T Beam

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Information about T Beam

Published on December 2, 2008

Author: hsrai



Slide 1: T-Beam INTRODUCTION : INTRODUCTION Design of reinforced concrete beams involves sizing and finding required quantity of steel based on the consideration of strength and serviceability requirements. It also involves detailing. The major consideration in the design of beams is bending moment. Hence first, beams are designed for bending moment and then the design for shear is taken up. Checks are applied for deflection and crack width. If the requirement for any limit state fails redesign is to be made. The detailing of reinforcement is to be made with neat sketches/drawings taking into account bond, cracking and durability considerations. Such kind of design consideration has been discussed. EFFICTIVE WIDTH OF THE FLANGE : Effective width instead of actual width of a flange beam is used for calculating the moment of resistance as in clause 23.1 of Code. The actual width of a flanged beam is equal to the distance between the middle points of the adjacent spans of the slab. The effective is used to allow for the fact that the compressive stress in the flange is higher above the rib than at some direction away from the rib. The effective width concept makes use of constant compressive stress over the entire effective width such that the force of compression remains same. A typical stress distribution in a T-beam is shown in fig. EFFICTIVE WIDTH OF THE FLANGE Slide 6: Clause 23.1.2 of the Code recommends that effective width of flange of T-beam may be taken as follow: bf = ____l_____ + bw (l/ b) + 4 bf = b     Where, bf = effective width of flange Bw = breath of the rib or web b = actual width of flange l = distance between points of zero moment in a beam; For continuous beams l may be taken as 0.7 times the Effective span. MINIMUM AND MAXIMUM REINFORCEMENT : Clause 26.5.1 of the Code requires that minimum tension reinforcement should not be less than that given by the following : Ao = 0.85 bwd fy   where, Ao = minimum area of tension reinforcement bw = breadth of the web of a flanged beam     Maximum area of tension steel or compression steel should not exceed 0.04 bw.D. Clause 23.1.1 requires that if the main reinforcement of the slab is parallel to the beam, transverse reinforcement must be provided as shown in the fig. Such reinforcement must be at least 60 % of the main reinforcement at mid span of the slab. MINIMUM AND MAXIMUM REINFORCEMENT TYPES OF PROBLEM : Types of problem on T-beam can be classified under the following four cases:   Case 1 Neutral axis lies in the flange In such cases, concrete below the neutral axis is assumed to have cracked. Thus, the beam is as good as a rectangular section having breadth B and effective depth d. Thus, all the expressions derived for a rectangular beam, singly reinforced as well as doubly reinforced. TYPES OF PROBLEM Slide 11: Case 2 T-section is balanced , singly reinforced and neutral axis lies in the web A T-beam can be idealized as shown in dig. Slide 12: If __Df__< or equal to 0.20 it is assumed that depth of d rectangular portion of the stress block is greater than that of the flange [ x2 = 3/7 of 0.48 d = 0.28d ]   Force of compression = force of compression in rectangle of size bw.d + force of compression in rectangle of size( bf – bw).Df = 0.36fckbwxm + 0.446fck(bf-bw)Df   Force of tension = 0.87fy At   Moment of resistance = 0.36fckbw xm(d – 0.42xm) + 0.446fck(bf– bw). Df(d – 0.5Df) Slide 13: If _Df_ >0.20, it is assumed that depth of rectangular d portion of the stress block is equal to yf.   Force of compression =0.36fckbw xm+ 0.446fck (bf - bw)y   Force of tension = 0.87 fy At   Moment of resistance = 0.36fckbw xm(d – 0.42xm)+ 0.446fck(bf – bw)yf(d – 0.5yf) Where, yf = (0.15xm+ 0.65Df) shouldn’t be greater than Df Slide 14: Case 3 T-section is under reinforced and neutral axis lies in the web   If __D__< or equal to 0.43 (= 3/7) it is assumed that depth of x rectangular portion of the stress block is greater than that of the flange as shown in fig.a   Force of compression = 0.36fck.x bw + 0.446fck (bf – bw)Df   Force of tension = 0.87fy At   Moment of resistance with respect to steel = 0.87fy At Z   Moment of resistance with respect to concrete = 0.36fckbwx(d- 0.42x) + 0.446fck(bf – bw)Df(d – 0.5Df) Slide 16: If _D_ > 0.43, it is assumed that depth of the rectangular X portion of the stress block is equal to y as shown in fig. b   Force of compression =0.36fckbw.x + 0.446fck (bf –bw)yf   Force of tension = 0.87fy At   Moment of resistance with respect to steel = 0.87fy At Z   Moment of resistance with respect to concrete = 0.36fck bw(d- 0.42x) + 0.446fck(bf – bw)yf(d – 0.5yf)   Where yf = [0.15x + 0.65 Df] shouldn’t e greater than Df Slide 18: Case 4 T-section is over reinforced and maximum depth of neutral axis lies in the web   If _Df_ < or equal to 0.20 , use eqs. of case 2 sub d part 1   If _Df__ > 0.20 , then use eqs. Of case 2 sub part 2 d Example 1 : A T-beam floor consists of 15cm thick RC slab monolithic with 30cm beams. The beams are spaced at 3.5 m c/c and their effective span is 6 m as shown in fig.If the superimposed load on the slab is 5 kN/m2, design an intermediate beam. Use M20 and Fe250. Example 1 Solution: : Dead load of slab = 0.15*25 = 3.75kN/m2   Superimposed load on slab = 5kN/m2   Total load on slab = 8.75kN/m2   Load per meter run of the beam = load on slab per unit area *c/c Distance between beams = 8.75*3.5 = 30.56kN/m   Effective width of flange ,bf =lo +bw +6Df = 600+30+6*15=220cm 6 6 Solution: Slide 22: Maximum value of flange width = 350cm > 220cm Therefore effective width of flange = 220 cm Let us adopt overall depth d of the beam equal to 40cm and effective cover equal to 40 mm so that effective depth is 36cm   Dead load on wed of beam = width of web*depth of web*concrete Density = 0.30*0.25*25 = 1.875kN/m   Total load on beam per meter run = 30.6 + 1.875= 32.475kN/m Slide 23: Factored maximum BM = wl2/8 = 1.5*32.475*62/8 = 219.2kNm   Let us assume that neutral axis lies in the flange, that is X = 0.87fyAt = 0.87*250At 0.36fckbf 0.36*20*2200 = 0.0137At   Factored BM = force of torsion * z 219.2*106 = 0.87*fyAt (d – 0.42x) = 0.87*250At(360 – 0.42*0.0137At) At = 2938mm2 And X = 40.25mm < 150mm   Therefore NA lies in the flange Slide 24: Use 5 – 28 mm bars (As =3079mm2 > 2938mm2)   Minimum area of tension steel , Ao = 0.85bwd / fy = 0.85*300*360 /250 = 367mm2 < 3079mm2 OK   Maximum area of tension steel, At =0.04bwD =0.04*300*400 = 4800mm2 > 3079mm2 OK

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