calculation template

1. 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 = 0.15×25kN/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 :
𝐼 𝑦
𝐼𝑥
= 4.675𝑚
2.699𝑚
= 1.732 ˂ 2.0
Slab 2 :
𝐼 𝑦
𝐼𝑥
= 3.802𝑚
3.475𝑚
= 1.094˂ 2.0
Slab 3 :
𝐼 𝑦
𝐼𝑥
= 5.698𝑚
2.939𝑚
= 1.94˂ 2.0
Action from Slab
Slab1 w1GK1 = 0.26×7.85kN/m2×0.803m = 1.639kN/m
w1QK1 = 0.26×3.0kN/m2×0.803m = 0.626kN/m
w2GK2 = 0.40×7.85kN/m2×1.896m = 5.953kN/m
w2QK2 = 0.40 ×3.0kN/m2×1.896m = 2.275kN/m
Slab2 w3GK3 = 0.36×7.85kN/m2×2.939m = 8.306kN/m
w3QK3 = 0.36×3.00kN/m2×2.939m = 3.174kN/m
Two
Way
Slab

3. Ref. Calculation Output
Action on Beam
Beam Self-weight = 0.20m × (0.5-0.15)m × 25 kN/m3
= 1.75kN/m
Span A – 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 – 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 – 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 – 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 – 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 – 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 – 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 = −
𝑤𝐿2
12
= −
[1.35 (14.5892 𝑘𝑁 𝑚⁄ )+1.50(4.2143𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -11.3448 kNm
FEMBC = -FEMCB = −
𝑤𝐿2
12
= −
[1.35 (17.6179 𝑘𝑁 𝑚⁄ )+1.50(5.3156𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -13.8480 kNm
FEMCD = -FEMDC = −
𝑤𝐿2
12
= −
[1.35 (17.2189 𝑘𝑁 𝑚⁄ )+1.50(5.1705𝑘𝑁 𝑚⁄ )](3.225𝑚)2
12
= -26.8694 kNm
FEMDE = -FEMED = −
𝑤𝐿2
12
= −
[1.35 (15.0966 𝑘𝑁 𝑚⁄ )+1.50(4.3988 𝑘𝑁 𝑚⁄ )](1.725𝑚)2
12
= -6.6899 kNm
FEMEF = -FEMFE = −
𝑤𝐿2
12
= −
[1.35 (15.3812 𝑘𝑁 𝑚⁄ )+1.50(4.5023𝑘𝑁 𝑚⁄ )](2.175𝑚)2
12
= -10.8481 kNm
FEMFG = -FEMGF = −
𝑤𝐿2
12

9. = −
[1.35 (13.2465 𝑘𝑁 𝑚⁄ )+1.50(3.7260𝑘𝑁 𝑚⁄ )](1.8𝑚)2
12
= -6.3374 kNm
FEMGH = -FEMHG = −
𝑤𝐿2
12
= −
[1.35 (21.0675 𝑘𝑁 𝑚⁄ )+1.50(6.5700𝑘𝑁 𝑚⁄ )](3.0𝑚)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 – B = [1.35(14.5892kN/m) + 1.50(4.2143kN/m)]
= 26.0169kN/m
Loading at Span B – C = [1.35(17.6179kN/m) + 1.50(5.3156kN/m)]
= 31.7575 kN/m
Loading at Span C – D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D – E = [1.35(15.0966kN/m) + 1.50(4.3988kN/m)]
= 26.9786 kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 26.9786 kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F – G = [1.35(13.2465kN/m) + 1.50(3.7260kN/m)]
= 23.4718 kN/m

12. Loading at Span G – H = [1.35(21.0675kN/m) + 1.50(6.5700kN/m)]
= 38.2961 kN/m

13. Ref. Calculation Output
Span A – B
+↻ ∑ 𝑀 𝐵 = 0
= 2.2875𝑚 𝑅 𝐴 − (26.0169 𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) −
0𝑘𝑁𝑚 + 9.8838𝑘𝑁𝑚
RA = 25.436kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RA - RB1 = 26.0169kN/m (2.2875m)
RB1 = 34.078kN
→ ∑ 𝐻𝐴 = ← ∑ 𝐻 𝐵1
= 0

14. Ref. Calculation Output
Span B – C
+↻ ∑ 𝑀 𝐶 = 0
= 2.2875𝑚𝑅 𝐵2 −
(31.7575 𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) −
9.8838𝑘𝑁𝑚 + 24.0739𝑘𝑁𝑚
RB2 = 30.119kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RB2 + RC1 = 31.7575 kN/m (2.2875m)
RC1 = 42.526kN
→ ∑ 𝐻 𝐵2 = ← ∑ 𝐻 𝐶1
= 0

15. Ref. Calculation Output
Span C – D
+↻ ∑ 𝑀 𝐷 = 0
= 3.225𝑚𝑅 𝐶2 − (31.0013 𝑘𝑁 𝑚⁄ )(3.225𝑚)(
3.225𝑚
2
) −
24.0739𝑘𝑁𝑚 + 20.7069𝑘𝑁𝑚
RC2 = 51.034kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 48.945kN
→ ∑ 𝐻 𝐶2 = ← ∑ 𝐻 𝐷1
= 0

16. Ref. Calculation Output
Span D - E
+↻ ∑ 𝑀 𝐸 = 0
= 1.725𝑚𝑅 𝐷2 − (26.9786 𝑘𝑁 𝑚⁄ )(1.725𝑚)(
1.725𝑚
2
) −
20.7069𝑘𝑁𝑚 + 6.9680𝑘𝑁𝑚
RD2 = 31.234kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RD2 + RE1 = 26.9786 kN/m (1.725m)
RE1 = 15.304kN
→ ∑ 𝐻 𝐷2 = ← ∑ 𝐻 𝐸1
= 0

17. Ref. Calculation Output
Span E – F
+↻ ∑ 𝑀 𝐸 = 0
= 2.175𝑚𝑅 𝐸2 − (27.5181 𝑘𝑁 𝑚⁄ )(2.175𝑚)(
2.175𝑚
2
) −
6.9680𝑘𝑁𝑚 + 7.0638𝑘𝑁𝑚
RE2 = 29.882kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 29.970kN
→ ∑ 𝐻 𝐸2 = ← ∑ 𝐻 𝐹1
= 0

18. Ref. Calculation Output
Span F – G
+↻ ∑ 𝑀 𝐹 = 0
= 1.8𝑚𝑅 𝐹2 − (23.4718𝑘𝑁 𝑚⁄ )(1.8𝑚)(
1.8𝑚
2
) −
7.0638𝑘𝑁𝑚 + 18.6536𝑘𝑁𝑚
RF2 = 14.686kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RF2 + RG1 = 23.4718kN/m (1.8m)
RG1 = 27.563kN
→ ∑ 𝐻 𝐹2 = ← ∑ 𝐻 𝐺1
= 0

19. Ref. Calculation Output
Span G – H
+↻ ∑ 𝑀 𝐺 = 0
= 3.0𝑚𝑅 𝐺2 − (38.2961 𝑘𝑁 𝑚⁄ )(3.0𝑚)(
3.0𝑚
2
) −
18.6536𝑘𝑁𝑚 + 0𝑘𝑁𝑚
RG2 = 63.662kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RG2 + RH = 38.2961kN/m (3.0m)
RH = 1.226kN
→ ∑ 𝐻 𝐺2 = ← ∑ 𝐻 𝐻
= 0

20. Ref. Calculation Output
Summary
At point A,𝑅 𝐴 = 25.436𝑘𝑁
↺ 𝑀𝐴 = 0𝑘𝑁𝑚
At point B,𝑅 𝐵 = 64.197𝑘𝑁
↺ 𝑀 𝐵 = −9.884𝑘𝑁𝑚
↻ 𝑀 𝐵 = +9.884𝑘𝑁𝑚
At point C,𝑅 𝐶 = 93.560𝑘𝑁
↺ 𝑀𝑐 = −24.074𝑘𝑁𝑚
↻ 𝑀𝑐 = +24.074𝑘𝑁𝑚
At point D,𝑅 𝐷 = 80.179𝑘𝑁
↺ 𝑀 𝐷 = −20.707𝑘𝑁𝑚
↻ 𝑀 𝐷 = +20.707𝑘𝑁𝑚
At point E,𝑅 𝐸 = 45.186𝑘𝑁
↺ 𝑀 𝐸 = −6.968𝑁𝑚
↻ 𝑀 𝐸 = +6.968𝑁𝑚
At point F,𝑅 𝐹 = 44.656𝑘𝑁

21. ↺ 𝑀 𝐹 = −7.064𝑘𝑁𝑚
↻ 𝑀 𝐹 = +7.064𝑘𝑁𝑚
At point G,𝑅 𝐺 = 91.225𝑘𝑁
↺ 𝑀 𝐺 = −18.654𝑘𝑁𝑚
↻ 𝑀 𝐺 = +18.654𝑘𝑁𝑚
At point H,𝑅 𝐻 = 51.226𝑘𝑁
↻ 𝑀 𝐻 = 0𝑘𝑁𝑚
Ref. Calculation Output
6.2 Load Distribution of Beam
Case 2: Alternative Loading
FEM
FEMAB = -FEMBA = −
𝑤𝐿2
12
= −
[1.35 (14.5892 𝑘𝑁 𝑚⁄ )+1.50(4.2143𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -11.3448 kNm
FEMBC = -FEMCB = −
𝑤𝐿2
12
= −
[1.35 (17.6179 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -10.3712 kNm

22. FEMCD = -FEMDC = −
𝑤𝐿2
12
= −
[1.35 (17.2189 𝑘𝑁 𝑚⁄ )+1.50(5.1705𝑘𝑁 𝑚⁄ )](3.225𝑚)2
12
= -26.8694 kNm
FEMDE = -FEMED = −
𝑤𝐿2
12
= =−
[1.35(15.0966 𝑘𝑁 𝑚⁄ )](1.725𝑚)2
12
= -5.0537 kNm
FEMEF = -FEMFE = −
𝑤𝐿2
12
= −
[1.35 (15.3812 𝑘𝑁 𝑚⁄ )+1.50(4.5023𝑘𝑁 𝑚⁄ )](2.175𝑚)2
12
= -10.8481 kNm
FEMFG = -FEMGF = −
𝑤𝐿2
12
= −
[1.35 (13.2465 𝑘𝑁 𝑚⁄ )](1.8𝑚)2
12
= -4.8283 kNm
FEMGH = -FEMHG = −
𝑤𝐿2
12
= −
[1.35 (21.0675 𝑘𝑁 𝑚⁄ )+1.50(6.5700𝑘𝑁 𝑚⁄ )](3.0𝑚)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 – B = [1.35(14.5892kN/m) + 1.50(4.2143kN/m)]
= 26.0169kN/m
Loading at Span B – C = [1.35(17.6179kN/m)]
= 23.7842kN/m
Loading at Span C – D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D – E = [1.35(15.0966kN/m)]
= 20.3804kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 26.9786 kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F – G = [1.35(13.2465kN/m)]
= 17.8828kN/m

25. Loading at Span G – H = [1.35(21.0675kN/m) + 1.50(6.5700kN/m)]
= 38.2961 kN/m

26. Ref. Calculation Output
Span A – B
+↻ ∑ 𝑀 𝐵 = 0
= 2.2875𝑚 𝑅 𝐴 − (26.0169 𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) +
7.3648𝑘𝑁𝑚
RA = 26.537kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RA - RB1 = 26.0169kN/m (2.2875m)
RB1 = 32.976kN
→ ∑ 𝐻𝐴 = ← ∑ 𝐻 𝐵1
= 0

27. Ref. Calculation Output
Span B – C
+↻ ∑ 𝑀 𝐶 = 0
= 2.2875𝑚𝑅 𝐵2 −
(23.7842𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) − 7.3648𝑘𝑁𝑚 +
22.4829𝑘𝑁𝑚
RB2 = 20.594kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RB2 + RC1 = 23.7842kN/m (2.2875m)
RC1 = 33.812kN
→ ∑ 𝐻 𝐵2 = ← ∑ 𝐻 𝐶1
= 0

28. Ref. Calculation Output
Span C – D
+↻ ∑ 𝑀 𝐷 = 0
= 3.225𝑚𝑅 𝐶2 − (31.0013 𝑘𝑁 𝑚⁄ )(3.225𝑚)(
3.225𝑚
2
) −
22.4829𝑘𝑁𝑚 + 20.5238𝑘𝑁𝑚
RC2 = 50.597kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 49.382kN
→ ∑ 𝐻 𝐶2 = ← ∑ 𝐻 𝐷1
= 0

29. Ref. Calculation Output
Span D - E
+↻ ∑ 𝑀 𝐸 = 0
= 1.725𝑚𝑅 𝐷2 − (20.3804 𝑘𝑁 𝑚⁄ )(1.725𝑚)(
1.725𝑚
2
) −
20.5238𝑘𝑁𝑚 + 6.0896𝑘𝑁𝑚
RD2 = 25.946kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RD2 + RE1 = 20.3804kN/m (1.725m)
RE1 = 9.210kN
→ ∑ 𝐻 𝐷2 = ← ∑ 𝐻 𝐸1
= 0

30. Ref. Calculation Output
Span E – F
+↻ ∑ 𝑀 𝐸 = 0
= 2.175𝑚𝑅 𝐸2 − (27.5181 𝑘𝑁 𝑚⁄ )(2.175𝑚)(
2.175𝑚
2
) −
6.0896𝑘𝑁𝑚 + 6.4726𝑘𝑁𝑚
RE2 = 9.750kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 30.102kN
→ ∑ 𝐻 𝐸2 = ← ∑ 𝐻 𝐹1
= 0

31. Ref. Calculation Output
Span F – G
+↻ ∑ 𝑀 𝐹 = 0
= 1.8𝑚𝑅 𝐹2 − (17.8828𝑘𝑁 𝑚⁄ )(1.8𝑚)(
1.8𝑚
2
) −
6.4726𝑘𝑁𝑚 + 17.7860𝑘𝑁𝑚
RF2 = 9.809kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RF2 + RG1 = 17.8828kN/m (1.8m)
RG1 = 22.380kN
→ ∑ 𝐻 𝐹2 = ← ∑ 𝐻 𝐺1
= 0

32. Ref. Calculation Output
Span G – H
+↻ ∑ 𝑀 𝐺 = 0
= 3.0𝑚𝑅 𝐺2 − (38.2961 𝑘𝑁 𝑚⁄ )(3.0𝑚)(
3.0𝑚
2
) −
17.7860𝑘𝑁𝑚
RG2 = 63.373kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RG2 + RH = 38.2961kN/m (3.0m)
RH = 51.516kN
→ ∑ 𝐻 𝐺2 = ← ∑ 𝐻 𝐻
= 0

33. Ref. Calculation Output
Summary
At point A,𝑅 𝐴 = 26.537𝑘𝑁
↺ 𝑀𝐴 = 0𝑘𝑁𝑚
At point B,𝑅 𝐵 = 53.571𝑘𝑁
↺ 𝑀 𝐵 = −7.365𝑘𝑁𝑚
↻ 𝑀 𝐵 = +7.365𝑘𝑁𝑚
At point C,𝑅 𝐶 = 84.409𝑘𝑁
↺ 𝑀𝑐 = −22.483𝑘𝑁𝑚
↻ 𝑀𝑐 = +22.483𝑘𝑁𝑚
At point D,𝑅 𝐷 = 75.328𝑘𝑁
↺ 𝑀 𝐷 = −20.524𝑘𝑁𝑚
↻ 𝑀 𝐷 = +20.524𝑘𝑁𝑚
At point E,𝑅 𝐸 = 38.960𝑘𝑁
↺ 𝑀 𝐸 = −6.090𝑘𝑁𝑚
↻ 𝑀 𝐸 = +6.090𝑁𝑚
At point F,𝑅 𝐹 = 39.911𝑘𝑁

34. ↺ 𝑀 𝐹 = −6.473𝑘𝑁𝑚
↻ 𝑀 𝐹 = +6.473𝑘𝑁𝑚
At point G,𝑅 𝐺 = 85.753𝑘𝑁
↺ 𝑀 𝐺 = −17.786𝑘𝑁𝑚
↻ 𝑀 𝐺 = +17.786𝑘𝑁𝑚
At point H,𝑅 𝐻 = 51.516𝑘𝑁
↻ 𝑀 𝐻 = 0𝑘𝑁𝑚
Ref. Calculation Output
6.3 Load Distribution of Beam
Case 3: Adjacent Loading
FEM
FEMAB = -FEMBA = −
𝑤𝐿2
12
= −
[1.35 (14.5892 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -8.5883 kNm
FEMBC = -FEMCB = −
𝑤𝐿2
12
= −
[1.35 (17.6179 𝑘𝑁 𝑚⁄ )+1.50(5.3156𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
12
= -13.8480 kNm

35. FEMCD = -FEMDC = −
𝑤𝐿2
12
= −
[1.35 (17.2189 𝑘𝑁 𝑚⁄ )+1.50(5.1705𝑘𝑁 𝑚⁄ )](3.225𝑚)2
12
= -26.8694 kNm
FEMDE = -FEMED = −
𝑤𝐿2
12
= =−
[1.35(15.0966 𝑘𝑁 𝑚⁄ )](1.725𝑚)2
12
= -5.0537 kNm
FEMEF = -FEMFE = −
𝑤𝐿2
12
= −
[1.35 (15.3812 𝑘𝑁 𝑚⁄ )+1.50(4.5023𝑘𝑁 𝑚⁄ )](2.175𝑚)2
12
= -10.8481 kNm
FEMFG = -FEMGF = −
𝑤𝐿2
12
= −
[1.35 (13.2465 𝑘𝑁 𝑚⁄ )+1.50(3.7260𝑘𝑁 𝑚⁄ )](1.8𝑚)2
12
= -6.3374 kNm
FEMGH = -FEMHG = −
𝑤𝐿2
12
= −
[1.35 (21.0675 𝑘𝑁 𝑚⁄ )](3.0𝑚)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 – B = [1.35(14.5892kN/m)]
= 19.6954kN/m
Loading at Span B – C = [1.35(17.6179kN/m) + 1.50(5.3156kN/m)]
= 31.7576kN/m
Loading at Span C – D = [1.35(17.2189kN/m) + 1.50(5.1705kN/m)]
= 31.0013 kN/m
Loading at Span D – E = [1.35(15.0966kN/m)]
= 20.3804kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181kN/m
Loading at Span E – F = [1.35(15.3812kN/m) + 1.50(4.5023kN/m)]
= 27.5181 kN/m
Loading at Span F – G = [1.35(13.2465kN/m) + 1.50(3.7260kN/m)]
= 23.4718kN/m

38. 38
Loading at Span G – H = [1.35(21.0675kN/m)]
= 28.4411kN/m

39. 39
Ref. Calculation Output
Span A – B
+↻
∑ 𝑀 𝐵
= 0
= 2.2875𝑚 𝑅 𝐴 −
(19.6954 𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) − 0𝑘𝑁𝑚 +
8.5565𝑘𝑁𝑚
RA = 18.786kNm
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RA - RB1 = 19.6954kN/m (2.2875m)
RB1 = 26.267kN
→ ∑ 𝐻𝐴 = ← ∑ 𝐻 𝐵1
= 0

40. 40
Ref. Calculation Output
Span B – C
+↻ ∑ 𝑀 𝐶 = 0
= 2.2875𝑚𝑅 𝐵2 −
(31.7576𝑘𝑁 𝑚⁄ )(2.2875𝑚)(
2.2875𝑚
2
) −
8.5565𝑘𝑁𝑚 + 24.5846𝑘𝑁𝑚
RB2 = 29.316kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RB2 + RC1 = 31.7576kN/m (2.2875m)
RC1 = 43.330kN
→ ∑ 𝐻 𝐵2 = ← ∑ 𝐻 𝐶1
= 0

41. 41
Ref. Calculation Output
Span C – D
+↻ ∑ 𝑀 𝐷 = 0
= 3.225𝑚𝑅 𝐶2 − (31.0013 𝑘𝑁 𝑚⁄ )(3.225𝑚)(
3.225𝑚
2
) −
24.5846𝑘𝑁𝑚 + 19.9097𝑘𝑁𝑚
RC2 = 51.439kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RC2 + RD1 = 31.0013kN/m (3.225m)
RD1 = 48.540kN
→ ∑ 𝐻 𝐶2 = ← ∑ 𝐻 𝐷1
= 0

42. 42
Ref. Calculation Output
Span D - E
+↻ ∑ 𝑀 𝐸 = 0
= 1.725𝑚𝑅 𝐷2 − (20.3804 𝑘𝑁 𝑚⁄ )(1.725𝑚)(
1.725𝑚
2
) −
19.9097𝑘𝑁𝑚 + 5.6838𝑘𝑁𝑚
RD2 = 25.825kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RD2 + RE1 = 20.3804kN/m (1.725m)
RE1 = 9.331kN
→ ∑ 𝐻 𝐷2 = ← ∑ 𝐻 𝐸1
= 0

43. 43
Ref. Calculation Output
Span E – F
+↻ ∑ 𝑀 𝐸 = 0
= 2.175𝑚𝑅 𝐸2 − (27.5181 𝑘𝑁 𝑚⁄ )(2.175𝑚)(
2.175𝑚
2
) −
5.6838𝑘𝑁𝑚 + 8.4135𝑘𝑁𝑚
RE2 = 28.671kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RE2 + RF1 = 27.5181kN/m (2.175m)
RF1 = 31.181kN
→ ∑ 𝐻 𝐸2 = ← ∑ 𝐻 𝐹1
= 0

44. 44
Ref. Calculation Output
Span F – G
+↻ ∑ 𝑀 𝐹 = 0
= 1.8𝑚𝑅 𝐹2 − (23.4718𝑘𝑁 𝑚⁄ )(1.8𝑚)(
1.8𝑚
2
) −
8.4135𝑘𝑁𝑚 + 14.2435𝑘𝑁𝑚
RF2 = 17.886kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RF2 + RG1 = 23.4718kN/m (1.8m)
RG1 = 24.363kN
→ ∑ 𝐻 𝐹2 = ← ∑ 𝐻 𝐺1
= 0

45. 45
Ref. Calculation Output
Span G – H
+↻ ∑ 𝑀 𝐺 = 0
= 3.0𝑚𝑅 𝐺2 − (28.4411 𝑘𝑁 𝑚⁄ )(3.0𝑚)(
3.0𝑚
2
) −
14.2435𝑘𝑁𝑚 + 0𝑘𝑁𝑚
RG2 = 47.410kN
↑ ∑ 𝐹𝑦 = ↓ ∑ 𝐹𝑦
RG2 + RH = 28.4411kN/m (3.0m)
RH = 37.914kN
→ ∑ 𝐻 𝐺2 = ← ∑ 𝐻 𝐻
= 0

46. 46
Ref. Calculation Output
Summary
At point A,𝑅 𝐴 = 18.786𝑘𝑁
↺ 𝑀𝐴 = 0𝑘𝑁𝑚
At point B,𝑅 𝐵 = 55.583𝑘𝑁
↺ 𝑀 𝐵 = −8.557𝑘𝑁𝑚
↻ 𝑀 𝐵 = +8.557𝑘𝑁𝑚
At point C,𝑅 𝐶 = 94.769𝑘𝑁
↺ 𝑀𝑐 = −24.585𝑘𝑁𝑚
↻ 𝑀𝑐 = +24.585𝑘𝑁𝑚
At point D,𝑅 𝐷 = 74.365𝑘𝑁
↺ 𝑀 𝐷 = −19.910𝑘𝑁𝑚
↻ 𝑀 𝐷 = +19.910𝑘𝑁𝑚
At point E,𝑅 𝐸 = 38.002𝑘𝑁
↺ 𝑀 𝐸 = −5.684𝑘𝑁𝑚
↻ 𝑀 𝐸 = +5.684𝑘𝑁𝑚
At point F,𝑅 𝐹 = 49.067𝑘𝑁
↺ 𝑀 𝐹 = −8.413𝑘𝑁𝑚

47. 47
↻ 𝑀 𝐹 = +8.413𝑘𝑁𝑚
At point G,𝑅 𝐺 = 71.773𝑘𝑁
↺ 𝑀 𝐺 = −14.243𝑘𝑁𝑚
↻ 𝑀 𝐺 = +14.243𝑘𝑁𝑚
At point H,𝑅 𝐻 = 37.914𝑘𝑁
↻ 𝑀 𝐻 = 0𝑘𝑁𝑚

48. 48
Ref. Calculation Output
Distribution Factor, DF
KAB = KBA =
4𝐸𝐼
𝐿
=
4𝐸𝐼
2.2875
= 1.7486 EI
KBC = KCB =
4𝐸𝐼
𝐿
=
4𝐸𝐼
2.2875
= 1.7486 EI
KCD = KDC =
4𝐸𝐼
𝐿
=
4𝐸𝐼
3.225
= 1.2403 EI
KDE = KDE =
4𝐸𝐼
𝐿
=
4𝐸𝐼
1.725
= 2.3188 EI
KEF = KF =
4𝐸𝐼
𝐿

49. 49
=
4𝐸𝐼
2.175
= 1.8391 EI
KFG = KGF =
4𝐸𝐼
𝐿
=
4𝐸𝐼
1.8
= 2.2222 EI
KGH = KHG =
4𝐸𝐼
𝐿
=
4𝐸𝐼
3.0
= 1.3333 EI
Ref. Calculation Output
DFAB = 𝐾 𝐴𝐵
𝐾 𝐴𝐵 +0
= 1.7468 𝐸𝐼
1.7468 𝐸𝐼+0
= 1
DFBA = 𝐾 𝐵𝐴
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
= 1.7468𝐸𝐼
1.7468 𝐸𝐼+1.7468 𝐸𝐼
= 0.5
DFBC = 𝐾 𝐵𝐶
𝐾 𝐵𝐴 + 𝐾 𝐵𝐶
= 1.7468 𝐸𝐼
1.7468 𝐸𝐼+1.7468 𝐸𝐼
= 0.5
DFCB = 𝐾 𝐶𝐵
𝐾 𝐶𝐵 + 𝐾 𝐶𝐷
= 1.7468 𝐸𝐼
1.7468 𝐸𝐼+1.2403 𝐸𝐼
= 0.585
DFCD = 𝐾 𝐶𝐷
𝐾 𝐶𝐵 +𝐾 𝐶𝐷
= 1.2403 𝐸𝐼
1.7468 𝐸𝐼+1.2403 𝐸𝐼
= 0.415
DFDC = 𝐾 𝐷𝐶
𝐾 𝐷𝐶+ 𝐾 𝐷𝐸
= 1.2403 𝐸𝐼
1.2403 𝐸𝐼+2.3188 𝐸𝐼
= 0.348
5
DFDE = 𝐾 𝐷𝐸
𝐾 𝐷𝐶+ 𝐾 𝐷𝐸
= 2.3188 𝐸𝐼
1.2403 𝐸𝐼+2.3188 𝐸𝐼
= 0.651
5
DFED = 𝐾 𝐸𝐷
𝐾 𝐸𝐷 + 𝐾 𝐸𝐹
= 2.3188 𝐸𝐼
2.3188 𝐸𝐼+1.8391 𝐸𝐼
= 0.557
7
DFEF = 𝐾 𝐸𝐹
𝐾 𝐸𝐷 + 𝐾 𝐸𝐹
= 1.8391 𝐸𝐼
2.3188 𝐸𝐼+1.8391 𝐸𝐼
= 0.442
3
DFFE = 𝐾 𝐹𝐸
𝐾 𝐹𝐸+ 𝐾 𝐹𝐺
= 1.8391 𝐸𝐼
1.8391 𝐸𝐼+2.2222 𝐸𝐼
= 0.452
8
DFFG = 𝐾 𝐹𝐺
𝐾 𝐹𝐸+ 𝐾 𝐹𝐺
= 2.2222 𝐸𝐼
1.8391 𝐸𝐼+2.2222 𝐸𝐼
= 0.547
2

50. 50
DFGF = 𝐾 𝐺𝐹
𝐾 𝐺𝐹 + 𝐾 𝐺𝐻
= 2.2222 𝐸𝐼
2.2222 𝐸𝐼+1.3333 𝐸𝐼
= 0.625
DFGH = 𝐾 𝐺𝐻
𝐾 𝐺𝐹 + 𝐾 𝐺𝐻
= 1.3333 𝐸𝐼
2.2222 𝐸𝐼+1.3333 𝐸𝐼
= 0.375
DFHG = 𝐾 𝐻𝐺
𝐾 𝐻𝐺 + 0
= 1.3333 𝐸𝐼
1.3333 𝐸𝐼+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 :
Øbar = 16mm
Ølink = 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 = 𝑎 𝑠𝑑 − ∅𝑙𝑖𝑛𝑘 − ∅ 𝑏𝑎𝑟 2⁄
= 39mm
Allowance in design for deviation, Δ Cdev = 10mm
Nominal cover, Cnom = Cmin + Δ Cdev
= 49mm Cnom
∴ Cnom = 51mm 51mm
4. Loading and Analysis
Loading on Slab 15:
Slab Thickness = 0.16m
Concrete Unit Weight = 25kN/m3
Self-weight = 25kN/m3 × 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
𝑙 𝑦
𝑙 𝑥
= 1.33 ˂ 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 ×
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
𝑙 𝑦
𝑙 𝑥
= 1.33 ˂ 2.0 (2 way slab)
Load on beam 2-3/D-F

57. 57

58. 58
Ref. Calculation Output
Characteristic permanent load, Gk
Beam self-weight,
βvy = 0.36 and 0.33
Beam self-weight
w0 = 0.2m × 0.35m × 25kN/m3
= 1.75 kN/m
w15 = βvynlx
= 0.36 x 5.5kN/m2 x 1.8m
= 3.564kN/m
w16 = βvynlx
= 0.33 x 6.75kN/m2 x 1.2m
= 2.673kN/m
= Brickwall
w1 = 2.6kN/m2 × 3.0m
= 7.8
Total GK = 15.787kN/m
Characteristic variable load, Qk
w15 = βvynlx
= 0.36 x 3.0kN/m2 x 1.8m
= 1.944kN/m
w16 = βvynlx
= 0.33 x 3.0kN/m2 x 1.2m
= 1.188kN/m

59. 59
Total QK = 3.132kNm

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 = 𝑊 𝑑 𝐿
2
V
= 26.01 𝑘𝑁 𝑚⁄ ×1.8𝑚
2
23.410
= 23.410kN kN
Bending Moment, 𝑀 = 𝑊 𝑑 𝐿2
8
M
= (26.01𝑘𝑁 𝑚⁄ )(1.8𝑚)2
8
10.534
= 10.534kNm kNm

61. 61
Ref. Calculation Output
Effective depth
d = h - Cnom - Ølink - Øbar
= 350mm-50mm-8mm-16mm d
= 278mm 278mm
d' = Cnom + Ølink +
∅ 𝑏𝑎𝑟
2
= 36mm
Design bending moment, Med = 7.153kNm
𝐾 = 𝑀
𝑏𝑑2 𝑓 𝑐𝑘
= 10 .534 ×106
𝑁𝑚𝑚2
(200𝑚𝑚 )(278𝑚𝑚)2(30𝑁/𝑚𝑚2 )
= 0.0227
Kbal = 0.167
K < Kbal ; Compression reinforcement is not required
z =
𝑑[0.5 + √0.25 −
𝐾 𝑏𝑎𝑙
1.134
]
z = 0.82𝑑
z = 227.96mm
Area of Tension steel, As
𝐴 𝑠 = 𝐾 𝑏𝑎𝑙 𝑓 𝑐𝑘 𝑏𝑑2
0.87𝑓 𝑦𝑘 𝑧
+ 𝐴 𝑠′
= (0.167)(30𝑁 𝑚𝑚2⁄ )(200𝑚𝑚)(278𝑚𝑚 )2
0.87 (500𝑁 𝑚𝑚2⁄ )(226.32𝑚𝑚)
+ 𝐴 𝑠′
= 780.925mm2

62. 62
Min and max reinforcement area, As,min
𝐴 𝑠,𝑚𝑖𝑛 = 0.26
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
𝑏𝑑
=
0.26
2.90𝑁 𝑚𝑚2⁄
500𝑁 𝑚𝑚2⁄
𝑏𝑑
= 0.001508 bd
Exceed 0.0013bd; use 0.0015bd
= 83.844mm2
Ref. Calculation Output
As,max = 0.04Ac
= 0.04 × 200mm × 350mm
= 2800mm2
5. Shear Reinforcement
Design shear force, Ved = 23.410kN
Concrete strut capacity
𝑉𝑅𝑑,𝑚𝑎𝑥 = 0.3𝑏 𝑤 𝑑𝑓𝑐𝑘(1−
𝑓 𝑐𝑘
250
)
(𝑐𝑜𝑡𝜃+𝑡𝑎𝑛𝜃)
= 0.3(200𝑚𝑚)(278𝑚𝑚)(30𝑁 𝑚𝑚2⁄ )(1−
30𝑁 𝑚𝑚2⁄
250
)
(𝑐𝑜𝑡𝜃+𝑡𝑎𝑛𝜃)
= 181.962𝑘𝑁 (𝜃 = 22°, 𝑐𝑜𝑡𝜃 = 2.5)
= 264.211𝑘𝑁 (𝜃 = 45°, 𝑐𝑜𝑡𝜃 = 1.0)
Ved < VRd,maxcot 𝜃= 2.5
Ved < VRd,maxcot 𝜃= 1.0

63. 63
∴ 𝜽 < 𝟐𝟐°
𝜃 = 0.5 sin−1
[
𝑉 𝐸𝑑
0.18𝑏 𝑤 𝑑𝑓𝑐𝑘(1−
𝑓 𝑐𝑘
250
)
]
= 0.5 sin−1
[
18.558×103 𝑁
0.18(200𝑚𝑚)(278𝑚𝑚)(30𝑁 𝑚𝑚2⁄ )(1−
30𝑁 𝑚𝑚2⁄
250
)
]
= 2.01°
Use 𝜃 = 22°
𝑡𝑎𝑛𝜃 = 0.40, 𝑐𝑜𝑡𝜃 = 2.5
Shear Link,
𝐴 𝑠𝑤
𝑠
=
𝑉 𝐸𝑑
(0.78𝑓𝑦𝑘 𝑑𝑐𝑜𝑡𝜃)
= 23.410×103 𝑁
(0.78)(500𝑁 𝑚𝑚2⁄ )(278𝑚𝑚)(2.5)
= 0.0864mm
Try link: H6 𝐴 𝑠𝑤=28.3mm2
Spacing, s = 28.3 𝑚𝑚2
0.0864 𝑚𝑚
= 327.67mm

64. 64
Ref. Calculation Output
Maximum spacing, = 0.75d
Smax = 0.75(278mm)
= 208.5mm
Use: H6 - 200 H6 - 200
Minimum links,
𝐴 𝑠𝑤
𝑠
= 0.08√ 𝑓𝑐𝑘
𝑏 𝑤
𝑓 𝑦𝑘
= 0.08√30 𝑁 𝑚𝑚2⁄
200𝑚𝑚
500𝑁 𝑚𝑚2⁄
= 0.175 mm
Spacing, s = 28.3 𝑚𝑚2
0.175 𝑚𝑚
= 161.454mm < 0.75d
Use: H6 - 175 H6 - 175
Minimum Links,
𝐴 𝑠𝑤
𝑠
= 0.08√ 𝑓𝑐𝑘
𝑏 𝑤
𝑓 𝑦𝑘
= 0.08√30 𝑁 𝑚𝑚2⁄
200𝑚𝑚
500𝑁 𝑚𝑚2⁄
= 0.175mm
Spacing, s = 28.3mm2/0.175mm
= 161.464mm ˂ 0.75d
Use: H6 - 175
Shear resistance of minimum links

65. 65
𝑉 𝑚𝑖𝑛 =
𝐴 𝑠𝑤
𝑠
(0.78𝑓𝑦𝑘 𝑑𝑐𝑜𝑡𝜃)
= 28.3𝑚𝑚2
200 𝑚𝑚
(0.78)(500𝑁 𝑚𝑚2⁄ )(278𝑚𝑚)(2.5)
= 43.833kN

66. 66
Ref. Calculation Output
Link Arrangement

67. 67
Ref. Calculation Output
6. Deflection
Percentage of required tension reinforcement,
𝜌
=
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 775.307𝑚𝑚2
(200𝑚𝑚)(278𝑚𝑚)
= 0.01405
Reference reinforcement ratio,
𝜌°
= √ 𝑓𝑐𝑘 × 10−3
= √30 𝑁 𝑚𝑚2⁄ × 10−3
= 0.00548
Percentage of required compression reinforcement,
𝜌′ = 0
𝜌 > 𝜌° ; Use equation:
𝑙
𝑑
= 𝐾[11 + 1.5√ 𝑓𝑐𝑘
𝜌°
𝜌−𝜌′ +
1
12
√ 𝑓𝑐𝑘√
𝜌′
𝜌°
𝑙
𝑑
= 1.0[11+ 1.5√30 𝑁 𝑚𝑚2⁄
0.00548
0.01405−0
+
1
12
√30 𝑁 𝑚𝑚2⁄ √
0
0.00548
]
𝑙
𝑑
= 1.0[11+ 0.5549+ 0]
𝑙
𝑑
= 11.555
Modification factor less than 7m = 1.00
Modification factor for steel area provided,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
=
943𝑚𝑚2
780.925𝑚𝑚2

68. 68
= 1.208 <1.5 OK!
Allowable span effective depth ratio = (l/d) allowable
= 11.555 × 1.208 × 1.00
= 13.954
Actual Span-effective depth, (l/d)
actual
=
1800 𝑚𝑚
278 𝑚𝑚
= 6.475 < (l/d) allowable OK!
Ref. Calculation Output
7. Cracking
Limiting crack width, wmax = 0.3mm
Steel stress,
𝑓𝑠
= 𝑓𝑦𝑘
1.15
[
𝐺 𝑘+0.3𝑄 𝑘
1.35𝐺 𝑘+1.5𝑄 𝑘
]
1
𝛿
= 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

69. 69
Bar spacing,
S
= 200mm- 2(50mm) -2(6mm)-16mm
= 72mm < 100mm OK!
8. Detailing

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm

71. 71
∅𝑙𝑖𝑛𝑘 = 6mm

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

73. 73

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

75. 75

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

77. 77

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

79. 79

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

81. 81

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
∅ 𝑏𝑎𝑟 1 = 12mm
∅ 𝑏𝑎𝑟 2 = 8mm
∅𝑙𝑖𝑛𝑘 = 6mm

83. 83

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 = 𝑎 𝑠𝑑 − ∅𝑙𝑖𝑛𝑘 −
∅ 𝑏𝑎𝑟
2
= 45mm – 6mm –
0.5(12mm)
= 33mm
Allowance in design for deviation, ∆Cdev = 10mm
Nominal cover
Cnom = Cmin + ∆Cdev
= 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(14.5892𝑘𝑁 𝑚⁄ ) +
0.10(4.2143 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
= 12.583kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(14.5892𝑘𝑁 𝑚⁄ ) +
0.106(4.2143 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
= 14.4306kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 26.0468 𝑘𝑁 𝑚⁄ × 2.2875𝑚
= 26.78kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 26.0468 𝑘𝑁 𝑚⁄ × 2.2875𝑚

86. 86
= 37.49kN

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(17.6179𝑘𝑁 𝑚⁄ ) +
0.10(5.3156 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
= 15.373kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(17.6179𝑘𝑁 𝑚⁄ ) +
0.106(5.3156 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
= 17.6147kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 31.7576 𝑘𝑁 𝑚⁄ × 2.2875𝑚
= 32.6904kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 31.7576 𝑘𝑁 𝑚⁄ × 2.2875𝑚
= 45.7666kN

88. 88

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(17.2189𝑘𝑁 𝑚⁄ ) +
0.10(5.1705 𝑘𝑁 𝑚⁄ )](3.225𝑚)2
= 29.8260kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(17.2189𝑘𝑁 𝑚⁄ ) +
0.106(5.1705 𝑘𝑁 𝑚⁄ )](3.225𝑚)2
= 34.1779kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 31.0013 𝑘𝑁 𝑚⁄ × 3.225𝑚
= 44.9906kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 31.0013 𝑘𝑁 𝑚⁄ × 3.225𝑚
= 62.9868kN

90. 90

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(15.0966𝑘𝑁 𝑚⁄ ) +
0.10(4.3988 𝑘𝑁 𝑚⁄ )](1.725𝑚)2
= 7.421kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(15.0966𝑘𝑁 𝑚⁄ ) +
0.106(4.3988 𝑘𝑁 𝑚⁄ )](1.725𝑚)2
= 8.5095kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 26.9786 𝑘𝑁 𝑚⁄ × 1.725𝑚
= 20.9422kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 26.9786 𝑘𝑁 𝑚⁄ × 1.725𝑚
= 29.319kN

92. 92

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(15.3812𝑘𝑁 𝑚⁄ ) +
0.10(4.5023 𝑘𝑁 𝑚⁄ )](2.715𝑚)2
= 12.035kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(15.3812𝑘𝑁 𝑚⁄ ) +
0.106(4.5023 𝑘𝑁 𝑚⁄ )](2.715𝑚)2
= 13.7988kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 27.5181 𝑘𝑁 𝑚⁄ × 2.715𝑚
= 26.9333kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 27.5181 𝑘𝑁 𝑚⁄ × 2.715𝑚
= 37.7066kN

94. 94

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(13.2465𝑘𝑁 𝑚⁄ ) +
0.10(3.726 𝑘𝑁 𝑚⁄ )](1.8𝑚)2
= 7.025kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(13.2465𝑘𝑁 𝑚⁄ ) +
0.106(3.726 𝑘𝑁 𝑚⁄ )](1.8𝑚)2
= 8.0611kNm
Shear Force
Outer Support, VED = 0.45 × wd × L
= 0.45 × 23.4718 𝑘𝑁 𝑚⁄ × 1.8𝑚
= 19.0121kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 23.4718 𝑘𝑁 𝑚⁄ × 1.8𝑚
= 26.6170kN

96. 96

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(21.0675𝑘𝑁 𝑚⁄ ) +
0.10(6.57 𝑘𝑁 𝑚⁄ )](3.0𝑚)2
= 31.907kNm
Support, MED = (0.106GK + 0.106QK)L2
= [0.106(21.0675𝑘𝑁 𝑚⁄ ) +
0.106(6.57 𝑘𝑁 𝑚⁄ )](2.2875𝑚)2
= 36.5345kNm
Shear Force
Outer Support,
VED
= 0.45 × wd × L
= 0.45 × 38.2961 𝑘𝑁 𝑚⁄ × 3.0𝑚
= 51.6998kN
Inner Support, VED = 0.63 × wd × L
= 0.63 × 38.2961 𝑘𝑁 𝑚⁄ × 3.0𝑚
= 72.3787kN

98. 98

99. 99
Ref. Calculation Output
4. Main Reinforcement
Effective Depth, d = ℎ − 𝐶 𝑛𝑜𝑚 − ∅𝑙𝑖𝑛𝑘 − 0.5∅ 𝑏𝑎𝑟
= 400mm – 60mm – 6mm – 0.5(12mm)
= 328mm
d’ = 𝐶 𝑛𝑜𝑚 + ∅𝑙𝑖𝑛𝑘 + 0.5∅ 𝑏𝑎𝑟
= 72mm

100. 100
Ref. Calculation Output
Span AB
Bending Moment = 12.583kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 12.583 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.013 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.013
1.134⁄ ]
= 0.9884d
= 0.9884 × 328mm
= 324.20mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 12.583 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9884 ×328𝑚𝑚
2H10
= 89.23mm2 157mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd

101. 101
= 0.0015 × 300mm × 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

102. 102
Ref. Calculation Output
Span BC
Bending Moment = 15.373kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 15.373 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.016 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.016
1.134⁄ ]
= 0.9857d
= 0.9857 × 328mm
= 324.20mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 15.373 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9857 ×328𝑚𝑚
2H10
= 109.31mm2 157mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd

103. 103
= 0.0015 × 300mm × 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

104. 104
Ref. Calculation Output
Span CD
Bending Moment = 29.826kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 29.826 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.031 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.031
1.134⁄ ]
= 0.9719d
= 0.9719 × 328mm
= 318.77mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 29.826 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9719 ×328𝑚𝑚
3H10
= 215.09mm2 236mm2
Minimum and maximum reinforcement area
As,min = 0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd

105. 105
= 0.0015 × 300mm × 328mm
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

106. 106
Ref. Calculation Output
Span DE
Bending Moment = 7.421kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 7.421 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.008 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.008
1.134⁄ ]
= 0.9929d
= 0.9929 × 328mm
= 325.67mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 7.421 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9929 ×328𝑚𝑚
2H10
= 52.38mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

107. 107
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

108. 108
Ref. Calculation Output
Span EF
Bending Moment = 12.035kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 12.035 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.012 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.012
1.134⁄ ]
= 0.9893d
= 0.9893 × 328mm
= 324.49mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 12.035 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9893 ×328𝑚𝑚
2H10
= 85.26mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

109. 109
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

110. 110
Ref. Calculation Output
Span FG
Bending Moment = 7.025kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 7.025 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.007 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.007
1.134⁄ ]
= 0.9938d
= 0.9938 × 328mm
= 325.96mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 7.025 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9938 ×328𝑚𝑚
2H10
= 49.54mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

111. 111
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

112. 112
Ref. Calculation Output
Span GH
Bending Moment = 31.907kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 31.907 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.033 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.033
1.134⁄ ]
= 0.9700d
= 0.9700 × 328mm
= 318.16mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 31.907 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9700 ×328𝑚𝑚
3H10
= 230.54mm2 236mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

113. 113
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

114. 114
Ref. Calculation Output
Support B
Bending Moment = 17.6147kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 17.6147 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.018 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.018
1.134⁄ ]
= 0.9839d
= 0.9839 × 328mm
= 322.71mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 17.6147 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9839 ×328𝑚𝑚
2H10
= 125.48mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

115. 115
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

116. 116
Ref. Calculation Output
Support C
Bending Moment = 34.1779kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 34.1779 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.035 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.035
1.134⁄ ]
= 0.9681d
= 0.9681 × 328mm
= 322.71mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 34.1779 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9681 ×328𝑚𝑚
4H10
= 247.44mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

117. 117
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

118. 118
Ref. Calculation Output
Support D
Bending Moment = 34.1779kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 34.1779 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.035 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.035
1.134⁄ ]
= 0.9681d
= 0.9681 × 328mm
= 322.71mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 34.1779 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9681 ×328𝑚𝑚
4H10
= 247.44mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

119. 119
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

120. 120
Ref. Calculation Output
Support E
Bending Moment = 13.7988kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 13.7988 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.014 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.014
1.134⁄ ]
= 0.9875d
= 0.9875 × 328mm
= 322.71mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 13.7988 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9875 ×328𝑚𝑚
2H10
= 97.94mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

121. 121
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

122. 122
Ref. Calculation Output
Support F
Bending Moment = 13.7988kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 13.7988 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.014 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.014
1.134⁄ ]
= 0.9875d
= 0.9875 × 328mm
= 322.71mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 13.7988 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9875 ×328𝑚𝑚
2H10
= 97.94mm2 157mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

123. 123
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

124. 124
Ref. Calculation Output
Support G
Bending Moment = 36.5345kNm
K = 𝑀 𝐸𝐷
𝑓 𝑐𝑘 𝑏𝑑2
= 36.5345 × 106
𝑁𝑚𝑚
30𝑁 𝑚𝑚2⁄ ×300𝑚 × (328𝑚𝑚)2
= 0.038 ˂ Kbal(0.167)
z = 𝑑 [0.5 + √0.25 − 𝐾
1.134⁄ ]
= 𝑑 [0.5 + √0.25 − 0.038
1.134⁄ ]
= 0.9653d
= 0.9653 × 328mm
= 316.61mm
Area of Tension Reinforcement
As = 𝑀 𝐸𝐷
0.87 × 𝑓 𝑦𝑘 ×𝑧
= 36.5345 × 106
𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.9653 ×328𝑚𝑚
4H10
= 265.26mm2 314mm2
Minimum and maximum reinforcement area
As,min =
0.26 (
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
) 𝑏𝑑
=
0.26 × (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) 𝑏𝑑
= 0.0015bd
= 0.0015 × 300mm × 328mm

125. 125
= 148.39mm2
As,max = 0.04Ac
= 0.04bh
= 0.04 × 300mm × 400mm
= 4800mm2

126. 126
Ref. Calculation Output
5. Shear Reinforcement
Concrete strut capacity
VRd,max = 0.36𝑏 𝑤 𝑑𝑓𝑐𝑘 (1−
𝑓𝑐𝑘
250⁄ )
cot𝜃+ tan 𝜃
= 322.48kN ( 𝜃 = 22° ,cot 𝜃 = 2.5 )
= 467.60kN( 𝜃 = 45° , cot 𝜃 = 1.0)
Support A and H
VED at A = 26.537kN
VED at H = 51.516kN
Shear links
𝐴 𝑠𝑤
𝑠
= 𝑉 𝐸𝐷
0.78𝑓 𝑦𝑘 𝑑 cot 𝜃
For Support A = 26 .537 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.083mm
For Support H = 51 .516 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.161mm
Try link H6 (Asw = 28.3mm2)
For Support A, s = 28.3𝑚𝑚2
0.083𝑚𝑚
= 341.05mm H6 - 250
For Support H, s = 28.3𝑚𝑚2
0.161𝑚𝑚
= 175mm H6 - 175

127. 127
Maximum Spacing = 0.75d
= 0.75 × 328mm
= 246mm
H6 - 250

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 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.107mm
For Support C = 51 .439 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.107mm
For Support D = 49.382 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.154mm

129. 129
For Support E = 29.882 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.093mm
For Support F = 31.181 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.098mm
For Support G = 63.662 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.199mm

130. 130
Ref. Calculation Output
Shear link for lowest shear using
For Support B = 30 .119 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.094mm
For Support C = 43.330 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.135mm
For Support D = 31.234 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.098mm
For Support E = 15.304 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.048mm
For Support F = 17.886 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.056mm
For Support G = 27.563 × 103
𝑁
0.78 ×500𝑁 𝑚𝑚2⁄ ×328𝑚𝑚 ×2.5
= 0.086mm
Try links H6 (Asw = 28.3mm2)
For highest shear using, spacing
For Support B, s = 28.3𝑚𝑚2
0.107𝑚𝑚
= 265.58mm H6 – 250
For Support C, s = 28.3𝑚𝑚2
0.161𝑚𝑚
= 175.94mm H6 – 150
For Support D, s = 28.3𝑚𝑚2
0.154𝑚𝑚
= 183.27mm H6 – 175

131. 131
For Support E, s = 28.3𝑚𝑚2
0.093𝑚𝑚
= 302.87mm H6 – 250
For Support F, s = 28.3𝑚𝑚2
0.098𝑚𝑚
= 290.25mm H6 – 250
For Support G, s = 28.3𝑚𝑚2
0.199𝑚𝑚
= 142.16mm H6 - 100

132. 132
Ref. Calculation Output
For lowest shear using, spacing
For Support B, s = 28.3𝑚𝑚2
0.094𝑚𝑚
= 300.49mm H6 – 250
For Support C, s = 28.3𝑚𝑚2
0.135𝑚𝑚
= 208.87mm H6 - 200
For Support D, s = 28.3𝑚𝑚2
0.098𝑚𝑚
= 289.76mm H6 – 250
For Support E, s = 28.3𝑚𝑚2
0.048𝑚𝑚
= 591.37mm H6 - 250
For Support F, s = 28.3𝑚𝑚2
0.056𝑚𝑚
= 506.00mm H6 - 250
For Support G, s = 28.3𝑚𝑚2
0.086𝑚𝑚
= 328.35mm H6 - 250
Minimum links
𝐴 𝑠𝑤
𝑠
= 0.08𝑓𝑐𝑘
1
2 𝑏 𝑤
𝑓𝑦𝑘
= 0.08 × (30 𝑁 𝑚𝑚2⁄ )
1
2 ×300𝑚𝑚
500𝑁 𝑚𝑚2⁄
= 0.262mm
Try link H6 (Asw = 28.3mm2)
Spacing, s = 28.3 𝑚𝑚2
0.262𝑚𝑚
= 108.02mm H6 - 100

133. 133
Shear resistance of minimum links
Vmin = (
𝐴 𝑠𝑤
𝑠
) (0.78𝑑𝑓𝑦𝑘 cot 𝜃)
= (28.3𝑚𝑚2
100 𝑚𝑚
)(0.78 × 328𝑚𝑚 × 500𝑁 𝑚𝑚2⁄ × 2.5)
= 90.50kN
Since the min. shear resistance is higher than every shear force
calculated,
∴ 𝑼𝒔𝒆 𝑯𝟔 − 𝟏𝟎𝟎 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒂𝒍𝒍 𝒔𝒑𝒂𝒏

134. 134
Ref. Calculation Output
6. Deflection
Check at Span AB
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 89.23𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0009
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 ×
10−3
= (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0009
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0009
−
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
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 157𝑚𝑚2
89.23𝑚𝑚2
= 1.75
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 14.4999 × 1.0 × 1.5
= 21.7499
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 2287.5𝑚𝑚
328𝑚𝑚
= 6.9741 OK
Ref. Calculation Output
Check at Span BC
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 109.31𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0011
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 ×
10−3
= (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0

136. 136
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0011
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0011
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 157𝑚𝑚2
109.31𝑚𝑚2
= 1.44
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 14.5318 × 1.0 × 1.44
= 20.9258
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 2287.5𝑚𝑚
328𝑚𝑚
= 6.9741 OK

137. 137
Ref. Calculation Output
Check at Span CD
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 215.09𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0022
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 × 10−3 = (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0022
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0022
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 236𝑚𝑚2
215.09𝑚𝑚2

138. 138
= 1.10
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 14.3688 × 1.0 × 1.10
= 15.8057
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 3225𝑚𝑚
328𝑚𝑚
= 9.8323 OK

139. 139
Ref. Calculation Output
Check at Span DE
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 52.38𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0005
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 × 10−3 = (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0005
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0005
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 157𝑚𝑚2
52.38𝑚𝑚2

140. 140
= 2.997
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 15.1416 × 1.0 × 1.5
= 22.7124
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 1725𝑚𝑚
328𝑚𝑚
= 5.26 OK

141. 141
Ref. Calculation Output
Check at Span EF
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 85.26𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0009
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 × 10−3 = (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0009
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0009
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 157𝑚𝑚2
85.26𝑚𝑚2

142. 142
= 1.84
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 14.6299 × 1.0 × 1.5
= 21.9449
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 2175𝑚𝑚
328𝑚𝑚
= 8.2774 OK

143. 143
Ref. Calculation Output
Check at Span FG
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 49.54𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0005
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 × 10−3 = (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0005
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0005
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 157𝑚𝑚2
49.54𝑚𝑚2

144. 144
= 3.17
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 15.1416 × 1.0 × 1.5
= 22.7124
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 1800𝑚𝑚
328𝑚𝑚
= 5.4878 OK

145. 145
Ref. Calculation Output
Check at Span GH
Percentage of required tension reinforcement
𝜌 =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 230.54𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0.0023
Reference reinforcement ratio
𝜌0 = ( 𝑓𝑐𝑘
)
1
2 ×
10−3
= (30 𝑁 𝑚𝑚2⁄ )
1
2 × 10−3
= 0.0055
Percentage of required compression reinforcement
𝜌′ = 𝐴 𝑠,𝑟𝑒𝑞
′
𝑏𝑑
= 0𝑚𝑚2
300𝑚𝑚 ×328𝑚𝑚
= 0
Factor for structural system, K = 1.3
𝜌 < 𝜌0
𝑙
𝑑
= 𝐾 [11 + 1.5 √ 𝑓𝑐 𝑘
𝜌0
𝜌
+ 3.2 √ 𝑓𝑐 𝑘 (
𝜌0
𝜌
− 1)
3
2
]
= 1.3 [11 + 1.5 √30𝑁 𝑚𝑚2⁄ × 10−3 0.0055
0.0023
+ 3.2 √30𝑁 𝑚𝑚2⁄ × 10−3 (0.0055
0.0023
− 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,
=
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 236𝑚𝑚2
230.54𝑚𝑚2

146. 146
= 1.02
Therefore allowable span-effective depth ratio
(
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 14.3632 × 1.0 × 1.02
= 14.6557
Actual span-effective depth
(
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
= 3000𝑚𝑚
328𝑚𝑚
= 9.1463 OK

147. 147
Ref. Calculation Output
7. Cracking
Limiting cracking width,
wmax
= 0.3mm
Steel Stress, fs = 𝑓𝑦𝑘
1.15
×
𝐺 𝑘+0.3𝑄 𝑘
(1.35𝐺 𝑘+1.50𝑄 𝑘)
1
𝛿
= 500𝑁 𝑚𝑚2⁄
1.15
×
21.0675𝑘𝑁 𝑚⁄ +0.30(6.57𝑘𝑁 𝑚⁄ )
38.2961𝑘𝑁 𝑚⁄
× 1.0
= 261.56mm2
Maximum allowing spacing = 150mm
Maximum allowing bar = 12mm
Bar spacing, s = 𝑏−2𝐶 𝑛𝑜𝑚−2∅𝑙𝑖 𝑛 𝑘− ∅ 𝑏𝑎𝑟
2
= 300𝑚−2(60𝑚𝑚)−2(6𝑚𝑚)−10𝑚𝑚
2
= 79mm

148. 148
Ref. Calculation Output
8. Detailing

149. 149

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: ∅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-∅bar/2 = 40mm-20mm/2
= 30mm
Allowance in design for deviation, ∆Cdev = 10mm
Nominal cover, Cnom = Cmin + ∆Cdev = 30mm + 10mm
= 40mm
∴ Cnom = 40mm
Ref. Calculation Output
4.0 Loading and Analysis
Slab self-
weight
= 0.14 × 25kN/m3 = 3.5kN/m2
Permanent load (excluding self –weight) = 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 ×
2m
= 26.55 kN/m
Shear Force, V = 𝑊 𝑑 𝐿
2
= (26.55 𝑘𝑁 𝑚⁄ ) × (3.125𝑚)
2
= 41.48 kN
Bending Moment, M = 𝑊 𝑑 𝐿2
8
= (26.55 𝑘𝑁 𝑚⁄ ) × (3.125𝑚)2
8
= 32.41 kNm
5.0 Main Reinforcement
Effective depth, d = h-Cnom-∅bar/2
= 140mm-40mm-(20mm/2)
= 90mm
Design bending moment, MEd = 32.41 kNm
= 𝑀 𝐸𝐷
𝑏𝑑2 𝑓 𝑐𝑘
= 32.41 × 106
𝑁𝑚𝑚
2000𝑚𝑚 × (90𝑚𝑚)2 ×30 𝑁 𝑚𝑚2⁄
= 0.032 < Kbal = 0.167
∴ Compression reinforcement is not required

153. 153
Ref. Calculation Output
z = d[0.5+√0.25 −
𝐾
1.134
]
z = d[0.5+√0.25 −
0.032
1.134
]
z = 0.97d > 0.95d
Use 0.95d 0.95d
As =
𝑀 𝐸𝐷
0.89𝑓𝑦𝑘 𝑧
= 32.41 × 106 𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.95 ×90𝑚𝑚
= 871.41 mm2
Main bar: H12-200 (566mm2/m) H12 - 200
Min. and max. reinforcement area,
As, min = 0.26(
𝑓𝑐𝑡𝑚
𝑓 𝑦𝑘
)bd
= 0.26x (
2.90 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
)bd
= 0.0015bd > 0.0013bd
Use 0.0015bd = 0.0015x2000mmx90mm
= 270mm2/m
As, max = 0.04Ac
= 0.04bh
= 0.04 × 2000mm × 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 ≤ 2.0
= 1 + (200/90)1/2
= 2.49 ≥ 2.0
ρ’ = As1/bd ≤ 0.02
= 566mm2/ (2000mm × 90mm)
= 0.0031 ≤ 0.02

155. 155
Ref. Calculation Output
VRd, c = [0.12 × 𝑘 × √100𝜌′ 𝑓𝑐𝑘
3
]bd
= [0.12 × 2.0 × √100 × 0.0031 × 30 𝑁 𝑚𝑚2⁄3
] × 2000mm × 90mm
= 90847.2 N = 90.8kN
Vmin = (0.35 × √𝑘23
× √ 𝑓𝑐𝑘 )bd
= (0.35 × √2.023
× √30 𝑁 𝑚𝑚2⁄ ) × 2000𝑚𝑚 × 90𝑚𝑚
= 975991.8N = 976.0 kN
VRd, c ; Vmin > VEd (OK)
7.0 Deflection
Percentage of required tension reinforcement,
𝜌 = As, req/bd
= 413.52mm2/ (2000m × 90mm)
= 0.0023
Reference reinforcement ratio
𝜌o = (fck)1/2x10-3
= (30 N/mm2)1/2x10-3
= 0.0055
Percentage of required compression reinforcemen
𝜌’ = As’, req/bd
= 0/ (2000mm × 90mm)
= 0

156. 156
Factor for structural system, K = 1.0
𝜌 < 𝜌o
𝑙
𝑑
= K[11+1.5√ 𝑓𝑐𝑘
𝜌o
𝜌
+3.2√ 𝑓𝑐𝑘(
𝜌o
𝜌
-1)3/2
]
𝑙
𝑑
= 1.0 (11 + 1.5 √30 𝑁 𝑚𝑚2⁄
𝜌0
𝜌
+ 3.2 √30 𝑁 𝑚𝑚2⁄ (√(
𝜌0
𝜌
− 1)
23
))
𝑙
𝑑
= 1.0 (11 + 19.6 + 28.8)
𝑙
𝑑
= 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 × 1.0 × 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 ≤ Smax, slabs = 400mm (OK)
Secondary bar,
Smax, slabs = 3.5h or 450mm = 490mm > 450mm
Max. bar spacing = 450m ≤ Smax, slabs = 450mm (OK)

158. 158

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
𝐿 𝑦
𝐿 𝑥
= 1.017m
Characteristic actions :
Permanent, Gk = 3kN/𝑚2
Variable, Qk = 3kN/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: Øbar = 20mm
2.0 Slab Thickness
Min thickness for fire resistance = 120mm
Estimated thickness of deflection control, h =
𝐿 𝑥
22
=
2250𝑚𝑚
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 = 𝑎 −
∅ 𝐵𝑎𝑟
2
= 20mm – 10mm
= 10mm
Use min. conc cover regard to durability due to higher value.
Allowance in design for deviation, ΔCdev = 10mm
Nominal cover, Cnom = Cmin+ΔCdev
= 55mm + 10mm
= 65mm
4.0 Loading and Analysis
Slab self-weight = 0.21m × 30 kN/m3
= 6.3 kN/m2
Permanent load = 3 kN/m2
Char. permanent action,
Gk
= 9.3 KN/𝑚2
Char. variable action, Qk = 3 KN/𝑚2
Design action, nd = 1.35GK + 1.5QK
= 1.35(9.3kN/m2)+1.5(3kN/m2)
= 17.06 kN/m2
Moment
Short span, Msx = αsx × n × Lx
2
= 0.062 x 17.06kN/m × (2.250m)2

162. 162
= 5.35 KNm/m
Long span, Msy = αsy × n × Ly
2
= 0.062 × 17.06kN/m × (2.288m)2
= 5.54kNm/m
5.0 Main Reinforcement
Effective depth, dx = h – Cnom - 0.5Øbar
= 210mm – 65mm - (0.5 x 20mm)
= 135mm

163. 163
Ref. Calculation Output
Effective depth, dx = h – Cnom - 0.5Øbar
= 210mm – 65mm - (1.5 x 20mm)
= 115mm
Short span
Msx = 5.35 kNm/m
k = 𝑀 𝑠𝑥
𝑏𝑑2 𝑓𝑐𝑘
= 5.35 × 106 𝑁𝑚𝑚
2250𝑚𝑚 × (135𝑚𝑚)2 ×30 𝑁 𝑚𝑚2⁄
= 0.0043
kbal = 0.167
K<𝐾𝑏𝑎𝑙 , compression is not required
z = d [0.5+√0.25 −
𝐾
1.134
]
= d [0.5+√0.25 −
0.01
1.134
]
= 0.99d
= 0.99 × 135mm
= 133.65mm/m
As = 𝑀 𝑠𝑥
0.87 × 𝑓𝑦𝑘 ×𝑧
= 5.35 × 106 𝑁𝑚𝑚
0.87 ×500 𝑁 𝑚𝑚2⁄ ×0.99 ×135𝑚𝑚
= 92.09mm2
H6-300(bot) (94𝒎𝒎 𝟐
/m)

164. 164
Long span
Msy = 5.54 kNm/m
k = 𝑀 𝑠𝑦
𝑏𝑑2 𝑓𝑐𝑘
= 5.54 × 106 𝑁𝑚𝑚
2288𝑚𝑚 × (115𝑚𝑚)2 ×30 𝑁 𝑚𝑚2⁄
= 0.006
K<𝐾𝑏𝑎𝑙 , compression is not required

165. 165
Ref. Calculation Output
z = d [0.5+√0.25 −
𝐾
1.134
]
= d [0.5+√0.25 −
0.014
1.134
]
= 0.99d
= 0.99 x 115mm
= 113.85mm/m
As =
𝑀 𝑠𝑦
0.87 × 𝑓𝑦𝑘 ×𝑧
= 5.54 × 106 𝑁𝑚𝑚
0.87 ×500𝑁 𝑚𝑚2⁄ ×0.99 ×115𝑚𝑚
= 108.03mm2
H6-250(bot) (113𝒎𝒎 𝟐
/m)
Min. and max. Reinforced area,
As,min = 0.26(
𝑓𝑐𝑡𝑚
𝑓𝑦𝑘
)bd
= 0.26 (
2.9 𝑁 𝑚𝑚2⁄
500 𝑁 𝑚𝑚2⁄
) bd
= 0.0015bd > 0.0013bd
= 0.0015 × 2288mm × 135mm
= 465.79mm2/m
As,max = 0.04Ac
= 0.04 × 2288mm × 210mm
= 19219.2mm2/m
6.0 Shear Reinforcement

166. 166
Max. Design shear force, 𝑉𝐸𝐷 = 𝑛 𝑑 ×
𝐿 𝑥
2
= 17.06 𝑘𝑁 𝑚⁄ ×
2.250𝑚
2
= 19.19kN
Design shear resistance, k = 1 + (200/d)1/2
≤ 2.0
= 1 + (200/135)1/2
= 2.22 ≥ 2.0

167. 167
Ref. Calculation Output
𝜌′ = 𝐴 𝑠
𝑏𝑑
≤ 0.02
= 92.02 𝑚𝑚2
2288𝑚𝑚 ×135𝑚𝑚
= 0.0003 ≤ 0.02
𝑉𝑅𝐷,𝑐 = (0.12 × 𝑘 × √100 × 𝜌′ × 𝑓𝑐𝑘
3
) 𝑏𝑑
= 0.12 × 2.0 × √100 × 0.012 × 30 𝑁 𝑚𝑚2⁄3
× 2288𝑚𝑚 × 135𝑚𝑚
= 244.78 kN
𝑉 𝑚𝑖𝑛 = (0.035𝑘3/2
𝑓𝑐𝑘1/2
)bd
= (0.035(2.0)3/2
(30 𝑁 𝑚𝑚2⁄ )1/2
)× 2288mm x 135mm
= 167.48kN
𝑽 𝑹𝑫: 𝑽 𝒎𝒊𝒏 ≥ 𝑽 𝒎𝒊𝒏
7.0 Reflection
Percentage of required tension reinforcement,
ρ =
𝐴 𝑠,𝑟𝑒𝑞
𝑏𝑑
= 108.03 𝑚𝑚2
2288𝑚𝑚 ×115𝑚𝑚
= 0.0004
Reference reinforcement ratio,
𝜌0 = (fck)1/2
x 10−3
= (30)1/2
x 10−3
= 0.0055

168. 168
Percentage of required compression reinforcement,
𝜌′ =
𝐴 𝑠′,𝑟𝑒𝑞
𝑏𝑑
=
0 𝑚𝑚
2288𝑚𝑚 ×135𝑚𝑚
= 0
Factor for structural system,
k = 1.0
p ≤ 𝑝0

169. 169
Ref. Calculation Output
1
𝑑
= 𝑘 [11+ 1.5 √ 𝑓𝑐𝑘
𝑝0
𝑝
+ 3.2 √ 𝑓𝑐𝑘 (
𝑝0
𝑝
-1)3/2
]
𝑙
𝑑
= 1.0 [11 + 1.5 √30
0.0055
0.0038
+ 3.2 √30 (
𝑝0
𝑝
-1)3/2
]
𝑙
𝑑
= 1.0 [11+ 11.9 + 5.24]
𝑙
𝑑
= 28.14
Modification factor for span less than 7m = 1.0
Modification factor for steel area provided =
𝐴 𝑠,𝑝𝑟𝑜𝑣
𝐴 𝑠,𝑟𝑒𝑞
= 108.03 𝑚𝑚2
108.03 𝑚𝑚2
= 1.0 ≤ 1.5
Therefore allowable span-effective depth
ratio
= (
𝑙
𝑑
)
𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒
= 28.14 × 1.0 × 2.288m
= 64.384m
Actual span-effective depth = (
𝑙
𝑑
)
𝑎𝑐𝑡𝑢𝑎𝑙
=
2288𝑚𝑚
115𝑚𝑚
= 19.86 ˂ 64.384
= OK
8.0 Cracking

170. 170
n = 210mm < 210mm
Main bar
δmax, slabs = 3h or 400mm
= 3 (210mm) or 400mm
= 630mm > 400
= OK

171. 171
Ref. Calculation Output
Max. bar spacing = 325mm ≤ δmax, slabs = 400mm
Secondary bar
δmax, slabs = 3.5 or 450mm
= 3.5 (210mm) or 400mm
= 735mm
= OK
Max. bar spacing = 350mm ≤ δ max, slabs = 450mm
9.0 Detailing
Reinforcement
01: H6 - 300

172. 172
02: H6 - 250

173. 173
CHAPTER 6
MATERIAL PROPOSED
6.1 Physical Damage of Offshore Building
1. Salt crystallisation
 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.
 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
 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.
 All the above types of physicals damage are mainly cosmetic although they do
reduce the cover depth to the reinforcement.
3. Marine growth
 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.
 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
∅ 𝑏𝑎𝑟 1 = 10mm
∅𝑙𝑖𝑛𝑘 = 6mm