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Structural Design 1 (Reinforced Concrete) Chapter 5: Shear and Diagonal Tension of Beams 1. Introduction on Shear Failure/Diagonal Tension Failure Introduction on Shear Failure/Diagonal Tension Failure Beams must also have an adequate safety margin against other types of failure, some of which may be more dangerous than flexural failure. This may be so because of the catastrophic nature of some other types of failure, should they occur. Shear Failure of reinforced concrete, more properly called Diagonal Tension Failure, is one example. Shear failure is difficult to predict accurately. In spite of many decades of experimental research, and the use of highly sophisticated analytical tools, it is not yet fully understood. Furthermore, if a beam without properly designed shear reinforcement is overloaded to failure; shear collapse is likely to occur suddenly, with no advance warning of distress. This is in strong contrast with the nature of flexural failure. Introduction on Shear Failure/Diagonal Tension Failure For typically under-reinforced beams, flexural failure is initiated by gradual yielding of the tension steel, accompanied by obvious cracking of the concrete and large deflections, giving ample warning and providing the opportunity to take corrective measures. Because of these differences in behavior, reinforced concrete beams are generally provided with special shear reinforcement (Stirrups) to ensure that flexural failure would occur before shear failure if the member should be severely overloaded. The shear stresses in most beams are far below the direct shear strength of the concrete. The real concern is with diagonal tension stress, resulting from the combination of shear stress and longitudinal flexural stress. 2. Types of Cracks Types of Cracks Cracks produced by shear/diagonal tension failure Types of Cracks a. Flexural cracks are vertical cracks that extend from the tension sides of beams up to the region of their neutral axes. Should beams have very deep webs, the cracks will be very closely spaced, with some of them coming together above the reinforcing and some disappearing there. These cracks may be wider up in the middle of the beam than at the bottom. b. Inclined cracks due to shear can develop in the webs of reinforced concrete beams either as independent cracks or as extensions of flexural cracks. Occasionally, inclined cracks will develop independently in a beam, even though no flexural cracks are in that locality. These cracks are called webshear cracks. Types of Cracks c. The usual type of inclined shear cracks are the flexure-shear cracks. They commonly develop in both prestressed and nonprestressed beams. d. Torsion cracks are quite similar to shear cracks except that they spiral around the beam. Should a plain concrete member be subjected to pure torsion, it will crack and fail along 45° spiral lines due to the diagonal tension corresponding to the torsional stresses. Although torsion stresses are very similar to shear stresses, they will occur on all faces of a member. As a result, they add to the shear stresses on one side and subtract from them on the other. Types of Cracks e. Bond cracks are due to bond stresses between the concrete and the reinforcement which will lead to a splitting along the bars f. Cracks can also occur in concrete members due to shrinkage, temperature change, settlements, and so on. 3. Shear Stresses in Concrete Beams Shear Stresses in Concrete Beams The average shearing stress, vave in concrete is calculated from: (applicable for rectangular sections.) where: vave = average shearing stress (kPa or MPa); Vu = factored shear force on the critical section at a distance “d” from the support of the beam; bw = width of the web; d = effective depth. Shear Stresses in Concrete Beams This stress is not equal to the diagonal tension stress. It merely serves as an indicator of the magnitude. Should this indicator exceeds a certain value, shear or web reinforcement (such as stirrups) will be necessary. The NSCP had presented basic shear equations that are in terms of shear forces and not shear stresses. In other words, the effective beam areas to obtain shear forces multiply the average shear stresses described in this section. 4. Shear Strength for Non-Prestressed Members Shear Strength for Non-Prestressed Members The nominal or theoretical shear strength of a member Vn is provided by the concrete and the steel reinforcement in accordance to NSCP 422.5.1.1. Vn = Vc + Vs The permissible shear strength Vu of a member is Vu = fVc + fVs Where: Vc = nominal shear strength provided by concrete in accordance with NSCP 422.5.5, 422.5.6, and 422.5.7; Vs = nominal shear strength provided by shear reinforcement in accordance with Section 422.5.10. 5. Shear Strength provided by Concrete for NonPrestressed Members, Vc Shear Strength provided by Concrete for Non-Prestressed Members, Vc Based on NSCP 422.5.3.1, the value of !"# used to calculate Vc, Vci, and Vcw for one-way shear shall not exceed 8.3MPa, unless allowed in Section 422.5.3.2. NSCP 422.5.5 - Vc for Non-Prestressed Members without Axial Force 1. Simplified Equation (NSCP 422.5.5.1) !" = 0.17()*′" ,- . This equation neglects the effect of moment Mu occurring simultaneously with Vu at the section considered. Shear Strength provided by Concrete for Non-Prestressed Members, Vc 2. Detailed / Rational Equation (NSCP 422.5.5.1) This value must be calculated separately for each point being considered in the beam. The purpose of his requirement is to limit Vc near the point of inflection. Shear Strength provided by Concrete for Non-Prestressed Members, Vc 2. Detailed / Rational Equation (NSCP 422.5.5.1) Where: Ø Vud/Mu should not be greater than 1 Ø λ = modification factor for lightweight concrete (NSCP 419.2.4) Ø rw = As/bwd Ø Mu = moment occurring at the point where Vu is critical Shear Strength provided by Concrete for Non-Prestressed Members, Vc NSCP 422.5.6 - Vc for Non-Prestressed Members with Axial Compression 1. Simplified Equation (NSCP 422.5.6.1) *+ !" = 0.17 (1 + / 012′" 45 6 14-. This equation neglects the effect of moment Mu occurring simultaneously with Vu at the section considered. Where: - Nu/Ag shall be expressed in MPa - Nu = factored axial load normal to cross section occurring simultaneously with Vu. It is positive for compression. - Ag = gross area of section. Shear Strength provided by Concrete for Non-Prestressed Members, Vc 2. Detailed / Rational Equation (NSCP 422.5.6.1) This equation takes into account the effects of the longitudinal reinforcing bars, moment Mu and shear Vu magnitudes. Shear Strength provided by Concrete for Non-Prestressed Members, Vc NSCP 422.5.7 - Vc for Non-Prestressed Members with Significant Axial Tension 1. Detailed / Rational Equation (NSCP 422.5.7.1) *+ 0 123′" 56 7 !" = 0.17 (1 + 3.5./ Where: - Nu is negative in tension - Nu/Ag shall be expressed in MPa - Vc shall not be less than zero 6. Shear Strength provided by Shear Reinforcement, Vs 6a. Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs When the factored shear Vu is high, it shows that serious cracks will develop unless some type of additional reinforcing is provided. This reinforcement is called as web reinforcement and takes the form of stirrups that enclose the longitudinal reinforcing faces of the beam. NSCP 422.5.10.5.1 describes them as follows: 1. Stirrups, ties, or hoops perpendicular to longitudinal axis of member; 2. Axial welded wire reinforcement with wires located perpendicular to longitudinal axis of member; 3. Spiral reinforcement. Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs For non-prestressed members, shear reinforcement shall be permitted to also consist of in accordance to NSCP 422.5.10.5.2: 1. Stirrups making an angle of 45 degrees or more with longitudinal tension reinforcement; 2. Longitudinal reinforcement with bent portion making an angle of 30 degrees or more with the longitudinal tension reinforcement; 3. Combination of stirrups and bent longitudinal reinforcement. Web Reinforcement Types of Stirrups Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs Bars called hangers or stirrup supports are placed on the compression side of the beam. The stirrups are passed around the tensile steel and to meet anchorage requirements are run as far into the compression side of the beam as practical and are hooked around the hangers. Bending the stirrups around the hangers reduces bearing stresses under the hooks. If these bearing stresses are too high, the concrete will crush and the stirrups will tear out. When a significant amount of torsion is present in a member, it will be necessary to use closed stirrups. Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs NSCP 420.2.2.4. Types of non-prestressed bars and wires to be specified for particular structural applications shall be in accordance with Table 420.2.2.4(a) for deformed reinforcement and Table 420.2.2.4(b) for plain reinforcement. Web Reinforcement 6b. Minimum Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs NSCP 409.6.3.1 states that the minimum area of shear reinforcement, Av(min), shall be provided in all regions where Vu exceeds 0.5ϕVc except for the cases in Table 409.6.3.1, where at least Av(min) shall be provided where Vu > ϕVc· Minimum Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs When shear reinforcing is required, the amount provided must fall between certain specified lower and upper limits. If the amount of reinforcing is too low, it may yield or even snap immediately after the formation of an inclined crack. As soon as diagonal cracks develop, the tension previously carried by the concrete is transferred to the web reinforcement. To prevent the stirrups from snapping at that time, their area is limited to the minimum value. The least amount of web reinforcement for any section of the beam is Minimum Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs NSCP 409.6.3.3. If shear reinforcement is required and torsional effects can be neglected according to Section 409.5.4.1, Av(min) shall be in accordance with Table 409.6.3.3. Minimum Web Reinforcement Shear Strength provided by Shear Reinforcement, Vs Bent-up Bar Web Reinforcing Minimum Web Reinforcement 6c. Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs The maximum shear Vu must not exceed the permissible shear capacity of the section fVn Vu ≤ fVn The purpose of stirrups is to minimize the size of the diagonal tension cracks or to carry the diagonal tension stress from one side on the crack to the other. Very little tension is carried by the stirrups until after a crack begins to form. Before the inclined cracks begin to form, the strain in the stirrups is equal to the strain in the adjacent concrete. Because this concrete cracks at very low diagonal tensile stresses, the stresses in the stirrups at that time are very small. Stirrups do not prevent inclined cracks until the cracks develop. Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs The nominal shear strength of the stirrups or shear reinforcement Vs crossing the crack can be calculated from the following expression Vs = Avfyn where - Vs = nominal shear strength provided by the shear reinforcement; - Av = cross sectional area of the stirrup bar; - n = number of stirrups crossing the crack. If a U stirrup is used, Av = 2 times the cross sectional area of the stirrup bar. If a UU stirrup is used, Av = 4 times the cross sectional area of the stirrup bar. Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs If it is conservatively assumed that the horizontal projection of the crack equals the effective depth d of the section (thus a 45° crack), the number of stirrups crossing the crack can be determined from the following: n = d/s where - s = center-to-center spacing of the stirrups. - d = effective depth. From the given equations: ( !" = $% &' " #$ %& ' != (! (NSCP 422.5.10.5.3) Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs Going through a similar derivation, the following can be determined for the required area for inclined stirrups (NSCP 422.5.10.5.4): #$ %& ' (!*+, + ./!,) != (! Where: - a = angle between the stirrups and the longitudinal axis. It shall not be less than 30° (NSCP 422.5.10.6.1). Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs NSCP 422.5.10.6.2 If shear reinforcement consists of a single bar or a single group of parallel bars having an area Av, all bent the same distance from the support, Vs shall be the lesser of (a) and (b): where a is the angle between bent-up reinforcement and longitudinal axis of the member. Spacing Limits for Shear Reinforcement Shear Strength provided by Shear Reinforcement, Vs Stirrups cannot resist appreciable shear unless they are crossed at an inclined crack. In order to make sure that each 45° crack is intercepted by at least one stirrup, the maximum spacing of vertical stirrups according to NSCP 409.7.6.2.2. But the desirable stirrups is 100 mm. spacing for Spacing Limits for Shear Reinforcement 7. Design of Reinforced Concrete Beam for Shear Design of Reinforced Concrete Beam for Shear The following steps summarize the design of vertical shear reinforcement (Stirrups): 1. Determine the design shear Vu that the member can sustain; 2. Draw the Vu diagram; 3. Calculate Vu at a distance d from the support; 4. Calculate fVc either considering the effects of Mu or not, depending on the given; 5. Determine the need for stirrups If Vu ≥ 0.5fVc then provide stirrups Design of Reinforced Concrete Beam for Shear The following steps summarize the design of vertical shear reinforcement (Stirrups): 6. Calculate the maximum theoretical spacing of the stirrups (fy ≤ 415 MPa) #$ %& ' != (! Design of Reinforced Concrete Beam for Shear The following steps summarize the design of vertical shear reinforcement (Stirrups): 7. Determine the maximum spacing to provide minimum area of shear reinforcement. 8. Compute maximum spacing. 9. For convenience, calculate the maximum theoretical spacing at 1-meter points along the span. Normally, the first stirrup is placed at a distance of d or s/2 from the face of the support. In practice, the practical minimum spacing is 75-100 mm. 8. Design Considerations Design Considerations In practice 10-12mm f stirrups are used. However, if the calculated design spacing are less than ¼ d, larger diameter stirrups can be used. Another alternative is to use UU stirrups instead of U stirrups. Different diameter stirrups should not be used in the same beam so as not to produce confusion. Design Considerations It is convenient to draw the Vudiagram and carefully label it with values of items such as fVc, ½ fVc, Vu, and Vu at a distance d from the support and to show the dimensions involved. The first stirrup may be placed Ø at a distance d from the face of the support. Ø at ½ s from the face of the support. Ø at 75 mm to 100 mm from the support. Examples (Shear and Diagonal Tension of Beams) Examples – Shear and Diagonal Tension of Beams 1. A rectangular reinforced concrete beam has the dimensions and reinforcement shown. If f’c = 20.70 MPa and fy = 415 MPa, what should be the ultimate shear capacity, Vu of the beam? Use the rational method of computing Vc. with Mu = 225 kNm. Examples – Shear and Diagonal Tension of Beams 2. A simply supported beam has a span of 6m and carries a uniformly distributed service dead load of 21.2 kN/m and a service live load of 51.10 kN/m. The weight of the beam is included in the service dead load. The beam is reinforced with a constant area of flexural steel equal to 3910 mm2. f‘c = 17.24 MPa, fy = 344.70 MPa, Width of beam is 400 mm with an effective depth of 560 mm. a. Compute the nominal shear strength of the concrete at a distance “d” from the support of the beam; b. Compute the maximum spacing of the 10mm stirrups and show the actual positions of the vertical stirrups; and c. Determine the location where stirrups are no longer required on either side of the midspan section. Examples – Shear and Diagonal Tension of Beams 3. A simply supported rectangular beam 400 mm wide having an effective depth of 550 mm carries a total factored load of 116 kN/m on a 6 m clear span. It is reinforced with 6360 mm2 of tensile steel, which continues uninterrupted into the supports. f‘c = 20.73 MPa a. Compute the design shear force which results from the application of the factored loads; b. Compute the shear capacity provided by concrete; and c. At what point is the web reinforcement no longer required measured from the support. Examples – Shear and Diagonal Tension of Beams 4. A rectangular beam has a width of 250 mm and an effective depth of 437.50 mm. It is reinforced with a tensile reinforcement having an area of 1875 mm2 placed at 62.5 mm above the bottom of the beam. The beam is subjected to a factored shear force Vu = 178 kN and moment Mu = 40 kN.m occurring at the point where Vu is critical. f‘c = 24.82 MPa. a. Compute the nominal shear strength provided by concrete (rational method); b. Compute the nominal shear strength provided by concrete with a tensile force of 44 kN (rational method); and c. Compute the nominal shear strength provided by concrete with an axial compression force of 44 kN. Any questions?