1 / 6

Chapter 9: Elasticity and Fracture

Chapter 9: Elasticity and Fracture. Christopher Chui. Statics: Forces in Equilibrium. 1 st condition for equilibrium: The sum of all forces is zero: S F x = 0, S F y = 0, S F z = 0 2 nd condition for equilibrium: The sum of all torques is zero: St = 0. Problem Solving in Statics.

luke
Download Presentation

Chapter 9: Elasticity and Fracture

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 9: Elasticity and Fracture Christopher Chui Chapter 9: Elasticity and Fracture - Christopher Chui

  2. Statics: Forces in Equilibrium • 1st condition for equilibrium: The sum of all forces is zero: SFx = 0, SFy = 0, SFz = 0 • 2nd condition for equilibrium: The sum of all torques is zero: St = 0 Chapter 9: Elasticity and Fracture - Christopher Chui

  3. Problem Solving in Statics • Choose one body at a time for consideration, and make a careful free-body diagram to show all forces acting on it • Choose a coordinate system and resolve the forces into their components • Using letters to represent unknowns, write down the equation for SFx = 0, SFy = 0, SFz = 0 • For St = 0 equation, choose any axis perpendicular to the xy plane. Pay attention to the sign of the torque • Solve these equations for the unknowns Chapter 9: Elasticity and Fracture - Christopher Chui

  4. Stability and Balance • If an object is displaced slightly, 3 possible outcomes: 1) the object returns to its original position—stable equilibrium; 2) the object moves even farther—unstable equilibrium; 3) the object remains in its new position—neutral equilibrium • A body whose CG is above its base of support will be stable if a vertical line projected downward from the CG falls within the base of support Chapter 9: Elasticity and Fracture - Christopher Chui

  5. Elasticity, Stress and Strain • Hooke’s law: DL is proportional to applied force • There is a limit of elasticity; plasticity follows; and finally breaking • DL =(1/E)(F/A)Lo E is elastic or Young’s modulus • Stress = force / area = F/A • Strain = change in length / original length = DL / Lo • E = stress / strain Chapter 9: Elasticity and Fracture - Christopher Chui

  6. Three Types of Stresses • Tensile stress • Compressive stress • Shear stress • Shear strain DL = (1/G)(F/A) Lo where G is shear modulus = ½ to 1/3 of the elastic modulus • DV/Vo= -(1/B) DP, where B is the bulk modulus Chapter 9: Elasticity and Fracture - Christopher Chui

More Related