Calculation of the Ultimate Limit State of shear exhaustion in reinforced concrete sections, according to EHE-08, Article 44 (4th edition, 2010).

• The calculation of the shear is a very complex issue, and this is reflected in the regulations, and therefore, in this program. As in most calculations specified in the regulations, the shear is calculated by verification, not by sizing, that is, from a starting data including a proposed transversal reinforcement, it is checked which maximum shear stress is admissible.
  
  
• For the shear calculation "... is established as a general method of calculation, the struts and ties." (EHE-08, 44.1).
  
  
• The shear calculation is applicable "exclusively at linear elements subjected to combined stresses of bending, shear and axial, and to plates or slabs working primarily in one direction". (EHE-08, 44.2).
  
  
• It has been specified the edition of the regulation used in this program because of the changes that this article has suffered between editions, and the detected errata in this edition, we trust remedied in later editions.
  
  
• In this program it is considered that the lower reinforcement is the tensioned one when the section is subjected to positive bending moments (as in the case of a double pinned beam). Therefore, if negative moments are introduced, the tensioned reinforcement will be at the top (as in the end of a fixed beam).
  
  

Parameters:
Materials:
Concrete: fck =	2530354045 505560708090100	N/mm2
Concrete deduction factor Yc =				Yc = 1.5 except in accidental situation (Yc = 1.3), fatigue or fire (EHE-08, 15.3)
Compression fatigue factor αcc =		(*1)		Usually αcc = 1 (at EHE-98 αcc = 0.85)
Indirect concrete control:	Sí
Steel: fyk =	.400500.	N/mm2		Ys = 1.15 except in accidental situation (Ys = 1.0), or intense control (Ys = 1.10) (see EHE-08, 15.3)
Steel deduction factor Ys =
Limit fyd to 400 N/mm2:	Sí			Recommendation at code comment of 40.2 for cracking control.
Section:
Minimum net width (2000 max.) b0 =		mm
Height (3000 max.) =		mm
Top nominal cover =		mm
Bottom and sides nominal cover =		mm
Is a column:	Sí
Shear reinforcement:
Vertical branches =		ø	.No hay681012 1416202532 40.	mm(*2)each		mm
Tilted branches =		ø	.No hay681012 1416202532 40.	mm each		mm
Angle with the member axis α =		º		The angle between the concrete compression struts and the member axis must be between 26.57º and 63.43º.
Ángulo de las bielas de compresión θ =		º (*3)
Longitudinal reinforcement:
Top reinforcement:		ø	.681012 1416202532 40.	mm	In flexural stressed members is considered compressed only the upper reinforcement. In columns is considered all the entered reinforcements.
Bottom reinforcement:		ø	.681012 1416202532 40.	mm
Lateral reinforcement (each face):		ø	.681012 1416202532 40.	mm(*4)
Stresses:	At one effective height		At support's edge		(Values at both columns should have the same sign.)
Compression value prestress included Nd =		kN		kN(*5)	(Compression is positive)
Moment prestress included Md =		mkN		mkN	Negative moments indicate that the tensions or lower compressions occur in the upper reinforcement. At columns is checked if compound compression exists (M/N < h/2).
Normal stress parallel to the shear stress σyd = (prestressed included)		N/mm2		(Tension positive. If the element is subject to compression parallel to the shear, for example, soil pressure in buried foundation elements).
(*1)The fatigue factor reflects this effect in the concrete, when subjected to high levels of compression due to long-term loads. It is generally adopted αcc = 1, although the author of the project should estimate its reduction to 0.85, if the permanent and total loads ratio is high, or depending on the characteristics of the structure (EHE-08, 39.4).
(*2)The code allows to reduce the bending radius in hoops or stirrups less than or equal to 12 mm diameter, which is recommended not use larger diameter styrrups, being preferable to place double or triple styrrups when required. (EHE-08, 69.3.4).
(*3)According to regulations, it looks like any value within the specified range can be adopted. The smaller this angle the more beneficial results are obtained; and the higher angle the more safety results. See author's notes.
(*4)The side reinforcement introduced is considered in columns as compressed. If you do not want to consider it, you should not introduce it into the program. However, it must be verified that reinforcement separations specified in the regulations are met (the "Aggregate and separations" utility can be used)
(*5)As the low axial values are favorable for the shear, two calculations should be made: one with total factored and combined stresses, and other only with dead loads stresses and with reduction factor.
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Results:
Materials:
Concrete compressive strength fcd =	20	N/mm2		(EHE-08, 39.4)
Concrete oblique compressive strength f1cd =	12	N/mm2		(EHE-08, 442.3.1 and 40.3.2 comments)
Average concrete tensile strength fct,m =	2.9	N/mm2		(EHE-08, 39.1)
Concrete effective shear strength (with shear reinf.) fcv =	30	N/mm2		(EHE-08, 44.2.3.2.2)
Design steel strength fyd =	400	N/mm2		(EHE-08, 38.3)
Web oblique compression:		(Calculated at support border)
Effective axial stress σ'cd (for Vu1) =	0	N/mm2 (*6)		(EHE-08, 44.2.3.1)
k =	1			(EHE-08, 44.2.3.1)
α average value =	90	º		(EHE-08, 44.2.3.1 comment)
Shear of exhaustion Vu1 =	1317.6	kN		(EHE-08, 44.2.3.1)
Tension at web:		(Calculated at one effective height from the support border)
Mechanical lever arm z =	494.1	mm (*7)		(EHE-08, 44.2.3.2.2)	z = 0,9 d (not flexural-compresion: Md/Nd > h/2)
Contribution of transversal reinforcement Vsu =	132.46	kN		(EHE-08, 44.2.3.2.2)
Stress parallel to the guideline σxd =	0	N/mm2		(tension is positive)
Cracks tilt angle cotg θe = (Between 0,5 and 2)	1			(EHE-08, 44.2.3.2.2)
β factor =	1			(EHE-08, 44.2.3.2.2)
ξ factor =	Value visible only to users with active subscription. (+info)
Tensioned reinforcement geometric ratio ρ1 =	5.4935	‰(*8)		(EHE-08, 44.2.3.2.1.2)
Average axial stress at web σ'cd =	0	N/mm2 (*9)		(EHE-08, 44.2.3.2.1.2)
Concrete contribution Vcu=	89.61	kN		(EHE-08, 44.2.3.2.2)
Minimum contribution Vu2=	122.12	kN(*10)		(EHE-08, 44.2.3.2.2 and 44.2.3.2.1.2)
Actual concrete contribution Vcu=	122.12	kN		(EHE-08, 44.2.3.2.2)
Exhaustion due to tension at web Vu2t=	254.58	kN(*11)		(EHE-08, 44.2.3.2.2)
Maximum allowable shear =	254.58	kN		(EHE-08, 44.2.3)
(*6) For the calculation of the effective axial stress, it can be considered that Ac is the total area, or the net area, without the reinforcements. In the latter case, the stress increases because the section reduces, being a favorable situation with low stresses and unfavorable with high streseses (that is why it is necessary to calculate with all the combinations of actions described in the code). This program always considers that the area is total (although not on the safety side), as reinforcements actually, being placed, will cause a reduction in the axial stress on the concrete.
(*7) Fixed a typo in the EHE so that the mechanical arm z must not exceed 0,9d instead of 0,9d'.
(*8)This geometric ratio is evaluated considering the effective height instead the total depth, as usual. The reinforcement to consider must be anchored to a minimum of one effective height from the study section.
(*9) Do not confuse with the effective axial stress, that the standard denoted by the same notation.
(*10) Corrected the EHE typo in 44.2.3.2.2: for f1/2 must say fcv1/2 and 44.2.3.2.1.2, where again there is another typo, since the brackets haven't been printed.
(*11) The EHE is ambiguous on this point, since it uses the term Vu2 to designate two different concepts: first the concrete resistance to shear stress, and secondly the resistance of the concrete and steel to the shear stress. Therefore it has been substituted in this formula the term Vu2 by Vu2t.

• In the calculation of the mechanical arm z in the case of flexural compression, it has been considered for the tensioned reinforcement distance to the axis z0, and for the mechanical capacity of the tensioned reinforcement Us, the less compressed reinforcement, since in principle, no tensioned reinforcement exists. The same happens with the geometric amount of tensioned reinforcement calculation ρ₁.
  
  
• The tilted reinforcements have to be placed in the tensions direction, that is, transverse to the compression struts. This way, for vertical loads, the top of the bars should be closer to the support. At columns where the shear stress can act in both directions (under wind or earthquake loads), the use of angled reinforcements may not be useful for not having symmetric behavior.
  
  
• The author does not understand how to interpret the regulation when having to choose an arbitrary value for the angle of inclination of the compression struts. For one, the smaller this value (26.57º) the most beneficial results are obtained, but in different situations it has to be choosed the worst, then it should take the largest angle (63.43º). However, the same code proposes in the comment of 44.2.3.1 a simplified formula wherein the selected strut angle is 45°, not being the most unfavorable situation. That's why the user should adopt the consideration that deems appropriate.
  
  
• For the tilted reinforcements, the more effective angle and produces higher shear strength is the perpendicular to the compression struts. In practice, angled rebars are not unusual and less with any angle, so when placed will be at 45º. The consequence is, as the compression struts cannot be horizontal, if slightly inclined styrrups are placed with the top closer to the support, would generate more effective transverse reinforcement, although the standard indicates that at least the minimum amount should form 90° with the beam axis (EHE-08, 44.2.3.4.1), without indicating limitation for columns. (In some cases with struts at 45°, to tilt 10º the styrrups will cause a 1.5% steel increasing, and a behavior improvement of 5%).
  
  
• The author has considered that the regulation has the indicated misprints, and considered as such. If such errors actually does not exists, the generated results from this program would not be correct, and so, be invalidated. It is the user who must determine the adequacy of the results, until the appearance of a new edition of the code.
  
  

Version 12/02/2013