JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHE ET
Made by:
Date:
Checked by:
Date:
Description
Reference
BS 8110
Data Data give iven Concrete Grade
Result
f cu cu
=
30
N/mm
f cu cu
=
30
f y
=
460 40
N/mm
f y
=
mm
cover =
460 40
Cover = Preliminary Sizing Assume Thickness of slab Diameter of bar
125 12
mm mm
Effective depth for the short span, d1 =
79
mm
d1 =
79
Effective depth for the long span, d2 =
67
mm
d2 =
67
(Residential Area)
qk =
4
gk =
4
n
12
Condition :
9
= =
(Worst Condition)
Four edges discontinuous Loading Total Live Load (Unfactored)
=
Screed+finishes (Unfactored)
= 0
Brickwall (Unfactor UDL) =
4 1 kN/m kN/m2 2
Total Dead Load ( Unfactored ) = 4 Design Load (Factored DL + Factored Factored LL) = Short Span, lx = Long Span, ly =
2
m
2
m
Table 3.15 Moment Coefficient
kN/m 2
kN/m
Heig Heigh ht of of bri brick ckwa walll = kN/m2 12
0
m
kN/m
=
lx = ly =
2
2 2
0.000 2
0.056 0.055
0
Main Reinforcement at MIDSPAN cl 3.5.3.4 Msx = Bsx*n*lx =
eqn 14
=
cl 3.4.4.4
z=
=
sx cu
77.8
Therefore, z =
kNm per me metre width
0.0142
k < 0.156. No compression reinforcement.
( 0.95d = 75.05 75.05
mm
88
mm
Asmin =
163
mm
Asmax =
5000
mm
10
at
As required =
Therefore, provide T
Ref. : JPA/P-063/Slab.xls/08-002
2.64
2
)
Provided As =
163
mm2 Main Reinforcement
As < As Asmin. min. Provided minimum reinforcement! mm 200 Area = 392
at MIDSPAN T at
10 200
page 1 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHEET
Made by:
Date:
Checked by:
Date:
Description
Reference
cl 3.4.6
Result
Check for Deflection 2
Table 3.11 Msx/b*d
=
0.4231
eqn 8
Fy
=
eqn 7
Modification factor =
N/mm
64.55 2
Table 3.10 Basic span/effective depth Modification
=
=
20
40
Deflection Check
Actual span/effective depth ratio =
25.32
OK!
OK!
Cracking Check max. spacing =
250
mm c/c
min. spacing =
100
mm c/c
Spacing provided =
200
mm c/c
Cracking Check OK!
OK!
Secondary Reinforcement at MIDSPAN cl 3.5.3.4 2
Msy = Bsy*n*lx =
eqn 14
cl 3.4.4.4 k = Msy/f cu*b*d = z=
65.47
Therefore,
z =
2.7
kNm per metre width
0.0201
k < 0.156. No compression reinforcement.
( 0.95d = 63.65 63.65
As required =
106
mm 2
mm
Asmin =
163
mm
Asmax =
5000
mm
Therefore, provide T
10
)
Provided As =
163
mm2
Secondary Reinforcement at MIDSPAN
As < Asmin. Provided minimum reinforcement!
at
200
Area =
392
T at
2
mm
10 200
Main Reinforcement at SUPPORT cl 3.5.3.4 2 eqn 14 Msx = Bsx*n*lx = 0.0 kNm per metre width 2
cl 3.4.4.4 k = Msx/f cu*b*d = z= Therefore,
79 z =
0
k < 0.156. No compression reinforcement.
( 0.95d = 75.05 75.05 mm
As required =
2
mm
Asmin =
163
mm
Asmax =
5000
mm
Therefore, provide T
Ref. : JPA/P-063/Slab.xls/08-002
0
10
)
at
Provided As =
163
mm2 Main
As < Asmin. Provided minimum reinforcement! 200
Area =
392
2
mm
Reinforcement at SUPPORT T 10 at 200
page 2 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHEET
Made by:
Date:
Checked by:
Date:
Description
Reference
Cracking Check max. spacing = min. spacing = Spacing provided =
250 100 200
Result
mm c/c mm c/c mm c/c
Cracking Check OK!
OK!
cl 3.5.3.7 Check for Shear eqn 19 Vsx = Bsx*n*lx = Table 3.16 v = Vsx/bd =
7.92
Table 3.9 100 As/bd
0.4963
Therefore,
N/mm 0.1003 < 0.8*sqrt(fcu), OK!
= Vc
Where Bsx =
=
0.7977
0.33
Shear Coefficient
0.33
0.33
( Table 3.16)
0.33
N/mm
0.33
Shear Check
Table 3.8 Therefore, V < Vc. No Shear Reinforcement required.
OK!
Second Reinforcement at SUPPORT cl 3.5.3.4 2 eqn 14 Msy = Bsy*n*lx = 0.00 kNm per metre width 2
cl 3.4.4.4 k = Msy/f cu*b*d = z= Therefore,
0
67 z =
k < 0.156. No compression reinforcement.
( 0.95d = 63.65 63.65 mm
As required =
0
2
mm
Asmin =
163
mm
Asmax =
5000
mm
Therefore, provide T Cracking Check max. spacing = min. spacing = Spacing provided =
10
250 100 200
)
Provided As =
163
mm2 Secondary Reinforcement
As < Asmin. Provided minimum reinforcement!
at
200
Area =
mm c/c mm c/c mm c/c
392
2
mm
at SUPPORT T at
10 200
Cracking Check OK!
OK!
cl 3.5.3.7 Check for Shear eqn 19 Vsy = Bsy*n*lx = Table 3.16 v = Vsx/bd =
7.92
Table 3.9 100 As/bd
0.5851
Therefore,
N/mm 0.1183 < 0.8*sqrt(fcu), OK!
= Vc
Where Bsy =
=
0.8781
0.33 N/mm
( Table 3.16) Shear Coefficient
0.33
Table 3.8 Therefore, V < Vc. No Shear Reinforcement required.
Ref. : JPA/P-063/Slab.xls/08-002
0.33
0.33 0.33
Shear Check OK!
page 3 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATION SHEET
Made by:
Date:
Checked by:
Date:
Description
Reference
Result
2 ex en rom the edge a minimum distance of 0.4 m
T 10 @ 200 T 10 200 T 10 @ 200 2
T @
10 200
T @
10 200
T 10 @ 200
Loading for Beam Design D/L L/L
1
Ref. : JPA/P-063/Slab.xls/08-002
4.00 4.00
0
2
kN/m 2 kN/m
1
Un factored
D/L L/L
1
4.00 4.00
2
kN/m 2 kN/m
1
page 4 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHEE T
Made by:
Date:
Checked by:
Date:
Description
Reference
BS 8110
Data given Concrete Grade
Result
f cu
=
30
N/mm
f cu
=
30
f y
=
460 40
N/mm
f y
=
mm
cover =
460 40
Cover = Preliminary Sizing Assume Thickness of slab Diameter of bar
100 12
mm mm
Effective depth for the short span, d1 =
54
mm
d1 =
54
Effective depth for the long span, d2 =
42
mm
d2 =
42
(Residential Area)
qk =
4
gk =
3.4
n
11.16
Condition :
9
= =
(Worst Condition)
Four edges discontinuous Loading Total Live Load (Unfactored)
=
Screed+finishes (Unfactored)
4
=
1
0
Brickwall (Unfactor UDL) =
kN/m2
Total Dead Load ( Unfactored ) = 3.4 Design Load (Factored DL + Factored LL) = Short Span, lx = Long Span, ly =
0.6
m
2
m
Table 3.15 Moment Coefficient
kN/m 2
kN/m
Height of brickwall = kN/m2 11.16
0
m
kN/m
=
lx = ly =
0.6
0.6 2
0.000 2
0.056 0
0
Main Reinforcement at MIDSPAN cl 3.5.3.4 Msx = Bsx*n*lx =
eqn 14
=
cl 3.4.4.4
z=
=
sx cu
54
Therefore, z =
kNm per metre width
0
k < 0.156. No compression reinforcement.
( 0.95d = 51.3 51.3
2
mm
Asmin =
130
mm
Asmax =
4000
mm
10
at
Therefore, provide T
)
mm
0
As required =
Ref. : JPA/P-063/Slab.xls/08-002
0.00
Provided As =
130
mm2 Main Reinforcement
As < Asmin. Provided minimum reinforcement! mm 200 Area = 392
at MIDSPAN T at
10 200
page 5 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHEE T
Made by:
Date:
Checked by:
Date:
Description
Reference
cl 3.4.6
Result
Check for Deflection 2
Table 3.11 Msx/b*d
=
0
eqn 8
Fy
=
eqn 7
Modification factor =
N/mm
0 2
Table 3.10 Basic span/effective depth Modification
=
=
20
40
Deflection Check
Actual span/effective depth ratio =
11.12
OK!
OK!
Cracking Check max. spacing =
250
mm c/c
min. spacing =
100
mm c/c
Spacing provided =
200
mm c/c
Cracking Check OK!
OK!
Secondary Reinforcement at MIDSPAN cl 3.5.3.4 2
Msy = Bsy*n*lx =
eqn 14
cl 3.4.4.4 k = Msy/f cu*b*d = z=
41.74
Therefore,
z =
0.3
kNm per metre width
0.0057
k < 0.156. No compression reinforcement.
( 0.95d = 39.9 39.9
As required =
19
mm 2
mm
Asmin =
130
mm
Asmax =
4000
mm
Therefore, provide T
10
)
Provided As =
130
mm2
Secondary Reinforcement at MIDSPAN
As < Asmin. Provided minimum reinforcement!
at
200
Area =
392
T at
2
mm
10 200
Main Reinforcement at SUPPORT cl 3.5.3.4 2 eqn 14 Msx = Bsx*n*lx = 0.0 kNm per metre width cl 3.4.4.4 k = Msx/f cu*b*d = z= Therefore,
54 z =
0
k < 0.156. No compression reinforcement.
( 0.95d = 51.3 51.3 mm
As required =
2
mm
Asmin =
130
mm
Asmax =
4000
mm
Therefore, provide T
Ref. : JPA/P-063/Slab.xls/08-002
0
10
)
at
Provided As =
130
mm2 Main
As < Asmin. Provided minimum reinforcement! 200
Area =
392
2
mm
Reinforcement at SUPPORT T 10 at 200
page 6 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATI ON SHEET
Made by:
Date:
Checked by:
Date:
Description
Reference
Cracking Check max. spacing = min. spacing = Spacing provided =
250 100 200
Result
mm c/c mm c/c mm c/c
Cracking Check OK!
OK!
cl 3.5.3.7 Check for Shear eqn 19 Vsx = Bsx*n*lx = Table 3.16 v = Vsx/bd =
0
Table 3.9 100 As/bd
0.726
Therefore,
N/mm 0 < 0.8*sqrt(fcu), OK!
= Vc
Where Bsx =
=
0.9959
0
( Table 3.16) Shear Coefficient
0.33
0
0
N/mm
0.33
Shear Check
Table 3.8 Therefore, V < Vc. No Shear Reinforcement required.
OK!
Second Reinforcement at SUPPORT cl 3.5.3.4 2 eqn 14 Msy = Bsy*n*lx =
0.00
2
cl 3.4.4.4 k = Msy/f cu*b*d = z= Therefore,
kNm per metre width
0
42 z =
k < 0.156. No compression reinforcement.
( 0.95d = 39.9 39.9 mm
As required =
0
2
mm
Asmin =
130
mm
Asmax =
4000
mm
Therefore, provide T Cracking Check max. spacing = min. spacing = Spacing provided =
10
250 100 200
)
Provided As =
130
mm2 Secondary Reinforcement
As < Asmin. Provided minimum reinforcement!
at
200
Area =
mm c/c mm c/c mm c/c
392
2
mm
at SUPPORT T at
10 200
Cracking Check OK!
OK!
cl 3.5.3.7 Check for Shear eqn 19 Vsy = Bsy*n*lx = Table 3.16 v = Vsx/bd =
2.21
Table 3.9 100 As/bd
0.9334
Therefore,
N/mm 0.0527 < 0.8*sqrt(fcu), OK!
= Vc
Where Bsy =
=
1.1531
N/mm
( Table 3.16) Shear Coefficient
0.33
0
Table 3.8 Therefore, V < Vc. No Shear Reinforcement required.
Ref. : JPA/P-063/Slab.xls/08-002
0.33
0 0.33
Shear Check OK!
page 7 of 8
JPA No.
Sheet
Rev.
2
Job Title: Subject:
Client: CALCULATION SHEET
Made by:
Date:
Checked by:
Date:
Description
Reference
Result
0.6 T 10 @ 200 T 10 200 T 10 @ 200 2
T @
10 200
T @
ex en rom the edge a minimum distance of 0.12 m
10 200
T 10 @ 200
Loading for Beam Design D/L L/L
0.3
Ref. : JPA/P-063/Slab.xls/08-002
1.02 1.20
1.4
2
kN/m 2 kN/m
0.3
Un factored
D/L L/L
0.3
1.02 1.20
2
kN/m 2 kN/m
0.3
page 8 of 8
