Calculating cracked deflections in Diamonds

A better knowledge of material properties and advanced analysis tools enable structural engineers to achieve more lightweight designs. As a result, comfort requirements are becoming increasingly important. For concrete beams and floors, the maximum deflection often becomes the decisive criterion. But then, new questions arise, such as “How should deflections be calculated?” and “What are the limit values for maximum deflections?”.

A few definitions up-front

The initial shape of a concrete beam or floor slab is the reference line (reference plane) of the load-bearing element in the unloaded condition.
The deflection is the displacement of the reference line (reference plane) under the influence of the applied loads compared to the initial shape.
The instantaneous deflection due to applied loads is the part of the deflection that occurs almost instantaneously after applying the loads.
The time-dependent deflection due to applied loads is the part of the deflection that occurs due to concrete creep after applying loads that are assumed to be constant through time.
The additional deflection from the point in time is the part of the deflection that occurs after . It consists of several components:
- The remaining part of the time-dependent deflection under the loads that have already been applied at the point in time ;
- The instantaneous and time-dependent deflection under the loads applied after the point in time $t_i$ .

Calculating the cracked deflection

The elastic deflection of reinforced concrete beams or floor slabs will only match the actual deflection if the beam’s or floor slab’s cracking moment is not exceeded. Because of the low tensile strength of concrete, this situation will usually not be applicable in practice. As soon as stresses exceed the concrete tensile strength, cracks will appear, and the cross-section’s stiffness will significantly decrease, resulting in more considerable deflections.

To calculate the deflection of reinforced concrete beams or floor slabs, we need to consider concrete cracking. The degree to which the concrete is cracked, will not only depend on its tensile strength. It will also depend on the actual amount of reinforcement.

The calculation method discussed hereafter (available in the Diamonds structural analysis software) is a logical extension of design rules for structural elements (such as beams) bearing loads in one direction only. After discussing its principles for the case of unidirectional load-bearing elements, we’ll examine how we can extend the scope of this calculation method to structural elements bearing loads in two directions.

For elements subject to bending, the following formula applies for the calculation of the vertical deflection $\delta$ :

(1) $\frac{d^2 \delta }{dx^2}=\frac{1}{r}=\frac{M}{E \cdot I}$

For reinforced concrete elements, the curvature value 1/r depends on whether the cross-section is cracked or not. A cross-section is cracked only when the bending moment exceeds the cracking moment $M_r$ . The cracking moment $M_r$ is determined by:

(2) $M_r=f_{ctm.fl} \cdot W$

in which:

$f_{ctm.fl}$ is the mean flexural tensile strength of concrete
$W$ is the moment of resistance of the uncracked fictitious cross-section that consists of the entire concrete cross-section augmented by $\alpha$ times the section of the reinforcement bars $\left( \alpha=\frac{E_s}{E_c} \right)$

For a cross-section in uncracked zones of the beam (for which $M < M_r$ ), the local curvature $\frac{1}{r_1}$ is calculated from:

(3) $\frac{1}{r_1}=\frac{M}{E \cdot I_1}$

in which:

$E$ is the Young’s modulus of concrete
$I_1$ is is the moment of inertia of the uncracked fictitious cross-section

For a fully cracked section (for which $M > M_r$ ), the local curvature $\frac{1}{r_2}$ is calculated from:

(3) $\frac{1}{r_2}=\frac{M}{E \cdot I_2}$

in which:

$E$ is the Young’s modulus of concrete
$I_2$ is the moment of inertia of the cracked fictitious cross-section, consisting only of the compressed concrete cross-section augmented by $\alpha$ times the section of the reinforcement bars $\left( \alpha=\frac{E_s}{E_c} \right)$ .

Figure 1: Distribution of cracked zones in a continuous beam.

Figure 1 shows distribution of cracked zones in a continuous beam. In the white zones, the bending moment $M$ remains smaller than the cracking moment $M_r$ . The beam is not cracked in those zones. In the light blue zones, the bending moment $M$ exceeds the cracking moment $M_r$ a little. While in the darker blue zones, the cracking moment $M_r$ is exceeded a lot more. The light blue zones will be less cracked than the darker blue zones.

So, we need an expression that takes the amount of cracking into account. The average local curvature $\frac{1}{r}$ for a cross-section in a cracked zone (for which $M>M_r$ ) is calculated as the weighted average between the uncracked and cracked curvature (EN 1992-1-1 §7.4.3):

(5) $\frac{1}{r}= (1-x) \frac{1}{r_1} + x \frac{1}{r_2}$

in which:

$x$ is the distribution coefficient equal to
(6) $1-0.5\left( \frac{M_r}{M_{ULS RC}} \right)^2$

Rather than calculating the deflection through a double integration of the local curvatures $\frac{1}{r}$ , we can divide the beam into several elements with variable stiffness $EI$ . Replacing $I$ & with $I_1$ for the uncracked zones and deriving $I$ for cracked zones from the relationship (5) in which curvatures $\frac{1}{r}$ are replaced by $\frac{M}{E \cdot I}$ .

We can easily extend the latter method to structural elements bearing loads in two directions. For each node of a finite element mesh, the moment $M$ and the cracking moment $M_r$ can be determined for both reinforcement directions. Formulae (5) and (6) then provide a way to estimate the stiffness $EI$ for both directions.

EXAMPLE: calculate cracked deformation in a beam

Given

C25/30, S500, $A_{top}=226mm^2$ , $A_{bottom}=942mm^2$ , $d_1=d_2= 40mm$ , $\phi =2$
dead load = 25kN/m + live load = 10kNm resulting in $M_{ULS RC} = 74kNm$

Calculation

Determine the cracking moment $M_r$ :

$M_r=f_{ctm.fl} \cdot W = 19.51kNm$

Determine the redistribution coefficient x:

$x=1-0.5\left( \frac{M_r}{M_{ULS RC}} \right)^2 =0.965$

The deformation of an uncracked cross-section $\delta_1$ equals:

$\delta_1 = \frac{5\cdot q\cdot L^4}{384\cdot E_c_0\cdot I_1}=2.50mm$

in which:

$E_c_0 = 31476MPa$
$I_1 = 156439cm^4$

The deformation of a fully cracked cross-section $\delta_2$ equals:

$\delta_2 = \frac{5\cdot q\cdot L^4}{384\cdot E_{c.\phi} \cdot I_2}=11.25mm$

in which:

$E_{c.\phi} = 10492MPa$
$I_2=104475cm^4$

The maximal cracked deflection $\delta$ in the beam equals:

$\begin{align*}\delta&= \left ( 1-x \right )\cdot \delta_1+x\cdot \delta_2\\&=\left ( 1-0.965\right )\cdot 2.50mm+0.965\cdot 11.25mm\\&=10.95mm\end{align*}$

Figure 2: Cracked deformation with creep in an ULS RC combination calculated by Diamonds

Looking for a more detailed example? Check our validation example.

Calculating additional deflection

The determination of additional deflection requires the time at which the loads act. The deflection as a function of time, can be calculated using superposition:

For each load group:
- calculate the instant cracked deflection using equations (1) to (6)
- calculate the delayed deflection due to creep using equations (1) to (5)
At each step in time take the sum of:
- the instant cracked deflection of the loads that are present
- the appropriate part of the delayed deflection due to creep

But that method comes with a few difficulties:

Cracking is non-elastic material behaviour. It does not allow superposition.
The moment of inertia of cracked cross-sections must be calculated instantaneously.
The creep factor for a given load depends on the age of the concrete at the time the load is applied: the older the concrete, the lower the creep factor for that load.

The following approach can be suggested (as implemented in Diamonds):

The superposition principle is applied by specifying which load combination should be considered decisive for concrete cracking.
This can be done separately for each load group (the amount of cracking is deemed to be constant through time for each load group). Usually, a rare load combination is recommended.
The creep factor is constant for each concrete quality and for all load case.

Calculating additional deflection: practical example in Diamonds

The above approach is illustrated through a practical example using Diamonds.
Figure 4 shows the geometry of a floor slab (from the building in Figure 3) for which the deflection is determined in three different ways.

Figure 4: Geometry model of one of the slabs from the Fonteynbrug residence on which the loads due to the partition walls act

Figure 5: Elastic deformation of the floor slab [mm]

Method 1: The elastic deflection (Figure 5) is only valid for an uncracked slab. This deflection depends on the modulus of elasticity and Poisson’s ratio of the concrete and the moment of inertia of the uncracked cross-section. This deflection is independent of the calculated reinforcement quantities. This deflection is only realistic for small loads. In all other cases, the elastic deflection underestimates the actual deflection.

Figure 6: Total cracked deflection on time infinity [mm]

Method 2: The total cracked deflection at time infinity (Figure 6) provides a conservative estimate of the deflection.
This deflection takes into account the reinforcement quantities, the cracked cross-section and the creep of the concrete. All loads are fictitiously applied at the same time. This deflection is conservative because it assumes that all loads (including the non-quasi-permanent part of the live load) cause creep.

Figure 7: Additional deflection due to the presence of the partition walls [mm]

Method 3: The cracked deflection over time takes into account the reinforcement quantities, the cracked cross-section, the creep of the concrete and the time at which the loads are applied.
In this example, it was assumed that the self-weight is present after 28 days, the dead loads after 60 days, the partition walls after 90 days and the live load after 120 days.

Time [days]	Deformation [mm]
t=-28	0,0
t=+28	-2,5
t=-60	-4,4
t=+60	-5,2
t= -90	-6,2
t= +90	-6,2
t=-120	-6,6
t= -120 RC	-7,3
t= ∞ RC	-10,5 (Figure 6)

The following table shows the increase in cracked deflection over time, calculated using Diamonds. The term ‘t=-90’ days refers to ‘just before the separation walls engage’. And ‘t=+90’ refers to immediately after the separation walls engage.

The deflection calculation over time has the advantage that it is possible to calculate additional deflection after applying specific loads. Figure 7 shows the additional deflection resulting from the installation of the partition walls.

Complying with design codes

Now that we have the maximum deflection calculated, to which value should it be limited? There is a distinction between the total deflection and the additional deflection.

The total deflection is generally limited to the span length divided by 250 (or L/125 for a cantilever) in an SLS QP (self-weight, dead loads and approximately 30% of the service loads).
By applying a counter-deformation, the deflection can be fully or partially compensated.
The additional deflection is generally limited to the span length divided by 500 (or L/250 for a cantilever) where damage to partition walls is to be avoided.

Conclusion

The non-linear material behaviour of concrete requires the calculation of deflection to consider cracking and creep. Advanced structural analysis software, like Diamonds, proves to be vital to estimate the total and additional deflections for 2D and 3D models.

If you want to learn more about the cracked deformation in Diamonds, we recommend this webinar.

Quiz

Which knowledge is essential to estimate the cracked deflection (select all possible answers).
1. Position of the stirrups.
2. The bending moment diagram.
3. The concrete grade.
4. The steel grade.
Which statement is true about the cracking moment ?
1. When the bending moment in a beam exceeds the cracking moment $M_r$ , the beam is fully cracked.
2. The cracking moment $M_r$ is the bending moment in a concrete beam or floor slab just before the tensile zone starts to crack.
3. When the bending moment in a beam exceeds the cracking moment $M_r$ , the curvature 1/r remains constant.
4. When the bending moment in a beam exceeds the cracking moment $M_r$ in one spot, the entire element is cracked.
The beam in the calculation example was loaded with a distributed load. Therefor the maximum bending moment was only found at mid span. Assume that we would load the beam so that the maximum bending moment is found over the entire span. Which statement would be true?
1. The total cracked deformation would be equal to $\delta_2$
2. The total cracked deformation would be equal to $\delta_1 + \delta_2$
3. The used formula for $\delta_1$ and $\delta_2$ are not applicable because the load distribution is different.
What is the strictest requirement that the cracked deflection t= ∞ RC (see table) satisfies? Use a span of 3m.
1. L/250
2. L/300
3. L/500

Calculating cracked deflections in Diamonds

A few definitions up-front

Calculating the cracked deflection

EXAMPLE: calculate cracked deformation in a beam

Given

Calculation

Calculating additional deflection

Calculating additional deflection: practical example in Diamonds

Complying with design codes

Conclusion

Quiz

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