higher than expected reinforcement in beam

[Anonymous]

Hi,

I have a beam in structure, the maximum moment is 167.3 kN.m . when I designed it with Robot ...as a 450 *600 mm beam with cover 50 mm .

I expected to see this results based on the Eurocode 2

K= M/bd^2*fck =167.3*10^6/(450*(532^2)*35)=0.038 <0.167 no compression reinforcement required

Lever arm z= d*(0.5+sqrt(0.25-k/1.134))

= 532*(0.5+sqrt(0.25-0.038/1.134))= 514 mm 

AS= M/(0.87*500*Z)= 167.3*10^6/(0.87*500*514) = 748.3 mm2

Robot results is 1084 mm2 far more than expectation. Even if u design the same beam with Tedds it gives lyou ower results than what Robot gives. 

[Rafal.Gaweda]

Robot file please

[Anonymous]

Hi

I am talking about beam number 35. However the continous beams 51-62 has the same problem. 

Regards

[Anonymous]

any comment?

[Artur.Kosakowski]

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[Anonymous]

There's a difference between Thoritical As and Provided As 
756 is the thoretical As reqired 
the other one in the Provided As which is the Area of the bars provided in beam reinforcement 

[Anonymous]

Thank you very much...that great this is the expected required area 756 mm2!  which can be acheived by providing 4f 16 bars , 

However, Robot provided double this value (with a message that the longitude reinforcement has been ogmented due to the cracks! ). I hope there is some justification for this high increase. 

can u please give me the equations used for the crack calculations and especially the amount of steel we need to add.

Thanks 

[Artur.Kosakowski]

I'n not sure if I understand you correctly. My assumption is that you calculated amount of required reinforcement for ULS:

K= M/bd^2*fck =167.3*10^6/(450*(532^2)*35)=0.038 <0.167 no compression reinforcement required

Lever arm z= d*(0.5+sqrt(0.25-k/1.134))

= 532*(0.5+sqrt(0.25-0.038/1.134))= 514 mm 

AS= M/(0.87*500*Z)= 167.3*10^6/(0.87*500*514) = 748.3 mm2

which similar to what Robot calculates for ULS only (4 fi 16). For SLS Robot follows the code provisions and as far as I can tell you haven't tried to calculate the required amount of reinforcement for SLS (cracks) by hand to compare the values obtained for both ULS and SLS combinations being included in the process.

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[Anonymous]

for ULS Perfect results but for the SLS:

Can I just see the formula robot used to calculate the reinforcement for cracks? because from one hands  it sounds so high to double the reinforcement for cracks ...and from the another hand I don't want to ignore the cracks.

[Artur.Kosakowski]

Hi Roland,

As I wrote Robot uses the formula from the design code which are applicable for calculations of crack width. I'm afraid it will be hard for me to look at the code and type all of them one by one on the forum. 

Considering your doubts about the amount of reinforcement - what about trying to calculate it by hand? If you find some discrepancy at some point just indicate where it is and then it will be much easier to investigate why this happened. 

You may also have a look atthe atatched verification example made by my French colleagues.

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.