Appunti Assessment of Historical Buildings

Name: Appunti Assessment of Historical Buildings
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Appunti del corso presi durante lezioneGrade:30L/30Syllabus1. Construction techniques of historical buildings. 2. Mechanical behaviour of masonry.3. Structural analysis techniques. 4. Masonry …

Esame Assessment of Historical Buildings

Facoltà Ingegneria civile ambientale e territoriale

Dal corso del Prof. Taliercio Alberto

Università Politecnico di Milano

A.A. 2022-2023

78 pagine

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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 types of masonry. It consists of: bricks + mortar joints

Let's consider a uniaxial compression test.

Comments

Bricks are stiffer & stronger than mortar
Bricks have a brittle behaviour
Mortar has a ductile behaviour
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:

Bricks & mortar obey Mohr-Coulomb criterion:

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

Mechanical behaviour of masonry

Masonry is a composite material made of components characterized by different mechanical properties.

Brickwork is one of the most common types of masonry. It consists of bricks and mortar joints.

Let's consider a uniaxial compression test.

Comments

Bricks are stiffer and stronger than mortar.
Bricks have a brittle behaviour.
Mortar has a ductile behaviour.
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:

Bricks & mortar obey Mohr-Coulomb criterion:

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

(σ_c + σ_a) sec φ = √E / 2√E / 2 + σ cot φ sec φ = √E / 2√E / 2 (sec φ - 1) = -2c cos φ / -1 + sec φ

σ_c = ^{2c cos φ} / _{1 - sec φ}

σ_a - σ_p sec φ = √E / 2 - (σ cot φ + √E / 2) sec φ = √E / 2c cos φ - √E / 2 sec φ - √E / 2 = 0

σ_c = ^{2c cos φ} / _{1 + sec φ}

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

2) Axisymmetric stress state

Thanks to this hypothesis we have a uniform stress state

σ^b₁₁ = σ^b₂₂ = σ^b₃₃

σ^m₁₁ = σ^m₂₂ = σ^m₃₃

σ₂₂ = σ₁₂ = σ₂₂ = -σ - compression

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

σ^B₁₁ h_B + σ^m₁₁ h_mu = 0 σ^B₁₁ - σ^m₁₁ ^h_mu / _{h_B} = -σ^m₁₁ - α_H

α_H = ^h_mu / _{h_B} (

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