Transversal reinforcement for RC beam by EC2

[Anonymous]

Hello,

Please could you help with an issue regarding transversal reinforcement in RC beam. I'm designing a simple one span beam by eurocode 2. Section of the beam is 450wx750d, tension reinforcement is 5d20, design shar force is about 250kN. Robot gives be required shear reinforcement of 5d6 legs at 140mm centres. It seems too much. I have done some simple calculations by hand and also using Tedds and i got much smaller amount of shear reinforcement. Just 3 legs of d6 is more than enough. Could you help me to identify the problem? I have alsoo attached robot file.

Thank you.

[Artur.Kosakowski]

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[Anonymous]

Thanks for response. I was talking about shear force near support. It is about 250kN only. If you will look at the provided number of legs and its spacing you will notice that it is definetely too much. I checked the required shear reinforcement near support by hand calcs and using Tedds and I've got different results. When I try to use British standards (BS 8110) to calculate the amount of shear reinforcement in Robot it gives me a good result, almost identical to a hand calculation and to results from Tedds. It seems there is a problem with shear reinforcement calculation using EC2. Please just try to compare the results for shear reinforcement to resist about 250kN for a given section and material properties with any other software, spreadsheet or hand caluclations. You can also compare results using another codes. And you will find a huge difference. I'm attaching a screenshoot with the result from Tedds. As you can see for the given conditions it's just enough to have 5 legs of d6 rebar near support at 300mm centres.

[Artur.Kosakowski · Accepted solution]

What is the value of

 

?

EC2:

Formula (6.8):

Vrds = Asw/s*z*fywd*cotTh

assuming 5d6 :

Asw = 5* 2.8274333882308137e-005 m2 = 0,00014135[m2]

Vrds = Asw/s*z*fywd*cotTh = 0,00014135 [m2]/0,14[m] * 0,579[m] * 434782[KPa] * 1.0 = 254 kN

If you find your post answered press the Accept as Solution button please. This will help other users to find solutions much faster. Thank you.

[Anonymous]

Problem is solved. Thank you for help.