1. Shear Force Formula
2. Compressive Force
3. Shear Force Fluid

Determine the Concrete Shear Capacity 17-2 Determine the Required Shear Reinforcing 17-2 Seismic and Nonseismic Spandrels 17-3 Seismic Spandrels Only 17-5 18 Wall Pier Boundary Elements Details of Check for Boundary Element Require-ments 18-1 Example 18-5 19 Input Data General 19-1 Using the Print Shear Wall Design Tables Form 19-1 Summary. Start studying Chapter 6. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The yield point of the material is exceeded B. The elastic limits of the material is not exceeded C. The elastic limits of the material is exceeded. Shear force at any section or point in a beam is the algebraic sum of all the.

Shear Strenth Of Reinforced Concrete Beams Per ACI-318-02.1.PDHonline Course S153 (4 PDH)Shear Strength of Reinforced ConcreteBeams per ACI 318-022012Instructor: Jose-Miguel Albaine, M.S., P.E.PDH Online PDH Center5272 Meadow Estates DriveFairfax, VA Phone & Fax: 703-988-0088www.PDHonline.orgwww.PDHcenter.comAn Approved Continuing Education Provider.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 1 of 24Shear Strength of Reinforced Concrete Beams per ACI 318-02Course Content1. IntroductionIn a simple beam subjected to bending, the fibers above the neutral axis arein compression, whereas tensile stresses occur in the fibers below this axis.The factors influencing shear strength and formation of inclined cracks areso numerous and complex that a definitive conclusion regarding the exactmechanism of inclined cracking resulting from high shear is difficult toestablish. The behavior of reinforced concrete beams at failure in shear isdistinctly different from their behavior in flexure. They fail abruptly withoutsufficient advanced warning, and the diagonal cracks that develop areconsiderably larger than the flexural cracks.A concrete beam reinforced with longitudinal steel only, will developdiagonal tensile stresses that tend to produce cracks, as indicated in Fig.

1.These cracks are vertical at the center of span and become more inclined asthey approach the supports, sloping at an angle approximately 45o. Thestresses that cause these cracks are known as diagonal tension. In order toprevent this type of failure additional reinforcing bars are added, which arecalled stirrups (perpendicular to the longitudinal axis of beam).A practical assumption is that the intensity of the vertical shear is consideredto be a measure of the intensity of the diagonal tension. Thus, when we usethe term shear, we really refer to the diagonal tension stresses. Since thestrength of concrete in tension is significantly lower than its strength incompression, design for shear is of major import in concrete structures.This course will use the strength design method or ultimate strength design(USD) to determine the shear strength of nonprestressed concrete beams.This method is consistent with the prevalent methodology endorsed by ACI318-02.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 2 of 242. Shear in Homogeneous Elastic BeamsHomogeneous elastic beams have stresses proportional to strains, and theshear stresses are obtained from the principles of classical mechanics ofmaterials, such as equation (1):v =I bV Q (Eq. 1)V = Shear force at section under considerationQ = statical moment about neutral axis of that portion of cross section lyingbetween a line through point in question parallel to neutral axis andnearest face (upper or lower) of beam.I = moment of inertia of cross section about neutral axisb = width of beam at given pointFigure No.

1.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 3 of 24The infinitesimal elements A1 and A2 of a rectangular beam in Figure 2 areshown with the tensile normal stress ft and shear stress v across plane a1 – a1,and a2 – a2 at a distance y from the neutral axis.dbN.A.aaA1A2Figure No. 2a1 a1a2 a2Cross Section Shear Stress DistributionvNeutral AxisyBending Stress DistributionfcftyFigure 3 shows the internal stresses acting on the infinitesimal element A2,and using Mohr’s circle, the principal stresses for element A2 in the tensilezone below the neutral axis can be computed as:ft (max) =ft2+ ft22+ v2principal tension (Eq. 2.a)fc (max) =ft2- ft22+ v2 principal compression (Eq. 2.b)The same procedure can be used to find the internal stresses on theinfinitesimal element A1.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 4 of 24VCTvvftftA2vvftftft (max)θStress state in element A2θTan 2222 max = vft / 2Figure No.

3The behavior of reinforced concrete beams is more complex since the tensilestrength of the concrete is approximately 10% of its compressive strength.For compressive element A1 the maximum principal stress is in compression,therefore cracking is prevented. On the other hand, for element A2 below theneutral axis the maximum principal stress is in tension, hence cracking maybe induced. For an element at the neutral axis near the support, the state ofstress would be that of pure shear thus having the maximum tensile andcompressive principal stress acting at a 45oangle from the plane of theneutral axis.Due to the complex mechanism of shear resistant mechanism in reinforcedconcrete beams (nonhomogeneous, non-elastic behavior), the ACI –ASCEJoint Committee 426 gives an empirical correlation of the basic conceptsdeveloped from extensive test results.3. Modes of Failure of Concrete Beams without Shear ReinforcementIn areas where the bending moments are large, cracks develop almostperpendicular to the axis of the beam. These cracks are termed flexuralcracks. In region of high shear due to the diagonal tension, the cracksdevelop as an extension of the flexural crack, and they are called flexureshear cracks. 4 shows the types of cracks expected in a.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 5 of 24reinforced concrete beam with or without effective diagonal tensionreinforcement.LcApplied LoadFlexural andFlexure-ShearDiagonalTensionFlexural andFlexure-ShearDiagonalTension(Web Shear) (Web Shear)ContinuousSupportSimpleEnd SupportFigure No.

4dIn the region of flexural failure, the cracks are mainly vertical in the middlethird of the beam span and perpendicular to the lines of principal stresses.The dominant cause of these cracks is large flexural stress f.Diagonal tension cracks may precipitate the failure of the beam if thestrength of the beam in diagonal tension is lower than its strength in flexuraltension. This type of crack starts with the development of a few verticalflexural at midspan.

Shear Force Formula

Thereafter, the bond between the reinforcing steel andthe surrounding concrete is destroyed at the support. Finally, two or threediagonal cracks develop at about 1 ½ to 2d distance from the face of thesupport. At this stage it is recognized that the beam has reached its shearstrength. The beams categorized as intermediate beam usually fall in thiscategory.The shear strength of deep beams is predominantly controlled by the effectof shear stress. These beams have a small shear span/depth ratio, a/d and arenot part of the scope of this work. ACI 318 section 11.8 addresses the shearstrength of deep beams.

Compressive Force

Shear Force Fluid

5 and Table No. 1 for classification ofbeams as a function of beam slenderness.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 6 of 24BeamCategoryFailureModeShear Span/Depth Ratio as aMeasure of SlendernessSlender FlexureConcentrated Load, a/d Distributed Load, Lc/dExceeds 5.5 Exceeds 16Intermediate Diagonal Tension 2.5-5.5 11-16Deep Shear Compression 1-2.5 1-5TABLE 1: Beam Slenderness Effect on Mode of FailureFor a uniformly distributed load, a transition develops from deep beam tointermediate beam effect.LcwPadFigure No. 5.www.PDHcenter.com PDH Course S153 www.PDHonline.orgPage 7 of 244. Shear Strength CriteriaThe ACI provisions of Chapter 11 considers the shear strength of beams onan average shear stress on the full effective cross section bwd. In a memberwithout shear reinforcement, shear is assumed to be resisted by the concreteweb.