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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