Estimating Bending Stress: Approximate Formula

Approximate calculation for the Reinforcement of a Section subjected to simple Bending Stress

Introduction

On some occasions it is necessary to calculate the reinforcement in the section of a concrete beam subjected to bending stress in an approximated form, either to verify calculations obtained in programs, to give an answer on the jobsite, for the preliminary dimensioning of a project before it is initiated, etc.

The formula that will be indicated later has been widely disclosed and is applied for the purposes indicated previously, only the question remains as to how precise and reliable its application is, as well as the range in which said formula is valid. 

Then, the purpose of this work is to precisely clarify the origin and validity of said equation, its reliability, and to define the range in which it can be applied.

The formula is:

Explanation of the origin of the approximated equation

A rectangular section subjected to simple bending as shown in Fig. 2 reaches the point of failure when the concrete in the most compressed fiber reaches a unitary deformation of 0.003, while the tensile steel closest to the tension edge has a deformation εs > 0.004.(Section 9.3.3.1 ACI 318 -14) This limit in the deformation of the reinforcement is intended to mitigate the fragile type of failure of the section in the case of an overload in the beam. 

The former leads to the section having a compressed upper zone whose resultant C has to be equal to the tensile force of the steel T in order to fulfill force equilibrium ΣF = 0

Since the deformation of the reinforcement is εs > εy = 0.002, the force in the steel will have a value of T = As.fy.

The resulting compression in the concrete will have a value of C = 0.85f’c.β1.c.b = 0.85f’c.β1.k.d let c = k.d.

From the equilibrium equation C = T , we have:

0.85f’c.β1.k.d.b = As.fy

Using mechanical volume ωR = As.fy/b.d.f’c, results:

.    ωR = 0.85β1.k

The moment equation ΣM = 0 will remain as:

Mn = 0.85f’c.β1.k.d.b(d – β1.k.d/2)

Using specific moment μR = Mn/b.d2.f’c and substituting ωR = 0.85β1.k

μn = ωR(1-ωR/1.7)

Moreover, the equation relates the nominal moment that a section can resist as the amount of reinforcement increases. 

Being Mu = ϕ Mn and μn = ϕ μR we have:

.    μu = Φ•ωR((1-ωR)/(1.7))

According to ACI 318 -14 the value of ϕ = 0.9 when εs < 0.005 (tension controlled) and decreases for lower values of εs. (Table 21.2.2)

This equation has a lower limit that is defined by the minimum amount and an upper limit that is defined by the smallest unitary deformation of the reinforcement ξs = 0.004.

ωRmin  = 0.00333x60000psi/4000psi = 0.05

kmax = 0.003/(0.003+0.004) = 0.4286 ωRmax = 0.85×0.85×0.4286 = 0.31

As a result, we can see in Fig. 3 the variation curve μu vs ωR for the complete range of ωR and a thicker line shows the range of values between the minimum mechanical volume ωR = 0.05 and the maximum mechanical volume ωR = 0.31.

In what follows, we will consider several simplifications, consisting of:

1. The concrete resistance will be set to f’c = 4000 psi. The reason is very simple, the compressed area of a bending section is small, and it is not economical to use high f’c values. In the case where f’c is greater, the equation is conservative. This leads to the assumption of β1 = 0.85

     2. The strength of the steel will be set at fy = 60000 psi

     3. We can assume that the maximum amount does not exceed 65% of the amount that generates the maximum capacity of the section, that is ωRmax = 0.31×0.65 = 0.2, which would be 4 times the amount of the minimum of ωRmin = 0.05

Evaluating the equation of μu for ωR = 0.2 we have:

μu = 0.9.0.2(1-0.2/1.7) = 0.16

The linear function that passes through the origin and through the point on the curve ωR = 0.2, μu = 0.16 will be:

μu/ωR = 0.16/0.2 = 0.8

μu = 0.8 ωR meaning:

Mu/b.d2.f’c = 0.8.As.fy/b.d.f’c where:

As = Mu/0.8.d.fy adjusting the units, gives:        As = Mu(kip.ft).12/0.8×60(ksi) which results:

 

.  As = (Mu)/(4•d)

  As = area of the reinforcement in in2

  Mu = factored moment in kip.ft      d = effective height in in.

With which the traditional equation is demonstrated, being completely valid for the stated conditions giving results on the safe side when the geometric amount is less than As/bd = 0.2×4000/60000 = 0.0133

Another way of saying it will be: when the geometric amount is less than 4 times de minimum geometric amount, that is, As/b.d < 0.00333.4 = 0.0133 the equation obtained is completely sound. 

Certainly, throughout the range of application of the formula, the amount of reinforcement obtained always exceeds that of the reinforcement calculated through more precise methods. If we apply the formula to larger amounts, then the formula gives lower values than what the section actually requires. 

Next, we will see two examples, the first within the valid range and the second outside of the range.

First example to verify the approximated formula:

 

1.) Section A 12″x24″ f’c = 4000 psi d = 22″ Mu = 155 kip.ft

Exact calculation: As = 1.66 in2

Applying approximated formula:

As = 155 kip.ft/4×22.in As = 1.76 in2 6 % mayor As/bd = 1.76 in2   /12 inx22 in = 0.0066  < 0.0133 O.K.

Note that in this case the geometric amount is approximately half of 0.0133, corresponding to the intermediate zone of validity that we have seen. In this zone is where the maximum difference between the real value and the approximated value should occur, always on the safe side. 

Second example to verify approximated formula: 

2.) Section B 12″x24″ f’c = 4000 psi d = 22″ Mu = 380 kip.ft

Exact calculation: As = 4.52 in2

Applying approximated formula: 

As = 380 kip.ft/4×22.in As = 4.32 in2

As/bd = 4.32 in2 /12 inx22 in = 0.0164 > 0.0133 Not valid.

Finally, concluding the initially stated objective of obtaining the simplified equation from the real function μu vs ωR of rectangular section beams, we find something interesting when transforming from the nominal equation μR vs ωR to the equation μu vs ωR. It must be clarified that the discussion has nothing to do with the approximated formula, but it is of interest, and we want to comment on it because of its importance.

In Fig. 4 we have enlarged the top right portion of Fig. 3 to show something very interesting and important. 

Consequently, when transforming from the nominal equation μR vs ωR to the equation μu vs ωR by multiplying μR by ϕ, from ωR = 0.271 (εt = 0.005) to ωR = 0.433 (εt =0.002) the value of ϕ decreases from 0.9 to 0.65. Resulting in μu remaining almost constant throughout this section. 

As a result, the capacity of the section does not increase even when the amount of reinforcement increase. 

We can deduce that it is not convenient to design the section with a reinforcement higher than ωR = 0.271. In other words, additional reinforcement would be an unnecessary waste since there would not be an increase in the capacity of the section. 

Therefore, to complete the evaluation range of the approximated formula, a new formula can be obtained that covers the section in which ωR > 0.2. As we can see in Fig. 3, we have obtain a new equation B in a similar manner for this section: 

    As :=   Mu

     3.8⋅d

which will always be on the safe side of the entire range above ωR > 0.2

Conclusions and recommendations.

3.1) The first step was to derive the approximated equation from the exact formulation of a rectangular section subjected to bending. It was verified that the approximated equation is a simplification of more complex calculations by assuming a linear function of the real behavior. 

3.2) The equation As = Mu/4xd is completely valid and safe within the range in which it can be applied. As long as the geometric amount is less than 4 times the minimum geometric amount, the formula is completely reliable. 

3.3) In the range of application shown, the difference due to excess does not exceed 6% of the exact calculated reinforcement. This occurs at about the midway point of the validity range. 

3.4) To broaden the application range of the approximated formula, we obtain a modification of the original formula As = Mu/3.8xd, which is also safe when ωR is greater than 0.2. 

3.5) Fig. 3 shows that the norm considers the range of ωR between 0.271 and 0.31. It is not convenient to design in this range because of the unnecessary waste of reinforcement without a significant increase in the capacity of the section. 

3.6) After completing this analysis, in the new version of the ACI 318 – 19 codes, a modification was introduced and now the beams will have a limit of εs < 0.005 so that the fragile failure limit coincides with the tension control limit for ϕ =0.9. Therefore, we resolve the problem that was in the previous section.

Translation By: Ivet Llerena (Eastern Engineering Group).

Structural Drawings By: Eastern Engineering Group.

© 2021 This article was written by Ernesto Valdes and published by Eastern Engineering Group. All rights reserved.