For many economic and aesthetic reasons, the design requirements for building floors are getting thinner and the span is getting larger. Concrete is a material that can withstand compression well but is weak in tension, so people have come up with the idea of inserting steel reinforcement to solve the tension problem. Tensile force causes the concrete to crack and the steel reinforcement starts working. Therefore, the view that concrete is not allowed to crack is extreme except in a few cases. Cracks increase the deflection, so the deflection of concrete cannot be calculated without taking into account cracking. Here is a quick way to calculate deflection according to Vietnamese Codes.
1. Basics of flexural behavior of reinforced concrete
Photo 1
Floor beams are the most common and basic structure. Due to their own weight and the load used above, the work is bending: in the beam, there is an internal bending moment M, the lower fiber is stretched while the upper fiber is shortened due to compression. The bending deformation has the form of a curve with radius R (Photo 1).
Note: E-Elastic module, I-moment of inertia, σ-stress, σ=My/I, y-height of the beam cross section to the neutral axis (N.A-where σ=0).
The basic bending formula in Photo 1 applies to homogeneous materials with 1 value of E. Reinforced concrete is not homogeneous, including 2 materials with different E: concrete (Eb) and reinforcement (Es). For the convenience of design engineers, the concept of converted cross section appears in the Standard. By converting the steel reinforcement to the equivalent concrete cross-sectional area to simplify the problem to a single material, concrete, according to the following basic formula, when the beam is still working in the elastic region:
α = Es/Eb ratio of elastic modulus of steel reinforcement and concrete.
Photo 2
– Tensile reinforcement: Area of reinforcement As converted to concrete cross-section α1.As
– Compressive reinforcement: Area of reinforcement A’s converted to concrete cross-section α2.A’s
Usually α2 > α1 because the actual compressive stress in concrete is greater than the elastic calculation due to the effects of shrinkage and creep.
Photo 3
The operation of reinforced concrete beams through stages
– M is still small (small load) M<Mcr, concrete has not cracked (Photo 3 blue segments on the M-ϕ diagram). The entire height D of the beam cross section is mobilized. The structural design problem is elastic up to this point (the diagram is a straight line).
This is the diagram for calculating deflection in beam segments without cracks. The participation of reinforcement in stiffness can be ignored.
– M increases, the fiber under the beam gradually increases tensile stress, until reaching the tensile strength fcr (in TC denoted as Rbt,ser), cracks appear (yellow segment on the diagram).
Photo 4
– Continue to increase M, the lower fiber tensile area cracks gradually, the crack gradually develops upwards, ignoring the height of the concrete in the crack when calculating the stiffness (Photo 4). The steel gradually bears the entire tensile force (red section) and reaches the steel strength Rs (corresponding to My). The naked eye cannot see the crack until the crack width increases by 0.005mm. The red section is a straight line, the cracked section still works elastically. There is a balance M=Cz=Tz, with C, T being the resultant force of the compressive and tensile areas on the stress diagram.
🔍This is the main working stage of reinforced concrete, using it to calculate the deflection at the locations where cracks appear.
Note that although the concrete in the cracked area is not included in the design consultation, it does not mean that it is completely useless, but still has the following functions:
– Withstands shear and torsion
– Holds the position of the reinforcement bars
– Protects the reinforcement from corrosion and fire
Photo 5
– M continues to increase, the upper compressed concrete area is increasingly narrowed and reaches the compressive strength limit Rb (corresponding to Mu), corresponding to the dark red curve on the M-ϕ diagram.
At this time, the beam is damaged, no longer able to bear the load, the structural design uses this diagram to calculate the bearing capacity of the beam according to the first limit state (TTGH), not calculating the deflection in this stage.
If there is too much steel, the stress in the steel has not reached the strength yet, the concrete in the compressed area has been compressed and broken due to exceeding the strength Rb. This is an unacceptable situation in construction design because of waste and danger due to brittle failure of the beam,
2. Deflection calculation
Familiar formula of the standard: f = f1 – f2 + f3
in which:
+ f1: deflection due to short-term effects of the entire load
+ f2: deflection due to short-term effects of long-term loads
+ f3: deflection due to long-term effects of long-term loads
Load = Long-term load + Short-term load
Long-term load: self-weight, long-term part of the load used (furniture, fixed equipment, etc., also known as Live load). The value is looked up in the table of TCVN 2737:1995, for example, equal to 20% of the load used for housing.
Short-term load: live load minus its long-term part; wind load, earthquake, …
Thus f = short-term deflection of short-term load + long-term deflection of long-term load
The reason for taking f1-f2 without including the deflection of short-term load is: Before the load is used, the structure has already borne most of the long-term load which is its own weight, cracks have appeared, making its stiffness no longer correspond to the small value of each short-term load.
In construction design, the deflection is calculated with standard load (not multiplying the overload). The reason is that if there is an overload, the deflection will only increase temporarily, and will decrease when the load returns to normal.
🔲The deflection of a beam subjected to bending is determined by the curvature (1/r) according to the theoretical formula:
f = ∫Ṁ (1/r)dx, integral along the length of the beam according to the coordinate x
The curvature is equal to the moment value divided by the stiffness D: (1/r) = M/D
Ṁ is not the moment M but the moment in the beam due to the concentrated force equal to 1 placed at the deflection location. The Ṁ diagram is in the form of a straight line.
Instead of calculating the integral, the design engineer can calculate the sum over the beam sections by multiplying the diagram on the beam sections with the moment of constant sign. In most practical cases, the cross-section of the floor beam remains unchanged. The curvature (1/r) on each section is allowed by the Standard to be determined proportionally according to M with the curvature (1/r)max at the location of the largest moment.
1/r = (1/r)max. M/Mmax
(1/r)max = Mmax/Dmax. Thus, the curve diagram (1/r) can be considered as the same as the moment diagram M
Depending on the value of Mmax, at this largest cross-section, there may or may not be a crack, from which the stiffness value Dmax corresponding to Mmax is determined.
Multiply the diagram f = area of the diagram (1/r) x ordinate of the diagram Ṁ corresponding to the center of gravity of the diagram (1/r)
In reality, the structure mainly bears evenly distributed loads, the M diagram has a parabolic shape. Proceed to multiply the diagram to calculate the deflection f, compare with the deflection fe to find the stiffness change coefficient k = f/fe
For example, some cases in construction design:
Photo 6
Simple beams
Photo 7
Cantilever
Photo 8
Continuous beams
The moment diagram in the continuous beam is also a parabolic segment similar to the simple beam (with positive moment in the middle of the span) and the cantilever beam (with negative moment on the support). So the problem is also reduced to the same way as the above two cases in structural design.
All cases of bending beams show that the stiffness change coefficient is equal to the value at the maximum moment position: K = Kmax. A simple way to practice is to use structural design software such as ETABS, SAFE with the elastic problem, only replace the elastic stiffness De=bh^3/12 of the floor beam by multiplying by the stiffness change coefficient Kmax=Dmax/De.
🔲Determine the stiffness D according to the standard:
D = Eb1. Ired
Eb1: deformation modulus of concrete
Ired: moment of inertia of the reinforced concrete cross-section
These values vary depending on whether the load is short-term or long-term, whether cracks appear or not, and are determined according to the formulas in the Standard.
When the load is long-term, there is an effect of creep on the concrete.
🔍Creeping is a phenomenon in which the load does not increase in value but is maintained for a long time (long-term load), causing the deformation of the structure to continue to increase over time.
3. Cracking
Customers often feel worried when they see cracks appear. Design consultants need to explain slowly: any type of concrete structure will crack, to be 100% sure that there will be no cracks, the only way is to use Prestressing. Therefore, TCVN allows cracks to appear when bearing loads in most cases:
Allows short-term cracks (due to static load + short-term temporary load) with a width not exceeding acrc1 and long-term cracks (due to static load + long-term temporary load) with a width not exceeding acrc2
The TC’s limit crack width values are intended to:
– Protect steel reinforcement from rust: acrc1 = 0.3mm; acrc2 = 0.4mm
– Limit seepage (for tanks, walls, tunnel roofs, …): acrc1 = 0.2mm; acrc2 = 0.3mm
The formula for calculating crack width according to TCVN is simple and clear. The largest crack width is at the position of the largest internal force, so design engineers do not have to do integration 😂.
4. Calculation example
🔲Cantilever
Structural design calculation diagram of a 2m cantilever reinforced concrete floor.
The largest elastic deflection at the end of the cantilever: fe = 1.49mm
Calculated with moment Mmax at the support (house wall), the stiffness reduction coefficient is 0.21. Long-term deflection: 6.59mm (on the stiffness reduction diagram)
On the nonlinear calculation diagram using SAFE software, with the following parameters:
Creep Coefficient: creep coefficient of concrete, taken from table 11 of the Standard, equal to 1.9
Shrinkage Strain: shrinkage strain of concrete, taken from clause 9.1.8 of the Standard, equal to 0.0002
Long-term deflection result (Long Term Cracked) 6.39mm.
Thus, the result of the elastic diagram applying the stiffness reduction coefficient is appropriate.
🔲Diagram of a conventional floor beam with mid-span deflection
Diagram of the structural design of an apartment building floor.
Maximum elastic deflection at mid-span: fe = 5.2mm
Stiffness reduction coefficient K at the position with moment Mmax in the middle of the span: Floor 0.13; Beam 0.30
Stiffness reduction diagram, maximum deflection at mid-span is f = 23.2mm
Nonlinear diagram using SAFE, maximum deflection at mid-span is f = 24.5mm.
Thus, the deflection result according to the elastic diagram with stiffness reduction coefficient is very suitable.
🔍Links File for example
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🍺Conclusion:
– The deflection of reinforced concrete must be calculated for cracks, cracks are a common working case.
– The structural stiffness is reduced (the deflection increases) several times due to the following reasons:
+ The appearance of cracks reduces the working height of the cross-section
+ Creep increases the deformation due to long-term loads
– When cracks appear, they often cause confusion, so it is necessary to patiently explain to the Customer that it is normal, not all cracks are dangerous, what are the standards for allowing cracks.
– Use construction design software with the stiffness reduction coefficient on the elastic diagram to simplify the calculation of the deflection instead of integrating like the TCVN formula👍.
– The values of the stiffness reduction coefficients calculated according to TCVN are close to the recommended ACI standard experience numbers: 0.25 for floors; 0.35 for beams.
– The stiffness reduction factor is only applied to the calculation diagram when calculating the deflection. The reason for the stiffness and cracking is calculated based on the maximum internal force Mmax of the elastic problem.
– Simplifying the calculation diagram and accurately determining the floor deflection is very important to quickly design the optimal floor solution, contributing greatly to optimizing construction costs 💸.
