Intro of Mechanics of Materials

Topics: Force, Continuum mechanics, Classical mechanics Pages: 3 (333 words) Published: May 10, 2013
Review of Chapter 1 and 2
• Reading work for this week
– Please go through entire chapter 1 and 2 and work out the example problems in the text without reading the solutions

Chapter 1
• Equilibrium of deformable body
– Free-body diagrams

• Concept of stress
– Normal and shear stress – Factor of safety – Design of connections

1.2 Equilibrium of a Deformable Body External Loads
• • • Surface forces are caused by direct contact Concentrated forces are applied to a point Linear Distributed forces w(s) [Force/length] are idealizations. The resultant force Fs is equivalent to the area under distributed loading curve and acts through centroid or geometric center of this area Body force is developed when one body exerts a force on another body without direct physical contact



Support Reactions
• Forces developed at supports or points of contact

Equations of equilibrium
ΣF=0 Σ Mo= 0

• Vector equations of equilibrium

• Scalar equations of equilibrium Σ Mx = 0 Σ My = 0 Σ Mz = 0 Σ Fx = 0 Σ Fy = 0 Σ Mo = 0

Σ Fx = 0 Σ Fy = 0 Σ Fz = 0

• Coplanar Forces

Internal Resultants

Internal Resultants

Internal Resultants for coplanar loading

Example 1-5

Free Body diagram

Concept of Stress

Stress Notations

General state of stress

Units
σ = force/unit area Normal or shear stress = Newtons/m2 1 Pascal (Pa) = 1 Newton/m2 1kPa (kilo Pascal) = 103 Newton/m2 1MPa (Mega Pascal) = 106 Newton/m2 1GPa (Giga Pascal) = 109 Newton/m2

Average shear stress

Τavg = V/A V = F/2 Simple or Direct Shear

Single (lap) shear

Double shear

Allowable Stress
• • • • Factor of Safety (F.S.) F.S. = Ffail / Fallow Always bigger than 1 F.S. = σfail / σallow F.S. = τfail / τallow

Chapter 2
• Deformation of a body is specified through strain • Concept of strain – Normal and shear strain – Examples

Normal Strain
• Єavg = (Δs’ – Δs)/(Δs) • Є = lim
B A along n

Δs’ – Δs Δs

• Δs’ ≈ (1 + Є) Δs

Shear strain...
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