DESIGN OF REINFORCED CONCRETE SUMBER UTAMA ----
Flexural and Strength Analysis of Beams and one-Way one-Way Slabs Harun Alrasyid
Department of Civil Engineering, Faculty of Civil Engineering and Planning
INSTITUT TEKNOLOGI SEPULUH NOPEMBER
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FLEXURAL ANALYSIS OF BEAMS Uncracke Uncr acked d Conc Concrete rete Stag Stage e At small loads when the tensile stresses are less than the modulus of rupture (the bending tensile stress at which the concrete begins to crac crack) k),, the the enti entire re cros crosss sect sectio ion n of the the beam beam resi resist stss bend bendin ing, g, wi with th compression on one side and tension on the other.
Concrete Cracked–Elastic Cracked–Elastic Stresses Stage As the load is increased after the modulus of rupture of the concrete is exceeded, cracks begin to develop in the bottom of the beam. The moment at which these cracks begin to form—that is, when the tensile stress in the bottom of the beam equals the modulus of rupture—is refe referr rre ed to as the the crac cracki king ng mom moment, nt, Mcr . As the load is further increased, these cracks quickly spread up to the vicinity of the neutral axis, and then the neutral axis begins to move upward.
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Uncrac Unc racked ked Con Concre crete te Stage Stage
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Concrete Cracked–Elastic Stresses Stage The cracks occur at those places along the beam where the actual moment is greater than the cracking moment, as shown in Figure (a). The stresses and strains for this range are shown in Figure (b). In this stage, the compressive stresses vary linearly with the distance from the neutral axis or as a straight line.
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BEAM FAILURE
Moment–curvature diagram for reinforced concrete beam with tensile reinforcing only.
Ultimate-Strength stage
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CRACKING MOMENT The area of reinforcing as a percentage of the total cross-sectional area of a beam is quite small (usually 2% or less), and its effect on the beam properties is almost negligible as long as the beam is uncracked. The stress in the concrete at any point a distance y from the neutral axis of the cross section can be determined from the following flexure formula in which M is the bending moment equal to or less than the cracking moment of the section and Ig is the gross moment of inertia of the cross section: The cracking moment is as follows:
Where: f r is the modulus of rupture of the concrete and yt is the distance from the centroidal axis of the section to its extreme fiber in tension. The “lambda” term is 1.0 for normal-weight concrete and is less than 1.0 for lightweight concrete,
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BEAM - EXAMPLE 1
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BEAM - EXAMPLE 1
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Ultimate or Nominal Flexural Moments
To obtain the nominal or theoretical moment strength of a beam, the simple steps to follow are illustrated 1. Compute total tensile force T = A s f y . 2. Equate total compression force C = 0.85f c ab to A s f y and solve for a. In this expression, ab is the assumed area stressed in compression at 0.85f c .The compression force C and the tensile force T must be equal to maintain equilibrium at the section. 3. Calculate the distance between the centers of gravity of T and C. (For a rectangular beam cross section, it equals d − a/2.) 4. Determine Mn, which equals T or C times the distance between their centers of gravity.
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Compression and tension couple at nominal moment.
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BEAM - EXAMPLE 2
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BEAM - EXAMPLE 2
Beam cross section
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STRENGTH ANALYSIS OF BEAMS ACCORDING TO ACI CODE
To accurately estimate the ultimate strength of a structure, it is necessary to take into account the uncertainties in material strengths, dimensions, and workmanship. This is done by multiplying the theoretical ultimate strength (called the nominal strength herein) of each member by the strength reduction factor , φ, which is less than 1. These values generally vary from 0.90 for bending down to 0.65 for some columns.
Derivation of Beam Expressions Nonlinear stress distribution at ultimate conditions.
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Some possible stress distribution shapes
The values of β1 given by the code (10.2.7.3) are intended to give this result. For fc values of 4000 psi or less, β1 = 0.85, and it is to be reduced continuously at a rate of 0.05 for each 1000-psi increase in fc above 4000 psi. Their value may not be less than 0.65. The values of β1 are reduced for high-strength concretes primarily because of the shapes of their stress–strain curves conditions.
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STRENGTH ANALYSIS OF BEAMS
For concretes with fc > 30 MPa, β1 can be determined with the following expression: β1 = 0.85 − 0.008 (fc − 30 MPa ) ≥ 0.65 The usable flexural strength of a member, φ Mn, must at least be equal to the calculated factored moment, Mu, caused by the factored loads φMn ≥ Mu
Beam internal forces at ultimate conditions.
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STRENGTH ANALYSIS OF BEAMS
For concretes with fc > 30 MPa, β1 can be determined with the following expression: β1 = 0.85 − 0.008 (fc − 30 MPa ) ≥ 0.65 The usable flexural strength of a member, φ Mn, must at least be equal to the calculated factored moment, Mu, caused by the factored loads φMn ≥ Mu
Figure 3. Beam internal forces at ultimate conditions.
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STRENGTH ANALYSIS OF BEAMS
For writing the beam expressions, reference is made to Figure 3. Equating the horizontal forces C and T and solving for a, we obtain
Because the reinforcing steel is limited to an amount such that it will yield well before the concrete reaches its ultimate strength, the value of the nominal moment, Mn , can be written as
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STRENGTH ANALYSIS OF BEAMS
If we substitute into this expression the value previously obtained for a (it was ρfyd/0.85fc ), replace As with ρbd , and equate φMn to Mu, we obtain the following expression:
Replacing As with ρbd and letting Rn = Mu/φbd2, we can solve this expression for ρ (the percentage of steel required for a particular beam) with the following results:
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STRAINS IN FLEXURAL MEMBERS
Furthermore, the code states that the maximum usable strain in the extreme compression fibers of a flexural member is to be 0.003. Finally, it states that for Grade 60 reinforcement and for all prestressed reinforcement we may set the strain in the steel equal to 0.002 at the balanced condition. Then, a value was derived for a, the depth of the equivalent stress block of a beam. It can be related to c with the factor β1 also given in that section:
Then the distance c from the extreme concrete compression fibers to the neutral axis is
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EXAMPLE 3
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STRENGTH REDUCTION or φ Factors
Strength reduction factors are used among these values are the following: 0.90 for tension-controlled beams and slabs 0.75 for shear and torsion in beams 0.65 or 0.75 for columns 0.65 or 0.75 to 0.9 for columns supporting very small axial loads 0.65 for bearing on concrete For ductile or tension-controlled beams and slabs where t ≥ 0.005, the value of φ for bending is 0.90. Should t be less than 0.005, it is still possible to use the sections if t is not less than certain values.
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STRENGTH REDUCTION or φ Factors
Variation of φ with net tensile strain t and c/dt for Grade 60 reinforcement and for prestressing steel.
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Minimum Percentage of Steel
for the minimum amount of flexural reinforcing can be written as a percentage, as follows:
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Balanced Steel Percentage
An expression was derived for depth of the compression stress block, a, by equating the values of C and T. This value can be converted to the neutral axis depth, c, by dividing it by β1: This is the balanced percentage, ρb:
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EXAMPLE 4
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EXAMPLE 4
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DESIGN OF RECTANGULAR BEAMS AND ONE-WAY SLABS
Load Factors
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Design of Rectangular Beams
Before the design of an actual beam is attempted, several miscellaneous topics need to be discussed. These include the following: 1. Beam proportions. 2. Deflections. 3. Estimated beam weight.
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Design of Rectangular Beams
Before the design of an actual beam is attempted, several miscellaneous topics need to be discussed. These include the following (continued):
4. Selection of bars. 5. Cover.
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Design of Rectangular Beams
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Design of Rectangular Beams
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ONE-WAY SLABS
Reinforced concrete slabs are large flat plates that are supported by reinforced concrete beams, walls, or columns; by masonry walls; by structural steel beams or columns; or by the ground. If they are supported on two opposite sides only, they are referred to as one-way slabs because the bending is in one direction only—that is, perpendicular to the supported edges.
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ONE-WAY SLABS
The thickness required for a particular one-way slab depends on the bending, the deflection, and shear requirements.
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EXAMPLE 5
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EXAMPLE 5
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EXAMPLE 5
