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

Width of beam = 300   

Depth of beam = 600    TotalLoad on Beam = 1.4g + 1.6q = (1.4 *25) + (1.6*14) = 22.4N

Required Longitudinal Reinforcement

Longitudinal reinforcement at tension and compression face (Min of two 12 mm diameter bar is required to be provided in tension) in single or multiple rows are provided.

b) Shear Force = 5/8 Pa L = 0.625 * 22.4 * 600 = 8400 = 8.4 KN

So, Area of Steel = 3*3.14/4 *30* 4.2 = 296 sqmm

Step one

Nominal shear stress

Tv = Vu/bd

Tv = 8.4 x 1000/(300 x 600) = 0.046N/mm2

Step two

Percentage of steel

Percent steel = Ast/bd x 100

Percent steel = (296 x 100)/ (300×600)

= 0.16%

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 greater than Tc

So 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

= 0.75 x 500

= 375mm

Provide 6mm diameter, 2-legged stirrups@100mm c/c.

Maximum spacing = 0.75d

= 0.75 x 230 = 172mm

Provide 6mm diameter 2-legged stirrups @ 100mmc/c

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