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Reinforced Concrete Beam Design · Acedra
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Reinforced Concrete Beam Design
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0:15
Concrete is a material that can handle a very small amount of tensile stress. If this stress is exceeded, the concrete fractures, making it unsuitable for applications where members are in tension only. In compression, however, concrete is very strong and can handle a large amount of stress before it eventually fractures. Beams are mainly subject to bending moments. Typically, due to gravity loads, the top of the
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beam is compressed while the bottom is in tension. Because concrete is weak in tension, the beam will fracture at the bottom. That is why reinforcement is required, mainly in the form of steel rebars. In the presence of steel rebars, when the beam is subject to bending moments with the rebars on the tension side, the top portion of the concrete will be in compression, and the tensile loads will be fully carried out by the steel rebars. This allows for an
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effective cross-section against bending that is not only stronger but also more ductile. This video aims to explain the ACI provisions for reinforced concrete beams, specifically for reinforced non-prestressed normal weight concrete beams subject primarily to bending moment and shear. It is highly advisable to watch this video to the end to know how to reinforce and dimension a concrete beam according to ACI under different conditions.
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We will also soon post a video with a full example, so please subscribe to not miss that video. Please note, however, that one example is never enough as each project is individual, so both videos combined will give the best possible insight. ACI specifies stability requirements for beams. If the beam is not continuously braced, say by a concrete slab, then the spacing of lateral bracing shall not exceed 50 times the width of the beam. Required strength shall be calculated
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in accordance with the factored load combinations in the following table.
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The minimum depth of the reinforced concrete beams can be determined referring to Table 9.3.1.1 of ACI. This table is applicable when using normal weight concrete and steel rebars with a yield strength of 60,000 psi. For fy other than 60,000 psi, the expressions in Table 9.3.1.1 shall be multiplied by the following factor. Simply supported means that the beam is supported at the very ends only. When a beam runs across
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multiple supports, the first span is considered one end continuous. Whenever the beam runs above two supports, as shown in the middle span, it has two continuous ends. Cantilever can either be taken as the overhang portion of a continuous beam or a beam that is fixed from one end only. Non-prestressed beams with axial forces less than 0.1 times the concrete compressive strength multiplied by the gross cross-sectional area shall be tension controlled as per Table 21.2.2,
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where the yield strain can be determined as follows. Strain in concrete and non-prestressed reinforcement shall be assumed to be proportional to the distance from the neutral axis. Equilibrium shall be satisfied at each section. The maximum strain at the extreme concrete compression fiber shall be assumed to be 0.003. Tensile strength of concrete is assumed to be nonexistent. Concrete stress of 0.85fc shall be assumed uniformly distributed over an equivalent
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compression zone bounded by edges of the cross-section and a line parallel to the neutral axis located at a distance “a” from the fiber of maximum compressive strain. A is equal to Beta one times c. This sets up an upper limit for the area of steel reinforcement. A minimum area of flexural reinforcement As,min shall be provided at every section where tension reinforcement is required by analysis. As,min shall be the larger of (a) and (b). The value of fy is limited to 80,000 psi.
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The width of a beam bw can be determined based on the flexural strength requirements by satisfying the following strength design equation where Mu is the maximum factored moment. At this stage, the required flexural reinforcement is not yet known, however, a range can be determined based on the required minimum and maximum As. For Grade 60 reinforcement and a specified concrete compressive strength of 4000 psi, as previously discussed,
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the minimum and maximum As can be determined as follows. The equation for maximum As is coming from the equilibrium of forces. Because the distance “a” is equal to Beta one times c, the expression can be written to solve for As such that the section is tension controlled. Using an initial estimate of 0.01bw d for the area of the reinforcement, which is about the average of max and min As, bw can be obtained by setting Mu equal to Phi Mn. Then bw can be determined
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as follows. The following are guidelines for beam economy when sizing beams: use whole inch increments for beam dimensions, use beam widths in multiples of 2 or 3 inches, use constant beam size from span to span and vary the reinforcement as required, use wide shallower beams rather than narrow deeper beams, use beam width at least 4 inches wider than the supporting column whenever possible. If not, then the same width. Avoid using beams narrower than the column as this increases
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the work required for the formwork because the columns extend along the depth of the beam. This also reduces the reinforcement congestion at the intersection between beam and column. After sizing the beam and approximating the area of steel required by taking the average of the minimum required and maximum allowed steel area, the number and size of bars can be determined to be equal to or greater than this number. After obtaining the exact steel reinforcement area,
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the strength of the member in flexure should be greater than the maximum magnified applied moment. The nominal flexural strength is based on moment equilibrium. The depth of the equivalent compressive stress block “a” can be found by setting the tension force in the reinforcement equal to the compressive force in the concrete block. A minimum area of shear reinforcement Av,min shall be provided at all regions where the
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following condition is met. Av is the area of two legs of a closed stirrup. Av,min divided by the shear reinforcement spacing s shall be the larger of (a) and (b) in Table 9.6.3.4.
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An exception is made for the following cases in Table 9.6.3.1 where Av minimum is required only if Vu is greater than Phi Vc. For normal weight concrete, lambda is equal to 1. The nominal one-way shear strength Vn is determined according to ACI 22.5, where there is a contribution from concrete and a contribution from steel. For non-prestressed members, Vc is determined according to the equations in Table 22.5.5.1. The following table contains
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these equations as well as the maximum permitted Vc according to ACI 22.5.5.1.1. The factor lambda “s” is required to account for the fact that the concrete shear strength does not increase proportionally to the depth. The term rho w is equal to the area of the flexural reinforcement As divided by bw times d. To minimize the likelihood of diagonal compression failure, the cross-sectional dimensions of the section must be selected to satisfy ACI Equation 22.5.1.2.
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Stirrups oriented perpendicular to the axis of the member and anchored to the longitudinal flexural reinforcement are the most commonly used type of shear reinforcement for beams in buildings. The nominal shear strength provided by stirrups is determined by ACI Equation 22.5.8.5.3. In general, Av is equal to Ab times the number of stirrup legs provided. Ab is the cross-sectional area of one
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stirrup leg. The maximum stirrup spacing across the width of the member is given in ACI Table 9.7.6.2.2 which translates to a minimum number of stirrup legs needed across the width of the beam. When more than two stirrup legs are required, the following configurations are to be used. The required area of transverse reinforcement can be determined with the following expression. The shear reinforcement requirements according to ACI are summarized here as follows. A minimum area
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of torsional reinforcement shall be provided in all regions where the torsion is greater than or equal to the threshold torsion. The threshold torsion can be extracted from Table 22.7.4.1(a). The minimum transverse reinforcement required will be the greater of (a) and (b), where At is the area of one leg of a closed stirrup. The minimum longitudinal reinforcement required will be the lesser of (a) and (b), where ph is the perimeter of the centerline of outermost
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closed transverse torsional reinforcement, and pcp is the outside perimeter of the concrete cross-section. As in most cases torsion in concrete members is ignored or not significant, the evaluation of the torsional strength will be explained in a separate video. The minimum concrete cover to be provided can be determined according to Table 20.5.1.3.1,
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which depends on the exposure to weather and humidity as well as the size of the used reinforcement. The minimum spacing of flexural reinforcing bars is shown as follows, where dagg is the nominal aggregate size in the concrete mixture, and Db is the overall diameter of the longitudinal reinforcement. The following equation can be used to determine the maximum number of reinforcing bars n_max that can fit in a single layer based on the minimum spacing
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limits. The value must be rounded down to the next whole number. Maximum center-to-center spacing of reinforcing bars is given in ACI 9.7.2.2 to control flexural cracking. The following equation can be used to determine the minimum number of reinforcing bars n_min that must be in a single layer to control flexural cracking. The value must be rounded up to the next whole number. For beams with an overall depth greater than 36 in., longitudinal skin reinforcement must
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be uniformly distributed on both faces of the beam for a distance equal to at least h/2 from the tension face to help control the width of the cracks that can form on these faces. The development length of deformed bars and deformed wires in tension is the embedment length required for the reinforcement to achieve the required tension without slipping from the concrete. The required development length is the greater of the lengths calculated in Table 25.4.2.3 and 12
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inches. To determine the modification factors in the equations, refer to Table 25.4.2.5. A reinforcement lap splice is a method of joining two reinforcement bars by overlapping them and wiring them together. For non-contact splices in flexural members, the transverse center-to-center spacing of spliced bars shall not exceed the minimum of one fifth of the required lap splice length and 6 inches. The lap splice length of deformed bars
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in tension is determined according to Table 25.5.2.1 and it depends on the development length shown earlier. Along development and lap splice lengths of longitudinal bars with yield strength greater than or equal to 80,000 psi, transverse reinforcement shall be provided such that Ktr is not smaller than 0.5 times the bar diameter. N is the number of longitudinal bars being developed and s is the spacing between transverse reinforcements. Atr
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is the total cross-sectional area of all transverse reinforcement within spacing s that crosses the potential plane of splitting through the reinforcement being developed. Shear reinforcement must be properly anchored and developed to be fully effective. Requirements for the development of stirrups are given in ACI 25.7.1 and are illustrated here. In general, the stirrups must extend as close to the compression and tension surfaces of the
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member as cover and other constraints permit and must extend at least a distance d from the extreme compression fiber. Each bend in the continuous portion of a single or multiple leg U stirrup and each bend in a closed stirrup must enclose a longitudinal bar. Minimum inside bend diameters and standard hook geometry for stirrups, ties, and hoops are given in ACI Table 25.3.2. Beams must have sufficient width to allow for bend radii at corners of stirrups.
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The following summarizes the recommended flexural reinforcement of continuous perimeter and non-perimeter beams. Perimeter beams are those located at the perimeter of the level in question. All remaining beams are non-perimeter beams. Last but not least, the deflection limits have to be met for serviceability requirements. For beams loaded such that the maximum bending moment at service loads Ma is less than or equal to
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two-thirds the cracking moment, the section is assumed to be uncracked. In this case, the gross moment of inertia Ig of the section (neglecting the longitudinal reinforcement) can be used in the deflection calculations. The cracking moment is determined by ACI Equation 24.2.3.5, where yt is the distance from the centroid of the gross section, neglecting the reinforcement. The modulus of rupture fr is calculated as follows.
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Once a section cracks, the cracked moment of inertia should be used instead. In general, it is determined by transforming the longitudinal steel reinforcement area to an equivalent area of concrete. The concrete below the neutral axis, which is located a distance equal to kd below the extreme compression fiber, is cracked and is ignored in the analysis. For a rectangular concrete beam with one layer of longitudinal reinforcement,
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it works out as follows, where n is the ratio of the steel elastic modulus to that of the concrete. In one-way flexural beams, creep and shrinkage due to sustained loads cause time-dependent deflections that can be 2 to 3 times greater than the immediate deflections. Ignoring them can be catastrophic. They can be calculated by multiplying the calculated immediate deflections due to sustained loads like dead loads by the factor lambda delta,
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which is determined by taking into account the area of the steel compression reinforcement if it exists. The value Xi depends on the duration of the load application.
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