Creo que hay una pequeña confusión

Creo que hay una pequeña confusión en lo que es resistencia al corte de una sección de concreto rectangular. Vc=k*(fc)^0.5*bw*d como verán la ecuación bw y d son unidades de longitud; (fc)^0.5 donde fc(F/L^2)=en unidades F^0.5/L(longitud); Como se sabe el cortante tiene unidad de medida de fuerza F. por tal razón k tiene que tener las dimensiones que hagan falta para que el cortante de en unidades de fuerza. escribiendo la ecuación nuevamente en unidades Vc -->F=k*(F^.5/L)*L*L entonces las unidades de K son F^0.5/L para que la ecuación este balanceada.
En conclusión la k varia dependiendo de las unidades que uses por ejemplo cuando usas (psi) y (in) k=2 lb^0.5/in; si usas (Mpa) (mm)
k=0.166 N^0.5/mm y 0.53kg^0.5/cm si usas Kg/cm2 y cm. pueden revisarlo en una calculadora hp o manualmente.
Cuando diseñamos tenemos que estar claros que unidades usamos porque si no podemos estar calculando mal los elementos esto cuando lo haces a mano o lo programas. Pero si usas software como etabs sap2000 y rob-ot etc no hay que preocuparse porque estos haces los correctivos en la las unidades de medida automáticamente.
En el problema anteriormente presentado no lo revise solo vi la imagen presentada y me pareció que el efecto de torsión en el elemento era alto.
Cuando este mensaje te sale (Error O/S #3 - Shear Stress exceeds maximum allowed) para cumplir según el código que estés aplicando hay varias alternativas. 1 incrementar la fuerza del concreto 2 incrementar la sección 3 analizar como disipas el efecto que te este dando el problema en este caso la torsión.
Si he cometido algún error me disculpo.
Saludos..
Pd No hay alternativa de incrementar acero porque el software a aplicado lo máximo que le permite el código que estés usando y aun así no le cumple.
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Results (English) 1:

I think there is a little confusion in what is resistance to cutting out a section of concrete rectangular. VC = k * (fc) ^ 0.5 * bw * d as the bw equation and d are units of length; (fc) ^ 0.5 where fc(F/L^2) = units F ^ 0.5 / L (length); As it is known the cutting edge has unit of measurement for force f. k therefore must have dimensions that make failure to shear of in force units. writing the equation again in Vc units - > F = k *(F^.5/L) * L * L K units are then F ^ 0.5 / L for which the equation to be balanced.In conclusion the various k depending on the units you use for example when you use (psi) and (in) k = 2 lb ^ 0.5 / in; If you are using (Mpa) (mm) k = 0.166 N ^ 0.5 / mm and 0.53kg^0.5/cm if you use Kg/cm2 and cm. can review it on an hp calculator or manually.When we design have to be clear that units use because if we cannot be calculating evil elements this when you do it by hand or programs it. But if you use software like etabs, sap2000 and rob-ot etc should not worry because they do the corrective measures in the measurement units automatically.In the above presented problem it check not only saw the presented image and it seemed to me that the effect of torque in the element was high.When you get this message (Error o/s #3 - Shear Stress exceeds maximum allowed) to meet according to the code you're using, there are several alternatives. 1 increase the strength of the concrete 2 increase section 3 analyze how you disipas the effect as you this giving the problem in this case the torque.I apologize if I've made a mistake.Best regards..PS there is no alternative of increasing steel because the software to applied the maximum that allows the code that you're using and still does not meet him.the_underdogNovice userNovice userProgress to next rank:2.6% Messages: 2Registered: Jue Abr 14, 2011 2:11 pmCountry: Panama (pa)Thanked: 0 timesThanked you: 0 times

Results (English) 2:

I think there is a little confusion in what is resistance to cutting a rectangular concrete section. Vc = k * (fc) ^ 0.5 * bw * d like bw see the equation d are units of length; (fc) ^ 0.5 where fc (F / L ^ 2) = 0.5 ^ F / L (length) units; As is known having the cutting force measurement unit F. k for that reason you have to have dimensions that are needed for the shear force units. writing the equation again in Vc units -.> F = k * (F ^ .5 / L) * L * L then the units are F ^ 0.5 K / L so that the equation is balanced
In conclusion k varies depending of the units you use for example when using (psi) and (in) k = 2 ^ 0.5 lb / in; if you use (Mpa) (mm)
k = 0.166 N ^ 0.5 / 0.5 mm ^ 0.53kg / cm if you use Kg / cm2 and cm. You can check it on a calculator hp or manually.
When we design we have to be clear that units use it because if we can not be miscalculating the elements when you do this by hand or programs. But if you use software such as ETABS SAP2000 and rob-ot etc not to worry because these make the corrections in the measurement units automatically.
On the issue previously presented not check only saw the image presented and I thought the twisting effect the item was high.
When this message goes like (Error O / S # 3 - Shear Stress exceeds maximum allowed) to meet according to the code you're using there are several alternatives. 1 increase the strength of concrete 2 section 3 increase analyze how dissipate the effect this giving you the problem here torsion.
If I made ​​a mistake I apologize.
Regards ..
Pd There is no alternative to increase steel because the software . applied the maximum that allows the code you are using and still not meet him
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Results (English) 3:

I think there is a Little confusion in what is Shear strength of a rectangular Section concrete. Vc = k ^ * (FC) 0.5 * * d like to see the equation BW and BW are units of length D; ^ 0.5 where FC (FC) (F / l ^ 2) = f ^ 0.5 units / l (Length); known as the Shear Force F has Unit of measure. For this Reason, K is to have the dimensions needed for cutting in units of Force. Writing the equation again in VC - > F = (f ^ * k * L * 5 / L) l then the units of K f ^ 0.5 / l for which the equation is balanced.In conclusion, The K varies depending on the units that you use for example when you use (PSI) and (in) K = 2 ^ 0.5 lb / in; if you use (MPA) (mm)0.166 n ^ k = 0.5 / 0.5 mm / cm ^ 0.53kg if you use kg / cm2 and cm. Can take a calculator HP or manually.We have to be clear that when designing units use because if we can"t be misjudging the elements this when you hand or programmes. But if you use software SAP2000 as ETABS and rob ot etc. do not worry because these do the Corrections in the units of measure automatically.The problem presented above do not check only saw the picture presented and it seemed to me that the effect of torsion in the element was high.When you get this message (error or # / S 3 - Shear stress exceeds maximum allowed) to serve as the code you"re using several alternatives. 1 increase the strength of Concrete increase 2 Section 3 analyze and disperses the effect is giving you the problem in this case torsion.I apologize if I have made a mistake.Greetings..PD, there is no alternative to increase steel because the software Applied the maximum allowed the code you"re using and still didn"t meet.The _ UnderdogNovice userNovice userProgress to the next level:2.6%Posts: 2Registered: Thu Apr 14, 2011 12:11 PMCountry: Panama (PA)Thanked: 0 timesI have thanked: 0 times