Aus AS3600 Deflection Presentation PDF

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Lecture 7 Serviceability deflections • Deflection limits • Deemed-to-comply conditions, simplified procedure and refined approach • Short and long term effects • Load combinations • Moment of inertia

There are 3 main types of deflection problem that may affect the serviceability of a structure: 1) Where excessive deflection causes either aesthetic or functional problems; 2) Where excessive deflection results in damage to either structural or nonstructural elements 3) Where dynamic effects cause discomfort to occupants. Examples case 1: Unacceptable sagging of slabs and beams. Problem overcome by limiting the magnitude of the final LONG-TERM (or TOTAL) deflection equal to the sum of the short-term (elastic) deflection plus time dependent deflections (caused by sustained load (creep) and shrinkage). Examples case 2: Deflection-induced damage to ceiling or floor finishes, cracking masonry walls and other brittle partitions. The deflection that occurs after the attachment of the nonstructural element in question, i.e. the INCREMENTAL deflection, must be limited. Examples case 3: vertical motion of floor. Very little quantitative info available in codes

Lef = min{Ln + D , L} D L

L

n

The design for serviceability is frequently the primary consideration when determining the cross-sectional dimensions of beams and floor slabs. This is particularly true in the case of slabs that are typically think in relation to their span and are therefore deflection sensitive! AS3600 specifies 3 basic approaches: •Deemed-to-comply conditions based on the span-to-depth ration (clause 8.5.4 for beams and clause 9.3.4 for slabs of AS3600) •Simplified approach based on the calculation of deflections using approximations (clause 8.5.3 for beams and clause 9.3.2 for slabs of AS3600) •Refined approach The simplified approach are simple to use and involve crude approximations of the complex effects of cracking, tension stiffening, concrete creep and shrinkage and load history. They often produce inaccurate of unconservative predictions when applied to lightly reinforced slabs.

The simplified approach Calculated simplified deflections smaller than deflection limits

Deflections must be calculated taking into account short-term mechanical effects (i.e. elastic deflections due to dead and live loads) and time-dependent effects that are function of the load history and temperature. The time-dependent deflections are difficult to calculate and the simplified method offers an easy solution. But, why are time-dependent effects so important and what exactly are they?

Time-dependent effects = creep and shrinkage

Creep deformations are dependent on LOAD and TIME

σ

Time

ε εplastic

εelastic

εelastic Time

Why is shrinkage so problematic?

Extra tensile strength between cracks called

Tension Stiffening

Shrinkage causes the gradual widening of existing cracks and a significant increase in deflections with time

Total deflection=elastic short-term deflection due to dead (G) and live (Q) loads + + long-term deflection due to sustained load (creep) and shrinkage Elastic deflection at mid-span=

CΔ

M L2ef EI

With M=maximum bending moment

With ψs from Table 4.1 of AS1170.0 (usually equal to 0.7)

In the absence of more accurate calculations, the additional long-term deflection of a reinforced beam due to creep and shrinkage may be calculated by multiplying the short-term (elastic) deflection caused by the sustained loads by a multiplier, Kcs, given by

k cs = [2 − 1.2( Asc / Ast ] ≥ 0.8

Clause 8.5.3.2

With ψL from Table 4.1 of AS1170.0 (usually equal to 0.4)

Total deflection

Δ total = Δ s + kcs Δ s .sus Elastic deflection due to dead load and short-term factorised live load

Elastic deflection due to dead load and long-term factorised live load

Clause 8.5.3.2

Moment of Inertia W

Δ=

L

WL3 EI

1 48

Mid-span deflection Δ

W=wL w L

Δ=

5 384

WL3 EI

Simply supported = Statically determined

Mid-span deflection Δ

Continuous beam = Statically undetermined

Ig Icr

Cracking Moment Mcr .......

A

B

1 1 I ef _midspan + Ief _ supportB 2 2

C

D

1 1 1 I ef _midspan + Ief _ supportB + Ief _ supportC 2 4 4

Deemed-to-comply conditions...