Q6 Design a Tbeam spanning 6 m supporting a oneway slab of t

Q.6. Design a T-beam spanning 6 m supporting a one-way slab of thickness 140 mm and subjected to a live load of 3.5 kN/m2 and a dead load (due to floor finish, partition, etc.) of 1.2 kN/m2, in addition to its self-weight. Assume fc 30 MPa and f 500 MPa and the centre-to-centre distance of beams as 4 m.

Solution

Loads:

Dead load due self weight of slab = 0.140 x 25 = 3.5 kN/m2

Superimposed dead load (Given) = 1.2 kN/m2

Live load (Given) =3.5 kN/m2

As per IS 456:2000 ( Indian standard for design of RCC)

Cl. 23.1.2. Effective width of flange (b_f ) = l₀ /6 + b_w + 6 D_f = 6000(span of beam)/6 + 300 (Assume) + 6 x 140 (Thickness of slab; Since it will act as flange)

b_f = 1000 + 300 + 6 x 140 = 2140 mm = 2140 mm (Consider)

Assuming simply supported beam , Span/effective depth ratio = 20

effective depth = 6000/20= 300 mm

Provide overall depth = 400 mm

Self weight of beam = (Depth of web - depth of flange) x width of web x 25

= (0.400-0.140)*0.300 x 25 = 1.95 kN/m

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Total load = 10.15 kN/m2

Since the slab is one way

Load per unit length of beam (w) = 10.15x (distance between the beams/2)x 2(Considering the beam to be intermediate one)

= 10.15x (4/2)*2

= 40.6 kN/m

Sagging Moment = w x span²/8 = (40.6 x 6²)/8 = 182.7 kN-m

Shear Force = w x span x 0.5 = 40.6 x6 x 0.5 = 121.8 kN

Factored forces (Assuming the 1.5 factor of safety)

Factored Design moment = 1.5 x 182.7 = 274.05 kN-m

Factored Shear Force = 1.5x 121.8 = 182.7 kN

Effective depth = 400 -25 -8(link)-16/2 = 359 mm

Neutral axis depth

x/d = 1.2 - (1.44 - (6.68 Mu/fckb_fd²)^0.5 = 1.2-(1.44-(6.68 x274.05 x 10⁶/30x2140x359²)^0.5

x/d = 0.096

x = 0.096 x 359 = 34.47 mm < D_f

Steel to be provided ( Design as a rectangular beaM SINCE x< D_f )

Ast = Mu/(0.87fy (d-0.42x) = 274.05 x 10⁶/ (0.87 x 500 (359-0.42x34.47))

= 1828.61 mm²

% steel provide base on width of web= 100 Ast/bd = 100x 1828.61/(300x359) = 1.69 %

Min steel specified = 0.85x100/fy = 0.85 x 100/500 = 0.17 % , Ast provided > Ast min

Provide 6 Nos H20 at bottom, Provide 2 NOs T12 bar top in flanged.

Design shear stress = Vu/bd = 182.7 x 10³/(300 x 359) = 1.69 N/mm²

Design shear strength of concrete for 1.69 % of steel, Table 19 , IS 456:2000

Tc=0.8 N/mm2

Shear taken by concrete Vuc = 0.8 x300x35910^-3 =86.16 kN

Shear to be taken by shear link = 182.7 -0.8 x300x359x10^-3 = 96. 54 kN

Proved H10 , 2 legged links @200 mm c/c spacing

Vus = 0.87 fy x Asv d/sv = 0.87x500x pi/4 x 10²x2x359/200 = 122589 N =122.58 kN

Total shear capacity of section =86.16+122.58 = 208.74 kN> 182.7 kN hence Safe