More Related Content Similar to Calculation template Similar to Calculation template (20) Calculation template1. Ref. Calculation Output
1.0 Analysis and Designof Beam
Chac.Variable Action, QK =3.0 kN/m2
Weight of Concrete = 25kN/m3
Action on Slab
Self-weight = 0.15Γ25kN/m3 = 3.75kN/m2
Finishes, ceiling and services = 1.5kN/m2
Brick wall = 2.6kN/m2
Chac. Permanent Action, GK = 7.85kN/m2
2. Ref. Calculation Output
Distribution of Actions from Slabs
Slab1 :
πΌ π¦
πΌπ₯
= 4.675π
2.699π
= 1.732 Λ 2.0
Slab 2 :
πΌ π¦
πΌπ₯
= 3.802π
3.475π
= 1.094Λ 2.0
Slab 3 :
πΌ π¦
πΌπ₯
= 5.698π
2.939π
= 1.94Λ 2.0
Action from Slab
Slab1 w1GK1 = 0.26Γ7.85kN/m2Γ0.803m = 1.639kN/m
w1QK1 = 0.26Γ3.0kN/m2Γ0.803m = 0.626kN/m
w2GK2 = 0.40Γ7.85kN/m2Γ1.896m = 5.953kN/m
w2QK2 = 0.40 Γ3.0kN/m2Γ1.896m = 2.275kN/m
Slab2 w3GK3 = 0.36Γ7.85kN/m2Γ2.939m = 8.306kN/m
w3QK3 = 0.36Γ3.00kN/m2Γ2.939m = 3.174kN/m
Two
Way
Slab
3. Ref. Calculation Output
Action on Beam
Beam Self-weight = 0.20m Γ (0.5-0.15)m Γ 25 kN/m3
= 1.75kN/m
Span A β B
Permanent Action, GK
= 6.6825kN/m + kN/m + 1.75kN/m
= 14.5892kN/m
Variable Action, QK
= 2.4300kN/m + 1.7843kN/m = 4.2143N/m
Span B β C
Permanent Action, GK
= 6.1256kN/m + 8.4923kN/m + 3.00kN/m
= 17.6179kN/m
Variable Action, QK
= 2.2275kN/m + 3.0881kN/m = 5.3156kN/m
5. Ref. Calculation Output
Span C β D
Permanent Action, GK
= 7.9819kN/m + 6.2370kN/m + 3.00kN/m
= 17.2189kN/m
Variable Action, QK
= 2.9025kN/m + 2.2680kN/m = 5.1705kN/m
Span D β E
Permanent Action, GK
= 4.6963kN/m + 7.4003kN/m + 3.00kN/m
= 15.0966kN/m
Variable Action, QK
= 1.7078kN/m + 2.6910kN/m = 4.3988kN/m
Span E β F
Permanent Action, GK
= 5.9214kN/m + 6.4598kN/m + 3.00kN/m
= 15.3812kN/m
Variable Action, QK
= 2.1533kN/m + 2.3490kN/m = 4.5023kN/m
6. Span F β G
Permanent Action, GK
= 4.9005kN/m + 5.3460kN/m + 3.00kN/m
= 13.2465kN/m
Variable Action, QK
= 1.782kN/m + 1.9440kN/m = 3.7260kN/m
7. Ref. Calculation Output
Span G β H
Permanent Action, GK
= 8.1675kN/m + 9.9000kN/m + 3.00kN/m
= 21.0675kN/m
Variable Action, QK
= 2.9700kN/m + 3.6000kN/m = 6.5700kN/m
8. Ref. Calculation Output
6.1 Load Distribution of Beam
Case 1: Maximum Loading
FEM
FEMAB = -FEMBA = β
π€πΏ2
12
= β
[1.35 (14.5892 ππ πβ )+1.50(4.2143ππ πβ )](2.2875π)2
12
= -11.3448 kNm
FEMBC = -FEMCB = β
π€πΏ2
12
= β
[1.35 (17.6179 ππ πβ )+1.50(5.3156ππ πβ )](2.2875π)2
12
= -13.8480 kNm
FEMCD = -FEMDC = β
π€πΏ2
12
= β
[1.35 (17.2189 ππ πβ )+1.50(5.1705ππ πβ )](3.225π)2
12
= -26.8694 kNm
FEMDE = -FEMED = β
π€πΏ2
12
= β
[1.35 (15.0966 ππ πβ )+1.50(4.3988 ππ πβ )](1.725π)2
12
= -6.6899 kNm
FEMEF = -FEMFE = β
π€πΏ2
12
= β
[1.35 (15.3812 ππ πβ )+1.50(4.5023ππ πβ )](2.175π)2
12
= -10.8481 kNm
FEMFG = -FEMGF = β
π€πΏ2
12
9. = β
[1.35 (13.2465 ππ πβ )+1.50(3.7260ππ πβ )](1.8π)2
12
= -6.3374 kNm
FEMGH = -FEMHG = β
π€πΏ2
12
= β
[1.35 (21.0675 ππ πβ )+1.50(6.5700ππ πβ )](3.0π)2
12
= -28.7221 kNm
10. Moment Distribution
Mem AB BA BC CB CD DC DE ED EF FE FG GF GH HG
CF 0 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0
DF 1 0.5 0.5 0.585 0.415 0.3485 0.6515 0.5577 0.4423 0.4528 0.5472 0.625 0.375 1
FEM -11.3448 11.3448 -13.8480 13.8480 -26.8694 26.8694 -6.6899 6.6899 -10.8481 10.8481 -6.3374 6.3374 -28.7221 28.7221
Dist 11.3448 1.2516 1.2516 7.6175 5.4039 -7.0326 -13.1470 2.3191 1.8392 -2.0425 -2.4683 13.9904 8.3943 -28.7221
Co 0.6258 0 3.8087 0.6258 -3.5163 2.7019 1.1595 -6.5735 -1.0212 0.9196 6.9952 -1.2341 0 4.1971
Dist -0.6258 -1.9044 -1.9044 1.6909 1.1995 -1.3457 -2.5157 4.2356 3.3591 -3.5838 -4.3310 0.7713 0.4628 -4.1971
Co -0.9522 0 0.8455 -0.9522 -0.6729 0.5998 2.1178 -1.2579 -1.7919 1.6796 0.3857 -2.1655 0 0.2314
Dist 0.9522 -0.4227 -0.4227 0.9507 0.6744 -0.9471 -1.7705 1.7009 1.3489 -0.9351 -1.1301 1.3534 0.8121 -0.2314
Co -0.2114 0 0.4753 -0.2114 -0.4735 0.3372 0.8504 -0.8852 -0.4676 0.6745 0.6767 -0.5651 0 0.4060
Dist 0.2114 -0.2377 -0.2377 0.4007 0.2842 -0.4139 -0.7737 0.7545 0.5984 -0.6118 -0.7394 0.3532 0.2119 -0.4060
Co -0.1188 0 0.2003 -0.1188 -0.2069 0.1421 0.3772 -0.3869 -0.3059 0.2992 0.1766 -0.3697 0 0.1059
Dist 0.1188 -0.1002 -0.1002 0.1906 0.1352 -0.1810 -0.3384 0.3864 0.3064 -0.2154 -0.2603 0.2311 0.1386 -0.1059
Co -0.0501 0 0.0953 -0.0501 -0.0905 0.0676 0.1932 -0.1692 -0.1077 0.1532 0.1155 -0.1302 0 0.0693
Dist 0.0501 -0.0476 -0.0476 0.0822 0.0583 -0.0909 -0.1699 0.1544 0.1225 -0.1217 -0.1471 0.0814 0.0488 -0.0693
0 9.8838 -9.8838 24.0739 -24.0739 20.7069 -20.7069 6.9680 -6.9680 7.0638 -7.0638 18.6536 -18.6536 0
11. Ref. Calculation Output
Uniform Distribution Loading
Loading at Span A β B = [1.35(14.5892kN/m) + 1.50(4.2143kN/m)]
= 26.0169kN/m
Loading at Span B β C = [1.35(17.6179kN/m) + 1.50(5.3156kN/m)]
= 31.7575 kN/m
Loading at Span C β D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D β E = [1.35(15.0966kN/m) + 1.50(4.3988kN/m)]
= 26.9786 kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 26.9786 kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F β G = [1.35(13.2465kN/m) + 1.50(3.7260kN/m)]
= 23.4718 kN/m
12. Loading at Span G β H = [1.35(21.0675kN/m) + 1.50(6.5700kN/m)]
= 38.2961 kN/m
13. Ref. Calculation Output
Span A β B
+β» β π π΅ = 0
= 2.2875π π
π΄ β (26.0169 ππ πβ )(2.2875π)(
2.2875π
2
) β
0πππ + 9.8838πππ
RA = 25.436kN
β β πΉπ¦ = β β πΉπ¦
RA - RB1 = 26.0169kN/m (2.2875m)
RB1 = 34.078kN
β β π»π΄ = β β π» π΅1
= 0
14. Ref. Calculation Output
Span B β C
+β» β π πΆ = 0
= 2.2875ππ
π΅2 β
(31.7575 ππ πβ )(2.2875π)(
2.2875π
2
) β
9.8838πππ + 24.0739πππ
RB2 = 30.119kN
β β πΉπ¦ = β β πΉπ¦
RB2 + RC1 = 31.7575 kN/m (2.2875m)
RC1 = 42.526kN
β β π» π΅2 = β β π» πΆ1
= 0
15. Ref. Calculation Output
Span C β D
+β» β π π· = 0
= 3.225ππ
πΆ2 β (31.0013 ππ πβ )(3.225π)(
3.225π
2
) β
24.0739πππ + 20.7069πππ
RC2 = 51.034kN
β β πΉπ¦ = β β πΉπ¦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 48.945kN
β β π» πΆ2 = β β π» π·1
= 0
16. Ref. Calculation Output
Span D - E
+β» β π πΈ = 0
= 1.725ππ
π·2 β (26.9786 ππ πβ )(1.725π)(
1.725π
2
) β
20.7069πππ + 6.9680πππ
RD2 = 31.234kN
β β πΉπ¦ = β β πΉπ¦
RD2 + RE1 = 26.9786 kN/m (1.725m)
RE1 = 15.304kN
β β π» π·2 = β β π» πΈ1
= 0
17. Ref. Calculation Output
Span E β F
+β» β π πΈ = 0
= 2.175ππ
πΈ2 β (27.5181 ππ πβ )(2.175π)(
2.175π
2
) β
6.9680πππ + 7.0638πππ
RE2 = 29.882kN
β β πΉπ¦ = β β πΉπ¦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 29.970kN
β β π» πΈ2 = β β π» πΉ1
= 0
18. Ref. Calculation Output
Span F β G
+β» β π πΉ = 0
= 1.8ππ
πΉ2 β (23.4718ππ πβ )(1.8π)(
1.8π
2
) β
7.0638πππ + 18.6536πππ
RF2 = 14.686kN
β β πΉπ¦ = β β πΉπ¦
RF2 + RG1 = 23.4718kN/m (1.8m)
RG1 = 27.563kN
β β π» πΉ2 = β β π» πΊ1
= 0
19. Ref. Calculation Output
Span G β H
+β» β π πΊ = 0
= 3.0ππ
πΊ2 β (38.2961 ππ πβ )(3.0π)(
3.0π
2
) β
18.6536πππ + 0πππ
RG2 = 63.662kN
β β πΉπ¦ = β β πΉπ¦
RG2 + RH = 38.2961kN/m (3.0m)
RH = 1.226kN
β β π» πΊ2 = β β π» π»
= 0
20. Ref. Calculation Output
Summary
At point A,π
π΄ = 25.436ππ
βΊ ππ΄ = 0πππ
At point B,π
π΅ = 64.197ππ
βΊ π π΅ = β9.884πππ
β» π π΅ = +9.884πππ
At point C,π
πΆ = 93.560ππ
βΊ ππ = β24.074πππ
β» ππ = +24.074πππ
At point D,π
π· = 80.179ππ
βΊ π π· = β20.707πππ
β» π π· = +20.707πππ
At point E,π
πΈ = 45.186ππ
βΊ π πΈ = β6.968ππ
β» π πΈ = +6.968ππ
At point F,π
πΉ = 44.656ππ
21. βΊ π πΉ = β7.064πππ
β» π πΉ = +7.064πππ
At point G,π
πΊ = 91.225ππ
βΊ π πΊ = β18.654πππ
β» π πΊ = +18.654πππ
At point H,π
π» = 51.226ππ
β» π π» = 0πππ
Ref. Calculation Output
6.2 Load Distribution of Beam
Case 2: Alternative Loading
FEM
FEMAB = -FEMBA = β
π€πΏ2
12
= β
[1.35 (14.5892 ππ πβ )+1.50(4.2143ππ πβ )](2.2875π)2
12
= -11.3448 kNm
FEMBC = -FEMCB = β
π€πΏ2
12
= β
[1.35 (17.6179 ππ πβ )](2.2875π)2
12
= -10.3712 kNm
22. FEMCD = -FEMDC = β
π€πΏ2
12
= β
[1.35 (17.2189 ππ πβ )+1.50(5.1705ππ πβ )](3.225π)2
12
= -26.8694 kNm
FEMDE = -FEMED = β
π€πΏ2
12
= =β
[1.35(15.0966 ππ πβ )](1.725π)2
12
= -5.0537 kNm
FEMEF = -FEMFE = β
π€πΏ2
12
= β
[1.35 (15.3812 ππ πβ )+1.50(4.5023ππ πβ )](2.175π)2
12
= -10.8481 kNm
FEMFG = -FEMGF = β
π€πΏ2
12
= β
[1.35 (13.2465 ππ πβ )](1.8π)2
12
= -4.8283 kNm
FEMGH = -FEMHG = β
π€πΏ2
12
= β
[1.35 (21.0675 ππ πβ )+1.50(6.5700ππ πβ )](3.0π)2
12
= -28.7221 kNm
23. Moment Distribution
Mem AB BA BC CB CD DC DE ED EF FE FG GF GH HG
CF 0.000 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.500 0.000
DF 1.000 0.500 0.500 0.585 0.415 0.349 0.652 0.558 0.442 0.453 0.547 0.625 0.375 1.000
FEM -11.3448 11.3448 -10.3712 10.3712 -26.8694 26.8694 -5.0537 5.0537 -10.8481 10.8481 -4.8283 4.8283 -28.7221 28.7221
Dist 11.3448 -0.4868 -0.4868 9.6514 6.8467 -7.6028 -14.2129 3.2316 2.5629 -2.7258 -3.2940 14.9336 8.9602 -28.7221
Co -0.2434 0.0000 4.8257 -0.2434 -3.8014 3.4234 1.6158 -7.1065 -1.3629 1.2814 7.4668 -1.6470 0.0000 4.4801
Dist 0.2434 -2.4129 -2.4129 2.3662 1.6786 -1.7561 -3.2830 4.7233 3.7460 -3.9612 -4.7870 1.0294 0.6176 -4.4801
Co -1.2064 0.0000 1.1831 -1.2064 -0.8781 0.8393 2.3617 -1.6415 -1.9806 1.8730 0.5147 -2.3935 0.0000 0.3088
Dist 1.2064 -0.5915 -0.5915 1.2194 0.8651 -1.1155 -2.0854 2.0200 1.6021 -1.0811 -1.3065 1.4959 0.8976 -0.3088
Co -0.2958 0.0000 0.6097 -0.2958 -0.5578 0.4325 1.0100 -1.0427 -0.5406 0.8010 0.7480 -0.6533 0.0000 0.4488
Dist 0.2958 -0.3049 -0.3049 0.4993 0.3542 -0.5027 -0.9398 0.8830 0.7003 -0.7014 -0.8476 0.4083 0.2450 -0.4488
Co -0.1524 0.0000 0.2497 -0.1524 -0.2514 0.1771 0.4415 -0.4699 -0.3507 0.3501 0.2041 -0.4238 0.0000 0.1225
Dist 0.1524 -0.1248 -0.1248 0.2362 0.1676 -0.2156 -0.4030 0.4577 0.3630 -0.2510 -0.3033 0.2649 0.1589 -0.1225
Co -0.0624 0.0000 0.1181 -0.0624 -0.1078 0.0838 0.2288 -0.2015 -0.1255 0.1815 0.1324 -0.1517 0.0000 0.0795
Dist 0.0624 -0.0591 -0.0591 0.0996 0.0706 -0.1089 -0.2037 0.1824 0.1446 -0.1421 -0.1718 0.0948 0.0569 -0.0795
0.0000 7.3648 -7.3648 22.4829 -22.4829 20.5238 -20.5238 6.0896 -6.0896 6.4726 -6.4726 17.7860 -17.7860 0.0000
24. Ref. Calculation Output
Uniform Distribution Loading
Loading at Span A β B = [1.35(14.5892kN/m) + 1.50(4.2143kN/m)]
= 26.0169kN/m
Loading at Span B β C = [1.35(17.6179kN/m)]
= 23.7842kN/m
Loading at Span C β D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D β E = [1.35(15.0966kN/m)]
= 20.3804kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 26.9786 kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F β G = [1.35(13.2465kN/m)]
= 17.8828kN/m
25. Loading at Span G β H = [1.35(21.0675kN/m) + 1.50(6.5700kN/m)]
= 38.2961 kN/m
26. Ref. Calculation Output
Span A β B
+β» β π π΅ = 0
= 2.2875π π
π΄ β (26.0169 ππ πβ )(2.2875π)(
2.2875π
2
) +
7.3648πππ
RA = 26.537kN
β β πΉπ¦ = β β πΉπ¦
RA - RB1 = 26.0169kN/m (2.2875m)
RB1 = 32.976kN
β β π»π΄ = β β π» π΅1
= 0
27. Ref. Calculation Output
Span B β C
+β» β π πΆ = 0
= 2.2875ππ
π΅2 β
(23.7842ππ πβ )(2.2875π)(
2.2875π
2
) β 7.3648πππ +
22.4829πππ
RB2 = 20.594kN
β β πΉπ¦ = β β πΉπ¦
RB2 + RC1 = 23.7842kN/m (2.2875m)
RC1 = 33.812kN
β β π» π΅2 = β β π» πΆ1
= 0
28. Ref. Calculation Output
Span C β D
+β» β π π· = 0
= 3.225ππ
πΆ2 β (31.0013 ππ πβ )(3.225π)(
3.225π
2
) β
22.4829πππ + 20.5238πππ
RC2 = 50.597kN
β β πΉπ¦ = β β πΉπ¦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 49.382kN
β β π» πΆ2 = β β π» π·1
= 0
29. Ref. Calculation Output
Span D - E
+β» β π πΈ = 0
= 1.725ππ
π·2 β (20.3804 ππ πβ )(1.725π)(
1.725π
2
) β
20.5238πππ + 6.0896πππ
RD2 = 25.946kN
β β πΉπ¦ = β β πΉπ¦
RD2 + RE1 = 20.3804kN/m (1.725m)
RE1 = 9.210kN
β β π» π·2 = β β π» πΈ1
= 0
30. Ref. Calculation Output
Span E β F
+β» β π πΈ = 0
= 2.175ππ
πΈ2 β (27.5181 ππ πβ )(2.175π)(
2.175π
2
) β
6.0896πππ + 6.4726πππ
RE2 = 9.750kN
β β πΉπ¦ = β β πΉπ¦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 30.102kN
β β π» πΈ2 = β β π» πΉ1
= 0
31. Ref. Calculation Output
Span F β G
+β» β π πΉ = 0
= 1.8ππ
πΉ2 β (17.8828ππ πβ )(1.8π)(
1.8π
2
) β
6.4726πππ + 17.7860πππ
RF2 = 9.809kN
β β πΉπ¦ = β β πΉπ¦
RF2 + RG1 = 17.8828kN/m (1.8m)
RG1 = 22.380kN
β β π» πΉ2 = β β π» πΊ1
= 0
32. Ref. Calculation Output
Span G β H
+β» β π πΊ = 0
= 3.0ππ
πΊ2 β (38.2961 ππ πβ )(3.0π)(
3.0π
2
) β
17.7860πππ
RG2 = 63.373kN
β β πΉπ¦ = β β πΉπ¦
RG2 + RH = 38.2961kN/m (3.0m)
RH = 51.516kN
β β π» πΊ2 = β β π» π»
= 0
33. Ref. Calculation Output
Summary
At point A,π
π΄ = 26.537ππ
βΊ ππ΄ = 0πππ
At point B,π
π΅ = 53.571ππ
βΊ π π΅ = β7.365πππ
β» π π΅ = +7.365πππ
At point C,π
πΆ = 84.409ππ
βΊ ππ = β22.483πππ
β» ππ = +22.483πππ
At point D,π
π· = 75.328ππ
βΊ π π· = β20.524πππ
β» π π· = +20.524πππ
At point E,π
πΈ = 38.960ππ
βΊ π πΈ = β6.090πππ
β» π πΈ = +6.090ππ
At point F,π
πΉ = 39.911ππ
34. βΊ π πΉ = β6.473πππ
β» π πΉ = +6.473πππ
At point G,π
πΊ = 85.753ππ
βΊ π πΊ = β17.786πππ
β» π πΊ = +17.786πππ
At point H,π
π» = 51.516ππ
β» π π» = 0πππ
Ref. Calculation Output
6.3 Load Distribution of Beam
Case 3: Adjacent Loading
FEM
FEMAB = -FEMBA = β
π€πΏ2
12
= β
[1.35 (14.5892 ππ πβ )](2.2875π)2
12
= -8.5883 kNm
FEMBC = -FEMCB = β
π€πΏ2
12
= β
[1.35 (17.6179 ππ πβ )+1.50(5.3156ππ πβ )](2.2875π)2
12
= -13.8480 kNm
35. FEMCD = -FEMDC = β
π€πΏ2
12
= β
[1.35 (17.2189 ππ πβ )+1.50(5.1705ππ πβ )](3.225π)2
12
= -26.8694 kNm
FEMDE = -FEMED = β
π€πΏ2
12
= =β
[1.35(15.0966 ππ πβ )](1.725π)2
12
= -5.0537 kNm
FEMEF = -FEMFE = β
π€πΏ2
12
= β
[1.35 (15.3812 ππ πβ )+1.50(4.5023ππ πβ )](2.175π)2
12
= -10.8481 kNm
FEMFG = -FEMGF = β
π€πΏ2
12
= β
[1.35 (13.2465 ππ πβ )+1.50(3.7260ππ πβ )](1.8π)2
12
= -6.3374 kNm
FEMGH = -FEMHG = β
π€πΏ2
12
= β
[1.35 (21.0675 ππ πβ )](3.0π)2
12
= -21.3308 kNm
36. Moment Distribution
Mem AB BA BC CB CD DC DE ED EF FE FG GF GH HG
CF 0 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0
DF 1 0.5 0.5 0.585 0.415 0.3485 0.6515 0.5577 0.4423 0.4528 0.5472 0.625 0.375 1
FEM -8.5883 8.5883 -13.8480 13.8480 -26.8694 26.8694 -5.0537 5.0537 -10.8481 10.8481 -6.3374 6.3374 -21.3308 21.3308
Dist 8.5883 2.6299 2.6299 7.6175 5.4039 -7.6028 -14.2129 3.2316 2.5629 -2.0425 -2.4683 9.3709 5.6225 -21.3308
Co 1.3149 0.0000 3.8087 1.3149 -3.8014 2.7019 1.6158 -7.1065 -1.0212 1.2814 4.6855 -1.2341 0.0000 2.8113
Dist -1.3149 -1.9044 -1.9044 1.4546 1.0319 -1.5047 -2.8130 4.5328 3.5949 -2.7018 -3.2651 0.7713 0.4628 -2.8113
Co -0.9522 0.0000 0.7273 -0.9522 -0.7524 0.5159 2.2664 -1.4065 -1.3509 1.7974 0.3857 -1.6325 0.0000 0.2314
Dist 0.9522 -0.3636 -0.3636 0.9972 0.7074 -0.9696 -1.8127 1.5378 1.2196 -0.9885 -1.1946 1.0203 0.6122 -0.2314
Co -0.1818 0.0000 0.4986 -0.1818 -0.4848 0.3537 0.7689 -0.9063 -0.4943 0.6098 0.5102 -0.5973 0.0000 0.3061
Dist 0.1818 -0.2493 -0.2493 0.3900 0.2767 -0.3912 -0.7314 0.7811 0.6195 -0.5071 -0.6128 0.3733 0.2240 -0.3061
Co -0.1246 0.0000 0.1950 -0.1246 -0.1956 0.1383 0.3906 -0.3657 -0.2536 0.3097 0.1867 -0.3064 0.0000 0.1120
Dist 0.1246 -0.0975 -0.0975 0.1874 0.1329 -0.1843 -0.3446 0.3454 0.2739 -0.2248 -0.2716 0.1915 0.1149 -0.1120
Co -0.0487 0.0000 0.0937 -0.0487 -0.0922 0.0665 0.1727 -0.1723 -0.1124 0.1369 0.0958 -0.1358 0.0000 0.0575
Dist 0.0487 -0.0468 -0.0468 0.0824 0.0585 -0.0833 -0.1558 0.1588 0.1259 -0.1054 -0.1273 0.0849 0.0509 -0.0575
0.0000 8.5565 -8.5565 24.5846 -24.5846 19.9097 -19.9097 5.6838 -5.6838 8.4135 -8.4135 14.2435 -14.2435 0.0000
37. 37
Ref. Calculation Output
Uniform Distribution Loading
Loading at Span A β B = [1.35(14.5892kN/m)]
= 19.6954kN/m
Loading at Span B β C = [1.35(17.6179kN/m) + 1.50(5.3156kN/m)]
= 31.7576kN/m
Loading at Span C β D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D β E = [1.35(15.0966kN/m)]
= 20.3804kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181kN/m
Loading at Span E β F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F β G = [1.35(13.2465kN/m) + 1.50(3.7260kN/m)]
= 23.4718kN/m
39. 39
Ref. Calculation Output
Span A β B
+β»
β π π΅
= 0
= 2.2875π π
π΄ β
(19.6954 ππ πβ )(2.2875π)(
2.2875π
2
) β 0πππ +
8.5565πππ
RA = 18.786kNm
β β πΉπ¦ = β β πΉπ¦
RA - RB1 = 19.6954kN/m (2.2875m)
RB1 = 26.267kN
β β π»π΄ = β β π» π΅1
= 0
40. 40
Ref. Calculation Output
Span B β C
+β» β π πΆ = 0
= 2.2875ππ
π΅2 β
(31.7576ππ πβ )(2.2875π)(
2.2875π
2
) β
8.5565πππ + 24.5846πππ
RB2 = 29.316kN
β β πΉπ¦ = β β πΉπ¦
RB2 + RC1 = 31.7576kN/m (2.2875m)
RC1 = 43.330kN
β β π» π΅2 = β β π» πΆ1
= 0
41. 41
Ref. Calculation Output
Span C β D
+β» β π π· = 0
= 3.225ππ
πΆ2 β (31.0013 ππ πβ )(3.225π)(
3.225π
2
) β
24.5846πππ + 19.9097πππ
RC2 = 51.439kN
β β πΉπ¦ = β β πΉπ¦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 48.540kN
β β π» πΆ2 = β β π» π·1
= 0
42. 42
Ref. Calculation Output
Span D - E
+β» β π πΈ = 0
= 1.725ππ
π·2 β (20.3804 ππ πβ )(1.725π)(
1.725π
2
) β
19.9097πππ + 5.6838πππ
RD2 = 25.825kN
β β πΉπ¦ = β β πΉπ¦
RD2 + RE1 = 20.3804kN/m (1.725m)
RE1 = 9.331kN
β β π» π·2 = β β π» πΈ1
= 0
43. 43
Ref. Calculation Output
Span E β F
+β» β π πΈ = 0
= 2.175ππ
πΈ2 β (27.5181 ππ πβ )(2.175π)(
2.175π
2
) β
5.6838πππ + 8.4135πππ
RE2 = 28.671kN
β β πΉπ¦ = β β πΉπ¦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 31.181kN
β β π» πΈ2 = β β π» πΉ1
= 0
44. 44
Ref. Calculation Output
Span F β G
+β» β π πΉ = 0
= 1.8ππ
πΉ2 β (23.4718ππ πβ )(1.8π)(
1.8π
2
) β
8.4135πππ + 14.2435πππ
RF2 = 17.886kN
β β πΉπ¦ = β β πΉπ¦
RF2 + RG1 = 23.4718kN/m (1.8m)
RG1 = 24.363kN
β β π» πΉ2 = β β π» πΊ1
= 0
45. 45
Ref. Calculation Output
Span G β H
+β» β π πΊ = 0
= 3.0ππ
πΊ2 β (28.4411 ππ πβ )(3.0π)(
3.0π
2
) β
14.2435πππ + 0πππ
RG2 = 47.410kN
β β πΉπ¦ = β β πΉπ¦
RG2 + RH = 28.4411kN/m (3.0m)
RH = 37.914kN
β β π» πΊ2 = β β π» π»
= 0
46. 46
Ref. Calculation Output
Summary
At point A,π
π΄ = 18.786ππ
βΊ ππ΄ = 0πππ
At point B,π
π΅ = 55.583ππ
βΊ π π΅ = β8.557πππ
β» π π΅ = +8.557πππ
At point C,π
πΆ = 94.769ππ
βΊ ππ = β24.585πππ
β» ππ = +24.585πππ
At point D,π
π· = 74.365ππ
βΊ π π· = β19.910πππ
β» π π· = +19.910πππ
At point E,π
πΈ = 38.002ππ
βΊ π πΈ = β5.684πππ
β» π πΈ = +5.684πππ
At point F,π
πΉ = 49.067ππ
βΊ π πΉ = β8.413πππ
47. 47
β» π πΉ = +8.413πππ
At point G,π
πΊ = 71.773ππ
βΊ π πΊ = β14.243πππ
β» π πΊ = +14.243πππ
At point H,π
π» = 37.914ππ
β» π π» = 0πππ
48. 48
Ref. Calculation Output
Distribution Factor, DF
KAB = KBA =
4πΈπΌ
πΏ
=
4πΈπΌ
2.2875
= 1.7486 EI
KBC = KCB =
4πΈπΌ
πΏ
=
4πΈπΌ
2.2875
= 1.7486 EI
KCD = KDC =
4πΈπΌ
πΏ
=
4πΈπΌ
3.225
= 1.2403 EI
KDE = KDE =
4πΈπΌ
πΏ
=
4πΈπΌ
1.725
= 2.3188 EI
KEF = KF =
4πΈπΌ
πΏ
49. 49
=
4πΈπΌ
2.175
= 1.8391 EI
KFG = KGF =
4πΈπΌ
πΏ
=
4πΈπΌ
1.8
= 2.2222 EI
KGH = KHG =
4πΈπΌ
πΏ
=
4πΈπΌ
3.0
= 1.3333 EI
Ref. Calculation Output
DFAB = πΎ π΄π΅
πΎ π΄π΅ +0
= 1.7468 πΈπΌ
1.7468 πΈπΌ+0
= 1
DFBA = πΎ π΅π΄
πΎ π΅π΄ + πΎ π΅πΆ
= 1.7468πΈπΌ
1.7468 πΈπΌ+1.7468 πΈπΌ
= 0.5
DFBC = πΎ π΅πΆ
πΎ π΅π΄ + πΎ π΅πΆ
= 1.7468 πΈπΌ
1.7468 πΈπΌ+1.7468 πΈπΌ
= 0.5
DFCB = πΎ πΆπ΅
πΎ πΆπ΅ + πΎ πΆπ·
= 1.7468 πΈπΌ
1.7468 πΈπΌ+1.2403 πΈπΌ
= 0.585
DFCD = πΎ πΆπ·
πΎ πΆπ΅ +πΎ πΆπ·
= 1.2403 πΈπΌ
1.7468 πΈπΌ+1.2403 πΈπΌ
= 0.415
DFDC = πΎ π·πΆ
πΎ π·πΆ+ πΎ π·πΈ
= 1.2403 πΈπΌ
1.2403 πΈπΌ+2.3188 πΈπΌ
= 0.348
5
DFDE = πΎ π·πΈ
πΎ π·πΆ+ πΎ π·πΈ
= 2.3188 πΈπΌ
1.2403 πΈπΌ+2.3188 πΈπΌ
= 0.651
5
DFED = πΎ πΈπ·
πΎ πΈπ· + πΎ πΈπΉ
= 2.3188 πΈπΌ
2.3188 πΈπΌ+1.8391 πΈπΌ
= 0.557
7
DFEF = πΎ πΈπΉ
πΎ πΈπ· + πΎ πΈπΉ
= 1.8391 πΈπΌ
2.3188 πΈπΌ+1.8391 πΈπΌ
= 0.442
3
DFFE = πΎ πΉπΈ
πΎ πΉπΈ+ πΎ πΉπΊ
= 1.8391 πΈπΌ
1.8391 πΈπΌ+2.2222 πΈπΌ
= 0.452
8
DFFG = πΎ πΉπΊ
πΎ πΉπΈ+ πΎ πΉπΊ
= 2.2222 πΈπΌ
1.8391 πΈπΌ+2.2222 πΈπΌ
= 0.547
2
50. 50
DFGF = πΎ πΊπΉ
πΎ πΊπΉ + πΎ πΊπ»
= 2.2222 πΈπΌ
2.2222 πΈπΌ+1.3333 πΈπΌ
= 0.625
DFGH = πΎ πΊπ»
πΎ πΊπΉ + πΎ πΊπ»
= 1.3333 πΈπΌ
2.2222 πΈπΌ+1.3333 πΈπΌ
= 0.375
DFHG = πΎ π»πΊ
πΎ π»πΊ + 0
= 1.3333 πΈπΌ
1.3333 πΈπΌ+0
= 1
51. 51
Ref. Calculation Output
6.3.1 Simply Supported Beam
1. Specification
Effective Span, L = 1.8m
Characteristic Action :
Finishes, gk = 1.5kN/m
Variable, qk = 3.0kN/m
Design life = 50 Years
Fire Resistance = R90
Exposure Classes = XS3
Materials :
Characteristic Strength of Concrete, fck = 30N/mm2
Characteristic Strength of Steel, fyk = 500N/mm2
Characteristic Strength of Link, fyk = 500N/mm2
52. 52
Unit Weight of Reinforced Concrete = 25kN/mm2
Assumed :
Γbar = 16mm
Γlink = 8mm
53. 53
Ref. Calculation Output
2. Size
Overall depth, h = L/13
= 1.8/13
= 0.138462m
= 138.462mm
Width, b = 0.4h
= 0.4(138.4615mm)
= 55.385mm
Use b x h = 200mm x 350mm 200mm
= 70000mm2 350mm
54. 54
Ref. Calculation Output
3. Durability, Fire and Bond Requirements
Min. concrete cover regard to bond, Cmin,b = 20mm
Min. concrete cover regard to durability,
Cmin,dur = 45mm
Min. required axis distance for R90 fire
resistance, asd = 45mm
asd = a + 10 = 55mm
Min. concrete cover regard to fire
Cmin = π π π β β
ππππ β β
πππ 2β
= 39mm
Allowance in design for deviation, Ξ Cdev = 10mm
Nominal cover, Cnom = Cmin + Ξ Cdev
= 49mm Cnom
β΄ Cnom = 51mm 51mm
4. Loading and Analysis
Loading on Slab 15:
Slab Thickness = 0.16m
Concrete Unit Weight = 25kN/m3
Self-weight = 25kN/m3 Γ 0.16m = 4.0kN/m
Finishing, ceiling and Services = 1.5kN/m
55. 55
Permanent Action, gk = 5.5kN/m
Variable Action, qk = 3.0kN/m
FS15
ly = 2.4m
lx = 1.8m
π π¦
π π₯
= 1.33 Λ 2.0 (2 way slab)
56. 56
Ref. Calculation Output
Loading on Slab 16:
Slab Thickness = 0.21m
Concrete Unit Weight = 25kN/m3
Self-weight = 25kN/m3 Γ
0.21m
= 5.25kN/m
Finishing, ceiling and Services = 1.5kN/m
Permanent Action, gk = 6.75kN/m
Variable Action, qk = 3.0kN/m
FS16
ly = 2.4m
lx = 1.8m
π π¦
π π₯
= 1.33 Λ 2.0 (2 way slab)
Load on beam 2-3/D-F
58. 58
Ref. Calculation Output
Characteristic permanent load, Gk
Beam self-weight,
Ξ²vy = 0.36 and 0.33
Beam self-weight
w0 = 0.2m Γ 0.35m Γ 25kN/m3
= 1.75 kN/m
w15 = Ξ²vynlx
= 0.36 x 5.5kN/m2 x 1.8m
= 3.564kN/m
w16 = Ξ²vynlx
= 0.33 x 6.75kN/m2 x 1.2m
= 2.673kN/m
= Brickwall
w1 = 2.6kN/m2 Γ 3.0m
= 7.8
Total GK = 15.787kN/m
Characteristic variable load, Qk
w15 = Ξ²vynlx
= 0.36 x 3.0kN/m2 x 1.8m
= 1.944kN/m
w16 = Ξ²vynlx
= 0.33 x 3.0kN/m2 x 1.2m
= 1.188kN/m
60. 60
Ref. Calculation Output
Design action, Wd = 1.35Gk + 1.5Qk Wd
= 1.35(15.787kN/m) + 1.5(3.132kN/m) 26.01
= 26.01 kN/m kN/m
Shear Force, V = π π πΏ
2
V
= 26.01 ππ πβ Γ1.8π
2
23.410
= 23.410kN kN
Bending Moment, π = π π πΏ2
8
M
= (26.01ππ πβ )(1.8π)2
8
10.534
= 10.534kNm kNm
61. 61
Ref. Calculation Output
Effective depth
d = h - Cnom - Γlink - Γbar
= 350mm-50mm-8mm-16mm d
= 278mm 278mm
d' = Cnom + Γlink +
β
πππ
2
= 36mm
Design bending moment, Med = 7.153kNm
πΎ = π
ππ2 π ππ
= 10 .534 Γ106
πππ2
(200ππ )(278ππ)2(30π/ππ2 )
= 0.0227
Kbal = 0.167
K < Kbal ; Compression reinforcement is not required
z =
π[0.5 + β0.25 β
πΎ πππ
1.134
]
z = 0.82π
z = 227.96mm
Area of Tension steel, As
π΄ π = πΎ πππ π ππ ππ2
0.87π π¦π π§
+ π΄ π β²
= (0.167)(30π ππ2β )(200ππ)(278ππ )2
0.87 (500π ππ2β )(226.32ππ)
+ π΄ π β²
= 780.925mm2
62. 62
Min and max reinforcement area, As,min
π΄ π ,πππ = 0.26
πππ‘π
ππ¦π
ππ
=
0.26
2.90π ππ2β
500π ππ2β
ππ
= 0.001508 bd
Exceed 0.0013bd; use 0.0015bd
= 83.844mm2
Ref. Calculation Output
As,max = 0.04Ac
= 0.04 Γ 200mm Γ 350mm
= 2800mm2
5. Shear Reinforcement
Design shear force, Ved = 23.410kN
Concrete strut capacity
ππ
π,πππ₯ = 0.3π π€ ππππ(1β
π ππ
250
)
(πππ‘π+π‘πππ)
= 0.3(200ππ)(278ππ)(30π ππ2β )(1β
30π ππ2β
250
)
(πππ‘π+π‘πππ)
= 181.962ππ (π = 22Β°, πππ‘π = 2.5)
= 264.211ππ (π = 45Β°, πππ‘π = 1.0)
Ved < VRd,maxcot π= 2.5
Ved < VRd,maxcot π= 1.0
63. 63
β΄ π½ < ππΒ°
π = 0.5 sinβ1
[
π πΈπ
0.18π π€ ππππ(1β
π ππ
250
)
]
= 0.5 sinβ1
[
18.558Γ103 π
0.18(200ππ)(278ππ)(30π ππ2β )(1β
30π ππ2β
250
)
]
= 2.01Β°
Use π = 22Β°
π‘πππ = 0.40, πππ‘π = 2.5
Shear Link,
π΄ π π€
π
=
π πΈπ
(0.78ππ¦π ππππ‘π)
= 23.410Γ103 π
(0.78)(500π ππ2β )(278ππ)(2.5)
= 0.0864mm
Try link: H6 π΄ π π€=28.3mm2
Spacing, s = 28.3 ππ2
0.0864 ππ
= 327.67mm
64. 64
Ref. Calculation Output
Maximum spacing, = 0.75d
Smax = 0.75(278mm)
= 208.5mm
Use: H6 - 200 H6 - 200
Minimum links,
π΄ π π€
π
= 0.08β πππ
π π€
π π¦π
= 0.08β30 π ππ2β
200ππ
500π ππ2β
= 0.175 mm
Spacing, s = 28.3 ππ2
0.175 ππ
= 161.454mm < 0.75d
Use: H6 - 175 H6 - 175
Minimum Links,
π΄ π π€
π
= 0.08β πππ
π π€
π π¦π
= 0.08β30 π ππ2β
200ππ
500π ππ2β
= 0.175mm
Spacing, s = 28.3mm2/0.175mm
= 161.464mm Λ 0.75d
Use: H6 - 175
Shear resistance of minimum links
65. 65
π πππ =
π΄ π π€
π
(0.78ππ¦π ππππ‘π)
= 28.3ππ2
200 ππ
(0.78)(500π ππ2β )(278ππ)(2.5)
= 43.833kN
67. 67
Ref. Calculation Output
6. Deflection
Percentage of required tension reinforcement,
π
=
π΄ π ,πππ
ππ
= 775.307ππ2
(200ππ)(278ππ)
= 0.01405
Reference reinforcement ratio,
πΒ°
= β πππ Γ 10β3
= β30 π ππ2β Γ 10β3
= 0.00548
Percentage of required compression reinforcement,
πβ² = 0
π > πΒ° ; Use equation:
π
π
= πΎ[11 + 1.5β πππ
πΒ°
πβπβ² +
1
12
β πππβ
πβ²
πΒ°
π
π
= 1.0[11+ 1.5β30 π ππ2β
0.00548
0.01405β0
+
1
12
β30 π ππ2β β
0
0.00548
]
π
π
= 1.0[11+ 0.5549+ 0]
π
π
= 11.555
Modification factor less than 7m = 1.00
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
=
943ππ2
780.925ππ2
68. 68
= 1.208 <1.5 OK!
Allowable span effective depth ratio = (l/d) allowable
= 11.555 Γ 1.208 Γ 1.00
= 13.954
Actual Span-effective depth, (l/d)
actual
=
1800 ππ
278 ππ
= 6.475 < (l/d) allowable OK!
Ref. Calculation Output
7. Cracking
Limiting crack width, wmax = 0.3mm
Steel stress,
ππ
= ππ¦π
1.15
[
πΊ π+0.3π π
1.35πΊ π+1.5π π
]
1
πΏ
= 500
1.15
[
9.75+0.3(3.0)
1.35(9.75)+1.5(3.0)
]1.0
= 288.813N/mm2
EC2: Max allowable bar spacing = 100mm
Table
4.2
70. 70
Ref. Calculation Output
5.4.2 Continuous Beam
1. Specification
Span AB
Effective Span, L = 2.2875m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 14.5892kN/m
Variable Loading, QK = 4.2143kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
72. 72
Ref. Calculation Output
Span BC
Effective Span, L = 2.2875m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 17.6179kN/m
Variable Loading, QK = 5.3156kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
74. 74
Ref. Calculation Output
Span CD
Effective Span, L = 3.225m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 17.2189kN/m
Variable Loading, QK = 5.1705kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
76. 76
Ref. Calculation Output
Span DE
Effective Span, L = 1.725m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 15.0966kN/m
Variable Loading, QK = 4.3988kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
78. 78
Ref. Calculation Output
Span EF
Effective Span, L = 2.175m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 15.3812kN/m
Variable Loading, QK = 4.5023kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
80. 80
Ref. Calculation Output
Span FG
Effective Span, L = 1.8m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 13.2465kN/m
Variable Loading, QK = 3.7260kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
82. 82
Ref. Calculation Output
Span GH
Effective Span, L = 3.0m
Dimension
Width = 300mm
Depth = 400mm
Characteristic Load
Permanent Loading, GK = 21.0675kN/m
Variable Loading, QK = 6.570kN/m
Design life = 50 Years
Fire Resistance = R120
Exposure Cement = XS3
Materials:
Unit Weight of Concrete = 25kN/m3
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 12mm
β
πππ 2 = 8mm
β
ππππ = 6mm
84. 84
Ref. Calculation Output
2. Durability, Fire and Bond Requirement
Min. Conc. Cover regard to bond, Cmin,b = 12mm
Min. conc. Cover regard to durability, Cmin,dur = 50mm
Min. required axis distance for R120 fire
resistance
asd = 35mm + 10mm
= 45mm
Min. concrete cover regard to fire,
Cmin = π π π β β
ππππ β
β
πππ
2
= 45mm β 6mm β
0.5(12mm)
= 33mm
Allowance in design for deviation, βCdev = 10mm
Nominal cover
Cnom = Cmin + βCdev
= 50mm + 10mm
= 60mm
85. 85
Ref. Calculation Output
3. Loading and Analysis
Span AB
Design Load, wd = 1.35GK + 1.50QK
= 1.35(14.5892kN/m) + 1.50(4.2143kN/m)
= 26.0468kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(14.5892ππ πβ ) +
0.10(4.2143 ππ πβ )](2.2875π)2
= 12.583kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(14.5892ππ πβ ) +
0.106(4.2143 ππ πβ )](2.2875π)2
= 14.4306kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 26.0468 ππ πβ Γ 2.2875π
= 26.78kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 26.0468 ππ πβ Γ 2.2875π
87. 87
Ref. Calculation Output
Span BC
Design Load, wd = 1.35GK + 1.50QK
= 1.35(17.6179kN/m) + 1.50(5.3156kN/m)
= 31.7576kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(17.6179ππ πβ ) +
0.10(5.3156 ππ πβ )](2.2875π)2
= 15.373kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(17.6179ππ πβ ) +
0.106(5.3156 ππ πβ )](2.2875π)2
= 17.6147kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 31.7576 ππ πβ Γ 2.2875π
= 32.6904kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 31.7576 ππ πβ Γ 2.2875π
= 45.7666kN
89. 89
Ref. Calculation Output
Span CD
Design Load, wd = 1.35GK + 1.50QK
= 1.35(17.2189kN/m) + 1.50(5.1705kN/m)
= 31.0013kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(17.2189ππ πβ ) +
0.10(5.1705 ππ πβ )](3.225π)2
= 29.8260kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(17.2189ππ πβ ) +
0.106(5.1705 ππ πβ )](3.225π)2
= 34.1779kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 31.0013 ππ πβ Γ 3.225π
= 44.9906kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 31.0013 ππ πβ Γ 3.225π
= 62.9868kN
91. 91
Ref. Calculation Output
Span DE
Design Load, wd = 1.35GK + 1.50QK
= 1.35(15.0966kN/m) + 1.50(4.3988kN/m)
= 26.9786kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(15.0966ππ πβ ) +
0.10(4.3988 ππ πβ )](1.725π)2
= 7.421kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(15.0966ππ πβ ) +
0.106(4.3988 ππ πβ )](1.725π)2
= 8.5095kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 26.9786 ππ πβ Γ 1.725π
= 20.9422kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 26.9786 ππ πβ Γ 1.725π
= 29.319kN
93. 93
Ref. Calculation Output
Span EF
Design Load, wd = 1.35GK + 1.50QK
= 1.35(15.3812kN/m) + 1.50(4.5023kN/m)
= 27.5181kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(15.3812ππ πβ ) +
0.10(4.5023 ππ πβ )](2.715π)2
= 12.035kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(15.3812ππ πβ ) +
0.106(4.5023 ππ πβ )](2.715π)2
= 13.7988kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 27.5181 ππ πβ Γ 2.715π
= 26.9333kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 27.5181 ππ πβ Γ 2.715π
= 37.7066kN
95. 95
Ref. Calculation Output
Span FG
Design Load, wd = 1.35GK + 1.50QK
= 1.35(13.2465kN/m) + 1.50(3.726kN/m)
= 23.4718kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(13.2465ππ πβ ) +
0.10(3.726 ππ πβ )](1.8π)2
= 7.025kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(13.2465ππ πβ ) +
0.106(3.726 ππ πβ )](1.8π)2
= 8.0611kNm
Shear Force
Outer Support, VED = 0.45 Γ wd Γ L
= 0.45 Γ 23.4718 ππ πβ Γ 1.8π
= 19.0121kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 23.4718 ππ πβ Γ 1.8π
= 26.6170kN
97. 97
Ref. Calculation Output
Span GH
Design Load, wd = 1.35GK + 1.50QK
= 1.35(21.0675kN/m) + 1.50(6.57kN/m)
= 38.2961kN/m
Design Moment
Span, MED = (0.09GK + 0.10QK)L2
= [0.09(21.0675ππ πβ ) +
0.10(6.57 ππ πβ )](3.0π)2
= 31.907kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(21.0675ππ πβ ) +
0.106(6.57 ππ πβ )](2.2875π)2
= 36.5345kNm
Shear Force
Outer Support,
VED
= 0.45 Γ wd Γ L
= 0.45 Γ 38.2961 ππ πβ Γ 3.0π
= 51.6998kN
Inner Support, VED = 0.63 Γ wd Γ L
= 0.63 Γ 38.2961 ππ πβ Γ 3.0π
= 72.3787kN
99. 99
Ref. Calculation Output
4. Main Reinforcement
Effective Depth, d = β β πΆ πππ β β
ππππ β 0.5β
πππ
= 400mm β 60mm β 6mm β 0.5(12mm)
= 328mm
dβ = πΆ πππ + β
ππππ + 0.5β
πππ
= 72mm
100. 100
Ref. Calculation Output
Span AB
Bending Moment = 12.583kNm
K = π πΈπ·
π ππ ππ2
= 12.583 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.013 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.013
1.134β ]
= 0.9884d
= 0.9884 Γ 328mm
= 324.20mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 12.583 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9884 Γ328ππ
2H10
= 89.23mm2 157mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
101. 101
= 0.0015 Γ 300mm Γ 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 Γ 300mm Γ 400mm
= 4800mm2
102. 102
Ref. Calculation Output
Span BC
Bending Moment = 15.373kNm
K = π πΈπ·
π ππ ππ2
= 15.373 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.016 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.016
1.134β ]
= 0.9857d
= 0.9857 Γ 328mm
= 324.20mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 15.373 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9857 Γ328ππ
2H10
= 109.31mm2 157mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
103. 103
= 0.0015 Γ 300mm Γ 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 Γ 300mm Γ 400mm
= 4800mm2
104. 104
Ref. Calculation Output
Span CD
Bending Moment = 29.826kNm
K = π πΈπ·
π ππ ππ2
= 29.826 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.031 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.031
1.134β ]
= 0.9719d
= 0.9719 Γ 328mm
= 318.77mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 29.826 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9719 Γ328ππ
3H10
= 215.09mm2 236mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
105. 105
= 0.0015 Γ 300mm Γ 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 Γ 300mm Γ 400mm
= 4800mm2
106. 106
Ref. Calculation Output
Span DE
Bending Moment = 7.421kNm
K = π πΈπ·
π ππ ππ2
= 7.421 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.008 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.008
1.134β ]
= 0.9929d
= 0.9929 Γ 328mm
= 325.67mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 7.421 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9929 Γ328ππ
2H10
= 52.38mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
108. 108
Ref. Calculation Output
Span EF
Bending Moment = 12.035kNm
K = π πΈπ·
π ππ ππ2
= 12.035 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.012 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.012
1.134β ]
= 0.9893d
= 0.9893 Γ 328mm
= 324.49mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 12.035 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9893 Γ328ππ
2H10
= 85.26mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
110. 110
Ref. Calculation Output
Span FG
Bending Moment = 7.025kNm
K = π πΈπ·
π ππ ππ2
= 7.025 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.007 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.007
1.134β ]
= 0.9938d
= 0.9938 Γ 328mm
= 325.96mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 7.025 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9938 Γ328ππ
2H10
= 49.54mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
112. 112
Ref. Calculation Output
Span GH
Bending Moment = 31.907kNm
K = π πΈπ·
π ππ ππ2
= 31.907 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.033 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.033
1.134β ]
= 0.9700d
= 0.9700 Γ 328mm
= 318.16mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 31.907 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9700 Γ328ππ
3H10
= 230.54mm2 236mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
114. 114
Ref. Calculation Output
Support B
Bending Moment = 17.6147kNm
K = π πΈπ·
π ππ ππ2
= 17.6147 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.018 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.018
1.134β ]
= 0.9839d
= 0.9839 Γ 328mm
= 322.71mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 17.6147 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9839 Γ328ππ
2H10
= 125.48mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
116. 116
Ref. Calculation Output
Support C
Bending Moment = 34.1779kNm
K = π πΈπ·
π ππ ππ2
= 34.1779 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.035 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.035
1.134β ]
= 0.9681d
= 0.9681 Γ 328mm
= 322.71mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 34.1779 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9681 Γ328ππ
4H10
= 247.44mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
118. 118
Ref. Calculation Output
Support D
Bending Moment = 34.1779kNm
K = π πΈπ·
π ππ ππ2
= 34.1779 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.035 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.035
1.134β ]
= 0.9681d
= 0.9681 Γ 328mm
= 322.71mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 34.1779 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9681 Γ328ππ
4H10
= 247.44mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
120. 120
Ref. Calculation Output
Support E
Bending Moment = 13.7988kNm
K = π πΈπ·
π ππ ππ2
= 13.7988 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.014 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.014
1.134β ]
= 0.9875d
= 0.9875 Γ 328mm
= 322.71mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 13.7988 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9875 Γ328ππ
2H10
= 97.94mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
122. 122
Ref. Calculation Output
Support F
Bending Moment = 13.7988kNm
K = π πΈπ·
π ππ ππ2
= 13.7988 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.014 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.014
1.134β ]
= 0.9875d
= 0.9875 Γ 328mm
= 322.71mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 13.7988 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9875 Γ328ππ
2H10
= 97.94mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
124. 124
Ref. Calculation Output
Support G
Bending Moment = 36.5345kNm
K = π πΈπ·
π ππ ππ2
= 36.5345 Γ 106
πππ
30π ππ2β Γ300π Γ (328ππ)2
= 0.038 Λ Kbal(0.167)
z = π [0.5 + β0.25 β πΎ
1.134β ]
= π [0.5 + β0.25 β 0.038
1.134β ]
= 0.9653d
= 0.9653 Γ 328mm
= 316.61mm
Area of Tension Reinforcement
As = π πΈπ·
0.87 Γ π π¦π Γπ§
= 36.5345 Γ 106
πππ
0.87 Γ500π ππ2β Γ0.9653 Γ328ππ
4H10
= 265.26mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
πππ‘π
ππ¦π
) ππ
=
0.26 Γ (
2.9 π ππ2β
500 π ππ2β
) ππ
= 0.0015bd
= 0.0015 Γ 300mm Γ 328mm
126. 126
Ref. Calculation Output
5. Shear Reinforcement
Concrete strut capacity
VRd,max = 0.36π π€ ππππ (1β
πππ
250β )
cotπ+ tan π
= 322.48kN ( π = 22Β° ,cot π = 2.5 )
= 467.60kN( π = 45Β° , cot π = 1.0)
Support A and H
VED at A = 26.537kN
VED at H = 51.516kN
Shear links
π΄ π π€
π
= π πΈπ·
0.78π π¦π π cot π
For Support A = 26 .537 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.083mm
For Support H = 51 .516 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.161mm
Try link H6 (Asw = 28.3mm2)
For Support A, s = 28.3ππ2
0.083ππ
= 341.05mm H6 - 250
For Support H, s = 28.3ππ2
0.161ππ
= 175mm H6 - 175
128. 128
Ref. Calculation Output
For highest shear using at the support
Support B = 34.078kN
Support C = 51.439kN
Support D = 49.382kN
Support E = 29.882kN
Support F = 31.181kN
Support G = 63.662kN
For lowest shear using at support
Support B = 30.119kN
Support C = 43.330kN
Support D = 31.234kN
Support E = 15.304kN
Support F = 17.886kN
Support G = 27.563kN
Shear links for highest shear using
For Support B = 34.078 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.107mm
For Support C = 51 .439 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.107mm
For Support D = 49.382 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.154mm
129. 129
For Support E = 29.882 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.093mm
For Support F = 31.181 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.098mm
For Support G = 63.662 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.199mm
130. 130
Ref. Calculation Output
Shear link for lowest shear using
For Support B = 30 .119 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.094mm
For Support C = 43.330 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.135mm
For Support D = 31.234 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.098mm
For Support E = 15.304 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.048mm
For Support F = 17.886 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.056mm
For Support G = 27.563 Γ 103
π
0.78 Γ500π ππ2β Γ328ππ Γ2.5
= 0.086mm
Try links H6 (Asw = 28.3mm2)
For highest shear using, spacing
For Support B, s = 28.3ππ2
0.107ππ
= 265.58mm H6 β 250
For Support C, s = 28.3ππ2
0.161ππ
= 175.94mm H6 β 150
For Support D, s = 28.3ππ2
0.154ππ
= 183.27mm H6 β 175
131. 131
For Support E, s = 28.3ππ2
0.093ππ
= 302.87mm H6 β 250
For Support F, s = 28.3ππ2
0.098ππ
= 290.25mm H6 β 250
For Support G, s = 28.3ππ2
0.199ππ
= 142.16mm H6 - 100
132. 132
Ref. Calculation Output
For lowest shear using, spacing
For Support B, s = 28.3ππ2
0.094ππ
= 300.49mm H6 β 250
For Support C, s = 28.3ππ2
0.135ππ
= 208.87mm H6 - 200
For Support D, s = 28.3ππ2
0.098ππ
= 289.76mm H6 β 250
For Support E, s = 28.3ππ2
0.048ππ
= 591.37mm H6 - 250
For Support F, s = 28.3ππ2
0.056ππ
= 506.00mm H6 - 250
For Support G, s = 28.3ππ2
0.086ππ
= 328.35mm H6 - 250
Minimum links
π΄ π π€
π
= 0.08πππ
1
2 π π€
ππ¦π
= 0.08 Γ (30 π ππ2β )
1
2 Γ300ππ
500π ππ2β
= 0.262mm
Try link H6 (Asw = 28.3mm2)
Spacing, s = 28.3 ππ2
0.262ππ
= 108.02mm H6 - 100
133. 133
Shear resistance of minimum links
Vmin = (
π΄ π π€
π
) (0.78πππ¦π cot π)
= (28.3ππ2
100 ππ
)(0.78 Γ 328ππ Γ 500π ππ2β Γ 2.5)
= 90.50kN
Since the min. shear resistance is higher than every shear force
calculated,
β΄ πΌππ π―π β πππ πππππππ πππ ππππ
134. 134
Ref. Calculation Output
6. Deflection
Check at Span AB
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 89.23ππ2
300ππ Γ328ππ
= 0.0009
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ
10β3
= (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0009
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0009
β
1)
3
2
]
= 1.3[11 + 0.0504 + 0.1034]
= 14.4999
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
135. 135
=
π΄ π ,ππππ£
π΄ π ,πππ
= 157ππ2
89.23ππ2
= 1.75
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 14.4999 Γ 1.0 Γ 1.5
= 21.7499
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 2287.5ππ
328ππ
= 6.9741 OK
Ref. Calculation Output
Check at Span BC
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 109.31ππ2
300ππ Γ328ππ
= 0.0011
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ
10β3
= (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
136. 136
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0011
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0011
β 1)
3
2
]
= 1.3[11 + 0.0375 + 0.1408]
= 14.5318
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 157ππ2
109.31ππ2
= 1.44
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 14.5318 Γ 1.0 Γ 1.44
= 20.9258
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 2287.5ππ
328ππ
= 6.9741 OK
137. 137
Ref. Calculation Output
Check at Span CD
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 215.09ππ2
300ππ Γ328ππ
= 0.0022
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ 10β3 = (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0022
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0022
β 1)
3
2
]
= 1.3[11 + 0.0206 + 0.0323]
= 14.3688
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 236ππ2
215.09ππ2
138. 138
= 1.10
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 14.3688 Γ 1.0 Γ 1.10
= 15.8057
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 3225ππ
328ππ
= 9.8323 OK
139. 139
Ref. Calculation Output
Check at Span DE
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 52.38ππ2
300ππ Γ328ππ
= 0.0005
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ 10β3 = (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0005
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0005
β 1)
3
2
]
= 1.3[11 + 0.0908 + 0.5566]
= 15.1416
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 157ππ2
52.38ππ2
140. 140
= 2.997
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 15.1416 Γ 1.0 Γ 1.5
= 22.7124
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 1725ππ
328ππ
= 5.26 OK
141. 141
Ref. Calculation Output
Check at Span EF
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 85.26ππ2
300ππ Γ328ππ
= 0.0009
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ 10β3 = (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0009
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0009
β 1)
3
2
]
= 1.3[11 + 0.0504 + 0.2034]
= 14.6299
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 157ππ2
85.26ππ2
142. 142
= 1.84
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 14.6299 Γ 1.0 Γ 1.5
= 21.9449
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 2175ππ
328ππ
= 8.2774 OK
143. 143
Ref. Calculation Output
Check at Span FG
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 49.54ππ2
300ππ Γ328ππ
= 0.0005
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ 10β3 = (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0005
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0005
β 1)
3
2
]
= 1.3[11 + 0.0908 + 0.5566]
= 15.1416
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 157ππ2
49.54ππ2
144. 144
= 3.17
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 15.1416 Γ 1.0 Γ 1.5
= 22.7124
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 1800ππ
328ππ
= 5.4878 OK
145. 145
Ref. Calculation Output
Check at Span GH
Percentage of required tension reinforcement
π =
π΄ π ,πππ
ππ
= 230.54ππ2
300ππ Γ328ππ
= 0.0023
Reference reinforcement ratio
π0 = ( πππ
)
1
2 Γ
10β3
= (30 π ππ2β )
1
2 Γ 10β3
= 0.0055
Percentage of required compression reinforcement
πβ² = π΄ π ,πππ
β²
ππ
= 0ππ2
300ππ Γ328ππ
= 0
Factor for structural system, K = 1.3
π < π0
π
π
= πΎ [11 + 1.5 β ππ π
π0
π
+ 3.2 β ππ π (
π0
π
β 1)
3
2
]
= 1.3 [11 + 1.5 β30π ππ2β Γ 10β3 0.0055
0.0023
+ 3.2 β30π ππ2β Γ 10β3 (0.0055
0.0023
β 1)
3
2
]
= 1.3[11 + 0.0197 + 0.0289]
= 14.3632
Modification factor for span greater than 7m
= 1.0
Modification factor for steel area provided,
=
π΄ π ,ππππ£
π΄ π ,πππ
= 236ππ2
230.54ππ2
146. 146
= 1.02
Therefore allowable span-effective depth ratio
(
π
π
)
πππππ€ππππ
= 14.3632 Γ 1.0 Γ 1.02
= 14.6557
Actual span-effective depth
(
π
π
)
πππ‘π’ππ
= 3000ππ
328ππ
= 9.1463 OK
147. 147
Ref. Calculation Output
7. Cracking
Limiting cracking width,
wmax
= 0.3mm
Steel Stress, fs = ππ¦π
1.15
Γ
πΊ π+0.3π π
(1.35πΊ π+1.50π π)
1
πΏ
= 500π ππ2β
1.15
Γ
21.0675ππ πβ +0.30(6.57ππ πβ )
38.2961ππ πβ
Γ 1.0
= 261.56mm2
Maximum allowing spacing = 150mm
Maximum allowing bar = 12mm
Bar spacing, s = πβ2πΆ πππβ2β
ππ π πβ β
πππ
2
= 300πβ2(60ππ)β2(6ππ)β10ππ
2
= 79mm
150. 150
Ref. Calculation Output
5.5 Slab Design
5.5.1 Simply Supported One Way Spanning Slab
1.0 Specification
Effective span, Leff = 5.4m
Characteristic actions
Permanent, Gk = 3.0kN/m2
Variable, qk = 3.0kN/m2
Design life = 50 years
Fire resistance = R120
Exposure classes = XS3
Materials
Characteristic strength of concrete, fck = 30N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Unit weight of reinforced concrete = 25kN/m3
Assumed: β
bar = 20mm
2.0 Slab Thickness
Min, thickness for fire resistance = 120mm
Thickness for deflection control, h = 120mm
3.0 Durability , Fire and Bond Requirements
151. 151
Min. concrete cover regard to bond, Cmin, b = 12mm
Min. concrete cover regard to durability, Cmin,
dur
= 20mm
Min. required axis distance for R120, a = 40mm
Min. concrete cover regard to fire
Cmin = a-β
bar/2 = 40mm-20mm/2
= 30mm
Allowance in design for deviation, βCdev = 10mm
Nominal cover, Cnom = Cmin + βCdev = 30mm + 10mm
= 40mm
β΄ Cnom = 40mm
Ref. Calculation Output
4.0 Loading and Analysis
Slab self-
weight
= 0.14 Γ 25kN/m3 = 3.5kN/m2
Permanent load (excluding self βweight) = 3.0kN/m2
Characteristic permanent action, Gk = 6.5kN/m2
Characteristic variable action, Qk = 3.0kN/m2
Design action, nd = 1.35Gk + 1.5Qk
= 1.35(6.5kN/m2) + 1.5(3.0kN/m2)
= 13.275kN/m2
152. 152
Consider 2m width, Wd = nd Γ
2m
= 26.55 kN/m
Shear Force, V = π π πΏ
2
= (26.55 ππ πβ ) Γ (3.125π)
2
= 41.48 kN
Bending Moment, M = π π πΏ2
8
= (26.55 ππ πβ ) Γ (3.125π)2
8
= 32.41 kNm
5.0 Main Reinforcement
Effective depth, d = h-Cnom-β
bar/2
= 140mm-40mm-(20mm/2)
= 90mm
Design bending moment, MEd = 32.41 kNm
= π πΈπ·
ππ2 π ππ
= 32.41 Γ 106
πππ
2000ππ Γ (90ππ)2 Γ30 π ππ2β
= 0.032 < Kbal = 0.167
β΄ Compression reinforcement is not required
153. 153
Ref. Calculation Output
z = d[0.5+β0.25 β
πΎ
1.134
]
z = d[0.5+β0.25 β
0.032
1.134
]
z = 0.97d > 0.95d
Use 0.95d 0.95d
As =
π πΈπ·
0.89ππ¦π π§
= 32.41 Γ 106 πππ
0.87 Γ500π ππ2β Γ0.95 Γ90ππ
= 871.41 mm2
Main bar: H12-200 (566mm2/m) H12 - 200
Min. and max. reinforcement area,
As, min = 0.26(
πππ‘π
π π¦π
)bd
= 0.26x (
2.90 π ππ2β
500 π ππ2β
)bd
= 0.0015bd > 0.0013bd
Use 0.0015bd = 0.0015x2000mmx90mm
= 270mm2/m
As, max = 0.04Ac
= 0.04bh
= 0.04 Γ 2000mm Γ 140mm
= 11200mm2/m
Secondary bar: H12-450 (1251mm2/m) H12 - 450
6.0 Shear Reinforcement
154. 154
Design shear force, VEd = 41.48kN
Design shear resistance,
k = 1 + (200/d)1/2 β€ 2.0
= 1 + (200/90)1/2
= 2.49 β₯ 2.0
Οβ = As1/bd β€ 0.02
= 566mm2/ (2000mm Γ 90mm)
= 0.0031 β€ 0.02
155. 155
Ref. Calculation Output
VRd, c = [0.12 Γ π Γ β100πβ² πππ
3
]bd
= [0.12 Γ 2.0 Γ β100 Γ 0.0031 Γ 30 π ππ2β3
] Γ 2000mm Γ 90mm
= 90847.2 N = 90.8kN
Vmin = (0.35 Γ βπ23
Γ β πππ )bd
= (0.35 Γ β2.023
Γ β30 π ππ2β ) Γ 2000ππ Γ 90ππ
= 975991.8N = 976.0 kN
VRd, c ; Vmin > VEd (OK)
7.0 Deflection
Percentage of required tension reinforcement,
π = As, req/bd
= 413.52mm2/ (2000m Γ 90mm)
= 0.0023
Reference reinforcement ratio
πo = (fck)1/2x10-3
= (30 N/mm2)1/2x10-3
= 0.0055
Percentage of required compression reinforcemen
πβ = Asβ, req/bd
= 0/ (2000mm Γ 90mm)
= 0
156. 156
Factor for structural system, K = 1.0
π < πo
π
π
= K[11+1.5β πππ
πo
π
+3.2β πππ(
πo
π
-1)3/2
]
π
π
= 1.0 (11 + 1.5 β30 π ππ2β
π0
π
+ 3.2 β30 π ππ2β (β(
π0
π
β 1)
23
))
π
π
= 1.0 (11 + 19.6 + 28.8)
π
π
= 59.4
Modification factor for span less than 7m = 1.0
157. 157
Ref. Calculation Output
Modification factor for steel area provided
= As, prov/As, req
= 566/413.52
= 1.37 < 1.5
Therefore allowable span-effective depth ratio
= (l/d)allowable
= 59.4 Γ 1.0 Γ 1.37
= 81.38
Actual span-effective depth
= (l/d)actual
= 3125mm/90mm
= 34.72 < (l/d)allowable
h = 140mm
Main bar,
Smax, slabs = 3h or 400mm = 420mm >
400mm
Max. bar spacing = 225mm β€ Smax, slabs = 400mm (OK)
Secondary bar,
Smax, slabs = 3.5h or 450mm = 490mm > 450mm
Max. bar spacing = 450m β€ Smax, slabs = 450mm (OK)
159. 159
Ref. Calculation Output
5.5.2 Simply Supported Two Way Spanning Slab
1.0 Specification
Long span, Ly = 2.288m
Short span, Lx = 2.250m
πΏ π¦
πΏ π₯
= 1.017m
Characteristic actions :
Permanent, Gk = 3kN/π2
Variable, Qk = 3kN/π2
Design life = 50 Years
Fire resistance = R120
Exposure classes = XS3
Materials :
Characteristic strength of concrete, fck = 30N/ππ2
Characteristic strength of steel, fyk = 500N/ππ2
Unit weight of reinforced concrete = 25KN/π3
Assumed: Γbar = 20mm
2.0 Slab Thickness
Min thickness for fire resistance = 120mm
Estimated thickness of deflection control, h =
πΏ π₯
22
=
2250ππ
22
Uses h
= 102mm 210mm
160. 160
Use h 210mm
3.0 Durability, Fire and Bond Requirements
Min. conc. cover regard to bond , Cmin = 20mm
Min. Conc. cover regard to durability,
Cmin,dur
= 55mm
Min. Required axis distance for R60, a = 20mm
161. 161
Ref. Calculation Output
Min. Conc. cover regard to fire, Cmin = π β
β
π΅ππ
2
= 20mm β 10mm
= 10mm
Use min. conc cover regard to durability due to higher value.
Allowance in design for deviation, ΞCdev = 10mm
Nominal cover, Cnom = Cmin+ΞCdev
= 55mm + 10mm
= 65mm
4.0 Loading and Analysis
Slab self-weight = 0.21m Γ 30 kN/m3
= 6.3 kN/m2
Permanent load = 3 kN/m2
Char. permanent action,
Gk
= 9.3 KN/π2
Char. variable action, Qk = 3 KN/π2
Design action, nd = 1.35GK + 1.5QK
= 1.35(9.3kN/m2)+1.5(3kN/m2)
= 17.06 kN/m2
Moment
Short span, Msx = Ξ±sx Γ n Γ Lx
2
= 0.062 x 17.06kN/m Γ (2.250m)2
162. 162
= 5.35 KNm/m
Long span, Msy = Ξ±sy Γ n Γ Ly
2
= 0.062 Γ 17.06kN/m Γ (2.288m)2
= 5.54kNm/m
5.0 Main Reinforcement
Effective depth, dx = h β Cnom - 0.5Γbar
= 210mm β 65mm - (0.5 x 20mm)
= 135mm
163. 163
Ref. Calculation Output
Effective depth, dx = h β Cnom - 0.5Γbar
= 210mm β 65mm - (1.5 x 20mm)
= 115mm
Short span
Msx = 5.35 kNm/m
k = π π π₯
ππ2 πππ
= 5.35 Γ 106 πππ
2250ππ Γ (135ππ)2 Γ30 π ππ2β
= 0.0043
kbal = 0.167
K<πΎπππ , compression is not required
z = d [0.5+β0.25 β
πΎ
1.134
]
= d [0.5+β0.25 β
0.01
1.134
]
= 0.99d
= 0.99 Γ 135mm
= 133.65mm/m
As = π π π₯
0.87 Γ ππ¦π Γπ§
= 5.35 Γ 106 πππ
0.87 Γ500 π ππ2β Γ0.99 Γ135ππ
= 92.09mm2
H6-300(bot) (94ππ π
/m)
164. 164
Long span
Msy = 5.54 kNm/m
k = π π π¦
ππ2 πππ
= 5.54 Γ 106 πππ
2288ππ Γ (115ππ)2 Γ30 π ππ2β
= 0.006
K<πΎπππ , compression is not required
165. 165
Ref. Calculation Output
z = d [0.5+β0.25 β
πΎ
1.134
]
= d [0.5+β0.25 β
0.014
1.134
]
= 0.99d
= 0.99 x 115mm
= 113.85mm/m
As =
π π π¦
0.87 Γ ππ¦π Γπ§
= 5.54 Γ 106 πππ
0.87 Γ500π ππ2β Γ0.99 Γ115ππ
= 108.03mm2
H6-250(bot) (113ππ π
/m)
Min. and max. Reinforced area,
As,min = 0.26(
πππ‘π
ππ¦π
)bd
= 0.26 (
2.9 π ππ2β
500 π ππ2β
) bd
= 0.0015bd > 0.0013bd
= 0.0015 Γ 2288mm Γ 135mm
= 465.79mm2/m
As,max = 0.04Ac
= 0.04 Γ 2288mm Γ 210mm
= 19219.2mm2/m
6.0 Shear Reinforcement
166. 166
Max. Design shear force, ππΈπ· = π π Γ
πΏ π₯
2
= 17.06 ππ πβ Γ
2.250π
2
= 19.19kN
Design shear resistance, k = 1 + (200/d)1/2
β€ 2.0
= 1 + (200/135)1/2
= 2.22 β₯ 2.0
167. 167
Ref. Calculation Output
πβ² = π΄ π
ππ
β€ 0.02
= 92.02 ππ2
2288ππ Γ135ππ
= 0.0003 β€ 0.02
ππ
π·,π = (0.12 Γ π Γ β100 Γ πβ² Γ πππ
3
) ππ
= 0.12 Γ 2.0 Γ β100 Γ 0.012 Γ 30 π ππ2β3
Γ 2288ππ Γ 135ππ
= 244.78 kN
π πππ = (0.035π3/2
πππ1/2
)bd
= (0.035(2.0)3/2
(30 π ππ2β )1/2
)Γ 2288mm x 135mm
= 167.48kN
π½ πΉπ«: π½ πππ β₯ π½ πππ
7.0 Reflection
Percentage of required tension reinforcement,
Ο =
π΄ π ,πππ
ππ
= 108.03 ππ2
2288ππ Γ115ππ
= 0.0004
Reference reinforcement ratio,
π0 = (fck)1/2
x 10β3
= (30)1/2
x 10β3
= 0.0055
168. 168
Percentage of required compression reinforcement,
πβ² =
π΄ π β²,πππ
ππ
=
0 ππ
2288ππ Γ135ππ
= 0
Factor for structural system,
k = 1.0
p β€ π0
169. 169
Ref. Calculation Output
1
π
= π [11+ 1.5 β πππ
π0
π
+ 3.2 β πππ (
π0
π
-1)3/2
]
π
π
= 1.0 [11 + 1.5 β30
0.0055
0.0038
+ 3.2 β30 (
π0
π
-1)3/2
]
π
π
= 1.0 [11+ 11.9 + 5.24]
π
π
= 28.14
Modification factor for span less than 7m = 1.0
Modification factor for steel area provided =
π΄ π ,ππππ£
π΄ π ,πππ
= 108.03 ππ2
108.03 ππ2
= 1.0 β€ 1.5
Therefore allowable span-effective depth
ratio
= (
π
π
)
πππππ€ππππ
= 28.14 Γ 1.0 Γ 2.288m
= 64.384m
Actual span-effective depth = (
π
π
)
πππ‘π’ππ
=
2288ππ
115ππ
= 19.86 Λ 64.384
= OK
8.0 Cracking
170. 170
n = 210mm < 210mm
Main bar
Ξ΄max, slabs = 3h or 400mm
= 3 (210mm) or 400mm
= 630mm > 400
= OK
171. 171
Ref. Calculation Output
Max. bar spacing = 325mm β€ Ξ΄max, slabs = 400mm
Secondary bar
Ξ΄max, slabs = 3.5 or 450mm
= 3.5 (210mm) or 400mm
= 735mm
= OK
Max. bar spacing = 350mm β€ Ξ΄ max, slabs = 450mm
9.0 Detailing
Reinforcement
01: H6 - 300
173. 173
CHAPTER 6
MATERIAL PROPOSED
6.1 Physical Damage of Offshore Building
1. Salt crystallisation
ο The surface of the concrete just above ground or water level is disrupted by the
growth of salts crystals in the pores of the concrete.
ο In temperature climates most of the evaporation takes places at surface the
crystals also form at the surface and do little damage and became worst in hotter
climates.
2. Abrasion
ο Occurs due to wave action carrying sand, shingle or other debris. Shipping
impact is another source of damage. The concrete needs to have sufficient
surface hardness to resist the abrasive forces.
ο All the above types of physicals damage are mainly cosmetic although they do
reduce the cover depth to the reinforcement.
3. Marine growth
ο Marine growth on concrete has generally been considered beneficial as it keeps
the concrete wet, thereby resisting diffusion of gases. Excessive growth can add
to the surface are of slender members such as piles, which could be important
when considering wave loading.
ο Concrete-eating Mollusca has been reported at one place in the Gulf area. They
have an affinity for limestone and therefore only attack concretes made with
limestone aggregate.
174. 174
6.2 Proposal of Material
Structure Material Advantages
Piling -Fibreglass Piling -Water proof
- Dislike by insect and
marine growth.
-Resist of salt water
-Pile cap is not needed
-Installed by vibratory
hammer with a sheet pile
clamp
Ground Beam -Concrete Class XS3 - Has longer durability
-Less moisture absorption
- Rust-resistant
reinforcement
- Resist to rust although
concrete crack
- Forms a stable film of
ferric oxide on its surface
due to alkalinity of
concrete.
Wall -Fly ash bricks -Less porous
-High compressive
strength
-Low thermal conductivity
-Lightweight, easy to
handle
175. 175
CHAPTER 7
CONCLUSION
In the end of the project, we had learnt the process of design planning. We feel thankful as a
chance is given to design the building and analyse it into a building that could be construct one day.
In this project, we had learn how to estimate the beam size and slab thickness, load distribution and
action, analyse the beam and slab from bar size to the deflection, cracking control and detailing.
Furthermore, guideline from Eurocode has eased our process of designing and analysing.
Overall, the structural analysis is needed to be carried out as deflection will occur due to
overload of beam and slab. Both permanent action and variable action followed by moment are
important to be determined as it will affect the choosing of bar size, spacing and the concrete unit. In
an addition, cracking may occur due to extremely hot and cold weather. Hence, choosing the right
material like concrete and steel bar is needed to be considered to reduce the risk of cracking. For
instance, building near to the sea is exposed to high salinity of sea water; choosing material that could
resist water should be put into the main priority.
In conclusion, a well-engineered structure will minimize the failure of structure and produce a
building that is stable and safe enough for living.
176. 176
REFERENCES
1. Al Nageim, H., Durka, F., Morgan, W. & Williams, D.T. 2010. Structural mechanics: loads,
analysis, materials and design of structural elements. 7th edition.
London, Pearson Education.
2. Amit A. Sathawane, R.S. Deotale. 2012. Analysis And Design Of Flat Slab
And Grid Slab And Their Cost Comparison.
3. Reinforced Concrete Design, 1990.Tata McGraw-Hill Publishing Company,
1st Revised Edition. Publisher of New Delhi.
4. Salvadori, M. & Heller, R. 1986. Structure in architecture: the building of
buildings. 3rd Edition. Englewood Cliffs, New Jersey, Prentice-Hall.
5. W.H.Mosley, J.H. Bungery & R. Husle. 1999. Reinforced Concrete Design
(5th Edition).Palgrave.
177. 177
WORK PROGRESSION
N
O
WORK PROGRESSION
WEEK
1 2 3 4 5 6 7 8 9
1
0
1
1
1
2
1
3
1
4
1 Forming group
2
Determine the suitable design to make a
model
3
Discussion on methodology and work
progression
4 Design a model in sketchup and autoCAD
5 Make a model follow by the real scale
5
Determinate the continues beam, simply
supported beam, one-ways slab, and two-
ways slab
6 Pre-presentation
7 Calculate the loads, and analyze the structure
8 Presentation
9 Final Report
178. 178
Ref. Calculation Output
5.4.2 Continuous Beam
9. Specification
Span AB
Effective Span, L = 3.235m
Dimension
Width = 200mm
Depth = 500mm
Characteristic Load
Action on slab
Selfweight=0.15x25=3.75 kN/m2
Finishes,ceiling and services=1.5 kN/m2
Brick wall =2.6 kN/m2
Permanent Loading, GK = 7.85kN/m
Variable Loading, QK = 3.0kN/m2
Design life = 50Years
Fire Resistance = R120
Exposure Cement = XC1
Materials:
Unit Weight of Concrete = 25kN/m3
179. 179
Characteristic strength of concrete, fck = 25N/mm2
Characteristic strength of steel, fyk = 500N/mm2
Characteristic strength of link, fyk = 500Nmm2
Assumed
β
πππ 1 = 10mm
β
ππππ = 6mm