = 1.00
7

steel area provided, = A s,prov = 226 = 2.58 > 1.5 use: 1.5
A s,req 87.7 115.64

So, allowable, l = 77.09 x 1 x 1.5 =
d

Actual , l = 3960 = 12.9 < 115.64 Ok !
d 307

7.3 CRACKING 160 300
200 250
Table 7.1N Limiting crack width, w max = 0.3 mm 240 200
Table 7.3N 280 150
Steel stress, 320 100
360 50
fs = f yk x Gk  0.3Qk 1

1.15 (1.35Gk  1.5Qk ) 

= 500 x 4.84 + (0.3x 0.3)) x 1.0
1.15 6.98

= 307 N/mm2

Max. allowable clear bar spacing = 100 mm s

Clear spacing, Ok !

s = [ 125 - 2(25) - 2(6) - 2(12) ] / 1 created by Mohd Rizalman
= 39 mm < 100 mm

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 7/7

Reference Calculation Output

DETAILING

RB 4 ( 125 x 350 )

created by Mohd Rizalman

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Output 1/7

Reference MUHAMMAD, MUKIM KUALA KUANTAN
Calculation

DESIGN CONTINUOUS 2-SPAN SUPPORTED BEAM RECTANGULAR SECTION

BEAM POSITION : RB 8 ( 150 x 450 )

LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

ESTIMATION OF LOADING FROM ROOF

gk (Roof):- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²
H.W Truss
Sgk = 0.30 kN/m²

= 1.00 kN/m² Sqk = 0.25 kN/m²
= 2.17 kN/m²
Action area from plan
gk (Ceiling level):- ESTIMATION OF LOADING FROM CEILING for beam RB8

Ceiling(Asbestos/Plaster) = 1.00 kN/m²
Services = 0.10 kN/m²
Sgk = 1.10 kN/m²

R7/R8/R9/R10

0.701m

0.85m

30°

1.20 m

gk (Roof) = 2.17 x 1.2 / (COS(30) = 3.01 kN/m

gk (Ceiling level) = 1.1 x 1.2 = 1.32 kN/m

Beam seflweight = 0.01.5125xx00..43550 x 25 = 1.69 kN/m

Sgk = 6.01 kN/m

qk (Roof) = 0.25 x 1.2 / (COS(30)) / 1.2 = 0.29 kN/m

Σgk = 6.01 kN/m Σgk = 6.01 kN/m
Σqk = 0.29 kN/m Σqk = 0.29 kN/m

3.96m 3.66m

(2) '(4) '(6)

Design Action W(4-6)max = 1.35gk + 1.5qk
W(2-4)max = 1.35gk + 1.5qk = (1.35 x 6.01) + (1.5 x 0.29)
= (1.35 x 6.01) + (1.5 x 0.29) = 8.55 kN/m
= 8.55 kN/m
W(2-4)min = 1.35gk W(4-6)min = 1.35gk

= (1.35 x 6.01) = (1.35 x 6.01)
= 8.11 kN/m = 8.11 kN/m

Beam Stiffness, K K(4-6) = 3EI/L = 3 / 3.66 = 0.82
K(2-4) = 3EI/L = 3 / 3.96 = 0.76

Distribution Factor, FA 0.76 / (0.76 + 0.82) = 0.48
FA(2-4) = 1 0.82 / (0.76 + 0.82) = 0.52
FA(4-2) = K(4-2) /(K(4-2) + K(4-6) =
FA(4-6) = K(4-6) /(K(4-2) + K(4-6) = 8.55x3.96²/12 = 11.17 kNm
FA(6-4) = 1 8.11x3.96²/12 = 10.60 kNm
8.55x3.66²/12 =
Fixed End Moment, FEM 8.11x3.66²/12 = 9.54 kNm
FEM(2-4)max = W(2-4)max .L(2-4)2 / 12 = 9.05 kNm
FEM(2-4)min = W(2-4)min .L(2-4)2 / 12 =
FEM(4-6)max = W(4-6)max .L(C4-6)2 / 12 =
FEM(4-6)min = W(4-6)min .L(4-6)2 / 12 =

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. 2/7
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

CASE 1; max,max. 8.55 kN/m
8.55 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

D 3.96m C 3.66m A
Span (m)
0.48 0.52 1.00
F.A 1.00 11.17 -9.54 9.54
FEM -11.17 -0.78 -0.85 -9.54
Balance 11.17
Distribute 5.59 -4.77 0.00
Balance -0.39 -0.42
15.59 -15.59
M on Support 0.00
(kNm)

15.59 kNm 8.55 kN/m
8.55 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.96R(2) + 15.59 - (8.55 x 3.96² / 2) = 0 3.66 R(6) + 15.59 - (8.55 x 3.66² / 2) = 0
3.96R(2) + 15.59 - 67.04 = 0 3.66 R(6) + 15.59 - 57.27 = 0 xy 0
3.96R(2) + -51.45 = 0 3.66 R(6) + -41.68 = 0 00 9.87
3.96R(2) = 51.45 3.66 R(6) = 41.68 0 12.99 -15.59
R(2) = 51.45 / 3.96 1.52 0 -15.59
R(2) = 12.99 kN R(6) = 41.68 / 3.66 3.96 -20.87 7.59
R(6) = 11.39 kN 3.96 19.9
6.29 0 0
∑Fy↑+ = 0 7.62 -11.39 0
R(2) + R(4) + R(6) - (8.55 x 3.96) - (8.55 x 3.66) = 0 0
12.99 + R(4) + 11.39 - (33.858) - (31.293) = 0 7.62 0
00
R(4)-40.77 = 0
R(4) = 40.77 kN 0 -4.33
1.52 -4.33
Shear Force Diagram 0.76 -4.33 1.52m

12.99 19.9 6.29 4.33
0 1.52m -20.87 7.62 4.33
6.95 4.33 1.33m

1.33m
0

-11.39

Bending Moment Diagram

-15.59

0 0
9.87 7.59

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. 3/7
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

CASE 2; max,min. 8.11 kN/m
8.55 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

DCA

Span (m) 3.96m 3.66m

F.A 1.00 0.48 0.52 1.00

FEM -11.17 11.17 9.05 9.05

Balance 11.17 -9.71 -10.52 -9.05

Distribute 5.59 -4.53
Balance -0.51 -0.55

M on Support 0.00 6.54 -6.54 0.00
(kNm)

6.54 kNm 8.11 kN/m
8.55 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.96R(2) + 6.54 - (8.55 x 3.96² / 2) = 0 3.66 R(6) + 6.54 - (8.11 x 3.66² / 2) = 0
3.96R(2) + 6.54 - 67.04 = 0 3.66 R(6) + 6.54 - 54.32 = 0 xy 0
3.96R(2) + -60.5 = 0 3.66 R(6) + -47.78 = 0 00 13.65
3.96R(2) = 60.5 3.66 R(6) = 47.78 0 15.28 -6.54
R(2) = 60.5 / 3.96 1.79 0 -6.54
R(2) = 15.28 kN R(6) = 47.78 / 3.66 3.96 -18.58 10.5
R(6) = 13.05 kN 3.96 16.63
6.01 0 0
∑Fy↑+ = 0 7.62 -13.05 0
R(2) + R(4) + R(6) - (8.55 x 3.96) - (8.11 x 3.66) = 0 0
15.28 + R(4) + 13.05 - (33.858) - (29.6826) = 0 7.62 0
00
R(4)-35.21 = 0
R(4) = 35.21 kN 0 -5.09
1.79 -5.09
Shear Force Diagram 0.89 -5.09 1.79m

15.28 16.63 6.01 5.09
7.62 5.09
1.61m 6.82 5.09 1.61m

0 1.79m 0

-13.05

-18.58

Bending Moment Diagram

-6.54

0 0
13.65 10.5

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. 4/7
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

CASE 3; min,max. 8.55 kN/m
8.11 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

Shear Force & Bending Moment Analysis ( maximum load for all span)

DCA

Span (m) 3.96m 3.66m

F.A 1.00 0.48 0.52 1.00

FEM -10.60 10.60 -9.54 9.54

Balance 10.60 -0.51 -0.55 -9.54

Distribute 5.30 -4.77
Balance -0.25 -0.27

M on Support 0.00 15.14 -15.14 0.00
(kNm)

15.14 kNm

8.11 kN/m 8.55 kN/m

3.96 m 3.66 m

R(2) R(4) R(6)

∑M4+ = 0 ∑M4+ = 0 S.F. data momen
3.96R(2) + 15.14 - (8.11 x 3.96² / 2) = 0 3.66 R(6) + 15.14 - (8.55 x 3.66² / 2) = 0
3.96R(2) + 15.14 - 63.59 = 0 3.66 R(6) + 15.14 - 57.27 = 0 xy 0
3.96R(2) + -48.45 = 0 3.66 R(6) + -42.13 = 0 00 9.22
3.96R(2) = 48.45 3.66 R(6) = 42.13 0 12.23 -15.14
R(2) = 48.45 / 3.96 1.51 0 -15.14
R(2) = 12.23 kN R(6) = 42.13 / 3.66 3.96 -19.89 7.75
R(6) = 11.51 kN 3.96 19.78
6.27 0 0
∑Fy↑+ = 0 7.62 -11.51 0
R(2) + R(4) + R(6) - (8.11 x 3.96) - (8.55 x 3.66) = 0 0
12.23 + R(4) + 11.51 - (32.1156) - (31.293) = 0 7.62 0
00
R(4)-39.67 = 0
R(4) = 39.67 kN 0 -4.08 1.51m
1.51 -4.08
Shear Force Diagram 0.754 -4.08

12.23 19.78 6.27 4.08
0 1.51m -19.89 7.62 4.08
6.95 4.08 1.35m

1.35m 0
-11.51

Bending Moment Diagram

-15.14

0 7.75 0

9.22 13.65 kNm Shear Maximum 15.28 kN
15.59 kNm 1. Support 2 = 20.87 kN
CONCLUSION 2. Support 4 = 13.05 kN
Moment Maximum 10.5 kNm 3. Support 6 =

1. Span 2-4 =
2. Support 4 =
3. Span 4-6 =

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. 5/7
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output
Table 4.1
3.1.2(3) DESIGN
3.2.2(2)
Materials : = 30 N/mm2 (Table A.3 BS 8500)
Table 4.2 Characteristic strength of concrete, f ck = 500 N/mm2
Table 4.4N Characteristic strength of steel, f yk = 500 N/mm2 (Table A.1 EN 1991-1)
Characteristic strength of link, f yk = 25 kN/m3
4.4.1.2 Unit weight of reinforced concrete
4.4.1.3
4.4.1.1(2) Assumed: f Tension = 12 ; f Compression = 12 mm ; f link = 6 mm

9.2.1.1 SIZE Use : b x h = 150 x 450 mm

6.1 DURABILITY, FIRE & BOND REQUIREMENTS

6.1 Min. conc. cover regard to bond, C min,b = 12 mm

6.1 Min. conc. cover regard to durability, C min,dur = 15 mm
#REF!
Table 5.5 EN 1992-1-2
a sd = 20 + 10 = 30 mm
Use:
Min. concrete cover regard to fire, C nom = 25 mm

C min = a sd - f link - f bar/2 = 30 - 6 - (12/2) = 18 mm
10 mm
Allowance in design for deviation, D0.C5 dev =

Nominal cover,

C nom = C min + DCdev = 15 + 10 = 25 mm

Effective depth, 450 - 25 - 6 - 12 = 407 mm d'
d = h - C nom - f link - f Tension = 25 + 6 + 12/2 = 37 mm
d' = Cnom + f link + f comp. /2 =

Minimum and maximum reinforcement area; dh
b
A s,min= 0.26 ( f ctm )bd = 0.26 ( 2.9 )bd = 0.0015 bd
f yk 500
101.3 mm2
A s,min= 0.0015 bd = 0.0015 x 150 x 450 = 2700 mm2

As,max= 0.04 bh = 0.04 x 150 x 450 =

MAIN REINFORCEMENT AT SPAN 2-4 Bending, Moment, M = 13.65 kNm

K = M / bd 2f ck
= 13.65 x 106 / (150 x 407 x 30) = 0.018 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.98d >0.95d

= 0.95 x 407 = 386.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = 13.7 x 106 / (0.87 x 500 x 386.7) = 81 mm2

MAIN REINFORCEMENT AT SUPPORT 4 Bending, Moment, M = 15.59 kNm

K = M / bd 2f ck
= 15.59 x 106 / (150 x 407 x 30) = 0.021 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.98d >0.95d

= 0.95 x 407 = 386.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = 15.6 x 106 / (0.87 x 500 x 386.7) = 93 mm2

MAIN REINFORCEMENT AT SPAN 4-6 Bending, Moment, M = 10.50 kNm

K = M / bd 2f ck
= 10.50 x 106 / (150 x 407 x 30) = 0.014 < K bal = 0.167

Compression reinforcement is not required

z = d [ 0.5 + 0.25 - K /1.134) ] = 0.99d >0.95d

= 0.95 x 407 = 386.7 mm

Area of tension steel Use : 2H 12
( 226 mm2)
A s = M / 0.87 f yk z = 10.5 x 106 / (0.87 x 500 x 386.7) = 62 mm2
created by Mohd Rizalman

Project: Design by: Date: Check by:
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Mohd Rizalman
Sheet No.
TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 6/7
Output
Reference MUHAMMAD, MUKIM KUALA KUANTAN
6.2 Calculation
6.2.3
SHEAR REINFORCEMENT V Ed = 20.9 kN

V Rd, max = 0.36b wdf ck(1 - f ck/250)
(cot q + tan q )

= 0.36 x 150 x 407 x 30(1 - 30/250) = 580219.2
(cot q + tan q ) (cot q + tan q )

V Rd, max ( q =22 o ) = 580219.2 }= 200.1 kN > cot 22 o = 2.5
(2.5 + 0.4) V Ed tan 22 o = 0.4
= 290.1 kN > cot 45 o = 1.0
V Rd, max ( q =45 o ) = 580219 tan 45 o = 1.0
(1 + 1)

Therefore angle q < 22°

Use : θ = 22.0 o tan q = 0.4 cot q = 2.5

Shear links

9.2.2 (6) Max. spacing, s max = 0.75d = 0.75 x 407 = 305 mm
Try link :
A sw = V ed = 20.9 x 10³ = 0.05
s 0.78f ykd cot q 0.78 x 500 x 407 x 2.5 = 1131 mm

H6 so, A sw = 57 mm2 : Spacing, s = 57 Use : H6 - 300
0.05

6.2.3(7) Additional longitudinal reinforcement

Additional tensile force,

DF td =0.5V Ed cot q = 0.5 x 20.87 x 2.5 = 26 kN Use : 1H 12
( 113 mm2)
A s req. = ΔF td = 26x10³ = 60 mm2
0.87f yk (0.87 x 500)

7.4 DEFLECTION

Percentage of required tension reinforcement,

r = A s,req = 81.2 = 0.001
bd 150 x 407

Reference reinforcement ratio,

r o =(f ck) 1/2 x 10-3 = (30) 1/2x10-3 = 0.0055

Percentage of required compression reinforcement,

r ' = A s ' ,req = 0 = 0.000
bd 150 x 407

Table 7.4N Factor for structural system, K = 1.3

r < ro Use equation --> (7.16a)

l  ro  3.2  ro 13 / 2 
= K 11  1.5 r r 
d  f ck f ck   (7.16a)

= 1.3 (11 + 33.85 + 96.60) = 183.89

Modification factor for :-

span less than 7 m > 7 = 1.00
7

steel area provided, = A s,prov = 226 = 2.79 > 1.5 use: 1.5
A s,req 81.2 275.84

So, allowable, l = 183.89 x 1 x 1.5 =
d

Actual , l = 3960 = 9.7 < 275.84 Ok !
d 407

7.3 CRACKING 160 300
200 250
Table 7.1N Limiting crack width, w max = 0.3 mm 240 200
Table 7.3N 280 150
Steel stress, 320 100
360 50
fs = f yk x Gk  0.3Qk 1

1.15 (1.35Gk  1.5Qk ) 

= 500 x 6.01 + (0.3x 0.3)) x 1.0
1.15 8.55

= 310 N/mm2

Max. allowable clear bar spacing = 100 mm s

Clear spacing, Ok !

s = [ 150 - 2(25) - 2(6) - 2(12) ] / 2 created by Mohd Rizalman
= 32 mm < 100 mm

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. 7/7
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

DETAILING

RB 8 ( 150 x 450 )

created by Mohd Rizalman

COLUMN DESIGN

Project: Design by: Date: Check by:
Mohd Rizalman
CADANGAN 1 UNIT RUMAH BANGLO 2 Ahmad bin Albab Sheet No.

TINGKAT DI ATAS LOT 2386/3, JLN TENGKU 02DKA20F1234 Output 1/13

Reference MUHAMMAD, MUKIM KUALA KUANTAN
Calculation

DESIGN OF COLUMN GL 2/J - DETERMINATION OF LOADING

ROOF LEVEL : BEAM RB 8 ( 150 x 450 ) GRIDLINE J/2-4

LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

ESTIMATION OF LOADING FROM ROOF

gk (Roof):- qk:-

Roof tiles = 0.67 kN/m² Roof with slope 30°
Insulated panel
H.W. Purlins = 0.20 kN/m²
H.W Truss
Sgk = 0.30 kN/m²

= 1.00 kN/m² Sqk = 0.25 kN/m²
= 2.17 kN/m²

gk (Ceiling level):- ESTIMATION OF LOADING FROM CEILING

Ceiling(Asbestos/Plaster) = 1.00 kN/m²
Services = 0.10 kN/m²
Sgk = 1.10 kN/m²

R7/R8/R9/R10
width = 9.45 m

0.701m
0.85m

30°

1.20 m

gk (Roof) = 2.17 x 1.2 / (COS(30) = 3.01 kN/m
1.32 kN/m
gk (Ceiling level) = 1.1 x 1.2 = 1.69 kN/m
6.01 kN/m
Beam seflweight = 0.01.5125xx00..43550 x 25 =
0.29 kN/m
Sgk =

qk (Roof) = 0.25 x 1.2 / (COS(30)) / 1.2 =

Σgk = 6.01 kN/m
Σqk = 0.29 kN/m

3.96m
(2) '(4)

ROOF LEVEL : BEAM RB 9 ( 150 x 450 ) GRIDLINE 2/J-H
LOCATION OF BEAM FROM ROOF BEAM TRUSSES LAYOUT:

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. 2/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

Consider T1/T8/T9/T10

30°

0.85m 3.96 m

2.83m

RB9 2.17 x 2.83 / (COS(30)) = 7.09 kN/m
gk (Roof) = 3.11 kN/m
gk (Ceiling level) = 1.1 x 2.83 = 1.69 kN/m
11.89 kN/m
Beam seflweight = 150 x 450 x 25 =
Sgk = 0.82 kN/m

qk (Roof) = 0.25 x 2.83 / (COS(30)) =

Σgk = 11.89 kN/m
Σqk = 0.82 kN/m

3.96m

(J) '(H)

1ST FLOOR LEVEL : BEAM FFB 8 ( 150 x 450 ) GRIDLINE J/2-4
FFB9 ( 150 x 450 ) GRIDLINE 2/J-H

LOCATION OF BEAM FROM 1ST FLOOR BEAM LAYOUT:

Slab GL J-H/2-4

Lx = 3.96 m

Ly = 3.96 m

Ly/Lx = 1.0 -------> 2 way slab

3 edge discontinuous

(1 long edge continuous)

From Table 3.15 BS8110

For Beam FFB8 GL J/2-4 β v = 0.30
For Beam FFB9 GL 2/J-H β v = 0.30

First Floor Beam (Loading from Slab area)

gk (Slab with 100 mm thk):- qk:(Imposed load on slab):-

Slab 0.100 x 25 = 2.50 kN/m² All area of slab

Rendering, Screed (50 mm thk) = 1.15 kN/m²

Clay floor tiles = 0.67 kN/m²

Ceiling(Asbestos/Plaster) = 1.00 kN/m²

Services = 0.10 kN/m²

gk = 5.42 kN/m² qk = 1.50 kN/m²
2750 mm
Loading by brickwall = 2.6 kN/m² high of wall = 3200 - 450 =

Loading For Beam FFB8 GL J/2-4 & FFB9 GL 2/J-H

gk:- qk:-
β v . n.Lx = 0.3x5.42x3.96 =
6.44 kN/m 0.3x1.5x3.96 = 1.78 kN/m

Brickwall = 2.6x2.75x25 = 7.15 kN/m

Beam Selfweight= 0.15x0.45x 25 = 1.69 kN/m

Total gk = 15.28 kN/m

Σgk = 15.28 kN/m Σgk = 15.28 kN/m
Σqk = 1.78 kN/m Σqk = 1.78 kN/m

3.96m 3.96m

(2) '(4) (J) '(H)

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. 3/13
MUHAMMAD, MUKIM KUALA KUANTAN

Reference Calculation Output

GROUND FLOOR LEVEL : BEAM GB 6 ( 150 x 325 ) GRIDLINE J/2-4
GB 7 ( 150 x 325 ) GRIDLINE 2/J-H

LOCATION OF BEAM FROM GROUND BEAM LAYOUT:

Ground Beam (Non Suspanded slab)

Loading by brickwall = 2.6 kN/m² high of wall = 3556 - 450 = 3106 mm

Loading For Beam GB6 GL J/2-4 & GB7 GL 2/J-H
gk:-
Brickwall = 2.6x3.106x25 = 8.08 kN/m

Beam Selfweight=0.15x0.325x 25 = 1.22 kN/m
Total gk = 9.29 kN/m

Σgk = 9.29 kN/m Σgk = 9.29 kN/m
Σqk = 0.00 kN/m Σqk = 0.00 kN/m

3.96m 3.96m

(2) '(4) (J) '(H)

Column Size : b x h = (150 x 300) mm
Actions:
Beam: GL J/2-4 = Roof 1st Fl. G. Fl.
Chac. Permanent action, Gk (kN/m) = 6.0 15.3 9.3
Chac. Variable action, Qk (kN/m) = 0.3 1.8 0.0
Design action, Wd max = 1.35Gk + 1.5Qk 8.5 23.3 12.5

Beam: GL 2/J-H Roof 1st Fl. G. Fl.
Chac. Permanent action, Gk (kN/m) = 11.9 15.3 9.3
Chac. Variable action, Qk (kN/m) = 0.8 1.8 0.0
Design action, Wd max = 1.35Gk + 1.5Qk = 17.3 23.3 12.5

Axial force on Column 2/J

- Roof to 1st Floor : High of column, L = 3200.00 mm

Action from beam GL J/2-4 = 8.5 x 3.96/2 = 16.83 kN

Action from beam GL 2/J-H = 17.3 x 3.96/2 = 34.25 kN

Column selfweight = 0.15 x 0.3 x 3.2 x 25 = 3.60 kN

Column axial force, N (Roof to 1st Floor) = 54.68 kN

- 1st Floor to Ground Floor : High of column, L = 3556.00 mm
Column axial force, N (Roof to 1st Floor) = 54.68 kN
Action from beam GL J/2-4 = 23.3 x 3.96/2 = 46.13 kN
Action from beam GL 2/J-H = 23.3 x 3.96/2 = 46.13 kN
Column selfweight = 0.15 x 0.3 x 3.556 x 25 = 4.00 kN

Column axial force, N (1st Floor to ground floor) = 150.95 kN

-Ground Floor to Footing : High of column, L = 1500.00 mm
Column axial force, N (Roof to 1st Floor) = 150.95 kN
Action from beam GL J/2-4 = 12.5 x 3.96/2 = 24.75 kN
Action from beam GL 2/J-H = 12.5 x 3.96/2 = 24.75 kN
Column selfweight = 0.15 x 0.3 x 1.5 x 25 = 1.69 kN

Column axial force, N (ground floor to footing) = 202.14 kN

created by Mohd Rizalman