• 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