A T-beam with flange width (B) 100, flange thickness (t-10), beam width b30, and

Question

A T-beam with flange width (B) 100, flange thickness (t-10), beam width b30, and depth (h) is 60cm is supported to a column at node A, and a beam at node B. Moment and shear diagrams are shown below. a) Calculate the required longitudinal reinforcement of the beam at span and at support. b) Calculate the design shear force for regions AC and BC. Design the shear reinforcement for the beam using TS-500 and TSC 2007 requirements. Ps-1.4g+1.6q Model 1 Column direct support) Ai I B (indirect support) 1515 15115m 570m 15 L=600cm 15 5/8 PAL 128

Explanation / Answer

Total Load = 1.4*25 + 1.6*14 = 35 + 22.4 = 57.4Kn

Hf = 60, 0.85 fck a*b = Asfy : ; As=0.85 *30 * 100*10 /500 ; As= 510sqcm

Considering 12mm size bar, unit weight = 0.87* 10 =8. 7 kg . So no. of bars = 5.8 or 6 nos. , 3nos. at top and 3nos. at bottom.

b. Shear Force at AC = 5/8*57.4*600, at BC = 3/8 *57.4 *600

   =21.525Kn at BC = 12.915

c. Step1 :- Nominal Shear Stress = 21525 /4000 = 5.3 N/mm2

Step two

Percentage of steel

Percent steel = Ast/bd x 100    ;( 510*100)/(300*100) = 0.17%

Step three

As per IS: 456: 2000

Tc = 0.48 + (0.56-0.48)/(0.75-0.5) (0.63 – 0.5)

Tc= 0.52 N/mm2

Therefore, Tv < Tc

No shear reinforcement required.

Step four

Provide minimum shear reinforcement;

As per IS : 456 : 2000

Asv/bsv = 0.4/(0.87 fy)

Assuming 6mm diameter, 2 – legged stirrups

Asv = (2 x 3.14 x 6 x 6)/4 = 56.54 mm2

Sv = (0.87fy.Asv)/0.4b

Sv = (0.87 x 250 x 56.54)/(0.4×300) = 102.47mm say 100mm

As per IS:456:2000,

Maximum spacing = 0.75d

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