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An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 meters and carries a U.D.L. of 3000 kg/m inclusive of its self weight. If the lever arm constant for the section is 0.865, the maximum intensity of shear stress, is

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Correct Answer: Option A
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4 Comments
Yugant Mankar
Yugant Mankar 5 years ago
Shear stress = F/(b x J x d).

= 0.5WL/(b x J x d).

= 0.5x30x600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].

= 9000/1081.25.

= 8.32
Yugant Mankar
Yugant Mankar 5 years ago
Shear stress = F/(b x J x d).

= 0.5WL/(b x J x d).

= 0.5x30x600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].

= 9000/1081.25.

= 8.32
Vishwajeet Patel
Vishwajeet Patel 6 years ago
Shear stress = F/(b x J x d).

= 0.5WL/(b x J x d).

= 0.5x30x600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].

= 9000/1081.25.

= 8.32
Kinjal Patel
Kinjal Patel 7 years ago
Shear stress = F/(b x J x d).

= 0.5WL/(b x J x d).

= 0.5x30x600/(25x50x0.865) [3000kg/m = 30 kg/cm and 6.00 m = 600 cm].

= 9000/1081.25.

= 8.32