Concrete Structure

REINFORCED CONCRETE BEAM DESIGN AND DETAILING 5.0 LOADING ON BEAMS

5.2 DESIGN LOADS

calculate load contribution from slab dead load from slab, Gslab = 25kN/m2 × 0.21 × (5.9+1.2) = 37.3 kN/m imposed load from slab, Qslab = 3.0 kN/m2 × 7.1 m = 21.3 kN/m 5.3 CAPACITY DESIGN

END SPAN BEAM B1 INTERIOR SPAN

for permanent and imposed loads, Fd = 1.2G + 1.5Q = 1.2 × (11.1 + 37.3) + 1.5 × 21.3 = 90.03 kN/m Beam load arrangement

END SPAN

5.4 ANALYSIS of BM and SF

use simplified method for beams 4 - continuous spans BMD 5.1 ANALYSIS OF BEAM B1

a. b. c.

d.

e. f. g. h.

i.

the beam supports the one-way slab in addition to its own self-weight typical L/d ratio varies from 10-20, according to spans and magnitude of imposed live loads try an L/d ratio of 16, L = 8400mm d = 8400 / 16 = 525 mm, although the beam is cast together with the slab and acts integrally with the slab as a T-beam in the positive bending region. For simplicity. We design the beam as rectangular section only. In the negative moment region, the beam cross section remains rectangular.

The beam reinforcements we choose is N24 bars and N12 @150mm ligatures for the quarterspans and N12 @ 300mm ligatures for the middle half-span the width of the beam is given as 1200mm from the earlier section, Cover = 20mm overall depth, D = d+0.5 diameter of bars + lig. diameter + cover = 525 + 12 + 12 + 20 D = 569 mm (round up to next 20mm, 580mm) d = 525 + 11 = 536mm dead load of beam (top portion of beam had been included as part of slab) G = (0.58 - 0.21) × 1.2 ×25 G = 11.1 kN/m

Moment = coefficient × FdL2 (AS3600 6.10.2.2) Approximate Moment Diagram for more than 2 spans SFD Shear = coefficient × FdL (AS3600 6.10.2.2) Approximate Shear Force Diagram for more than 2 spans

END SUPPORT

END SPAN

at end A & E, negative moment, M- = 1/16 × FdL2 = 1/16 × 90.03 × 8.4 × 8.4 = 397.03 kNm from HB71 chart 4.1 M/bd2 = 397.03 × 106 / (1200 × 536 × 536) = 1.15 Ast = 0.003 × 1200 × 536 = 1929.6mm2 provide 5 N24 2250mm2

at end span AB & DE, positive moment, M+ = 1/11 × FdL2 = 1/11 × 90.03 × 8.4 × 8.4 = 577.50 kNm from HB71 chart 4.1 M/bd2 = 577.50 × 106 / (1200 × 536 × 536) = 1.68 Ast = 0.0044 × 1200 × 536 = 2830.08mm2 provide 7 N24 3150mm2 `

5.5 DEFLECTION CHECKS - using deemed-to-comply span-to-depth ratios

END SUPPORT & INTERIOR SPAN

END SPAN

where

0.0044

INTERIOR SUPPORT

Lef = effective span = 8400mm Lef/d = span-to-depth ratio ̗/Lef = deflection limit selected = 1/500 EC = standard strength of concrete after 28 days = 32800 for 40MPa concrete Fd.ef = effective design load per unit area = (1.0 + kcs) G + (ψs + kcs × ψl) Q kcs= [2.0 - 1.2 × (Asc/Ast)] ≥ 0.8 (Asc is the area of steel in the compression zone) = 2.0 (given that Asc = 0, AS3600 8.5.3.2) β = bef / bw ≥ 1.0 (assume rectangular section) =1200/1200 = 1.0 p = Ast/befs at end span (mid) = 3150/1200×536 = 0.0049 if p ≥ 0.005, k1 = 0.02 +2.5p if p ≤0.005, k1 = 0.1-13.5p in this case, p < 0.005, k1 = 0.03385 k2 = 2.4/384 for an end span (0.0063); 1.5/384 for an interior span (0.0039) (AS/NZS 1170) ψs = 0.7 = short term load combination factors (AS/NZS 1170.0) ψl = 0.4 = long term load combination factors (AS/NZS 1170.0) Fd.ef = (1.0 +2.0) G + (0.7 + 2.0 × 0.4) Q = 3.0 G + 1.5 Q = 3.0 × (11.1+37.3) + 1.50 ×21.3 Fd.ef = 177.15 kN/m2 Fd.ef /b = 177.15/1.2 = 147.625 kN/m2

INTERIOR SUPPORT

at support B,C & D, negative moment, M- = 1/10 × FdL2 = 1/10 × 90.03 × 8.4 × 8.4 = 635.25kNm from HB71 chart 4.1 M/bd2 = 635.25 × 106 / (1200 × 536 × 536) = 1.84 Ast = 0.0048 × 1200 × 536 = 3087.36mm2 provide 7 N24 3150mm2

from HB71 chart 4.28 for ̗/Lef = 1/500, Lef/d ≤ 13.3 our initial L/d = 16 >13.3, not acceptable need to recalculate If ̗/Lef = 1/500, mid span deflection = 16.8mm

INTERIOR SPAN at span BC&CD, positive moment, M- = 1/16 × FdL2 = 1/16 × 90.03 × 8.4 × 8.4 = 397.03 kNm from HB71 chart 4.1 M/bd2 = 397.03 × 106 / (1200 × 536 × 536) = 1.15 Ast = 0.003 × 1200 × 536 = 1929.6mm2 provide 5 N24 2250mm2

After 3 rounds of calculation, get to the acceptable L/d ratio with in 0.5 margin. To save space, only show final round of recalculation.

0.0048

13.3

5.1* ANALYSIS OF BEAM B1

END SUPPORT & INTERIOR SPAN

END SPAN

try an L/d ratio of 14.8, L = 8400mm d = 8400 / 14.8 = 568 mm, overall depth, D = d+0.5 diameter of bars + lig. diameter + cover = 568 + 12 + 12 + 20 D = 612 mm (round up to next 20mm, 620mm) d = 568 + 8 = 576mm dead load of beam (top portion of beam had been included as part of slab) G = (0.62 - 0.21) × 1.2 ×25 G = 12.3 kN/m

5.2* DESIGN LOADS

calculate load contribution from slab dead load from slab, Gslab = 25kN/m2 × 0.21 × (5.9+1.2) = 37.275 kN/m imposed load from slab, Qslab = 3.0 kN/m2 × 7.1 m = 21.3 kN/m

0.00383

0.00275

5.3* CAPACITY DESIGN

for permanent and imposed loads, Fd = 1.2G + 1.5Q = 1.2 × (12.3 + 37.275) + 1.5 × 21.3 Fd = 91.44 kN/m Beam load arrangement

5.4* ANALYSIS of BM and SF

use simplified method for beams 4 - continuous spans BMD Moment = coefficient × FdL2 (AS3600 6.10.2.2) Approximate Moment Diagram for more than 2 spans SFD Shear = coefficient × FdL (AS3600 6.10.2.2) Approximate Shear Force Diagram for more than 2 spans

END SUPPORT & INTERIOR SPAN at end A & E, negative moment, M- = 1/16 × FdL2 = 1/16 × 91.44 × 8.4 × 8.4 = 403.25 kNm from HB71 chart 4.1 M/bd2 = 403.25 × 106 / (1200 × 576 × 576) = 1.01 Ast = 0.00275 × 1200 × 576 = 1900.8 mm2 provide 5 N24 2250mm2 0.00425

END SPAN at end span AB & DE, positive moment, M+ = 1/11 × FdL2 = 1/11 × 91.44 × 8.4 × 8.4 = 586.55 kNm from HB71 chart 4.1 M/bd2 = 586.55 × 106 / (1200 × 576 × 576) = 1.47 Ast = 0.00383 × 1200 × 576 = 2647.30mm2 provide 6 N24 2700mm2 `

INTERIOR SUPPORT at support B,C & D, negative moment, M- = 1/10 × FdL2 = 1/10 × 91.44 × 8.4 × 8.4 = 645.20kNm from HB71 chart 4.1 M/bd2 = 645.20 × 106 / (1200 × 576 × 576) = 1.62 Ast = 0.00425 × 1200 × 576 = 2937.6 mm2 provide 7 N24 3150mm2

5.5* DEFLECTION CHECKS - using deemed-to-comply span-to-depth ratios

where Lef = effective span = 8400mm Lef/d = span-to-depth ratio ̗/Lef = deflection limit selected = 1/500 EC = standard strength of concrete after 28 days = 32800 for 40MPa concrete Fd.ef = effective design load per unit area = (1.0 + kcs) G + (ψs + kcs × ψl) Q kcs= [2.0 - 1.2 × (Asc/Ast)] ≥ 0.8 (Asc is the area of steel in the compression zone) = 2.0 (given that Asc = 0, AS3600 8.5.3.2) β = bef / bw ≥ 1.0 (assume rectangular section) =1200/1200 = 1.0 p = Ast/befs at end span (mid) = 2700/1200×576 = 0.0039 if p ≥ 0.005, k1 = 0.02 +2.5p if p ≤0.005, k1 = 0.1-13.5p in this case, p < 0.005, k1 = 0.047 k2 = 2.4/384 for an end span (0.0063); 1.5/384 for an interior span (0.0039) (AS/NZS 1170) ψs = 0.7 = short term load combination factors (AS/NZS 1170.0) ψl = 0.4 = long term load combination factors (AS/NZS 1170.0) Fd.ef = (1.0 +2.0) G + (0.7 + 2.0 × 0.4) Q = 3.0 G + 1.5 Q = 3.0 × (12.3+37.275) + 1.50 ×21.3 Fd.ef = 180.675 kN/m2 Fd.ef /b = 180.675/1.2 = 150.56 kN/m2 From the formula above, end span: Lef/d ≤ [(0.047× 1/500 × 1200×32800)/(0.0063 × 180.675)]^1/3 = 14.841 our initial L/d = 14.8 < 14.841, acceptable and within 0.5 margin could stop here From the formula above, interior span: p = Ast/befs at end span (mid) = 2250/1200×576 =0.0033 if p ≥ 0.005, k1 = 0.02 +2.5p if p ≤0.005, k1 = 0.1-13.5p in this case, p < 0.005, k1 = 0.056 k2 = 2.4/384 for an end span (0.0063); 1.5/384 for an interior span (0.0039) (AS/NZS 1170) Lef/d ≤ [(0.056× 1/500 × 1200×32800)/(0.0039 × 180.675)]^1/3 = 18.4 our initial L/d = 14.8 < 18.4, acceptable but beyond 0.5 margin may recalculate until it gets within 0.5 margin, but for easy construction, interior span may stay same with end span.

If L/d = 18.4 for interior span, d = 457mm D = 501mm (need to round up to next 20mm, 520mm) d = 476mm If ̗/Lef = 1/500, mid span deflection = 16.8mm

Final solution: Beam: D = 620mm, d = 576mm, Steel: 6 N24 end span, 7 N24 interior support, 5 N24 end support and interior span.