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Estratto del documento

MECHANICAL BEHAVIOUR OF MASONRY

Masonry is a COMPOSITE MATERIAL made of components characterized by DIFFERENT MECHANICAL PROPERTIES.

  • BRICKWORK is one of the most common type of masonry. It consist of:BRICKS & MORTAR JOINTS

Let's consider a UNIAXIAL COMPRESSION TEST

COMMENTS:

  1. bricks are STIFFER & STRONGER than mortar
  2. bricks have a BRITTLE BEHAVIOUR
  3. mortar has a DUCTILE BEHAVIOUR
  4. MASONRY has an intermediate behaviour between brick and mortar

COMPRESSIVE STRENGTH OF MASONRY

* To determine the COMPRESSIVE STRENGTH of masonry let's make the following hypothesis:

  1. BRICKS & MORTAR obey MOHR - COULOMB criterion:

|τ| ≤ c + σ tan φ - internal friction angle

* We can assume that the real masonry specimen can be substitute by a layered medium

2) Axisymmetric Stress State

* Thanks to this hypothesis we have a uniform stress state

σb11 = σm11 σb22 = σm22 σb33 = σm33 σzz = σ12 = -σ → compression

* Since there are not horizontal stresses acting on the element, the horizontal stress average must be zero:

σm 11 hm + σb 11 hb = 0 σm 11 = -σb 11 hm / hb = -σm 11 αh αh = hm / hb ( M ⋅ (Σ ⋅ m) ∈ M (t ⋅ m)

M (Σ - t) M ≤ 0 the eigenvalue of the tensor must be NON-POSITIVE

* this inequality reads that Σ - t MUST BE NEGATIVE SEMI-DEFINITE

b) COMPRESSIVE PRINCIPAL STRESS ONLY ALONG 1 DIRECTION (II=0)

! THE CRACKING STRAIN TENSOR IS ORTHOGONAL TO THE STRESS TENSOR

  • εcr > 0 → non-negative
  • σi:εcr = 0 → orthogonality
  • σ & εcr COLLINEAR → same principal directions

* From a physical point of view, the ORTHOGONALITY CAN between σ & ξ means that there is NO POWER DISSIPATED in CRACKING

FAILURE MODES OF MASONRY

1) UNIAXIAL COMPRESSION: failure usually occurs by cracking and sliding in the head and/or bed joints.

2) UNIAXIAL TENSION:

The AVERAGE SHEAR AT WHICH CRUSHING OCCURS is

Tc = λcF/A PB/2FH(1 + 2λ/B √F/8t)

Tc = P/2H B/B(1 + 2λ/B √F/8t) = σB/2H(1 + 2λ/B √F/8t)

Comparing Tc & Tot we have that Tot/Tc is a REDUCTION FACTOR that takes into account the FAILURE COMPRESSIVE STRENGTH of MASONRY

Normally, in masonry walls, σ 0

e = N/2

e ≤ b/2

(1)

Mo - Nc - mob/ℓ + λkxF a/2 ≤ b/2 (P - 2Mo/ℓ)

λkx = 2Pb/FH

* in this case NO CRACKS develops so NO PLASTIC

HINGES will form.

MACROSCOPIC BEHAVIOUR of MASONRY

* To analyze a masonry building it is possible to use:

① MICROMECHANICAL APPROACH

  • Takes into account of the heterogeneous nature of the building
  • Bricks and mortar are modelled separately
  • Cannot be used to analyze big buildings because it requires a huge mesh

② MACROMECHANICAL APPROACH:

  • Masonry is replaced by a homogeneized material
  • Rough mesh required
  • It is necessary to determine the homogeneized properties of the material

MACROSCOPIC ELASTIC BEHAVIOUR of MASONRY

* Linear elasticity can be used to analyze masonry walls at service conditions, when the presence of cracks can be neglected

① MODEL NEGLECTING HEAD JOINTS:

* Hp: {Head joints neglected Poisson’s ratio neglected νhm ≈ pbo ≈ 0

* Masonry is modelled as a layered medium

* Let’s define: Ebx = σ⁄Eb Emx = σ⁄Em

* The MACROSCOPIC VERTICAL STRAIN of the layered medium is:

E = Ebho + Emihm⁄Rhohm= σ⁄Eb + σ⁄Em= σ⁄E

* the MICROSCOPIC STRAIN FIELD (ξ(x)) must be PERIODIC, then

ξ(x) = E x + μper(x)

ξ(x) = E + ξper(x)

∮ ξper(x) = ξ + φ

* if J(x) and ξ(x) are PERIODIC then the HILL'S MACRO-HOMOGENEITY holds:

⟨J(x) : ξ(x)⟩ = ⟨J(x)⟩ : ⟨ξ(x)⟩ = ζ : E

it states the EQUIVALENCE of the VIRTUAL WORK at the MICRO & MACRO SCALE

METHOD 1: 2-STEP HOMOGENIZATION APPROACH:

• STEP 1

* Let's consider a LAYERED MEDIUM

with a 3D stress state

⎧ k = Σ11 + Ỹ11

⎪ 22 = Σ22 + Ỹ22

⎨ 12 = Σ12 + Ỹ12

⎩

⎧ 33 = E33 + Ę33

⎪ 13 = E13 + Ę13

⎨ 23 = E23 + Ę23

⎩

Ỹj: auxiliary stress

Ęj: auxiliary strain

→ the remaining components are:

⎧ 13 = Σ13

⎪ 23 = Σ23

⎨ 33 = Σ33

⎩

⎧ Ęk = E11

⎪ Ę22 = E22

⎨ Ę12 = E12

⎩

Macroscopic Strength of Masonry

Homogenization Theory

* Let's consider a masonry wall subjected to the maximum load PE that the wall can sustain.

* We substitute the real material with an homogenized medium and we want to determine the strength domain Ghom.

* Ghom must be such that: lim PE = phom when RVE, we have E → 0 that PE = phom

Static Definition of Ghom (Suquet)

Ghom = { ∫ 1 Σ = < U(zξ) > div U(zξ) = 0 ∀ x ε V t = U(xm) antiperiodic over ∂V[[ U(x) ]] Mτ = 0 across ⊂ SτU(x) ε G(x) ∀ z ε V }

  1. div U(x) = 0 → because we are neglecting body forces
  2. t antiperiodic over ∂V → at 2 opposite points, the stress vector must equal and opposite due to the periodicity of the geometry of the real body
  3. [[ U(x) ]] Mτ = 0 across any ⊂ Sτ → the stress field in the RVE can be discontinuous but, the stress vector must be continuous along Sτ

no jump of the stress field across Sτ

discontinuity surface

Dettagli
Publisher
Elebi1
A.A. 2022-2023
78 pagine
SSD Ingegneria civile e Architettura ICAR/08 Scienza delle costruzioni
I contenuti di questa pagina costituiscono rielaborazioni personali del Publisher Elebi1 di informazioni apprese con la frequenza delle lezioni di Assessment of Historical Buildings e studio autonomo di eventuali libri di riferimento in preparazione dell'esame finale o della tesi. Non devono intendersi come materiale ufficiale dell'università Politecnico di Milano o del prof Taliercio Alberto.