Deflection controlWhy limit deflections?1. To avoid failures in supported partitions (not

load bearing wall structures).

Help to limit over-deflectionsThe help can be so called precambering (opposite upward deflection) of shuttering of horizontal spanning member (usually a slab). Enclosed figure shows relation between loads and deflections. Max allowed precambering according to Eurocode is span/250. It is suitable especially for protecting partitions.

On this figure long term effects are not respected

2. To avoid bad appearance and/or generally upsetting feeling of ‘liveliness’ in the structure

2. To avoid bad appearance and/or generally upsetting feeling of ‘liveliness’ in the structure. „Virtual Inverse deflection“.

3. To protect collapse of the support structure. (Seldom, by special circumstances, only.)

How to ensure acceptable deflections?

a) Calculate the deflection of designed structure/member and compare it with a suitable limit.

b) Respect recommended arrangement of designed structure/member.Another special „detailing“ → → → span/depth ratio.

a) Technique based on calculation of structural’sdeflection.The condition of reliability concerning deflection is:

f ≤ flim

There can be many various limits there. Eurocode 2

gives only following two:

Under the quasi-permanent load, the long term

deflection should not exceed span/250, in order to

avoid impairment of appearance and general utility.

Under the quasi-permanent load, the deflection after

removing temporary supports should be limited to

span/500 to avoid damage to adjacent parts of the

structure. (Partition walls as was presented.)

a) Technique based on calculation of structural’s deflection

The calculation of the actual value of deflection (marking f, inthis lecture) is considered as not very accurate and the limitsand the technique are not obligatory. However, it is safe andimportant for understanding of all related processes andcircumstances.

If RC will be homogenous and elastic material, the calculation of deflection will be simple according to structural mechanics –theory of elasticity (Mohr theorem).e.g. simple supported beam with the span “l” loaded with uniform load “p” of material with the modulus of elasticity “E” and Cross-sectional parameters represented by “J”, has deflection in the middle of the span:

� =�

��� .

� .��

� .�[1]

But RC is not “macroscopically” homogenous material, itcracks and more of its properties are changing during thetime due to more influences. In the following chart thepotential problems with structural RC are defined and it isnoticed relevant solution.Will be explained in the following lecture.

Difference ideal mass

x real RC

Solution by Eurocode

Macroscopic inhomogeneity:

concrete + reinforcement

Transformed CS and it’s stiffness and

its parameters are used.

Concrete cracks under load at

different way.

Model of not cracked and fully

cracked transformed CS and

interpolation in it.

Concrete changes its

properties during the time.

Unique models for the influence of

creep and of drying shrinkage,

separately.

Macroscopic heterogeneity of RC

This fact is by RC solved with usage of so called “Transformed” or ”Idealized” cross – section.This is a CS where the sectional area of reinforcement As is substituted with ae multiple of it. Where:

ae = Es/Ecm

Where modules of elasticity Es = 200 GPa = const.Ecm depends on concrete strength class and is noticeably smaller.

Notice: by transformed CS is RC CS transformed into concrete CS.

Commonly

In the equation [1] is the multiple E.J stiffness of the deflected CS The

technique how to express the difference between RC and ideal homogenous

mass is in the Eurocode based on the substitution of ideal stiffness E.J with

the real (more real) one. Common marking of the stiffness is B. By

identification of the real B booth E and J are modified. See following.

By transformed CS is the substitutive area Ast. ae virtual only, and has the centroid equal with the centroid of reinforcement. Distance of the neutral axis xl and xll is identical with the position of centroid axis of the same cross-section (in case of bended CS).

*) Characteristics of Transformed CS commonly are market with the index „i“, e.g. Ji, xi etc. Usually b= bi= bl= bII

Characteristics of Transformed CS without crack are market with the index „l“, e.g. Jl, xl etc.) Characteristics of Transformed CS with crack are market with the index

„ll“, e.g. Jll, xll etc. Usually (by rectangle) b= bi= bl= bII

How RC cracksFirstly, it is important to understandthe process of RC cracking.For the simplicity, it is explained onRC member in pure tension. Forbending, it is in principle the same.There are three important phases,there:Phase I uncracked - fig a) bothconcrete and reinforcement actfully in tension and there are nocracks in concrete. Of course, actingforce F1 is relatively small. (sc<fctk)Phase II cracked (partly) - fig b) firstcracks occur and with increasing ofF2 occurs new an new cracks untilthe situation on fig. c) is reached.From this point, no new crackoccurs and with increasing F3 growsthe strain of reinforcement incracks, only. It is phase III. -inelastic (fully cracked)(F1< F2< F3).

More details of stress – strain conditions surrounding a crack in the phase 2. or 3. le ≈ S0 ≈ anchoring length

Demonstration of previous cracking sequence in L-D diagram

Transformation of L-D diagram into L-D common relation according to Eurocode 2 NOT ESSENTIAL

Influence of acting bending moment M on stiffness B of RC cross-section(By flexural members)

sr

By the real service load of RC structures/members phase ll is presumed

equivalent stiffness B must be interpolated. The formula is: 1

�=

1

��. (1 − ) +

1

��� . [2]

Where:

BI = Ecm . JI and BIl = Ecm . JIl

= 1 − . (��

�) 2

interpol. coefficient (for tensional fastening)

�� is stress in the reinforcement after the first crack occurs

� is stress in the reinforcement for load considered. Stress can be substituted with M:

= 1 − . (���

��) 2

�� acting M of loading (loading case) considered is coefficient for load duration

= 1,0 for short term load

= 0,5 for long term variable or permanent load

By real structures the most stressed CS are usually in phase II so usage of interpolation is actual. The question is, which quantity to interpolate for expression the actual stiffness of related CS? Usually curvature (1/r) or directly the stiffness B is used. Interpolation formula for stiffness B is:

ME > Mcr by phase II !

Deflections and proper loading combination

Finding of relevant load combination for deflection calculation is

little bit complicated.

As the cracking should be presumed as irreversible process, for

calculation of Mcr and all stress values concerning stiffness B, Bl, Bll

characteristic load combination should be used (equation 6.14b by

EN 1990).

For calculation of the deflection itself, frequent or quasi-

permanent combination should be used, in relation to the type of

calculated deflection.

For short-term deflection (without influence of creep and

shrinkage) frequent combination should be used.

For long-term deflection (with influence of creep and shrinkage)

quasi-permanent combination should be used.

Mcr (cracking moment) should be calculated for the ideal/transformed CS without crack! Basic theory of elasticity should be used and stress in marginal fibres in tension is presumed to be equal to fct,k; 0,05.

Long-term deflection and long-term effects

The long-term deflections are significantly greater when compared to short-term

ones (about 2-3-times). The responsible processes of the increase are creep of

concrete and shrinkage.

Creep influence

Is expressed simply by

modification of the concrete’s

modulus of elasticity. Instead of

Ecm is Ecm,eff used. Hold true:

Ecm,eff = Ecm/(1+,t0)

The value of Ecm,eff with help of a

nomogram in the next page

figure can be found.

The unpleasant thing is that all calculation must be done twice, of stiffness Bl

and Bll and their interpolation with the Ecm,eff and e,eff. It is not enough to put Ecm,eff it into deflection calculation/formula, only.

e,eff = Es/Ecm,eff

In the diagram it is h0 = 2AC/u, where: AC is the sectional area of the CS and uthe perimeter exposed to drying.

Example: Calculate by the slab with h=300 mm (dried from both surfaces), concrete C30/37 with CEM “N” (normal setting speed) the value of ,t0, the relative humidity of surrounding air is 50%. 1. step: to choose proper diagram ( there are more diagrams in EN, there]

The value of ,t0 = 2,85

Shrinkage influence to deflection (curvature)Concrete shrinkage has the influence to deflection of members due to not uniform drying or/and not symmetrical reinforcement. Examples:

In the case of symmetrical CS and also

reinforcement is shrinkage symmetrical

to centroid axis of the member, which

shrinks but there is no curvature.

In the case of not symmetrical CS

and/or reinforcement shrinkage o

concrete lead to curvature –

deflection of the member. Pay

attention to the orientation of the

curvature! The deflection due to

external load can be increased or

decreased!

Shrinkage formulas Curvature due to shrinkage:

1/rcs = cs . e,eff. S/J

Where:

cs is the magnitude of free concrete shrinkage in actual humidity temperature and time.(Formulas in the EN 1992-1-1.)

�

�=

��

��. (1 − ) +

���

���

.

Deflection due to shrinkage can be obtained as value of bending moment M of the same CS of member loaded with the relevant curvature. E.g. for s. s. beam the max deflection in the middle is:

��ℎ =1

8.

1

��ℎ . �2

Note: deflection due to shrinkage should be about one order smaller when compared to the creep deflection.

Stiffness of the whole members

In the previous text the calculation of

stiffness in relation to the level of stress in

actual CS. By members stressed with

bending moment M is the value of

stiffness valid just in this CS. By other CS

wit lover M is the stiffness equally higher

– see enclosed figure and M-B relation.

Full respect to this fact will complicate the

calculation of deflection a lot. Fortunately

is by the EN 1992-1-1 allowed safe

simplification – to presume by the

member alongside the whole span

constant stiffness equal to the minimum

found in the most stressed CS.

Task: analyse yourself from the same point of view shape of stiffness line alongside of primary beam of frame structure.

Deflection control due to comparison with max span/depth ratio

This technique of indirect deflection

control is in the Eurocode recommended

over the previous one. The technique is

based on calculation of member’s

span/depth ratio and comparing them

with recommended limit. The calculation

of an actual ratio l/d is not complicated.

More complication can be presumed by usage of complicated formulas for

calculation of the limit (max allowed) l/d – see the following slide do not memorize!!This is the reason why by various code companion literature instead of these graphs have been presented. See over next slide.

Do not memorize formulas and following graphs!

max

To be honest, the values of limiting span/depth ratio must be multiplied with the values of k given in Table 7.4N in Eurocode .

Practical loading history

Identified quantities without creep influence

END OF SLS