Question

The continuous beam shown in Fig. 1 is part of a beam-and-slab floor system of an office building and is typical of a series of beams spaced at 4 m intervals. In addition to the dead load, the live load on the floor is 3 kPa. Using f'c = 32MPa, N32 bars for main reinforcement, a) Find the maximum positive and negative moments using the AS3600 simplified method. b) Design the beam cross-section for the maximum amount of tensile steel required at (i) interior supports, and (ii) mid-span. In both cases use singly reinforced design. M 13.0 m 13.0 m 13.0 m 0.45 m 0.45 m 0.45 m 0.45 m Fig. 1a (Continuous beam profile) Beam Section details Flange thickness: Width of web: Depth of Beam below slab: Depth to centre of steel in tension: 150 mm 500 mm 600 mm 50 mm Fig. 1b (Typical Section)

Answer 1

1. 1ooo Continums Beam doading i) shele:- 24x15. 3.60 kN/m2 V c = 24 kNm3 11) L. L = 3 kN/m2 Using AS 360 Wid=1:283.62+1, 5X3 = 4.32+45=(8.82 Kallm? en dood on beam = (spacing 11) Self with beam (fastired) = 1:24 24 x 500 6600 Toll we beam) = 35.28+8.64=193.9 2 KN]»? 35.28 8-82x = 8.64 kaum 100x1000 2 Coefficients that Simplified Method ]. 1 + T6 to 2 E) 742.25KNM Cat guleus Suppat) (midspan) // At the suppe Mmaxt ve=742-254 2. Three Span beau In= 13m b=500mm M=xx Edx [n d = 600+ 15o-50 We know from AS 3600 moment d=700mm to ha ++ @so a) So maximum nepture moment 43.92X13 -X43.92813= +674.77 kNm o Golvèn b= Scomm, d= Fleum; fc= 32 Nmm, fy = Sou n rosmo 85X32 2353x10x742.25 19x500x1700)*32 fz.os As=:00 68X500X700 = 238mm As minimum 132 x500x70 = 1980mm > 238 mm2 D So formile As = 1980mm - As minimemal vi Inteur Suppril Smice at nid Spon M+ = 674.77 kam which is < Mo=742.25 Therefore, have a lo fonde As - 1980 mmt Singhy Rein freed lilsis N32 bars 1980x4 = 13 ban! 3 banli Price 3N32 bars i Top at Intews Support Amel 3N32 bourat Bottem at Mid-Span. S = { 1-1 500 0.054 1-1 1-0.025 E.O$4X 0.013 = 0.00068 2x50 han